Rotating Frames of Reference

1 Rotating Frames of Reference Phil Lucht Rimrock Digital Technology, Salt Lake City, Utah 84103 last update: February 9, 2017 Maple code is available upon request. Comments and errata are welcome. The material in this document is copyrighted by the author. The graphics look ratty in Windows Adobe PDF viewers wh en not scaled up, but look just fine in this excellent freeware viewer: https://www.tracker-software.co m /product/pdf-xchange-editor . The table of contents has live links, and u se of a wide Bookmarks pane is recommended. Overview .................................................................................................................................................. 4 Summary.................................................................................................................................................. 5 1. Notation, important role of the Prime Symbol, and other Preliminaries ...................................... 8 1.1 The basis vectors e n and e'n and two ways in which they are related ............................................. 8 1.2 Expansions of a vector a nd use of primes and parentheses ............................................................ 9 1.3 Special case where a' i is unambiguous ......................................................................................... 11 1.4 When are two vectors equal? ................................................................................................ ........ 13 1.5 The small rotation of a vector about an axis ............................................................................... ..14 1.6 The time rate of cha nge of a rotating vector ............................................................................... ..16 1.7 Rate of change of the basis vectors........................................................................................ ....... 17 1.8 Notations for the many time derivatives of vectors r, r', b and L.................................................. 18 1.9 Angular momentum ........................................................................................................... ........... 21 1.10 No frame label is needed for d/dt of a scalar function ................................................................ 22 1.11 When do operations d/dt and taking a component "commute" ? ................................................ 23 2. The G Rule for arbitrary vector a and its derivation .................................................................... 25 3. The Apparatus and its Observ er at Rest in Frame S'.................................................................... 28 4. The Relationship between the Two Frames S and S'..................................................................... 30 4.1 Explanation of Fig (4.1.1): Frame S in the plane of paper........................................................... 30 4.2 Explanation of Fig (4.2.1) : Vector ω pointing directly out of paper........................................... 31 4.3 Comments on b• S and b• S'............................................................................................................. 31 4.4 Special Case #1 : ω axis through Frame S origin......................................................................... 32 4.5 Special Case #2 : ω axis through Frame S' origin........................................................................ 33 4.6 The Turntable................................................................................................................................ 34 4.7 The Earth....................................................................................................................................... 35 4.8 The Flying Camera Platform......................................................................................................... 36 5. The Goal of the next two sections ........................................................................................... ......... 37 6. Determination of velocities................................................................................................. .............. 38 6.1 Velocity v S'................................................................................................................................... 38 6.2 Velocity v ≡ vS.............................................................................................................................. 38 2 6.3 Velocity v' S.................................................................................................................................... 39 6.4 Velocity Summary ........................................................................................................... ............. 39 6.5 Velocities fo r Special Cases............................................................................................... ........... 39 6.6 Comments ..................................................................................................................................... 40 7. Determination of accelerations ........................................................................................................ 42 7.1 Acceleration a' S............................................................................................................................. 42 7.2 Acceleration a ≡ aS........................................................................................................................ 43 7.3 Acceleration a S'............................................................................................................................ 44 7.4 Accelerati on Summary.................................................................................................................. 44 7.5 Relation between b•• S and b•• S'........................................................................................................ 44 8. The Fictitious Forces....................................................................................................... .................. 46 8.1 Development of the Fictitious Forces ....................................................................................... ....46 8.2 Interpretation of the Centrifugal and Euler Fictitious Forces ....................................................... 46 8.3 Interpretations of the Coriolis Fictitious Force ........................................................................... ..49 8.4 Special Case #1 Problems................................................................................................... .......... 53 8.5 Problems on the su rface of the Earth ....................................................................................... .....54 8.6 Tethered satellites and Tidal Forces....................................................................................... ....... 56 8.7 Special Case #3 ............................................................................................................ ................. 60 8.8 Tides on the Earth ......................................................................................................... ................ 62 9. Notation comparison with Marion (1970) and Thornton & Marion (2003) ................................ 84 10. Notation comparison with Goldstein (1950) and Goldstein, Poole and Safko (2001) ............... 86 11. Angular Momentum and Fictitious Torques; the Reynolds Transport Theorem..................... 88 11.1 Introduction................................................................................................................................. 88 11.2 Expression of L(c) and L•(c) in terms of Frame S' objects .......................................................... 90 11.3 Fictitious Torques and Newton's Rotational Law in a non-inertial frame .................................. 92 11.4 Application: Fictitious Torques in Fluid Dynamics................................................................... 93 11.5 Application: Fictitious Forces in Fluid Dynamics ..................................................................... 95 11.6 Comments on the Reynolds Transport Theorem ........................................................................ 96 12. Summary of the Forward Problem Solution .............................................................................. 100 12.1 Summary of the Forward Problem equations (non-swap notation) .......................................... 100 12.2 Summary of the Forward Problem equations (swap notation).................................................. 102 13. The Inverse Problem........................................................................................................ ............. 104 13.1 Brute Force Method ........................................................................................................ .......... 104 13.2 Swap Rules Method .................................................................................................................. 105 13.3 Summary of the Inverse Problem Equations (non -swap notation) ........................................... 107 13.4 Summary of the Inverse Probl em Equations (swap notation)................................................... 108 13.5 Why the Swap Rules Work....................................................................................................... 109 14. Rotating Frames in Curvilinear Coordinates............................................................................. 112 15. Ant on Turntable Problems .................................................................................................. ....... 115 Kinematics common to all Ant Problems ......................................................................................... 115 15.1 Problem 1: Ant crawls at constant speed V to the Origin of Frame S'..................................... 119 15.2 Problem 2: Ant spirals in at constant V and Ω to the Origin of Frame S'................................ 128 3 15.3 Problem 3: Inverse Problem: Ant fli es in Fram e S at constant velocity V ............................. 135 15.4 Problem 4: The Projectile Problem of Section 8.3................................................................... 144 Appendix A: Derivation of R( ξ) and Properties of Rotation Matrices.......................................... 150 Appendix B: The G Rule for a Tensor of Rank n ........................................................................... 154 Appendix C: The Foucault Pendulum............................................................................................. .160 C.1 Drawings, Notation, Coordinates and Basis Vectors ................................................................. 161 C.2 Qualitative Solution....................................................................................................... ............. 164 C.3 Equations of Motion for the Foucault Pendulum (Spherical Coordinates) ................................ 164 C.4 The Simple Pendulum ................................................................................................................ 167 C.5 The Spherical and Foucault Pendulums ..................................................................................... 174 C.6 Equations of Motion for the Foucault Pendulum (Cartesian Coordinates) ................................ 189 C.7 Verification of the Cartesian equa tions of motion and string tension ........................................ 191 C.8 Numerical solutions of the equations of motion (Cartesian Coordinates).................................. 195 Appendix D: Center of gravity and torque for a tethered satellite................................................ 204 D.1 Definition of Center of Gravity............................................................................................ ......204 D.2 Center of Gravity fo r a 2-mass Dum bbell Satellite.................................................................... 207 D.3 Dumbbell Satellite Center of Gravity with the Far Approximation: Numerical Examples ....... 213 D.4 Dumbbell Satellite Center of Gravity for equal masses and no approximation......................... 216 D.5 Center of Gravity fo r a single-sphe re satellite ........................................................................... 221 Appendix E: Spherical Coordinate Unit Vectors and Particle Kinematics.................................. 226 E.1 Angle Conventions ..................................................................................................................... 226 E.2 Matrix Approach ........................................................................................................................ 226 E.3 The motion of a particle in spherical coordinates....................................................................... 230 E.4 Curvilinear coordinates approach........................................................................................... ....231 E.5 Polar Coordinates ......................................................................................................... ............. 232 E.6 The Affine Connection ..................................................................................................... ......... 233 Appendix F : The Dumbbell (Tethered) satellite as an example of rotating frame analysis........ 236 F.1 Kinematics of the satel lite in rotating Frame S........................................................................... 237 F.2 Angular momentum of the satellite and its time derivative in Frame S ..................................... 240 F.3 The torque on the Dum bbell Satellite in Frame S'...................................................................... 242 F.4 The fictitious torque on the satellite in Frame S......................................................................... 246 F.5 Equations of Motion for the satellite in Frame S (Spherical Coordinates)................................. 249 F.6 Force analysis of the satellite in Frame S (Spherical Coordinates) ............................................ 253 F.7 Numerical solutions of the equations of motion (Spherical Coordinates) .................................. 260 F.8 Force analysis of the satellite in Frame S (Cartesi an Coordinates) ............................................ 264 F.9 Verification of the Cartesian equa tions of motion and stick tension .......................................... 268 F.10 Numerical solutions of the equations of motion (Cartesian Coordinates) ................................ 273 References............................................................................................................................................ 284 Overview 4 Overview This m onograph presents an extended discussion of doing physics in non-inertial frames of reference. The first chapters present the general theory, while the latter chapters and appendices contain examples concerning ant paths on turntables, tides, pendulums, fluid flows and tethered/dumbbell satellites. The presentation is entirely self-contained with all support material provided. Where possible, connections are made to external sources on this subject. Both linear (force) and rotational (torque) viewpoints are considered. Maple is used extensivel y to plot particle trajectories, to obtain and plot numerical solutions of differential equations (dsolve), and to verify complicated equalities. Our general context is an Apparatus containing a Par ticle observed from two fram es of reference called S and S'. Frame S' is rotating and translating in some arbitrary manner with respect to Frame S as indicated in this drawing, Fig 1 The Particle is located at position r relative to the Frame S origin, and at position r' relative to the Frame S' origin. These vectors are related by r = r' + b where b is a dynamic vector connecting the frame origins. Rotation is about some possibly moving axis with some angular velocity ω which might be changing in both direction and magnitude. We describe the relationship between the properties of the Particle as measured in these two frames of reference. The entire discussion takes place in a non-relativistic framework where time is the same in the two frames. Even in this limited cont ext, things are fairly complicated. An important subtopic of the rotating frames di scussion might be called "Newtonian mechanics in non-inertial frames" where one considers the fate of F = m a in a non-inertial frame. This is where the famous fictitious forces and less-fa mous fictitious torques appear. Most mechanics textbooks which treat rotating frames, having a multitude of other topics to address, spend 10-20 pages on the subject with the following itinerary: state the G Rule (see below), use it to derive an inter-frame velocity and/or acceleration relation, discuss fictitious forces in a rotating frame with emphasis on the Coriolis force, do a few basic problems, and end up treating the Foucault pendulum. Overview 5 A notable exception is the book of Taylor which devotes 40 pages to the subject including a nice discussion of the tides. (In his book, our frames S and S' are called S 0 and S.) In this document, having the luxury of no space limitations, we try to probe more deeply into the technical nuts and bolts of rotating frames analysis. Almo st all calculations are done in line for the reader to see. The "papal we" mode of presentation is often used below, as if this paper had multiple authors who seem to own the equations, drawings and experiments as their personal possessions. The approved mode of course is to use passive or impersonal sentence constructions as if the author did not exist. Interestingly, Taylor uses the "I mode" which is perhaps more honest and is certainly refreshing. Summary Section 1 sets up the notation used through out the document. As shown in Fig 1, a decision was made that Frame S' is the rotating frame, even though this conflicts with the choice made by many textbook authors. We refer to this notation as our non-swap notation and all our development work is done in this notation. One can imagine another version of Fig 1 which has S ↔ S', which is then our swap notation . Often it is more convenient to have Frame S be the rotating frame to a void an avalanche of primes in the equations of interest, and in that case the "swap" not ation is more useful. All key results are summarized in Sections 12 and 13 in both "non-swap" and "swap" notations. Unit basis vectors are ei and e 'i for Frame S and Frame S'. A generic vector a can be expanded on either basis. The presence of two frames of reference often associates with a vector a another vector a' which leads to the need for a compact no tation which distinguishes the components (a') i and (a)' i which are often different. The prime symbol ' plays a central role in the notation and is never used to indicate time differentiation (overdots are used for that purpose). After considering the meaning of equality for two vectors, we develop the notion of "conical rotational motion" according to a• = ω x a and then apply that notion to the basis vectors. The need for putting a frame label on the time derivative of a vector (but not a scalar or vector component) is demonstrated. It is then shown that the Particle in Fig 1 has four distinct velocities and eight distinct accelerations, reinforcing the need for a precise notation. Finally, it is noted that the operations of computing a time derivative and taking a component generally do not commute. Section 2 derives and discusses what we call " the G Rule ", namely, (d a/dt) S = (da/dt)S'+ ω x a where Frame S' rotates at rate ω relative to Frame S and a is an arbitrary vector. Section 3 describes an Observer and an Apparatus containing a Particle, all of which are in the rotating Frame S'. Section 4 explains the relationship between Frames S and S' for a general placemen t of the instantaneous rotation axis about which Frame S' rotates at vector angular frequency ω. Two special cases are identified for the location of this axis, and then three applications are roughly outlined. Section 5 (which is a very brief) states the goal of subsequent sections whic h is basically to determine how, given Particle properties in Frame S', one may determine these properties in Frame S. The results are eventually summarized in Section 12. Overview 6 Section 6 derives the relationships among the four velocities mentioned above. Section 7 derives the relationship among three of the eight accelerations . This and the preceding section are as dry as dust (mud's thirsty sister), but the results are of key importance, so everything is done step by step. Section 8 addresses the traditional subject of fictitious forces and interprets them. First the centrifugal and Euler forces are interpreted with the aid of some drawings, then comes the Coriolis force. An arm- waving interpretation of this force is provided for a simple four-projectile problem (on a turntable), and an exact solution to this problem later appears in Section 15. Three applications involving fictitious forces are then considered, but not fully analyzed: problems with moving objects near the surface of the Earth, tethered satellites , and ocean tides . Section 9 relates our notation to that used by Marion (1970) and Thornton & Marion [T&M] (2003). It is found that the Marion texts are very close to our "swap" notation. Section 10 then relates our notation to that used by Goldstein (1950) and Goldstein, Poole and Safko [GPS] (2001). These texts assume that both reference fra mes have the same origin which simplifies their presentations. Section 11 comments on the angular momentum vector L and its time derivative, and establishes how these vectors are related in the two frames. The notion of fictitious torques is introduced. It is demonstrated how both fictitious forces and torques are applied in fluid dynamics . Brief comments are made concerning fluid material and control volumes and the Reynolds Transport Theorem. Section 12 summarizes the set of equations which fulfill the goal stated in Section 5: given properties in Frame S', what are they in Frame S? The results are given in both "non-swap" and "swap" notation. Section 13 then considers the Inverse Problem: given properties in Frame S, what are they in Frame S'? The inverse equations are first obtained by laboriously inverting those presented in Section 12.1, and are then reobtained by a simple symmetry operation. The Inverse Problem equations are then summarized in both "non-swap" and "swap" notation. Section 14 briefly adds the complica tion of having a different orthogonal curvilinear coordinate system in each of the Frames S and S'. Up to this poi nt, only Cartesian coordinates have been used. Section 15 treats three "ant on turntable" problems in some detail. In the first two problems, the ant crawls in a certain manner on the turntable as it rota tes (Frame S') and the ant's position, velocity and acceleration are computed in inertial Frame S. In the third problem, the ant flies in a straight line in Frame S just over the turntable surface, and the ant's positi on, velocity and acceleration are computed in Frame S'. Many (hopefully entertaining) Maple trajectory plots are presented, along with the very simple code for these plots. In the final section, the projectile problem of Section 8.3 is solved using the third problem results, and it is noted that there are hard ways to solve rotation problems that can be avoided. Overview 7 Appendix A computes a certain matrix R(ξ ) which relates spherical unit vectors to Cartesian ones. Appendix B derives the G Rule for a general tensor of rank-n. Appendix C contains a detailed discussion of the plane ( simple ) pendulum , the spherical pendulum , and the Foucault mode of the spherical pendul um, including details of the Airy precession. A compact summary may be found at the start of Appendix C. Appendix D defines the notion of a center of gravity (as distinct from a center of mass) and computes the location of the center of gravity for various or ientations of a dumbbell satellite. Certain intuitive notions regarding the location of the center of gravity are seen to be not fully accurate. Appendix E summarizes useful facts about spherical coordinates . An emphasis is on relations involving the spherical unit vectors including their spa tial and time derivatives. The position, velocity and acceleration of a point particle are expressed in terms of these spherical unit vectors. A small section provides similar information for polar coordinates. Appendix F undertakes a detailed study of the motion of a dumbbell or tethered satellite in circular orbit around the Earth. The equations of motion are ob tained both from a fictitious torque analysis and a fictitious force analysis. Numerical solutions are presented for satellite librations and more general motions. A compact summary may be found at the start of Appendix F. References are then given for all works mentioned. Section 1: Preliminaries 8 1. Notation, important role of the Prime Symbol, and other Preli minaries Note : When an equation is repeated, its equation number is put in italics. 1.1 The basis vectors e n and e'n and two ways in which they are related Frame S has Cartesian basis unit vectors en, while Frame S' has Cartesian basis unit vectors e'n . The two sets of basis vectors are related by some rotation we shall call R (repeated indices have implied sums), e n = R e'n n = 1,2,3 or ( en)i = Rij(e'n)j (1.1.1) where on the right the component subscripts refer to components in Frame S. The above relation between e n and e'n can also be written in this manner e n = (R-1)nm e'm or e'n = Rnm em (1.1.2) a fact proven below in a short series of steps. Notice that the right equation in (1.1.1) involves a sum of basis vector components, whereas the equations in (1.1.2) involve a sum of basis vectors. Proof of (1.1.2): Step 1: The orthonormal Cartesian basis vectors have these properties δ n,k = en • ek = e'n • e'k and ( en)k = δn,k (1.1.3) where ( en)k = ek • en denotes the kth component of e n in Frame S. The basis vectors en are axis-aligned in Frame S, while the e'n are axis-aligned in Frame S'. The dot product is a • b = b • a ≡ aibi . Step 2: Note that (1.1.1) (1.1.3) (1.1.3) ( e'n)m = em • e'n = em • [R-1en] = ( em)i (R)-1 ij(en)j = δm,i (R)-1 ij δn,j = (R-1)mn = (RT)mn = Rnm ( 1 . 1 . 4 ) where in the last step we use the fact that R -1 = RT (or RRT = RTR = 1) for any rotation. Such a matrix R is said to be "real orthogonal" (T means transpose). Step 3: We wish to show (1.1.2) that e' n = Rnmem. We shall verify this by showing that all three components of each side are equal in Frame S, and to do this we dot both sides into ek : LHS = ek • e'n = (e'n)k = Rnk // using (1.1.4) RHS = e k • [Rnm em] = Rnm ek • em = Rnm δk,m = Rnk . QED Section 1: Preliminaries 9 Time dependence of the R ij If Frame S is fixed and Frame S' is rotating, then we really have e n = R(t)e 'n(t) where the R matrix is a function of time, so we have R ij(t). Similarly, if Frame S is moving and Frame S' is fixed, en(t) = R(t) e 'n and again one has R ij(t). Only in the case where there is no rotation between the frames are the R ij independent of time. This means that ω = 0 in Fig 1. Since our document is about "rotating frames of reference" we exclude this no-rotation case from consideration. Comment : In the above it was convenient to represent all three Cartesian unit vectors in Frame S and Frame S' by ei and e'i. Later when we do particular problems, we shall often revert to the following more standard physics notation: x 1, x2, x3 = x, y, z e1, e2, e3 = x^, y^, z^ e' 1, e'2, e'3 = x^', y^', z^' (1.1.5) Then for example δ n,k = en • ek = e'n • e'k for n = 1 and k = 2 says 0 = x^• y^ = x^'• y^'. 1.2 Expansions of a vector and use of primes and parentheses Note : We write (a) i as a component of vector a , but ( en)i as a component of en . In the first case the a in (a)i is not bolded, but since en is decorated with a label n, it gets bolded. It is just our convention. Any vector a can be expanded on either set of basis vectors so that, with implied summation on i, a = ai ei = a'i e'i . a i = a • ei a' i = a • e'i . (1.2.1) If some other vector named a' is lurking in the wings, one might want to be more careful labeling components. A safe method would be this: a = (a) i ei = (a)'i e'i (a)i = a • ei (a)' i = a • e'i a' = (a')i ei = (a')'i e'i (a')i = a' • ei (a')' i = a' • e'i . (1.2.2) Here, a prime inside a parentheses is part of the vector name, whereas a prime outside a parentheses denotes a vector component in Frame S' (whereas no prime outside means a component in Frame S). Unless the relationship between vectors a and a' has a certain simple form, it is very likely that (a') i ≠ (a)'i . In this case the notation a' i would be ambiguous since one doesn 't know whether it refers to (a') i or (a)'i. It is true that the notation a i could be unambiguously identified with (a) i, but we shall maintain the parentheses just to be uniform. Example 1. As an example of the four expansions above, let us consider a = en : e n = (en)i ei => ( en)i = δn,i // by inspection e' n = (e'n)'i e'i => ( e'n)'i = δn,i // by inspection Section 1: Preliminaries 10 en = (en)'i e'i => ( en)'i = (R-1)ni // using (1.1.2) that en = (R-1)ni e'i e' n = (e'n)i ei => ( e'n)i = Rni . // using (1.1.2) that e'n = Rni ei We can now restate these results showing the dot pr oducts which represent each expansion coefficient, and in this way we obtain expr essions for all the dot products, en = (en)i ei => ( en)i = ei • en = δn,i e' n = (e'n)'i e'i => ( e'n)'i = e'i • e'n = δn,i e n = (en)'i e'i => ( en)'i = e'i • en = (R-1)ni = RT ni = Rin e' n = (e'n)i ei => ( e'n)i = ei • e'n = Rni . (1.2.3) Notice that all these results are consistent with claims made earlier. In passing, we note that if R is any rotation (so therefore RTR = 1), then a • b = [R a] • [Rb] . ( 1 . 2 . 4 ) One line proof : [Ra] • [Rb] = [R a]k[Rb]k = RkiaiRkjbj = (RT)jk Rkiaibj = (RTR)jiaibj = δj,iaibj = aibi = a • b Matrix Notation to show how components are related. Consider the following expansion of vector a on the basis vectors e'j, a = (a)' j e'j // (1.2.2) = (a)' j { Rji ei } // (1.1.2) = (a)' j { (R-1)ij ei } // R = (R-1)T real orthogonal rotation = { (R -1)ij(a)'j )ei . // reorder (1.2.5) Comparing this to a = (a) i ei of (1.2.2) we conclude that, since ei is a complete basis, (a)i = (R-1)ij(a)'j ⇔ ⎝⎜⎜⎛ ⎠⎟⎟⎞ (a)1 (a)2 (a)3 = R-1 ⎝⎜⎜⎛ ⎠⎟⎟⎞ (a)'1 (a)'2 (a)'3 (1.2.6) where R-1 is a 3x3 rotation matrix. Section 1: Preliminaries 11 We can repeat the above discussion replacing a with a' with this result, (a')i = (R-1)ij(a')'j ⇔ ⎝⎜⎜⎛ ⎠⎟⎟⎞ (a')1 (a')2 (a')3 = R-1 ⎝⎜⎜⎛ ⎠⎟⎟⎞ (a')'1 (a')'2 (a')'3 . (1.2.7) These matrix equations are convenient for computing the components of a vector on the ei basis if they are known in the e'i basis (and vice versa) . 1.3 Special case where a' i is unambiguous We shall now examine the type of relationship between a' and a in which (a') i = (a)'i and therefore we can use the notation a' i without ambiguity. First, the components (a)' i and (a) i are related in the following simple manner, using (1.2.3), (a)'n = e'n • a = (e'n)m (a)m = Rnm(a)m . (1.3.1) Now suppose we define a new vector a' in this way, a' ≡ Ra . ( 1 . 3 . 2 ) If a vector a transforms into a' according to (1.3.2), we say it is a "vector under rotations" which means it "transforms as a vector under rotations". When written in Frame S components this says (a') n = Rnm(a)m . ( 1 . 3 . 3 ) Comparison of (1.3.1) and (1.3.3) shows that (a)'n = (a')n ( 1 . 3 . 4 ) and therefore in this case we can use a' n ≡ (a)'n = (a')n . ( 1 . 3 . 5 ) Thus, if the vectors a and a' are related by a' = R a where R is the rotation appearing in e n = R e'n , then we can dispense with the parentheses as s hown in (1.3.4). We still have (a')' i which requires parentheses. Example 2: Consider equation (1.1.1) , en = R e'n . Section 1: Preliminaries 12 Since this is not of the form a' ≡ Ra, we may not dispense with the parentheses. In fact from (1.2.3) we have ( en)'i = Rin (e'n)i = Rni ( 1 . 3 . 6 ) and these are not the same because rotation matrices ar e not symmetric. Unlike normal vectors for which one writes the transform a' = Ra , the basis vectors are "back-rotated" so e'n = (R-1)en. Active and Passive One can think of a' = R a as an active rotation of vector a into another vector a' within Frame S. In this case, the components of a' are (a') i. The alternative is to think of vector a as not moving at all in Frame S, but the basis vectors are back-rotated from en to e'n taking us to Frame S'. In this back-rotated basis the components of a are (a)'i. This is the passive view of a rotation and is in f act the view we take throughout this document because we want to observe activities in Frame S from Frame S' and vice versa. Each view has its usefulness and we have just shown that if a' = R a, then (a)' n = (a')n. In the active view, the "apparatus" is rotated and the axes stay put, while in the passive view the apparatus stays put but the axes are back-rotated. There is a th ird view in which the apparatus and the axes are both rotated in the same direction, and this view is usef ul in the discussion of covariance of equations in Lucht Tensor . One can regard the three views as three "exper iments" one might perform. As noted, we shall work with the passive view exclusively in this document. Example 3: Soon we shall be dealing with the Fig 1 equation r' = r - b. Since this is not of the form r' ≡ Rr , we may not dispense with the parentheses, and we expect that (r') i and (r)'i will be different. Footnote : More generally, if R is the linearized version of some general transformation x' = F(x) at a point x, so that d x' = R( x) dx, then (1.3.2) that a' = R a says that a "transforms as a contravariant vector with respect to the underlying transformation F ". In general R( x) is a combination of rotation and stretch and is a function of location. In our current document we d eal only with R(x) = R = a rotation that is the same at all points in space. It turns out however that the notation a' i is unambiguous in the general case as well as we shall now show. Lucht Tensor uses a different notation for basis vectors, and to make the connection between our current document and Tensor one must take { en,e'n} → { un,en} Frame S = { en} → Frame S = { un} en = R e'n → un = Rei Frame S' = { e'n} → Frame S' = { en} . Then in the language of Tensor where • is the "covariant dot product", one has (implied summations), (a')n = a' • un = (R a) • (Ren) = (R a)i(Ren)i = Rij(a)j Ri k(en)k = (RijRi k) (a)j(en)k = δ j k(a)j(en)k = (a)j(en)j = a • en = (a)'n . Section 1: Preliminaries 13 In this general case, u n are still axis-aligned basis vectors, but en are generally not axis-aligned and are generally not unit vectors. For example, in spherical coordinates e1 = r^, e2 = r θ^ and e3 = rsinθ φ^ as shown in (E.6.8). 1.4 When are two vectors equal? This topic will probabl y seem strange and unnecessary, but it has been a constant annoyance to the author so here are some words on the subject. When we say two vectors A and B are the same or are equal, we mean that the two vectors have the same components in the same coordinate system and we write A = B. This does not require that vectors A and B coincide. It might be that B is a translated copy of A. To be really fussy, we could define a stronger equality A =• B to mean that not only do the vectors ha ve the same components in the sense of A = B, but the vectors actually coincide with each other. We shall have no use for A =• B in this document. For us, two vectors are "the same" even if translated from one another. In light of this interpretation of vectors bein g equal, we can examine the meaning of certain statements. For example, we normally say "a particle is located at r in Frame S ". This really means the particle is at point r in Frame S which has coordinates (x,y,z). What this means in terms of the graphic vector r is that if the vector r is translated so that its tail is at the origin of Frame S, then its tip will be at the particle location. The vector r can be drawn anywhere in a picture. It describes the displacement of a particle in Frame S from the origin in Frame S. Example 4: When we say en = R e 'n as in (1.1.1) above, it is understood that the tails of all vectors involved (the en and the e'n) are at a common location, as in this picture (1.4.1) even though, in our application of Fig 1, the en are drawn with their tails at the origin of Frame S while the e'n are drawn with their tails at the origin of Frame S'. Example 5: In the expansion r = (r)iei we normally think of vector r having its tail at the origin of Frame S, while in the expansion r = (r')'ie'i one would be inclined to think of vector r as having its tail at the origin of Frame S'. In our stricter sense of coincidence noted above, we might say (r) i ei =•/ (r')'i e'i but this is not of interest. Wh at we care about is that (r) i ei = (r')'i e'i in the sense A = B above and we don't care if the vectors A and B are translated relative to one another. What we care about is that the vectors have the same components in any given Frame. Section 1: Preliminaries 14 1.5 The small rotation of a vector about an axis We do this first algebraically and second geometrically. In the algebrai c (linear algebra) approach, the reader must accept a few facts about rotation matric es. The 3x3 matrix which rotates a vector by angle φ about rotation axis n^ according to the right hand rule is obtaine d by exponentiating another 3x3 matrix, Rn^(φ) = exp(-i φ n^ • J) , ( 1 . 5 . 1 ) where the (J) k are 3x3 matrices known as the rotation generator matrices: J1 = ⎟⎟⎟ ⎠⎞ ⎜⎜⎜ ⎝⎛ − 0 i 0i 0 00 0 0 J 2 = ⎟⎟⎟ ⎠⎞ ⎜⎜⎜ ⎝⎛ − 0 0 i0 0 0i 0 0 J 3 = ⎟⎟⎟ ⎠⎞ ⎜⎜⎜ ⎝⎛− 0 0 00 0 i0 i 0 . The numbers in these three matrices can be summarized in this single statement, (Jk)ij = - i εkij ( 1 . 5 . 2 ) where ε is the totally antisymmetric permutation tens or defined by εabc = +1 if abc is an obtained from 123 by an even number of pairwise swaps (such as 312) εabc = -1 if abc is an obtained from 123 by an odd number of pairwise swaps (such as 213) εabc = 0 otherwise (ie, when two or more indices have the same value such as 122 or 333) (1.5.3) Note that ε abc = - εbac regardless of index value. A cross product component can be expressed in terms of the permutation tensor as [ a x b]i = εijkajbk with implied sums on j and k. It is convenient to define a vector rotation angle in this manner φ ≡ φ n^ ( 1 . 5 . 4 ) and then the rotation (1.5.1) may be written in these new ways, R n^(φ) = R(φ) = exp(-i φ • J) . (1.5.5) For a small rotation d φ = dφ n^, this may be approximated as (using ex = 1 + x + ... but applied to x = matrix) R(dφ) = exp(-i d φ • J) ≈ 1 - i dφ • J . (1.5.6) With these preliminary remarks out of the way, we can consider the rotation of a vector a by a small amount as we move from time t to time t + dt , a(t+dt) = R(d φ) a( t ) ( 1 . 5 . 7 ) Section 1: Preliminaries 15 where dφ ≡ dφ n^ is in some arbitrary direction n^ which is unrelated to the direction of the vector a(t) . The change in vector a is given by d a = a(t+dt) - a(t) = R(d φ) a(t) - a(t) ≈ [1 - i dφ • J] a(t) - a(t) = - i d φ • J a(t) = - i d φk Jk a(t) . Taking the ith component of the above equation we get dai = - i dφk[Jka]i = - i dφ k(Jk)ijaj = - i dφ k( -i εkij ) aj = - εkij dφk aj = + εikj dφk aj and going back to vector notation we find that d a = dφ x a ( 1 . 5 . 8 ) which is our main result. It tells us the change in a vector a under a small rotation d φ. Here then is a graphical derivation of this same fact for a limited geometry. Consider this picture ( 1 . 5 . 9 ) where the small rotation vector d φ points out of the plane of paper, and where a(t) happens to lie in the plane of paper (hence this picture does not cover th e most general case). Using the right hand rule for cross products, we can see that d a (shown on the right) lies in the direction of d φ x a, so we can write that da = C dφ x a where C is some constant. We can determine C from the equation |d a| = C |dφ x a| . Since dφ and a are at right angles, we know that |d φ x a| = |dφ| |a| = dφ a. But from the picture on the right it seems quite clear that |d a| ≈ a dφ. Therefore |d a| = C |dφ x a| => d φ a = C dφ a => C = 1 (1.5.10) so we end up with d a = dφ x a which agrees with (1.5.8). In the case that a does not lie in the plane of paper, the geometric derivation takes more work, and in this case we just rely on the algebraic result. We have made free use of the notion that translated vectors are "equal". Notice the fact that d a = dφ x a does not depend on the distance D between the rotation axis and the tail of vector a! The result is true even if this dist ance is 0 so that the tail of vector a lies right on the rotation axis. In this case the pair of arrows in the left picture coincides with the pair of arrows in the right picture. Section 1: Preliminaries 16 Footnote: The equation R( φ) = exp(-i φ • J) can be interpreted as a rotation in N dimensions with a set of three appropriate NxN generator matrices (J k)ij. Only when N = 3 is (1.5.2) valid or even meaningful. In group theoretic language, these NxN matrices J k form an N-dimensional irre ducible representation of the Lie Algebra so(N) which is [J a, Jb] = iεabcJc. When J is exponentiated as shown in (1.5.5), the rotations R(φ) are then elements of the Lie Group known as the Rotation Group SO(N). The meaning is Special (det = +1), Orthogonal (as in RTR= 1), and N dimensions. The vector φ then has N components. For N = 2, the generators are J k = σk/2 where σ k are the so-called 2x2 Pauli matrices associated with "spin 1/2". 1.6 The time rate of change of a rotating vector In the previous section we found that, d a = dφ x a . (1.5.8) Dividing by dt gives (d a/dt) = ω x a where ω ≡ dφ/dt . (1.6.1) This equation is of fundamental importance in this document. It describes the rotation of vector a at rate ω about an axis parallel to ω. The equation can be represented by e ither of these "cone pictures" : (1.6.2) (a) (b) Comments: 1. In (b), the tip of vector a traverses a circular path and so does the tail. The tail is always distance D from the rotation axis. One usually sees the simple r picture (a) where D = 0. Recall from above that distance D of the tail from the rotation axis is irrelevant. The motion of the vector a is exactly the same in both these cone pictures. ( Recall the comments in S ection 1.4 about "when are two vectors equal". ) 2. It is probably best to describe the rotating motion of vector a as a conical motion rather than a "circular motion", even though the tip of vector a travels in a circular moti on. In the special case that ψ = π/2, meaning the tip of the cone has moved to the center of its circular end face (and ω • a = 0), then vector a would swing around in a true "circular motion". Section 1: Preliminaries 17 3. Note that one might have ω = ω(t) so the vector ω could be changing in both direction and magnitude as time progresses. But at time t we have a definite ω^(t), and this is sometimes referred to as the instantaneous direction of rotation at time t, and ω(t) the instantaneous angular velocity. In everything below, we always think of ω in this instantaneous sense, even though we might draw guide circles and cones to show the instantaneous motion at some instant of time t. 4. Knowledge of vector ω does not in fact say where the rotation axis is located. Again we invoke the comments of Section 1.4 above. There are two translati onal degrees of freedom in that the rotation axis can be translated parallel to itself in two dimensi ons. Comparing (a) and (b) above, one sees an example of the same ω but two different rotation axes, one translated re lative to the other. In (b) the red and black ω vectors are the same vector. One normally draws the vector ω on the rotation axis as done in red. 5. As noted, in (a) and (b) the conical motion of vector a is the same and is independent of the placement of the rotation axis as long as it is parallel to ω. However , in Fig 1 there is a large change in the physical relationship between Frame S and Frame S' if the rotation axis is translated and ω stays the same. 6. In the case that ω = constant, a simple solution to (1.6.1) a• = ω x a can be obtained. This equation is a system of three coupled first-order linear ordinary differential equations. Writing a = axx^ + ayy^ + azz^ and ω = ωz^, the three equations become a •x = -ωay, a•y = ωax, and a •z = 0. Thus a••x = -ω2ax and we end up with this general solution, ax(t) = Acos( ωt) – Bsin( ωt) a x(0) = A // a x2 + ay2 = A2 + B2 = circle ay(t) = Asin( ωt) + Bcos( ωt) a y(0) = B az(t) = az(0) = C = constant a z(0) = C . (1.6.3) The vector a starts out at some a(0) = (A,B,C) and does the conical mo tion drawn above, not just in the instantaneous sense, but in the full sense so that a really goes around the entire cone. 1.7 Rate of change of the basis vectors We can apply our rate of change rule (1.6.1) to the basis vectors e'n to obtain (d e'n/dt ) = ω x e'n . We are implicitly computing this derivative while "standing" in Frame S. That is to say, looking at Fig 1 in the Introduction above, we can regard Frame S as being fixed to the paper and Frame S' is rotating and its basis vectors e'n are changing as stated above. It is extrem ely important to make this fact explicit, so we now add a label S showing this fact, (d e'n/dt)S = ω x e'n . (1.7.1) Were we to compute this same deriva tive standing in Frame S', we would get (d e'n/dt)S' = 0 ( 1 . 7 . 2 ) Section 1: Preliminaries 18 because in Frame S' the basis vectors e'n are not rotating at ω, but are just sitting there frozen. By the same argument, we know that (d en/dt)S = 0 . ( 1 . 7 . 3 ) What about the fourth possible derivative (d en/dt )S' ? We will show in (2.5) below that in fact (d en/dt)S' = – ω x en . ( 1 . 7 . 4 ) It seems at least reasonable that if the e'n are rotating relative to the en by ω, then the en are rotating relative to the e'n by -ω. Main Conclusion : We thus arrive at the notion that, when dealing with multiple frames of reference and vectors represented in terms of their respective basis v ectors, we absolutely must indicate with a label the frame in which a time derivative is being calculated. 1.8 Notations for the many time d eri vatives of vectors r, r', b and L In the Overview we described r and r' as the position vectors of a Particle with respect to Frames S and S'. We can associate with these two vectors four different time derivatives. (d r/dt)S (d r/dt)S' ( d r'/dt)S (d r'/dt)S' (1.8.1) In the usual manner, we represent a time derivativ e of a vector by an over-dot, and a second time derivative by an over-double-dot. The first time derivative of position r is velocity v, and the second is acceleration a. The same for r', v' and a' , so v = r• v' = r•' a = v• = r•• a' = v•' = r••' . (1.8.2) In order to save space, we can define these two operators ∂S ≡ (d/dt)S ∂S' ≡ (d/dt)S' ( 1 . 8 . 3 ) Although the ∂ symbol is used in partial differentiation, our use here is exactly as defined above and so our ∂ is not a partial derivative (except in a very abstract sense). The list of derivatives above can be written now in several ways. Within each column below, all objects are exactly the same thing, just written in different notations: ( r in Frame S appears as r' in Frame S' ) Section 1: Preliminaries 19 Table of first derivatives of r Table of first derivatives of r' (d r/dt)S (d r/dt)S' ( d r'/dt)S (d r'/dt)S' r•S ≡ r• r•S' r•'S r•'S' ≡ r•' vS ≡ v vS' v'S v'S' ≡ v' ∂Sr ∂S'r ∂ Sr' ∂S'r' . (1.8.4) For further clutter reduction, we have added the four new notations shown in red according to the following rule: When a vector and all its derivatives (here only one) are all in (or associated with) the same frame, we suppress the frame subscript and just let the prime or lack of it "do the talking", and refer to such as object as being "natural". There is no notational ambiguity in so doing. The velocities shown in the center two columns above, vS' = (dr/dt)S' and v'S = (dr'/dt)S, shall be referred to as "cross velocities", as dis tinct from the two "natural velocities". What about second derivatives? Things ar e more complicated now because vector r can have four distinct second derivatives, and so can vector r'. Just as done above, we construc t a table for each, and within each column of each table, all objects are the same object : ( r in Frame S appears as r' in Frame S' ) Table of second derivatives of r ∂S∂Sr ∂S∂S'r ∂S'∂Sr ∂S'∂S'r r••SS ≡ r••S ≡ r•• r••SS' r••S'S r••S'S' ≡ r••S' v•SS ≡ v•S ≡ v• v•SS' v•S'S v•S'S' ≡ v•S' aSS ≡ aS ≡ a aSS' aS'S aS'S' ≡ aS' (1.8.5) Table of second derivatives of r' ∂S∂Sr' ∂S∂S'r' ∂S'∂Sr' ∂S'∂S'r' r••'SS ≡ r••'S r••'SS' r••'S'S r••'S'S' ≡ r••'S' ≡ r••' v•'SS ≡ v•'S v•'SS' v•'S'S v•'S'S' ≡ v•'S' ≡ v•' a 'SS ≡ a'S a'SS' a'S'S a'S'S' ≡ a'S' ≡ a' (1.8.6) Here we again use the rule mentioned above that when a vector and all its derivatives are in the same frame, we let the overall prime or lack of it do the ta lking. We have introduced a second rule as well, which says that whenever both frame subscripts are the same, we suppress one of them to save space. In the tables above we then have two "natural" accelerations a and a', and six "cross accelerations". So now we have the following "natural" vectors havi ng minimal (that is, no) frame subscript clutter: r• v r•• v• a natural in Frame S r•' v' r••' v•' a ' natural in Frame S' (1.8.7) Section 1: Preliminaries 20 Recall from the Introduction that the vector b connects the origins of two frames. We can make a table of first and second derivatives for this vector as well. Although b• is a velocity and b•• is an acceleration, we shall not make up separate symbol names for these objects (though some authors do). Also, note that there is no vector in Fig 1 called b', we just have r = r' + b. Here are the corresponding tables for first and second derivatives of b, where we use only the second rule above that when two frame subscripts are the same we suppress one of them. Table of first derivatives of b . (Items are the same within each column.) (d b/dt)S (d b/dt)S' b• S b• S' ∂Sb ∂S'b ( 1 . 8 . 8 ) Table of second derivatives of b ∂S∂Sb ∂S∂S'b ∂S'∂Sb ∂S'∂S'b b•• SS ≡ b•• S b•• SS' b•• S'S b•• S'S' ≡ b•• S' (1.8.9) Comment: Just as a reminder, any derivative in the above section can be expanded onto either Frame S or Frame S' basis vectors, so any derivative has Frame S and Frame S' components. This statement would be true for any vector and we just remind the reader that the notation like (d r/dt)S does not mean the components are evaluated only in Frame S. Just as an example, this first derivative has these two expansions where the expansion coefficien ts (components) are shown on the right, (d r/dt)S = [(d r/dt)S]i ei [(d r/dt)S]i = (d r/dt)S • ei (d r/dt)S = [(d r/dt)S]'i e'i [(d r/dt)S]'i = (d r/dt)S • e'i . (1.8.10) Section 1: Preliminaries 21 1.9 Angular momentum The drawing of interest is a reoriented Fig 1, to which we have added arbitrary points c and c', (1.9.1) Whereas the linear momentum p = mv of a particle does not require a reference point, angular momentum L does require such a reference point. For example, th e Particle in the above figure has many different values of L in Frame S, some of which we might denote as follows, L(0) = r x mv // L in Frame S with respect to Frame S origin L(c) = (r-c) x mv // L in Frame S with respect to Frame S point c L(b) = (r-b) x mv // L in Frame S with respect to Frame S point b L(b) = r' x m v . // L in Frame S with respect to Frame S' origin (1.9.2) The last two lines are exactly the same since b is a vector between the two origins so r'= r-b. Using the general form L(c), we may identify the following two "natural" angular momenta in Frames S and S', L(c) S ≡ (r-c) x mvS = ( r-c) x mv = ( r-c) x p ≡ L(c) L'(c') S' ≡ (r'-c') x mv'S' = (r'-c') x mv' = (r'-c') x p' ≡ L'(c') . (1.9.3) These definitions are analogous to p ≡ mv and p' ≡ mv' in the linear momentum world. The time derivative of the first of these objects is given by L•(c) = L•(c) S = ∂S L(c) S = ∂S [ (r-c) x mvS] = (vS - c•S) x mvS + (r-c) x ∂S(mvS) = – c•S x mvS + (r-c) x maS = – c• x mv + (r-c) x ma . Section 1: Preliminaries 22 A similar result is obtained by priming everything on the above line, so we end up with these four equations: L(c) = ( r-c) x mv (1.9.4) L•(c) = ( r-c) x ma – c• x mv ( 1 . 9 . 5 ) L'(c') = (r'-c') x mv' ( 1 . 9 . 6 ) L•'(c') = (r'-c') x ma' – c•' x mv' . ( 1 . 9 . 7 ) If one thinks of L(c) = (r-c) x mv just as a cross product of two vectors to get a third vector, one can apply our Theorem (A.20) to write ( L(c))' = ( r'-c') x mv' and therefore ( L(c))' = L'(c'). It is just another notation. The discussion of angular momentum continues in Section 11. 1.10 No frame label is needed for d/dt of a scalar function If one is differentiating a specific component of a vector, such as a i(t) = 3t2, there is no need to add the frame label since the derivative of this function is 6t no matter what frame it is computed in. This is true when differentiating any component of any tensor. That is to say, if T ij..(t) is a component of a tensor, then (dT ij..(t)/dt)S = (dTij..( (t)/dt)S' = (dTij..(t)/dt) . (1.10.1) We would like simply to say that the d/dt derivative of any scalar function does not need a label S or S', but the word "scalar" has multiple meanings. In one meaning, any single function f(t) is a scalar function since it is a 1-tuple of functions, but in another mean ing (tensorial scalar), only a function which is a rotational scalar is a scalar function, and this would rule out the component of a vector as being a scalar function. We refer to the first meani ng in the title of this subsection. Section 1: Preliminaries 23 1.11 When do operations d/dt and taking a component "commute" ? We claim the following theorem (to be proved below) : Commutation Theorem: ( 1 . 1 1 . 1 ) If the time derivative ( ∂X) is computed in the same frame (Frame X) in which components are taken, then the operations of taking the time derivative and taking the component can be done in either order with the same result, so these operations are said to commut e. Otherwise the operations do not commute. Thus (∂ Sa)j = ∂S(a)j but ( ∂Sa)'j ≠ ∂S(a)'j (∂S'a)'j = ∂S'(a)')j (∂S'a)j ≠ ∂S'(a)j . In each of these four equations, on the left we compute ∂ Xa and then we take a component, while on the right side we take a component and then compute ∂X on that component. Goldstein makes a point about the non-commuting cases as a "word of caution" on the bottom of page 133 with an example on the top of page 134. In Goldstein, Poole and Safko the caution is stated on page 173 below (4.86), but the example has been removed. As a preliminary to proving the theorem, we note that for a scalar u and vector v one can write, (d[u v]/dt)S = (du/dt) v + u (d v/dt)S S (d[u v]/dt)S' = (du/dt) v + u (d v/dt)S' S' or ∂ S(uv) = (∂Su) v + u (∂Sv) S ∂S'(uv) = (∂S'u) v + u (∂S'v) S ' ( 1 . 1 1 . 2 ) where recall that ∂ S = ∂S' = ∂t when applied to a scalar like u. Within either frame this is just the calculus Leibniz product rule applied to as function u(t) times a vector function v(t), ∂ t(uv) = (∂tu) v + u (∂tv) , so (1.11.2) is just a statement of this known rule in the two frames. One can trivially generalize (1.11.2) to a sum of the form u ivi (implied sum on i) to get ∂ S(uivi) = (∂Sui) vi + ui (∂Svi) S ∂S'(uivi) = (∂S'ui) vi + ui (∂S'vi) S ' ( 1 . 1 1 . 3 ) Reader Exercise : Show that ∂X(a x b) = (∂Xa) x b + a x (∂Xb) for X = S or S' (more Leibniz). (1.11.4) Proof of the Commutation Theorem: In the proof we use all the following facts developed above and gathered below for convenience. Notice that it is not assumed that a' = R a and in fact the vector a' does not appear anywhere. Section 2: The G Rule 24 (R-1)ij = (RT)ij = Rji rotation is real orthogonal (1.1.4) (1.11.5) Rij(t) = function of time so ( ∂tRij(t)) ≠ 0 and (∂t(R-1)ij(t)) ≠ 0 (ω ≠ 0) "rotating frames" a = (a)i ei = (a)'i e'i expansions in the two frames (1.2.2) ( en)i = δn,i ( e'n)'i = δn,i ( en)'i = (R-1)ni ( e'n)i = Rni basis vector components (1.2.3) (a)'i = Rij(a)j . components of a in the two frames (1.3.1) ∂Sen = 0 and ∂S'e'n = 0 frame definitions (1.7.3), (1.7.2) ∂S f(t) = ∂S' f(t) = ∂ t f(t) time derivative of a component (1.10.1) ∂S(uivi) = (∂Sui) vi + ui (∂Svi) product rule in frame S ∂S'(uivi) = (∂S'ui) vi + ui (∂S'vi) product rule in frame S' (1.11.3) Here then is the detailed proof of the above theorem. First, consider ∂S with a = (a)i ei : (∂Sa)j = (∂S[(a)iei])j = (∂S(a)i) (ei)j = (∂t(a)i) δi,j = ∂t(a)j ∂S(a)j = ∂t(a)j ⇒ ∂S(a)j = (∂Sa)j // commutes ( ∂S with Frame S components) (∂Sa)'j = (∂S[(a)iei])'j = (∂S(a)i) (ei)'j = (∂t(a)i) Rji = Rji(∂t(a)i) (∂S(a)'j) = ∂S[Rji(a)i] = Rji(∂S(a)i) + (∂SRji) (a)i = Rji(∂t(a)i) + (∂tRji) (a)i ⇒ (∂S(a)'j) = (∂Sa)'j + (∂tRji) (a)i ⇒ (∂S(a)'j) ≠ (∂Sa)'j // does not commute ( ∂S with Frame S' components) Next consider ∂ S' with a = (a)'i e'i : (∂S'a)j = (∂S'[(a)'ie'i])j = (∂S'(a)'i) (e'i)j = (∂S'(a)'i) Rij = (R-1)ji (∂t(a)'i) (∂S'(a)j) = ∂S'[(R-1)ji(a)'i] = (∂t(R-1)ji)(a)'i + (R-1)ji(∂t(a)'i) ⇒ (∂S'(a)j) = (∂S'a)j + (∂t(R-1)ji)(a)'i ⇒ (∂S'(a)j) ≠ (∂S'a)j // does not commute ( ∂S' with Frame S components) (∂S'a)'j = (∂S[(a)'ie'i])'j = (∂S'(a)'i) (ei')'j = (∂S'(a)'i) δi,j = (∂S'(a)'j) = (∂t(a)'j) ∂S'(a)'j = ∂t(a)'j ⇒ ∂S'(a)'j = (∂S'a)'j // commutes ( ∂S' with Frame S' components) Section 2: The G Rule 25 2. The G Rule for arbitrary vector a and its derivation In this section, a is a generic vector -- it could be any vector. The "G Rule for vector a" is this (written various equivalent ways) ∂Sa = ∂S'a + ω x a (d a/dt)S = (da/dt)S' + ω x a a•S = a•S' + ω x a ( a•S)i = ( a•S')i + iεijkωjak . // components (2.1) In the first line we use the abbreviations ∂X = (d/dt) X where X is a frame of re ference. G is in honor of Herbert Goldstein since the Rule appears on page 133 of his classic 1950 book Classical Mechanics (and he uses the vector G instead of our a). Perhaps another name for this rule would be "the ru le which relates the time derivatives of a vector taken in two frames of reference S and S' where Frame S' is rotating relative to Frame S at instanta neous vector angular velocity ω about some unspecified rotation axis which is parallel to the vector ω". Equation (2.1) goes by a few obscure names, some people calling it a "transport theorem" or a "basic kinematic equation", but most authors who use it give it no name. Comment : The content of (2.1) must have been known to Coriolis (see Gaspard-Gustave) in 1835, but the cross product notation was not in use at that time. According to Crowe, the cross product idea arose gradually from the work of Hamilton (quaternions) a nd Grassmann (vector areas) as early as the 1840's, and later from work of Gibbs in the 1880's, with th e possible involvement of a certain Reverend O'Brien in 1852. In any event, the cross product notation (a nd bolded vector notation in general) was first introduced to the textbook-reading public when Gibbs ga ve his student Wilson the task of publishing and improving his notes. Their book Vector Analysis , published in 1901, was reprinted 7 times and was then made a Dover book in 1960 and can be found online (see Refs). It happens, however, that this book makes no mention of the G Rule or Coriolis forces or "frames of reference". It is a math book, not a physics book. Since the G Rule applies to any vector a, sometimes the vector is left out and one writes this operator equation, (d/dt) S = (d/dt) S' + ω x . (2.2) Here is where and how the G rule appears in some popular mechanics texts: (d G/dt)space = (dG/dt)body + ω x G // Goldstein p 133 (4-100) // Goldstein Poole Safko p 172 (4.82) (d Q/dt)fixed = (dQ/dt)rotating + ω x Q // Marion p 342 (11.7) // Thornton Marion p 390 (10.12) (d Q/dt)S0 = (dQ/dt)S + ω x Q // Taylor p 342 (9.30) (d/dt)space = (d/dt) body + ω x // Goldstein p 133 (4-102) (d/dt) s = (d/dt) r + ω x // Goldstein Poole Safko p 173 (4.86) Section 2: The G Rule 26 Proof of the G Rule : We want to show that ∂Sa = ∂S'a + ω x a . (2.1) Start with the known fact (1.7.1), ∂Se'i = ω x e'i . (1.7.1) Then using the above equation and facts collected in (1.11.5) we find, ∂Sa = ∂S[(a)'ie'i] = (∂S(a)'i)e'i + (a)'i(∂Se'i) = (∂S(a)'i)e'i + (a)'i ω x e'i = ( ∂S(a)'i)e'i + ω x [(a)'ie'i] = (∂S(a)'i)e'i + ω x a . (2.3) But, ∂S'a = ∂S'[(a)'ie'i] = (∂S'(a)'i) e'i since ∂S'e'i = 0 . (2.4) Inserting this into (2.3) then gives ∂Sa = ∂S'a + ω x a which is the desired G rule (2.1). QED Comment : The G Rule involves two frames of reference calle d Frame S and Frame S'. It is true that in Fig 1 and Fig (1.9.1) we think of Frame S as being "glued to the paper", but there is nothing that says the paper might not be rotating about some other axis ω'. In other words, even if Frame S and Frame S' are both non-inertial frames, the G Rule is still valid because nothing in its derivation requires that either frame be inertial. Inertial only matters when we later start talking about F = m a. The G Rule involves Frame S' rotating by ω relative to Frame S. An implication is that any equation obtained below by use of the G Rule is also valid even if both Frame S and Frame S' are non-inertial. Example 1: (reversing the above) Suppose a = e'i . Then our rule (2.1) says (d e'i/dt)S = (d e'i/dt)S' + ω x e'i . But we noted in (1.7.2) the obvious fact that (d e'i/dt)S' = 0, so the above becomes (d e'i/dt)S = ω x e'i which agrees with (1.7.1). Example 2: Suppose a = ei . Then rule (2.1) says (d ei/dt)S = (d ei/dt)S' + ω x ei . But we noted in (1.7.3) the obvious fact that (d ei/dt)S = 0, so the above becomes Section 2: The G Rule 27 (d ei/dt)S' = – ω x ei ( 2 . 5 ) and we have now derived the claim made in (1.7.4). Example 3: Suppose a = ω. Then rule (2.1) says (dω/dt)S = (dω/dt)S' + ω x ω = (dω/dt)S' so for this one vector ω both derivatives are the same and we write (dω/dt)S = (dω/dt)S' ≡ dω/dt = ω• . (2.6) Example 4. Using the dot notation of (1.8.2), and assuming there are two related vectors a and a', the G Rule states a•S = a•S' + ω x a that is ∂Sa• = ∂S'a• + ω x a a•'S = a•'S' + ω x a' that is ∂Sa•' = ∂S'a•' + ω x a' . (2.7) Example 5. We can apply this rule to any of the vectors listed in Section 1.8. Here are a few examples: b• S = b• S' + ω x b ( 2 . 8 ) r•S = r•S' + ω x r or vS = vS' + ω x r r•'S = r•'S' + ω x r' o r v'S = v'S' + ω x r' (2.9) v•S = v•S' + ω x vS o r aS = aS' + ω x vS v•'S = v•'S' + ω x v'S or a'S = a'S' + ω x v'S . (2.10) Question : Are there any restrictions on the vector a for which the G Rule applies? The only fact used above about a is that a can be expanded on Cartesian basis vectors as a = a'i e'i and that the component functions a' i(t) are differentiable. If the tail of vector a does not lie at the origin of Frame S', we just translate a such that this is the case, as per Section 1.4. Significance of the G Rule: In all the computations below, basically the only operations done are these: • Apply the G Rule to some vector • Apply ∂S to both sides of some equation • Apply ∂S' to both sides of some equation Note: In Appendix B we generalize the G Rule to tensors of arbitrary rank, so the G Rule of this Section is the general case applied to rank 1 tensors (vectors). Section 3: Rotating Observer 28 3. The Apparatus and its Observer at Rest in Frame S' In our general "experiment" to be de scribed below, Frame S' is rotating with respect to Frame S. This does not necessarily imply that Frame S is "at rest", but Frame S is at rest with respect to the paper on which we draw Fig 1. If Frame S is truly at rest with respect to the stars, then Frame S is called an inertial frame, and in such a frame Newton's 2nd Law F = ma is valid. We do not in general assume that S is such an inertial frame. Imagine now that we have some Apparatus sitti ng in Frame S' which contains a Particle which undergoes some motion. The Particle might be a mass on one or more springs, or it might be a Particle in ballistic flight, or it might be a Particle of matter in a gear wheel which is turning in some complicated machine, or it might be a Particle of a fluid or of an elastic solid. An Observer also sitting (at rest) in Frame S' ha s some measurement equipment, can see the axes of Frame S' of course, and does certa in measurements on the Particle while Frame S' is rotating with respect to Frame S . The Particle is located at position r in Frame S and position r' in Frame S'. The Observer can see Frame S and is aware of both r and r' and can make measurements of them both. For example, although r is the position vector of the Particle relative to the Frame S origin, our Observer measures its components in Frame S' this way r = (r)'ie'i (r)' i = r • e'i Meanwhile, measurements of r' reveal these different components r' = (r')'ie'i (r')' i = r' • e'i . However, the Observer can only measure frame-S' derivatives. From our list we then select some items : vS' = r•S' = (d r/dt)S' // cross velocity v'S' = r•'S' = (d r'/dt)S' ≡ v' // natural (1.8.4) aS' = v•S' = (d vS'/dt)S' = (d2r/dt2)S' // cross acceleration a'S' = v•'S' = (d v'S'/dt)S' = (d2r'/dt2)S' ≡ a' // natural (1.8.6) b• S' = (d b/dt)S' rate of change of b as viewed from Frame S' . (1.8.9) For example, the Observer might measure the frame-S Particle location r(t) at time t, wait one time tick dt, then measure it again to get r(t+dt). Then (the components of r(t) are [ r(t)]'i ), vS'(t) = [ r(t+dt) – r(t)]/(dt) or vS'(t) = (d r/dt)S' // cross velocity Alternatively, the Observer could measure r'(t) and r'(t+dt) and get v'S'(t) = [ r'(t+dt) – r'(t)]/(dt) or v'S'(t) = (d r'/dt)S' = v' . // natural Section 3: Rotating Observer 29 Waiting another dt tick, one could measure r(t+2dt) and r'(t+2dt) and then deduce vS'(t+dt) and v'S'(t+dt). From these the Observer could determine aS'(t) = [ vS'(t+dt) - vS'(t)]/(dt) or aS'(t) = (d vS'/dt)S' // cross a'S'(t) = [ v'S'(t+dt) - v'S'(t)]/(dt) or a'S'(t) = (d v'S'/dt)S' = a' . // natural As a final example, here the Observer measures a cross angular momentum and a natural one : L(c) = (r-c) x mv (1.9.4) L•(c) S' = ∂S'L(c) = [L(c+dt)(t+dt) - L(c)(t)]/(dt) // cross L'(c') = (r'-c') x mv' (1.9.6) L•'(c') S' = ∂S'L'(c') = [L'(c'+dt)(t+dt) - L'(c')(t)]/(dt) = L•'(c') . // natural Section 4: Relationship between S and S' 30 4. The Relationship between the Two Frames S and S' 4.1 Explanation of Fig (4.1.1): Frame S in the plane of paper The relation between the two frames is shown in this picture, a snapshot at some time t : ( 4 . 1 . 1 ) There is a lot to be said about this picture (whi ch is the same as Fig 1 in the Overview) . The axes e2 and e3 of Frame S lie in the plane of paper and ar e "aligned with paper" as shown and remain fixed relative to paper, so the Frame S origin lies in th e plane of paper. The origin of Frame S' is displaced by amount b from the origin of Frame S, and this S' origin does not lie in the plane of paper. The location of the Particle does not lie in the plane of paper, so vectors r, r' and b , although coplanar with each other, are each not in the plane of paper. Similarly, the ω rotation axis does not lie in the plane of paper nor is it parallel to it. Frame S' is instantaneously rotating about some axis indicated by ω(t). Each of the basis vectors e'n is rotating according to (1.7.1), (d e'n/dt)S = ω x e'n, as the Frame S' moves rigidly in rotation about the ω rotation axis. The origin of Frame S' is instantaneously rotating along the green circle of some instantaneous radius r ω which has its center on the rotation axis at a green dot. This green dot, meanwhile, is moving at some velocity vωpar in the plane of the green circle re lative to Frame S. This indicates the motion of the rotation axis parallel to itself. Suppose Frame S' contains a rigid object fixed relative to the Frame S' axes. If we were to select some point P in that rigid object, that point P would be instantaneously rotating about the ω(t) axis along a circle similar to the one shown a bove, but which has a different radius and a different center point along the same rotation axis. If ω^ were constant in time and if vωpar were 0, the origin of Frame S' really would move along the full green circle shown, but we have in mind that ω = ω(t) and this varies in time, both in magnitude and direction. Thus, the green circle is itself tilting to stay in a plane perpendicular to ω(t). Viewed from Frame S, the unit vectors of Frame S' are oriented and move according to several equations we have already dealt with, Section 4: Relationship between S and S' 31 e'n(t) = R-1(t) en (1.1.1) e'n(t) = Rnm(t) em (1.1.2) (d e'n/dt)S = ω(t) x e'n( t ) . (1.7.1) 4.2 Explanation of Fig (4.2.1) : Vector ω pointing directly out of paper We now draw Fig (4.1.1) from a different perspective. The reader will hopefully forgive the "artist" for not attempting to draw Fig (4.2.1) as a precision 3D rotated version of Fig (4.1.1), but hopefully the general features of the drawing are sufficient for our purposes below. ( 4 . 2 . 1 ) Frame S remains fixed relative to paper as time varies , but is now rotated relative to Fig (4.1.1) and the S origin no longer lies in the plane of paper. At time t for which the picture is drawn, the origin of Frame S' and the green circle and its center do lie in the plane of paper. The ω vector points straight out at the viewer as indicated by the circled dot. Vectors r, r' and b do not lie in the plane of paper. Fig (4.1.1) is meant to give the general lay of th e land, but Fig (4.2.1) is the one we will work with below. 4.3 Comments on b• S and b• S' The quantity b• S = (d b/dt)S describes the instantaneous velocity of the Frame S' origin relative to that of Frame S. When ω = 0, Frame S' merely translates relative to Frame S with position b, velocity b• S and acceleration b•• S . When ω ≠ 0, however, for a general placement of the rotation axis one can regard b, Section 4: Relationship between S and S' 32 b• S and b•• S as quantities derived from the location and move ment of that axis and from the value of ω, all of which one imagines are controlled by some mechan ical outside agency. An exception is Special Case #2 below where the rotation axis passes through the Frame S' origin. The quantity b• S' = (d b/dt)S' is a "cross velocity" in the sense of Se ction 1.8 and is difficult to interpret. In Special Case #1 below, however, the vector b is effectively glued to the Frame S' axes and b• S' = 0, causing simplification of several equati ons which will be obtained below. The connection between b• S and b• S' is provided by the G Rule (2.1) applied to vector b, b• S = b• S' + ω x b . ( 4 . 3 . 1 ) In Section 7.5 we derive the re lationship between accelerations b•• S = ∂Sb• S and b•• S' = ∂S'b• S' . 4.4 Special Case #1 : ω axis through Frame S origin The rotation axis always passes through the origin of Frame S, so Fig (4.2.1) appears as follows, Special Case #1 ( 4 . 4 . 1 ) The Frame S' origin and the green circle are still in the plane of paper, but the Frame S origin is in general not, so vector b is not in the plane of paper. In this situation, the vector b and the vectors e'n all rotate Section 4: Relationship between S and S' 33 together as if they were thin metal rods soldered together. If b" = R b and e"n = Re'n then angles are fixed, as indicated by b" • e"n = [R b] • [Re'n] = b • e'n using (1.2.4) and R = R ω(dφ). See Fig (1.5.9). This has the immediate implication that, just as (d e'n/dt)S' = 0 in (1.7.2), here we have b• S' ≡ (db/dt)S' = 0 . ( 4 . 4 . 2 ) Therefore, from (4.3.1), we have instantaneous conical motion for b as seen in Frame S, b• S ≡ (db/dt)S = ω x b . // Special Case #1 (4.4.3) (4.4.4) This says that, as seen in Frame S, the change in b is always perpendicular to b, so the length of b does not change. We still allow ω = ω(t) and, as ω^(t) changes, the tip of vector b (which is the origin of Frame S') describes some path on the surface of a sphe re of radius b in Frame S. Here we allow ω^(t) to change, but the rotation axis must always pass through the Frame S origin. 4.5 Special Case #2 : ω axis through Frame S' origin Here the rotation axis always passes through the origin of Frame S' : Special Case #2 ( 4 . 5 . 1 ) In this case our Fig (4.2.1) green circle has shrunk around the Frame S' origin. In this situation, we can think of the "driving parameters" being the three parameters of b and the three parameters of ω giving the 6 independent (Galilean) parameters defining the inst antaneous relationship between the frames. Think of Section 4: Relationship between S and S' 34 Frame S' as a camera platform which is supported on a boom b(t) and which independently controls its own orientation and rotation ω(t). Then b• S and b•• S are determined by b(t), and b• S' is given by (4.3.1) as b• S' = b• S – ω x b . ( 4 . 5 . 2 ) We now look at three sample applications. The first two fall into the Special Case #1 category, while the third is Special Case #2. 4.6 The Turntable Although obsolete, the phonograph turntable continues to provide an excellent visualization of rotating frames. It turns slowly enough for on e to actually see it turn, it is fairly large, and the surface is not shiny and completely featureless. It is believed that throughout history such turntables normally rotated clockwise in both hemispheres of the Earth, but in our drawings below we shall think of our turntable as turning counterclockwise relative to the little spindle sticking up. (4.6.1) An ant (Particle) is crawling around on such a turntable which is rotating with some ω(t). The Frame S' is set up some distance b = | b(t)| from the spindle, and has its e'1 axis pointing "to the right" and its e'2 axis pointing to the spindle. That is to say, e'1 = θ^ and e'2 = -r^ if we think of r,θ as polar coordinates for fixed Frame S. In this application, ω(t) = ω(t)e3 and the rotation axis passes though the origin of Frame S, so this is a Special Case #1 situation with ω^(t) = e3 being constant. Conversely, vector b(t) maintains its magnitude b, but b^(t) rotates about the spindle at angular rate ω(t) as seen from Frame S which is at rest. Section 4: Relationship between S and S' 35 (4.6.2) This type of Frame S' designation would be familiar to merry-go-round riders hanging on and facing the center. While once in Sydney, the author rode "The Roto r" for which Frame S' can be regarded as being at the cylindrical wall to which the rider is glued by centrifugal force (the circular floor then drops away). ( w i k i ) ( 4 . 6 . 3 ) In Section 15 we shall solve three "ant on turntable" problems using the above geometry. 4.7 The Earth The non-rotating Cartesian Frame S is located at Earth center with e3 pointing to the North Pole. The e1 axis points out through a point on the equator to some fixed distant star (Star 1), and then e2 is the third Cartesian axis which points out through a different poin t on the equator to Star 2. If we ignore the Earth's orbiting around the Sun, and the motion of the Orion spiral galactic arm containing the Sun, and the motion of the Milky Way galactic center away from the "Dipole repeller" and so on, then Frame S is an inertial frame. Section 4: Relationship between S and S' 36 Rotating Frame S' has its origin at some arbitrar y fixed point on the surface of the Earth. For this system, the e'3 axis points "up", meaning in the r^ direction for Frame S spherical coordinates. The e'1 axis points to the east, and the e'2 axis points north. The Earth rotates at some ω = ωe3 with ω > 0. Since the ω axis passes through the Frame S origin, this is another Special Case #1 application. ( 4 . 7 . 1 ) Figure (4.7.1) and the discussion of this section are in the "non-swap" notation mentioned in the Section 1 summary at the start of this document. In pr actice, one usually uses the "swap" notation S ↔S' for Earth problems in order to avoid the appearance of primes in equations. This will be done in Appendix C concerning the Foucault pendulum. 4.8 The Flying Camera Platform Some Apparatus is located in Frame S instead of S' , and Frame S' is a "camera platform" which flies around in some complicated way and observes th e activity in Frame S. In this case, both ω(t) and b(t) would be under the command of the pilot of the camera platform. One would set this up as a Special Case #2 situation, so ω passes through the Frame S' origin and b• S is the velocity of the platform origin and b(t) is its location relative to Frame S. One would use the "inverse problem" equations of Section 13 to get the primed quantities in terms of the unprimed ones. Section 5: Goal 37 5. The Goal of the next two sections The symbols appearing here are defined in Section 1.8. An Observer in Frame S' measures various properties of a Particle in motion, r' v' a' L'(c') L•'(c') We want to know how these properties of the Particle appear in Frame S, r v a L(c) L•(c) and we want to know the various other v and a forms of Section 1.8 in terms of the basic Frame S' objects listed above. Later in Section 13 we will want to know how to solve "the inverse problem" of finding the primed quantities if the unprimed ones are known. Section 6: Velocities 38 6. Determination of velocities The notations used here are described in Section 1.8. There are four distinct ve locities, and we want to express three of them in terms of the fourth which is the natural velocity in Frame S', (d r'/dt)S' = v'S' ≡ v' . We shall make frequent use of the Fig 1 relationship between r and r', r = r' + b ( 6 . 1 ) as well as the G Rule for vector b, as in (4.3.1) b• S = b• S' + ω x b . ( 6 . 2 a ) Inserting (6.1) as b = r - r' into (6.2a) gives identity b• S + ω x r' = b• S' + ω x r . ( 6 . 2 b ) 6.1 Velocity v S' Apply (d/dt) S' to (6.1) to get (6.3a), then use (6.2a) to get (6.3b) : vS' = v' + b• S' ( 6 . 3 a ) vS' = v' + b• S – ω x b . ( 6 . 3 b ) 6.2 Velocity v ≡ vS Apply (d/dt) S to (6.1) to get ∂Sr = ∂Sr' + ∂Sb or v = v'S + b• S . ( 6 . 4 ) Now use the G Rule (2.1) for vector r', ∂Sr' = ∂S'r' + ω x r', to get v'S = v' + ω x r' . ( 6 . 5 ) Insert (6.5) into (6.4) to get the first line below. Th e second and third lines make use of (6.2a) and (6.1) . v = v' + ω x r' + b• S ( 6 . 6 a ) v = v' + ω x r' + b• S' + ω x b ( 6 . 6 b ) v = v' + ω x r + b• S' . ( 6 . 6 c ) Section 6: Velocities 39 6.3 Velocity v' S Solve (6.4) for v'S, v'S = v – b• S . ( 6 . 7 ) Then solve (6.6a) for v - b• S and install into (6.7) to get the first line below, v'S = v' + ω x r' ( 6 . 8 a ) v'S = v' + ω x r – ω x b ( 6 . 8 b ) v'S = v' + ω x r + b• S' - b• S . ( 6 . 8 c ) The remaining two lines come from using (6.1) and (6.2a). 6.4 Velocity Summary v = v' + ω x r' + b• S = v' + ω x r + b• S' (6.6a,c) vS' = v' + b• S – ω x b = v' + b• S' (6.3b,a) v'S = v' + ω x r' (6.8a) (6.9) Thus we have expressed the three velocities vS = v, vS' and v'S in terms of v'S' = v'. 6.5 Velocities for Special Cases These cases were discussed above in Section 4.4 and 4.5. For Special Case #1 , where the ω axis passes through the Frame S origin, in any equations above set b• S' = 0 // (4.4.2) b• S = ω x b = ω x(r – r') = ω x r – ω x r' . // (4.4.3) (6.10) Then the Section 6.4 summary becomes v = v' + ω x r Special Case #1 only (6.6a) vS' = v' Special Case #1 only (6.3b) v'S = v' + ω x r' . general (6.8a) (6.11) For Special Case #2 , where the ω axis passes through the Frame S' origin, b• S is a driving parameter, so we select just the b• S forms from the summary Section 6: Velocities 40 v = v' + ω x r' + b• S general (6.6a) vS' = v' + b• S – ω x b general (6.3b) v'S = v' + ω x r' . general (6.8a) (6.12) 6.6 Comments 1. Consider these two results from a bove (picked more or less at random) r = r' + b (6.1) v = v' + ω x r + b• S' . (6.6c) Either equation can be "evaluated" in either Fr ame S or Frame S'. Evaluation in Frame S gives (r) i = (r')i + (b)i (v)i = (v')i + εijk(ω)j(r)k + (b• S')i while evaluation in Frame S' gives (r)'i = (r')'i + (b)'i (v)'i = (v')'i + εijk(ω)'j(r)'i + (b• S')'i . // ( ω)'j = (ω )j This is a situation where we fully expect to have (r') i ≠ (r)'i and (v') i ≠ (v)'i as mentioned in Section 1.2, so the careful placement of primes is important. This is just a reminder. 2. Based on the equations above , it is clear that we have r' ≠ Rr v' ≠ Rv where R is the rotation appearing in (1.1.1) which relates our two frames, en = R e'n . Therefore, in general the pairs of vectors ( r,r') and ( v,v') are not "vectors under rotations" in the sense of (1.3.2). 3. On the other hand, the vectors ω and b appearing in these formulas are normal "vectors under rotations" in the sense of (1.3.2), since we just define b' and ω' by these equations, b' = R b ω' = Rω so (b')i = Rij(b)j = (b)'i = b'i ( ω')i = Rij(ω)j = (ω)'i = ω'i (6.13) since, according to (1.3.4), we can dispense with pa rentheses for such vectors. These equations then give us the components of b and ω in Frame S'. For example, e'i• b = (b)'i = b'i. Section 6: Velocities 41 4. In deriving the various velocity rela tions above, we have basically only used r = r' + b and the G Rule. Both the fact that r = r' + b and the G Rule are valid even if bot h Frame S and Frame S' are non-inertial frames, so one could think of "the paper" to which Frame S is glued in Fig (4.2.1) as possibly rotating. The implication is that the velocity relations above do not require that either frame be inertial. This same comment applies to the acceleration re lations derived in Section 7 below. It is only when we later add the equation F = ma to the story in Section 8 that we have to regard Frame S as being an inertial frame. Section 7: Accelerations 42 7. Determination of accelerations The notations used here are describe d in Section 1.8. There are eight distinct accelerations of interest, and we could express seven of them in terms of the eight h which is the natural acceleration in Frame S', ( d a'/dt)S' = a'S' ≡ a' . We shall only express the three accelerations aS = a, aS' and a'S in terms of a'S' = a'. Below (6.9) one sees that this is exactly what we did with the velocities. Here, however, we examine a'S first. As noted in the Section 7 Summary, "dry as dust", and our apologies. 7.1 Acceleration a' S The G Rule (2.1) for v'S says ∂ Sv'S = ∂S'v'S + ω x v'S or a'S = a'S'S + ω x v'S . ( 7 . 1 ) Notice the unusual cross derivative a'S'S which involves both S and S'. This is one those cross accelerations appearing in (1.8.6). To compute this, we must go back to the G Rule (2.1) for r', v'S = v' + ω x r' . // v'S = ∂Sr', v' = ∂S'r' (6.8a) Apply ∂ S' ≡ (d/dt)S' to both sides to get ∂S'v'S = ∂S'v' + ∂S'(ω x r') = ∂S'v' + ω• x r' + ω x (∂S'r') // see (1.11.4) and (2.6) or a'S'S = a' + ω• x r' + ω x v' . ( 7 . 2 ) Here we used (2.6) that ∂ Sω = ∂S'ω = ω• . Now install (7.2) for a'S'S into (7.1) to get a'S = [a' + ω• x r' + ω x v'] + ω x v'S . ( 7 . 3 ) Now replace v'S in the last term using the G Rule for r' [ (6.8a) a few lines above ] a'S = [a' + ω• x r' + ω x v'] + ω x [v' + ω x r' ] or a 'S = a' + ω• x r' + 2 ω x v' + ω x (ω x r' ) . ( 7 . 4 ) The famous "Coriolis factor of 2" has now appeared and will be trivially transferred into aS in the next section. It is useful to review the steps above to see where this factor of 2 comes from : Section 7: Accelerations 43 1. Write the G Rule for v'S a'S = a'S'S + ω x v'S 2. Write the G Rule for r' v'S = v' + ω x r' 3. Insert 2 into 1 a'S = a'S'S + ω x v' + ω x (ω x r') // 1st ω x v' term 4. Apply ∂S' to 2 to get a'S'S = a' + ω• x r' + ω x v' // 2nd ω x v' term, 5. Install 4 into 3: a'S = [a' + ω• x r' + ω x v'] + ω x v' + ω x (ω x r') and now we have two ω x v' terms and only natural Frame S' objects r', v' and a'. 7.2 Acceleration a ≡ aS Start with (6.4) which is ∂S applied to r = r' + b , v = v'S + b• S . (6.4) Apply ∂S again to get a = a'S + b•• S . ( 7 . 5 ) Then insert (7.4) for a'S into (7.5) to get the same result as (7.4) but with b•• S tacked on, a = a' + ω• x r' + 2 ω x v' + ω x (ω x r') + b•• S . (7.6a) S S' Euler Coriolis centripetal frame We have attached a name to each contribution to a and will discuss these terms below. Since we really want all primed objects on the right side, we can anticip ate the result (7.12) derived in Section 7.5 below, b•• S = b•• S' + ω• x b + 2ω x b• S' + ω x (ω x b) (7.12) to get this alternate form for a , a = a' + ω• x r' + 2 ω x v' + ω x (ω x r') + [b•• S' + ω• x b + 2ω x b• S' + ω x (ω x b) ] = a' + ω• x r + 2 ω x v' + ω x (ω x r) + 2ω x b• S' + b•• S' (7.6b) where we think here of r as just a shorthand for b + r' to reduce the number of terms. Section 7: Accelerations 44 7.3 Acceleration a S' Start with (6.3a) which is ∂S' applied to r = r' + b, vS' = v' + b• S' . (6.3a) Apply ∂S' again to get ∂S'vS' = ∂S'v' + b•• S' or aS' = a' + b•• S' . ( 7 . 7 ) 7.4 Acceleration Summary a = a' + ω• x r' + 2 ω x v' + ω x (ω x r') + b•• S (7.6a) a 'S = a' + ω• x r' + 2 ω x v' + ω x (ω x r' ) (7.4) aS' = a' + b•• S' (7.7) (7.8) 7.5 Relation between b•• S and b•• S' Write the G Rule (2.1) for b• S' then solve it for b•• S' ∂Sb• S' = b•• S' + ω x b• S' b•• S' = ∂Sb• S' – ω x b• S' . (7.9) Now apply ∂ S to (6.2a) then solve for ∂Sb• S', b•• S = ∂Sb• S' + ∂S (ω x b) ∂Sb• S' = b•• S – ∂S (ω x b) . ( 7 . 1 0 ) Insert (7.10) into (7.9) to get the first line be low, then use (6.2a) to get the second line, b•• S' = [ b•• S – ∂S (ω x b)] – ω x b• S' = b•• S – ∂S (ω x b) – ω x [b• S – ω x b] = b•• S – ω• x b – 2ω x b• S + ω x (ω x b) . ( 7 . 1 1 ) The inversion of this equation may be found by using (6.2a), b• S = b• S'+ ω x b, Section 7: Accelerations 45 b•• S' = b•• S – ω• x b – 2ω x [b• S' + ω x b] + ω x (ω x b) = b•• S – ω• x b – 2ω x b• S' – ω x (ω x b) so then b•• S = b•• S' + ω• x b + 2ω x b• S' + ω x (ω x b) . (7.12) In a Special Case #1 problem we have b• S' = 0 from (6.10), hence b•• S' = 0, so that b•• S = ω• x b + ω x (ω x b) . // Special Case #1 (7.13) Section 8: Fictitious Forces 46 8. The Fictitious Forces 8.1 Development of the Fictitious Forces We start with (7.6a) which says a = a' + ω• x r' + 2 ω x v' + ω x (ω x r') + b•• S . (7.6a) (8.1.1) S S' Euler Coriolis centripetal frame Now for the first time we assume that Frame S is an inertial reference frame. Newton's law in inertial Frame S says ( m = mass of the Particle) F = ma ( 8 . 1 . 2 ) with a given as above in (8.1.1), so that, reordering the 5 terms, F = ma = mb•• S + ma' + mω x (ω x r') + 2m ω x v' + mω• x r' . (8.1.3) Now suppose we imagine an "effective" version of Newton's Law that works in rotating Frame S', F'eff = ma' . ( 8 . 1 . 4 ) Solving (8.1.3) for m a' tells us that (second line uses (8.1.2)) m a' = F'eff = F – m b•• S – mω x (ω x r') – 2m ω x v' – mω• x r' (8.1.5) m a' = F'eff = m a – m b•• S – mω x (ω x r') – 2m ω x v' – mω• x r' . (8.1.6) We can write the second equality in (8.1.5) as F'eff = F + F'fict w h e r e ( 8 . 1 . 7 ) F'fict = – m b•• S – mω x (ω x r') – 2m ω x v' – mω• x r' . (8.1.8) frame centrifugal Coriolis Euler Here F'fict represents "fictitious" forces ("pseudo" forces) that mysteriously have to be added to "real forces" F to make our bogus (8.1.4) "Newton's Law" F'eff = ma' be valid in Frame S'. We put a prime on F'fict as a reminder that it is a force which appears in non-inertial Frame S' 8.2 Interpretation of the Centrifugal and Euler Fictitious Forces To emphasize the long and short vector idea in our interpretation of the centrifugal and Euler fictitious forces, we take the Earth scenario of Fig (4.7.1) rath er than a more general situation like Fig (4.2.1) : Section 8: Fictitious Forces 47 (8.2.1) Suppose in the above picture v' = 0 so r' is static in Frame S'. Then (8.1.8) becomes F'fict = – m b•• S – mω x (ω x r') – mω• x r' . (8.1.8) with v' = 0 (8.2.2) frame centrifugal Euler In this expression it is the short vector r' which appears in both the centrifugal and Euler terms. Since Fig (8.2.1) is a Special Case #1 situation, we know from (7.13) that, b•• S = ω• x b + ω x (ω x b) . // Special Case #1 (7.13) (8.2.3) Inserting this into (8.2.2) gives, F'fict = – m [ ω x (ω x b) + ω• x b] – mω x (ω x r') – mω• x r' = – m ω x (ω x [b+ r']) – mω• x [b+ r'] = – m ω x (ω x r) – mω• x r . centrifugal Euler (8.2.4) Now the frame term – m b•• S is gone and it is the long vector r which appears in the two fictitious force terms. Equation (8.2.4) is the conventional form for F'fict. The centrifugal term may be written Ax(AxC)=(A•C)A - A2C – mω x (ω x r) = – mω2[ z^ x (z^ x r)] = – mω2 [ (z^• r)z^ - r ] = – mω2 [ z - r ] = + m ω2ρ ( 8 . 2 . 5 ) Section 8: Fictitious Forces 48 which is the usual way one states this centrifugal force. The vector r in Fig (8.2.1) is doing conical motion about the ω vector, and the tip of r executes circular motion with radius vector ρ. If the Earth rotation were varying only in magnitude, one would have for the Euler force in (8.2.4), – mω• x r = – mω•r z^ x r^ = – mω•r sinθ φ^ // (E.2.15) = – mω•ρ φ^ = – maφ φ^ // (E.3.6) (8.2.6) and this is the expected fictitious linear force directed in the - φ^ direction. Accelerate your Persian Rug to the right, you feel a fictitious force to the left. We end up then with F'fict = + mω2ρ – maφ φ^ . ( 8 . 2 . 7 ) centrifugal Euler Looking at the expression (8.2.3) for the frame acceleration component b•• S, we can break it down in similar fashion to get (see Fig (8.2.1), – m b•• S = – m ω x (ω x b) – mω• x b = m ω2ρb – mω•ρb φ^ . (8.2.8) centrifugal Euler where these terms are for the origin of Frame S' located at b in Frame S. The vector b in Fig (8.2.1) is doing conical motion about the ω vector, and the tip of b executes circular motion with radius vector ρb. Finally, we return to the two short-vector terms in (8.2.2), ΔF'fict = – mω x (ω x r') – mω• x r' = m ω2ρ' – mω•ρ' φ^ . (8.2.9) centrifugal Euler These terms describe the two fictitious forces relative to the Frame S' origin. The vector r' in Fig (8.2.1) is doing conical motion "about the ω vector", and the tip of r' executes "circular motion" with radius vector ρ'. To see this conical motion more clearly, we s how on the left below how the static (in Frame S') vector r' moves on the blue lampshade surface as the Earth rotates. When the tails of the r' vectors on this lampshade are translated to a common point, one obtains the conical motion shown on the right. The angle ψ is determined by r' • z^ = r'cosψ. Section 8: Fictitious Forces 49 ( 8 . 2 . 1 0 ) The point of this interpretative section is that, since b + r' = r, the centrifugal and Euler forces are additive: those of the Frame S' origin ( b) plus those relative to the Frame S' origin (short r') add up those relative to the center of the Earth (long r). 8.3 Interpretations of the Coriolis Fictitious Force Qualitative Arm-Waving Interpretation of the Coriolis Force The Coriolis fictitious term is always the main topic of any textbook or web page which deals with motion in rotating frames, so we won't have much to say about it other than to give a popular qualitative explanation of the direction of the effect. We did show very carefully how the factor of 2 arises in the derivation of the term which is F'cor = -2m ω x v', and indeed, the expression itself was derived in full. The careful reader will notice a certain amount of "free play" in the workings of what follows. Consider the pictures below where we launch four co lored projectiles horizontally from a launch platform that is screwed to a frictionless turntable surface (perhaps the projectiles are hockey pucks). Fixed Frame S and rotating Frame S' have a common origin at the spindle. The launching is done in rotating Frame S' and the projectiles are sent off in the four directions of the compass at equal speeds V. These velocities are represented by the four black equal-length arrows in the right side picture below. The colored arrows on the left show the initial Frame S velocities of these projectiles. In Frame S the initial v's are not the same size because the turntable adds an upward tangential amount vt to each (v t = aω). Since Frame S is an inertial frame, we can regard the colored arrows on the left as also representing the straight-line trajectories of the projectiles in Frame S. Section 8: Fictitious Forces 50 So, each projectile is launched with the same initial adder vt in Frame S and, being in free flight, maintains that vt during its flight. Notice that three of the projectiles move into to a region of larger radius, while the orange one heads to a smaller radius region. When a projectile moves to a larger radius, the particles of the turntable move faster CCW than the projectile's vt causing a velocity differential between the turntable and the projectile. We want now to examine this differential in the four cases. The short black arrows on the left show the motion of the turntable particles relative to the projectile, as will now be reviewed. ( 8 . 3 . 1 ) For the black projectile in mid flight, the turntable particles under the projectile are moving to the northwest relative to the projectile, so the projectile is seen by the turntable particles to be drifting to the right. Hence the curved black trajectory path on the right of Fig (8.3.1). On the left below is a crude strobe picture where a turntable particle's path in Frame S is shown in red and the arrows are then transferred to the right with a common tail to show what projectile motion the turntable particle sees in its rest frame. black blue (8.3.2) Section 8: Fictitious Forces 51 For the blue projectile in mid flight, the turntable partic les under the projectile are moving to the north relative to the projectile, so the projectile is seen to be drifting south. Hence the curved blue trajectory path on the right of Fig (8.3.1). The right strobe picture above shows this effect. For the red projectile in mid flight, the turntable particles under the projectile are moving to the northeast relative to the projectile, so the projectile is seen to be drifting to the left. Hence the curved red trajectory path on the right. For the orange projectile in mid flight, the turntable partic les under the projectile are moving to the south relative to the projectile (these particles are at a smaller radius and move more slowly than v t), so the projectile is seen to be drifting to the north. Hen ce the curved orange trajectory path on the right. Viewed from the direction of launch in Frame S', all four trajectories drift "to the right". This is in agreement with the right hand rule applied to our expression F'cor = -2m ω x v' = +2m v' x ω. What is not particularly obvious from the above qualitative discussion is that the | F'cor| is exactly the same for all four projectiles, and indeed for a projectile launched in any direction. Comments: 1. If the turntable were going CW instead of CCW, the drift directions would all be reversed, both by the qualitative argument, and by the F'cor expression. Projectiles would drift to the left instead of to the right. 2. One can think of Fig (8.3.1) as a view of the Earth from the North Pole. In this case, the vectors do not lie in the plane of paper, but qualitatively the conclusi on is the same: projectiles drift to the right in the Northern Hemisphere. A view from the South Pole woul d then show drift to the left in the Southern Hemisphere since ω is reversed. 3. The colored trajectories on the right in Fig (8.3.1), when applied to air masses moving into a region of Low pressure, look like this (8.3.3) and explain why lower pressure regions are CCW cyclon ic in the Northern Hemisphere. Since lows often drift to the east in the western US, warm Mexican air is felt prior to the low's arrival, and cool Canadian air is felt afterwards. Section 8: Fictitious Forces 52 Superposition Interpretation of the Coriolis Force Recall now the fictitious forces seen by the projectiles in Fig 8.5, F'fict = – m b•• S – mω x (ω x r') – 2m ω x v' – mω• x r' . (8.1.8) frame centrifugal Coriolis Euler Since the Frame S and Frame S' origins align, b = 0 and b•• S = 0 . If we assume ω = constant, and express the centrifugal term in the simpler form of (8.2.9), then the projectiles are controlled by F'fict = mω2r' – 2m ω x v' . (8.3.4) centrifugal Coriolis [To obtain the first term, set ψ = 90o in the cone picture of Fig (8.2.10) so then ρ' = r'.] One can think of the above as the superpositi on of two problems. The centrifugal term alone accelerates the projectiles radially outward in proportion to their distances from the turntable center, so the effect of this term is different for the four pr ojectiles as they progress in flight. The Coriolis term alone causes each projectile to deflect to its right ( ω > 0) along a path that is part of a circle as we now show. Recall the bogus Newton's Law F'fict = m a' from (8.1.4) where a' = (d v'/dt)S'. If F'fict = – 2m ω x v' alone, then we have a' = -2ω x v' or (d v'/dt)S' = Ω x v' where Ω = (-2ω) . (8.3.5) According to (1.6.1) and Fig (1.6.2) vector v' must rotate on a cone whose axis is Ω. For the turntable situation, however, v' is always in the plane of the turntable, so that cone must be flat, so vector v' goes in a circle at rate Ω. The trajectory r'(t) is then also circular so the Corio lis deflection is part of a circle. Here is some Maple code illustrating this fact for the blue projectile. We enter the circular v' (called v) and integrate to get the trajectory. We set parameters ω = +1, a = 1,V = 1 with initial condition v'(0) = V x^' and then create a plot, Section 8: Fictitious Forces 53 (8.3.6) The projectile is deflected "to the right" in this case since ω > 0. The actual deflection of the four projectiles is th en a superposition of the centrifugal and Coriolis motions and is therefore not perfectly circular. We shall solve this problem exactly in Section 15.5 and plot the all four projectile trajectories. 8.4 Special Case #1 Problems Recall from (8.1.5) that F'eff = F – mb•• S – mω x (ω x r') – 2m ω x v' – mω• x r' . (8.1.5) frame centrifugal Coriolis Euler If the rotation axis passes through the center of Frame S, we know from (7.13) that b•• S = ω• x b + ω x (ω x b) Special Case #1 (7.12) so that, using r = r' + b (6.1) in the third line below, F'eff = F – m[ω• x b + ω x (ω x b)] – mω x (ω x r') – 2m ω x v' – mω• x r' = F – m ω x (ω x [b +r' ] ) – 2m ω x v' – mω• x [b + r'] // = F + F'fict = F – mω x (ω x r) – 2m ω x v' – mω• x r . Special Case #1 (8.4.1) so F'fict = – mω x (ω x r) – 2m ω x v' – mω• x r Special case #1 (8.4.2) This is similar to (8.2.4) but now we have included the Coriolis term. Here r is the long vector, but v' is the rate of change of the short vector r'. Section 8: Fictitious Forces 54 Problems involving the motion of objects on or near the Earth's surface, or of objects in orbit around the Earth, fall into Special Case #1. In the first problem class, Frame S' can be defined as shown in Fig (4.7.1). 8.5 Problems on the surface of the Earth Let F0 = mg0 where go = -g0r^ is a vector pointing to the center of the Earth, and g 0 = GME/RE2. Then for problems involving motions of objects near the surface of the Earth, one has these real forces, F = mg0 + possible other real forces . (8.5.1) Possible other real forces might include air friction, wind, the action of magnetic fields on charged particles, etc. Let T = 24*60*60 = 86400 ~ 10 5 sec be the nominal period of a day. The day has a seasonal variation in its duration of roughly (see e.g. https://www.iers.org/IERS/EN/Science/EarthRotation/LODplot.html ) dT/dt ~ 0.5 msec/day ~ 10-3 * 10-5 ~ 10-8 // rms (8.5.2) with smaller short-term variations. From this we compute a rough value for ω•, |ω•| = |d(2π/T)/dt| = 2 πT-2 (dT/dt) ~ 6 * 10-10 * 10-8 ~ 10-17 sec-2 . (8.5.3) For activities on the surface of the Earth, r ≈ RE ~ 107 m so, ω• x r ~ 10-17 * 107 ~ 10-10 ~ 10-9 g . // max value at the equator (8.5.4) Thus in (8.4.1) we neglect the Euler term – m ω• x r to get F'eff = (m g0 + possible other real forces) – m ω x (ω x r) – 2m ω x v' . (8.5.5) Carrying out a static experiment ( v' = 0) to measure the g vector at some location on the Earth (no other forces in this experiment), one finds from (8.5.5), F'eff = m g0 – mω x (ω x r) r = REr^ ≡ mg , ( 8 . 5 . 6 ) where g is then the local gravity vector (which does not quite point to Earth center). In terms of this g, one then has F'eff = (m g + possible other real forces) – 2m ω x v' (8.5.7) Section 8: Fictitious Forces 55 so only the Coriolis fictitious force is left -- the centrifugal term has been absorbed into m g. By how much do g0 and g differ? We can write g - g0 = - ω x (ω x r) = ω2r cosθ ρ^ r = R E (8.5.8) where θ is latitude measured from the equator and ρ^ is the usual cylindrical unit vector pointing away from the rotation axis of the Earth at our point on the surface. So we s ee the direction of the difference to be ρ^ for any θ. The magnitude is roughly ω = 7.27 x 10-5 sec-1 // ωE = 2π radians / [ 24*60*60 sec ] r ≈ 6.37 x 10 6 // = R E ( 8 . 5 . 9 ) ω2r cosθ = (7.27)2(6.37) 10-4cosθ = 337 x 10-4 cosθ ~ 3 x 10-2 m/sec2 cosθ ~ (3/1000)g 0 cosθ so the difference is small but not zero. Presumably the local surface of the Earth is perpendicular to g and not g0, and one would certainly expect this to be true for a quiet ocean surface. We now repeat the above discussion in the "swap" notation mentioned at the start of the Summary. Swap Notation. The meaning of "swap notation" is that, in Fig 1, the vectors ω and b stay put, but all other vectors undergo V↔V'. This latter group includes basis vectors ei ↔ e'i, r ↔ r' , v ↔ v', a ↔ a' and of course Frame S ↔ Frame S'. This is nothing more than a cha nge of the way things are labeled. If a non-swap equation has the number (x.x.x), then th e corresponding equation in swap notation will be given the number (x.x.x) s ( 8 . 5 . 1 0 ) Here we rewrite equations (8.5.5), (8.5.6), (8.5.7) and (8.5.8) in swap notation. Feff = (m g0 + possible other real forces) – m ω x (ω x r') – 2m ω x v . (8.5.5) s Carrying out a static experiment ( v = 0) to measure the g vector at some location on the Earth (no other forces in this experiment), one finds from (8.5.10), Feff = m g0 – mω x (ω x r') r' = REr^' ≡ mg , ( 8 . 5 . 6 ) s where g is the local gravity vector (which does not quite point to Earth center). In terms of this g, one then has Feff = (m g + possible other real forces) – 2m ω x v (8.5.7) s so only the Coriolis fictitious force is left -- the centrifugal term has been absorbed into m g. Finally, g - g0 = - ω x (ω x r') = ω2r' cosθ' ρ^' . r' = R E (8.5.8) s Section 8: Fictitious Forces 56 8.6 Tethered satellites and Tidal Forces Consider a satellite, consisting of a pair of radially-aligned tethered masses m 1 and m2, which orbits the Earth at angular frequency ω. Inertial Frame S is fixed with its origin at the center of the Earth, while Frame S' is fixed to the tethered satellite and so Frame S' rotates at ω as the satellite moves around the Earth. The tethered masses are assumed to be at rest in Frame S' and are radially aligned with r 1 > r2. One can show (see Appendix F) that such a system is stable due to a restoring torque and, in the stable position, the line between the masses points to the center of the Earth as the satellite orbits. It might take some damping effort to achieve this stable configuration. This restoring torque plus damping over time is what caused the slightly distorted Moon to present its same face to the Earth as it orbits (apart from a small residual libration). Since we have a Special Case #1 situation (rotation axis through Frame S origin) we use (8.4.1) for the effective force acting on a mass m in Frame S' ( F is the real force), F'eff = F – mω x (ω x r) – 2m ω x v' – mω• x r . Special Case #1 (8.4.1) real centripetal Coriolis Euler where r is a long vector from the center of the Earth to mass m. In their stable positions, the two masses have vi' = 0 so there is no Coriolis term. The orbit has a constant ω, so there is no Euler term. Thus the effective fo rce on a static mass m in Frame S' is given by, F'eff = F – mω x (ω x r) = F – m[(ω•r)ω - ω2r] = F + mω2r r^ . ( 8 . 6 . 1 ) The real force F is the force of gravity plus the force of the tether T > 0. We then write for our two tethered masses, recalling that r 1 > r2, F'eff,1 = - (Gm Em1/r12) r^ – Tr^ + m1ω2r1 r^ = [ - (Gm Em1/r12) + m1ω2r1 - T ] r^ F'eff,2 = - (Gm Em2/r22) r^ + Tr^ + m2ω2r2 r^ = [ - (Gm Em2/r22) + m2ω2r2 + T ] r^ (8.6.2) where M E = mass of Earth, G = gravitational c onstant. It is convenient to define, f(r,m) ≡ -mM EG/r2 + mω2r // f '(r,m) = 2mMG/r3 + mω2 (8.6.3) so F'eff,1 = [ f(r1, m1) - T] r^ F'eff,2 = [ f(r2,m2) + T ] r^ . ( 8 . 6 . 4 ) Since both masses are at rest in Frame S', their accelerations are a'i = 0 so F'eff,i = mia'i = 0, and we conclude that the tether tension is given by Section 8: Fictitious Forces 57 T = f(r 1, m1) = - f(r2.m2) . ( 8 . 6 . 5 ) We now borrow a picture from page 120 of the Tethers in Space Handbook (Cosmo and Lorenzini) which shows the tethered satellite situ ation. The upper mass is at r 1, the lower at r 2, and the "center of gravity" is at r0, all measured from the center of the Eart h. We refer to the two masses as m 1 and m2 and to ω0 as ω. (8.6.6) If the entire satellite were suddenly compressed to a point mass M = m 1+ m2 and if this point mass were placed at the center-of-gravity locatio n, that point mass M would continue to orbit the Earth at radius r 0 and frequency ω and and nothing exciting happens. For either of these "systems" we have the following balance between gravitational force and centrifugal force, (GM EM/r02) = Mω2r0 or GM E/r02 = ω2r0 . (8.6.7) Comment : There is a very small distance between the sate llite's center of mass and its center of gravity. See Appendix D on this subject and in particul ar the numerical examples of (D.3.16). At this point to make things simple we shall assume m 1 = m2 = m. Then from (8.6.5), Section 8: Fictitious Forces 58 T = f(r 1) = - f(r2) f(r) ≡ -mMEG/r2 + mω2r // gravitational force + centrifugal force f '(r) = 2mMG/r3 + mω2 . (8.6.8) Note that f(r 0) = 0 according to (8.6.7). Setting r 1 = r0 + Δr in (8.6.8), T = f(r 0 + Δr) = -mMG/ (r 0+Δr)2 + mω2(r0+Δr) = -(mMG/r 02) (1 + Δ r/ r0)-2 + mω2(r0+Δr) // assume Δr << r0 ≈ -(mMG/r 02) (1 - 2 Δ r/ r0) + mω2(r0+Δr) // (1+x)n ~ 1 + nx for small x = - m ω2r0 (1 - 2 Δ r/ r0) + mω2(r0+Δr) // using (8.6.7) = 3 m ω2(Δr) = 3 ( m M G / r 03)(Δr ) . ( 8 . 6 . 9 ) The real gravitational force gives 2/3 of this result, while the fictitious centrifugal term gives 1/3 of the result, a fact we will note again several times below. If ρ represents an arbitrary vertical displacement away from r 0, so r = r 0 + ρ, then the above says, f(r0 + ρ) = 3mω2ρ ( 8 . 6 . 1 0 ) which looks like this near ρ= 0 (8.6.11) This f(r0+ρ) represents the radial force on any untethered Particle that might be present in Frame S' located a radial distance ρ from r0. A particle at r > r 0 is pushed up in Frame S', while a particle at r < r 0 is pushed down. Again, this force in Frame S' is due to the combination of gravitational (2/3) and centrifugal (1/3) forces. Another way to view the above calculation is this, T = [f(r 0 + Δr) - f(r0)] + f(r0) ≈ f '(r0) Δr + f(r0) // f '(r 0) is the gradient of the force f(r) at r=r 0 = f ' ( r 0) Δr // f(r 0) = 0 from (8.6.7) = ( 2 m M G / r 03 + mω2) Δr // f '(r 0) from (8.6.8) = ( 2 m ω2 + mω2) Δr // (8.6.7) again = 3 m ω2 Δr . ( 8 . 6 . 1 2 ) Section 8: Fictitious Forces 59 So the tension in the tether due to tidal force is equal to Δr times the derivative of f(r) evaluated at r = r 0, T ≈ Δr f '(r0) = Δ r 3mω2 = 3 Δr (mGM/r 03) . (8.6.13) Again, the tidal force acting up the upper mass is T pushing up, and the tidal force acting on the lower mass is T pulling down, all in Frame S'. One might then write tidal force per unit mass = ± 3 Δr (GM/r03) . // tidal acceleration (8.6.14) Once again, 2/3rds of the tidal force arises from the gravitational gradient at the satellite while 1/3 arises from the centrifugal gradient. The result (8.6.14) appears in Cosmo and Lorenzini p 123 with Δr=L and ω=ω0. Comments: 1. One may regard tension T as an example of a "tidal force" which tends to "rip apart" objects in orbit around a central force. In our example, the tidal force is T = f '(r 0) Δr = 3 (mMG/r 03) Δr. For objects in orbit around the Earth, this is a small or moderate force, but for objects orbiting massive black holes, the force is strong enough to rip apart all known material s, a process called "spaghettification", a sort of cosmic disposal. 2. According to (8.6.14), an untethered water droplet on the surface of the upper mass will migrate to the upper extremity of that mass, while a water droplet on the surface of the lower mass will migrate to the lower extremity of that mass. A crude intuition would seem to say that the Earth ought to pull both water drops to the lower extremity of each mass, but that is not what happens. One must get into the rotating Frame S' to see what happens. This is essentially why the usual tides on the Earth "bulge" on the side facing the Moon and on the side facing away from the Moon. The Sun also has its smaller effect. This subject is considered in the next section. 3. See comments below (8.8.26) for a discussion of why the tether tidal force (8.6.14) has a leading factor of 3, whereas the lunar Earth tide (8.8.26) has a leading factor of 2. 4. Appendix F presents a more complete analysis of the tether or "dumbbell" satellite, allowing for an arbitrary 3D orientation of the masses and analyz ing their motion when not in the stable vertical orientation. For example, a general dynamic expressi on for the tether tension T appears in (F.6.26). For the tether vertically aligned and at rest the limit of th is expression is (F.6.30) which is the same as (8.6.12) above. The reader is warned that Appendix D and F are both written in "swap" notation where prime and noprime labels are swapped relative to the no-swap notation used above. This is done to reduce the annoying number of primes that would othe rwise be present in many expressions. 5. Real tethered satellites have seen action since th e 1960's (including snapped tethers), see the brief history of Chen et al. [2013]. Tether applications include: (1) Power generation and thrust using a conducting tether in solar and planetary magnetic fiel ds and plasmas; (2) lifting, stabilizing and moving objects around in space (et, reboosting tired satellites); (3 ) research in controlled-gravity environments (distance down the tether see (8.6.12)); (4) gravity-wave and conventional ante nna experiments. Tether lengths have varied from tens of meters to tens of kilometers. Section 8: Fictitious Forces 60 8.7 Special Case #3 Consider the following situation which we shall call Special Case #3 where we have replaced the rotation vector ω by Ω where now Ω = dφ/dt : (8.7.1) This resembles Special Case #1 becau se the rotation axis passes through the origin of Frame S. However, in this figure we intend that the axes of Frame S' al ways line up with those of Frame S, so the only thing that varies is the vector b(t). The axes e'i are no longer "soldered" to the b vector, and e'i = ei at all times. Although the axes of Frame S' do not rotate relative to those of Frame S, we still put this case into our "rotating frames" basket because the origin of Frame S' is instantaneously rotating about the Ω axis. In Special Case #3 there is no distinction between ∂Sa and ∂S'a for any vector a : ∂Sa = ∂S[aiei] = (∂Sai) ei = (∂tai) ei ∂ S'a = ∂S'[aiei] = (∂S'ai) ei = (∂tai) ei . (8.7.2) One can interpret ∂Sa = ∂S'a as being the G Rule ∂Sa = ∂S'a + ω x a with ω= 0. Thus we can just write a• = ∂Sa = ∂S'a . The complicated analysis of Sections 6,7 and 8 is now much simpler: r = r' + b r• = r•' + b• ⇒ v = v' + b• v• = v•' + b•• ⇒ a = a' + b•• . ( 8 . 7 . 3 ) Newton's Law in Frame S' now has only one fictitious force, Section 8: Fictitious Forces 61 m a' = m a - m b•• F'eff = F - m b•• F'fict = – m b•• . // Special Case #3 (8.7.4) Recall that our general case fictional force expression was F'fict = – m b•• S – mω x (ω x r') – 2m ω x v' – mω• x r' . (8.1.8) frame centrifugal Coriolis Euler In Special Case #3 only the "frame" portion of the fictiti ous force exists and we can interpret this as being the full fictional force in which we set ω = 0 and ω• = 0. Regardless of how the Frame S' origin moves throug h space, we can always interpret its motion as an instantaneous rotation, as s uggested by this drawing, (8.7.5) At time t the Frame S' origin is rotating along the green circle shown. Comment : Historical authors have compared the behavior of Frame S' to a frying pan in the fast-moving hands of a busy cook who always maintains the pan's orientation. We are instead reminded of the behavior of a gimbaled alcohol stove often used in sa ilboats, or better yet, a gimbaled sailboat compass. Section 8: Fictitious Forces 62 8.8 Tides on the Earth The basic picture In general, the orbital pattern of a binary gravitatio nal system has this planar appearance, where each object traverses its own ellipse (8.8.1) http://abyss.uoregon.edu/~js/ast122/lectur es/lec10.html In this section, however, we restrict our interest to a special case where each mass traces out a circular path, not an elliptical one. The picture is this, where we define R 1,R2, M1,M2, r12,R,R' as shown: ( 8 . 8 . 2 ) Each object is assumed to be a spherically symmetric mass distribution and can thus be treated as a point mass at its center for gravitational purposes (see Sec tion D.5). An inertial Frame S has its origin at the center-of-mass point, and the binary system rota tes in the plane of paper at angular frequency Ω about this Frame S origin. Frame S' is attached to sphere 2 and we assume that sphere 2 maintains its orientation "relative to the stars" so that at some later time we have this picture Section 8: Fictitious Forces 63 ( 8 . 8 . 3 ) The situation with Frame S and Frame S' is then exactly that of Special Case #3 described in the previous section, where b = R and the only fictitious force in Frame S' is F'fict = – m b•• . Points A and B are two fixed points on the surface of sphere 2. To apply Fi g (8.8.3) to the Moon-Earth system, we temporarily turn off the rotation of the Earth (M 2) about its axis (not shown), so the this new Earth has a fixed orientation relative to the stars. Distance R can be found from the usual center-of-mass equation (viewed from Frame S), 0 = [M2Rr^ - M1R' r^]/(M1+M2) R+R' = r 12 ( 8 . 8 . 4 ) which reports out the obvious fact that R = r 12 [M1/(M1+ M2)] R' = r 12 [M2/(M1+ M2) ] . ( 8 . 8 . 5 ) We collect here some data on the Sun, Earth and Moon : M S = 1.989 x 1030 kg R S = 695,700 km G = 6.674 x 10-11 m3/(kg-sec2) ME = 5.972 x 1024 kg R E = 6371 km (8.8.6) MM = 7.342 x 1022kg R M = 1737 km T M = 27.32 days (sidereal) r ES = 1.496 x 108 km // average e = .0167 rME = 384,400 km // average e = .0549 We then compute, Section 8: Fictitious Forces 64 1 ------- 2 Earth-Sun R/R 2 = r12 [M1/(M1+ M2)] /R2 = rES [ME/(ME+ MS)] /RS = .000646 Moon-Earth R/R 2 = r12 [M1/(M1+ M2)] /R2 = rME [MM/(MM+ ME)] /RE = 0.737 (8.8.7) So for the Earth-Sun system, the center of mass is ba sically at the center of the Sun, while for the Moon- Earth system, the center of mass lies at a point 3/4 the radius of the Earth from the center. We could redraw our figure for these two cases, but the ki nematics does not change, so we won't bother. Force Equations Newton's Law for a mass m in non-inertial Frame S' is, according to (8.7.4), F'eff = ma' where F'eff = F – m b•• ( 8 . 8 . 8 ) and F is the sum of all Frame S forces acting on mass m. It is clear from Fig (8.8.3) that (since R is a constant) b = Rr^ d r^/dt = Ωθ^ d θ^/dt = - Ωr^ // see (E.5.6) b• = R d r^/dt = RΩθ^ b•• = RΩ dθ^/dt = - RΩ2r^ . ( 8 . 8 . 9 ) Therefore (8.8.8) becomes m a' = F'eff = F + m RΩ2r^ . ( 8 . 8 . 1 0 ) For a Particle in or on the Earth, the real forces are F = Fg1 + Fg2 + Fng where Fg1 = the gravitational force due to sphere 1 (the Moon) Fg2 = the gravitational force due to sphere 2 (the Earth) Fng = any non-gravitational forces (8.8.11) Section 8: Fictitious Forces 65 and then (8.8.10) may be written m a' = Fg1 + Fg2 + Fng + m RΩ2r^ . ( 8 . 8 . 1 2 ) This is the effective Newton's Law for a mass m in Frame S'. If a mass m is at rest on the surface of the Earth (in Frame S') , then a' = 0 and we find Fng = - Fg1 - Fg2 - m RΩ2r^ . (8.8.13) If there are no other non-gravitational forces affecting mass m, then Fng is just the force of the Earth's surface pushing up on mass m to hold it in place so it has a' = 0 . The relation between r 12 and Ω If we were to replace sphere 2 (the Earth) with a point mass M 2 at its center, nothing would change in our orbiting picture. This point mass does a circular orb it around the binary center of mass with radius R and angular frequency Ω. The usual rule for circular motion of a point particle says that the gravitational force balances the centrifugal force, so M 1M2G/r122 = M2Ω2R or M1G/r122 = Ω2R . ( 8 . 8 . 1 4 ) Now we do a thought experiment. We imagine the Eart h's core to be solid and we grind out a small spherical cavity around the Earth's center point. We take the ground-out material and compress it into a point mass m and we place that mass m at the center of the cavity. The orbit of the Earth-Moon system is unaffected by this alteration. In eff ect we now have two point masses in identical orbits with the Moon: the Earth of mass M 2-m and the central particle of mass m. For each we have M 1G/r122 = Ω2R. The claim then is that the point mass m simply floats in the center of the cavity. In Frame S' there is no total force acting on this mass m to cause it to move from its position. This total force of 0 is the sum of the gravitational force pulling it to the Moon, and the centrifugal force pushing it away from the Moon, 0 = - ( Fg1 + m RΩ2r^) or mM 1G/r122 = m RΩ2 or M 1G/r122 = Ω2R which is the same as (8.8.14) above. We can replace the R in (8.8.14) with the R of (8.8.5) to get M1G/r122 = Ω2 r12 [M1/(M1+ M2)] or (M 1+ M2)G/r123 = Ω2 (8.8.15) Section 8: Fictitious Forces 66 and this is the relationship between r 12 and Ω for given masses M 1 and M2. Using (8.8.14) in (8.8.12) one finds, m a' = Fg1 + Fg2 + Fng + (mM1G/r122) r^ . (8.8.16) Tidal Force at an arbitrary point on the Earth Now consider a particle of mass m at some arbitrary location C on the surface of the Earth. We define angles θ and β as shown, where β is very small, ( 8 . 8 . 1 7 ) Here we show a brand new r^ and θ^ which have nothing to do with those used in Fig (8.8.3). The old r^ in (8.8.16) is now d^0, a unit vector to the right in (8.8.17). Vector d points to point C from the center of sphere 1 while vector d0 links the two object centers, so d 0 = r12. We can then write Fg1 = - (M1mG/d2) d^ ( 8 . 8 . 1 8 ) and so (8.8.16) is recast once again as m a' = Fg1 + Fg2 + Fng +( mM1G/r122) d^0 = - (M 1mG/d2) d^ + Fg2 + Fng + (mM1G/d02) d^0 = Fg2 + Fng + mM1G ( d^0/d02 – d^/d2) = Fg2 + Fng + Ftid, ( 8 . 8 . 1 9 ) where Ftid ≡ mM1G ( d^0/d02 – d^/d2) . // tidal force (8.8.20) For a mass m at the center of the Earth, d = d0 so Ftid = 0 in agreement with our thought experiment above. It is the fact that sphere 1's gravitational field va ries slightly (in direction and magnitude) at different points on sphere 2 (as shown in (8.8.18)) which results in the tidal force. Equation (8.8.20) appears in Taylor p 332 as equation (9.12) and in Butikov as equation (2 ). We have tried to match Taylor's notation. Section 8: Fictitious Forces 67 Comment: If the rotating Moon-Earth system in Fig (8.8.17) were replaced by a static non-rotating system in which the Earth and Moon were held apart by a very long, stiff (1020 N) rod, would the tidal force be the same as shown in (8.8.20)? Or would the water bulge only on the side of the Earth facing the Moon? In this case we have in inertial Frame S' the fo llowing force on a mass m on the surface of the Earth F' = m a' = Fg1 + Fg2 + Fng // Earth, Moon and Stick Since the gravitational force Fg1 of M1 on a mass m at point A is larger than at point B in Fig (8.8.17), it seems likely that water would only bulge on the A side of the Earth. So it is not just the non-uniformity of the gravitational field that causes the double-bulge tide on the real Earth, it is this non-uniformity in combination with the balance provided by the rotation which causes there to be zero force on a particle at the center of the Earth. Tidal force in Cartesian coordinates It is useful to express Ftid in both Cartesian and polar coordi nates with the approximation that R2/d0 << 1. In that case, from the Law of Cosines and Fig (8.8.17), d 2 = R22 + d02 - 2d0R0cos(π-θ) = R22 + d02 + 2d0R0cosθ = d02[ 1 + (R2/d0)2 + 2 (R2/d0)cosθ ] ≈ d 02[ 1 + 2 (R 2/d0)cosθ ] ( 8 . 8 . 2 1 ) so that d-3 ≈ d0-3 [ 1 + 2(R 2/d0)cosθ]-3/2 ≈ d0-3 [ 1 - 3(R 2/d0)cosθ ] . (8.8.22) Armed with this fact, we next write, again looking at Fig (8.8.17), d0 = d0x^ d = (d0 + R2 cosθ) x^ + R2sinθ y^ . ( 8 . 8 . 2 3 ) We assume for now the usual x axis to the right and y axis up, though this will be changed below. Then, Ftid = mM 1G { d0/d03 – d/d3 } = m M 1G { d0x^/d03 – [(d0 + R2 cosθ) x^ + R2sinθ y^] d0-3 [ 1 - 3(R 2/d0)cosθ ]} = ( m M 1G/d02) { x^ – [(1 + (R2/d0) cosθ) x^ + (R2/d0) sinθ y^] [ 1 - 3(R 2/d0)cosθ ] } = ( m M 1G/d02) { x^ – x^ + 3(R2/d0)cosθ x^ - (R2/d0) cosθ) x^ - (R2/d0) sinθ) y^ } + O((R 2/d0))2 ≈ (mM 1G/d02) {2(R2/d0)cosθ x^ - (R2/d0) sinθ) y^ } Section 8: Fictitious Forces 68 = ( m M 1G/d03) {2R2cosθ x^ - R2sinθ) y^ } = ( m M 1G/d03) ( 2x x^- y y^ ) . (8.8.24) Using this simple form, it is easy to plot the tidal force field Ftid(x,y) in the region of the Earth, (8.8.25) We can evaluate Ftid in (8.8.24) at the left, right, top and bottom of the Earth: left: A (x,y) = (-R 2,0) Ftid = -2R2(mM1G/d03)x^ points left right: B (x,y) = (R 2,0) Ftid = 2R2(mM1G/d03)x^ points right top: (x,y) = (0,R 2) Ftid = -R2(mM1G/d03)y^ points down bottom: (x,y) = (0,-R 2) Ftid = R2(mM1G/d03)y^ points up (8.8.26) and these results seem in agreement with the a bove field plot (zero force at Earth center). 2 versus 3 Question : For the Earth tide situation we have just shown that at point B farthest from the Moon there is a total force | Ftid| = (2mM1G/d03)R2. For the tether analysis of (8.6.9) setting Δr = R2 produces a tidal force of | Ftid| = T = ( 3mMG/r03)R2 . Why are the two red integers different? Section 8: Fictitious Forces 69 Answer : For the tether satellite, the axes of the satellite Frame S' are rotating so the fictitious force set includes both the frame and "local" centrifugal contributions – m b•• S – mω x (ω x r') of (8.1.5). It was shown that the centrifugal term accounts for 1/3 of the tid al force. But for the Ea rth tide analysis, the axes of Frame S' do not rotate, so only – m b•• S exists so that extra 1/3 is missing. We might expect that extra 1/3 to reappear if we did an analysis of tides on an Earth which is phase locked to always have the same face pointing toward the Moon (Reader Exercise). On the other hand, for a tethered satellite that is dropping straight down toward the Earth, we would e xpect the 1/3 to be missing (Reader Exercise). One gets the general impression that the water surfa ce might have the following shape, where the black arrows show the magnitude of the tidal force at various locations, in agreement with (8.8.25) and (8.8.26), (8.8.27) The tidal acceleration is very weak compared to the local gravitational force on the Earth. For the lunar tidal case, continuing the Maple code above we find for the tidal acceleration at point B, aB = 2R2(M1G/d03) = 2(M 1G/d02)(R2/d0) = 2(M MG/rM-E2)(RE/rM-E) (8.7.28) so basically the tidal acceleration is 10 -7 the size of g = 9.8. It is rather amazing what such a small force can do when it is differentially applied to a lot of water. Section 8: Fictitious Forces 70 Tidal force in polar coordinates To express the tidal force in polar coordinates, we use (E.5.4) with ρ→ r = R2 and φ→θ to get x^ = cosθ r^ – sinθ θ^ r = R 2 y^ = sinθ r^ + cosθ θ^ . ( 8 . 8 . 2 9 ) Then, Ftid = (mM 1G/d03) { 2x x^- y y^ } = (mM 1G/d03) { 2rcosθ [cosθ r^ – sinθ θ^] - rsinθ [sinθ r^ + cosθ θ^] = (mM 1G/d03) { (2cos2θ - sin2θ)r^ + ( -3cos θsinθ)θ^ } . (8.8.30) The coefficients of the unit vectors can be simplified, 2cos 2θ - sin2θ = 2(1/2)*(1+cos2 θ) - (1/2)(1-cos2 θ) = (1/2) [ 2 + 2cos2 θ - 1 + cos2θ ] = (1/2) [ 1 + 3cos2 θ] = (3/2) [ cos2 θ +1/3 ] -3cos θsinθ = (-3/2) 2sin θcosθ = -(3/2)sin2 θ . (8.8.31) Therefore the tidal force in the polar coor dinates appearing in Fig (8.8.17) is, Ftid = (3/2) (mM 1G R2/d03) [ (cos2θ +1/3) r^ - sin2θ θ^ ] . (8.8.32) Since the term with the 1/3 is radially symmetric ar ound the Earth, it really has no effect on tides and is usually just dropped. This last result appears in Butikov as (5), (6) and (7). Equation of the water surface Here we follow the development of Butikov. We fi rst make an ansatz that the water surface on an idealized non-rotating water-covered Earth is de scribed by the following simple equation, r(θ ) = R2 + a cos2θ . a > 0 (8.8.33) The constant term is R 2 so that <r( θ)> = R2. One chooses the angle 2 θ to get the shape suggested in Fig (8.8.27). The problem then is to dete rmine the constant a which will be << R 2. This is not the true equation of an ellipse, but we shall call it an ellipse anyway. Consider the following picture of the water on the Earth, Section 8: Fictitious Forces 71 (8.8.34) Here nc is normal to the black circle at angle θ , while ne is normal to the red ellipse. In polar coordinates, we know that n^c = r^, and we wish to compute n^e. Our motivation is to compute angle α which will then lead us to an expression for a. So far we know that cos α = n^c• n^e = r^• n^e . Define f(r,θ) = r - R2 - acos2θ and the ellipse (8.8.33) is then given by f(r, θ) = 0. We know that the normal to a 2D surface is given by its gradient, so in polar coordinates we have ne = ∇f(r,θ) = (∂rf)r^ + (1/r)(∂θf)θ^ = r^ + 2(a/r)sin2θ θ^ |ne| = (1 + [2(a/r)sin2 θ]2 )1/2 1/| ne| = (1 + [2(a/r)sin2 θ]2 )-1/2 ≈ 1 - (1/2) [2(a/r)sin2 θ]2 (8.8.35) cosα = n^c• n^e = r^• n^e = r^• ne / |ne| = r^• [ r^ + 2(a/r)sin2 θ θ^] / |ne| = 1/| ne| ≈ 1 - (1/2) [2(a/r)sin2 θ]2 ≈ 1 - α2/ 2 . ( 8 . 8 . 3 6 ) Therefore using r = R 2, α = 2(a/R 2)sin2θ . ( 8 . 8 . 3 7 ) Having found a geometric value for α, we now seek another expression for α based on physics. Consider, Section 8: Fictitious Forces 72 (8.8.38) Since the affected water surface is assumed stable, the total force on the particle of water at the dot must be normal to the red surface. Therefore |F tid,θ | / mg = tan α ≈ α where g = GM 2/R22 (8.8.39) From (8.8.32) we then write α ≈ |Ftid,θ | / mg = (3/2) (mM 1G R2/d03) sin2θ / (mGM 2/R22) = (3/2) (M 1/M2) (R2/d0)3 sin2θ . (8.8.40) Comparing this α to that of (8.8.37) one finds, 2(a/R 2)sin2θ = (3/2) (M 1/M2) (R2/d0)3 sin2θ so a = (3/4) R 2 (M1/M2) (R2/d0)3 . ( 8 . 8 . 4 1 ) Thus the shape "ansatz" (8.8.33) was a good one. Th is result for a appears in Butikov as (9) and (11). Taylor obtains the same result in his (9.18) (h=2 a) by treating the water surface as an equipotential surface (see Footnote at the end of this section). The variation between low and high tides is H = 2a and we may compute this from (8.8.41) for both the Moon-Earth and Sun-Earth systems (in km), Therefore, H lunar_tide = 53.49 cm ~ 1.8 feet . Hsolar_tide = 24.58 cm ~ 0.8 feet . (8.8.42) A non-inlander will recognize these as reasonable ballp ark values for ocean tides, lending credence to the model at hand. Section 8: Fictitious Forces 73 Comment : The red "ellipse" shown in (8.8.34) is the cross section in the plane of paper of a red ellipsoid formed by rotating the ellipse about the z axis. This el lipsoid then specifies the water level at all places on the Earth. Tidal patterns for an arbitrary rotation axis of the Earth We now turn the rotation of the Earth back on (we turn ed if off earlier). For the real Earth, there are many complications that arise. There are land masses. Lake water has nowhere to go. There is friction between the water and the land which slows down the Earth's rotation slightly over time. There is weather and there are ocean tidal currents which do not flow infin itely fast. We shall not attempt to analyze this general situation. Instead, we imagine an idealized Earth covered wi th water and the Earth turns under the water with no "friction", and an Observer just stands in the wate r and measures the tide height as a function of time. What does that Observer see? It of course depends on where the Earth's axis of rotation is located relative to our picture. Consider, ( 8 . 8 . 4 3 ) We have now redefined the Earth frame on the right to be Frame S (formerly it was Frame S') and we have drawn new x,y,z axes for this new Frame S so the z axis points away from mass M 1. In Frame S then the angle θ is the usual spherical-coordinates polar angle. We now assume that the Earth rotates about some axis z^' (new Frame S') which is obtained by rotating the z^ axis by angles θ1 and φ1 as follows (see (E.2.2) for the matrix), z^' = Rz(φ1) Ry(θ1) z^ ≡ R1 z^ = ⎝⎜⎜⎛ ⎠⎟⎟⎞ cosθ1cosφ1 -sinθ1 sinθ1cosφ1 cosθ1sinφ1 cosφ1 sinθ1sinφ1 -sinθ1 0 cos θ1 z^ = ⎝⎜⎜⎛ ⎠⎟⎟⎞ cosθ1cosφ1 -sinθ1 sinθ1cosφ1 cosθ1sinφ1 cosφ1 sinθ1sinφ1 -sinθ1 0 cos θ1 ⎝⎜⎛ ⎠⎟⎞0 0 1 = ⎝⎜⎜⎛ ⎠⎟⎟⎞ sinθ1cosφ1 sinθ1sinφ1 cosθ1 . (8.8.44) This rotation is sufficient to put the z^' in any desired direction ( θ1.φ1). The rotation R 1 moves all vectors r to new vectors r' = R1r. Thus we may write Section 8: Fictitious Forces 74 r' = ⎝⎜⎛ ⎠⎟⎞x' y' z' = ⎝⎜⎜⎛ ⎠⎟⎟⎞ cosθ1cosφ1 -sinθ1 sinθ1cosφ1 cosθ1sinφ1 cosφ1 sinθ1sinφ1 -sinθ1 0 cos θ1 ⎝⎜⎛ ⎠⎟⎞x y z = R1 r and r = ⎝⎜⎛ ⎠⎟⎞x y z = ⎝⎜⎛ ⎠⎟⎞ cosθcosφ cosθsinφ -sinφ -sinφ cosφ 0 sinθ cosφ sinθsinφ cosθ ⎝⎜⎛ ⎠⎟⎞x' y' z' = R1-1 r' = R1T r' (8.8.45) where the second matrix is the transpose of th e first since rotations are real orthogonal R-1 = RT. If we define spherical coordinates (r, θ,φ) for (x,y,z) and (r'. θ',φ') for (x',y',z') (of course r' = r) then the above may be written, cancelling the r factors, ⎝⎜⎛ ⎠⎟⎞ sinθcosφ sinθsinφ cosθ = ⎝⎜⎜⎛ ⎠⎟⎟⎞ cosθ1cosφ1 cosθ1sinφ1 -sinθ1 -sinφ1 cos φ1 0 sinθ1cosφ1 sinθ1sinφ1 cosθ1 ⎝⎜⎛ ⎠⎟⎞ sinθ'cosφ' sinθ'sinφ' cosθ' (8.8.46) which is three scalar equations. The third equation is this cosθ = sinθ 1cosφ1sinθ'cosφ' + sinθ1sinφ1sinθ'sinφ' + cosθ1cosθ' . (8.8.47) If the Earth turns at rate ω so φ' = ω t, we then have cosθ = sinθ1cosφ1sinθ'cosωt + sinθ1sinφ1sinθ'sinωt + cosθ1cosθ' ( 8 . 8 . 4 8 ) cos2θ = 2cos 2θ - 1 . Recall the equation of the water surface from (8.8.33), r(θ ) = R 2 + a cos2θ . a > 0 (8.8.33) This implies a tide height of h(θ) = a cos2 θ . ( 8 . 8 . 4 9 ) Therefore on our idealized Earth which rotates about an axis ( θ 1,φ1) relative to Fig (8.8.43) we obtain the following tide height during the day h(t) = a cos2 θ(t) = a [ 2cos2θ(t) - 1 ] = a [ 2 ( s i n θ 1cosφ1sinθ'cosωt + sinθ1sinφ1sinθ'sinωt + cosθ1cosθ')2 - 1 ] . (8.8.50) Here θ' indicates the line of latitude at which our Observer is positioned. Section 8: Fictitious Forces 75 Example 1 : (z^' = y^) Suppose the Earth's rotation axis were in the y^ direction in Fig (8.8.43) (pointing out of the plane of paper). In that case one has θ1= π/2 and φ1= π/2, since z^' = Rz(π/2) Ry(π/2) z^ = ⎝⎜⎜⎛ ⎠⎟⎟⎞ sinθ1cosφ1 sinθ1sinφ1 cosθ1 = ⎝⎜⎛ ⎠⎟⎞ 0 1 0 = y^ . (8.8.51) Then from (8.8.50), h(t) = a [ 2(sin θ 1cosφ1sinθ'cosωt + sinθ1sinφ1sinθ'sinωt + cosθ1cosθ')2 - 1 ] = a [ 2 ( s i n θ'sinωt) 2 - 1 ] = a [ 2 s i n 2θ'sin2ωt - 1 ] . ( 8 . 8 . 5 2 ) If the Observer were at the Earth's equator θ' = π/2 (which is in the plane of paper of Fig (8.8.43)), one would have h(t) = a [ 2sin2ωt - 1 ] = a sin(2 ωt) = a cos(2 ωt - π/2) . (8.8.53) Comparison with (8.8.49) shows that 2 θ = 2ωt - π/2 so θ = ωt - π/4. The Observer sees a full amplitude swing of ±a in the tide. As this Observer moves toward the pole so θ' decrease, the amplitude of the tide decreases as (8.8.52) shows. At the pole, where θ' = 0, one finds h(t) = -a (a constant) all the time, which seems reasonable since θ = π/2 all the time and so h( θ) = a cos2 θ = a cosπ = -a. Here is a Maple rendition of this Example where we set a = 1 and ω = 1 so one day lasts T = 2 π. In this code we refer to θ' as θ2: Section 8: Fictitious Forces 76 (8.8.54) The top trace is for θ' = 90o (equator) which has the full tide amplit ude, and then as one approaches the pole in steps of 20o this amplitude decreases ending up with h(t) ≈ -a for θ ' = 10o. On this Earth there are always two equal high tides per day. Example 2 : (tipping z^' toward the x^ direction) Suppose however that the Earth's rotation axis points in the z^ direction of (8.8.43) so θ1= 0 and φ1= 0. In this case we expect to have no tides at all since for any θ' latitude line θ = θ' is constant. This is borne out in the above Maple code Maple where we again plot h(t), (8.8.55) At the equator the Observer is stuck low tide all the time (bottom trace). Conversely, an Observer at the pole is stuck at high tide all the time (top trace is θ' = 10o). As we start increasing θ1 away from 0, the nature of the tidal traces changes. Here is a set of trace sets for various θ1 always with φ1= 0: Section 8: Fictitious Forces 77 θ 1 = 0 θ1 = 5o θ1 = 10o θ1 = 20o θ 1 = 40o θ 1 = 60o θ1 = 70o θ1 = 80o ( 8 . 8 . 5 6 ) θ1 = 90o When θ1= 0o there are no tides at all since the rotation is in the z^ direction in Fig (8.8.43). As we gradually tip the Earth's rotation axis toward the x^ direction, things change. Up to about θ1 = 40o there is one tide per day, but beyond this point there are tw o unequal tides per day and they become equal when θ1 reaches 90o. The point here is that many tidal patterns are possible depending on the Earth's direction of rotation. Tidal patterns for the rotating Earth (but still a water world in which water flows instantly) Relative to Fig (8.8.43) the actual axis z^' of the Earth's rotation varies over time, as suggested by this picture from wiki, (8.8.57) https://upload.wikimedia.org/wiki pedia/co mmons/4/43/Earth-Moon.PNG Section 8: Fictitious Forces 78 We transcribe the situation depicted above into a drawing more compatible with Fig (8.8.43), (8.8.58) Here the black arrow is z^' (Earth's rotation axis) and it is located at θ1= 118.6o and is in the plane of paper so φ1 = 0. The intersection of the two orbital planes is called the line of nodes and for the time indicated in the picture, that line is perpendicular to the plane of paper. This situation of maximum tilt 28.58o occurs once every 18.6 years, a time called the "major lunar standstill". At a time 9.3 years later than the above drawing, the Earth's rotation axis in effect move s to the right edge of the green cone and then the 28.58o = 23.44+5.14 gets replaced by 18.30o = 23.44-5.14 which is the "minor lunar standstill". The half- angle of the green cone is 23.44o. Near the time depicted in the picture, the Earth' s rotation axis in effect moves around a different cone once per month as indicated in blue in this picture (blue cone half-angle = 28.58o) (8.8.59) Thus, a half month later than the configuration shown, one will have θ1 = 90 - 28.58 = 61.42o. At times in between, θ 1 lies in the range ( 61.4o, 118.6o) and φ1 takes small values with | φ1| ≤ 28.58o . Here we repeat our plot set above for θ1 = 100o and φ1 = 20o just as an example: Section 8: Fictitious Forces 79 (8.8.60) At the equator (red) there are two small equal tides pe r day. At other latitudes there are still two tides per day, but they are unequal. As the Earth rotates, it is true that at any latitude θ' there is an outward-pointing centrifugal force of equal magnitude all around the Earth, but we expect this not to affect the tides. The high tides are nominally 12 hours apart. In fact the Moon moves with a 27.3 day period in the same direction the Earth rotates, so when 12 hours has passed, the Moon has moved ahead 12/27.3 = 0.44 hours = 26.4 minutes, so one has to wait another 26 mi nutes for the next lunar high tide, so the time between high tides is about 12 hours 26 minutes. This cau ses the time of high tide to move relative to a wall clock in any location, which is why we have tide tables and tide clocks. Roughly the solar tides have half the influence of th e lunar ones as indicated in (8.8.42). They add and cancel depending on the position of the Sun and Moon. This nice picture of D.J. Jeffery shows the extremal situations (the word spring does not mean the season Spring) (8.8.61) So the maximal spring tides are about 2 weeks apart and the same is true for the intervening minimal neap tides. Section 8: Fictitious Forces 80 A good discussion of the above tidal model is give n in Taylor's textbook p 330-336. A more detailed discussion is presented in the excellent (and downloa dable) paper by Butikov. Both sources are very readable. Tides on the real Earth We leave the reader with this perhaps disappointing picture of actual tides on the Earth. http://www.nauticed.org/sailing-blog/how-the-tides-work/ (8.8.62) (We are unable to locate the original source of this graphic.) Presumably these are measured long-term averages of tidal high-low differences and one sees how they generally range from 0 cm to 140 cm. A wave resonance at the Bay of Fundy can cause a 1700 cm (50 foot) high/low tide difference there. Recall our toy model calculation that H lunar_tide = 53.49 cm ~ 1.8 feet . Hsolar_tide = 24.58 cm ~ 0.8 feet . (8.8.42) At least the values in the map are in the same ballp ark as the toy model! The sailing author at the above link notes that tides are almost non-existent in the Caribbean and Mediterranean seas, which are dark blue in the figure. As Butikov points out, the simple model in which the Earth rotates under a static ellipsoid of water in a frictionless manner is very far from reality. One mu st consider the dynamical aspects of the problem which involve the massive water currents which attempt to maintain the tidal ellipsoid and the interaction of such currents with land masses and the ocean botto m and with themselves (water has some viscosity). Section 8: Fictitious Forces 81 The currents of course are subject to Coriolis forces. Water flow velocity is not infinite and is affected by ocean depth. One result is that there is a delay betw een the Moon being at local meridian (transit) and the occurrence of high tide (the "local lunitidal interval"). This delay is extremely variable, ranging from a few minutes to ~20 hours. The white lines in (8.8.2) ar e loci on which a high tide occurs at the same time (equal tidal phase) and the white line nodes have no tid es at all (known as amphidromes) for a particular tidal "component" like the main one called M 2. Tides flow around these points. The problem is one of wave dynamics and forced oscillation on a rotating ob ject and is well beyond the scope of our document (but is treated by Butikov and in his references). Footnote: Equation of the Wate r Surface by the Potential Method There are many ways to set up the potential method for determining the idealized tidal shape of Fig (8.8.34), see Taylor and Butikov for alternatives to our method. In the following presentation we omit some detailed steps which involve small- ε approximations. We seek a potential V(x,y) which solv es this equation (no minus sign in ∇V= F) ∇V(x,y) = Fg2 + Ftid = - (GM 2m/r2) r^ + (mM 1G/d03) ( 2x x^- y y^ ) (8.8.63) where we use (8.8.24) for Ftid. The exact solution for V is the following (by inspection) : V(x,y) = (GM 2m/r) + (GM 1m/d03)( x2 - y2/2) . // ∇(1/r) = -(1/r2)r^, r > 0 (8.8.64) Of interest are surfaces on which V is a constant (since we expect the water surface to be such a surface) so we let k be a constant and write (M 2/r) + (M 1/d03)( x2 - y2/2) = k dim(k) = M/L (8.8.65) which is the same as (M 2/k)2 = (x2+y2)(1 - 2εx2 + εy2) = r2 (1 - 2εr2cos2θ + εr2sin2θ) ε ≡ (M 1/kd03) . d i m ( ε) = L-2 (8.8.66) This is a quartic equation whose shape is very simila r to the "ellipse" of (8.8.33). For example, for M 2/k = 1 and ε = .05 we may compare the quartic (red) to the unit circle (blue) : Section 8: Fictitious Forces 82 (8.8.67) To zeroth order in ε we can identify (M 2/k) ≈ R2, the radius of the Earth. We can use this within quantities already of order ε, but more accuracy is needed for the standalone (M 2/k). Equation (8.8.66) is a quadratic in r2 and we solve it for r2(θ) and then r( θ) to get, always for small ε, r(θ) ≈ (M 2/k)[ 1 - (1/8)ε(sin2θ - 2cos2θ)(M2/k)2] . (8.8.68) We then demand that <r( θ)> = R 2 (averaged over θ), just as was done in (8.8.33), so R2 = <r(θ)> = (M2/k)[ 1 - (1/8)ε( 1 2 - 2 12 )(M2/k)2] = (M2/k)[ 1+ (1/16)(M 2/k)2ε ] . (8.8.69) This is then solved for (M 2/k) to order ε with the result (M 2/k) ≈ R2[1- (1/16) εR22] ( 8 . 8 . 7 0 ) which shows the first order correction from just using R 2. The locus of the quartic (8.8.66) is now, R 22[1-(1/16)εR22]2 = (x2+y2) [1 -ε(2x2 - y2) ] ε ≈ (M 1/M2)R2/d03 . ( 8 . 8 . 7 1 ) We then examine the quartic at its right edge (high tide) and its top edge (low tide) to find |Δ|high = (15/16)(M 1/M2)R24/d03 |Δ|low = (9/16)(M 1/M2)R24/d03 H = |Δ| high + |Δ|low = (M1/M2)R24/d03 { (15/16) + (9/16) } (8.8.72) = (3/2)(M 1/M2)R24/d03 . But H = 2a in terms of (8.8.33) so we get Section 8: Fictitious Forces 83 a = (3/4)(M 1/M2)R24/d03 ( 8 . 8 . 7 3 ) in agreement with (8.8.41). Comment : Requiring <r( θ)> = R2 (averaged over θ) both here and implicitly in (8.8.33) is a reasonable approximation, but in the true 3D problem one should really average over θ,φ so that the water volume displaced by low tides equals that piled up by high tides. Section 9: Marion 84 9. Notation comparison with Marion (1970) and Thornton & Marion (2003) In this Section we compare our notation for rotatin g-frame kinematics and non-i nertial-frame physics to that of Marion (1970) and Thornton and Marion (T&M 2003). Their notation is close to our "swap" notation, so before making any comparisons, we restate various of our equations in swap notation. A swap notation eq uation has an s subscript on the equation number and is obtained from the corresponding non-swap equation by prime ↔ noprime ( b and ω do not change) : r' = b + r (6.1)s v' = v + ω x r + b• S' (6.6a)s F' = ma' = mb•• S' + m a + mω x (ω x r) + 2m ω x v + mω• x r (8.1.3)s m a = Feff = F' - m b•• S' – mω x (ω x r) – 2m ω x v – mω• x r (8.1.5)s In the above swap notation equations, Frame S' is fixed (f) and Frame S is the rotating frame (r) : We now compare these equations with those of the Marion series authors: r' = b + r (6.1)s r' = R + r // Marion p 341 (11.1) // T&M p 388 (10.1) v' = v + ω x r + b• S' (6.6a)s vf = vr + ω x r + V // Marion p 344 (11.12) // T&M p 391 (10.17) F' = ma' = mb•• S' + m a + mω x (ω x r) + 2m ω x v + mω• x r (8.1.3)s F = m af = mR•• f + m ar + mω x (ω x r) + 2m ω x vr + mω• x r // Marion p 344 (11.17) // T&M p 392 (10.23) m a = Feff = F' - m b•• S' – mω x (ω x r) – 2m ω x v – mω• x r (8.1.5)s m ar = Feff = F - m R•• f – mω x (ω x r) – 2m ω x vr – mω• x r // Marion p 344 (11.19) // T&M p 392 (10.25) Based on these comparisons, we make the following translation table: Section 9: Marion 85 our swap Marion ( 9 . 1 ) notation authors S r name of the rotating frame (r = rotating) S' f name of the fixed frame (f = fixed) ∂ S (d/dt) rotating time derivative in the rotating frame ∂S' (d/dt) fixed time derivative in the fixed frame r r position in rotating frame v vr velocity in rotating frame a ar acceleration in rotating frame r' r' position in fixed frame v' vf velocity in fixed frame a' af acceleration in fixed frame F' F force in fixed frame = true force in rotating frame Feff Feff total effective force in the rotating frame Ffict total fictitious force in the rotating frame b R location of the rotating frame orig in (measured in the fixed frame) b• S' R• f,V velocity of the rotating frame or igin (measured in the fixed frame) b•• S' R•• f acceleration of the rotating frame origin (measured in the fixed frame) Section 10: Goldstein 86 10. Notation comparison with Goldstein (1950) and Goldstein, Poole and Safko (2001) In this Section we compare our notation for rotatin g-frame kinematics and non-i nertial-frame physics to that of Goldstein on (1950) and Goldstein, Poole and Safko (GFS 2001). These texts are also close to our "swap" notation. Howe ver, in neither text is the reader informed of the location of the origin of the inertial frame or of the rotating frame. In Goldstein we are told that r is a "vector from the origin of the terrestrial system" and that "terrestrial measurements are usually made with respect to a coordinate system fixed in the earth, which therefore rotates with a constant angular velocity ω relative to the inertial system ". Having studied their rotational equa tions, it is our conclusion that the two frames of reference must have their origins co-sited at the center of the Earth, (10.1) One would likely align e3 with e'3 and call it z^. Most of the follow-on discussion is about the Coriolis force which involves velocity and not r, so the location of the origin is not important. In terms of our Fig 1, this means that b = 0 and so r' = r and we are simultaneously Special Case #1 and Special Case #2 since the rotation axis passes through both origins. For this situation, in our velocity table (1.8.4) the left and right sides are exactly the sa me. Similarly acceleration table (1.8.6) is identical to table (1.8.5). For example, with b = 0 we have, vS' = (∂ r/∂t)S' = (∂ r'/∂t)S' = v'S' = v' aS' = (∂2r/∂t2)S' = (∂2r'/∂t2)S' = a'S' = a' . ( 1 0 . 2 ) Of course b•• S' = 0 when b = 0. We now gather up some of our equations in swap notation and state them for the case that b = 0, making use of (10.2) above: r' = r // b = 0 (6.1)s v' = v + ω x r (6.5)s a ' = a + ω• x r + 2 ω x v + ω x (ω x r) (7.4)s m a = Feff = F' – mω x (ω x r) – 2m ω x v – mω• x r . (8.1.5)s In the above swap notation equations, Frame S' is fi xed (space) and Frame S is the rotating frame (r) : We start our comparison with equation (6.5) s : v' = v + ω x r (6.5)s vs = vr + ω x r . // Goldstein p 135 (4-104) Section 10: Goldstein 87 // GPS p 175 (4.88) The next comparison is (we add ω• x r to their equations), a ' = a + ω• x r + 2 ω x v + ω x (ω x r) (7.4)s as = ar + ω• x r + 2 ω x vr + ω x (ω x r) . // Goldstein p 135 (4-105) // GPS p 175 (4.89) And finally, m a = Feff = F' – mω x (ω x r) – 2m ω x v – mω• x r (8.1.5)s m ar = Feff = F – mω x (ω x r) – 2m ω x vr – mω• x r . // Goldstein p 135 (4-106,7) // GPS p 175 (4.90,1) Based on these comparisons, we construct the following translation table: our swap Goldstein ( 1 0 . 3 ) notation authors b = 0 so r = r' S r name of the rotating frame (r = rotating) S' s name of the fixed frame (s = space) ∂ S (d/dt) r time derivative in the rotating frame ∂S' (d/dt) s time derivative in the fixed frame r r position in rotating frame v vr velocity in rotating frame a ar acceleration in rotating frame r' r position in fixed frame v' vs velocity in fixed frame (s = space) a' as acceleration in fixed frame F' F force in fixed frame = true force in rotating frame Feff Feff total effective force in the rotating frame Ffict total fictitious force in the rotating frame b 0 location of the rotating frame orig in (measured in the fixed frame) b• S' 0 velocity of the rotating frame or igin (measured in the fixed frame) b•• S' 0 acceleration of the rotating frame origin (measured in the fixed frame) Since b = 0 is assumed, the Goldstein and GPS texts only treat a special case of the "rotating frames of reference" scenario we depict in Fig 1 in which b(t) is a general dynamic vector. Section 11: Angular Momentum 88 11. Angular Momentum and Fictitious To rques; the Reynolds Transport Theorem 11.1 Introduction Nomenclature is an issue for this subject, a nd here is a vague partial table of usage: "physics" continuum mechanics r x p angular momentum L moment of momentum (moment of p) r x F torque N or τ moment of force (moment of F), moment ( M or m) r⊥ = rsinθ moment arm moment arm Σi ri x Fi with ΣiFi= 0 a sum of torques a couple, a torque, a pure moment (11.1.1) The moment arm r ⊥ is the component of r which is perpendicular to F (or p) as in the drawings below. We shall use the "physics" terminology, so : A Particle located at position r and having linear momentum p is said to have angular momentum L = r x p about the origin, in reference to the origin, or with respect to the origin. (11.1.2) A Particle located at position r which is acted upon by a force F is said to experience a torque N = r x F about the origin, in reference to the origin, or with respect to the origin. (11.1.3) In other words, the tail of vector r is at the origin of the coordinate system and L and N as above are both shown with respect to that point, as in the picture on the left below. Note that a Particle could refer to an actual point particle, or to a small piece of a rigid body, or to a small chunk of a fluid or an elastic solid. Sometimes we want to use L and N with respect to some other reference point, call it c, which is not the origin of the coordinate system. This is shown on the right where the origin of the picture on the left has been translated slightly down and to the right, (11.1.4) The angular momentum and torque with respect to point c are given by L(c) = (r-c) x p L(0) = r x p N(c) = (r-c) x F N(0) = r x F . (11.1.5) Section 11: Angular Momentum 89 We therefore introduce a label " c" to indicate the point of reference for an L or an N. If the reference point is the origin, then we write things as shown on the right above. In the work below, a certain amount of complexity is introduced by allowing c ≠ 0, but the intention is to "do the general case". This then brings up the significance of the last line in (11.1.1) above, and suggests the benefits of dealing with "a couple" when possible. Consider: Theorem: If the sum of a set of forces acting on an object is 0, then the sum of the torques associated with those forces (acting on that same object) is independent of the point c chosen as the reference point for all the torques. (11.1.6) proof: N(c) = Σi(ri-c) x Fi = Σi rix Fi - c x (ΣiFi) = Σi rix Fi - 0 = N(0) . Example: Imagine a cylindrical steel bar in a state of to rsional strain due to e qual and opposite torques applied to the ends of the bar through small gra bber chucks, one at each end (no gravity). So N1 = -N2 = N and each torque is twisting the ba r counterclockwise as seen looking at each end. The effect of a chuck on the bar can be represented as a continuous sum of tangential forces acting on the thin band of surface of the bar under the chuck, which forces add up to zero (think pairwise). Therefore, since the total force of a chuck on the bar is zero, the torque of a chuck on the bar is independent of reference point. Since the same is true for each end, one can pick some arbitrary point c (such as c = 0) and reference both torques to that point, and then add them "l egally" to conclude that the total torque on the bar is 0. The bar thus shows no angular acceleration. (11.1.7) In this example, each chuck represents a couple or pure moment acting on the bar. Section 11: Angular Momentum 90 11.2 Expression of L(c) and L•(c) in terms of Frame S' objects We replicate Fig (1.9.1) which now includes a single torque reference point which is c is Frame S and c' in Frame S'. (11.2.1) The following three equations are obvious from the drawing, r - r' = b ( 1 1 . 2 . 2 ) c - c' = b (11.2.3) r - c = r '- c' . ( 1 1 . 2 . 4 ) For arbitrary reference point c one has from (1.9.4-7), L(c) = ( r-c) x mv (1.9.4) (11.2.5) L•(c) = ( r-c) x ma – c• x mv (1.9.5) (11.2.6) L'(c') = (r'-c') x mv' (1.9.6) (11.2.7) L•'(c') = (r'-c') x ma' – c•' x mv' . (1.9.7) (11.2.8) The reference point c might be moving. We know from Sections 6 and 7 how r,v,a and r',v',a' are related, r = r' + b (6.1) (11.2.9) v = v' + ω x r' + b• S or p = p' + m(ω x r' + b• S) (6.6a) (11.2.10) a = a' + ω• x r' + 2 ω x v' + ω x (ω x r') + b•• S . (7.6a) (11.2.11) The middle equation above can be written, Section 11: Angular Momentum 91 r• = r•' + ω x r' + b• S . (11.2.12) The torque reference point c in Fig (11.2.1) is just like the particle point r, so the above for c becomes c• = c•' + ω x c' + b• S . (11.2.13) Part of the goal of Section 5 is to express L(c) and L•(c) in terms of Frame S' quantities. In the following all algebra is shown to provide an easily traceable pa th since there won't be any result verifications: L(c) = ( r-c) x p = ( r'-c') x [ p' + m(ω x r' + b• S)] // (11.2.4), (11.2.10) = ( r'-c') x p' + m( r'-c') x (ω x r' + b• S) = L'(c') + m( r'-c') x [ (ω x r') + b• S] . // (11.2.7) Now take the ∂S time derivative of L(c) to get, L•(c) = m( r-c) x a – mc• x v // (11.2.6) = m( r'-c') [ a' + ω• x r' + 2 ω x v' + ω x (ω x r') + b•• S] – mc• x v // (11.2.4), (11.2.11) = [m( r'-c') x a' – mc•' x v'] + m( r'-c') x [ ω• x r' + 2 ω x v' + ω x (ω x r') + b•• S] – m c• x v + mc•' x v' = L•'(c') + m( r'-c') x [ ω• x r' + 2 ω x v' + ω x (ω x r') + b•• S] // (11.2.8) – m c• x v + mc•' x v' = L•'(c') + m( r'-c') x [ ω• x r' + 2 ω x v' + ω x (ω x r') + b•• S] – m (c•' + ω x c' + b• S) x ( v' + ω x r' + b• S) + m c•' x v' . // (11.2.13), (11.2.10) These results may now be summarized: L(c) = L'(c') + m( r'-c') x [ (ω x r') + b• S] (11.2.14) L•(c) = L•'(c') + m( r'-c') x [ ω• x r' + 2 ω x v' + ω x (ω x r') + b•• S] – m (c•' + ω x c' + b• S) x ( v' + ω x r' + b• S) + m c•' x v' (11.2.15) These equations express L(c) and L•(c) entirely in terms of Frame S' objects, for the general case where the angular momentum reference point c is arbitrarily selected. The second equation can be rewritten Section 11: Angular Momentum 92 L•'(c') – L•(c) = (r'-c') x Ffict + m(c•' + ω x c' + b• S) x ( v' + ω x r' + b• S) – mc•' x v' (11.2.16) F'fict = – m b•• S – mω x (ω x r') – 2m ω x v' – mω• x r' . (8.1.8) Comment : Just as with the v and a equations of Section 6, the above equations involving angular momentum and its time derivative are valid even if bo th Frame S and Frame S' are non-inertial. This is so because the derivations above are based on the G Rule and those v and a equations, both of which do not require that either Frame be inertial. 11.3 Fictitious Torques and Newton's Rotational Law in a non-inertial frame We now assume that Frame S is an inertial frame and we compare Newton's (2nd) Law for linear motion with that for circular motion, both in Frame S, F = p• = ma = m r•• // p• and r•• "natural" in Frame S (11.3.1) N(c) = L•(c) = I(c) α = I(c) ψ••z^ = (mr2)ψ••z^ . // L•(c) and ψ•• "natural" in Frame S (11.3.2) Here is a drawing showing the various parameters ( I(c) is the moment of inertia about point c ). (11.3.3) Although the general motion of the Particle in Fig (11.2. 1) could be construed as an instantaneous circular motion about some obscure rotation axis, the above equa tions are not really helpful in the general case, and we work instead with torque in terms of angular momentum. L(c) is the angular momentum of our Particle in Frame S relative to reference point c, while N(c) is some externally applied torque about that same reference point acting on the Particle. In Frame S' we want to find some effective Newton's Rotational Law, N'(c') eff = L•'(c') ( 1 1 . 3 . 4 ) N'(c') eff = N(c) + N'(c') fict // defines N'(c') fict (11.3.5) where N'(c') fict is the fictitious torque that mysteriously appears in non-inertial Frame S', so then Section 11: Angular Momentum 93 N(c) + N'(c') fict = L•'(c') . (11.3.6) We can then apply this bogus Newton's Rotational La w in non-inertial Frame S'. This is in complete analogy with the use of fictitious forces as reviewed in Section 8. Solve (11.3.6) for N'(c') fict and then replace N(c) by L•(c) in (11.3.2) to get N'(c') fict = L•'(c') - L•(c) . (11.3.7) Replacing the right side of the above using (11.2.16) gives N'(c') fict = (r'-c') x F'fict + m(c•' + ω x c' + b• S) x ( v' + ω x r' + b• S) – mc•' x v' (11.3.8) where F'fict = – m b•• S – mω x (ω x r') – 2m ω x v' – mω• x r' . (8.1.8) (11.3.9) frame centrifugal Coriolis Euler Writing this out in full, N'(c') fict = - ( r'-c') x [ mb•• S +mω x (ω x r') + 2m ω x v' + mω• x r'] + m (c•' + ω x c' + b• S) x ( v' + ω x r' + b• S) – mc•' x v' . (11.3.10) In the special case that b = 0 and c = 0 we know from (11.2.3) that c' = 0 as well. In this case, (11.3.8) simplifies to become N'(c') fict = r' x F'fict . (11.3.11) 11.4 Application: Fictitious Torques in Fluid Dynamics The object of interest is a blob of fluid contained in a moving volume V m. The boundary S m of this volume V m moves and changes shape such that every point on S m moves at a velocity which matches the local fluid flow velocity v(r,t). As a result, no particles of fluid either enter or leave the blob volume V m as it moves. One implication is that the mass M of the blob V m remains constant. This blob, of some fixed mass M, is our "object of mechanical interest" (later called "the system"). At time t, we imagine that V m sheds a snake skin V c which then remains frozen in time. Then V m(t) aligns with V c at time t and probably at no other time. Whereas V m moves, V c is fixed. V m is called a "material volume" since it flows with the material, while V c is called a "control volum e". In general, all volume integrals except those being differentiated in time are expressed as integrals over Vc. Here is Newton's Rotational Law N = L• (11.3.2) for the this blob of mass M in inertial Frame S, Section 11: Angular Momentum 94 N(0) = ∫Sc r x t dS + ∫Vc r x ρB dV = (d/dt) S [ ∫Vm r x ρv dV ] . // Lai (7.9.1) (11.4.1) This equation appears in Lai et al. p 430 as equation (7.9.1). Here t is a possible external "surface traction" (force per area) acting on the blob's surface S c, B is some possible "body force" per unit mass acting on the blob's interior (perhaps gravity g), and ρ is the mass density of the fluid. Notice that ∫Vm r x ρv dV is an integral of d L = r x dp over the blob, where d p = dm v and dm = ρdV, so this integral L(0) blob is referenced to the point c = 0. Frame S' is some rotating frame whose orig in is aligned with that of Frame S, so b = 0, and the angular momentum reference point in Frame S' is chosen as c' = 0 (recall (11.2.3) c - c' = b). The integrals for N(0) can be evaluated in any frame one likes , and evaluating them in rotating Frame S' gives N(0) = ∫Sc' r' x t dS' + ∫Vc' r' x ρB dV' . (11.4.2) Here the integration point r' runs over surface S c' in the first integral, and over volume V c' in the second. The body force B and the surface traction t are unchanged but are now expressed in terms of r'. What does equation (11.4.1) look like in non-inertial Fram e S' ? According to (11.3.4) and (11.3.5) it is this: N(0) + N'(0) fict = L•'(0) = (d/dt) S'L'(0) . ( 11.3.4),(11.3.5) Using (11.4.2) this may be written as, ( ∫Sc' r' x t dS' + ∫Vc' r' x ρB dV' ) + N'(0) fict = (d/dt ) S' [ ∫Vm' r' x ρv' dV'] . (11.4.3) Here everything is computed in rotating Frame S', but t and B are the same as in Frame S since they are not affected by the fact that the frame S' is rotating. Also, ρ' = ρ since this is mass per volume and the differential volume element is not affected by a rotation. The control volume V c' and its boundary S c' are fixed in Frame S'. On the right, V m' aligns with V c' at time instant t, and ∫Vm' r' x ρ'v' dV' = L'(0) blob . (11.4.4) The question remains: what is the fictitious torque N'(0) fict appearing in (11.4.3)? For a particle of mass dm' at location r' in the blob the contribution is, from (11.3.11), (terms reordered on second line) d N'(0) fict = r' x dF'fict = r' x [– b•• S – ω x (ω x r') – 2 ω x v' – ω• x r'] dm' = r' x [– b•• S – ω• x r' – ω x (ω x r') – 2 ω x v'] dm' . (11.4.5) Section 11: Angular Momentum 95 When this is integrated over the blob, one finds N'(0) fict = ∫Vc'r' x [– b•• S – ω• x r' – ω x (ω x r') – 2 ω x v'] dm' // Lai (7.9.9) (11.4.6) = – ( ∫Vc'r' dm') x b•• S – ∫Vc'r' x(ω• x r'] dm' – ∫Vc'r' x [ω x (ω x r')]dm' - 2 ∫Vc'r' x (ω x v')dm' . frame Euler centrifugal Coriolis This expression appears in Lai et al. as p 431 (7.9.9). Equation (11.4.3) with (11.4.6) is applied on Lai et al. pp 431-432 to a conventional rotating sprinkler. The sprinkler is at translational re st and its horizontal watering tube of length 2r 0 rotates at ω. The control volume V c' is the interior of this rotating watering tube (angled ends are very short), so V c' is fixed in rotating Frame S'. The solution to the problem is shown to be ω = -(Q/A)sin θ/r0 where A is the area of the orifice on each end of the watering tube, Q is the volume of water flow delivered to the sprinkler per unit time, and θ is the angle of each tube-e nd opening viewed from above. (11.4.7) 11.5 Application: Fictitious Forces in Fluid Dynamics A similar equation applies for the fictitious forces (rather than torques) in this same fluid dynamics example. Perhaps this should have appeared in S ection 8, but the groundwork has been laid here. The corresponding equations are F = dp/dt or F = (∫Sc t dS + ∫VcρB dV) = (d/dt) S [ ∫Vm r x ρv dV ] (11.5.1) and F + F'fict = (d/dt ) S' [ p' ] or ( ∫Sc t dS + ∫Vc ρB dV ) + F'fict = (d/dt) S' [ ∫Vm' ρv' dV'] . (11.5.2) Again using (11.3.9) integrated over the blob with dm' = ρdV', F'fict = – b•• S ∫V' dm' – ω x (ω x ∫Vc'r' dm' ) – 2 ω x ∫Vc' v' dm' – ω• x ∫Vc'r' dm' . ( 1 1 . 5 . 3 ) Section 11: Angular Momentum 96 Setting M ≡ ∫V'dm' = total mass of the blob, and combining the last two equations while reordering terms, one gets (d/dt) S' [ ∫Vm' ρv' dV'] = ( ∫Sc t dS + ∫Vc ρB dV ) – [ M b•• S + 2 ω x ∫Vc' v' dm' + ω• x ∫Vc'r' dm' + ω x (ω x ∫Vc'r' dm' ) . / / L a i ( 7 . 7 . 1 4 ) ( 1 1 . 5 . 4 ) This appears as equation (7.7.14) on p 429 of Lai et al. (they use a0 = b•• S and m = M). Here is a translation table relating our notati on to that of Lai et al. (figure on page 428) Lai u s F1 S f i x e d f r a m e F2 S' rotating frame (moving frame) r r position in fixed frame x r' position in rotating frame (d r/dt)F1 = vF1 (d r/dt)S = vS = v velocity in fixed frame (d x/dt)F2 = vF2 (d r'/dt)S' = v'S' = v' velocity in rotating frame R0 b vector linking frame origins r = R 0 + x r = b + r' (D/Dt)F1 = (D/Dt) (d/dt) S derivative in fixed frame (a0) F1 = a0 b•• S (11.5.5) In Lai et al. equations (7.9.1) through (7.9.8) symbol v means vF1 which is our v. In Lai et al. equation (7.9.9) show ing the fictitious torque, symbol v means vF2 which is our v' . 11.6 Comments on the Reynolds Transport Theorem Although a bit off our path, it seems useful to tie this topic in with the previous section. 1. The operation ∂ /∂t differs from the operation d/dt when applie d to an "Eulerian" function of space and time, in which case d/dt is called a material derivativ e and is written D/Dt in fluid dynamics notation, df( r,t)/dt = ∂f/∂t + ∇ f • dr/dt = ∂ f/∂t + v • (∇f) ≡ Df(r,t)/Dt . (11.6.1) In an Eulerian function, the position coordinate r is the current position of a Pa rticle of fluid as one would expect. (In a Lagrangian function, the position argument is the position at which a Particle started out at some earlier time t 0. ) In general, any property of a fluid f( r,t) (such as temperature or density or velocity) varies with r, so the term v • (∇f) does not in general vanish. Section 11: Angular Momentum 97 2. The right side of equation (11.4.1) or (11.5.1) shows the total time derivative of an integral over a material volume V m, which integral represents a mechanical prope rty of our blob object of interest. It is always possible to replace such a time derivativ e with a set of control volume and control surface integrals using a rather elegant theorem known as the Reynolds Transport Theorem (1903) , (d/dt) [ ∫Vm T dV ] = ∫Vc (∂T/∂t) dV + ∫Sc T(v•n) dS = ∫Vc [(dT/dt)+ T div v] dV (11.6.2) where T = T( r,t) is any reasonable function. For example, T could be a scalar like ρ, or a component of a vector like r x ρv in (11.5.1), or a component of any tensor T ijk... . This theorem appears as (7.4.1) and (7.4.2) in Lai et al. page 418 and a proof is given on the next page. [ n is a unit normal to the surface ] Notice that each term has the units of T times volume/sec. If T = ρ , the left expression is dM/dt = 0 and the far right integral being 0 for any V c requires that (dρ/dt) + ρ div v = 0 which is a form of the continuity equation ∂ρ/∂t + div(ρ v) in which ρ is mass density and J = ρv is the mass-current density. 3. Although we write V c = Vm(t) at time t, it is understood that V c is independent of time -- it is that shed snake skin referred to above. Therefore, one regards ( ∂Vc/∂t) = (dV c/dt) = 0, and then in (11.6.2) we can write ∫Vc (∂T(r,t)/∂t) dV = (∂/∂t) [∫Vc T(r,t) dV ] = (d/dt) [ ∫Vc T(r,t) dV ] (11.6.3) where in the last step we use the fact that the integral is a function only of time, since r is integrated out. So The Reynolds Transport Theorem (11.6.2) can be written this way (d/dt) [∫Vm TdV] = (d/dt) [ ∫Vc TdV ] + ∫Sc T(v•n) dS = ∫Vc [(dT/dt)+ T div v] dV . (11.6.4) 4. Writing T = ρb where b is some extensive blob property per unit mass, (11.6.4) becomes (d/dt) [∫Vm b ρdV] = (d/dt) [ ∫Vc b ρdV ] + ∫Sc bρ(v•n) dS = ∫Vc [(d(ρb)/dt)+ (ρb) div v] dV . (11.6.5) In this case, one can regard ∫Vm b ρdV = ∫Vm b dm as the "total amount of b" in the moving fluid blob. This moving blob which recall maintains all its partic les is sometimes called "the system", and the total amount of b in the system might be called B sys. Then (11.6.5) can be written as (dB sys/dt) = (dB CV/dt) + ∫CS bρ(v•n) dS = ∫CV [(d(ρb)/dt)+ (ρb) div v] dV (11.6.6) Section 11: Angular Momentum 98 where CV (or C.V.) is a traditional notation for V c, the control volume, and CS is S c, the control surface. For example, here is a typical web appearance of the Reynolds Transport Theorem in the form of the left equality in (11.6.6) and (11.6.5), (11.6.7) which points out another common nota tion: V with a horizontal bar ( V ¯¯) refers to volume, to distinguish it from V without a slash which refers to velocity. We solved this problem by using lower case v for velocity. When the bar is short, one gets V ¯ which looks a bit like an upside down A, and in fact the logic "for all" symbol ∀ is sometimes used. 5. Conceivably, the vague similarity between the left equation in (11.6.6) and the G Rule (2.1) might be the reason some people refer to the G Rule as a tr ansport theorem. This does seem far fetched. 6. Applying the left equality of (11.6.4) to T = r x ρv gives, in Frame S, (d/dt) S [ ∫Vm r x ρv dV ] = (d/dt) S [ ∫Vc r x ρv dV ] + ∫Sc r x ρv (v•n) dS , (11.6.8) so that (11.4.1) may be written N(0) = ( ∫Sc r x t dS + ∫Vc r x ρB dV ) = (d/dt) [ ∫Vc r x ρv dV ] + ∫Sc (r x ρv) (v•n) dS . // Lai (7.9.8) (11.6.9) This says that the total torque on a fluid blob eq uals the rate of change of the angular momentum contained in the frozen control volume V c plus the rate of outflow of angular momentum from that volume. This equation appears as (7.9.8) in Lai et al. p 431. When working in a rotating frame, we have to add to the left side of (11.6.9) the fictitious torques stated in (11.4.6) and which Lai et al. states as (7.9.9). Here then are a few quotes from Lai et al. (p 430-431) which use the continuum mechanics terminology as s hown in the table at the start of this Section : --------------- Section 11: Angular Momentum 99 The equation (7.9.8) and its italicized interpretation express the familiar law of conservation of angular momentum. In Chapter 7 Lai et al. have similar se ctions for conservation of mass, energy, and linear momentum, and a final section on the inequality of entropy, each of these being a "principle". Each section uses the Reynolds Transport Theorem to replace its D/Dt [ ∫Vm ...] object, and each section ends up with an equation like (7.9.8) with an italicized interpretation. Section 12: Forward Problem Summary 100 12. Summary of the Forward Problem Solution 12.1 Summary of the Forward Problem equations (non-swap notation) We now summarize the results of Sections 6, 7, 8 and 11. The first set of equations below is valid regardless of whether either of these frames is inerti al (they could both be non-inertial). The second set of equations involving fictitious forces and torques assume s that Frame S is inertial (and Frame S' is not). (12.1.1) Definitions and Equations r, v, a position, natural veloc ity and natural acceleration in Frame S (12.1.2) r', v', a' position, natural velocity and natural acceleration in Frame S' ω angular velocity of Frame S' relative to Frame S b vector directed from origin of Frame S to origin of Frame S' r = b + r' (6.1) (a) v = v' + ω x r' + b• S (6.6a) (b) v = v' + ω x r + b• S' (6.6c) (c) a = a' + ω• x r' + 2 ω x v' + ω x (ω x r') + b•• S (7.6a) (d) S S' Euler Coriolis centripetal frame a = a' + ω• x r + 2 ω x v' + ω x (ω x r) + 2ω x b• S' + b•• S' (7.6b) (e) L(c) = L'(c') + m( r'-c') x [ (ω x r') + b• S] (11.2.14) (f) L•(c) = L•'(c') + m( r'-c') x [ ω• x r' + 2 ω x v' + ω x (ω x r') + b•• S] – m (c•' + ω x c' + b• S) x ( v' + ω x r' + b• S) + m c•' x v' (11.2.15) (g) Section 12: Forward Problem Summary 101 Fictitious Forces (Section 8) For using fictitious forces we have (for these equations Frame S is inertial) F = ma // true Newton's Law in inertial Frame S (8.1.2) F'eff = ma' // fake Newton's Law in rotating Frame S' (8.1.4) F'eff = F + F'fict . (8.1.7) (12.1.3) For the general case, the fictitious forces can be expressed as F'fict = – m b•• S – mω x (ω x r') – 2m ω x v' – mω• x r' . (8.1.8) (12.1.4) frame centrifugal Coriolis Euler For Special Case # 1 problems ( ω axis passes through Frame S origin), we have F'fict = – mω x (ω x r) – 2m ω x v' – mω• x r . Special Case #1 (8.4.2) (12.1.5) centrifugal Coriolis Euler Fictitious Torques (Section 11) // Frame S is inertial N'(c') fict = ( r'-c') x F'fict + m(c•' + ω x c' + b• S) x ( v' + ω x r' + b• S) – mc•' x v' (11.3.8) or N'(c') fict = - ( r'-c') x [ mb•• S + mω x (ω x r') + 2m ω x v' + mω• x r'] (12.1.6) + m (c•' + ω x c' + b• S) x ( v' + ω x r' + b• S) – mc•' x v' (11.3.10) Section 12: Forward Problem Summary 102 12.2 Summary of the Forward Problem equations (swap notation) As a reminder, we quote from above: Swap Notation. The meaning of "swap notation" is that, in Fig 1, the vectors ω and b stay put, but all other vectors undergo V↔V'. This latter group includes basis vectors ei ↔ e'i, r ↔ r' , v ↔ v', a ↔ a' and of course Frame S ↔ Frame S'. This is nothing more than a cha nge of the way things are labeled. If a non-swap notation equation has the number (x.x.x), then the corresponding equation in swap notation will be given the number (x.x.x) s . (8.5.10) We are now going to translate everything in Section 12.1 into swap notation. Fig (12.2.1) is the same as Fig (12.1.1) but the primes are swapped as just discussed. The first set of equations below is valid regardless of whether either of these frames is inertial (they could both be non-inertial). The second set of equations involving fictitious fo rces assumes that Frame S' is inertial (and therefore Frame S is not). (12.2.1) Definitions and Equations r', v', a' position, natural veloc ity and natural acceleration in Frame S' (12.2.2) r, v, a position, natural velocity and natural acceleration in Frame S ω angular velocity of Frame S relative to Frame S' b vector directed from origin of Frame S' to origin of Frame S r' = b + r (6.1)s (a) v' = v + ω x r + b• S' (6.6a)s (b) v' = v + ω x r' + b• S (6.6c)s (c) a' = a + ω• x r + 2 ω x v + ω x (ω x r) + b•• S' (7.6a)s (d) S' S Euler Coriolis centripetal frame Section 12: Forward Problem Summary 103 a' = a + ω• x r' + 2 ω x v + ω x (ω x r') + 2ω x b• S + b•• S (7.6b)s (e) L'(c') = L(c) + m( r-c) x [ (ω x r) + b• S'] (11.2.14) s (f) L•'(c') = L•(c) + m( r-c) x [ ω• x r + 2 ω x v + ω x (ω x r) + b•• S'] – m (c• + ω x c + b• S') x ( v + ω x r + b• S') + m c• x v (11.2.15) s (g) Fictitious Forces (Section 8.1) For using fictitious forces we ha ve (Frame S' is inertial) F' = ma' // true Newton's Law in inertial Frame S' (8.1.2)s Feff = ma // fake Newton's Law in rotating Frame S (8.1.4)s Feff = F' + Ffict (8.1.7)s (12.2.3) For the general case, the fictitious forces can be expressed as Ffict = – m b•• S' – mω x (ω x r) – 2m ω x v – mω• x r (8.1.8)s (12.2.4) frame centrifugal Coriolis Euler For Special Case # 1 problems ( ω axis passes through Frame S' origin), we have Ffict = – mω x (ω x r') – 2m ω x v – mω• x r' Special Case #1 (8.4.1)s (12.2.5) centrifugal Coriolis Euler Fictitious Torques (Section 11.3) N(c) fict = ( r-c) x Ffict + m(c• + ω x c + b• S') x ( v + ω x r + b• S') – mc• x v (11.3.8)s or N(c) fict = - ( r-c) x [ mb•• S' + mω x (ω x r) + 2m ω x v + mω• x r] (12.2.6) + m (c• + ω x c + b• S') x ( v + ω x r + b• S') – mc• x v (11.3.10) s Section 13: The Inverse Problem 104 13. The Inverse Problem Consider these two problems which concern the exact same physical situation (non-swap notation): Forward Problem: given: r', v', a', L'(c'), L•'(c') find: r, v, a, L(c), L•(c) // summarized in Section 12.1 Inverse Problem: given: r, v, a, L(c), L•(c) find: r', v', a', L'(c'), L•'(c') // to be summarized in Section 13.3 below Recall that equations involving just the above quanti ties are valid even if both Frames S and S' are non- inertial. On the other hand, equations involving F'fict and N'(c) fict require that Frame S be inertial. If Frame S is inertial for the Forward Problem, it is also inertial for the Inverse Problem and in that case we know that there is no fictitious force or torque in Frame S, so these items are not included on the list of quantities shown above. We shall first compute the inverse equations by brute force, then at the end show how they can be obtained by a set of simple swap rules. 13.1 Brute Force Method Looking at the Section 12.1 summary, equation (6.1) is easily inverted r = b + r' => r' = r - b ( 1 3 . 1 . 1 ) Similarly for (6.6a), where the third line below uses identity (6.2b), v = v' + ω x r' + b• S (6.6a) v' = v – ω x r' – b• S ( 1 3 . 1 . 2 ) v' = v – ω x r – b• S' . ( 1 3 . 1 . 3 ) Equation (7.6a) requires a bit more effort to invert. We first solve (7.6a ) for a' a = a' + ω• x r' + 2 ω x v' + ω x (ω x r') + b•• S (7.6a) a' = a – ω• x r' – 2 ω x v' – ω x (ω x r') – b•• S . (13.1.4) Replace v' using (13.1.2), a' = a – ω• x r' – 2 ω x [v – ω x r' – b• S] – ω x (ω x r') – b•• S Section 13: The Inverse Problem 105 = a – ω• x r' – 2 ω x v + 2 ω x (ω x r') + 2 ω x b• S – ω x (ω x r') – b•• S = a – ω• x r' – 2 ω x v + ω x (ω x r') + 2 ω x b• S – b•• S . (13.1.5) With r' = r – b we can regard the RHS of (13.1.5) as be ing expressed entirely in terms of Frame S objects. Now, by first shuffling terms in (7.11), b•• S' = b•• S – ω• x b – 2ω x b• S + ω x (ω x b) = > (7.11) 2 ω x b• S – b•• S = – ω• x b + ω x (ω x b) – b•• S' we can replace the last two terms in (13.1.5) to get a' = a – ω• x r' – 2 ω x v + ω x (ω x r') – ω• x b + ω x (ω x b) – b•• S' = a – ω• x r – 2 ω x v + ω x (ω x r) – b•• S' (13.1.6) Inversion of the L(c) and L•(c) can also be done by this brute force method, but we spare the reader and instead quote the results later after es tablishing the "swap rules method". Summary of the inverse problem results obtained by brute force: r' = r - b (13.1.1) v' = v – ω x r' – b• S (13.1.2) v' = v – ω x r – b• S' (13.1.3) a' = a – ω• x r' – 2 ω x v + ω x (ω x r') + 2 ω x b• S – b•• S (13.1.5) a' = a – ω• x r – 2 ω x v + ω x (ω x r) – b•• S' (13.1.6) 13.2 Swap Rules Method Without any justification yet, let us postulate that we can obtain our inverse problem equations directly from the forward problem equations (and vice versa) using this set of Swap Rules : Section 13: The Inverse Problem 106 forward problem equations ← swap rules → inverse problem equations r ↔ r' b ↔ – b ω ↔ – ω L(c) ↔ L'(c') (13.2.1) v ↔ v' b• S ↔ – b• S' L•(c) ↔ L•'(c') a ↔ a' b•• S ↔ – b•• S' // Swap Rules This set of rules is different from our rules for going between swap and no-swap notation, which are these r ↔ r' b ↔ b ω ↔ ω L(c) ↔ L'(c') S ↔ S' (13.2.2) v ↔ v' b• S ↔ b• S' L•(c) ↔ L•'(c') a ↔ a' b•• S ↔ b•• S' // rules for going between swap and no-swap notation. The big difference is that with the Swap Rules we are negating the vectors b (and its derivatives) and ω. Later in Section 13.5 we will justify the set of ru les (13.2.1). Whereas the change from non-swap notation to swap notation is just a cosmetic relabeling of a problem's variables, application of the Swap Rules changes a problem which is the Forward Problem into a different problem which is the Inverse Problem defined at the start of Section 13.1. Since the inverse problem equations computed by brut e force are sitting just above, let's apply the Swap Rules (13.2.1) to them and see what we get: r = r' + b (13.1.1)swapped v = v' + ω x r + b• S' (13.1.2)swapped v = v' + ω x r' + b• S (13.1.3)swapped a = a' + ω• x r + 2 ω x v' + ω x (ω x r) + 2 ω x b• S' + b•• S' (13.1.5)swapped a = a' + ω• x r' + 2 ω x v' + ω x (ω x r') + b•• S . (13.1.6)swapped And now we directly quote from the summary from Section 12.1 above r = b + r' (6.1) v = v' + ω x r + b• S' (6.6c) v = v' + ω x r' + b• S (6.6a) a = a' + ω• x r + 2 ω x v' + ω x (ω x r) + 2ω x b• S' + b•• S' (7.6b) a = a' + ω• x r' + 2 ω x v' + ω x (ω x r') + b•• S (7.6a) Since the last two sets of equations are identical, we have demonstrated that the Swap Rules presented in (13.2.1) do indeed convert either set of equations into the other. Section 13: The Inverse Problem 107 13.3 Summary of the Inverse Problem Equations (non-swap notation) As noted above, we obtain the Inverse Problem equations by applying the Swap Rules to the Forward Problem equations. But the Swap Rules can be thought of as having two steps: (1) do the swap one usually does to go from swap to non-swap notation; (2) then take b→ -b (including derivatives) and ω→ -ω (including derivatives ). We have step (1) already carried out in Section 12.2, so to get the equations below we need only carry out step (2) on the Section 12.2 equations. Here is a repeat of Fig (12.1.1) which of course applies to both the Forward Problem and the Inverse Problem : (13.3.1) r', v', a' position, natural veloc ity and natural acceleration in Frame S' (13.3.2) r, v, a position, natural velocity and natural acceleration in Frame S ω angular velocity of Frame S' relative to Frame S b vector directed from origin of Frame S to origin of Frame S' r' = - b+ r ( a ) v' = v – ω x r – b• S' // all these equations are from (12.2.2) with ω and b negated (b) v' = v – ω x r' – b• S ( c ) a' = a – ω• x r – 2 ω x v + ω x (ω x r) – b•• S' ( d ) a' = a – ω• x r' – 2 ω x v + ω x (ω x r') + 2 ω x b• S – b•• S (e) L'(c') = L(c) + m( r-c) x [ -(ω x r) - b• S'] (f) L•'(c') = L•(c) + m( r-c) x [- ω• x r - 2 ω x v + ω x (ω x r) - b•• S'] – m (c• - ω x c - b• S') x ( v - ω x r - b• S') + m c• x v (g) Section 13: The Inverse Problem 108 13.4 Summary of the Inverse Problem Equations (swap notation) The results of this section are those of Section 13.3 but with S ↔ S' and v'↔v for all vectors except b and ω. Again, this is just a change of labeling. We rep eat Fig (12.2.1) which shows the swap notation case : (13.4.1) r, v, a position, natural veloc ity and natural acceleration in Frame S (13.3.2) r', v', a' position, natural velocity and natural acceleration in Frame S' ω angular velocity of Frame S relative to Frame S' b vector directed from origin of Frame S' to origin of Frame S r = - b+ r' ( a ) v = v' – ω x r' – b• S (b) v = v' – ω x r – b• S' ( c ) a = a' – ω• x r' – 2 ω x v' + ω x (ω x r') – b•• S ( d ) a = a' – ω• x r – 2 ω x v' + ω x (ω x r) + 2 ω x b• S' – b•• S' (e) L(c) = L'(c') + m( r'-c') x [ -(ω x r') - b• S] (f) L•(c) = L•'(c') + m( r'-c') x [- ω• x r' - 2 ω x v' + ω x (ω x r') - b•• S] – m (c•' - ω x c' - b• S) x ( v' - ω x r' - b• S) + m c•' x v' (g) Section 13: The Inverse Problem 109 13.5 Why the Swap Rules Work Preview : We are going to show here that if one starts with an initial picture of the physical situation between two frames of references a nd a Particle, and if one applies th e Swap Rules (13.2.1) to that picture, one ends up with a final picture which is exactly the same as the initial picture. Therefore, if the initial physical picture is described by a set of equa tions, then applying the Swap Rules to those equations gives new equations which also apply to the initial pictur e (since it is the same as the final picture). Thus, if the initial set of equations is valid, so is the fi nal set of equations obtained via these Swap Rules. The equations obtained by application of the Swap Ru les provide the "answers" for our Inverse Problem defined at the start of Section 13. Start with Fig (12.1.1), (13.5.1) Apply the Swap Rules of (13.2.1) to get (13.5.2) Rather than take b → -b as a label, we have flipped the arrow on the b vector to achieve the same result. Now observe the above scenario from a camera plat form which is rotating counterclockwise at rate ω with its rotation axis the same as that shown in the figure. Viewed from this camera's rotating frame, Frame S is at rest, and Frame S' is rotating counterclockwise. That camera-viewed picture is then, Section 13: The Inverse Problem 110 (13.5.3) This is the picture one sees if one observes things fr om Frame S. Our equations of interest remain valid despite the camera's rotation because these equations are based on the G Rule which is a function only of the relative relationship between frames, and making the camera rotate above does not change this relationship. Moreover, these equations are the sa me regardless of where one puts the ω axis, regardless of where one places the Particle, and regardless of where and how one orients the two Reference frames. So let's move these things around a bit in the above picture to get this new drawing which has the same equations, (13.5.4) Now we rotate this picture about an axis near the center and perpendicular to the plane of paper. Such a rotation again does not change the equations a ssociated with the picture. We then have (13.5.5) Section 13: The Inverse Problem 111 But this is the same as the picture we started with above (apart from colors and text orientation), (13.5.1) Since the Swap Rules, along with various equation-invariant reorientations, produce a final picture which is the same as the initial picture, when those Swap Rules are applied to a valid set of equations which apply to the initial picture, the resulting equations are also valid for the initial picture since this is the same as the final picture. Section 14: Curvilinear Coordinates 112 14. Rotating Frames in Curvilinear Coordinates The solution equations for our Forward Problem ar e summarized in Section 12.1 and 12.2 above, and those for the Inverse Problem are summarized in Secti on 13.3 and 13.4. All equations are stated in bolded vector notation. Such equations may be projected ont o (dotted with) any complete set of basis vectors, such as the r^, θ^, φ^ used in spherical coordinate s. Every orthogonal curvilinear coordinate system has such a set of orthonormal unit basis vectors which we shall call e^i, orthonormality meaning e^i • e^j = δi,j. In general, curvilinear basis vectors like e^i= r^, θ^, φ^ are different at different points in space, so one can think of them as e^i(r). It is appropriate then to use them as basis vectors for a vector field V(r) or for a vector associated with a discrete Particle located at position r such as the velocity or acceleration of that Particle. We might want to use one curvilinear system of coordinates ξi with basis unit vectors e^i for Frame S, and an entirely different system ξ'i with basis unit vectors e^'i for Frame S'. We might, for example, have ξi be spherical coordinates and ξ'i be toroidal coordinates. He re then is the situation, Cartesian coords and basis vectors Curvilinear coords and basis vectors Frame S r i ei ξi e^i Frame S' (r') i e'i ( ξ')i e^'i (14.1) ei• ej = e'i• e'j = e^i • e^j = e^'i • e^'j = δi,j // orthonormality of all bases (14.2) There must exist some matrix R( ξ) such that e^i(r) = R(ξ)ei for any given curvilinear system. Recall now equations (1.1.1) and (1.1.2), en = R e'n n = 1,2,3 or ( en)i = Rij(e'n)j (1.1.1) en = (R-1)nm e'm or e'n = Rnm em . (1.1.2) In these equations replacing en → e^n and e'n → en and R → R(ξ) gives, e^n = R(ξ ) en n = 1,2,3 or ( e^n)i = R(ξ )ij(en)j e^n = (R(ξ)-1)nm em or en = R(ξ )nm e^m (14.3) which we summarize in the first line below. The second line is for then some other curvilinear coordinate system in Frame S'. e^n = R(ξ) en => en = R(ξ)nme^m or e^n = [R(ξ)]-1 nm em e^'n = R'(ξ') e'n => e'n = R'(ξ ')nme^'m or e^'n = [R'(ξ ')]-1 nm e'm . (14.4) Section 14: Curvilinear Coordinates 113 Example of an R( ξ) matrix. In spherical coordinates with ordering 1,2,3 = r, θ,φ, where θ is the polar angle and φ the azimuth, the matrix R( ξ) is given by (note that R-1 = RT), R(ξ) = ⎝⎜⎛ ⎠⎟⎞ cosφsinθ cosφcosθ -sinφ sinφ sinθ sinφcosθ cosφ cosθ -sinθ 0 [R( ξ)]-1 = ⎝⎜⎛ ⎠⎟⎞ cosφsinθ sinφsinθ cosθ cosφcosθ sinφ cosθ -sinθ -sinφ cosφ 0 (14.5) as shown in (A.9), (A.11). We can then use (14.3) that e^n = [R(ξ) ]-1 nm em to write (in matrix notation), ⎝⎜⎜⎛ ⎠⎟⎟⎞ e^1 e^2 e^3 = ⎝⎜⎜⎛ ⎠⎟⎟⎞ r^ θ^ φ^ = ⎝⎜⎛ ⎠⎟⎞ cosφsinθ sinφsinθ cosθ cosφcosθ sinφ cosθ -sinθ -sinφ cosφ 0 ⎝⎜⎜⎛ ⎠⎟⎟⎞ x^ y^ z^ = [R( ξ) ]-1 ⎝⎜⎜⎛ ⎠⎟⎟⎞ x^ y^ z^ = [R( ξ) ]-1 ⎝⎜⎜⎛ ⎠⎟⎟⎞ e1 e2 e3 (14.6) or r^ = cosφsinθ x^ + sinφsinθ y^ + cosθ z^ θ^ = cosφcosθ x^ + sinφcosθ y^ – sinθ z^ φ^ = –sinφ x^ + cosφ y^ . ( 1 4 . 7 ) Expansions and naming. If V is an arbitrary vector, we then have these four expansions of interest : V = Viei V i = V • ei V = (V)'i e'i (V)' i = V • e'i V = (V)i e^i ( V)i = V • e^i V = (V)'i e^'i ( V)'i = V • ei ( 1 4 . 8 ) where we use italics to denote curvilinear vector co mponents. It is common practice, once a curvilinear system is selected, to make these replace ments so the italics are no longer needed, ( V)i → Vξi ( V)'i → Vξi' ( 1 4 . 9 ) In cylindrical coordinates r, θ,z and r',θ',z' this would mean, for example, ( V)1 → Vr ( V)'1 → Vr' ( V)2 → Vθ ( V)'1 → Vθ' ( V)3 → Vz ( V)'2 → Vz' . (14.10) Equation Example . Consider now this equation taken from the Section 12.1 summary, v = v' + ω x r + b• S' . (6.6c) (14.11) We can view such an equation in any of our four bases as just discussed above, Section 14: Curvilinear Coordinates 114 (v)i = (v')i + εijk(ω)j(r)k + (b• S')i components in basis ei (v)'i = (v')'i + εijk(ω)'j(r)'k + (b• S')'i components in basis e'i ( v)i = ( v')i + εijk(ω)j(r)k + ( b• S')i components in basis e^i ( v)'i = (v')'i + εijk(ω)'j(r)'k + (b• S')'i . components in basis e^'i (14.12) For example, in r, θ,z cylindrical coordinates if we have ω = ω z^, then (ω)j = δj3ω , so in the third line above we get εijk(ω)j(r)k = εijk ω δj3 (r)k = ω εi3k(r)k = - ω εik3(r)k so that line becomes ( v)i = (v')i - ω εik3(r)k + (b• S')i components in basis e^i or ( v)1 = (v')1 - ω ε123(r)2 + (b• S')1 ( v)2 = (v')2 - ω ε213(r)1 + (b• S')2 ( v)3 = (v')3 - ω ε3k3(r)k + (b• S')3 = ( v')3 + (b• S')3 . (14.13) This then translates into (since r = rr^ + zz^ = rrr^ + rzz^ and rθ = 0) vr = v'r - ω rθ + (b• S')r = v' r + (b• S')r vθ = v'θ + ω rr + (b• S')θ = v' θ + ω r + ( b• S')θ vz = v'z + (b• S')z . ( 1 4 . 1 4 ) For any Special Case #1 problem (see Section 4.4) one has b• S' = 0 and the above equations become extremely simple vr = v'r vθ = v'θ + ω r vz = v'z . ( 1 4 . 1 5 ) Section 15: Ant on Turntable 115 15. Ant on Turntable Problems The main purpose of the following four "ant problem" examples is to exercise the results summarized in Sections 12 and 13 above and to demonstrate the use of non-Cartesian coordinates as outlined in Section 14. Problems 1 and 2 are "forward" problems, while Problems 3 and 4 are "inverse" problems. Problem 4 concludes the analysis of the 4-projectile problem begun in Section 8. Everything is done in "no-swap" notation where Frame S' is the rotating frame. Some secondary purposes are to provide the read er with many examples of manipulating basis vectors, using the bulletproof vector component not ation of Section 1, and applying simple matrix methods. Kinematics common to all Ant Problems Consider a turntable occupied by an ant as shown in this drawing. Here Frame S is a fixed frame with origin at the turntable center, while Frame S' (glu ed to the turntable surface) is a rotating frame. (15.1) In Frame S, the vector r has coordinates (r, θ) in standard polar coordinates. In Frame S', the vector r' has coordinates (r', θ') in standard polar coordinates. When φ = 0, red Frame S' lies directly under black Frame S and the axes line up. For any angle φ one has b = -b e'2. Since the rotation axis goes through the origin of Frame S, the turntable problems fall into Special Case #1 of Section 4.4. Basis vectors e3 = e'3 (not labeled) point to the viewer as does the ω vector for ω>0. The relation between the three angles θ, θ' and φ is complicated and can be indirectly obtained by writing the laws of sines and cosines for the triang le shown on the right above. The left and bottom internal triangle angles are obvious. The top one is then π - (θ + π/2 - φ ) - (π /2 - θ ') = θ '-θ+φ ( 1 5 . 2 ) None of this angle detail will be needed below (except in a Reader Exercise). Section 15: Ant on Turntable 116 In the first two Problems considered below, an ant executes some crawling motion on the turntable as described by certain r', v', and a' in Frame S'. Our task in each pr oblem is to use our Section 12.1 summary results to compute r, v, and a as seen in Frame S and to plot some trajectories r(t). In the third problem, the ant becomes a flying ant doing a straight-line fly-by at constant velocity in Frame S just over the turntable surface, a fly-by described by a certain r, v, and a. This is an example of the Inverse Problem discussed in Section 13 and our goal here is to compute r', v', and a' in Frame S' using the equations provided in Section 13.3. In each frame we define Cartesian and cylindr ical coordinates and unit vectors as follows: Frame S r i = x,y,z basis vectors ei = x^, y^, z^ ξi = r,θ,z basis vectors e^i = r^, θ^, z^ . Frame S' r' i = x',y',z' basis vectors e'i = x^', y^', z^' ξ'i = r',θ',z' basis vectors e^'i = r^', θ^', z^' (15.3) We are thus providing a specific example of (14.1) concerning general curvilinear coordinates in two frames of reference. What do we know about all the basis vectors? In order to illustrate some of the work of Section 1.1, we provide the reader with a complete set of socket wrenches even though only a few of these tools will act ually be used below. All of the following relations can be obtained by inspection from the above figure: Within Frame S we have e^i = Rz(θ)ei . for example r^ = Rz(θ) x^ (a) A corresponding equation applies in Frame S' , e^'i = Rz(θ')e'i . for example r^' = Rz(θ') x^' (b) The relation between the Frame S and Frame S' Cartesian unit vectors is e'i = Rz(φ)ei . for example x^' = Rz(φ) x^ (c) The relation between the Frame S and Frame S' cylindrical unit vectors is e^'i = Rz(φ)e^i . for example r^' = Rz(φ) r^ (d) Relations (d) and (a) can be combined to get e^'i = Rz(φ+θ)ei . for example r^' = Rz(φ+θ) x^ (e) (15.4) Section 15: Ant on Turntable 117 Writing the basis vector relations in matrix notation Recall the following theorem of (1.1.1) and (1.1.2) : (we use dummy basis vector names an and a'n ) , an = R a'n ⇔ an = Σm(R-1)nm a'm (15.5) On the left, we rotate vector a'n to get vector an. On the right, we express an as a linear combination of the basis vectors a'n. Remember that the subscripts on the a and a' are labels, not components! Suppose we take the kth component of the equation on the right of (15.5), [ an]k = Σm(R-1)nm [a'm]k . ( 1 5 . 6 a ) One can write this as Ank = Σm(R-1)nm A'mk where A nk = [an]k and A'nk = [a'n]k . (15.6b) For a matrix A nk one knows that n is the row index and k is the column index. Therefore, saying A nk = [an]k is the same as saying that the vector an is the nth row of matrix A. Thus we can write (15.6a) in this manner, ⎝⎜⎜⎛ ⎠⎟⎟⎞ a1 a2 a3 = R-1 ⎝⎜⎜⎛ ⎠⎟⎟⎞ a'1 a'2 a'3 ⇔ an = R a'n n = 1,2,3 . (15.7) All our rotations of interest in (15.4) are z-rotations which, from (A.1), have the form Rz(ψ) = ⎝⎜⎛ ⎠⎟⎞ cosψ -sinψ 0 sinψ cosψ 0 0 0 1 . (A.1)z Example 1 : Apply (15.7) to (15.4e) which says e^'i = Rz(φ+θ)ei : ⎝⎜⎜⎛ ⎠⎟⎟⎞ e^'1 e^'2 e^'3 = Rz(-θ' - φ) ⎝⎜⎜⎛ ⎠⎟⎟⎞ e1 e2 e3 = ⎝⎜⎛ ⎠⎟⎞ cos(θ '+φ) sin(θ'+φ) 0 -sin(θ'+φ) cos(θ '+φ) 0 0 0 1 ⎝⎜⎜⎛ ⎠⎟⎟⎞ e1 e2 e3 or ⎝⎜⎜⎛ ⎠⎟⎟⎞ r^' θ^' z^' = ⎝⎜⎛ ⎠⎟⎞ cos(θ '+φ) sin(θ'+φ) 0 -sin(θ'+φ) cos(θ '+φ) 0 0 0 1 ⎝⎜⎜⎛ ⎠⎟⎟⎞ x^ y^ z^ . Writing out the linear combinations, one gets Section 15: Ant on Turntable 118 r^' = cos(θ'+φ) x^ + sin(θ'+φ) y^ θ^' = - sin(θ'+φ) x^ + cos(θ '+φ) y^ z^' = z^ . ( 1 5 . 8 ) Example 2 : Apply (15.8) to (15.4c) which says e'i = Rz(φ)ei: ⎝⎜⎜⎛ ⎠⎟⎟⎞ e'1 e'2 e'3 = Rz(-φ) ⎝⎜⎜⎛ ⎠⎟⎟⎞ e1 e2 e3 = ⎝⎜⎛ ⎠⎟⎞ cosφ sinφ 0 -sinφ cosφ 0 0 0 1 ⎝⎜⎜⎛ ⎠⎟⎟⎞ e1 e2 e3 or ⎝⎜⎜⎛ ⎠⎟⎟⎞ x ^' y^' z^' = ⎝⎜⎛ ⎠⎟⎞ cosφ sinφ 0 -sinφ cosφ 0 0 0 1 ⎝⎜⎜⎛ ⎠⎟⎟⎞ x^ y^ z^ . Writing out the linear combinations, one gets x^' = cosφ x^ + sinφ y^ y^' = -sinφ x^ + cosφ y^ // e'2 = - sinφ e1 + cosφ e2 z^' = z^ . ( 1 5 . 9 ) We could reduce our 3x3 matrix work to 2x2 for the turntable examples, but other problems require the full 3x3 notation so we maintain it throughout. Relation between Frame S and Frame S' Assume at time t = 0 we have φ = φ0 in Fig (15.1). If the rotation follows some angular velocity profile ω = ω(t), and since ω = dφ/dt, one has dφ/dt = ω(t) => φ(t) = φ0 + ∫0 t ω(τ)dτ . (15.10) For simplicity, we shall assume constant ω in which case, φ(t) = φ 0 + ωt . ( 1 5 . 1 1 ) Motion of vector b From Fig (15.1) and from (15.9) one finds, b(t) = -b e'2 = -b [- sin φ x^ + cosφ y^] = bsinφ x^ – b cosφ y^ . (15.12) Section 15: Ant on Turntable 119 15.1 Problem 1: Ant crawls at constant speed V to the Origin of Frame S' (15.1.1) Ant's Motion in Frame S'. Assume the ant starts at some (r' 0,θ'0) at t = 0 and crawls with constant speed V toward the S' origin, v' = -V r^' . ( 1 5 . 1 . 2 ) We can integrate this within Frame S' (where r^' is fixed) to get r'(t) = r'0 -Vt r^' where r'0 = r'0r^' = (r'0,θ'0) . (15.1.3) The magnitude of r'(t) is given by r' = r'0 - V t ( 1 5 . 1 . 4 ) since we shall only be interested in times small enough so r' > 0. The angle θ' never changes, so θ' = θ'0 . ( 1 5 . 1 . 5 ) Finally, since V = constant, the acceleration is a' = 0 . ( 1 5 . 1 . 6 ) Thus, in line with our Forward Problem statement, these are the given quantities in Frame S' , r' = r'0 – Vt r^' v' = –V r^' a' = 0 . ( 1 5 . 1 . 7 ) Section 15: Ant on Turntable 120 Using (15.8) for r^', we can write v' = -V r^' as v' = –Vcos( θ'+φ) x^ – Vsin(θ'+φ) y^ . (15.1.8) Our goal is to compute r, v and a as seen in Frame S. Trajectory r(t) of the ant in Frame S Above we found that b(t) = bsin φ x^ – b cosφ y^ (15.12) r^' = cos(θ'+φ) x^ + sin(θ'+φ) y^ . (15.8) Recall from (12.2.1a) that, r = b + r' = b + r' r^' . (12.2.1a) Therefore from (15.12) and (15.8) quoted just above we can write r(t) = ( bsinφ x^ – b cosφ y^ ) + r' (cos(θ'+φ) x^ + sin(θ'+φ) y^ ) = [bsin φ + r'cos(θ'+φ)] x^ + [– b cos φ + r'sin(θ'+φ)] y^ . Setting r' = (r'0 - Vt) by (15.1.4), and thinking of φ = φ(t) (ie, a function of time) as in (15.10,11), r(t) = [ bsin φ + (r'0 - Vt)cos( θ'+φ)] x^ + [ –b cos φ + (r'0 - Vt) sin( θ'+φ)] y^ or r(t) = x x^ + y y^ w h e r e ( 1 5 . 1 . 9 ) x = bsin φ + (r'0 – Vt)cos( θ'+φ) y = – bcos φ + (r'0 – Vt)sin( θ'+φ) . This r(t) then is the trajectory of the ant in Frame S. Velocity v(t) of the ant in Frame S Since the turntable falls into our Special Case #1 of Section 4.4 ( ω through origin of Frame S), we know that b• S' = 0 (vector b is soldered to the Frame S' unit vectors). From (12.1.2b) we have, v = v' + ω x r + b• S' (12.1.2b) which then says, setting b• S' = 0, Section 15: Ant on Turntable 121 v = v' + ω x r . ( 1 5 . 1 . 1 0 ) The first term v' we will replace by (15.1.8); the second we compute, v' = –Vcos( θ'+φ) x^ – Vsin(θ'+φ) y^ (15.1.8) ω x r = [ωz^] x [x x^ + y y^ ] = ωxy^ – ωyx^ . (15.1.11) Therefore (15.1.10) says v = [ –Vcos( θ'+φ) x^ – Vsin(θ'+φ) y^] + ωxy^ – ωyx^ = [–V cos( θ'+φ) – ω y ] x^ + [–V sin(θ '+φ) + ω x] y^ or v = vx x^ + vy y^ w h e r e ( 1 5 . 1 . 1 2 ) v x = –Vcos( θ'+φ) – ω y vy = –Vsin( θ'+φ) + ω x where x,y are given in (15.1.9). We can go ahead and insert x and y from there to get – ωy = -ω[– bcosφ + (r'0 – Vt)sin( θ'+φ)] = ω bcosφ – ω(r'0 – Vt)sin( θ'+φ) ω x = ω[ bsinφ + (r' 0 – Vt)cos( θ'+φ)] = ωbsinφ + ω(r'0 – Vt)cos( θ'+φ) so v x = –Vcos(θ '+φ) + ωbcosφ – ω(r'0 – Vt)sin( θ'+φ) vy = –Vsin( θ'+φ) + ωbsinφ + ω(r'0 – Vt)cos( θ'+φ) . (15.1.13) This v(t) then is the velocity of the ant in Frame S. Acceleration a(t) of the ant in Frame S From (12.1.2e) we find that a = a' + ω• x r + 2 ω x v' + ω x (ω x r) + 2ω x b• S' + b•• S' (12.1.2e) but in this Special Case #1 problem we have b• S' = 0 and b•• S' = 0 so a = a' + ω• x r + 2 ω x v' + ω x (ω x r) . (15.1.14) We shall ponder the terms one at a time. Section 15: Ant on Turntable 122 As noted in (15.1.6), a' = 0. Our turntable is restricted to have ω• = ω• z^ so, similar to (15.1.11) above, we find ω• x r = ω•xy^ – ω•yx^ . (15.1.15) Next, we install (15.1.8) for v' to get ω x v' = [ωz^] x [–Vcos( θ'+φ) x^ – Vsin(θ'+φ) y^] = -ωV cos(θ'+φ) y^ + ωV sin(θ '+φ) x^ = ωVsin(θ '+φ) x^ – ωVcos(θ'+φ) y^ . (15.1.16) With (15.1.11) the last term of (15.1.14) becomes ω x (ω x r) = [ωz^] x [ωxy^ – ωyx^] = -ω2xx^ – ω2yy^ . // = - ω2 r, centripetal accel. (15.1.17) Combining all the terms then gives a = a' + ω• x r + 2 ω x v' + ω x (ω x r) = 0 + (ω•xy^ – ω•yx^) + 2ωVsin(θ '+φ) x^ - 2ωVcos(θ'+φ) y^ -ω2xx^ – ω2yy^ = [ – ω•y + 2ωVsin(θ '+φ) – ω2x] x^ + [ω•x – 2ω Vcos(θ'+φ) – ω2y]y^ or a = ax x^ + ay y^ w h e r e ( 1 5 . 1 . 1 8 ) a x = – ω•y + 2ωVsin(θ '+φ) – ω2x a y = ω•x – 2ω Vcos(θ'+φ) – ω2y where x,y are given by (15.1.9). This a(t) then is the acceleration of the ant in Frame S. Section 15: Ant on Turntable 123 Summary of the Solution to Problem 1 r(t) = x x^ + y y^ w h e r e (15.1.9) x = bsin φ + (r'0 – Vt)cos( θ'+φ) y = – bcos φ + (r'0 – Vt)sin( θ'+φ) v = vx x^ + vy y^ w h e r e (15.1.12) v x = – Vcos( θ'+φ) – ω y vy = – Vsin( θ'+φ) + ωx a = ax x^ + ay y^ w h e r e (15.1.18) a x = – ω•y + 2ωVsin(θ '+φ) – ω2x a y = ω•x – 2ω Vcos(θ'+φ) – ω2y and φ = φ(t) = φ0 + ∫0 t ω(τ)dτ = φ0 + ωt for constant ω . (15.10) (15.1.19) The x and y in equations (15.1.19) are given by (15.1.9), and θ' = θ'0 by (15.1.5). Selected Plots We set φ0 = 0 so Frame S' starts directly below Frame S and is aligned with it, so then φ = ωt. We set θ'0 = θ' = 0 so our ant approaches the Frame S' origin along the e'1 axis : (15.1.20) With these assumptions (15.1.9) becomes r(t) = x x^ + y y^ w h e r e (15.1.21) x = bsin( ωt) + (r'0 – Vt)cos( ωt) y = – bcos( ωt) + (r'0 – Vt)sin( ωt) . Section 15: Ant on Turntable 124 Each plot is finite because the trip is ove r when the ant reaches the S' origin at t max = r'0/V. We set b = 0.5, r' 0 = 1, V = 0.1. The ant therefore starts at (x,y) = (r' 0,-b) = (1,-0.5). Here are trajectory plots for various values of ω : ω = 1 / 2 ω = 1 ω = 3 b = 0.5, r' 0 = 1, V = 0.1 (15.1.22) The middle plot was generated by the following Maple code based on (15.1.21), where R ≡ r'0, Similar code is used to make all the other plots below. Setting b = 0, we get these more traditional plots (Frame S and Frame S' origins now coincide) : ω = 1/2 ω = 1 ω = 3 b = 0, r' 0 = 1, V = 0.1 (15.1.23) Section 15: Ant on Turntable 125 Next we plot the velocity v from (15.1.12) only for the middle ω = 1 case above on the right below, with the corresponding trajectory plot r on the left: (15.1.24) ω = 1 plot of r(t) ω = 1 plot of v(t) For seven different (but unknown) times, we draw the ve locity vector on the right and then transfer it to where we think it ought to go on the trajectory plot on the left. Things at least seem reasonable. A proper visual check would require a program to automate the above process. Next we plot the acceleration a on the right below using (15.1.18), again for ω = 1, with the corresponding trajectory plot r on the left: ω = 1 plot of r(t) ω = 1 plot of a(t) (15.1.25) Section 15: Ant on Turntable 126 For the same seven (still unknown) times plus one more, we draw the acceleration vector on the right and transfer it to where we think it ought to go on the left. Again, this is just a sa nity check to make sure things seem reasonable. Reader Exercise: From (15.1.2) one has v' = -V r^' so that v ' r = v' • r^ = -V r^' • r^ = -V cos(θ'-θ+φ) according to Fig (15.1), where v' r is the radial component of the ant velocity v' in polar coordinates. On the other hand, the radial component of v is given in (15.1.13) as, v r = r^ • v = r^ • [vx x^ + vy y^] = vxr^ • x^ + vyr^ • y^ = vx cosθ + vysinθ where v x = –Vcos(θ '+φ) + ωbcosφ – ω(r'0 – Vt)sin( θ'+φ) v y = –Vsin( θ'+φ) + ωbsinφ + ω(r'0 – Vt)cos( θ'+φ) . (15.1.13) Looking at the above expressions for v' r and vr, it seems unlikely that they could be equal since v r involves terms linear in time t and is a function of ω, b, r'0 whereas v' r does not seem to involve these terms and parameters at all. Yet equation (14.15), which applies to any Special Case #1 problem like Problem 1, claims v r = v'r . The Exercise is to demonstrate that in fact v r = v'r . Hints: (1) Set r' 0 – Vt = r' and show that v r = ωb cos(φ -θ) - ωr' sin(θ'+φ-θ) - V cos( θ'+φ-θ) . (2) Show that the first two terms cancel due to a Law of Sines for Fig (15.1). QED. Results expressed in matrix notation In the case that b = 0, ω• = 0, φ0 = 0, φ = ω t and θ'0 = θ' = 0 ( triplet of plots in (15.1.23) ) we can summarize our results as follows: r(t) = x x^ + y y^ w h e r e (15.1.9) (a) x = (r' 0 – Vt)cos( ωt) y = (r' 0 – Vt)sin( ωt ) v = vx x^ + vy y^ w h e r e (15.1.12) + (15.1.13) (b) v x = –Vcos(ω t) – ω (r'0 – Vt)sin( ωt) v y = –Vsin( ωt) + ω (r'0 – Vt)cos( ωt) a = ax x^ + ay y^ w h e r e (15.1.18) (c) a x = + 2ωVsin(ωt) – ω2 (r'0 – Vt)cos( ωt) a y = – 2ωVcos(ωt) – ω2 (r'0 – Vt)sin( ωt) . (15.1.26) Section 15: Ant on Turntable 127 These equations can be written in matrix notation as follows, ⎝⎜⎛ ⎠⎟⎞ x y z = ⎝⎜⎛ ⎠⎟⎞ cosωt -sinω t 0 sinωt cosωt 0 0 0 1 ⎝⎜⎜⎛ ⎠⎟⎟⎞ r'0-Vt 0 z = Rz(ωt) ⎝⎜⎜⎛ ⎠⎟⎟⎞ r'0-Vt 0 z ⎝⎜⎜⎛ ⎠⎟⎟⎞ vx vy vz = ⎝⎜⎛ ⎠⎟⎞ cosωt -sinω t 0 sinωt cosωt 0 0 0 1 ⎝⎜⎜⎛ ⎠⎟⎟⎞ -V ω(r'0-Vt) vz = Rz(ωt) ⎝⎜⎜⎛ ⎠⎟⎟⎞ -V ω(r'0-Vt) vz ⎝⎜⎜⎛ ⎠⎟⎟⎞ ax ay az = ⎝⎜⎛ ⎠⎟⎞ cosωt -sinω t 0 sinωt cosωt 0 0 0 1 ⎝⎜⎜⎛ ⎠⎟⎟⎞ -ω2(r'0-Vt ) -2ωV az = Rz(ωt) ⎝⎜⎜⎛ ⎠⎟⎟⎞ -ω2(r'0-Vt ) -2ωV az . (15.1.27) Using more systematic notation, we rewrite the above three matrix equations as, r = (r)iei where ⎝⎜⎜⎛ ⎠⎟⎟⎞ (r)1 (r)2 (r)3 = Rz(ωt) ⎝⎜⎜⎛ ⎠⎟⎟⎞ r'0-Vt 0 (r)3 v = (v)iei where ⎝⎜⎜⎛ ⎠⎟⎟⎞ (v)1 (v)2 (v)3 = Rz(ωt) ⎝⎜⎜⎛ ⎠⎟⎟⎞ -V ω(r'0-Vt) (v)3 a = (a)iei where ⎝⎜⎜⎛ ⎠⎟⎟⎞ (a)1 (a)2 (a)3 = Rz(ωt) ⎝⎜⎜⎛ ⎠⎟⎟⎞ -ω2(r'0-Vt ) -2ωV (a)3 . (15.1.28) The components shown in these matrix equations de scribe the spiral solution path, velocity and acceleration of our Problem 1 ant in Frame S. It happens that (r) 3 = (v)3 = (a)3 = 0. Section 15: Ant on Turntable 128 15.2 Problem 2: Ant spirals in at constant V and Ω to the Origin of Frame S' Ant's Motion in Frame S' In order to challenge our formalism a bit, the ant now crawls on the turntable in a more complicated manner. The ant in Frame S' starts at r'0 = r'0 x^' and crawls in a spiral path toward the S' origin. This spiral path is the output of Problem 1 with the Problem 1 parameters set to b = 0, ω• = 0, φ0 = 0 and θ '0 = 0. The ant moves at constant radial speed V toward th e S' origin while at the same time rotating CCW at constant Ω about that origin. In order to find r', v' and a' for this problem, we merely adjust the results stated in (15.1.28) by taking ω → Ω and priming appropriate objects. r' = (r')'ie'i where ⎝⎜⎜⎛ ⎠⎟⎟⎞ (r')'1 (r')'2 (r')'3 = Rz(Ωt) ⎝⎜⎜⎛ ⎠⎟⎟⎞ r'0-Vt 0 (r')'3 v' = (v')'ie'i where ⎝⎜⎜⎛ ⎠⎟⎟⎞ (v')'1 (v')'2 (v')'3 = Rz(Ωt) ⎝⎜⎜⎛ ⎠⎟⎟⎞ -V Ω(r'0-Vt) (v')'3 a' = (a')'ie'i where ⎝⎜⎜⎛ ⎠⎟⎟⎞ (a')'1 (a')'2 (a')'3 = Rz(Ωt) ⎝⎜⎜⎛ ⎠⎟⎟⎞ -Ω2(r'0-Vt ) -2ΩV (a')'3 . (15.2.1) This ant path in Frame S' has the following general appearance (depending on parameters), (15.2.2) We first wish to know the components of r', v' and a' on the en basis vectors. This problem was addressed in (1.2.7) which we quote Section 15: Ant on Turntable 129 (a')i = (R-1)ij(a')'j ⇔ ⎝⎜⎜⎛ ⎠⎟⎟⎞ (a')1 (a')2 (a')3 = R-1 ⎝⎜⎜⎛ ⎠⎟⎟⎞ (a')'1 (a')'2 (a')'3 . (1.2.7) w h e r e ( 1 5 . 2 . 3 ) en = R e'n n = 1,2,3 or ( en)i = Rij(e'n)j . (1.1.1) In our application here, we know from (15.4c) that , e'n = Rz(φ)en ⇒ en = Rz(-φ)e'n ⇒ R = R z(-φ) ⇒ R-1 = Rz(φ) . (15.2.4) Therefore, first setting a' = r' , we find from (15.2.3) and (15.2.1) that ⎝⎜⎜⎛ ⎠⎟⎟⎞ (r')1 (r')2 (r')3 = Rz(φ) ⎝⎜⎜⎛ ⎠⎟⎟⎞ (r')'1 (r')'2 (r')'3 = Rz(φ)Rz(Ωt) ⎝⎜⎜⎛ ⎠⎟⎟⎞ r'0-Vt 0 (r')'3 = Rz(Ωt+φ) ⎝⎜⎜⎛ ⎠⎟⎟⎞ r'0-Vt 0 (r')'3 = ⎝⎜⎛ ⎠⎟⎞ cos(Ωt+φ) -sin(Ωt+φ) 0 sin(Ωt+φ) cos(Ωt+φ) 0 0 0 1 ⎝⎜⎜⎛ ⎠⎟⎟⎞ r'0-Vt 0 (r')'3 r' = (r')iei . (15.2.5) Consider next a' = v' and a' = a' and use the column vectors on the right in (15.2.1) to get, ⎝⎜⎜⎛ ⎠⎟⎟⎞ (v')1 (v')2 (v')3 = ⎝⎜⎛ ⎠⎟⎞ cos(Ωt+φ) -sin(Ωt+φ) 0 sin(Ωt+φ) cos(Ωt+φ) 0 0 0 1 ⎝⎜⎜⎛ ⎠⎟⎟⎞ -V Ω(r'0-Vt) (v')'3 v' = (v')iei ⎝⎜⎜⎛ ⎠⎟⎟⎞ (a')1 (a')2 (a')3 = ⎝⎜⎛ ⎠⎟⎞ cos(Ωt+φ) -sin(Ωt+φ) 0 sin(Ωt+φ) cos(Ωt+φ) 0 0 0 1 ⎝⎜⎜⎛ ⎠⎟⎟⎞ -Ω2(r'0-Vt ) -2ΩV (a')'3 a' = (a')iei . (15.2.6) Then do the matrix multiplication to obtain, (r') 1 = (r'0-Vt)cos(Ωt+φ) (r')2 = (r'0-Vt)sin(Ωt+φ) (v') 1 = -Vcos( Ωt+φ) - Ω (r'0-Vt)sin(Ωt+φ) (v')2 = -Vsin(Ωt+φ) + Ω (r'0-Vt)cos(Ωt+φ) (a') 1 = -Ω2(r'0-Vt )cos(Ωt+φ) + 2ΩVsin(Ωt+φ) (a')2 = -Ω2(r'0-Vt )sin(Ωt+φ) - 2ΩVcos(Ωt+φ) . (15.2.7) Using our less formal notation we have now shown that the trajectory of our spiraling ant in Frame S' can be expressed in terms of Frame S basis vectors as follows ( later we will set φ = ωt ) , Section 15: Ant on Turntable 130 r'(t) = (r') x x^ + (r')y y^ w h e r e ( a ) (r') x = (r'0 – Vt)cos( φ + Ωt ) (r') y = (r'0 – Vt)sin( φ + Ωt) v' = (v')x x^ + (v')y y^ w h e r e ( b ) ( v ' ) x = –Vcos(φ + Ωt) – Ω (r'0 – Vt)sin( φ + Ωt) ( v ' ) y = –Vsin( φ + Ωt) + Ω (r'0 – Vt)cos( φ + Ωt) a ' = (a')x x^ + (a')y y^ w h e r e ( c ) ( a ' ) x = 2Ω Vsin(φ + Ωt) – Ω2(r'0 – Vt)cos( φ + Ωt) ( a ' ) y = – 2ΩVcos(φ + Ωt) – Ω2(r'0 – Vt)sin( φ + Ωt) . (15.2.8) Trajectory r(t) of the ant in Frame S According to (15.2.8a), r'(t) = (r'0 – Vt) [ cos( φ + Ωt) x^ + sin(φ + Ωt) y^ ] . (15.2.8a) (15.2.9) From (12.1.2a) and then from (15.12) we have r = b + r' (12.1.2a) b(t) = bsin φ x^ – b cosφ y^ . (15.12) Therefore, installing (15.2.9) for r' and just above for b, r(t) = x x^ + y y^ w h e r e ( 1 5 . 2 . 1 0 ) x = bsinφ + ( r'0 – Vt)cos( φ + Ωt) y = – bcos φ + (r'0 – Vt)sin( φ + Ωt) . The Ω = 0 limit of this result agrees with the θ' = 0 limit of (15.1.9), the Problem 1 trajectory. This r(t) then is the trajectory of the ant in Frame S. Velocity v(t) of the ant in Frame S Start with two equations used in Problem 1, v = v' + ω x r (15.1.10) ω x r = [ωz^] x [x x^ + y y^ ] = ωxy^ – ωyx^ . (15.1.11) Section 15: Ant on Turntable 131 Then use v' from (15.2.8b) in (15.1.10) just above to get v = vx x^ + vy y^ w h e r e ( 1 5 . 2 . 1 1 ) v x = –Vcos(φ + Ωt) – Ω (r'0 – Vt)sin( φ + Ωt) – ωy v y = –Vsin( φ + Ωt) + Ω (r'0 – Vt)cos( φ + Ωt) + ω x . Finally, insert (15.2.10) for x and y so that, v x = –Vcos(φ + Ωt) – Ω (r'0 – Vt)sin( φ + Ωt) – ω[– bcosφ + (r'0 – Vt)sin( φ + Ωt)] v y = –Vsin( φ + Ωt) + Ω (r'0 – Vt)cos( φ + Ωt) + ω [bsinφ + ( r'0 – Vt)cos( φ + Ωt)] or v x = –Vcos(φ + Ωt) – (ω + Ω) (r'0 – Vt)sin( φ + Ωt) + ωbcosφ v y = –Vsin( φ + Ωt) + (ω + Ω) (r'0 – Vt)cos( φ + Ωt) + ωbsinφ . (15.2.12) The Ω = 0 limit of these last equations gives the θ' = 0 limit of (15.1.13). This v(t) then is the velocity of the ant in Frame S. Acceleration a(t) of the ant in Frame S Start again with (15.1.14), a = a' + ω• x r + 2 ω x v' + ω x (ω x r) . (15.1.14) (15.2.13) The first term is given by (15.2.8c) a ' = (a')x x^ + (a')y y^ w h e r e (15.2.8c) ( a ' ) x = 2Ω Vsin(φ + Ωt) – Ω2(r'0 – Vt)cos( φ + Ωt) ( a ' ) y = – 2ΩVcos(φ + Ωt) – Ω2(r'0 – Vt)sin( φ + Ωt) . The 2nd and 4th terms we obtain by quoting these results from the previous problem, ω• x r = ω•xy^ – ω•yx^ (15.1.15) ω x (ω x r) = [ωz^] x [ωxy^ – ωyx^] = -ω2xx^ – ω2yy^ . // = - ω2 r, centripetal accel. (15.1.17) The third term of (15.2.13) is 2ω x v' = 2[ωz^] x [(v')x x^ + (v')y y^] = 2ω(v')x y^ – 2ω (v')y x^ . Adding these terms one can rewrite (15.2.13) as, Section 15: Ant on Turntable 132 a = ax x^ + ay y^ w h e r e ( 1 5 . 2 . 1 4 ) ax = 2Ω Vsin(φ + Ωt) – Ω2(r'0 – Vt)cos( φ + Ωt) – ω•y – ω2x – 2ω (v')y ay = – 2ΩVcos(φ + Ωt) – Ω2(r'0 – Vt)sin( φ + Ωt) + ω•x – ω2y + 2ω(v')x where x = bsinφ + ( r'0 – Vt)cos( φ + Ωt ) (15.2.10) y = – bcos φ + (r'0 – Vt)sin( φ + Ωt) and (v') x = –Vcos(φ + Ωt) – Ω (r'0 – Vt)sin( φ + Ωt) (v')y = –Vsin( φ + Ωt) + Ω (r'0 – Vt)cos( φ + Ωt ) . (15.2.8b) This a(t) then is the acceleration of the ant in Frame S. The result is admittedly a bit complicated, but the point is that we were able to obtain the result using our Section 12 summary equations without too much effort . [ See "The Hard Way" in Section 15.4 below. We have not dealt with any differential e quations in obtaining the above results. ] Trajectory Plots We set φ 0 = 0 so Frame S' starts directly below Frame S and is aligned with it, and then φ = ωt. Equation (15.2.10) then reads r(t) = x x^ + y y^ w h e r e ( 1 5 . 2 . 1 5 ) x = bsin( ωt) + ( r'0 – Vt)cos( ωt + Ω t) y = – bcos( ωt) + (r'0 – Vt)sin( ωt + Ω t) . Each plot is finite because the trip is ove r when the ant reaches the S' origin at t max = r'0/V. We set b = 3, r' 0 = 2, V = 0.4 and ω = 1. The ant therefore starts at (x,y) = (r' 0,-b) = (2,-3). Here are trajectory plots for various values of Ω : Section 15: Ant on Turntable 133 Ω = 6 Ω = 10 Ω = 20 b = 3, r' 0 = 2, V = 0.4, ω = 1 (15.2.16) The blue circles have radius b = 3 (origin of Fr ame S'), and the green circles have radius b+r' 0 = 5. These trajectories should seem quite reasonable to the reader, knowing what that ant is up to in Frame S', shown generically in Fig (15.2.2). The middle plot was generated by the following Maple code based on (15.2.15), where R ≡ r' 0, : ( 1 5 . 2 . 1 7 ) When ω and Ω have opposite sign, things can look quite different, Ω = -2 Ω = -3 Ω = -3.6 ( 1 5 . 2 . 1 8 ) Finally, here are some trajectories with V = 0 : Section 15: Ant on Turntable 134 r'0 = 0.7, b = 3, V = 0, ω = 1 and : Ω = 3.5 Ω = -7 Ω = 25.1 ( 1 5 . 2 . 1 9 ) Using the parameters of the third plot, if one were to space 10,000 ants evenly on the turntable, and have each one carry a grain of sand on its back, one woul d have constructed a ra ndom orbital sander. The closure of the plots occurs whenever Ω/ω is a ratio of integers, but that could take many revolutions if those integers are large. Section 15: Ant on Turntable 135 15.3 Problem 3: Inverse Problem: Ant flies in Frame S at constant velocity V A flying ant starting at position r0 flies just above the turntable surf ace in Frame S in a straight line at constant velocity V at angle θ relative to the x axis. First state r,v,a and then compute r',v',a' and plot the trajectory r' of the particle as seen in Frame S'. Use the sa me Frame S / Frame S' setup as in the previous problems. In this problem continue to use both notations for the unit vectors as in (1.1.5), x 1, x2, x3 = x, y, z e1, e2, e3 = x^, y^, z^ e' 1, e'2, e'3 = x^', y^', z^' (1.1.5) Ant's Motion in Frame S The flying ant starts at location r0 and has velocity V = Vn^ with V constant, so v = Vn^ r = Vt n^ + r0 a = 0 . ( 1 5 . 3 . 1 ) As noted, the ant flies on a line which has angle θ relative to the e1 axis, so n^ = Rz(θ) e1 = Rz(θ) x^ . ( 1 5 . 3 . 2 ) This angle θ is defined in the usual polar sense: it is counterclockwise from the positive x axis. Trajectory r'(t) of the ant in Frame S' First, we need to write n^ in Frame S' basis vectors. Recall (15.4c) which says e'i = Rz(φ) ei for example e'1 = Rz(φ) e1 . (15.4c) Therefore n^ = Rz(θ) e1 = Rz(θ) Rz(-φ) e'1 = Rz(θ-φ) e'1 . (15.3.3) Second, how does the point r0 in Frame S appear in Frame S' ? Using (15.4c) just above gives r0 = (r0)i ei = (r0)i Rz(-φ)e'i . (15.3.4) Third, from (15.12) ( or just looking at Fig (15.1) ) we know that b = -b e'2 . (15.12) Now for the trajectory we start with (13.3.2a), Section 15: Ant on Turntable 136 r' = r - b (13.3.2a) = (Vt n^ + r0 ) + b e'2 // from (15.3.1) and (15.12) = Vt R z(θ-φ) e'1 + (r0)i Rz(-φ)e'i + b e'2 // from (15.3.3) and (15.3.4) so r' = Vt R z(θ-φ) e'1 + Rz(-φ) [(r0)i e'i] + b e'2 . (15.3.5) We now use the following notations (these are all "natural" components in sense of Section 1.8 ) (r0)1 = x0 (r')' 1 = x' (r0)2 = y0 (r')'2 = y' . In Frame S', e'1 = (1,0,0), so we write (15.3.5) in matrix notation in Frame S' as follows : ⎝⎜⎛ ⎠⎟⎞ x' y' z' = Vt ⎝⎜⎛ ⎠⎟⎞ cos(θ -φ) -sin(θ-φ) 0 sin(θ-φ) cos(θ -φ) 0 0 0 1 ⎝⎜⎛ ⎠⎟⎞ 1 0 0 + ⎝⎜⎛ ⎠⎟⎞ cosφ sinφ 0 -sinφ cosφ 0 0 0 1 ⎝⎜⎜⎛ ⎠⎟⎟⎞ x0 y0 0 + b ⎝⎜⎛ ⎠⎟⎞ 0 1 0 or x' = Vt cos( θ-φ) + x0cosφ + y0sinφ y' = Vt sin( θ-φ) – x0sinφ + y0cosφ + b z' = 0 . The conclusion is this : r'(t) = x' x^' + y' y^' w h e r e ( 1 5 . 3 . 6 ) x ' = V t c o s ( θ-φ) + x0cosφ + y0sinφ y ' = V t s i n ( θ-φ) – x0sinφ + y0cosφ + b where φ = φ0 + ωt . This then is the trajectory r' of the flying ant as seen in Frame S'. Trajectory r'(t) of the ant in Frame S': Alternate Method Since we are going to have a discrepancy with Thornton and Marion below, it seems healthy to confirm (15.3.6) by an alternate derivation that does not make use of the matrix notation used above. We start as before above (15.3.5), r' = r - b = (Vt n^ + r0 ) + b e'2 . r' = trajectory of ant in Frame S' Using the theorem (1.1) + (1.2) we find from (15.3.3) and (15.3.4) that, Section 15: Ant on Turntable 137 n^ = Rz(θ-φ) e'1 = Rz(φ-θ)11 e'1 + Rz(φ-θ)12 e'2 = cos(φ-θ) e'1 - sin(φ-θ) e'2 r0 = (r0)i ei = (r0)i Rz(-φ)e'i = x0 Rz(-φ)e'1 + y0 Rz(-φ)e'2 = x 0 { Rz(φ)11 e'1 + Rz(φ)12 e'2 } + y0 { Rz(φ)21 e'1 + Rz(φ)22 e'2 } = x 0 { cos(φ ) e'1 – sin(φ) e'2 } + y0 {sin(φ) e'1 + cos(φ ) e'2 } = [ x 0 cos(φ ) + y0 sin(φ)] e'1 + [– x0 sin(φ) + y0 cos(φ )] e'2 . Therefore, r' = Vt n^ + r0 + b e'2 = Vt [cos( φ-θ) e'1 - sin(φ-θ) e'2] + [x0 cos(φ ) + y0 sin(φ)] e'1 + [- x0 sin(φ) + y0 cos(φ )] e'2 + b e'2 = [Vt cos( φ-θ) + x 0 cos(φ ) + y0 sin(φ) ]e'1 + [- Vt sin( φ-θ) - x0 sin(φ) + y0 cos(φ ) + b] e'2 = [Vt cos( θ-φ) + x 0 cos(φ ) + y0 sin(φ) ]e'1 + [ Vt sin( θ-φ) - x0 sin(φ) + y0 cos(φ ) + b] e'2 and this does agree with (15.3.6). Velocity v'(t) of the ant in Frame S' From the Inverse Problem equations of Section 13.3 we have from (13.3.2b), v' = v – ω x r – b• S' . (13.3.2b) Since this is a Special Case #1 problem, we have b• S' = 0 and then v' = v – ω x r = v – ω x (r'+b) = v – ω x r' – ω x b // (13.3.2a) that r' = r - b = V Rz(θ-φ) x^' – ω x [ x' x^' + y' y^'] – ω x [ -b y^'] // (15.3.1) v, (15.3.3) n^ and (15.12) b = V Rz(θ-φ) x^' –ωx' y^' + ω y' x^' – ωb x^' . // ω = ω z^' Using the following unofficial notations, ( v')'1 = v'x' ( v')'2 = v'y' the above equation in matrix notation in Frame S' is Section 15: Ant on Turntable 138 ⎝⎜⎜⎛ ⎠⎟⎟⎞ v'x' v'y' v'z' = V ⎝⎜⎛ ⎠⎟⎞ cos(θ -φ) -sin(θ-φ) 0 sin(θ-φ) cos(θ -φ) 0 0 0 1 ⎝⎜⎛ ⎠⎟⎞ 1 0 0 + ⎝⎜⎛ ⎠⎟⎞ ωy'-ωb -ωx' 0 or v'x' = V cos(θ-φ) + ω (y'-b) v'y' = V sin(θ-φ) - ωx' v'z' = 0 . Then using (15.3.6) for x' and y' we get, v'(t) = v'x' x^' + v'y' y^' w h e r e ( 1 5 . 3 . 7 ) v ' x' = V cos(θ-φ) + ω [Vt sin(θ-φ) – x0sinφ + y0cosφ] v ' y' = V sin(θ-φ) – ω [Vt cos(θ-φ) + x0cosφ + y0sinφ] where φ = φ0 + ωt . This then is the velocity v' of the flying ant as seen in Frame S'. Acceleration a'(t) of the ant in Frame S' From the Inverse Problem equations in Section 13.3 we have from (13.3.2d), a' = a – ω• x r – 2 ω x v + ω x (ω x r) – b•• S' . (13.3.2d) Setting b•• S'= 0 for our Special Case #1 problem, and using (15.3.1) for r, v and a (a = 0) we get a' = – ω• x [Vt n^ + r0] – 2 ω x [Vn^] + ω x (ω x [Vt n^ + r0]) . We shall stop here, but the calculation can be continued in a manner similar to that for r' and v'. Trajectory Plots Recall (15.3.6) from above r'(t) = x' x^' + y' y^' w h e r e (15.3.6) x ' = V t c o s ( θ-φ) + x0cosφ + y0sinφ y ' = V t s i n ( θ-φ) – x0sinφ + y0cosφ + b where φ = φ0 + ωt . For plotting purposes, we set φ0 = 0 so φ = ωt. Since b merely offsets plots vertically by amount b, we lose no interest by setting b=0, causing the Frame S and Frame S' origins to coincide. Then from (15.3.6), Section 15: Ant on Turntable 139 x' = Vt cos( θ-ωt) + x0cos(ωt) + y0sin(ωt) y' = Vt sin( θ-ωt) – x0sin(ωt) + y0cos(ωt) . (15.3.8) In all plots below we set ω = 1. In the first three plots the flying ant starts out at r0 = (-1/2,0) and flies north so θ = π/2. We superpose a green circle of radius R = 1 and take note of the time T it takes for the ant to reach the circle. The Maple code used for the left plot is this (x' = xp) : Here then are plots for three decreasing values of V. As the ant flies more slowly, the turntable turns more radians before the ant reaches a distan ce R = 1 from the turntable center. V = .85 , T = 1.03 V = 0.3, T = 2.9 V = .061, T = 14 ( 1 5 . 3 . 9 ) These plots may be compared to those appearing on on page 395 of Thornton and Marion, ( 1 5 . 3 . 1 0 ) Section 15: Ant on Turntable 140 Their plotting method is to compute the fictitious force acceleration a' = – 2 ω x v + ω x (ω x r) and then to numerically integrate a' twice to get v' and then r' which is then plotted. [ Again, see "The Hard Way" in Section 15.4 below.] Although our plots are close to theirs in appearance, our numbers for V and T differ significantly from theirs. Here are th e three plots our code generates using the numbers specified in the T&M images above, V = 1.5 , T = 0.86 V = 0.8, T = 2.9 V = 0.45, T = 17.3 ( 1 5 . 3 . 1 1 ) For the next three plots, we still have ω = 1 and the flying ant starts in the same place r0 = (-1/2,0), but now the ant flies southeast at speed V so θ = -π/4. Interestingly, we see that it is possible for the ant (as seen in Frame S') to execute a loop, a cusp, or a bump soon after taking flight. The lower set of figures show blowups of the parts of the upper paths, V = 0.42, T = 3.1 V = 0.35, T = 3.7 V = 0.31, T = 4.2 (15.3.12) Section 15: Ant on Turntable 141 ( 1 5 . 3 . 1 3 ) It is not easy to intuitively explain these plots. We do know that for all plots v t = ωr = 1*(1/2) = 0.5 at time t = 0. So even though the ant is flying southeas t in Frame S, in Frame S' there is at t = 0 a v t adder of 0.5 upwards (north) due to the motion of the turntable so in Frame S' the ant starts off going northeast. The ant moves to a smaller radius so v t is reduced so the and moves more toward the east and in the left case to the south as well, but then Frame S' whic h started under Frame S is rotating up to the right The left pair of plots in (15.3.12) may be compared to two other plots appearing in T&M on page 395, (15.3.14) Again the plots are similar, but the numbers ar e different (but in the same ball park). Using (15.3.7) we plot the velocity that goes with the second trajector y shown in the first triplet above, and at least things seem reasonable: Section 15: Ant on Turntable 142 r(t) for V = 0.3, T = 2.9 v(t) for V = 0.3, T = 2.9 (15.3.15) Finally, we move the starting position to (-1 ,-1) and have the ant fly northeast so θ = π/4. This path takes him over the origin of Frame S (and Frame S'), so we expect to see the Frame S' trajectory touch the origin at one point along the trajectory (e xcept in the left plot where V = 0) V = 0, T = 5 V = 0.11, T = 25 V = 0.2, T = 15 ( 1 5 . 3 . 1 6 ) Since the turntable is rotating counterclockwise at ω=1, these trajectories run clockwise at all times. When V = 0, the apparent motion of the static fly is ci rcular in Frame S'. In the middle plot we see that the ant spirals in, reaches the origin, the spirals out. In the right plot he does the same thing, but more quickly. In the 1960's some excellent frames-of-reference movies were pr oduced. One of them involves a frictionless puck moving on a smooth table mounted to a large wooden frame which is rotated (no doubt by students). Two affable "doctors" are rotating on that frame with the table. Doctor #1 on the left launches the puck, but Doctor #2 has nothing to do since the puck just returns to Doctor #1. Section 15: Ant on Turntable 143 ( 1 5 . 3 . 1 7 ) We show two possible Frame S' paths for a puck launched from (x 0, y0) = (-1,0) and ω = 1. V = 0.6, T = 3.2, θ = 0 V = 0.2, T = 5, θ = π/4 (15.3.18) This classic movie is archived at http://www.youtube.com/watch?v=3ug23VTMies . (The next film in this series treats the Foucault Pendulum which we describe in Appendix C.) Section 15: Ant on Turntable 144 15.4 Problem 4: The Projectile Problem of Section 8.3 It will be recalled that in Section 8.3, as a demonstr ation of the Coriolis force, four projectiles are fired horizontally in four directions as in Figure (8.3.1). Since each projectile is like one of our flying ants, we already have a complete solution to this problem wh ich we shall plot below. But first, it is very enlightening to approach this problem "the hard way" and then to appreciate the power of the equations which directly relate particle pr operties in Frame S and Frame S'. The Hard Way It was noted in (8.3.4) that the projectiles (or our flying ant) experience the following fictitious forces, F'fict = mω2r' – 2m ω x v' . (8.3.4) centrifugal Coriolis Using bogus Newton's Law (8.1.4) that F'eff = ma', the above equation can be written, (d v'/dt)S' = ω2r' – 2ω x v' . (15.4.1) Since ω is a constant, and working in Fram e S', we differentiate once to get (d2v'/dt2)S' = ω2 (dr'/dt)S' – 2ω x (dv'/dt)S' . In the abbreviated "natural" notation of Section 1.8 this says v••' + 2ω x v•' - ω2v' = 0 . (15.4.2) Now for the rest of this section, we temporarily drop all primes just to reduce clutter. Then the above becomes v•• +2ω x v• - ω2v = 0 . (15.4.3) Expanding the vectors of interest, v = vxx^ + vyy^ + vzz^ v• = v•xx^ + v•yy^ + v•zz^ v•• = v••xx^ + v••yy^ + v••zz^ ω x v• = [ω z^] x [v•xx^ + v•yy^ + v•zz^] = ωv•xy^ - ωv•yx^ . Inserting these expansions into (15. 4.3) and isolating the coefficients of the three unit vectors, we get Section 15: Ant on Turntable 145 v••x - 2ωv•y - ω2vx = 0 v••y + 2 ωv•x - ω2vy v••z - ω2vz = 0 . (15.4.4) The third equation has obvious solutions, one of which is v z = 0 which is what applies to our turntable problems. That leaves the first two equations, v••x - 2ωv•y - ω2vx = 0 v••y + 2ωv•x - ω2vy = 0 . (15.4.5) This is a system of two coupled, second-order, linear ODE's with constant coefficients. It takes some amount of work to solve such an equation and we will start down that path. Apply a Laplace Transform, [ s2Vx(s) - s vx (0) - v•x (0) ] - 2 ω[ s Vy(s) - vy(0) ] - ω2 Vx(s) = 0 [ s2Vy(s) - s vy (0) - v•y (0) ] + 2 ω[ s Vx(s) - vx(0) ] - ω2 Vy(s) = 0 (15.4.6) (s 2-ω2)Vx(s) – 2ωs Vy(s) = s v x (0) + v•x (0) – 2ω vy(0) // get V x and Vy on the left (s2-ω2)Vy(s) + 2ωs Vx(s) = s v y (0) + v•y (0) + 2ω vx(0) ⎝⎛ ⎠⎞s2-ω2 -2ωs 2ωs s2-ω2 ⎝⎛ ⎠⎞Vx(s) Vy(s) = ⎝⎜⎛ ⎠⎟⎞s vx (0) + v•x (0) – 2ω vy(0) s vy (0) + v•y (0) + 2ω vx(0) // write as matrix equation ⎝⎛ ⎠⎞Vx(s) Vy(s) = ⎝⎛ ⎠⎞s2-ω2 2ωs -2ωs s2-ω2 ⎝⎜⎛ ⎠⎟⎞s vx (0) + v•x (0) – 2ω vy(0) s vy (0) + v•y (0) + 2ω vx(0) / (s2+ω2)2 // Maple assists Vx(s) = { (s2-ω2) [s vx (0) + v•x (0) – 2ω vy(0)] +2ωs [s vy (0) + v•y (0) – 2ω vx(0)] }/(s2+ω2)2 Vy(s) = { -2 ωs [s vx (0) + v•x (0) – 2ω vy(0)] + (s2-ω2) [s vy (0) + v•y (0) – 2ω vx(0)] }/(s2+ω2)2 One can then look up the inverse Laplace transforms of all these functions, s3/(s2+ω2)2 c o s ( ωt) - (1/2)ωt sin(ωt) s2/(s2+ω2)2 [ s i n ( ωt) + ω tcos(ωt)]/(2ω) s/(s2+ω2)2 t sin(ωt)/(2ω) 1/(s2+ω2)2 [sin(ωt) - ωtcos(ωt)]/(2ω3) (15.4.7) and the problem is solved including the initial conditions. But, we don't have to solve this coupled system of differen tial equations because we already know the solution, and we didn't have to even look at a differential equation to find it! The solution is (15.3.7) which we quote Section 15: Ant on Turntable 146 v'(t) = v'x' x^' + v'y' y^' w h e r e (15.3.7) v ' x' = V cos(θ-φ) + ω [Vt sin(θ-φ) – x0sinφ + y0cosφ] v ' y' = V sin(θ-φ) – ω [Vt cos(θ-φ) + x0cosφ + y0sinφ] where φ = φ0 + ωt . Changing this to our temporary no- primes de-cluttered notation, v(t) = vx x^ + vy y^ where v x = V cos(θ-φ) + ω [Vt sin(θ-φ) – x0sinφ + y0cosφ] v y = V sin(θ-φ) – ω [Vt cos(θ-φ) + x0cosφ + y0sinφ] where φ = φ0 + ωt . We shall now use Maple to verify that these f unctions solve the coupled equations (15.4.5): ( 1 5 . 4 . 8 ) Sometimes, for a given problem in rotational motion, there is an easy way and a hard way to solve the problem. Another example of a hard way is the num eric integration mentioned after Fig (15.3.10). At this point we restore the primes on Frame S' quantities. Section 15: Ant on Turntable 147 The Four Projectiles Recall Fig (8.3.1) which shows the deflecting projectiles on the right. (8.3.1) In Frame S' at t = 0 the four projectiles are launched in the four Frame S' axis directions and each has the same speed V' (black arrows on the right). This V' wa s called V in Section 8.3, but since it is a speed in Frame S' we now call it V'. We now examine the initial speeds and launch angles θ of the four projectiles as seen in Frame S: black (upper) V' y^ + vty^ θ = π/2 V = V' + v t red (lower) -V' y^ + vty^ θ = -(π /2)*sign(V'-v t) V = |V' - v t| orange (left) -V' x^ + vty^ θ = π - Δθ V = V'2+vt2 blue (right) V' x^ + vty^ θ = Δθ V = V'2+vt2 Here vt = aω is the turntable upward speed where the launch occurs. The angle θ is the usual polar azimuthal angle measured CCW from the x^ axis which points to the right, and Δθ = tan-1(vt/V') . We treat each of these Frame S velocities as the velocity of a Problem 3 ant flyover. The Frame S' trajectories are then given by (15.3.8) where the Fr ame S and Frame S' origins coincide at the spindle (b=0) and have aligned axes at time t = 0, x' = Vt cos( θ-ωt) + x 0cos(ωt) + y0sin(ωt) y' = Vt sin( θ-ωt) – x0sin(ωt) + y0cos(ωt) . (15.3.8) At t=0 the four projectiles start at (x 0,y0) = (x'0,y'0) = (a,0) so things simplify a bit more, x'(t) = Vt cos( θ-ωt) + acos(ω t) y'(t) = Vt sin( θ-ωt) – asin( ωt) . (15.4.9) Section 15: Ant on Turntable 148 These are very simple expressions indeed, consid ering the coupled differential equations above. It remains only to have Maple plot the projectile tr ajectories with the initial velocities installed : (15.4.10) Here then are some plots. The first four all have V' = 5 and but have increasing duration. The notion of the deflections being roughly circular (end of Section 8.3) is not too bad for tmax ≤ 1 : V' = 5: tmax = 0.2 tmax = 0.5 tmax = 1.0 tmax = 4 ( 1 5 . 4 . 1 1 ) In the next plots we fix tmax = 3 but take V' to ever-decreasing values : Section 15: Ant on Turntable 149 tmax=3: V' = 1 V' = 0.5 V' = 0.2 V' = 0.05 ( 1 5 . 4 . 1 2 ) As V' → 0, all four curves eventually become the same curve. The blue and orange curves line up early because, with small V', both projectiles accelerate radially outward in Frame S' with no tangential velocity. The orange one first goes to the left, then reverses and goes to the ri ght and follows the blue particle in nearly the same trajectory. Appendix A: Spherical Coordinates 150 Appendix A: Derivation of R( ξ) and Properties of Rotation Matrices Here we derive equations (14.4) thr ough (14.6) for spherical coordinates. First, just for reference, here are the three active rotation matrices used below : R x(θ) = ⎝⎜⎛ ⎠⎟⎞ 1 0 0 0 cosθ -sinθ 0 sinθ cosθ Ry(θ) = ⎝⎜⎛ ⎠⎟⎞ cosθ 0 sinθ 0 1 0 -sinθ 0 cosθ Rz(θ) = ⎝⎜⎛ ⎠⎟⎞ cosθ -sinθ 0 sinθ cosθ 0 0 0 1 . (A.1) These are called "active" si nce they rotate a vector forward (counter clockwise) relative to fixed axes by amount θ according to the right hand rule when the thumb is aligned with the axis in question. From the usual picture of spherical coordinates, (A.2) one can see (by staring hard enough) that r^ = Rz(φ) Ry(θ) z^ = Rz^ R ≡ Rz(φ) Ry(θ) θ^ = Rz(φ) Ry(θ) x^ = Rx^ φ^ = Rz(φ) Ry(θ) y^ = Ry^ // R y(θ) does nothing here (A.3) which we rewrite as e^1 = R e3 where e^1 = r^ e3 = z^ e^2 = R e1 e^2 = θ^ e 1 = x^ e^3 = R e2 e^3 = φ^ e2 = y^ . (A.4) We can repair the ordering of the basis vectors en on the right using R 2 = (z^,x^,y^) as follows e3 = R2 e1 ⎝⎜⎛ ⎠⎟⎞ 0 0 1 = ⎝⎜⎛ ⎠⎟⎞ 0 1 0 0 0 1 1 0 0 ⎝⎜⎛ ⎠⎟⎞ 1 0 0 R 2 = ⎝⎜⎛ ⎠⎟⎞ 0 1 0 0 0 1 1 0 0 = ( z^,x^,y^) Appendix A: Spherical Coordinates 151 e1 = R2 e2 ⎝⎜⎛ ⎠⎟⎞ 1 0 0 = ⎝⎜⎛ ⎠⎟⎞ 0 1 0 0 0 1 1 0 0 ⎝⎜⎛ ⎠⎟⎞ 0 1 0 e2 = R2 e3 ⎝⎜⎛ ⎠⎟⎞ 0 1 0 = ⎝⎜⎛ ⎠⎟⎞ 0 1 0 0 0 1 1 0 0 ⎝⎜⎛ ⎠⎟⎞ 0 0 1 . (A.5) Maple tells us that R 2-1 = R2T and det(R 2) = 1, confirming that R 2 is a rotation. Putting (A.5) into (A.4), e^1 = R R2 e1 e^2 = R R2 e2 e^3 = R R2 e3 ( A . 6 ) or e^i = R(ξ) ei . ( A . 7 ) Recalling (14.4), e^n = R(ξ) en => en = R(ξ)nme^m or e^n = [R(ξ)]-1 nm em e^'n = R'(ξ') e'n => e'n = R'(ξ ')nme^'m or e^'n = [R'(ξ ')]-1 nm e'm . (14.4) we have therefore found the sought-after matrix R( ξ) appearing in the top left of (14.4). R(ξ) = R R2 . ( A . 8 ) Specific evaluation gives R = Rz(φ) Ry(θ) = ⎝⎜⎛ ⎠⎟⎞ cosφ -sinφ 0 sinφ cosφ 0 0 0 1 ⎝⎜⎛ ⎠⎟⎞ cosθ 0 sinθ 0 1 0 -sinθ 0 cosθ = ⎝⎜⎛ ⎠⎟⎞ cosφcosθ -sinφ cosφsinθ sinφ sinθ cosφ sinφsinθ -sinθ 0 cos θ (A.9) and then R(ξ) = R R2 = ⎝⎜⎛ ⎠⎟⎞ cosφcosθ -sinφ cosφsinθ sinφ sinθ cosφ sinφsinθ -sinθ 0 cos θ ⎝⎜⎛ ⎠⎟⎞ 0 1 0 0 0 1 1 0 0 = ⎝⎜⎛ ⎠⎟⎞ cosφsinθ cosφcosθ -sinφ sinφ sinθ sinφcosθ cosφ cosθ -sinθ 0 (A.10) from which we find [R( ξ)]-1 = [R(ξ)]T = ⎝⎜⎛ ⎠⎟⎞ cosφsinθ sinφsinθ cosθ cosφcosθ sinφ cosθ -sinθ -sinφ cosφ 0 . (A.11) From the top right equation of (14.4) (quoted above) we can write e^i = [R(ξ) ]-1 ij ej = [R(ξ) ]-1 i1 e1 + [R(ξ) ]-1 i2 e2 + [R(ξ) ]-1 i3 e3 . (A.12) Appendix A: Spherical Coordinates 152 This can be written in this matrix notation: ⎝⎜⎜⎛ ⎠⎟⎟⎞ e^1 e^2 e^3 = ⎝⎜⎛ ⎠⎟⎞ cosφsinθ sinφsinθ cosθ cosφcosθ sinφ cosθ -sinθ -sinφ cosφ 0 ⎝⎜⎜⎛ ⎠⎟⎟⎞ e1 e2 e3 = [R(ξ)]-1 ⎝⎜⎜⎛ ⎠⎟⎟⎞ e1 e2 e3 (A.13a) or ⎝⎜⎜⎛ ⎠⎟⎟⎞ r^ θ^ φ^ = ⎝⎜⎛ ⎠⎟⎞ cosφsinθ sinφsinθ cosθ cosφcosθ sinφ cosθ -sinθ -sinφ cosφ 0 ⎝⎜⎜⎛ ⎠⎟⎟⎞ x^ y^ z^ = [R( ξ)]-1 ⎝⎜⎜⎛ ⎠⎟⎟⎞ x^ y^ z^ (A.13b) or r^ = cosφsinθ x^ + sinφsinθ y^ + cosθ z^ θ^ = cosφcosθ x^ + sinφcosθ y^ - sinθ z^ φ^ = -sinφ x^ + cosφ y^ ( A . 1 3 c ) which are the well known expressions for the spherical unit vectors in terms of the Cartesian ones. The inverse of the last three equations can be obtai ned in the following manner (matrix from (A.10)) ⎝⎜⎜⎛ ⎠⎟⎟⎞ e1 e2 e3 = [R(ξ)] ⎝⎜⎜⎛ ⎠⎟⎟⎞e^1 e^2 e^3 = ⎝⎜⎛ ⎠⎟⎞ cosφsinθ cosφcosθ -sinφ sinφ sinθ sinφcosθ cosφ cosθ -sinθ 0 ⎝⎜⎜⎛ ⎠⎟⎟⎞e^1 e^2 e^3 (A.14a) or ⎝⎜⎜⎛ ⎠⎟⎟⎞ x^ y^ z^ = [R(ξ)] ⎝⎜⎜⎛ ⎠⎟⎟⎞ r^ θ^ φ^ = ⎝⎜⎛ ⎠⎟⎞ cosφsinθ cosφcosθ -sinφ sinφ sinθ sinφcosθ cosφ cosθ -sinθ 0 ⎝⎜⎜⎛ ⎠⎟⎟⎞ r^ θ^ φ^ (A.14b) or x^ = cosφsinθ r^ + cosφcosθ θ^ - sinφ φ^ y^ = sinφsinθ r^ + sinφ cosθ θ^ + cosφ φ^ z^ = cosθ r^ - sinθ θ^ . (A.14c) Appendix E contains further inform ation on spherical coordinates. Some Properties of Rotation Matrices A rotation matrix R is always real orthogonal which means R-1 = RT. It follows that RRT = 1 ⇒ ΣkRik(RT)kj = δi'j ⇒ ΣkRikRjk = δi,j RTR = 1 ⇒ Σk(RT)ikRkj = δi'j ⇒ ΣkRkiRkj = δi,j . (A.15) Also RRT = 1 ⇒ det(RRT) = det(1) ⇒ [det(R)]2 = 1 For a rotation matrix, det(R) = +1. (A.16) Appendix A: Spherical Coordinates 153 The determinant of R may be written using a standa rd expansion for the determinant of a matrix, 1 = det(R) = ΣijkεijkRi1Rj2Rk3 , ( A . 1 7 ) where the permutation tensor was described in (1.5.3). This can be generalized to read εabc = ΣijkεijkRiaRjbRkc for any a,b,c (A.18) where (A.17) is the particular case with abc = 123. We leave the proof of (A.18) as a Reader Exercise. Now apply Σ aRna to both sides and sum on a to get ΣaRna εabc = ΣaRna ΣijkεijkRiaRjbRkc = Σijkεijk [ ΣaRnaRia] RjbRkc = Σijkεijk [δn,i] RjbRkc // (A.15) = Σjk εnjkRjbRkc . We have just derived the following rarely-sta ted property of any 3x3 rotation matrix R : ΣaRnaεabc = ΣjkεnjkRjbRkc for any n,b,c . (A.19) Theorem : If A' = R A, B' =RB and C' = R C, then C = A x B ⇔ C' = A' x B' . (A.20) Proof ⇐ : C' = A' x B' ⇒ C'n = Σjkεnjk A'jB'k ⇒ (R C)n = Σjkεnjk (RA)j(RB)k ⇒ (ΣaRnaCa) = Σjkεnjk(ΣbRjbAb)(ΣcRkcBc) = = Σbc [ ΣjkεnjkRjbRkc] AbBc = Σ bc [ ΣaRnaεabc] AbBc // (A.19) = Σ aRna [ΣbcεabcAbBc ] = Σ a Rna [A x B ]a . In vector notation we have just shown that R C = R( AxB). Apply R-1 from the left to conclude that C = A x B, QED. Run the steps in reverse to prove ⇒ . This theorem confirms the intuitive fact that if A,B,C all transform as normal vectors under R, then if C = AxB in Frame S, then C' = A'xB' in rotated Frame S' . The fact that A•B = A'•B' is more obvious and requires only (A.15) : Σj(A')j(B')j = Σj(ΣbRjbAb)(ΣcRjcBc) = Σbc δbcAbBc = ΣbAbBb . Appendix B: Rank-n G Rule 154 Appendix B: The G Rule for a Tensor of Rank n In this section we continue to us e these shorthand operator notations, ∂S ≡ (d/dt)S ∂S' ≡ (d/dt)S' ∂t ≡ (d/dt) . (1.8.3) (B.1) Recall from (1.10.1) that for a scalar function or for a component of any tensor one has, (∂STijk... ) = (∂S'Tijk... ) = (∂tTijk... ) , (1.10.1) (B.2) and from (2.5) that ∂ S'ei = – ω x ei . (2.9) (B.3) So far we know all about the G Rule for tens ors of rank 0 and 1 (scalar and vector), ∂SA = ∂S'A // = ∂tA; A is a scalar function of t (B.2) ∂ SA = ∂S'A + ω x A . // A is a vector (2.1) (B.4) What happens for tensors of rank 2 or more? To expl ore this question, we first collect a few facts. A tensor of rank-n has the following e xpansion (where q represents the n th letter of the alphabet) T = Σabc..q. Tabc..q (ea⊗eb⊗ec.....⊗ eq) . (B.5) Here T has n subscripts and there is a tensor product of n basis vectors ei. This expansion is a generalization of the expansion of a vector, V = ΣiViei . ( B . 6 ) The basis vectors are axis-aligned unit vectors, so we have ( en)i = δn,i . (1.1.3) (B.7) The meaning of the ⊗ symbol is nothing more than the following, [A⊗B⊗....⊗Q ]abc...q = AaBbCc....Qq . ( B . 8 ) This particular tensor A ⊗B⊗....⊗Q happens to be the direct product of the n vectors A,B,C...Q, but there are of course general tensors T which cannot be written as such a direct product. As a simple tensor product example, ( ea⊗eb)ij = (ea)i (eb)j = δa,i δb,j . ( B . 9 ) Appendix B: Rank-n G Rule 155 Here is the expansion of the direct product tensor T = A⊗B, A⊗B = Σab [A⊗B]ab ea⊗eb = Σab AaBb ea⊗eb . ( B . 1 0 ) [ For more details on such tensor products and expansions, see Lucht Tensor Products with ei→ ui] . Before continuing, we develop two very similar Lemmas: Lemma 1: ∂S(A⊗B) = (∂SA) ⊗ B + A⊗(∂SB) ( B . 1 1 ) where A and B are vectors whose components are functions of time. Proof of Lemma 1: ∂S(A⊗B) = ∂S( Σab AaBb ea⊗eb ) // expansion (B.10) = Σ ab ∂S(AaBb) ea⊗eb / / s i n c e ∂Sen = 0 by (1.7.3) = Σab ∂t(AaBb) ea⊗eb / / ( B . 2 ) = Σab [(∂tAa)Bb + Aa(∂tBb)] ea⊗eb // regular calculus product rule = Σ ab [(∂SAa)Bb + Aa(∂SBb)] ea⊗eb // restore ∂S using (B.2) = Σ ab [(∂SA)aBb + Aa(∂SB)b] ea⊗eb /// commutation rule (1.11.1) = Σ ab [(∂SA)aBb] ea⊗eb + Σab Aa(∂SB)b ea⊗eb // write as two terms = ( ∂ SA) ⊗ B + A⊗(∂SB) . // expansion (B.10) QED Recall from (1.2.1) that one can expand a vector A on either the ei or the e'i basis, A = ΣiAiei = ΣiA'ie'i . (1.2.1) (B.12) Similarly, one can expand the tensor T of (B.5) on the e'i basis to get T = Σabc..q. T'abc..q (e'a⊗e'b⊗e'c.....⊗e'q) ( B . 1 3 ) with this special case A⊗B = Σab [A⊗B]'ab e'a⊗e'b = Σab A'aB'b e'a⊗e'b . ( B . 1 4 ) We then have Lemma 2 which is the same as Lemma 1 but with S →S' , Appendix B: Rank-n G Rule 156 Lemma 2: ∂S'(A⊗B) = (∂S'A) ⊗ B + A⊗(∂S'B) ( B . 1 5 ) The proof exactly follows that of Lemma 1 but we show it anyway: ∂ S'(A⊗B) = ∂S'( Σab A'aB'b e'a⊗e'b ) // expansion (B.14) = Σab ∂S'(A'aB'b) e'a⊗e'b / / s i n c e ∂S'e'n = 0 by (1.7.2) = Σab ∂t(A'aB'b) e'a⊗e'b // (B.2) = Σ ab [(∂tA'a)B'b + A'a(∂tB'b)] e'a⊗e'b // regular calculus product rule = Σ ab [(∂S'A'a)B'b + A'a(∂S'B'b)] e'a⊗e'b // restore ∂S' using (B.2) = Σ ab [(∂S'A)'aB'b + A'a(∂S'B)'b] e'a⊗e'b // commutation rule (1.11.1) = Σ ab [(∂S'A)'aB'b] e'a⊗e'b + Σab A'a(∂S'B)'b e'a⊗e'b // write as two terms = ( ∂ S'A) ⊗ B + A⊗(∂S'B) . // expansion (B.14) QED G Rule for a Rank-2 Tensor Recall from (B.5) that, T = Σ abTab (ea⊗eb) . (B.5) (B.16) Then apply ∂S, using the facts ∂ Sen = 0 and (B.2), (∂ST) = Σab(∂STab) (ea⊗eb) = Σab(∂tTab) (ea⊗eb) . (B.17) On the other hand, (∂S'T) = Σab (∂S'Tab) (ea⊗eb) + ΣabTab ∂S'(ea⊗eb) = Σ ab (∂tTab) (ea⊗eb) + ΣabTab ∂S'(ea⊗eb) // (B.2) = ( ∂ ST) + ΣabTab ∂S'(ea⊗eb) . // (B.17) (B.18) But, ∂S'(ea⊗eb) = (∂S'ea) ⊗ eb + ea⊗(∂S'eb) // by Lemma 2 (B.15) = – [ ( ω x ea) ⊗ eb + ea⊗ (ω x eb) . // (B.3) (B.19) Appendix B: Rank-n G Rule 157 Inserting this last result into (B.18) and swapping sides then gives the G rule for a rank-2 tensor, (∂ ST) = (∂S'T) + ΣabTab [ (ω x ea) ⊗ eb + ea⊗ (ω x eb) ] . (B.20) We can write the G rule for a vector T (rank-1 tensor) in a similar form, (∂ST) = (∂S'T) + ω x T = ( ∂ S'T) + ω x (ΣaTaea) = ( ∂ S'T) + ΣaTa ω x ea . ( B . 2 1 ) G Rule for a Rank-n Tensor It is easy to show that the above general pattern appl ies to tensors of rank 3 and higher, and one ends up with the following lowest tensor G rules, where the tensor rank is shown on the left . 0 (∂ST) = (∂S'T) 1 (∂ST) = (∂S'T) + ΣaTa ω x ea 2 (∂ST) = (∂S'T) + ΣabTab [ (ω x ea) ⊗ eb + ea⊗ (ω x eb) ] 3 (∂ST) = (∂S'T) + ΣabcTabc [ (ωxea)⊗eb⊗ec + ea⊗(ωxeb)⊗ec + ea⊗eb⊗(ωxec) ] (B.22) These G rule results may also be expressed in components. For example: 1 (∂ ST)i = (∂S'T)i + ΣaTa (ω x ea)i = ( ∂S'T)i + ΣaTa εirs ωr (ea)s = ( ∂S'T)i + ΣaTa εirs ωr δa,s = ( ∂S'T)i + εirs ωrTs . ( B . 2 3 ) 2 (∂ST)ij = (∂S'T)ij + ΣabTab [ (ω x ea) ⊗ eb + ea⊗ (ω x eb) ]ij = ( ∂S'T)ij + ΣabTab [ (ω x ea)i (eb)j + ( ea)i(ω x eb)j ] = ( ∂S'T)ij + [ Σab Tab (ω x ea)i δb,j + Σab Tab δa,i (ω x eb)j ] = ( ∂S'T)ij + [ Σa Taj (ω x ea)i + Σb Tib (ω x eb)j ] = ( ∂ S'T)ij + [ Σa Taj εirsωr(ea)s + Σb Tib εjrsωr(eb)s ] Appendix B: Rank-n G Rule 158 = ( ∂S'T)ij + [ Σa Taj εirsωrδa,s + Σb Tib εjrsωrδb,s ] = ( ∂ S'T)ij + [ Tsj εirsωr+ Tis εjrsωr ] = ( ∂S'T)ij + εirsωr Tsj + εjrsωrTis . ( B . 2 4 ) We shall do one more case to es tablish the general pattern, 3 (∂ ST)ijk - (∂S'T)ijk = ΣabcTabc [ (ωxea)⊗eb⊗ec + ea⊗(ωxeb)⊗ec + ea⊗eb⊗(ωxec) ]ijk = ΣabcTabc [ (ωxea)i(eb)j(ec)k + ( ea)i(ωxeb)j(ec)k + (ea)i(eb)j(ωxec)k ] = Σ abcTabc(ωxea)iδb,jδc,k + ΣabcTabcδa,i(ωxeb)jδc,k + ΣabcTabcδa,i δb,j(ωxec)k = ΣaTajk(ωxea)i + ΣbTibk(ωxeb)j + ΣcTijc(ωxec)k = ΣaTajkεirsωr (ea)s + ΣbTibkεjrsωr (eb)s + ΣcTijcεkrsωr (ec)s = Σ aTajkεirsωr δa,s + ΣbTibkεjrsωr δb,s + ΣcTijcεkrsωrδc,s = T sjkεirsωr+ Tiskεjrsωr + Tijsεkrsωr so we conclude that (∂ST)ijk = (∂S'T)ijk + εirsωrTsjk + εjrsωrTisk + εkrsωrTijs . (B.25) Looking at the last two results we can infer the rank-4 result (∂ ST)ijkl = (∂S'T)ijkl + εirsωrTsjkl + εjrsωrTiskl + εkrsωrTijsl + ε lrsωrTijks . (B.26) Summary of G rule for all ranks (tensor form) ( B . 2 7 ) 0 (∂ST) = (∂S'T) 1 (∂ ST) = (∂S'T) + ΣaTa ω x ea 2 (∂ST) = (∂S'T) + ΣabTab [ (ω x ea) ⊗ eb + ea⊗ (ω x eb) ] 3 (∂ ST) = (∂S'T) + ΣabcTabc [ (ωxea)⊗eb⊗ec + ea⊗(ωxeb)⊗ec + ea⊗eb⊗(ωxec) ] 4 (∂ ST) = (∂S'T) + ΣabcdTabcd [ (ωxea)⊗eb⊗ec⊗ed + ea⊗(ωxeb)⊗ec⊗ed + ea⊗eb⊗(ωxec)⊗ed + ea⊗eb⊗ec⊗(ωxed) ] etc. Appendix B: Rank-n G Rule 159 Summary of G rule for all ranks (component form) ( B . 2 8 ) 0 (∂ST) = (∂S'T) 1 (∂ST)i = (∂S'T)i + εirs ωrTs = (∂S'T)i + (ω x T)i 2 (∂ST)ij = (∂S'T)ij + εirsωr Tsj + εjrsωrTis 3 (∂ST)ijk = (∂S'T)ijk + εirsωrTsjk + εjrsωrTisk + εkrsωrTijs 4 (∂ST)ijkl = (∂S'T)ijkl + εirsωrTsjkl + εjrsωrTiskl + εkrsωrTijsl + ε lrsωrTijks etc. In the main body of our document we deal only with vector quantities r, v and a , but one can easily imagine "frames of reference problems" which deal also with tensors, such as the inertia tensor. Applications involving continuum mechanics abound in tensors. Appendix C: Foucault Pendulum 160 Appendix C: The Foucault Pendulum This is a long appendix so we provide an overview: The opening section reviews the traditional brief treatment of the Foucault pendulum precession. Section C.1 sets up the kinematics of the spherical pendulum. Section C.2 gives a quick qualitative explanation of the Foucault precession. Section C.3 derives the Foucault pendulum angular equa tions of motion, stated in (C.3.9). Section C.4 derives the exact solution to the simple (plane) pendulum in (C.4.20) or (C.4.26). Section C.5 reviews the exact solution of spherical pendulum, and also gives qualitative hints as to the nature of the pendulum motion. After stating the si mple conical motion solution, the motion of thin ellipses is reviewed and the Airy precession appears. After reviewing the Foucault mode of the spherical pendulum, the interference between th e Airy and Foucault precessions is analyzed. A detailed analysis of the Foucault pendulum at the Pantheon in Paris is presented. Finally, it is shown how to numerically obtain 2D and 3D plots of arbitrary motions of a spherical pendulum using Maple. Section C.6 derives the Foucault equations of motion in Cartesian coordinates in (C.6.9). Section C.7 shows that the x,y,z equations of motion of Section C.6 are entirely equivalent to the θ,φ equations of motion of Section C.3. Section C.8 studies numerical solutions of both the Foucault and Spherical pendulums. For the latter the Airy precession is demonstrated. Nutshell Analysis of the Foucault Pendulum This subject is usually treated in the small oscillati on limit in Cartesian coordinates (e.g. Taylor, Marion, Thornton and Marion). Applying Newton's Law F = ma including the Coriolis force one quickly obtains a pair of coupled linear ODE's which, in "top view" (x,y) notation, are x•• – 2ωcosβ y• + Ω2x = 0 ω = Earth rotation rate, β = polar angle (colatitude) from North Pole y•• + 2ωcosβ x• + Ω2y = 0 Ω = g/l = swing rate, ω << Ω, l = string length, g=gravity . (C.1) If the second equation is multiplied by i and added to the first, one gets η•• + 2iωcosβη• = - Ω2η . η ≡ x + i y ( C . 2 ) We ask Maple to try out the following candidate solution to (C.2), keeping in mind that ω << Ω, η(t) = A e -i(ωcosβ)t cos(Ω t ) . ( C . 3 ) Appendix C: Foucault Pendulum 161 ( C . 4 ) Thus the equation (C.2) is solved by the candidate solution (C.3) to order ( ω/Ω)2 and so is a good approximate solution. Writing out x = Re η and y = Im η one finds that y/x = tan φ where φ = -ω cosβ t φ• = -ω cosβ ( C . 5 ) which indicates a slow clockwise rotation of the swing axis for cos β > 0 (Northern Hemisphere). This is in agreement with one's knowledge that, in said He misphere, on each swing the pendulum is deflected a little to the right by the Coriolis force (s ee Fig (C.2.1) or Fig (C.8.7) below ). Our treatment below does not assume small oscilla tion. The pendulum is treated as a full-blown "spherical pendulum" which is influenced by the rotation of the Earth. We derive the pendulum equations of motion first in spherical coordinates in Section C.3, then later in Cartesian coordinates in Section C.6, since these are better for the numerical work of Secti on C.8. We show in (C.6.14) that equations (C.1) above are in fact the small-oscillation lim it of the general equations of motion. C.1 Drawings, Notation, Coordinates and Basis Vectors Fig (4.7.1) showed a typical Earth problem kinematic scenario in the "non-swap" notation where Frame S' is the rotating frame. Here, we choose instead to us e the "swap" notation where Frame S is the rotating frame, since this eliminates the need for scores of prime symbols which would otherwise clutter the equations. So now Frame S' is at the center of the Earth and is an inertial fra me, and Frame S is on the surface of the Earth and is a non-inertial frame. This use of primes is then in accordance with the Goldstein and Marion references di scussed in Sections 9 and 10. In addition to this S ↔S' swap, we make a few other cha nges compared to Fig (4.7.1). First, we select the basis vectors en differently. Since we are going to be dealing with a spherical pendulum, we would like the angle between the pendulum string and local vertical to be the polar angle of a spherical coordinate system for Frame S. This means that we want the z^ = e3 axis associated with this spherical system to be pointed toward the center of the Earth, rather than pointing "up". This then suggests that we take x^ = e1 as pointing "north", and y^ = e2 as pointing "east" and then the resulting vectors en form a right-handed coordinate system. Second, we place the origin of Frame S at a height l above the surface of the Earth, where l is the length of the spherical pendulum's string. Then the length of vector b is b = R e + l, where R e is the radius of the Earth. The Frame S origin is then at the pendulum pivot point. The new kinematic picture is then, Appendix C: Foucault Pendulum 162 (C.1.1) The next step is to use spherical coordinates in both Frames S and S'. For Frame S' , with origin at the center of the Earth and assumed fixed relative to distant stars, we select spherical coordinates, so r' = (r',θ',φ'). We shall assume that the origin of Frame S is located at b = (b, β, α) where β is the polar angle measured down from the North Pole (colatitude, range 0 to π), α is the origin's azimuth (longitude), and b = R e+ l where R e is the radius of the Earth and l is the length of the pendulum string (picture coming soon). Parameters α and b will play no role in what follows. Note that ω = ωe'3 with ω > 0 since the Earth rotates counterclockwise as viewed from above the North Pole, causing the Sun to rise in the east. Th e same Earth, viewed looking upward from beneath the South Pole, appears to rotate clockwis e (and the Sun still rises in the east). For Frame S (red, shown in the Northern Hemisphere) we select another set of spherical coordinates r = (r,θ,φ) as will be described momentarily. The Cartesian unit basis vectors en of Frame S can be expressed in te rms of the Frame S' spherical unit basis vectors r^', θ^', φ^' evaluated at the position r' = b = (b, β, α) as follows, e1 = -θ^' = "north" = x^ e2 = φ^' = "east" = y^ e3 = - r^' = "down" = z^ . According to the picture above, ω = ωe'3 can be expanded on these Frame S basis vectors, ω = - ωcosβ e3 + ωsinβ e1 . (C.1.2) Appendix C: Foucault Pendulum 163 As noted, the origin of Frame S is located distance l above the surface of th e Earth. If we position ourselves at this origin and gaze down onto the Earth's surface we see what is shown on the left below. A side view is presented on the right, and a 3D view below: (C.1.3) The meaning of the Frame S spherical coordinates r, θ,φ should be clear. These are the standard coordinates one would use to study a spherical pe ndulum in the absence of the Coriolis force. The Cartesian basis vectors en for Frame S are related to the spherical ones for Frame S as follows (see (A.14c) or (E.2.7)), e1 = x^ = sinθcosφ r^ + cosθcosφ θ^ - sinφ φ^ e2 = y^ = sinθsinφ r^ + cosθsinφ θ^ + cosφ φ^ e3 = z^ = cosθ r^ - sinθ θ^ . (C.1.4) Appendix C: Foucault Pendulum 164 C.2 Qualitative Solution On each swing the pendulum veers "a little to the right" in the Northern Hemisphere due to the Coriolis fictitious force -2m ω x v, so the problem is to compute the time for a full 360 degree rotation. Here we show the first four swings of the pendulum (viewed from above) (C.2.1) The net effect is that the plane of the swinging pendulum precesses clockwise. The amount of veering shown in the drawing is highly exaggerated since we know that when the pendulum is located at the North Pole the period will be one sidereal day ( ≈ 23 hours 56 minutes). If each swing takes 10 seconds, that would be about 24*3600/10 = 8,640 swings for a full revolution, so each swing would show only 360/8640 ≈ .04 degrees of precession. Away from the North Pole we know the precession rate will be even slower, at the equator it will be zero, and in the Southern Hemisphere the pendulum will precess in the opposite direction. These qualitative facts may be deduced from the direction and magnitude of the Coriolis force as discussed in Section 8.3. The gene ral pattern was noted for the small-angle limit of the theory in the solution (C.3). C.3 Equations of Motion for the Foucault Pendulum (Spherical Coordinates) In our "swap" notation context, the bogus Newton's law for non-inertial Frame S is, Feff = ma . (12.2.3) (C.3.1) The effective force Feff consists of real forces and fictitious fo rces. At the end of Section 8.5 in (8.5.7) s we showed that, for surface-of-the-Earth problems (which are Special Case #1), and in the "swap" notation, Feff = (m g + possible other real forces) – 2m ω x v , (8.5.7)s where force m g already incorporates the effect of the fictitious centrifugal force. In the spherical Foucault pendulum problem, we have "possible other real forces" = T, the tension of the massless pendulum string pulling on the pendulum of mass m. Thus, Appendix C: Foucault Pendulum 165 Feff = m g + T – 2m ω x v . (C.3.2) Although g does not point to the exact center of the Earth, we shall assume that it does, so g = gz^. The directional error in making this assumption is on the order of Δg/g0 ~ .003 ρ^ radians (see (8.5.8,9) ) which is insignificant in our analysis of the Foucault pendulum. The error is probably even less than this when one considers that the Earth's surface is perpendicular to g and not g0, but we don't want to get involved with such fine details. Using g = gz^ in (C.3.2) and using the latter in (C.3.1) we find m a = mg z^ + T – 2m ω x v ( C . 3 . 3 ) which is our vector "equation of motion" for the pe ndulum. Our task now is to evaluate the terms in (C.3.3) and then to balance the components on both sides to come up with three scalar equations of motion. The ω vector (C.1.2) can be expanded, using (C.1.4) for e3 and e1, ω = - ωcosβ e3 + ωsinβ e1 = - ωcosβ[cosθ r^ - sinθ θ^] + ωsinβ[cosφsinθ r^ + cosφcosθ θ^ - sinφ φ^] = ω(-cosβcosθ + sinβcosφsinθ) r^ + ω(cosβsinθ + sinβcosφcosθ ) θ^ + ω(-sinβ sinφ) φ^ . ( C . 3 . 4 ) The string tension vector may be written as follows, where T > 0, T = - T r^ . ( C . 3 . 5 ) If one wants to consider unusual in itial conditions for the pendulum wh ich would cause negative string tension, such as starting it at position r = - lz^ with some tiny velocity, one can imagine the string to be replaced with a massless rigid stick of length l. Such a stick pendulum then works for either sign of tension T and then matches the equations of motion. It remains to find expressions for v and a expanded on the spherical unit vectors. These are developed in (E.2.11) from which we quote: (dr^/dt)S = r^• = θ• θ^ + sinθ φ• φ^ (dθ^/dt)S = θ^• = - θ• r^ + cosθ φ• φ^ (dφ^/dt)S = φ^• = - sinθ φ• r^ - cosθ φ• θ^ . (E.2.11) (C.3.6) Using (E.3.6) for acceleration a one then has, Appendix C: Foucault Pendulum 166 r = l r^ v = r• = l r^• = l [θ• θ^ + sinθ φ• φ^] a = v• = l (- θ•2- sin2θ φ•2) r^ + l (θ•• - sinθcosθ φ•2) θ^ + l ( 2cosθ θ• φ• + sinθ φ••) φ^ (C.3.7) so we now have a viable v and a to use in (C.3.3). The Coriolis cross product can now be computed from (C.3.4) and (C.3.7), ω x v = [ω(-cosβ cosθ + sinβcosφsinθ) r^ + ω(cosβ sinθ + sinβ cosφcosθ ) θ^ + ω(-sinβ sinφ) φ^] x [l [θ• θ^ + sinθ φ• φ^] or ω x v /(ωl) = (-cos βcosθ + sinβcosφsinθ) r^ x [θ• θ^ + sinθ φ• φ^] + ( cosβsinθ + sinβ cosφcosθ ) θ^ x [θ• θ^ + sinθ φ• φ^] + (-sin βsinφ) φ^ x [θ• θ^ + sinθ φ• φ^] = (-cosβcosθ + sinβcosφsinθ) [θ• φ^ - sinθ φ• θ^] // see (E.2.12) + ( cosβsinθ + sinβcosφcosθ ) [sinθ φ• r^] + (-sin β sinφ) [- θ• r^] = [ (cosβsinθ + sinβcosφcosθ )sinθ φ• + sinβ sinφ θ• ] r^ – [ (-cosβcosθ + sinβcosφsinθ)sinθ φ•] θ^ + [ (-cosβcosθ + sinβcosφsinθ) θ• ] φ^ . (C.3.8) We can now assemble the pieces from (C.1.4), (C.3.5), (C.3.7) and (C.3.8) to write the equation of motion (C.3.3) divided by m, a = g z^ + T/m – 2 ω x v , (C.3.3) as l(- θ•2- sin2θ φ•2) r^ + l(θ•• - sinθcosθ φ•2) θ^ + l( 2cosθ θ• φ• + sinθ φ••) φ^ = g ( c o s θ r^ - sinθ θ^) - (T/m) r^ – 2ω l [(cosβsinθ + sinβcosφcosθ )sinθ φ• + sinβ sinφ θ• ] r^ + 2 ωl [(-cosβcosθ + sinβcosφsinθ)sinθ φ•] θ^ – 2ω l [(-cosβcosθ + sinβcosφsinθ) θ• ] φ^ . Matching components gives these three scalar equations of motion (the unit vectors on the right are just reminders of the origin of the equations), Appendix C: Foucault Pendulum 167 l(- θ•2- sin2θ φ•2) = gcosθ – (T/m) – 2ω l [(cosβsinθ + sinβcosφcosθ )sinθ φ• + sinβsinφ θ• ] r^ l(θ•• - sinθcosθ φ•2) = - g sin θ + 2ωl [(-cosβcosθ + sinβcosφsinθ)sinθ φ•] θ^ l( 2cosθ θ• φ• + sinθ φ••) = – 2ωl [(-cosβcosθ + sinβ cosφsinθ) θ• ] φ^ which we can simplify slightly to get θ•2 + sin2θ φ•2 = -(g/ l)cosθ + T/(ml) + 2ω [(cosβsinθ + sinβcosφcosθ )sinθ φ• + sinβsinφ θ• ] θ•• - sinθcosθ φ•2 = - (g/ l) sinθ - 2ωsinθ φ• (cosβcosθ - sinβcosφsinθ) 2cosθ θ• φ• + sinθ φ•• = 2ωθ• (cosβcosθ - sinβcosφsinθ) . (C.3.9) These are the equations of motion for a spherical pendulum operating on the rotating Earth with no approximations other than the modest ones that g does not vary over the small distance l and that g = gz^. Of course there is no air friction and the string is massless and lossless where it attaches. Reader Exercise : Can equations (C.3.9) be obtained from the conventional Lagrangian or Hamiltonian formalisms? Are these approaches valid in non- inertial frames, or are modifications needed? C.4 The Simple Pendulum In this section we temporarily turn off the rotation of the Earth to study the behavior of pendulum without that complication. Setting ω = 0 in (C.3.9) the pendulum equations of motion become θ• 2 + sin2θ φ•2 = -(g/ l)cosθ + T/(ml) r^ θ•• - sinθcosθ φ•2 = - (g/ l)sinθ θ^ 2cosθ θ• φ• + sinθ φ•• = 0 . φ^ (C.4.1) We now seek a solution of (C.4.1) for which φ• = 0. In this case the last equation requires φ•• = 0 and we are left with just two equations, θ•2 = -(g/ l)cosθ + T/(ml) r^ θ•• = - (g/ l)sinθ . θ^ ( C . 4 . 2 ) These equations describe a pendulum that seems to swing in the plane φ = constant and so this is an example of a plane pendulum, also known as a simp le pendulum. One can solve the second equation for θ(t), and then the first equation gives the tension T(t). If the pendulum is started at some polar angle θ0 and released perfectly so φ = φ0 and φ• = 0, the pendulum swings in a plane, but at it swings through θ = 0 the coordinate φ = φ0 discontinuously jumps to φ = φ0+π, so there is a continuity issue for φ at θ = 0. We could treat this technically using Heaviside Appendix C: Foucault Pendulum 168 and delta functions, but shall not go dow n that road. It is a simple fact that, for spherical coordinates, points on the z-axis where θ = 0 have an undefined value of φ . For small angles θ, one finds for the above initial condition that θ•• +(g/ l)θ = 0 ⇒ θ(t) = θ0cos(Ωt) Ω ≡ g/l {θ(0) = θ0 .θ•(0) = 0} (C.4.3) On the other hand, if the pendulum is initialized with θ = 0 and some velocity sufficient to take it up to a maximum angle θ0 we get θ•• +(g/ l)θ = 0 ⇒ θ(t) = θ0sin(Ωt) Ω ≡ g/l {θ(0) = 0, θmax = θ0 } (C.4.4) Exact Solution for the Simple Pendulum We provide this detailed solution because it generally does not app ear in textbooks. For larger θ 0 the simple pendulum equation θ•• + (g/ l)sinθ = 0 falls into a class of second order non-linear ODE's which have the form x•• = f(x) where x•• means ∂ t2x and we seek x(t). The solution is not hard to obtain, as we now outline. The first step is to define v ≡ x• : v ≡ x• ⇒ x•• = (dv/dt) = (dv/dx)(dx/dt) = (dv/dx)v ⇒ (dv/dx)v = f(x) ⇒ vdv = f(x)dx ⇒ (v2/2) = ∫ x f(x')dx' + C ⇒ (dx/dt) = 2 ⇒ dt = 1 2 dx ∫ x f(x')dx' + C ⇒ t(x) = ( 1 2 ∫ x dx" ∫ x" f(x')dx' + C ) + C' (C.4.5) where integration constants C and C' are determined by initial conditions. This solution gives t = t(x) which one must then "invert" to obtain x = x(t). If we take f(x) = - Ω 2sinx and then x → θ the (v2/2) result of (C.4.5) becomes (θ•2/2) = ∫ θ f(θ')dθ' + C = - Ω2 ∫ θ sin(θ')dθ' + C = Ω2cosθ + C . (C.4.6) Appendix C: Foucault Pendulum 169 We set the zero of potential energy for the pendulum at the bottom, θ = 0. At angle θ the pendulum has risen a height h = l - lcosθ so the potential energy at θ is then V = mgh = mg l(1-cosθ). We shall now apply the boundary conditions of (C.4.4). At t = 0 the pendulum is at θ = 0 and we give it a kick in the + θ direction with some initial velocity lθ•(0). We assume that this causes the pendulum to rise up to some max angle θ0 < π. A too-large kick results in over-t he-top behavior which we exclude. At t = 0 the total pendulum energy is (1/2)m[ lθ•(0)]2 . At the top of the swing the total energy is mg l(1-cosθ0). Therefore from energy conservation, (1/2)m[ lθ•(0)]2 = mg l(1-cosθ0) or (1/2)[ θ•(0)] 2 = (g/ l)(1-cosθ0) = Ω2(1-cosθ0) so θ•(0) = 2 Ω 1-cosθ0 = 2Ωsin(θ0/2) . (C.4.7) Evaluating (C.4.6) at t = 0 gives (1/2)[ θ•(0)] 2 = Ω2cos(0) + C = Ω2 + C . Comparing this last equation with the middle line of (C.4.7) gives, C = -Ω 2cosθ0 . ( C . 4 . 8 ) From (C.4.5) the solution for t( θ) is then ⇒ t(θ) = ( 1 2 ∫ θ dθ" ∫ θ" f(θ')dθ' + C ) + C' = ( 1 2 ∫ θ dθ" Ω2cosθ" + C ) + C' = ( 1 2 Ω ∫ θ dθ" cosθ" - cosθ0 ) + C' (C.4.9) The integral of interest appears on page 179 of GR7 as 2.571.4 (where we shall use the second form), (C.4.10) Appendix C: Foucault Pendulum 170 and where F( φ,k) is "the elliptic integral of the first kind". In our case b = 1 and a = -cos θ0 so r = 2 a+1 = 2 1-cosθ0 = 1 sin(θ0/2) γ = sin-1[ 1-cosθ 1-cosθ0 ] = sin-1 [ sin(θ/2) sin(θ0/2) ] . (C.4.11) Using integral evaluation (C.4.10) in (C.4.9) one finds that, t(θ) = 1 Ω F( sin-1 [ sin(θ/2) sin(θ0/2) ], sin(θ0/2) ) + C' = 1 Ω F(φ, k) + C' where φ = sin-1 [ sin(θ/2) k ] , k = sin( θ0/2) . (C.4.12) The function F is defined in GR7 page 860 as 8.111.2, (C.4.13) from which we see that F(0,k) = 0. Recall our boundary condition that θ(0) = 0 with some θ•(0) > 0. From (C.4.12) we find at t = 0 and θ = 0 that 0 = 1 Ω F(0, k) + C' = 0 + C' ⇒ C' = 0 . (C.4.14) Our final solution before inversion is then t(θ) = 1 Ω F(φ, k) where φ = sin-1 [ sin(θ/2) k ] , k = sin( θ0/2), Ω ≡ g/l . (C.4.15) The Jacobi elliptic function sn(u) is usually defined in this rather obscure manner, ( C . 4 . 1 6 ) If one writes sn(u) = sin( φ), then the right sides of (C.4.13) and (C.4.16) are the same, so F(φ ,k) = u = sn-1(sinφ) or sn(F(φ ,k)) = sinφ . (C.4.17) Appendix C: Foucault Pendulum 171 We now apply this last equati on to (C.4.15) as follows: sn(F(φ ,k)) = sinφ or sn(Ω t) = sin { sin-1 [ sin(θ/2) k ] } = sin(θ/2) k so sin(θ/2) = k sn( Ωt) k = sin( θ0/2) and θ(t) = 2 sin-1( k sn(Ωt) ) k = sin( θ0/2) (C.4.18) and the inversion of (C.4.15) is now complete. Fo llowing the convention of GR7, since sn(u) has an implicit parameter k, we make it explicit by writin g sin(u) = sin(u,k) and then our solution is sin(θ/2) = k sn( Ωt,k) k = sin( θ0/2) θ(t) = 2 sin-1( k sn(Ωt,k) ) k = sin( θ0/ 2 ) ( C . 4 . 1 9 ) or sin(θ/2) = sin( θ0/2) sn(Ω t, sin(θ0/2)) // boundary conditions below ⇒ θmax = θ0 θ(t) = 2 sin-1[ sin(θ0/2) sn(Ω t, sin(θ0/2)) ] for θ(0) = 0, θ•(0) = 2Ωsin(θ0/2) . (C.4.20) When k is not too large, one has sn(x,k) ≈ sin(x). If θ0 is small, then so is θ and the first line of (C.4.20) reads, θ/2 ≈ (θ0/2) sin(Ω t) ⇒ θ(t) = θ0 sin(Ωt ) ( C . 4 . 2 1 ) which is the correct small-angle solution for θ(0) = 0 and max angle θ0 as was stated in (C.4.4). Here is a Maple plot of θ(t) from (C.4.20) with a selection of peak angles θ0 with Ω = 1 so that the small- angle period is T = 2 π/Ω = 2π : Appendix C: Foucault Pendulum 172 (C.4.22) For small θ0 the function θ(t) is very sine-like with period T = 2 π, but for larger θ0 the period increases and the top flattens out. For θ0 = 177o the top is quite flat, indicating a long "hang time" when the pendulum (on its rigid massless stick) lingers in the near-vertical position. The complete elliptic integral of the first kind is defined by (some sources write K as K), K(k) ≡ F(π/2,k) = ∫0 π/2 dα 1-k2sin2α // see (C.4.13), and note K(0) = π/2 (C.4.23) Since sn(u) = sin( φ) as shown above (C.4.17), we know that the peak value of sn(u) is 1 and this occurs one quarter of a wave into the sn(u) wa veform. From (C.4.17) we know that sn(F(φ ,k)) = sinφ ⇒ sn(F(π /2,k)) = 1 ⇒ sn(K(k)) = 1 (C.4.24) Thus, a quarter period of the function sn(u,k) along the real u axis is just K(k) and so the full period is then T = 4K(k). Here is a plot of 4K(k) showing how it increases from 2 π to larger values slowly approaching the limit 4K( ∞) = ∞ : (C.4.25) Appendix C: Foucault Pendulum 173 Since a quarter period of sn(x,k) is K(k), a quarter period of sn( Ωt,k) is then K(k)/Ω . For the more standard boundary conditions shown in (C.4.3) where θ(0) = θ0 and .θ•(0) = 0, the solution is obtained by shifting (C.4.20) ahead by a quarter period, which means taking t → t + K(k)/ Ω or Ωt → Ωt + K(k) . The result is then, with k = sin( θ0/2), sin(θ/2) = sin( θ0/2) sn(Ω t + K(sin( θ0/2)), sin(θ0/2)) θ(t) = 2 sin-1[ sin(θ0/2) sn(Ω t + K(sin( θ0/2)), sin(θ0/2)) ] θ(0) = θ0 , θ•(0) = 0 . (C.4.26) Now for small angles the first line gives θ/2 ≈ θ 0/2 sin(Ω t+π/2) = θ0/2 cos(Ω t) ⇒ θ (t) = θ0 cos(Ωt) (C.4.27) in agreement with (C.4.3) Conventions : We have adopted the notation for K(k) and sn(u,k) which is used by the 2010 NIST Handbook of Mathematical Functions , by Gradshteyn and Ryzhik 7th edition (2007), and by the Bateman Manuscript Project (1953), though the last two sources write K as K. The reader is warned that there is another common convention which appears in the precu rsor to the NIST Handbook, namely the heavily- used 1964 Handbook of Mathematical Functions of Abramowitz and Segun (A S). The connection is this K(k) = K AS(k2) / / k2 = m sn(u,k) = sn AS(u,k2) = sn(u | k2) // sometimes = sn(u; k2) (C.4.28) In using these functions one must be very careful to learn the convention used by a given source. For example, our ancient Maple V states that (notice the k2 in the integral but k in the argument), // am(z,k) = sin-1sn(z,k) So this version of Maple is using our "m odern" NIST 2010 notation for argument k. Appendix C: Foucault Pendulum 174 C.5 The Spherical and Foucault Pendulums Having studied the simple pendulum, we now return the more general spherical pendulum and at the end we look at the Foucault mode of this pendulum. Since there are many subsections below, here is a list of their headings: (a) The equations of motion for the Spherical Pendulum (b) Lz as constant of the motion (c) E as another constant of the motion (d) Exact solution to the Spherical Pendulum (outline) (e) The nature of the general solution for the Spherical Pendulum (f) The Conical Motion solution of the Spherical Pendulum (g) The thin ellipse scenario for the Spherical Pendulum (h) The Intrinsic Airy Precession of the Spherical Pendulum (i) The Foucault Mode of the Spherical Pendulum (j) Interference between the Airy and Foucault precession (k) The Foucault Pendulum at the Pantheon in Paris (l) General Numerical solutions of the Spherical Pendulum The French name Foucault is pronounced f oo-ko' with accent on the ko (sounds like so). (a) The equations of motion for the Spherical Pendulum As before, we turn off the rotation of the Earth by setting ω = 0, so equations (C.3.9) become θ• 2 + sin2θ φ•2 = -(g/ l)cosθ + T/(ml) r^ θ•• - sinθcosθ φ•2 = - (g/ l)sinθ θ^ 2cosθ θ• φ• + sinθ φ•• = 0 . φ^ (C.5.1) These are the equations of motion for a spherical pendulum in the presence of a uniform gravitational field of strength g. It is useful to know something a bout the solution of these equations before we turn the Earth's rotation back on. Operationally, in the sense of a numerical solution, one can regard the last two equations of (C.5.1) as a pair of coupled second-orde r non-linear ODE's for functions θ(t) and φ(t) subject to initial conditions θ 0, φ0, θ• 0 and φ•0. Once these equations are solved (physically we know a unique solution must exist), the first equation may be used to determine T(t), the string tension. (b) Lz as constant of the motion The last equation in (C.5.1) may be written in this form d/dt ( sin2θ φ• ) = 0 ( C . 5 . 2 ) Appendix C: Foucault Pendulum 175 which says that sin2θ φ• must be a "constant of the motion". To understand the meaning of this constant, we first compute the torque about the string pivot point due to the gravitational force on the mass m, N = r x mg = r x mg z^ = mg l r^ x [cosθ r^ - sinθ θ^] = -mgl sinθ φ^ . (C.5.3) Since this torque lies in a plane normal to the z axis, as in (C.1.3), we may conclude that the Cartesian torque component N z = 0. The angular version of Newton's Law says N = dL/dt , and therefore we expect that the quantity L z will be a constant of the motion. Direct calculation shows that Lz = L • z^ = m ( r x v) • z^ = m ( z^ x r) • v = m lsinθ φ^ • (l [θ• θ^ + sinθ φ• φ^]) = m l2sin2θ φ• ( C . 5 . 4 ) where we have made use of (C.1.4) for z^ and (C.3.7) for v. Thus, we see that our third equation of motion in (C.5.1), rewritten as in (C.5.2), is just the statement that dL z/dt = 0. In the Lagrangian formulation one finds that L z is one of the canonical momenta which is constant because φ is a cyclic coordinate, meaning it does not appear in the Lagrangian (see Comment below). In terms of the spherical unit vector s, one finds (again using (C.3.7) for v) that the vector angular momentum of the pendulum mass is given by L = m r x v = ml2 [θ• φ^ - sinθ φ• θ^] . (C.5.5) Then application of N = dL/dt, with N as in (C.5.3) and unit-vector change rates as in (C.3.6), simply reproduces the last two equations of motion in (C.5.1). So we conclude that : h ≡ L z/(ml2) = sin2θ φ• ( C . 5 . 6 ) is a constant of the motion of the spherical pendulum. Comment : In Lagrangian dynamics one has L = KE-PE = (1/2)mv2+mlcosθ with v2 = l2(θ•2+ sin2θφ•2). The Euler Lagrange equations d t(∂L/∂θ•) = (∂L/∂θ) and dt(∂L/∂φ•) = (∂L/∂φ) produce the last two equations of (C.5.1). Since φ is cyclic (does not appear in L) the second equation says d t(∂L/∂φ•) = 0 which says d tLz = 0. The first equation in (C.5.1) does not appear since it is an equation of constraint. (c) E as another constant of the motion We have shown in (C.5.6) that φ• = h/sin2θ. If this is substituted into the second equation of (C.5.1) one obtains θ•• - h2cosθ /sin3θ + (g/ l)sinθ = 0 ( C . 5 . 7 ) Appendix C: Foucault Pendulum 176 and we just put this equation on hold for a mome nt, noting that it is a second-order non-linear ODE. Another constant of the motion is the total energy E which may be regarded as E = T + V = kinetic energy + potential energy (here the zero point of potential is put at the pendulum pivot point), E = (1/2)mv 2 - mg lcosθ = (1/2) m l2(θ•2+ sin2θφ•2) – mg lcosθ (C.5.8) where v2 = l2(θ•2+ sin2θ φ•2) according to (C.3.7) for v. By rescaling the energy, we can take this constant of the motion to be (note that E and E could have either sign), E = θ•2/2 + sin2θφ•2/2 – (g/ l)cosθ . E = m l2E (C.5.9) Using (C.5.6) to eliminate φ• gives, E = θ•2/2 + (h2/2sin2θ) – (g/ l)cosθ o r ( C . 5 . 1 0 ) E = θ•2/2 + Ve(θ) where V e(θ) ≡ (h2/2sin2θ) – (g/ l)cosθ . Unlike (C.5.7), equation (C.5.10) is a first-order ODE so we prefer it to (C.5.7). In fact, if one multiplies (C.5.7) by θ• and notes that θ• θ•• = (1/2) ∂t(θ•2) and does a few easy integrals, one obtains an equation of the form ∂t[stuff] = 0 so then stuff = constant. That "stuff" is the right side of (C.5.10) and we have then provided an interpretation for the constant (energy) [ θ• is an "integrating factor" for (C.5.7) ]. (d) Exact solution to the Spherical Pendulum (outline) Equation (C.5.10) is a first order non-linear ODE for θ (t) which we know how to solve: θ• = (2 E - 2Ve(θ) )1/2 => d θ = (2 E - 2Ve(θ) )1/2dt => dt = (2 E - 2Ve(θ) )-1/2 dθ boundary condition => t( θ) = ∫θ0 θ dθ' 1 2E - (h2/sin2θ') + (2g/ l)cosθ' . // t(θ 0) = 0 so θ (0) = θ0 (C.5.11) Letting z' = cos θ', so dz' = -sin θ' dθ' = - 1-z'2 dθ' one finds dθ' 2E - (h2/sin2θ') + (2g/ l)cosθ' = - dz' 1-z'2 1 2E - (h2/[1-z'2]) + (2g/ l)z' = - dz' 2E [1-z'2] - h2 – (2g/ l)z'[1-z'2] = - dz' 2E -2Ez'2 - h2 – (2g/ l)z'+ (2g/ l) z'3] Appendix C: Foucault Pendulum 177 = - 1 (2g/l) [ z'3 - (El/g) z'2 - z' + ( l/g) (E- h2/2 ) ] . (C.5.12) Therefore t(θ) = ∫z z0 dz' 1 (2g/l) [ z'3 - (El/g) z'2 - z' + ( l/g) (E- h2/2 ) ] // z = cosθ , z0 = cosθ0 = l/(2g) ∫z z0 dx 1 (x-a)(x-b)(x-c) ( C . 5 . 1 3 ) where a,b,c are the roots of the cubic equation x3 - (El/g) x2 - x + ( l/g) (E-h2/2 ) = 0. The dimensionless integral appearing on the last line can be evaluated in closed form using this indefinite integral, . (C.5.14) Here EllipticF is the incomplete elliptic integral of th e first kind as was defined in (C.4.13) and (C.4.17), EllipticF(sin( φ),k) = ∫0 sin(φ) dt 1 (1-t2)(1-k2t2) = F(φ,k) = sn-1(sinφ,k) . (C.5.15) There are standard (albeit complicated) formulas for th e roots of a cubic. Therefore, we have a closed form result for t( θ) which can then be "inverte d" to obtain a solution for θ(t). Then from (C.5.6) we get φ• = h/sin2(θ(t)) => φ(t) = φ0 + h ∫0 t dt' /sin2(θ(t')) (C.5.16) and the problem of the spherical pendulum is completely solved more or less in closed form. The string tension is then determined by the fi rst equation in (C.5.1). The solution θ(t) is a function of θ0, l, g, m and the two constants of the motion E = m l2E and Lz = m l2h. Somehow it must describe both low-amplitude pendulum motions (E < 0) as well as violent hi gh-speed over-the-top maneuvers (large E > 0). Appendix C: Foucault Pendulum 178 (e) The nature of the general solution for the Spherical Pendulum Recall (C.5.10) where E is the total scaled energy for the pendulum mass, E = θ•2/2 + Ve(θ) where V e(θ) ≡ (h2/2sin2θ) – (g/ l)cosθ . (C.5.10) The "effective potential" V e(θ) has the following shape (plotted here for h2/2 = .3 and (g/ l) = 1) ( C . 5 . 1 7 ) Since θ•2/2 = ( E - Ve(θ)) must be positive, one must have V e(θ) ≤ E which is valid only between θmin and θmax as shown. These are the "turning points" where θ• = 0. Differentiation of (C.5.10) tells us that, 0 = θ• θ•• + Ve'(θ) θ• // V e'(θ) ≡ dVe(θ)/dθ = slope of V e(θ) or θ•• = – Ve'(θ) . ( C . 5 . 1 8 ) At the right turning point the slope V e'(θ) is positive so θ•• < 0 which accelerates the particle to the left. At the left turning point the slope V e'(θ) is negative so θ•• > 0 which accelerates the particle to the right. The particle therefore bounces back and forth between th ese two turning points in some manner. Thus the motion of the pendulum is constrained on the spherical surface r = l between two horizontal circles of angles θmin and θmax and hits both these angles once per "oscillation". Meanwhile, from (C.5.16) the action in the φ dimension of the problem is controlled by φ• = h/sin2(θ(t)) => φ(t) = φ0 + h ∫0 t dt' /sin2(θ(t')) (C.5.16) (C.5.19) so as θ(t) does the oscillation just discussed between θmin and θmax, φ(t) increases as shown in its own complicated functional manner. Appendix C: Foucault Pendulum 179 A quick tour of web animations of the spheri cal pendulum shows the amazing complexity of the possible motions. If the string is replaced with a ma ssless stiff rod, over-the-top motions are included. Some examples (search youtube if these are dead links) : http://www.youtube.com/watch?v=6hCLkTENfSA . http://www.youtube.com/watch?v=VS1dU5HpfOM&feature=relmfu Both animations demonstrate the θmin ≤ θ ≤ θmax idea, though it takes a while to see in the first animation. (f) The Conical Motion solution of the Spherical Pendulum The spherical pendulum has an obvious simple solution where θ = θ0 = constant, so the string motion traces out a cone. In this case the equations of motion (C.5.1) become sin2θ0 φ•2 = -(g/ l)cosθ0 + T/(ml) r^ cosθ0 φ•2 = (g/ l) θ^ φ•• = 0 . φ^ ( C . 5 . 2 0 ) The second equation says ω φ ≡ φ• = (g/l)secθ0 . Since this is a constant, the third equation is satisfied as well, and then the first says T = mg sec θ0. In the small angle limit for θ0, ωφ slows down to its smallest possible value ωφ ≡ g/l = Ω which is the frequency of a small-a ngle plane pendulum. Conversely, as we try to achieve θ0 → π/2, secθ0→ ∞ and both ωφ and tension T become infinite, which seems pretty reasonable. This solution can of course be obtained by elementary methods as well. Reader Exercise : For the conical motion solution, θ• = 0 so E = Ve(θ) in (C.5.10). How does this fit in with Fig (C.5.17)? Is the pendulum stuck at a turning point? Are there two conical solutions for any given energy E ? Can one have a small conical motion with θ near π ( string = stick)? (g) The thin ellipse scenario for the Spherical Pendulum We now return to the general spherical pendulum equations θ• 2 + sin2θ φ•2 = -(g/ l)cosθ + T/(ml) r^ θ•• - sinθcosθ φ•2 = - (g/ l)sinθ θ^ 2cosθ θ• φ• + sinθ φ•• = 0 . φ^ (C.5.1) (C.5.21) In the plane pendulum analysis of Section C.4 we set φ• = 0 to simplify these equations. However, once the plane swing path is opened even sli ghtly into a thin elliptical path, the φ (t) function becomes just as active as the θ(t) function. For such a thin ellipse, θ bounces between the θmax and θmin turning points and θmin will be very small. Roughly speaking, for each full elliptical swing cycle of θ, φ(t) wraps 2 π Appendix C: Foucault Pendulum 180 about the origin, so the functions have simila r "frequencies". From (C.5.16) the velocity φ• = h/sin2(θ(t)) is very uneven and will have large peaks when θ is near θmin. It is useful to look at an actual simulation to see the action of the two angle variables. We start by entering the last two equations in (C.5.21) using Ω2 = g/l : (C.5.22) Then we set in some initial conditions and create the simulation (numerical solution). Here we start at θ(0) = 90o and some small amount of φ•(0) = 0.4 which results in an ellipse that is thin but not very thin : ( C . 5 . 2 3 ) Next we extract the four solution functions of interest which are θ(t),φ(t),θ•(t) and φ•(t) [ see our Maple User's Guide ] , ( C . 5 . 2 4 ) We are now ready to make plots: Appendix C: Foucault Pendulum 181 θ = red φ = black φ• = blue (C.5.25) We see in red the expected θ(t) bouncing between θmax= π/2 and θmin ≈ 0.3. The blue φ•(t) has peaks when θ is small, as noted above. The black φ just winds around to ever-increasing φ as the pendulum bob goes in its elliptical path (not an exact ellipse). We conjecture that for a thin elliptical orbit, the behavior of the pendulum for small or large θmax is very similar to what we saw for the plane pendul um in Section C.4. In particular if we set θmax very close to π, we expect to see the top of the red θ curve become flat, corresponding to the long hang time mentioned in that Section. Here we set θ max = 0.995* π : θ = red φ = black φ• = blue (C.5.26) The period of the θ motion is correspondingly increased. We have no similar conjecture to make about the φ(t) behavior of the pendulum! Appendix C: Foucault Pendulum 182 (h) The Intrinsic Airy Precession of the Spherical Pendulum Consider again the two equations of motion. θ•• - sinθcosθ φ• 2 = - Ω2 sinθ θ^ 2cosθ θ• φ• + sinθ φ•• = 0 . φ^ (C.5.21) (C.5.27) A pair of equivalent equations was noted earlier, θ•• - h 2cosθ /sin3θ + Ω2sinθ = 0 (C.5.7) φ• = h/sin2(θ(t)) (C.5.6) (C.5.28) With this second pair, one solves the first for θ(t) and uses that in the second to get φ(t). In all these equations the terms are of similar size, so nothing can be neglected in an attempt to make any approximation for a "thin orbit". For this reas on it is difficult to come up with an arm-waving explanation of the fact that the orbit processes due to the relation between the θ and φ behavior. It turns out that, when θ max is not too large, for each period of the θ motion, the azimuth φ wraps around a little less than 2 π and this causes the orbit to precess as the pendul um swings (this has nothing to do with Earth rotation which is turned off). Here we demonstrate this effect using θ(0) = 1 radian and doing mod( φ,2π) in the plot of φ: θ = red mod( φ,2π) = black φ• = blue (C.5.29) The vertical edge of the black φ curve is slipping to the left relative to the red and blue curves. In his 1851 paper Airy (see Refs.) derived an expr ession for this "intrinsic apsidal precession rate" of the spherical pendulum doing thin elliptical orbits wh ich our numerical solution above demonstrates. His formula is (now called "the Airy precession"), ω airy/ωswing = Tswing /Tairy = (3/8)(ab/ l2) = (3/8π)(πab/l2) = (3/8π)(A/l2) . (C.5.30) Appendix C: Foucault Pendulum 183 In this approximate formula, a and b are the semimajor and semiminor axes of the narrow ellipse which is slowly precessing, and l is the length of the string (A = πab is the area of the ellipse). This formula is most accurate for small oscillations and gets less pr ecise for larger ones, requiring correction terms. The orbit precesses in the same direction that the bob rotates around the ellipse (see Fig. (C.8.14) ). The derivation in Airy's paper is quite involved. An alternate derivation appears in the text of Synge and Griffith, pp 373-381. The result appears on p 381 in the form δφ = (3A/4 l2) where δφ is the amount of precession during one full swing of the pendulum. Then it takes N = δφ/2π swings to get 2 π of precession and so 1/N = 2 π/δφ = (3/8π)(A/l2) in agreement with (C.5.30) above. We shall return to this precession in the next secti on and later when we plot orbits in Section C.8. (i) The Foucault Mode of the Spherical Pendulum We now turn the Earth's rotation back on and are f aced with the full equations of motion from (C.3.9), θ•2 + sin2θ φ•2 = -(g/ l)cosθ + T/(ml) + 2ω [(cosβsinθ + sinβcosφcosθ )sinθ φ• + sinβsinφ θ• ] θ•• - sinθcosθ φ•2 = - (g/ l) sinθ - 2ωsinθ φ• (cosβcosθ - sinβcosφsinθ) 2cosθ θ• φ• + sinθ φ•• = 2ωθ• (cosβcosθ - sinβcosφsinθ) . (C.3.9) We seek a planar-like solution where ω, φ• and φ•• are all very small. It is still true that φ must wind roughly 2π for each θ swing, but most of the time (away from the low point) φ• is very small since the assumed orbit is nearly planar. The equations then reduce to the following, θ•2 = -(g/ l)cosθ + T/(ml) + 2ω [sinβ sinφ θ• ] θ•• = - (g/ l) sinθ 2cosθ θ• φ• = 2ωθ• (cosβcosθ - sinβcosφsinθ) . (C.5.31) The second equation is the standard equation for the general-amplit ude plane pendulum. This problem was exactly solved in Section C.4 for two different in itial conditions. Here is one of those solutions, θ(t) = 2 sin-1[ sin(θ0/2) sn(Ω t + K(sin( θ0/2)), sin(θ0/2)) ] θ(0) = θ0 θ•(0) = 0 . (C.4.26) Dividing the third equation of (C.5.31) by 2cos θ gives, φ• = ω (cosβ - sinβcosφtanθ) . ( C . 5 . 3 2 ) For small oscillations (in θ) we drop the second term to get φ• = ωcosβ ( C . 5 . 3 3 ) Appendix C: Foucault Pendulum 184 In our arrangement of the spherical coordinates with z ^ pointing down, this indicates a clockwise precession of the pendulum when it is viewed from above, and so is consistent w ith our initial Foucault calculation (C.5) . (j) Interference between the Airy and Foucault precession We have found that for small-angle motion of a sphe rical pendulum on the Earth's surface (magnitudes), ω F = 2ω|cosβ| // Foucault precession, ω = rotation rate of the Earth ωI ≈ Ω(3/8π)(A/l2) // Intrinsic precession, Ω = g/l = pendulum swing frequency where ω ≈ Ω = g/l is the small-angle pendulum frequency. The ratio is then ωI/ωF = Ω(3/8π)(A/l2) / (2ω|cosβ|) = (3/16 π) (g/l /ω) (A/ l2) |secβ| // dimensionless = (3/16π ) g A ω-1 l-5/2 |secβ| . (C.5.34) We would like the intrinsic precession rate to be much less than the Foucault rate so that one can then ignore the intrinsic effect. To make this ratio small, one of course wants to make the ellipse area A as small as possible, one wants to stay away from the Earth's equator where β = π/2 and sec β = ∞, and one wants a large l . This subject gets a good treatment in Schumacher and Tarbet (2009), where the Airy formula appears as equation (2). These authors propose an electronic device to neutralize the intrinsic precession to allow for a much shorter string on a Foucault pendulum. (k) The Foucault Pendulum at the Pantheon in Paris According to Google Maps the Pantheon is located at latitude 48.8468 degrees, so β = 90-48.8468= 41.1532 degrees. The length of the pendulum is purported to be 67 meters which we assume is the exact distance from the pivot point to the center of mass of the swinging weight. Wiki https://en.wikipedia.org/wiki/Gravity_of_Earth says g = 9.81 m/sec2 in Paris. According to https://en.wikipedia.org/wiki/ Sidereal _time the Earth's sidereal (relative to the stars) rotation period is 23.9344699 hours. We now have Maple do a few calculations. Firs t we compute the pendulum period (w used for ω , and the function evalf means "evaluate to a floating point number") : Appendix C: Foucault Pendulum 185 (C.5.35) The full-cycle period is therefore T swing = 16.42 seconds. Stopwatch measurements for the first full swing in this video https://www.youtube.com/watch?v=59phxpjaefA were 16.37, 16.38. 16.36 which average to 16.37, pretty close. One is never sure that video speeds are reproduced in real time. Next we compute the Foucault rotation period using ωFouc = ωEarth cosβ from (C.5.33) or (C.5) : ( C . 5 . 3 6 ) The period is then 31.787 hours or 31 hours and 47.2 minutes. The site https://en.wikipedia.org/wiki/List_of_Foucault_pendulums#France claims 31 hours and 50 minutes, but that is probably a calculated and not a measured value. Says m = 28 kg so this pendulum does not blow around much in the breeze. Next we estimate the effect of the Airy precessi on, assuming the swing is about 2 meter long and the ellipse width is 2 cm (hopefully it is smaller th an that). The ratio_Airy is taken from (C.5.30) Appendix C: Foucault Pendulum 186 ( C . 5 . 3 7 ) Thus suggests that we get one Airy precession for every 171.76 Foucault precessions. The direction of the Airy precession depends on which way the ellipse path is traversed. The error induced in the Foucault period is then about ± 11 minutes, assuming the a a nd b values shown. One would think that with a careful "burned string" launch of the pendulum, one could get "a" down to maybe 1 mm, which would reduce the 32 hour period error to ± 1 minute. ( l) General Numerical Orbits of the Spherical Pendulum In Section C.6 below we derive the spherical pe ndulum equations of motion directly in Cartesian coordinates and then in Section C.8 we plot various pendulum trajectories in Cartesian space. As long as one avoids hitting the singular points θ = 0 and θ = π, one can do this directly from the angular equations of motion as we now show. This method has an advantag e over that of Section C.8 in that negative values of z here are allowed, meaning motion of the pe ndulum bob in the upper he misphere is permitted. The spherical pendulum equations of motion are given in (C.5.1), θ• 2 + sin2θ φ•2 = -(g/ l)cosθ + T/(ml) r^ θ•• - sinθcosθ φ•2 = - (g/ l)sinθ θ^ 2cosθ θ• φ• + sinθ φ•• = 0 . φ^ (C.5.1) (C.5.38) We enter into Maple the last two equations as was show n in (C.5.22), and have ds olve find solutions as in (C.5.23). Here is an example where we give the pendulum a good kick at t = 0, θ(0) = 0.5 φ(0) = 0 θ•(0) = 1.7 φ•(0) = 1.0 (C.5.39) We then extract θ(t) and φ(t) and plot them : red θ bounces and black φ winds around, Appendix C: Foucault Pendulum 187 θ (t) = red φ(t) = black (C.5.40) Alternatively, one can plot the trajectory in ( θ,φ) space where θ bounces and φ winds up vertically, (C.5.41) We then generate Cartesian coordinates ( l = 1), (C.5.42) and plot the trajectory in these coordinates (recall that z^ points down), Appendix C: Foucault Pendulum 188 (C.5.43) x(t) = red y(t) = black z(t) = blue We can make a 3D plot of the trajectory as follows ( z^ axis points down, up is up) (C.5.44) Appendix C: Foucault Pendulum 189 Solving the first equation in (C.5.38) for T, and using m = 1 kg, l = 1 m, and g = 9.8 m/s2, we can plot the string tension for the above trajectory, (C.5.45) Since the string tension goes negative at the three high spots in the motion shown in (C.5.44), the pendulum must have a massless stick in place of a st ring in order to realize the trajectories of this example. C.6 Equations of Motion for the Foucault Pendulum (Cartesian Coordinates) We start with the vector equation (C.3.3), wh ich is Newton's Law in non-inertial Frame S, m a = mg z^ + T – 2m ω x v . (C.3.3) (C.6.1) where z^ points down. From (C.1.2) ω can be written ω = - ωcosβ e3 + ωsinβ e1 = ω[- cosβ z^ + sinβ x^] ( C . 6 . 2 ) so – 2m ω x v = -2mω [ - cosβ z^ + sinβ x^] x [vxx^ + vyy^ + vzz^ ] = -2mω [ - cosβ(vxy^ - vyx^) + sinβ(vyz^ - vzy^) ] = -2mω [ cosβvyx^ - (cosβvx+ sinβvz) y^ + sinβvyz^ ] . (C.6.3) From (C.3.5) the string tension is, Appendix C: Foucault Pendulum 190 T = - T r^ = -(T/ l)(xx^ + yy^ + zz^) . // l = x2+y2+z2 (C.6.4) The three equations of motion from (C.6.1) are then ma x = -(T/ l)x - 2mωcosβvy may = -(T/ l)y + 2mω (cosβvx+ sinβvz) maz = mg - (T/ l)z - 2mωsinβvy ( C . 6 . 5 ) or x•• = -(T/m l)x - 2ωcosβy• y•• = -(T/m l)y + 2ω (cosβx•+ sinβz•) z•• = g -(T/m l)z - 2ωsinβy• . ( C . 6 . 6 ) Eliminate T from the first two equations : yx•• = -(T/m l)xy - 2ωycosβy• xy•• = -(T/m l)xy + 2ωx(cosβx•+ sinβz•) yx•• - xy•• = - 2ω ycosβy• - 2ωx(cosβx•+ sinβz•) . (C.6.7) Do this again for the first and third equations of (C.6.6) : zx•• = -(T/m l)xz - 2ωzcosβy• xz•• = xg -(T/m l)xz - 2ωxsinβy• zx•• - x z•• = -xg - 2 ωzcosβy• + 2ωxsinβy• . (C.6.8) We then have these three equations of interest 1 yx•• - xy•• = - 2ω ycosβy• - 2ωx (cosβx•+ sinβz•) 2 zx•• - x z•• = - xg - 2 ωy•(zcosβ - xsinβ ) 3 x 2+y2+z2 = l2 ( C . 6 . 9 ) where l is the length of the pendulum string. These are the equations of motion for the Foucault pendulum in Cartesian coordinates. We can in theory solve for x(t), y(t) and z(t) given some initial conditions. Once the three equations are solved, we ca n find the tension T(t) from (say) the first equation of (C.6.6) : x•• = -(T/m l)x - 2ωcosβy• ⇒ (T/m l)x = - 2ωcosβy• - x•• ⇒ T = - m l(2ωcosβy• + x••) / x . ( C . 6 . 1 0 ) Appendix C: Foucault Pendulum 191 Small oscillation limit For very small pendulum swings one has T ≈ mg ⇒ T/m l ≈ (g/l) x• and y• are small z• ≈ 0 ( C . 6 . 1 1 ) In zeroth order the first two equations of (C.6.6) then say (r = l = length of string) x•• + (g/ l)x ≈ 0 y•• + (g/ l)y ≈ 0 ⇒ Ω = g/l ( C . 6 . 1 2 ) and the pendulum swings as x.y ~ sin( Ωt) with Ω = g/l . To first order we add back the small velocity terms in (C.6.6) to get x•• + Ω2x + 2ωcosβy• ≈ 0 y•• + Ω2y - 2ωcosβx• ≈ 0 . // bottom view (C.6.13) In our system z^ points down, but if it were to point up we could take x →x and y -y as coordinates one would use when viewing the pendulum from above. The equations are then x•• + Ω2x - 2ωcosβy• ≈ 0 y•• + Ω2y + 2ωcosβx• ≈ 0 // top view (C.6.14) These equations appear in (C.1). C.7 Verification of the Cartesian equations of motion and string tension The algebra above is quite complex so we want to be sure that our Cartesian equations of motion are correct. First, here are the last two angul ar equations of motion from (C.3.9) with l set to r, eq1 θ•• - sinθcosθ φ•2 = - (g/r) sin θ - 2ωsinθ φ• (cosβcosθ - sinβcosφsinθ) // r = l eq2 2cos θ θ• φ• + sinθ φ•• = 2ωθ• (cosβcosθ - sinβcosφsinθ) . (C.2.9) (C.7.1) Meanwhile, here are the Cartesian equations of motion from (C.6.9) (in reverse order), eq3 zx•• - x z•• = - xg - 2 ωy•(zcosβ - xsinβ ) eq2 yx•• - xy•• = -2ω ycosβy• - 2ωx(cosβx•+ sinβz•) . (C.6.9) (C.7.2) Appendix C: Foucault Pendulum 192 Below we shall show that : Task (a): eq2 of (C.7.2) ⇒ eq2 of (C.7.1) Task (b): [eq3 + (cos θsinφ)*eq2] of (C.7.2) ⇒ eq1 of (C.7.1) That is to say, angular eq1 of (C.7.1) is a certain linear combination of eq3 and eq2 of (C.7.2). If we can show Task (a) and Task (b) above, then we have shown that (C.7.2) ⇔ (C.7.1), and this then serves as verification of (C.7.2). Maple must replace x,y,z and derivatives with r, θ,φ and derivatives. For coordi nates and first derivatives, (C.7.3) The second derivatives are messier, but Maple is happy to do the calculations, (C.7.4) Appendix C: Foucault Pendulum 193 Task (a): Show that eq2 of (C.7.2) ⇒ eq2 of (C.7.1) We enter eq2 of (C.7.2) and do some manipulations, suppressing the output except for the last step : (C.7.5) We manually transcribe the resulting equation, 2cosθθ•φ• + sinθ φ•• -2ωcosβcosθ θ• + 2ωsinθcosφsinβθ• = 0 or 2cosθθ•φ• + sinθ φ•• = 2ωcosβcosθ θ• - 2ωsinθcosφsinβθ• or 2cosθθ•φ• + sinθ φ•• = 2ωθ•(cosβcosθ - sinθcosφsinβ) . (C.7.6) This last equation is a match for eq2 of (C .7.1) so we have accomplished Task (a). Task (b): eq3 + (cos θsinφ)eq2 of (C.7.2) ⇒ eq1 of (C.7.1) The code continues from that shown above. Equation eq2 is already entere d, so we now enter eq3, form the linear combination for eq1, then process the results with a series of typical tortuous Maple steps, (C.7.7) We again manually transcribe the resulting equation, -cosθsinθ φ• 2 + (2ωsinθcosθcosβ - 2ωsin2θcosφsinβ)φ• + sinθ(g/r) + θ•• = 0 or Appendix C: Foucault Pendulum 194 θ•• - cosθsinθ φ•2 = – sinθ(g/r) - 2ωsinθ(cosθcosβ - sinθcosφsinβ)φ• . (C.7.8) This is a match for eq1 of (C.7.1) so we have accomplished Task (b). Tension equation verification Using the angular equations of motion (C.7.1), we now show that the followi ng two tension expressions are equivalent (the second is Ca rtesian (C.6.10) while the first is angular (C.3.9)), where r = l , T/mr = θ•2 + sin2θ φ•2 + (g/r)cos θ - 2ω [(cosβsinθ + sinβcosφcosθ )sinθ φ• + sinβsinφ θ• ] T/mr = - (2 ωcosβy• + x••)/x . (C.7.9) Our task of showing that the above equations have equal right sides is the same as showing that x{θ•2 + sin2θ φ•2 + (g/r)cos θ - 2ω [(cosβsinθ+sinβcosφ cosθ )sinθ φ• + sinβsinφ θ• ]} = -(2ωcosβy• + x••) or L H S = R H S . ( C . 7 . 1 0 ) We first get the complicated left hand side LHS entered: (C.7.11) We then compute RHS = -(2 ωcosβy• + x••), Appendix C: Foucault Pendulum 195 (C.7.12) Notice that RHS contains second derivatives θ•• and φ••. We shall eliminate these derivatives by manually solving the angular equations of motion (C.7.1) for Tdd = θ•• and Pdd = φ•• : ( C . 7 . 1 3 ) To show that LHS = RHS, we define d = LHS-RHS and show that d = 0: (C.7.14) Thus d = 0 and LHS = RHS and the two expressions for T in (C.7.9) are equivalent. C.8 Numerical solutions of the equations of motion (Cartesian Coordinates) Our task is to solve the set of equations (C.6.9) (eq1 now has a new meaning) : eq1 x 2+y2+z2 = r2 / / r = l eq2 yx•• - xy•• = -2ω ycosβy• - 2ωx(cosβx• + sinβz•) eq3 zx•• - x z•• = -xg -2 ωy•(zcosβ - xsinβ ) . (C.6.9) (C.8.1) We enter eq2 and eq3 writing de rivatives for example as x•• = xdd : Appendix C: Foucault Pendulum 196 ( C . 8 . 2 ) At this point zd = z• and zdd = z•• are unspecified. We use eq1 of (C.8.1) to compute z• and z•• in terms of x and y, ( C . 8 . 3 ) When these expressions are installed, eq2 and eq3 becomes these formidable-looking equations which contain two unknown functions x(t) and y(t) and constants r = l, β,g and ω = w : ( C . 8 . 4 ) Appendix C: Foucault Pendulum 197 The Foucault Pendulum at the Pantheon in Paris (revisited) For our first plot, we again consider the Foucau lt pendulum set up in the Pantheon. The numbers are discussed above (C.5.35) : (C.8.5) The Maple code to invoke a solution is as follows: ( C . 8 . 6 ) We have initialized the pendulum at x(0) = 1m. Here is a plot of the first few swings, (C.8.7) Notice that the vertical scale is highly magnified, so really these swings are very close to the x axis. We want to view the orbits from above, not from below. Since the z axis points down, and since we are viewing from above, we negate the original y axis to get a new y axis appropriate for our plots viewed from the top. This negation is done in the odeplot call seen above. Appendix C: Foucault Pendulum 198 As expected, and as shown earlier in (C.2.1), the Cori olis force deflects each half-swing to the right (as seen from above). In order to plot functions like x(t) more generally , we extract our functions of interest from the Maple dsolve environment as follows, ( C . 8 . 8 ) Here functions like X(t) are taken directly fro m the dsolve listprocedure output structure while unavailable derivatives such as Xdd = x•• are manually approximated. For details on how this works and other information on dsolve (including a debugger's gui de), see the author's Maple User Guide. We can now make a plot of the three functions x(t),x•(t) and x••(t) for the first swing, x = red x• = black x•• = blue (C.8.9) The shape of red x(t) appears to be sinusoidal as predicted by the small angle model. The black curve vx(t) = x•(t) seems to cross the x axis around 16.4 suggesting that this is the period of the first swing of the pendulum, in agreement with (C.5.35). We can zo om in on the crossing to get a better view, Appendix C: Foucault Pendulum 199 ( C . 8 . 1 0 ) This indicates that the period is about 16.4211 sec. Foucault Pendulum on a rotating platform We now stop the Earth's rotation, transport the entire Pantheon to the North Pole ( β = 0o) and mount it on a sturdy platform which rotates once every 77 seconds. We do this just to explore the pendulum orbit trajectories that might result. With equally-scaled axes we get this Foucault path for a duration of 90.3 seconds, (C.8.11) Next, when the pendulum is released at x = 1m, it is given a small v y velocity of -0.1 m/s, which means that vytop = +0.1 m/s : Appendix C: Foucault Pendulum 200 (C.8.12) The sharp cusps of the previous orbit are now smoothed out. Conversely, if we apply v y = +0.1 m/s so that vytop = -0.1 m/s : (C.8.13) Appendix C: Foucault Pendulum 201 Desktop Spherical Pendulum Here we set l = 1 m and turn off the Earth's rotation completely with ω = 0, so our Foucault pendulum becomes a spherical pendulum. We start it moving with these initial conditions : (C.8.14) In 30 seconds the orbit precesses as shown, a phenomenon known as the intrinsic apsidal Airy precession discussed near (C.5.30). The orbit precesses in the same directional sense in which it orbits. This effect has nothing to do with the Earth's rotation, and turning ω back on makes no visible change in the above picture. Reader Exercise: (1) Construct the above pendulum with 1 meter of thread and a small weight hung from a ceiling lamp fixture. With x ~ 1/2m start an elliptical motion of the shape shown above. Count the number of swings it takes for the orbit to precess 90 degrees and compare to the above figure. If we change the initial position to x(0) = 0.9 to ge t more swing, here is the orbit out to t = 10: Appendix C: Foucault Pendulum 202 (C.8.15) It appears that the higher starting point has increased the precession rate. Here are some 3D plots of the above solution trajectory : ( C . 8 . 1 6 ) In the left image, the camera is looking down from the pivot point of the pendulum so the previous 2D picture is roughly replicated. Then the camera moves down to lower viewing angles. We already discussed in (C.5.30) Airy's 1851 form ula for the precession of the elliptical orbit, ω airy/ωswing = Tswing /Tairy = (3/8)(ab/ l2) = (3/8π)(πab/l2) = (3/8π)(A/l2) . (C.5.30) Appendix C: Foucault Pendulum 203 As noted there, in this approximate formula, a and b are the semimajor and semiminor axes of the narrow ellipse which is slowly precessing, and l is the length of the string (A = πab is the area of the ellipse). This formula is most accurate for small oscillations and gets less precise for larger ones, requiring correction terms. We can apply it to our two examples above. A tick mark in (C.8.14) is .04m, so st aring the picture we can estimate, since l = 1 m, a = 0.50 m b = 2.25 ticks = 2.25 *.04 = 0.09 m 1/N = (3/8)*(0.5)*(.09) = .016875000 // Maple N = 59.25925926 N/4 = 14.81481482 (C.8.17) The Airy prediction for Fig (C.8.14) is that there are 14.8 ellipses per quarter turn of the precession. A count of ellipses in the figure gives about 13.5, fair ly close given that these are not really small narrow ellipse oscillations. For (C.8.15) the numbers are a = 0.90 m b = 2 ticks = 2 *.04 = 0.08 m 1/N = (3/8)*(0.9)*(.08) = .027000000 N = 37.03703704 N/4 = 9.259259260 (C.8.18) The formula predicts 9.26 ellipses per quarter turn, while Fig (C.8.15) shows about 6.5. We expect a worse result since this is definitely not a small osc illation. But both results are in the ballpark, and Airy predicts that the second figure has fewer ellipses per quarter turn than the first figure. Appendix D: Center of Gravity 204 Appendix D: Center of gravity and torque for a tethered satellite D.1 Definition of Center of Gravity The phrase "center of gravity" is often used as a synonym for "center of mass" which complicates searching for information about the former concept. The "center of mass" of a system of particles is well known to be rcms = Σimiri Σimi = 1 M Σimiri M = Σimi // discrete rcms = ∫dVρ r ∫dVρ = 1 M ∫dV ρ r M =∫dV ρ // continuous . (D.1.1) We use "cms" to mean center of mass even though "com" might be more reasonable. Notice that rcms is measured with respect to th e same origin used for the ri or r. For a rigid object, the center of mass is a definite point that does not move around relative to th e object. It is completely determined by the spatial mass distribution of the object. The center of gravity is a completely different anim al, though it happens to align with the center of mass for a uniform gravitational field. For that reason one never deals with a distinct center of gravity concept for human-scale engineering objects on the Earth's surface. Consider a system of masses m i each of which experiences some force Fi. We define F = ΣiFi N(R) = Σi (ri-R) x Fi N(0) = Σiri x Fi . (D.1.2) Here F is the sum of the forces acting on all the masses m i, N(R) is the total torque on the system with respect to some arbitrary point R, and N(0) is the total torque with resp ect to the selected origin. An obvious theorem is that N(R) = Σi (ri-R) x Fi = Σiri x Fi - R x ΣiFi = N(0) - R x F ( D . 1 . 3 ) which shows how the two torques are related. If F = 0, which is often the case, then the torque is the same with respect to any point. One characteristic of a center of gravity point rcog is that the total torque on a system measured with respect to point rcog vanishes, so N(rcog) = 0. From (D.1.3) we see that this is the same as saying N(0) = rcog x F , so N(rcog) = 0 ⇔ N(0) = rcog x F rcog = "center of gravity" . (D.1.4) Appendix D: Center of Gravity 205 The significance of rcog is that it is a point which allows the relation between the total system torque N(0) and the total system force F to have the same form as Ni = ri x Fi for a single point particle. It is not obvious that such a point rcog exists for some arbitrary system of particles. So far this notion of "center of gravity" has nothing to do specifically with gravity, but below we shall add an additional part of the definition which does bring in gravity. Unlike the center of mass, the center of gravity (i f it exists) may not be unique and it generally moves around in an object as the object changes orientat ion in an external force field If we dot N(0) = rcog x F with the vector F we find N(0)• F = 0 . ( D . 1 . 5 ) As shown above, N(0) and F are well-defined computable quantities and if N(0)• F ≠ 0, then rcog cannot possibly exist (because if it did exist one must have N(0)• F = 0). So (D.1.5) is a condition for the existence of rcog. Here is a method for locating rcog, a variation of Symon p 258 which we present in the context of an asteroid of mass M near the Earth. This description includes the second characteristic of the center of gravity point which is that it is point at whic h gravitational action is effectively focused. One first computes the total gravitational force F = ΣiFi due to the Earth (summed over all points in the asteroid) and one then knows the direction of the sum vector F. One creates a line along F and then translates that line parallel to F until the line passes through the center of the Earth. The center of gravity of the asteroid lies on that translated line a distance r cog from the center of the Earth where r cog is determined by, GMEM/rcog2 = F ⇒ r cog = GMEM/F rcog = rcog (-F^) F = F F^ . ( D . 1 . 6 ) Here is an illustration, (D.1.7) Appendix D: Center of Gravity 206 From the Earth's point of view, one could replace th e entire asteroid with a point mass M at location rcog and the Earth would feel the same gravitational pull from that point mass as it does from the asteroid. Furthermore, if the asteroid were in a circular or bit around the Earth keeping its same aspect facing the Earth (not very likely), then the orbiting characteris tics of the asteroid would be the same as for its replacement point mass, and one would have for example GM EM/rcog2 = Mω2rcog as the balance between gravitational and centrifugal force. If the astero id does not rotate or tumbles in some manner, at any point in its orbit rcog will lie on the orbit shown, but the location of that point within the asteroid changes so that the new r cog computed for a new position and orienta tion still equals the orbit radius. In this case the rcog position will change relative to the asteroid, whereas the cms point always has the same position relative to the asteroid. Here we enhance th e above drawing by adding another position of the tumbling asteroid in its orbit, (D.1.8) One implication is that, for any orientation of the aste roid in orbit, there exists a position of the asteroid such that the total force F will have the same magnitude as at any other position, since r cog is the same. One might wonder what Symon's operational prescription for computing rcog has to do with our opening section about torque. First, for each point in the asteroid we know that ri x Fi = 0 ( ri tails are at Earth center) because each ri to mass m i is parallel to the force Fi on mass m i. Thus according to (D.1.2) we have N(0) = 0. Therefore the condition (D.1.5) that N(0)• F = 0 is trivially satisfied. Then rcog exists and is a vector which satisfies the center of gravity definition (D.1.4) that N(0) = rcog x F. This equation is satisfied by Symon's computed rcog because rcog is parallel to F, so the equation says 0 = 0. From (D.1.4) we then see that the N(Rcog) = 0 so rcog is a point with respect to which the total torque on the asteroid (due to gravitational force from the Earth) is 0. In this example it happens that any point R along the rcog line satisfies N(0) = R x F and any such point R therefore is a point of zero total torque N(R) = 0. But only one point on this line gives the concentration point of the gravitational force such that GMEM/rcog2 = F . Appendix D: Center of Gravity 207 We shall now consider a 2-mass tethered sate llite as a relatively simple example where we can compute the cog point and then watch it move relative to the cms point. D.2 Center of Gravity for a 2-mass Dumbbell Satellite To head off a proliferation of primes, we now switch to "swap notation" in which Frame S is the rotating frame while Frame S' is the fixed inertial frame. Rather than assume an arbitrary dumbbell orientation, we assume for this section that the dumbbell lies in the plane of paper: (D.2.1) Frame S' has its origin at the center of the Earth and is assumed fixed relative to the stars. Frame S has its origin at the center of mass of the satellite as shown. The orbit of the satellite lies in the plane of paper. The vector ω describes the angular velocity of the satellite and has nothing to do with Earth's rotation. Moreover, at t = 0 the axes of both frames line up: x^ = x^', y^ = y^' and z^ = z^' . The two masses m 1 and m2 are connected by a massless rigid stick of length s = r 1+r2 . Since the masses are assumed to be unequal, mass m 1 is restricted to lie on a sphere of radius r 1 in Frame S', while mass m 2 lies on a different sphere of radius r 2 (to be computed below). For purposes of the drawing, we have m 2 > m1 so that r 2 < r1. Appendix D: Center of Gravity 208 The polar angle in Frame S of m 1 (and the stick) is θ. The angles α 1 and α2 are positive. Assumption : We will show below that the center of gravit y is very close to the center of mass for any angle θ. For example, if the stick is 10 m long, the distance between these two centers for a low-Earth orbit is about 3 microns. For this reason, we shall assume that it is the center of mass point rcms which goes in a circular orbit around the Earth, even though we know from Section D.1 that it is really rcog that does this. So as the satellite goes around the Earth and θ varies in time, the point rcog moves a very small amount relative to rcms, and we shall just ignore this tiny motion. The satellite center of mass point then rotates around the Earth at some rate ω = 2π/T. For a low-Earth orbit, T is on the order of 88 minutes. As the sate llite moves in this orbit, Frame S rotates with the satellite while Frame S' stays fixed relative to the stars. This means that the Frame S axis z' always points away from the center of the Earth. One can generalize the discussion for an elliptical orbit, but things are complicated enough for a circular orbit so we stick with that simplification. In Appendix F we allow the dumbbell to move out of the plane of paper, but here for simplicity we assume it is in the plane of paper for any θ. With the above assumption, we may identify rcms with b, our usual vector connecting the two Frame origins, so b = rcms ( D . 2 . 2 ) and then from the figure, b + r1 = r'1 b + r2 = r'2 . ( D . 2 . 3 ) The Law of Cosines gives, for the right and left triangles, r' 12 = r12 + b2 - 2r1b cos(π -θ) = r12 + b2 + 2r1b cosθ r' 22 = r22 + b2 - 2r2b cosθ . ( D . 2 . 4 ) Center of Mass From (D.1.1) we know that the center of mass of the satellite is given by, rcms = m1r1+m2r2 m1+m2 = 0 Frame S (D.2.5) r'cms = m1r'1+m2r'2 m1+m2 = b . Frame S' (D.2.6) Due to the constraint of the massless stick one has, Appendix D: Center of Gravity 209 r^2 = - r^1 ( D . 2 . 7 ) so (D.2.5) says m 1r1 = - m2r2 ⇒ m1r1 = m2r2 , r2/r1 = m1/m2 , r^2 = – r^1 . (D.2.8) Angles Looking at the right triangles from the center of the Earth to the two dashes lines, one sees that sinα 1 = (r1sinθ) / r'1 ⇒ r' 1sinα1 = r1sinθ sinα2 = (r2sinθ) / r'2 ⇒ r' 2sinα2 = r2sinθ . (D.2.9) The same triangles reveal that cosα1 = (b + r 1cosθ)/r'1 ⇒ r' 1cosα1 = b + r1cosθ cosα2 = (b - r 2cosθ)/r'2 ⇒ r' 2cosα2 = b - r2cosθ . (D.2.10) We shall need the following dot products, r^'1 • z^ = cosα1 r^'1 • y^ = cos(π/2-α1) = sinα1 r^'2 • z^ = cosα2 r^'2 • y^ = cos(π/2+α1) = -sinα2 ( D . 2 . 1 1 ) r^'1 • r^'2 = cos(α1+α2) = cosα1 cosα2 - sinα1 sinα2 = (r' 1r'2)-1[ (r'1cosα1) (r'2cosα2) - (r'1sinα1) (r'2sinα2) ] = (r' 1r'2)-1[ ( b + r1cosθ) ( b - r2cosθ) - (r1sinθ) (r2sinθ) ] // (D.2.9) and (D.2.10) = (r' 1r'2)-1[b2 +b(r1-r2)cosθ - r1r2cos2θ - r1r2sin2θ ] = (r' 1r'2)-1[b2 +b(r1-r2)cosθ - r1r2 ] . (D.2.12) Where is Center of Gravity of the Satellite? The gravitational forces acting on the two satellite masses due to the Earth are. F1 = - (GM Em1/r'12) r^'1 = - (GM Em1/r'13) r'1 F2 = - (GM Em2/r'22) r^'2 = - (GM Em2/r'23) r'2 . ( D . 2 . 1 3 ) Note the direction of the "long" vectors r'1 and r'2 in the drawing. Appendix D: Center of Gravity 210 Following the prescription of Section D.1 for finding th e location of the center of gravity, we compute the total gravitational force on the satellite, F = - (GM Em1/r'12) r^'1 - (GMEm2/r'22) r^'2 = (-GM E) [ (m1/r'12) r^'1 + (m2/r'22) r^'2 ] . (D.2.14) Taking components: F y = F • y^ = (-GM E) [ (m1/r'12) r^'1• y^ + (m2/r'22) r^'2• y^ ] = (-GM E) [ (m1/r'12)sinα1 – (m2/r'22) sinα2 ] / / ( D . 2 . 1 1 ) = (-GM E) [ (m1/r'13) r'1sinα1 – (m2/r'23) r'2 sinα2 ] = (-GM E) [ (m1/r'13) r1sinθ – (m2/r'23) r2sinθ ] // (D.2.9) = (-GM E) [ (m1/r'13) r1sinθ – (m1/r'23) r1sinθ ] // (D.2.8) = (-GM E) (m1r1sinθ) [ (1/r'13) – (1/r'23) ] Fz = F • z^ = (-GM E) [ (m1/r'12) r^'1• z^ + (m2/r'22) r^'2• z^ ] = (-GM E) [ (m1/r'12)cosα1 + (m2/r'22)cosα2 ] / / ( D . 2 . 1 1 ) = (-GM E) [ (m1/r'13)r'1cosα1 + (m2/r'23)r'2cosα2 ] = (-GM E) [ (m1/r'13)( b + r1cosθ) + (m2/r'23)(b - r2cosθ) ] // (D.2.10) = (-GM E) [ b {(m 1/r'13) + (m2/r'23)} + cosθ { (m1r1/r'13) - (m2r2/r'23) } ] = (-GM E) [ b {(m 1/r'13) + (m2/r'23)} + cosθ { (m1r1/r'13) - (m1r1/r'23) } ] // (D.2.8) = (-GM E) [ b {(m 1/r'13) + (m2/r'23)} + m1r1cosθ { (1/r'13) - (1/r'23) } ] so Fy = (-GM E) (m1r1sinθ) [ (1/r'13) – (1/r' 23) ] F z = (-GM E) [ b {(m 1/r'13) + (m2/r'23)} + m1r1cosθ { (1/r'13) - (1/r'23) } ] . (D.2.15) Recall now from (D.1.6) the rule for obtaining rcog : Appendix D: Center of Gravity 211 GMEM/rcog2 = F ⇒ r cog = GMEM/F rcog = rcog (-F^) F = F F^ . (D.1.6) (D.2.16) Just to have a picture, we now add rcog to Fig (D.2.1), picking a graphic location for the point which makes things easy to draw (we don't know yet where it actually lies), (D.2.17) The center of gravity point lies a distance r cog (to be computed) along a line which makes some angle δ relative to the vertical axis. Since r^cog = - F^, we may write using (D.2.15), tan(δ) = - Fy/Fz = - (m 1r1sinθ) (1/r'13) – (1/r'23) b [ (m1/r'13) + (m2/r'23) ] + m1r1 cosθ [ (1/r'13) - (1/r'23)] = - (m 1r1sinθ) r'23 – r'13 b [ (m1r'23) + (m2r'13) ] + m1r1 cosθ [ r'23 - r'13] . (D.2.18) If the masses are vertically aligned, meaning θ = 0 or π, then sinθ = 0 and we see that δ = 0, as expected. Since for general masses and angle θ one does not have r'1 = r'2, one may conclude: Fact 1: In general the center of gravity does not lie on the line between the center of the Earth and the c e n t e r o f m a s s . ( D . 2 . 1 9 ) Appendix D: Center of Gravity 212 Earlier we conjectured that this was the case, a nd here we see it in our dumbbell satellite example. Next we wish to compute the distance r cog to the center of gravity. Using (D.2.14) and (D.2.12), F2 = (GME)2 [ (m1/r'12)2 + (m2/r'22)2 + (m1/r'12) (m2/r'22) r^'1 • r^'2 ] // (D.2.14) = (GM E)2 [(m1/r'12)2 + (m2/r'22)2 + (m1/r'12)(m2/r'22)(r'1r'2)-1(b2 +b(r1-r2)cosθ - r1r2) ] // (D.2.12) = (GM E)2 [(m1/r'12)2 + (m2/r'22)2 + (m1/r'13)(m2/r'23)(b2 +b(r1-r2)cosθ - r1r2) ] so F = GM E (m1/r'12)2 + (m2/r'22)2 + 2 (m1/r'13) (m2/r'23)(b2 + b(r1-r2)cosθ - r1r2) . (D.2.20) Then, using M = (m 1+m2) and mi = μiM, we find from (D.2.16) that, rcog = GMEM/F = 1 [ (μ1/r'12)2 + (μ2/r'22)2 + 2 (μ1/r'13) (μ2/r'23)(b2 + b(r1-r2)cosθ - r1r2) ]1/4 (D.2.21) r ' 12 = r12 + b2 + 2r1b cosθ r ' 22 = r22 + b2 - 2r2b cosθ // μi ≡ mi/M (D.2.4) which is somewhat more complicated than our expression for r cms, r cms = b . (D.2.2) We are then led to: Fact 2: In general the center of gravity does not lie th e same distance from the center of the Earth as the c e n t e r o f m a s s . ( D . 2 . 2 2 ) Quick check on (D.2.21) : If m 1= 0, then μ1 = 0 and μ2 = 1 and the result is r cog = r'2 which is correct. Reader Exercise : If the two masses are vertically aligned with θ = 0, then (a) Show that: r 2 = b - r'2 r'1+r'2 = 2b r 1 = r'1 - b (b2 +b(r1-r2)cosθ - r1r2) = r'1r'2 ( D . 2 . 2 3 ) (b) Show that r cog = 1 (μ1/r'12) + (μ2/r'22) = m1+m2 (m1/r'12) + (m2/r'22) (D.2.24) (c) Show that r cog < rcms , where r cms = b = (r'1+r'2)/2 and μ1+μ2= 1 Appendix D: Center of Gravity 213 D.3 Dumbbell Satellite Center of Gravity with the Far Approximation: Numerical Examples A practical tethered satellite is not going to have Earth-scale dimens ions so we now make the obvious assumption that r' 1, r'2 and b are much larger than r 1 and r2. We then define smallness parameters, ε1 ≡ (r1/b) (D.2.8) ε2 ≡ (r2/b) = (r 2/r1) (r1/b) = (m 1/m2)ε1 = (μ1/μ2)ε1 . (D.3.1) Then (b 2 + b(r1-r2)cosθ - r1r2) = b2 (1 + (ε1-ε2)cosθ - ε1ε2) (D.3.2) so that from (D.2.21), rcog = 1 [ (μ1/r'12)2 + (μ2/r'22)2 + 2 (μ1/r'13) (μ2/r'23)b2(1 + (ε1-ε2)cosθ - ε1ε2) ]1/4 . (D.3.3) From (D.2.4) we know also that r' 12 = r12 + b2 + 2r1bcosθ = b2 [ 1 + 2(r 1/b)cosθ + (r1/b)2] = b2 [ 1 + 2ε1cosθ + ε12] r' 22 = r22 + b2 - 2r2b cosθ = b2 [ 1 – 2(r 2/b)cosθ + (r2/b)2] = b2 [ 1 – 2ε2cosθ + ε22] so then r' 1 = b 1 + 2ε1cosθ + ε12 r'2 = b 1 – 2ε2cosθ + ε22 . ( D . 3 . 4 ) So far everything is exact, but we shall now expand r cog as a series in our small parameters. Since ε 2 = (μ1/μ2)ε1, we replace ε 2 by this expression so there is then only one small parameter ε1. Maple is ready to carry out this task. We first enter our expressions of interest, using rp1 for r' 1 etc : Appendix D: Center of Gravity 214 (D.3.5) We then instruct Maple to expand rcog in a power series around ε1 = 0 and we ask for the first four terms of the expansion, (D.3.6) A perhaps unexpected result is that there is no term linear in ε1, regardless of the masses. We leave it to the energetic reader to concoct a theoretical explanation of this fact (not all odd terms vanish, just the first odd term). Our conclusion then is, to second order in smallness parameter ε1, rcog ≈ b [ 1 + 3 4 μ1 1-μ1 (-3cos2θ +1) ε12 ] = b [ 1 + 34 m1 m2 (-3cos2θ +1) ε12 ] = b [ 1 + 3 4 m1 m2 ( 1 - 3 cos2θ ) r12 b2 ] = b [ 1 + 34 r2 r1 ( 1 - 3 cos2θ ) r12 b2 ] // (D.2.8) = b [ 1 + 3 4 ( 1 - 3 cos2θ ) r1r2 b2 ] . (D.3.7) Then r cog - rcms = rcog - b ≈ [ 3 4 ( 1 - 3 cos2θ ) r1r2 b2 ] b (D.3.8) and Appendix D: Center of Gravity 215 rcog - rcms rcms ≈ 3 4 ( 1 - 3cos2θ ) r1r2 b2 . (D.3.9) Fact 3: If θ = ±54.7o, we get r cog = rcms through order ε 12, since this angle has cos θ = 1/ 3 . But the line to the center of gravity does not in general have δ = 0 since in general r' 1 ≠ r'2 at this angle. (D.3.10) Fact 4 : If the masses are vertically aligned, meaning θ = 0 or π , then cos2θ = 1 and we find that rcog - rcms rcms = 3 4 ( 1 - 3 ) r1r2 b2 = - 3 2 r1r2 b2 . ( D . 3 . 1 1 ) Since the right side is negative, we have r cog < rcms and the center of gravity in this case is closer to the Earth than the center of mass and of course lies on the line to the center of mass. Fact 5 : If the masses are horizontally aligned, meaning θ = ±π/2, then cos θ = 0 and we find that rcog - rcms rcms = 3 4 ( 1 - 0 ) r1r2 b2 = + 3 4 r1r2 b2 . (D.3.12) The center of gravity in this case is farther from the Earth than the center of mass. The line to the center of gravity does not have δ = 0 unless m 1 = m2. The above cases show the extremes of the factor ( 1 - 3 cos2θ ) and hence of r cog - rcms . For a general angle θ the result lies between the two limiting cases. Numerical Examples From (8.8.6) one has R E = 6371 km. If we put our satellite in orbit 200 km above the Earth's surface, then b = 6371+200 = 6571 km. A higher orbit of course give s a larger b and a smaller offset between r cms and rcog. We take the maximum displacement between r cog and rcms from Fact 5 to obtain | rcog - rcms rcms |max = 3 4 r1r2 b2 . ( D . 3 . 1 3 ) For a given mass separation (tether length) s = r 1 + r2 the product r 1r2 is maximized when r 1= r2 (easy to show) and this in turn means m 1 = m2, so to get the worst case we set r 1 = r2 = s/2 to get | rcog - rcms rcms |max = 3 4 14 (s/b)2 = (3/16)(s/b)2 s = stick length = tether length (D.3.14) Note the quadratic dependence on the ratio s/b. Regardi ng the left side of this equation as dr/b, we have Maple compute dr for three cases: Appendix D: Center of Gravity 216 ( D . 3 . 1 5 ) For a s = 10 m tether length, the max offset is 2.8 mi crons which we feel pretty comfortable ignoring. For a s = 1 km tether length, the max offset is 2.8 cm, still pretty small. For a s = 50 km tether length, the max offset is about 70 m. (D.3.16) The quantities | r cog - rcms rcms |max for each case are shown as dr/b : 10-12, 10-8 and 10-5. This justifies our approximation that it is essentially the center of mass point which orbits the Earth, and the variation between rcog and rcms is generally relatively small. Howeve r, in the last case if one really deployed a 50 km tether from a small spacecraft, the center of gravity point would lie 70 meters out onto the tether. (The YES2 satellite in 2007 depl oyed a 30 km tether, see Chen et al.). D.4 Dumbbell Satellite Center of Gravity for equal masses and no approximation (a) Equal Masses and general θ Recall the general no-approximation re sults (D.2.18) and (D.2.21), tan(δ) = - (m 1r1sinθ) (1/r'13) – (1/r'23) b [ (m1/r'13) + (m2/r'23) ] + m1r1 cosθ [ (1/r'13) - (1/r'23)] (D.2.18) rcog = 1 [ (μ1/r'12)2 + (μ2/r'22)2 + 2 (μ1/r'13) (μ2/r'23)(b2 + b(r1-r2)cosθ - r1r2) ]1/4 (D.2.21) where Appendix D: Center of Gravity 217 r'12 = r12 + b2 + 2r1b cosθ r'22 = r22 + b2 - 2r2b cosθ . (D.2.4) (D.4.1) Setting m 1 = m2 (which implies r 1 = r2 = s/2 and also μ1 = μ2 = 1/2) gives these simpler forms, tan(δ) = - (r1sinθ) r'23 – r'13 b (r'23 + r'13) + r1 cosθ (r'23 - r'13) rcog = 2 r'1r'2 [ r'24 + r'14 + 2r'1r'2(b2 - r12) ]1/4 . (D.4.2) (b) Equal Masses and θ = 0 (vertically aligned) If we further specify that θ = 0 we get δ = 0 and moreover, r'1 = (b+r1) r' 2 = (b- r1) ( D . 4 . 3 ) b2 = r'12 + r12 - 2r'1r1 ⇒ b2 - r12 = r'1(r'1-2r1) (D.4.4) rcog = 2 r'1r'2 [ r'24 + r'14 + 2r'1r'2(b2 - r12) ]1/4 = 2 (b+r1)(b- r1) [ (b- r1)4 + (b+r1)4 + 2(b+r 1)(b- r1)(b2 - r12) ]1/4 = 2 (b+r1)(b- r1) [ 4(b2 + r12)2 ]1/4 = 2 (b+r1)(b- r1) 2 b2+ r12 = (b2 - r12) b2+ r12 = b2 b (1 - (r1/b)2) 1+ (r1/b)2 = b 1 - (r1/b)2 1+ (r1/b)2 . ( D . 4 . 5 ) The exact results are then (equal masses and θ = 0): tan(δ) = 0 ⇒ δ = 0 ( D . 4 . 6 ) rcog = b 1 - (r1/b)2 1+ (r1/b)2 . // m 1= m2 and θ = 0 (D.4.7) We then find that rcog - rcms = b [ 1 - (r1/b)2 1+ (r1/b)2 - 1 ] r cog - rcms rcms = 1 - (r1/b)2 1+ (r1/b)2 - 1 . ( D . 4 . 8 ) Appendix D: Center of Gravity 218 The first term is less than 1, so th e right side is negative and thus r cog < rcms. So in this vertically aligned case the center of gravity is closer to the Earth than the center of mass. This figure summarizes our result (with r cms - rcog exaggerated) (D.4.9) Looking back at the force equation (D.2.14), we can confirm result (D.4.7) fairly quickly : F = - (GM Em1/r'12) r^'1 - (GMEm2/r'22) r^'2 = - [(GM Em1/r'12) + (GMEm2/r'22) ] z^ // θ = 0 F = GM Em1[ (1/r'12) + (1/r'22)] = GM Em1[r'22 + r'12] / [r'1r'2]2 // m 1= m2 = GM Em1[(b-r1)2 +(b+r1)2] / [(b+r1)(b-r1)]2 = = GM Em1[2b2 + 2r12] / (b2-r12)2 = 2 GM Em1 (b2 + r12) / (b2-r12)2 = 2 GM Em1 b-2 (1 + (r1/b)2) / (1-(r1/b)2)2 so from (D.2.16), rcog = GME(2m1)/F = b (1-(r 1/b)2) / 1 + (r1/b)2 , (D.4.10) in agreement with (D.4.7). If we now further assume (r1/b) << 1 then (D.4.7) becomes rcog ≈ b (1 - (r 1/b)2) (1 - (1/2) (r 1/b)2 ) = b [ 1 - (3/2)(r 1/b)2] . rcog - rcms rcms = rcog - b b = - (3/2)(r 1/b)2 ( D . 4 . 1 1 ) in agreement with (D.3.11). Appendix D: Center of Gravity 219 (c) Equal Masses and θ = π/2 (horizontally aligned) We start again with (D.4.2) for equal masses, tan(δ) = - (r 1sinθ) r'23 – r'13 b (r'23 + r'13) + r1 cosθ (r'23 - r'13) rcog = 2 r'1r'2 [ r'24 + r'14 + 2r'1r'2(b2 - r12) ]1/4 (D.4.2) r'12 = r12 + b2 + 2r1b cosθ r' 22 = r22 + b2 - 2r2b cosθ . (D.4.1) When θ = π/2 we not only have r 1 = r2 but also r' 1 = r'2 along with cos θ = 0. This at once implies that tan(δ) = 0, and for r cog we find r'12 = r12 + b2 ⇒ - r 12 = b2 - r'12 ⇒ [b2 - r12] = 2b2 - r'12 rcog = 2 r'12 [ 2r'14 + 2r'12(b2 - r12) ]1/4 = 2 r'12 [ 2r'14 + 2r'12(2b2 - r'12) ]1/4 = 2 r'12 { 4r'12b2 }1/4 = [r'16 b2 ]1/4 = [r'13 b ]1/2 = r'13 b ( D . 4 . 1 2 ) = [r'16 b2 ]1/4 = [(b2+ r12)3 b2 ]1/4 = b [1 + (r 1/b)2 ]3/4 . (D.4.13) The exact results are then (equal masses and θ = π/2) tan(δ) = 0 r cog = r'13 b = b [1 + (r 1/b)2 ]3/4 rcog - rcms = b { [1 + (r 1/b)2 ]3/4 – 1 } rcog - rcms rcms = [1 + (r 1/b)2 ]3/4 – 1 (D.4.14) Since the first term on the right is greater than 1, we find r cog > rcms so for the horizontally-aligned equal-mass satellite the center of gravity is farther from the center of the Earth than the center of mass, in agreement with our earlier conclu sion based on approximation. This figure summarizes our result (with r cog - rcms exaggerated), Appendix D: Center of Gravity 220 (D.4.15) We have added in blue a portion of a circle centered at Earth cen ter which passes through the masses. The point rcog lies below this circle, as we now verify, rcog < r'1 ? r'13 b < r'1 ? r'13 < r'12 b ? r' 1 < b ? yes . (D.4.16) Looking back at the force equations (D.2.15), we can confirm result (D.4.12) very quickly, Fz = -2GM Em1(b/r'13) Fy = 0 rcog = GME(2m1)/F = r'13 b . (D.4.17) If we now further assume (r1/b) << 1 the result (D.4.13) becomes r cog = b [1 + (r 1/b)2 ]3/4 ≈ b [1 + (3/4) (r 1/b)2 ] so rcog - rcms rcms = rcog - b b = + (3/4)(r 1/b)2 ( D . 4 . 1 8 ) in agreement with (D.3.12). Appendix D: Center of Gravity 221 D.5 Center of Gravity for a single-sphere satellite For the vertically aligned dumbbell satellite, we found that r cog < rcms. People argue that this is so because gravity acts more strongly on the mass closer to the Earth. This argument does not help much, however, for the horizontally aligned satellite where instead one has r cog > rcms and neither mass is closer than the other to the Earth. One wonders if one can assemble a rigid satellite fro m a finite number of masses such that these two opposite effects cancel out, resulting in rcog = rcms, at least for some orientation of the masses. Could such a solution be found that works for any orientati on of the rigid satellite? We leave these questions to the reader and return instead to the above "argument". The argument applied to a sphere gives a wrong an swer. One would argue for a spherical satellite that the near half of the sphere is closer to the Earth (w here gravity is stronger) than the far half, so the center of gravity should be offset toward the Earth from the center of mass. As the reader no doubt knows, a uniform sphere or spherical shell is in fa ct a "rigid assembly of masses" for which rcog = rcms. So in such an object, the two effects found for the vertically and horizontally aligned satellites do in fact exactly cancel out. Here we demonstrate that rcog = rcms for a uniform thin shell by two basic methods. Once this is shown, the result then applies for an y symmetric assembly of shells such as a sphere or a thick spherical shell. Method A. In electrostatics one of Maxwell's equations says div E = 4πρ (cgs units). One applies the integral form of this law ∫E • dA = 4π∫ρ dV = 4πQ to show that the electric field outside a uniform spherical shell of charge is independent of the radius of the shell and thus is the same as if the charge were all concentrated at the center of the shell. Th e argument is that, due to rotational invariance (or "symmetry"), the direction of the electric field can only be radial, so then ∫E • dA = E * 4πR2 and one then finds that E = Q/R2 which is indeed the electric field of a point charge Q at distance R. The same argument can be applied to a uniform shell of mass so that the gravitational field must be in the radial direction and is independent of the shell radius R, so the shell acts as a point mass at its center. If this shell is a satellite then the Earth cannot determ ine from gravity alone the radius of that satellite, so one can replace the satellite by a point mass at its center. The Earth never knows, and nothing changes in the orbit. But for the point mass certainly r cog = rcms , so this fact applies as well to any uniform spherical shell or sphere satellite. Method B (the ever-popular brute force method). This me thod is more direct, and is found in high-school physics texts, though the integral we obtain below is not. The drawing below is upside-down relative to our earlier drawings, to be more compatible with usua l spherical coordinate notation. So the center of the Earth is now at the top and the center of mass of the satellite is located a di stance b from Earth center, as earlier. This particular satellite is a spherical shell of radius r and very thin thickness dr. The mass of the shell is then M shell = [(4πr2)dr]ρ where ρ is the shell's uniform mass density. Appendix D: Center of Gravity 222 (D.5.1) Quantity dm is a tiny chunk of mass on the spherical shell of radius r with dm = ρ(dA)dr = ρ (r 2dΩ)dr = ρ (r2sinθdθdφ)dr . (D.5.2) The force experienced by this mass dm due to mass M E is given by d F = -(GM Edm/r12)r^1 = -(GM E/r13) dm r1 . (D.5.3) As usual in spherical coordinate s, and as shown in (E.2.6), r = rsinθcosφ x^ + rsinθsinφ y^ + rcosθ z^ ( D . 5 . 4 ) so then r1 = b + r = -b z^ + rsinθcosφ x^ + rsinθsinφ y^ + rcosθ z^ = r s i n θcosφ x^ + rsinθ sinφ y^ + (-b+ rcos θ)z^ . (D.5.5) Therefore, d F = -(GM E/r13) dm r1 = -(GM E/r13) dm [rsinθ cosφ x^ + rsinθ sinφ y^ + (-b+ rcos θ)z^ ] . (D.5.6) Then, Appendix D: Center of Gravity 223 dFx = dF • x^ = -(GM E/r13) ρ (r2sinθdθdφ)dr (rsinθcosφ) dFy = dF • y^ = -(GM E/r13) ρ (r2sinθdθdφ)dr (rsinθsinφ) dFz = dF • z^ = -(GME/r13) ρ (r2sinθdθdφ)dr (-b+rcos θ) . (D.5.7) Now integrate d φ from 0 to 2 π to get the force on a ring of mass of angular width dθ on the shell at fixed θ. But sinφ has zero integral over this range and so does cos φ, so the first two terms integrate to nothing, while the integral of d φ in the last line gives 2 π. We are left with this resulting force, all in the z direction, dFring = (GM E/r13) ρ (r2sinθdθ 2π) dr (b-rcosθ) . (D.5.8) Now integrate over all rings of the shell to get a total force F experienced by the spherical shell satellite, F = ∫0 π dθ (GME/r13) ρ (r2sinθdθ 2π) dr (b-rcosθ) = 2 πρGME r2dr ∫0 π dθ sinθ (b-rcosθ) (r2+b2-2rbcosθ)3/2 . (D.5.9) This is a famous discontin uous integral that we evaluate below, but for now we use Maple. Assuming that r < b which means the Earth center lies outside the shell, Maple says, ( D . 5 . 1 0 ) where 2/b2 is the value of the integral. Thus we have shown that F = 2πρGME r2dr ∫0 π dθ sinθ (b-rcosθ) (r2+b2-2rbcosθ)3/2 = 2πρGME r2dr (2/b2) = G M E (4πr2dr)ρ /b2 = GMEMshell / b2 . ( D . 5 . 1 1 ) We now use the center of gravity definition from (D.1.6) to find, r cog = GMEMshell / F = b ( D . 5 . 1 2 ) Since rcms = b as well, we conclude that for a spherical shell satellite one has r cog = rcms . Appendix D: Center of Gravity 224 If the mass center M E were inside the shell so b < r, Maple gives a different answer , ( D . 5 . 1 3 ) showing that the force experienced by the uniform shell due to a mass M E inside the shell is zero. It seems unsportsmanlike not to actual do the integral to see why it is discontinuous at r = b. Here is a brief tour: 1. Change variables from θ to r 1 and look at the endpoints in the new variable, r12(θ=0) = r2 + b2-2rb(1) = (b-r)2 ⇒ r1(θ=0) = |b-r| ≡ α r12(θ=π) = r2 + b2-2rb(-1) = (b+r)2 ⇒ r1(θ=π) = b+r ≡ β It is the absolute value that causes the discontinuous behavior at b = r, as we shall see. 2. Since r 12 = r2 + b2 - 2rbcosθ one has sin θdθ = r1dr1/(rb) . (D.5.14) 3. The factor (b - rcosθ ) = (b 2-r2+r12)/(2b), while the denominator of the integral is just r 13. 4. The integral reduces to two elementary power integrals to give I(r) = ∫0 π dθ sinθ (b-rcosθ) (r2+b2-2rbcosθ)3/2 = [ (b2-r2) ∫α β dr1/r12 + ∫α β dr1 1 ] / (2rb2) = [ (b-r) {(b+r) - |b-r| |b-r| } + (b+r) - |b-r| ] / (2rb2) . (D.5.15) For b > r one has |b-r| = b-r and the value comes out I(r) = [ 2r + 2r ] / (2rb2) = 2/b2 . ( D . 5 . 1 6 ) For b < r one has |b-r| = r-b and the value is instead Appendix D: Center of Gravity 225 I(r) = [ -2b + 2b ] / (2rb2) = 0 . ( D . 5 . 1 7 ) The full result is then I(r) = (2/b 2)H(b-r) // Heaviside function (D.5.18) which has this appearance, ( D . 5 . 1 9 ) On can argue that I(b) = (1/b2) which is the average of the values at the discontinuity. Appendix E: Spherical Unit Vectors and Kinematics 226 Appendix E: Spherical Coordinate Unit Vectors and Particle Kinematics E.1 Angle Conventions Here is our "physics" convention for spherical coordinate angles θ and φ : (E.1.1) The reader is advised that the results belo w are often quoted in the literature with θ↔φ which is the "math" convention (eg, Wolfram, left below). Sometimes after doing θ↔φ some sources replace polar angle φ with latitude π/2 - φ (right below) : Two sphe rical angle conventions we do not use. In the right picture, φ and θ might be longitude and latitude, or right ascension and declination. If you are in charge of getting a spacecraft to Vul can, please pay attention to these conventions. E.2 Matrix Approach The usual matrices for actively rotating a vector about the x, y, or z axis are these: Rx(θ) = ⎟⎟⎟ ⎠⎞ ⎜⎜⎜ ⎝⎛ − θ θθ θ cos sin 0sin cos 00 0 1 Ry(θ) = ⎟⎟⎟ ⎠⎞ ⎜⎜⎜ ⎝⎛ − θ θθ θ cos 0 sin0 1 0sin 0 cos Rz(θ) = ⎟⎟⎟ ⎠⎞ ⎜⎜⎜ ⎝⎛− 1 0 00 cos sin0 sin cos θ θθ θ . ( E . 2 . 1 ) The Cartesian unit vectors x^, y^. z^ can be rotated into the spherical unit vectors r^, θ^, φ^ as follows Appendix E: Spherical Unit Vectors and Kinematics 227 θ^ = Rz(φ) Ry(θ) x^ = ⎟⎟⎟ ⎠⎞ ⎜⎜⎜ ⎝⎛ −− θ θϕ θ ϕ ϕ θϕ θ ϕ ϕ θ cos sinsin sin cos sin coscos sin sin cos cos 0⎝⎜⎛ ⎠⎟⎞1 0 0 = ⎝⎜⎛ ⎠⎟⎞ cosθcosφ cosθsinφ -sinθ = ⎝⎜⎛ ⎠⎟⎞θ^•x^ θ^•y^ θ^•y^ φ^ = Rz(φ) Ry(θ) y^ = ⎟⎟⎟ ⎠⎞ ⎜⎜⎜ ⎝⎛ −− θ θϕ θ ϕ ϕ θϕ θ ϕ ϕ θ cos sinsin sin cos sin coscos sin sin cos cos 0⎝⎜⎛ ⎠⎟⎞0 1 0 = ⎝⎜⎛ ⎠⎟⎞ -sinφ cosφ 0 r^ = Rz(φ) Ry(θ) z^ = ⎟⎟⎟ ⎠⎞ ⎜⎜⎜ ⎝⎛ −− θ θϕ θ ϕ ϕ θϕ θ ϕ ϕ θ cos sinsin sin cos sin coscos sin sin cos cos 0⎝⎜⎛ ⎠⎟⎞0 0 1 = ⎝⎜⎛ ⎠⎟⎞ sinθcosφ sinθ sinφ cosθ . (E.2.2) One can interpret the elements of the vectors on the right as dot products as we show on the first line. Notice that these three vectors on the right are just the columns of the matrix shown, so one can say ( θ^ φ^ r^ ) = ⎟⎟⎟ ⎠⎞ ⎜⎜⎜ ⎝⎛ −− θ θϕ θ ϕ ϕ θϕ θ ϕ ϕ θ cos sinsin sin cos sin coscos sin sin cos cos 0 = Rz(φ) Ry(θ) ≡ R . (E.2.3) It is then obvious that ⎝⎜⎛ ⎠⎟⎞ x^•θ^ x^•φ^ x^•r^ y^•θ^ y^•φ^ y^•r^ z^•θ^ z^•φ^ z^•r^ = ⎟⎟⎟ ⎠⎞ ⎜⎜⎜ ⎝⎛ −− θ θϕ θ ϕ ϕ θϕ θ ϕ ϕ θ cos sinsin sin cos sin coscos sin sin cos cos 0 = Rz(φ) Ry(θ) = R (E.2.4) from which one can read off any desired dot produc t. As verification, consider an example, Example : y^ • θ^ = y^ • (Rx^) = Σi (y^)i(Rx^)i = Σij(y^)iRij(x^)j = Σijδ2iRijδ1j = R21. Looking at the last equation of (E.2.2) multiplied by r, since r = r r^ one sees that x = rsin θcosφ y = rsin θsinφ z = rcos θ ( E . 2 . 5 ) which is the "inverse transformation" associated with spherical coordinates. See Section E.4 below. The three equations (E.2.2) can be trivially wr itten out as follows (changing to a standard r, θ,φ order) r^ = sinθ cosφ x^ + sinθsinφ y^ + cosθ z^ θ^ = cosθcosφ x^ + cosθsinφ y^ - sinθ z^ φ^ = -sinφ x^ + cosφ y^ ( E . 2 . 6 ) Appendix E: Spherical Unit Vectors and Kinematics 228 which can be inverted to give x^ = sinθcosφ r^ + cosθcosφ θ^ - sinφ φ^ y^ = sinθsinφ r^ + cosθsinφ θ^ + cosφ φ^ z^ = cosθ r^ - sinθ θ^ . ( E . 2 . 7 ) For those not wanting to invert (transpose) the 3x3 matrix R, (E.2.7) can be quickly verified by looking at the unit vector dot products in (E.2.4). Eq. (E.2.6) can be similarly verified . The spherical unit vectors are orthogonal due to their construction in (E.2.3). Example: r^ • θ^ = (R z^ ) • (R x^ ) = z^ • (RTRx^) = z^ • x^ = 0 Example: r^ • r^ = (R z^ ) • (R z^ ) = z^ • (RTRz^) = z^ • z^ = 1 Here R ≡ Rz(φ) Ry(θ) has the property RTR = 1 since it is a rotation. So r^ • θ^ = r^ • φ^ = θ^ • φ^ = 0 . (E.2.8) The spatial derivatives of the spherical unit vectors are easy to compute ( ∂θr^ means ∂r^/∂θ ), ∂rr^ = 0 ∂θr^ = θ^ ∂φr^ = sinθ φ^ ∂rθ^ = 0 ∂θθ^ = - r^ ∂φθ^ = cosθ φ^ ∂rφ^ = 0 ∂θφ^ = 0 ∂φφ^ = -sinθ r^ -cosθ θ^ . (E.2.9) Example : ∂θθ^ = ∂θ ⎝⎜⎛ ⎠⎟⎞ cosθcosφ cosθsinφ -sinθ = ⎝⎜⎛ ⎠⎟⎞ -sinθcosφ -sinθsinφ -cosθ = - r^. In Fig (E.1.1) one can see for example that ∂θr^ = θ^. Often one draws a little triangle to verify an equation like this: ( E . 2 . 1 0 ) dr^ = dr θ^ ≈ 1*dθ θ^ = dθ θ^ ⇒ dr^ = dθ θ^ ⇒ θ^ = dr^/dθ = ∂θr^ . Time derivatives of the unit vectors ar e then obtained by the chain rule ( ∂tr^ means d r^/dt ), Appendix E: Spherical Unit Vectors and Kinematics 229 ∂tr^ = θ• θ^ + φ• sinθ φ^ ∂tθ^ = – θ• r^ + φ• cosθ φ^ ∂tφ^ = – φ• sinθ r^ – φ• cosθ θ^ . ( E . 2 . 1 1 ) Example : ∂tr^ = (∂rr^) r• + (∂θr^) θ• + (∂φr^) φ• = 0 + θ^θ• + sinθ φ^φ• . Note : If we regard r,θ ,φ as coordinates of Frame S, then ∂tr^ = (d r^/dt)S = ∂Sr^ as in Section 1.7 and 1.8. The cross products of the spherical unit vectors follow the right hand rule, so looking at Fig (E.1.1): r^ x θ^ = φ^ θ^ x φ^ = r^ φ^ x r^ = θ^ ( E . 2 . 1 2 ) The first cross product has ordering r, θ,φ and the last two are cyclic permutations. Here are some other cross products: x^ x r^ = [ sinθcosφ r^ + cosθcosφ θ^ - sinφ φ^] x r^ = - cos θcosφ φ^ - sinφ θ^ x^ x θ^ = [ sinθcosφ r^ + cosθcosφ θ^ - sinφ φ^] x θ^ = sinθcosφ φ^ + sinφ r^ x^ x φ^ = [ sinθcosφ r^ + cosθcosφ θ^ - sinφ φ^] x φ^ = -sinθ cosφ θ^ + cosθcosφ r^ (E.2.13) y^ x r^ = [ sinθsinφ r^ + cosθsinφ θ^ + cosφ φ^] x r^ = -cosθsinφ φ^ + cosφ θ^ y^ x θ^ = [ sinθsinφ r^ + cosθsinφ θ^ + cosφ φ^] x θ^ = sinθsinφ φ^ - cosφ r^ y^ x φ^ = [ sinθsinφ r^ + cosθsinφ θ^ + cosφ φ^] x φ^ = - sinθsinφ θ^ + sinθsinφ r^ (E.2.14) z^ x r^ = [ cosθ r^ - sinθ θ^ ] x r^ = sinθ φ^ z^ x θ^ = [ cosθ r^ - sinθ θ^ ] x θ^ = cosθ φ^ z^ x φ^ = [ cosθ r^ - sinθ θ^ ] x φ^ = -cosθ θ^ - sinθ r^ ( E . 2 . 1 5 ) Appendix E: Spherical Unit Vectors and Kinematics 230 E.3 The motion of a particle in spherical coordinates The equations below are easily derived from (E.2.11): r = r r^ / / p o s i t i o n ( E . 3 . 1 ) v = vr r^ + vθθ^ + vφ φ^ // velocity v r = r• v θ = r θ• v φ = r φ• sinθ ( E . 3 . 2 ) a = ar r^ + aθθ^ + aφ φ^ // acceleration a r = r•• - rθ•2 – r φ•2 sin2θ a θ = 2 r• θ• + rθ•• - r φ•2 sinθ cosθ a φ = 2 r• φ• sinθ + 2 r θ• φ• cosθ + rφ••sinθ . (E.3.3) Example : v = r• = ∂t(rr^) = r•r^ + r(∂tr^) = r•r^ + r[ θ• θ^ + φ• sinθ φ^ ] = r•r^ + rθ•θ^ + rφ•sinθφ^ . Notice that r•• = v•r ≠ ar and similarly for other components. If for some reason particle motion is restricted to a spherical surface of radius r (such as in our dumbbell satellite), we can set r• = r•• = 0 in the above equations to get r = r r^ / / p o s i t i o n ( E . 3 . 4 ) v = vθθ^ + vφφ^ // velocity is tangent to the sphere : r • v = 0 v θ = r θ• v φ = r φ• sinθ ( E . 3 . 5 ) a = ar r^ + aθθ^ + aφ φ^ // acceleration a r = - rθ•2 – r φ•2 sin2θ a θ = rθ•• - r φ•2 sinθ cosθ a φ = 2 r θ• φ• cosθ + rφ••sinθ . ( E . 3 . 6 ) From these equations many other useful results can be obtained, for example, r^ x v = vθ r^ x θ^ + vφ r^ x φ^ = vθ φ^ – vφ θ^ r^ x a = aθ r^ x θ^ + aφ r^ x φ^ = aθ φ^ – aφ θ^ ( E . 3 . 7 ) Appendix E: Spherical Unit Vectors and Kinematics 231 E.4 Curvilinear coordinates approach Spherical coordinates are defined by the "inve rse transformation" on the left below: inverse transformation transformation x = rsin θcosφ r = + x2+y2+z2 0 ≤ r < ∞ y = rsin θsinφ cos θ = z/r 0 ≤ θ ≤ π -1 ≤ cosθ ≤ 1 z = rcos θ s i n θ = + 1-(z/r)2 0 ≤ sinθ ≤ 1 s i n φ = y/(rsinθ) -1 ≤ sinφ ≤ 1 c o s φ = x/(rsinθ) 0 ≤ φ < 2π -1 ≤ cosφ ≤ 1 (E.4.1) We quote now a set of results from our Tensor document which won't be needed but which we include for completeness. Equation numbers (...)T refer to that document. Spherical coordinates are just an example of curvilinea r coordinates which fit into a general formalism: x = (x1, x2, x3 ) = (x,y,z) // x-space coordinates x' = (x1', x2',x3') = (r,θ,φ) // x'-space coordinates x = F-1(x') ↔ x = rsin θcosφ // a non-linear inverse transformation y = r s i n θsinφ z = rcos θ . (1.6)T The linearized transformation local to a point defines certain R and S matrices (the "differentials") : d x' = R( x) dx R ik(x) ≡ (∂x'i/∂xk) R = S-1 // dx' i = Rij dxj d x = S( x') dx' Sik(x') ≡ (∂xi/∂x'k) S = R-1 // dx i = Sij dx'j . (2.1.6)T S = ⎝⎜⎛ ⎠⎟⎞ sinθ cosφ rcosθcosφ -rsinθsinφ sinθ sinφ rcosθsinφ rsinθ cosφ cosθ -rsinθ 0 // compute from above definition of S ij (3.4.4)T g¯' = STS = ⎝⎜⎜⎛ ⎠⎟⎟⎞ 1 0 0 0 r2 0 0 0 r2sin2θ , det(g ¯') = r4sin2θ // metric tensor and its determinant (5.13.14) T J(r,θ,φ) = det(S) = det(g¯') = r2sinθ . // Jacobian (5.13.16) T (ds)2 = Σijg¯'ij dx'i dxj = (dr)2 + r2(dθ)2 + r2sin2(dφ)2 // distance (5.13.18) T Because the metric tensor g ¯' is diagonal, the coordinates r, θ.φ are orthogonal. Footnote : Above we use the temporary developmental notation of Tensor where all indices are down and covariant objects get an overbar. Some rules for c onversion to the Standard Notation are these: Appendix E: Spherical Unit Vectors and Kinematics 232 g¯ij → gij covariant metric tensor v ¯i → vi covariant vector gij → gij contravariant metric tensor v i → vi contravariant vector (dx i → dxi) dx'i = Rij dxj → dx'i = Ri j dxj E.5 Polar Coordinates Here we use a notation common for two of the cylindrical coordinates ( ρ,φ,z), (E.5.1) x = ρcosφ y = ρsinφ ( E . 5 . 2 ) ρ^ = cosφ x^ + sinφ y^ ρ^ = Rz(φ) x^ φ^ = -sinφ x^ + cosφ y^ φ^ = Rz(φ) y^ (E.5.3) x^ = cosφ ρ^ – sinφ φ^ x^ = Rz(-φ) ρ^ y^ = sinφ ρ^ + cosφ φ^ y ^ = Rz(-φ) θ^ (E.5.4) ρ^ • x^ = cosφ ρ^ • y^ = sinφ φ^ • x^ = -sinφ φ^ • y^ = cosφ ( E . 5 . 5 ) ρ^• = φ• φ^ φ^• = - φ• ρ^ ( E . 5 . 6 ) Proof of (E.5.6) : ρ^• = dρ^/dt = d/dt(cos φ x^ + sinφ y^) = -sinφ φ• x^ + cosφ φ• y^ = -sinφ φ• [cosφ ρ^ – sinφ θ^ ] + cosφ φ• [ sinφ ρ^ + cosφ θ^] = φ• φ^ φ^• = dθ^/dt = d/dt( -sin φ x^ + cosφ y^) = -cosφ φ• x^ - sinφ φ• y^ = -cosφ φ•[cosφ ρ^ – sinφ θ^ ] - sinφ φ• [ sinφ ρ^ + cosφ θ^] = - φ• ρ^ Appendix E: Spherical Unit Vectors and Kinematics 233 E.6 The Affine Connection Recall from above the claim that ∂ rr^ = 0 ∂θr^ = θ^ ∂φr^ = sinθ φ^ ∂rθ^ = 0 ∂θθ^ = - r^ ∂φθ^ = cosθ φ^ ∂rφ^ = 0 ∂θφ^ = 0 ∂φφ^ = -sinθ r^ -cosθ θ^ . (E.2.9) (E.6.1) We wish to put these equations into a more general framework which is an extension of the discussion of Section E.4. In the notation of our Tensor document, for an arbitrary curvilinear coordinate system x' the derivatives of the tangent base vectors (called en in that document) are written ∂'jen = ΣiΓ 'i jn ei . (E.6.2) This just says that the change in a basis vector obtained by moving a small amount in some direction is (and of course must be) some linear combination of the basis vectors. The coe fficients of the linear combination are known as the affine connection (or Levi-Civita connection ) Γ 'c ab = Γ 'c ba. The reason for the primes is that Cartesia n coordinates are thought of as x, while curvilinear ones of some particular type are x'. In Cartesian x-space one has Γc ab = 0 because basis vectors don't vary with position in that space. Then Γ'c ab is the affine connection in x'-space. For exam ple, as noted in Section E.4, in spherical coordinates one has x = (x1,x2,x3) = (x,y,z) x ' = (x'1,x'2,x'3) = (r,θ,φ) ( E . 6 . 3 ) and the inverse transformation (E.2.5) is x = F-1(x'). As shown below, e1(x') = e1(r,θ,φ) = r^ and then as an example of (E.6.2) we write (for j=2 and n=1, a nd the linear combination has only one non-zero term): ∂'2e1 = ∂x2'e1 = ∂θe1(r,θ,φ) = ∂θr^(r,θ,φ) = θ^ = (1/r) eθ = (1/r)e2 = Γ' 2 21 e2 = Γθ θr eθ . (E.6.4) The curvilinear unit basis vectors e^n are related to the tangent base vectors by e^n = (1/h'n) en where the h'n = |en| are the so-called scale factors. The derivatives of the e^n are then given by : ∂' je^n = ∂'j(h'n-1en) = (∂'jh'n-1)en + h'n-1(∂'jen) = - h' n-2 (∂'jh'n) en + h'n-1( ΣiΓ 'i jn ei) = - h' n-1 (∂'jh'n) e^n + h'n-1( ΣiΓ 'i jn h'i e^i) Appendix E: Spherical Unit Vectors and Kinematics 234 = ( 1 / h ' n) [Σi h'i Γ 'i jne^i – (∂'jh'n) e^n] . ( E . 6 . 5 ) In Tensor the Cartesian basis vectors of x-space are called ui, but in this document they are called ei so we need a different symbol ei for the tangent base vectors. In Section 14 we use these ei with e^i as the unit tangent base vectors, ξ = (ξ1,ξ2,ξ3) in place of x' = (x'1,x'2,x'3), and Γ' = Γ with no prime since ξ has no prime (trying not to confuse Γ = 0 of x-space with Γ ≠ 0 of ξ-space). In this notation (E.6.5) would appear as (∂e^n/∂ξj) = (1/hn) [Σi(hiΓ i jne^i) - (∂ hn/∂ξj)e^n] . (E.6.6) Going back to the Tensor notation, the affine connection for a coordinate system is related to the system's metric tensor g according to Γ d ab = (1/2) g dc [ ∂ag¯bc + ∂bg¯ca – ∂cg¯ab] = 0 since g ¯ij = δij x-space (Cartesian) Γ' d ab = (1/2) g' dc [ ∂'ag¯'bc + ∂'bg¯'ca – ∂ 'cg¯'ab] . x'-space (E.6.7) For spherical coordinates, one has (using r, θ,φ = 1,2,3 where θ = polar, φ = azimuth) 1 2 3 h r = 1 h θ = r h φ = rsinθ // scale factors (h' 1= hr) er = r^ e θ = r θ^ e φ = rsinθ φ^ // tangent base vectors ( e1 = er) e^r = r^ e ^θ = θ^ e ^φ = φ^ // curvilinear unit basis vectors g¯'ab = ⎝⎜⎜⎛ ⎠⎟⎟⎞ 1 0 0 0 r2 0 0 0 r2sin2θ g' ab = inverse(g ¯'ab) // metric tensor (E.6.8) where for example g ¯'33 = gφφ = r2sin2θ. For spherical coordinates only 9 of the 27 elements of Γ'd ab are non-zero (computed from (E.6.7)) : Γ' 1 22 = - r Γ' 2 12 = Γ' 2 21 = 1/r // notation example: Γ' 1 22 = Γr θθ Γ' 1 33 = - r sin2θ Γ' 3 13 = Γ' 3 31 = 1/r Γ' 2 33 = - cosθsinθ Γ' 3 23 = Γ' 3 32 = cotθ . (E.6.9) Example done in Section 14 notation with (E.6.6) : (∂e^n/∂ξj) = (1/hn) [Σi(hiΓ i jne^i) - (∂ hn/∂ξj)e^n] (∂ e^3/∂ξ2) = (1/h3) [Σi(hiΓ i 23e^i) - (∂ h3/∂ξ2)e^3] (∂φ^/∂θ) = (1/rsin θ) [(h3Γ 3 23e^3) - (∂ [rsinθ]/∂θ)e^3] Appendix E: Spherical Unit Vectors and Kinematics 235 = (1/rsin θ) [(rsinθ * cotθ * φ^) - rcosθ * φ^] = (1/rsin θ) [rcosθ φ^ - rcosθ φ^] = 0 ( E . 6 . 1 0 ) and with some effort we have verified that ∂ θφ^ = 0 as appears in (E.6.1). This fact is of course obvious just looking at Fig (E.1.1), but things can be less obvious in obscure coordinate systems. Footnote : To be consistent with the Footnote at the end Section E.4 we really should use ∂¯j to indicate the derivative in (E.6.2) since it transforms as a covariant vector, but it just adds confusion to do so. Appendix F: The Dumbbell Satellite 236 Appendix F : The Dumbbell (Tethered) satellite as an example of rotating frame analysis This is a very long appendix so we provide an overview: Section F.1 lays out kinematic details of the coordinates we use and discusses simple geometric facts of the satellite. We use the "swap notation" wherein Frame S' is the inertial frame and Frame S is the rotating frame. The masses m 1 and m2 can be equal or different. Section F.2 computes the angular momentum L(0) of the satellite and its time derivative L•(0) in rotating Frame S where the origin of Frame S is used as a reference point. Section F.3 computes the true torque N'(b) on the satellite due to the Earth's gravitational attraction acting on the two masses. The resulting torque is computed in inertial Frame S' and has only a φ^ component as shown in (F.3.7). This computation is done in two extra ways to verify the result. In the far approximation, the expression for N'(b) simplifies to that shown in (F.3.13). Section F.4 then computes the fictitious torque which appears in rotating Frame S. This torque N(0) fict is stated in (F.4.11) and has φ^ and θ^ components. An interpretation is provided. Section F.5 then uses "Newton's Angular Law" L•(0) = N'(b) + N(0) fict to obtain the angular equations of motion for the satellite as stated in (F .5.7) using the far approximation. Certain special-case solutions are extracted which demonstrate the notion of in-plane and out-of-plane small- θ libration of the satellite with frequencies given in (F.5.10) and (F.5.13). Section F.6 basically starts over working only with for ces, not torques. The two angular equations of motion obtained in Section F.5 are again obtained, and a third equation determines the tension T in the dumbbell stick. One can interpret this tension T as a tidal force acting on mass m 1 which is distance r 1 from the Frame S center of mass. An equal and opposite tidal force acts on mass m 2 at the other end of the stick. When the satellite is ve rtically aligned, one finds T = 3m 1 ω2r1 in (F.6.30). The tension T does not appear in the angular analysis of Section F.5 si nce T makes no contribution to torque about the Frame S origin, being aligned with the stic k. Tension T is part of the radial force equation (F.6.19) of Section F.6, but there is no radial equation in Section F.5. There the stick is regarded just as a constraint, and it is typical for the forces of constraint not to be determined in the simplest analysis. Section F.7 provides some Maple numerical solutions of the far-approximation θ ,φ equations of motion of the satellite stated in (F.5.7). The two librati on modes are verified, and a more general case is examined. Section F.8 reworks the satellite force analysis entirely in Cartesian coordinates. The equations of motion for the variables x,y,z are shown in (F.8.15) with tension T provided by (F.8.20). Section F.9 shows that the x,y,z equations of motion of Section F.8 are entirely equivalent to the θ,φ equations of motion of Section F.5. Appendix F: The Dumbbell Satellite 237 Section F.10 examines various numerical solutions for the dumbbell satellite, working now in Cartesian coordinates. The libration modes are again examined, a nd a more general solution is studied. There is no "conical solution". F.1 Kinematics of the satellite in rotating Frame S As with Appendix D, this section uses the "swap notation" described in the Summary for Section 1 where the rotating frame is Frame S and the non-rotating frame is Frame S'. This is done to reduce the number of primes since most activity will be in the rotating satellite Frame S. We now place the dumbbell satellite in a more genera l orientation than it was in Appendix D : (F.1.1) Description of the Figure Half the battle is having a clear picture of what is going on and we shall expend many words to describe the above drawing. It shows the dumbbell satellite in or bit in a completely arbitrary orientation. The two masses m 1 and m2 are connected by a massless stick (not shown) of length r 1+r2 = s . If we find that this stick is always in tension for some situation, we can replace it in that situation with a non-stretching massless tether . If the stick were to go into compression, th e replacement tether would lose its linear shape and we don't want to deal with that problem. As shown much later in Section F.10, the stick is always in tension for normal situations. The gray-filled triangle is a part of the plane φ = constant which has normal vector φ^. This plane is not in the plane of paper. The fill region contains two non-right triangles shown in blue and red. The red Appendix F: The Dumbbell Satellite 238 triangle is on the viewer's side of paper, while the blue one lies behind the paper. Each of these triangles contains the vector b as an edge. Frame S' is an inertial frame whose center is the center of the Earth and which is assumed fixed with respect to the stars. Frame S is a non-inertial frame whose center is locat ed at the satellite center of mass point. We make the assumption discussed above (D.2.2) that we can i gnore the tiny offset between the center of mass and center of gravity of the satellite, so we then regard the Frame S origin as travelling in a circular orbit around the Earth. Recall from (D.3.16) that this o ffset is about 3 microns for s = 10 meters. At time t = 0 shown in the figure, the axes of both Frames align with each other. At any time, axes y,z and y',z' are in the plane of paper. The unit vector z^ always points from the center of the Earth to the origin of Frame S. so b = bz^ at any time. The axes x,x' always point directly out of the plane of paper so x^ = x^' . The orbital rotation vector ω = ωx^ also points out of the plane of paper and recall that ω = 2π/T where T is about 88 minutes for a low-Earth orbit. We have in mind an orbit at any altitude, but we do assume a circular orbit. The circled dot at the bottom is the center of the Earth (mass M E) and we have drawn the Earth surface in green. The blue circle is the orbit of the cen ter of mass of the satellite (the Frame S origin) and it lies in the plane of paper. Note that ω is for the orbit and has nothing whatsoever to do with the rotation of the Earth. The above picture is the same whether or not the Earth rotates at its 24 hour ωE rate about some obscure ω^E axis (not shown). One could try to treat the satellite as a reduced-mas s single-particle system as is done for planetary orbits, but even if possible this would obscure details we want to be visible. We really have here a 3-body problem where the three bodies are point masses, but ther e is a constraint (the stick), so perhaps it is a 2 1/2-body problem. If the dumbbell were treated as a rigid object, it is then a 2-body problem where one body is not a point mass. Naming of coordinates In Frame S mass m 1 has spherical coordinates (r 1,θ1,φ1) while mass m 2 has coordinates (r 2,θ2,φ2). The reader is now forewarned about ou r upcoming slipshod notation. We define (θ,φ) ≡ (θ1,φ1). Thus Fig (F.1.1) shows θ,φ and not θ1,φ1. The three unit vectors of spherical coordinates for mass m 1 will always be called r^, θ^, φ^ and never r^1, θ^1, φ^1. These spherical coordinates are in the "physics" convention: angle θ is the polar angle down from the "vertical" z axis, φ is the azimuthal angle measured from the x axis toward the y axis. See Appendix E.1 regarding conventions. Because r1 and r2 are collinear, we know that θ2 = π -θ and φ2 = π + φ. Since the Frame S origin is the center of mass, we also know that r 2 = (m1/m2)r1. Our strategy is to avoid the subscript-2 coordinates whenever possible and express everything in terms of the mass m 1 coordinates (r 1,θ,φ). The mass m 2 unit vectors are related to the m 1 unit vectors by r^2 = - r^, θ^2 = + θ^, and φ^2 = - φ^. Each unit vector points toward increasing parameter va lue for its associated mass. To see these last relations, it helps to stare at Fig (F.1.1) and think of the gray triangle as being in the plane of paper. We shall only make use of φ^2 = - φ^ below. See Appendix E regarding the unit vectors r^, θ^ and φ^. Appendix F: The Dumbbell Satellite 239 To summarize : r1 = [x1,y1,z1] = (r1,θ,φ) = coordinates of mass m 1 velocity = v1 r2 = [x2,y2,z2] = (r2,θ2,φ2) = coordinates of mass m 2 velocity = v2 θ2 = π-θ φ2 = φ+π r2 = (m1/m2) r1 r^2 = - r^1 = - r^ . (F.1.2) In Frame S mass m 1 is constrained to lie on a sphere of radius r 1, while mass m 2 is constrained to lie on a sphere of radius r 2. The picture assumes m 1 < m2 so r1 > r2. If m1 lies at a point on its sphere, m 2 lies on the inverse point but on its sphere, as the spherical coordinates above show. If the masses are the same, the two spheres coincide. Due to these constraints on the vectors r1 and r2, the corresponding velocity vectors v1 and v2 of the two masses must be tangential to their respec tive spheres. Thus, for example, for mass m 1 we can write v1 = v1θ θ^ + v1φ φ^ , whereas r1 = r1 r^ . (F.1.3) Some Basic Kinematic Facts The internal angles of the red triangle are β1, α1 and π-θ at the Frame S origin. Thus we know from a Law of Cosines that r'12 = b2 + r12 - 2br1cos(π-θ) = b2 + r12 + 2br1cosθ . The internal angles of the blue triangle are β 2, α2 and θ at the Frame S origin. Thus, r' 22 = b2 + r22 - 2br2cosθ . The three Laws of Sines for the two triangles tells us that sinβ1 b = sinα1 r1 = sin(π-θ) r1' = sinθ r1' // red triangle sinβ2 b = sinα2 r2 = sinθ r'2 . // blue triangle Looking at the drawing it is clear that b + r1 = r'1 and b + r2 = r2', so one can write r'2 - r2 = r'1 - r1 = b . Recall the center of mass condition from (D.2.8) that m 1r1 = - m2r2 ⇒ m1r1 = m2r2 , r2/r1 = m1/m2 , r^2 = – r^ . (D.2.8) Appendix F: The Dumbbell Satellite 240 Applying the Frame S time derivative ∂S gives similar results for r → v and r → a, see below (velocity and acceleration). Summary: r' 12 = b2 + r12 + 2br1cosθ ( F . 1 . 4 ) r'22 = b2 + r22 - 2br2cosθ ( F . 1 . 5 ) sinβ1 b = sinα1 r1 = sinθ r1' // red triangle (F.1.6) sinβ2 b = sinα2 r2 = sinθ r'2 // blue triangle (F.1.7) r'2 - r2 = r'1 - r1 = b r'1 = b + r1 r'2 = b + r2 (F.1.8) m1r1 = - m2r2 ⇒ m1r1 = m2r2 , r 2/r1 = m1/m2 , r^2 = – r^1 (F.1.9) m 1v1 = - m2v2 ⇒ m1v1 = m2v2 , v 2/v1 = m1/m2 , v^2 = – v^1 (F.1.10) m1a1 = - m2a2 ⇒ m1a1 = m2a2 , a 2/a1 = m1/m2 , a^2 = – a^1 (F.1.11) F.2 Angular momentum of the satellite and its time derivative in Frame S Our upcoming path is to obtain the equati ons of motion for the satellite in terms of θ, φ coordinates. As a demonstration of the method of Section 11.3 we sha ll do this first using Newton's rotational second law with fictitious torques. Later in Section F.6 we will do this again using Newton's linear second law with fictitious forces as a demonstrati on of the method of Section 8.1. For angular momentum (and later torque) we take our reference point to be c = 0 in Frame S (the origin) and thus c' = b in Frame S'. Then (we show all detail in this first calculation), L(0) = r1 x p1 + r2 x p2 = m1 r1 x v1 + m2 r2 x v2 // dim = ML2/T = m1r1 x v1 + m2[-(m1/m2)r1] x [-(m1/m2)v1] // (F.1.9) and (F.110) = m1r1 x v1 + [m1r1] x [(m1/m2)v1] = m1{ r1 x v1 + (m1/m2) r1 x v1 } = m 1[ 1 + (m1/m2) ] r1 x v1 = m1[ (m2+m1)/m2 ] r1 x v1 = ( m 1/m2) M r1 x v1 = (m1/m2) Mr1 r^ x v1 // M ≡ m1+m2 (F.2.1) = ( m 1/m2) Mr1 r^ x (v1θθ^ + v1φφ^) = (m1/m2) Mr1 [ v1θ r^ x θ^ + v1φ r^ x φ^ ] // (F.1.3) Appendix F: The Dumbbell Satellite 241 = (m 1/m2) M r1 (v1θ φ^ – v1φ θ^) // (E.2.12) = (m 1/m2) M r12 ( θ• φ^ – φ• sinθ θ^) . // (E.3.5) dim=ML2/T 3 The dumbbell has no angular momentum around the r^ axis which seems very reasonable since it consists of two point masses aligned with r^. (A real tethered satellite has non -point masses and one could imagine undesired torsion oscillation being a problem.) Taking a ∂ S time derivative gives the rate of change of angular momentum in Frame S (all variables here are the "natural" ones in Frame S ), L•(0) = ∂S{ (m1/m2) M r1 x v1 } // (F.2.1) = (m 1/m2) M ( v1 x v1 + r1 x a1 ) = (m 1/m2) Mr1 ( r^ x a1 ) = (m1/m2) Mr1 r^ x [ ar r^ + aθθ^ + aφ φ^ ] // (E.3.6) = (m1/m2) Mr1 [ aθ φ^ - aφ θ^ ] // (E.2.12) and then (E.3.6) for next line = (m 1/m2) Mr12 [ (θ••- φ•2 sinθ cosθ) φ^ – (2θ• φ• cosθ + φ••sinθ) θ^ ] . Here are the conclusions so far: L(0) = (m1/m2) M r12 ( θ• φ^ – φ• sinθ θ^) ( F . 2 . 2 ) L•(0) = (m1/m2) Mr12 [ (θ••- φ•2 sinθ cosθ) φ^ – (2θ• φ• cosθ + φ••sinθ) θ^ ] . (F.2.3) One obvious statement can be made looking at these equations: there is no angular momentum about the r^ axis and this vanishing angu lar momentum never changes. The equation of motion for the satellite within Fram e S is given by (11.3.4), but converted to swap notation, N(0) eff = L•(0) . (11.3.4)s (F.2.4) Our next task then is to compute the total effectiv e torque on the satellite in Frame S which from (11.3.5) is (again converted to swap notation), N(0) eff = N'(b) + N(0) fict . (11.3.5)s (F.2.5) Appendix F: The Dumbbell Satellite 242 Here N'(b) is the Frame S' torque on the satellite relative to point b in Frame S', and N(0) fict is the fictitious torque that arises because Fram e S is a rotating frame of reference. F.3 The torque on the Dumbbell Satellite in Frame S' The torque (in Frame S') of the Earth on the satellite ( relative to the Frame S origin) is given by N'(b) = r1 x F1 + r2 x F2 / / d i m = L3 M /T (F.3.1) where all four of these vectors are shown in Fig (F.1.1). Recall from (D.2.13) that F1 = - (GM Em1/r'12) r^'1 = - (GM Em1/r'13) r'1 F1 ≡ |F1| = (GMEm1/r'12) F2 = - (GM Em2/r'22) r^'2 = - (GM Em2/r'23) r'2 F2 ≡ |F2| = (GMEm2/r'22) . (F.3.2) // The stick tension T exerts no torque on the satellite masses since T is collinear with r1 and r2. Therefore, r1 x F1 = r1 x [ - (GMEm1/r'13) r'1 ] = - (GM Em1/r'13) r1 x r'1 r2 x F2 = r2 x [ - (GMEm2/r'23) r'2 ] = - (GM Em2/r'23) r2 x r'2 . (F.3.3) Using (F.1.8) we evaluate the cross products making use of the first line of (E.2.15) for r^ x z^ , r1 x r'1 = r1 x (b + r1) = r1 x b = r1b r^ x z^ = r1b (-sinθ φ^) = -r1bsinθ φ^ r2 x r'2 = r2 x (b + r2) = r2 x b = [-r2r^] x [b z^] = -r2br^ x z^ = -r2b(-sinθ φ^) = r2bsinθ φ^ . (F.3.4) Then inserting (F.3.4) into (F.3.3), r1 x F1 = (GM Em1/r'13) r1b sinθ φ^ ( F . 3 . 5 ) r2 x F2 = - (GM Em2/r'23) r2bsinθ φ^ = - (GM Em1/r'23) r1bsinθ φ^ . // (F.1.9) (F.3.6) Thus the Earth's torque on the satellite in Frame S' is, N'(b) = r1 x F1 + r2 x F2 = (GM Em1/r'13) r1b sinθ φ^ - (GMEm1/r'23) r1bsinθ φ^ = (GM Em1br1sinθ) [1/r'13 - 1/r'23] φ^ . (F.3.7) Appendix F: The Dumbbell Satellite 243 Note that this equation uses r1 and r2 and not r'1 and r'2 because the torque reference point is point b. We can confirm the two torque contributions by co mputing them geometrically from Fig (F.1.1) using these right-hand-rule helper drawings: (F.3.8) Then: r1 x F1 = r1F1 sin(π-β1) φ^ = r1F1sinβ1φ^ = r1F1(bsinθ/r'1) φ^ // (F.1.6) then (F.3.2) = r1(GMEm1/r'12)(bsinθ/r'1) φ^ = (GM Em1/r'13) r1b sinθ φ^ // agrees with (F.3.5) r2 x F2 = r2F2 sin(π-β2) φ^2 = r2F2 sinβ2 φ^2 = r2F2 sinβ2 [ -φ^ ] = r 2F2 (bsinθ/r'2) [ -φ^ ] = r2(GMEm2/r'22) (bsinθ/r'2) [ -φ^ ] // (F.1.7) and (F.3.2) = r 1(GMEm1/r'22) (bsinθ/r'2) [ -φ^ ] // (F.1.9) = - (GM Em1/r'23) r1bsinθ φ^ // agrees with (F.3.6) where we have used the fact that φ^2 = - φ^1 = - φ^ . Appendix F: The Dumbbell Satellite 244 There is a third way to compute the above torque, ba sed on the torque theorem (D.1.3) which says that N(R) = N(0) - R x F . In our current context of wo rking in Frame S' this reads N'(b) = N'(0) - b x F . ( F . 3 . 9 ) The torque N'(0) relative to the Frame S' origin is exactly 0 N'(0) = r' x F1 + r'2 x F2 = 0 + 0 = 0 since r'1 is collinear with F1 and r'2 is collinear with F2 as shown in Fig (F.1.1). We should include the stick tension/compression T, but ( r'1 - r'2) x T = (r1 - r2) x T = 0 - 0 = 0 so T can be ignored. Thus we find from (F.3.9) that N'(b) = - b x (F1+ F2) = - b x[ - (GM Em1/r'13) r'1 - (GMEm2/r'23) r'2 ] = (GM Em1/r'13) b x r'1 + (GMEm2/r'23) b x r'2 = (GM Em1/r'13) (r'1-r1) x r'1 + (GMEm2/r'23) (r2' - r2) x r'2 // (F.1.8) = - (GM Em1/r'13) r1 x r'1 - (GM Em2/r'23) r2 x r'2 = - (GM Em1/r'13)[-r1bsinθ φ^] - (GM Em2/r'23) [ r2bsinθ φ^ ] // (F.3.4) then (F.1.9) = ( G M Em1br1sinθ) [1/r'13 - 1/r'23] φ^ ( F . 3 . 1 0 ) in agreement with (F.3.7). Here are some quick checks on (F.3.10) : • If the satellite is vertically aligned, θ = 0,π , then sinθ=0 and N'(b) = 0 as expected (no moment arms). • If the satellite is horizontally aligned and m1= m2, then r'1 = r'2 so N'(b) = 0 (balanced moment arms). • If m 2> m1, then r2< r1 so for horizontal alignment one has r' 2<r'1 so [ (1/r' 13) - (1/r'23) ] < 0. Then if m 1 is on the right, we have θ = π/2 and sin θ = 1 and then (F.3.10) has N(b) = -(positive) φ^. But for this orientation φ^ = φ^1 = - x^ so N(b) = (positive) x^ : Appendix F: The Dumbbell Satellite 245 (F.3.11) In this case the lever arms balance in the sense that m 2r2 = m1r1, but m2 is closer to Earth center so it feels the stronger force and so we expect N'(b) = (positive) x^ . Far Approximation If we now assume as before that r 1',r'2,b >> r1, r2 we can approximate [ (1/r1'3) - (1/r'23) ] by adding on to the Maple code shown in (D.3.5) to get where recall e1 = ε1 ≡ (r1/b). This time there is a leading linear term in ε1 so (1/r'13) - (1/r'23) ≈ 3 cosθ b3(-μ2) (r1/b) = - 3r 1cosθ/ (μ2b4) . (F.3.12) Installing this result into (F.3.7) gives N'(b) = GM Em1b r1sinθ [ (1/r'13) - (1/r'23)] φ^ (F.3.7) ≈ - GM Em1b r1sinθ ( 3r1cosθ/ (μ2b4) ) φ^ = - 3GME(m1/μ2) b-3r12sinθcosθ φ^ . ( F . 3 . 1 3 ) Appendix F: The Dumbbell Satellite 246 To this order of approximation, the torque N(b) vanishes when the satellite is horizontally aligned as well as when it is vertically alig ned, and this is due to r 1' ≈ r'2 in the horizontal case. F.4 The fictitious torque on the satellite in Frame S Recall the general fictitious torque expression given in (11.3.10), acting on a single particle of mass m, N'(c') fict = - ( r'-c') x [ mb•• S +mω x (ω x r') + 2m ω x v' + mω• x r' ] + m (c•' + ω x c' + b• S) x ( v' + ω x r' + b• S) – mc•' x v' . (11.3.10) Converted to swap notation this says, N(c) fict = - ( r-c) x [ mb•• S' +mω x (ω x r) + 2m ω x v + mω• x r ] + m (c• + ω x c + b• S') x ( v + ω x r + b• S') – mc• x v . (11.3.10) s (F.4.1) Here ω is the angular rotation rate of the satellite about the Earth. Our application has the torque center at c = 0 (and c' = b) so this simplifies somewhat to N(0) fict = - m r x [ b•• S' +ω x (ω x r) + 2 ω x v + ω• x r] + b• S' x ( mv + ω x [m r] ) . (F.4.2) frame centrifugal Coriolis Euler Recall from (8.1.8) that the square bracket in the above is - Ffict/m so we can trace the origin of the terms. Now write (F.4.2) separately for each of the two masses of the satellite : N(0) f,1 = - m1r1 x [ b•• S' + ω x (ω x r1) + 2ω x v1 + ω• x r1] + b• S' x ( m1v1 + ω x[m1r1]) N(0) f,2 = - m2r2 x [ b•• S' + ω x (ω x r2) + 2ω x v2 + ω• x r2] + b• S' x ( m2v2 + ω x [m2r2]) . (F.4.3) In the second line, use (F.1.9) to replace m 2r2 = - m1r1 and (F.1.10) to replace m 2v2 = - m1v1 : N(0) f,2 = + m1r1 x [ b•• S' + ω x (ω x r2) + 2ω x v2 + ω• x r2] + b• S' x ( -m1v1 + ω x [-m1r1]) . (F.4.4) Next, add the two torques to get the total fictiti ous torque on the satellite seen in Frame S, Appendix F: The Dumbbell Satellite 247 N(0) fict = N(0) f,1 + N(0) f,2 = - m 1r1 x [ b•• S' + ω x (ω x r1) + 2ω x v1 + ω• x r1] + b• S' x ( m1v1 + ω x[m1r1]) + m 1r1 x [ b•• S' + ω x (ω x r2) + 2ω x v2 + ω• x r2] + b• S' x ( -m1v1 + ω x [-m1r1]) = - m 1r x { ω x (ω x [r1- r2]) + 2ω x [v1-v2] + ω• x [r1- r2] } centrifugal Coriolis Euler = - m 1r x { ω x (ω x [r1 + m1 m2 r1]) + 2ω x [v1 + m1 m2 v1] + ω• x [r1 + m1 m2 r1] } = - m 1 (1+m1 m2 ) r1 x { ω x (ω x r1) + 2ω x v1 + ω• x r1 } = - m1 m2 M r1 x { ω x (ω x r1) + 2ω x v1 + ω• x r1 } . (F.4.5) All terms involving b• S' and b•• S' have cancelled out. For the Earth orbit we have ω• = 0 and ω = ωx^. Now consider this vector identity: C x [A x (A x C)] = C x [ ( A•C)A - A2C ] = (A•C) C x A = – (A•C) A x C . (F.4.6) Then r1 x [ ω x (ω x r1) ] = – (ω•r1) (ω x r1) = - ω2r12 ( x^ • r^)( x^ x r^) // use (E.2.4) for x^ • r^ and (E.2.13) for x^ x r^ = - ω2r12 ( sinθcosφ )(- cosθcosφ φ^ - sinφ θ^ ) = ω 2r12 sinθcosφ (cosθcosφ φ^ + sinφ θ^ ) . (F.4.7) Next, the Coriolis term in (F.4.5) involves, r1 x (ω x v1) = ( r1•v1)ω - (r1•ω)v1 // A x (B x C) = ( A•C)B - (A•B)C . (F.4.8) But v1 is tangent to the radius-r 1 sphere to which m 1 is constrained, so ( r1•v1) = 0. Then r1 x (ω x v1) = - ( r1•ω)v1 = - r1ω (r^•x^)v1 = - r1ω sinθcosφ v1 . (F.4.9) We now have this somewhat complicated expression for N(0) fict : Appendix F: The Dumbbell Satellite 248 N(0) fict = - (m1/m2)M r1 x [ ω x (ω x r1) + 2ω x v1 ] = - ( m 1/m2)M { r1 x [ω x (ω x r1)] + 2 r1 x (ω x v1) } = - ( m 1/m2)M { ω2r12 sinθcosφ [ cosθcosφ φ^ + sinφ θ^ ] - 2r1ω sinθcosφ v1 } (F.4.10) centrifugal Coriolis Now using (E.3.5), v1 = v1θθ^ + v1φφ^ = r1 θ• θ^ + r1 sinθ φ• φ^ (E.3.5) we then have, centrifugal Coriolis N(0) fict = - (m1/m2)M{ω2r12sinθcosφ [cosθcosφ φ^ + sinφ θ^] - 2r1ω sinθcosφ (r1θ• θ^ + r1 sinθ φ• φ^ ) } = - ( m 1/m2)M { (ω2r12 sinθcosφsinφ - 2r12ωsinθcosφθ•) θ^ + ( ω2r12 sinθcosφ cosθcosφ - 2r12ωsinθcosφsinθφ• ) φ^ } = - ( m 1/m2)M { ωr12 sinθ cosφ (ωsinφ -2θ•) θ^ +ωr12sinθcosφ (ω cosθ cosφ - 2sinθ φ•) φ^ } = - ( m 1/m2)Mωr12sinθcosφ [ (ωsinφ -2θ•) θ^ + (ω cosθ cosφ - 2sinθ φ•) φ^ ] . (F.4.11) How might one interpret this simple result? We examine the pieces of N(0) fict= 0 as they appear in (F.4.11). Terms involving velocities θ• and φ• arise from the Coriolis term, the other terms come from the centrifugal term. (a) the "frame effects" due to the motion of b (b• S' and b•• S') are equal and opposite for the two masses because the origin of Frame S is at the center of mass causing m 2r2 = - m1r1, as shown in (F.1.9). (b) within Frame S, the (F.4.5 ) centrifugal acceleration term ω x (ω x r1) = -ω2r1 tries to push m 1 to a larger radius. But r1 is constrained to lie on a sphere of radius r 1 so m1 cannot go to a larger radius. This centrifugal acceleration is neutralized by part of the tension in the stick which we avoided talking about. This centrifugal term, by the way, involves "the short vector" r1 and not the long vector r'1. We discussed this situation in Section 8.2 for an Earth-ba sed Frame S'. In our current context, the picture that corresponds to Fig (8.2.10) is the following Appendix F: The Dumbbell Satellite 249 (F.4.12) The arrow in each location during the orbit represents the position vector r1 of mass m 1 where we assume that other effects are turned off so r1 stays fixed in Frame S. When th ese arrows are transferred to the picture on the right with common tails, we see that the tip of r1 does in fact go around in a circle of radius r1 and that is why the corresponding centrifugal force acting on m 1 is -ω2r1 . (c) Within Frame S, the Coriolis force - 2m 1 ω x v1 tries to deflect mass m 1 "to the right" in Fig (F.1.1). But again, r1 is constrained to lie on a sphere of radius r 1 so m1 cannot deflect to a different radius. This Coriolis force is neutralized by the rest of the tension in the stick. F.5 Equations of Motion for the satellite in Frame S (Spherical Coordinates) After much effort, we have arrived at th is set of results for the satellite : L(0) = (m1/m2) M r12 ( θ• φ^ – φ• sinθ θ^) (F.2.2) L•(0) = (m1/m2) Mr12 [ (θ••- φ•2 sinθ cosθ) φ^ – (2θ• φ• cosθ + φ••sinθ) θ^ ] . (F.2.3) N(b) = (GM Em1br1sinθ) (1/r'13 - 1/r'23) φ^ (F.3.7) r'12 = b2 + r12 + 2br1cosθ (F.1.4) r'22 = b2 + r22 - 2br2cosθ (F.1.5) N(0) fict = - (m1/m2)Mωr12 sinθcosφ [ (ωsinφ -2θ•) θ^ + (ω cosθ cosφ - 2sinθ φ•) φ^ ] . (F.4.11) Appendix F: The Dumbbell Satellite 250 Using the orbit equation (8.6.7 ) applied to the satellite, GME = ω2b3 , // larger b means smaller ω ( F . 5 . 1 ) we can rewrite the true torque above as N'(b) = (ω2b4m1r1sinθ) (1/r'13 - 1/r'23) φ^ . (F.5.2) Writing L•(0) = N'(b) + N(0) fict then gives the satellite vector equation of motion, (m1/m2) Mr12 [ (θ••- φ•2 sinθ cosθ) φ^ – (2θ• φ• cosθ + φ••sinθ) θ^ ] = ( ω2b4m1r1sinθ) (1/r'13 - 1/r'23) φ^ - ( m 1/m2)Mωr12 sinθcosφ [ (ωsinφ -2θ•) θ^ + (ω cosθ cosφ - 2sinθ φ•) φ^ ] . (F.5.3) Divide all three terms by the factor (m 1/m2)Mr12. The coefficient of the first term on the right becomes (ω 2b4m1r1sinθ) / [ (m1/m2)Mr12] = (1/r1) (ω2b4(m2/M)sinθ) so the vector equation of motion is then, [ (θ••- φ•2 sinθ cosθ) φ^ – (2θ• φ• cosθ + φ••sinθ) θ^ ] = ( 1 / r 1)(ω2b4μ2sinθ) (1/r'13 - 1/r'23) φ^ - ωsinθcosφ [ (ωsinφ -2θ•) θ^ + (ω cosθ cosφ - 2sinθ φ•) φ^ ] . (F.5.4) Using now the far approximation (F.3.12) that (1/r' 13) - (1/r'23) = -3r 1cosθ/ (μ2b4), we get [ (θ••- φ•2 sinθ cosθ) φ^ – (2θ• φ• cosθ + φ••sinθ) θ^ ] = - 3 ω2sinθcosθ φ^ - ω sinθcosφ [ (ωsinφ -2θ•) θ^ + (ω cosθ cosφ - 2sinθ φ•) φ^ ] (F.5.5) As a reminder, the first line above is L•(0), the second line is true torque N'(b), and the last line is the fictitious torque N(0) fict created by the fact that Frame S is a rotating frame of reference, and we display the fictitious contributions in blue to keep track of them for a while below. Appendix F: The Dumbbell Satellite 251 Comment : Notice that the equation of motion is independent of m 1 and m2 and hence of r 1 and r2. If we were to vary the ratio m 1/m2, we just "slide the stick" in Fig (F.1.1) so the Frame S origin remains at the center of mass point. The equa tion does depend on b through ω, since ω2 = GME/b3. Moving all terms to the left side, (F.5.5) becomes (θ••- φ•2 sinθ cosθ) φ^ – (2θ• φ• cosθ + φ••sinθ) θ^ (F.5.6) + 3 ω2sinθcosθ φ^ + ωsinθcosφ (ωsinφ -2θ•) θ^ + ωsinθcosφ (ω cosθ cosφ - 2sinθ φ•) φ^ = 0 . The component equations are then θ••- φ•2 sinθ cosθ + 3ω2sinθcosθ + ωsinθcosφ (ω cosθ cosφ - 2sinθ φ•) = 0 // φ^ – (2θ• φ• cosθ + φ••sinθ) + ω sinθcosφ (ωsinφ -2θ•) = 0 / / θ^ where the fictitious torque terms are shown in blue . Changing the sign of the second equation and making a few adjustments we get, θ•• + sinθcosθ(3ω2 - φ•2) + ωsinθcosφ (ω cosθ cosφ - 2sinθ φ•) = 0 // φ^ φ•• + 2θ• φ• cotθ – ωcosφ (ωsinφ -2θ•) = 0 / / θ^ (F.5.7) After much effort using the fictitious tor que method we have finally arrived at the spherical equations of motion for the dumbbell satellite! These are two ordinary differential equations with ti me t as the variable. The equations are 2nd order in both θ(t) and φ(t) and they are non-linear due to terms like θ•φ• and φ•2 and sinθ . Finally, the equations are coupled, so they form a system of two 2nd order, coupled, non-linear ODE's. This system of two non- linear 2nd order ODE's can be trivially replaced with an equivalent system of four non-linear 1st order ODE's as follows, v• θ + sinθcosθ(3ω2 - φ•2) + ωsinθcosφ (ω cosθ cosφ - 2sinθ φ•) = 0 // φ^ v•φ + 2 cotθ θ• φ• – ω cosφ (ωsinφ -2θ•) = 0 / / θ^ θ• = vθ φ• = vφ ( F . 5 . 8 ) We mention this only because this is the first step Maple takes (internally) when it sets about solving a pair of 2nd order ODE's with its num erical dsolve command (coming soon). Right now, we make an anzats that (F.5.7) has a solution for which φ = π/2 (so cos φ = 0) and φ does not change, so that both φ• and φ•• = 0 at all times. For such a solution, the dumbbell lies in the plane of paper Appendix F: The Dumbbell Satellite 252 of Fig (F.1.1) (in the plane of the orbit) and the only Frame S motion is in the θ degree of freedom. In this case the two equations in (F.5.7) simplify to θ•• + 3ω2sinθcosθ = 0 / / φ^ 0 = 0 . // in-plane // θ^ (F.5.9) For small θ, the first equation becomes θ•• + 3ω2θ = 0 which indicates sinusoidal oscillation in θ of the dumbbell about θ = 0 with frequency ωosc2 = 3ω2 so ωosc1 = 3 ω T osc1 = (1/ 3 ) T ≈ 0.58 T // in-plane libration (F.5.10) where recall that ω is the orbital rotation rate of the satellite. For a low-Earth orbit with T = 88 minutes, the dumbbell initialized to φ = π/2 and a small angle θ would have an oscillation period of 88*.58 = 51 minutes. Next we look for a solution with φ = φ• = 0 where the satellite at t = 0 is in a plane perpendicular to the plane of paper in Fig (F.1.1) (perp to the orbit plane). In this case the equations (F.5.7) become, θ•• + 3ω2sinθcosθ + ωsinθ(ω cosθ) = 0 / / φ^ sinθ φ•• – ω sinθ ( -2θ•) = 0 / / θ^ or θ•• +(2ω) 2sinθcosθ = 0 / / φ^ φ•• +2ωθ• = 0 . / / out-of-plane // θ^ (F.5.11) For small angles θ we have then θ•• +(2ω)2θ = 0 / / φ^ φ•• = -2ωθ• . / / θ^ (F.5.12) The first equation implies θ oscillation at frequency ωosc2 = 2ω T osc2 = 0.5T // out-of-plane libration (F.5.13) while the second equation shows that th e dumbbell cannot remain very long at φ = 0 since φ•• ≠ 0, so this oscillation solution is only a temporary solution and the dumbbell is not stable in the plane φ = 0. Both libration frequencies appear on page 126 of Cosm o and Lorenzini with a reference (on C&L p 169) to the following item, Appendix F: The Dumbbell Satellite 253 2. Beletskii, V. V. and Levin, E. M., "Dynamics of Space Tether Systems", Advances in the Astronautical Sciences , Vol. 83. (Univelt, Inc. 1993) Reader Exercise : Is the in-plane libration solution stable against small perturbations? F.6 Force analysis of the satellite in Frame S (Spherical Coordinates) Having obtained the dumbbell satellite θ,φ equations of motion in (F.5.7) using the fictitious torques method. we now set out to rederive these same equa tions using Newton's linear second law with fictitious forces . This time we also obtain an expression for the stick tension T which enables us to comment on the tidal forces for a static satellite positioned at θ = 0. Obtain the three equations of motion The only true forces on a dumbbell mass are gravity and stick tension T. From (F.3.2) we then write F'1 = - (GM Em1/r'13) r'1 - T r^1 = - (GM Em1/r'13)( b + r1) - T r^1 F'2 = - (GM Em2/r'23) r'2 - T r^2 = - (GM Em2/r'23)( b + r2) - T r^2 . (F.6.1) As noted earlier, the center of gravity is not quite at the center of mass in Fig (F.1.1), but the above equations are exact despite this fact. The forces are primed because they are forces in the inertial Frame S'. The reader is reminded that we are using the "swap notation" where prime ↔noprime relative to the non-swap notation. In order to use Newton's Law in rotating Frame S, we must include the fictitious forces. We translate the result of (8.1.8) to swap notation to obtain Ffict,1 ≈ – m1b•• S' – m1ω x (ω x r1) – 2m1 ω x v1 – m1ω• x r1 Ffict,2 ≈ – m2b•• S' – m2ω x (ω x r2) – 2m2 ω x v2 – m2ω• x r2 . (F.6.2) frame centrifugal Coriolis Euler In these equations, the b acceleration is given by the swap version of (7.13) which then states b•• S' = ω• x b + ω x (ω x b) . // Special Case #1 (7.13)s (F.6.3) and the vectors b and ω are given as in Fig (F.1.1) by ω = ωx^ b = bz^ . ( F . 6 . 4 ) Finally we may state Newton's Law for each mass, Appendix F: The Dumbbell Satellite 254 Feff,1 = m1 a1 ( F . 6 . 5 ) ≈ - (GMEm1/r'13)r'1 - T r^1 – m1b•• S' – m1ω x (ω x r1) – 2m1 ω x v1 – m1ω• x r1 Feff,2 = m2a2 (F.6.6) ≈ - (GMEm2/r'23)r'2 - T r^2 – m2b•• S' – m2ω x (ω x r2) – 2m2 ω x v2 – m2ω• x r2 where we have now a set of six scalar equations. Using (F.1.9) through (F.1.11), (F.6.6) can be rewritten, Feff,2 = - m1a1 (F.6.7) ≈ - (GMEm2/r'23)r'2 + T r^1 – m2b•• S' + m1ω x (ω x r1) + 2m1 ω x v1 – m1ω• x r1 . Adding (F.6.5) and (F.6.7) gives 0 = - (GM Em1/r'13) r'1 - (GMEm2/r'23)r'2 - (m1+m2)b•• S' or (m1+m2)b•• S' = - (GM Em1/r'13) r'1 - (GMEm2/r'23)r'2 (F.6.8) This equation is just F = m a in inertial Frame S' for the total satellite where b is the center of mass. Ignoring the small offset between center of mass a nd center of gravity, the three equations (F.6.8) describe the circular orbit of the satellite around the Ea rth. We may then regard the equation (F.6.5) as a set of three scalar equations for the three unknowns θ ,φ and T where recall r1 = (r1,θ,φ) in the spherical coordinates of Fig (F.1.1). Our next task is to write vector equation (F.6.5) in spherical coordinates to obtain the three equations of motion. After expanding the left side, we then consid er the right side of (F.6.5) one term at a time: Feff,1 = m1 a1 (F.6.5) ≈ - (GMEm1/r'13)r'1 - T r^1 – m1b•• S' – m1ω x (ω x r1) – 2m1 ω x v1 – m1ω• x r1 1 2 3 4 5 6 Left side of (F.6.5): m 1 a1 = m1(ar r^ + aθθ^ + aφ φ^ ) (E.3.6) = m 1r1[( - θ•2 - φ•2 sin2θ) r^1 + ( θ•• - φ•2 sinθ cosθ) θ^ + (2 θ• φ• cosθ + φ••sinθ) φ^ ] (F.6.9) Term 1: - (GM Em1/r'13)r'1 = - (GM Em1/r'13)(b + r1) // (F.1.8) = - (GM Em1/r'13)(bz^ + r1r1^) // (F.6.4) = - (GM Em1/r'13)(bcosθ r^1 - b sinθ θ^ + r1r1^) // (E.2.7) = - (GM Em1/r'13)[ (bcosθ +r1) r^1 - b sinθ θ^ ] ( F . 6 . 1 0 ) Appendix F: The Dumbbell Satellite 255 Term 2: - T r^1 a s i s ( F . 6 . 1 1 ) Term 3: – m 1b•• S' = – m1ω• x b - m1 ω x (ω x b) // (F.6.3) = - m 1ω x (ω x b) // satellite in circular orbit, ω• = 0 = - m 1(ω•b)ω + m1ω2b //- A x (A x C) = -( A•C)A + A2C = m 1ω2b = m1ω2bz^ // (F.6.4) = m 1ω2b [ cosθ r^1 - sinθ θ^] // (E.2.7) (F.6.12) Term 4: – m 1ω x (ω x r1) = -m1(ω•r1)ω + m1ω2r1 // identity shown above = - m 1ω2r1(x^•r^1)x^ + m1ω2r1 r^1 // (F.6.4) = - m 1ω2r1sinθcosφx^ + m1ω2r1 r^1 // (E.2.4)) = - m 1ω2r1sinθcosφ[sinθcosφ r^1 + cosθcosφ θ^ - sinφ φ^] + m1ω2r1 r^1 // (E.2.7) = - m 1ω2r1 [ (sin2θcos2φ - 1) r^1 + (sinθ cosθcos2φ) θ^ + (- sinθcosφsinφ) φ^ (F.6.13) Term 5: -2m 1 ω x v1 = -2m1 [ωx^] x ( vθθ^ + vφφ^) // (E.3.5) = - 2 m 1ω [vθ x^ x θ^ + vφ x^ x φ^ ] = - 2 m 1ω [vθ (sinθcosφ φ^ + sinφ r^1)+ vφ(-sinθ cosφ θ^ + cosθcosφ r^1)] // (E.2.13) = - 2 m 1ω [ (vθsinφ + vφcosθcosφ)r^1 + (-vφsinθ cosφ) θ^ + (vθsinθcosφ) φ^ ] = - 2 m 1ωr1 [ (θ• sinφ + φ• sinθ cosθcosφ)r^1 + (-φ• sin2θ cosφ) θ^ + (θ• sinθcosφ) φ^ ] // (E.3.2) ( F . 6 . 1 4 ) Term 6: – m 1ω• x r1 = 0 because we assume ω• = 0 (F.6.15) Having all the bits and pieces, we now assemble the three component equations of (F.6.5). Appendix F: The Dumbbell Satellite 256 Feff,1 = m1 a1 (F.6.5) ≈ - (GMEm1/r'13)r'1 - T r^1 – m1b•• S' – m1ω x (ω x r1) – 2m1 ω x v1 – m1ω• x r1 1 2 3 4 5 6 r^1: m1r1( - θ•2 - φ•2 sin2θ) = - (GM Em1/r'13)(bcosθ +r1) - T + m 1ω2b cosθ - m1ω2r1(sin2θcos2φ - 1) 1 2 3 4 - 2 m 1ωr1 (θ• sinφ + φ• sinθ cosθcosφ) ( F . 6 . 1 6 ) 5 θ^: m1r1 ( θ•• - φ•2 sinθ cosθ) = + (GM Em1/r'13) b sinθ - m1ω2bsinθ 1 3 - m 1ω2r1 sinθcosθcos2φ + 2m1ωr1φ• sin2θcosφ ( F . 6 . 1 7 ) 4 5 φ^: m1r1(2 θ• φ• cosθ + φ••sinθ) = + m1ω2r1sinθcosφsinφ - 2m1ωr1(θ• sinθcosφ) (F.6.18) 4 5 We now rewrite the three equations dividing by m 1 and using (F.5.1) that GM E = ω2b3 : r^1: r1( - θ•2 - φ•2 sin2θ) = - ( ω2b3/r'13)(bcosθ +r1) - T/m1 + ω2b cosθ - ω2r1(sin2θcos2φ - 1) - 2 ωr1 (θ• sinφ + φ• sinθ cosθcosφ) ( F . 6 . 1 9 ) θ^: r1 ( θ•• - φ•2 sinθ cosθ) = + (ω2b3/r'13) b sinθ - ω2bsinθ - ω2r1 sinθcosθcos2φ + 2ωr1φ• sin2θcosφ ( F . 6 . 2 0 ) φ^: (2 θ• φ• cosθ + φ••sinθ) = + ω2sinθcosφsinφ - 2ω(θ• sinθcosφ) (F.6.21) If one uses (F.1.4) that r' 12 = b2 + r12 + 2br1cosθ in (F.6.20), the pair of equations (F.6.20) and (F.6.21) can in theory be solved for θ (t) and φ(t), given appropriate initial cond itions. The solutions can then be inserted into (F.6.19) to obtain a result for the stick tension T(t). Verify the angular equations of motion We can rewrite(F.6.21) as φ^: sinθ φ•• + 2 θ• φ• cosθ - ωsinθcosφ (ωsinφ - 2θ•) ( F . 6 . 2 2 ) which matches the θ^ torque equation (F.5.7), sinθ φ•• + 2θ• φ• cosθ – ω sinθcosφ (ωsinφ -2θ•) = 0 . // θ^ (F.5.7) Next, the θ^ equation (F.6.20) may be rewritten Appendix F: The Dumbbell Satellite 257 r1 ( θ•• - φ•2 sinθ cosθ) = + (ω2b3/r'13) b sinθ - ω2bsinθ - ω2r1 sinθcosθcos2φ + 2ωr1φ• sin2θcosφ or r 1 ( θ•• - φ•2 sinθ cosθ) = +ω2bsinθ [(b/r'1)3 - 1] - ω2r1 sinθcosθcos2φ + 2ωr1φ• sin2θcosφ or θ•• - φ•2 sinθ cosθ = +ω2bsinθ [(b/r'1)3 - 1]/r1 - ω2sinθcosθcos2φ + 2ωφ• sin2θcosφ . (F.6.23) Now assume the far approximation where r' 1, b >> r. Recall that r' 12 = b2 + r12 + 2br1cosθ (r' 1/b)2 = 1 + (r 1/b)2 + 2(r1/b)cosθ (r' 1/b)3 = [ 1 + (r 1/b)2 + 2(r1/b)cosθ ]3/2 (b/r' 1)3 = [ 1 + (r 1/b)2 + 2(r1/b)cosθ ]-3/2 ≈ 1 + (-3/2) [(r 1/b)2 + 2(r1/b)cosθ ] ≈ 1 + (-3/2) 2(r 1/b)cosθ = 1 - 3(r 1/b)cosθ so [(b/r' 1)3 - 1] ≈ - 3(r1/b)cosθ . ( F . 6 . 2 4 ) Then (F.6.23) becomes θ•• - φ• 2 sinθ cosθ = +ω2bsinθ [- 3(r1/b)cosθ]/r1 - ω2sinθcosθcos2φ + 2ωφ• sin2θcosφ or θ•• - φ•2 sinθ cosθ = -3ω2sinθcosθ - ω2sinθcosθcos2φ + 2ωφ• sin2θcosφ or θ•• - φ• 2 sinθ cosθ + 3ω2sinθcosθ + ω2sinθcosθcos2φ -2ωφ• sin2θcosφ = 0 or θ•• + 3ω 2sinθcosθ - φ•2 sinθ cosθ + ωsinθcosφ(ωcosθcosφ - 2φ•sinθ) = 0 (F.6.25) which matches the φ^ torque equation (F.5.7), θ•• + 3ω2sinθcosθ - φ•2 sinθ cosθ + ωsinθcosφ (ω cosθ cosφ - 2sinθ φ•) = 0 // φ^ (F.5.7) At this point we have derived the same angular equations of motion (F.5.7) in two different ways. Appendix F: The Dumbbell Satellite 258 Obtaining the tension in the stick (or tether) Finally we come to the r^1 equation (F.6.19), r^1: r1( - θ•2 - φ•2 sin2θ) = - ( ω2b3/r'13)(bcosθ +r1) - T/m1 + ω2b cosθ - ω2r1(sin2θcos2φ - 1) - 2 ωr1 (θ• sinφ + φ• sinθ cosθcosφ) . (F.6.19) Inserting the far approximation (F.6.24) that (b/r' 1)3 ≈ 1 - 3(r1/b)cosθ into the above gives r1( - θ•2 - φ•2 sin2θ) = -ω2[1 - 3(r1/b)cosθ] (bcosθ +r1) - T/m1 + ω2b cosθ - ω2r1(sin2θcos2φ - 1) - 2 ωr1 (θ• sinφ + φ• sinθ cosθcosφ) r1( - θ•2 - φ•2 sin2θ) = -ω2(bcosθ +r1) + 3ω2(r1/b)cosθ(bcosθ +r1) - T/m1 + ω2b cosθ - ω2r1(sin2θcos2φ - 1) - 2ωr1 (θ• sinφ + φ• sinθ cosθcosφ) r1( - θ•2 - φ•2 sin2θ) = -ω2bcosθ - ω2r1 + 3ω2r1cosθ(cosθ +r1/b) - T/m 1 + ω2b cosθ - ω2r1sin2θcos2φ + ω2r1 - 2ωr1 (θ• sinφ + φ• sinθ cosθcosφ) r 1( - θ•2 - φ•2 sin2θ) = 3ω2r1cosθ(cosθ +r1/b) - T/m 1 - ω2r1sin2θcos2φ - 2ωr1(θ• sinφ + φ• sinθ cosθcosφ) ( - θ• 2 - φ•2 sin2θ) ≈ 3ω2cos2θ - T/(m1r1) - ω2sin2θcos2φ - 2ω(θ• sinφ + φ• sinθ cosθcosφ) ( - θ• 2 - φ•2 sin2θ) ≈ ω2(3cos2θ -sin2θcos2φ) - T/(m 1r1) - 2ω(θ• sinφ + φ• sinθcosθcosφ) . The stick tension is then given by (this will later be verified in Section F.9), T = m 1r1[ ω2(3cos2θ - sin2θcos2φ) + θ•2 + φ•2 sin2θ - 2ω(θ• sinφ + φ• sinθcosθcosφ) ] . (F.6.26) If the angular velocities are very small such that | θ• | << ω and | φ• | << ω, the result becomes T ≈ m1ω2r1(3cos2θ - sin2θcos2φ) . // small velocities (F.6.27) In Cartesian coordinates this becomes, T ≈ (m 1ω2/r1)(3r12cos2θ- r12sin2θcos2φ) = (m 1ω2/r1)(3z2-x2) . // small velocities (F.6.28) In this small-velocity limit, the tensi on in the stick is positive as long as Appendix F: The Dumbbell Satellite 259 |x| < 3 |z| ⇒ T > 0 // small velocity limit (F.6.29) so for small x displacements T is always positive. As noted, in general one must solve (F.6.20) and (F.6.21) for θ (t) and φ(t), then (F.6.26) gives T(t). Tidal Force If the dumbbell is static at θ = 0, we see from (F.6.26) or (F.6.27) that T = 3m 1ω2r1 ( F . 6 . 3 0 ) which we associated with a "tidal force". The factor of 3 arises from (F.6.24) which in turn arises from the power 3 in the gravitational force factor in (F.6.1), (GM Em1/r'13) = (ω2b3 m1/r'13) = ω2m1 (b/r'1)3 . It happens that in Frame S' the gradient of the radial gravitational field at mass m 1 is ∂r1'(-GMEm1/r'12) = ∂r1'(-ω2b3m1/r'12) = -ω2b3m1 ∂r1' (r'1-2) = 2ω2b3m1/r'13 = 2 ω2m1(b/r'1)3 ( F . 6 . 3 1 ) so one can associate the factor of 3 in (F.6.30) with this gradient. However, the result (F.6.30) really comes from a sum of several terms in (F.6.5) for a static dumbbell, as we now review (the r^1 equation): Feff,1 = m1 a1 (F.6.5) ≈ - (GMEm1/r'13)r'1 - T r^1 – m1b•• S' – m1ω x (ω x r1) – 2m1 ω x v1 – m1ω• x r1 1 2 3 4 5 6 r^1: m1r1( - θ•2 - φ•2 sin2θ) = - (GM Em1/r'13)(bcosθ +r1) - T + m 1ω2b cosθ - m1ω2r1(sin2θcos2φ - 1) 1 2 3 4 - 2 m 1ωr1 (θ• sinφ + φ• sinθ cosθcosφ) (F.6.16) 5 m1r1( 0 - 0) = - (GM Em1/r'13)(b +r1) - T + m 1ω2b - m1ω2r1(0 - 1) - 2m 1ωr1 (0 + 0) 1 2 3 4 5 0 = - m 1(ω2b3/r'13)(b +r1) - T + m 1ω2b + m1ω2r1 1 2 3 4 0 = {- m 1ω2[1 - 3(r1/b)](b +r 1)} - T + m 1ω2b + m1ω2r1 1 2 3 4 Appendix F: The Dumbbell Satellite 260 0 = { - m 1ω2 - m1ω2b +3m1ω2r1 } - T + m 1ω2b + m1ω2r1 1 2 3 4 0 = 3m 1ω2r1 - T ( F . 6 . 3 2 ) 1,3,4 2 Thus in rotating Frame S the expression (F.6.30) for the tidal force T has contributions from the gravitational gradient (term 1) and from the "frame" term – m 1b•• S (term 3) which is the centrifugal contribution due to the acceleration of Frame S toward the Earth, and finally from the "local centrifugal term" – m 1ω x (ω x r1) (term 4). F.7 Numerical solutions of the equations of motion (Spherical Coordinates) The angular equations of motion for mass m 1 of the dumbbell (or tether) sa tellite are stated in (F.5.7) which we replicate here, θ•• + sinθcosθ(3ω 2 - φ•2) + ωsinθcosφ (ω cosθ cosφ - 2sinθ φ•) = 0 φ•• + 2θ• φ• cotθ – ωcosφ (ωsinφ -2θ•) = 0 . (F.5.7) (F.7.1) The first step is to enter these two equations into Maple: ( F . 7 . 2 ) For illustration purposes we have se t the satellite orbit frequency to ω = 1 sec -1 so ω = 1 ⇒ T = 2 π = 6 . 2 8 s e c Verification of in-plane libration The initial conditions are taken to be ( see Fig (F.1.1) to see that φ = π/2 is the in-plane situation ) θ = 0.2 φ = π/2 θ• = 0 φ• = 0.01 . (F.7.3) Appendix F: The Dumbbell Satellite 261 Here θ = 0.2 = 11.5o is a fairly small angle. We now call Mapl e's ODE solver routine dsolve and plot 40 times θ(t) in red and Mod2Pi( φ) in black. The peaks of the red curve are thus at 40*.2 = 8. Without this mod routine, the black φ curve just winds up without limit. (F.7.4a) red = 40*θ black = Mod2p( φ) Rather than plot a cosine wave, dsolve keeps θ positive all the time and has φ jump by π each time the solution passes through θ = 0. In spherical coordinates the figure shows the expected θ curve! In order to avoid exactly hitting θ = 0 which is a singular point in spherical coordinates ( φ is undefined there), we have added a small φ• = .001 to cause the dumbbell to slightly miss the z axis. One can see from the second equation in (F.7.1) that the numerical integrator is faced with φ•• = 2θ• φ• cotθ + stuff, and cotθ blows up at θ = 0 (and generates an error message in odepl ot). We can expand the region t = (2,4) : (F.7.4b) From (F.5.10) for in-plane libration one predicts, T ocs1 = T/ 3 = 6.28/1.73 = 3.63 sec (F.7.5) and this value is verified by the vertical line in the above figure (each tick is .04) Appendix F: The Dumbbell Satellite 262 Verification of out-of-plane libration The initial conditions are now taken to be θ = 0.2 φ = 0 θ• = 0 φ• = 0 (F.7.6) We then rerun the above code with a different set of "inits" : (F.7.7a) Now the swing misses θ = 0 of its own accord. One can see that the black φ curve starts moving away from φ = 0 at about t = 0.5 and in general the black φ curve has smooth rises near small θ. We again expand the region t = (2,4) : (F.7.7b) From (F.5.13) for out-of-plane libration one predicts, T ocs2 = T/2 = 6.28/2 = 3.14 (F.7.8) while the figure shows about 3.16, close enough. Maple can also plot θ•(t) and φ•(t) and here is a plot showing all four curves : Appendix F: The Dumbbell Satellite 263 (F.7.9) red = 10*θ black = φ green = 10* θ• blue = φ• When red θ nears θ = 0, blue φ• has major action since φ is quickly changing by π . And the green θ• of spherical coordinates has to make a radical change since it is in e ffect suddenly reversing course. A more general solution Here in addition to starting with θ(0) = 0.2 and φ(0) = 0, we provide a push in the azimuthal direction so that mass m 1 of the dumbbell satellite then swings around in azimuth while it oscillates in θ : θ = 0.2 φ = 0 θ• = 0 φ• = 2 . (F.7.10) (F.7.11) Notice that the azimuthal velocity slows down near the peaks of red θ(t). Energy is transferred back and forth between the θ and φ degrees of freedom in this system. One can use the same odeplot routine used above to make "orbital" plots in angle space, Appendix F: The Dumbbell Satellite 264 (F.7.12) Rather than produce more plots here, we shall defer to Section F.9 where we shall plot x(t) and y(t) instead of θ(t) and φ(t). F.8 Force analysis of the satellite in Frame S (Cartesian Coordinates) Section F.6 used the force analysis to develop equations of motion for the dumbbell satellite in spherical coordinates. Here we repeat that de velopment but in Cartes ian coordinates where things are in many ways simpler. We follow Section F.6 down Newton's Law (F.6.5) , Feff,1 = m1 a1 (F.6.5) (F.8.1) ≈ - (GMEm1/r'13)r'1 - T r^1 – m1b•• S' – m1ω x (ω x r1) – 2m1 ω x v1 – m1ω• x r1 . 1 2 3 4 5 6 As a reminder, this is Newton's Law ( Feff = m a) for mass m 1 of the satellite in Frame S where fictitious forces are included. Although this mass has coordinate r1 in Frame S, we shall refer to its components without 1 subscripts, so r1 = (x,y,z). This is similar to how we used r1 = (r1,θ,φ) in spherical coordinates where θ and φ had implied "1" subscripts. We also write v1 = v and a1 = a to reduce clutter. The quantity T is the tension in the stick (or tether). We now evaluate the six terms of (F.8.1) in Cartesia n coordinates, mimicking (F.6.9) through (F.6.14): Left side of (F.8.1): m 1a1 = m1a = m1(axx^ + ayy^ + azz^ ) (F.8.2) Term 1: - (GM Em1/r'13)r'1 = - (GM Em1/r'13)(b + r1) = - (GM Em1/r'13) [ bz^ + xx^ + yy^ + zz^ ] Appendix F: The Dumbbell Satellite 265 = - (GM Em1/r'13) [ xx^ + yy^ + (b+z) z^ ] ( F . 8 . 3 ) Term 2: - T r^1 = -(T/r 1)r1 = -(T/r 1) [ x x^ + yy^ + zz^] (F.8.4) Term 3: – m 1b•• S' = – m1ω• x b - m1 ω x (ω x b) // (F.6.3) = - m 1ω x (ω x b) // satellite in circular orbit, ω• = 0 = - m 1(ω•b)ω + m1ω2b //- A x (A x C) = -( A•C)A + A2C = m 1ω2b = m1ω2bz^ ( F . 8 . 5 ) Term 4: – m 1ω x (ω x r1) = -m1(ω•r1)ω + m1ω2r1 // identity shown above = - m 1ω2(x^•r1)x^ + m1ω2r1 = -m1ω2(x)x^ + m1ω2 [ xx^ + yy^ + zz^] = m 1ω2 [ yy^ + zz^] ( F . 8 . 6 ) Term 5: -2m 1 ω x v1 = -2m1 [ωx^] x ( vxx^ + vyy^ + vzz^) = - 2 m 1ω [ vyz^ - vzy^] ( F . 8 . 7 ) Term 6: – m 1ω• x r1 = 0 because we assume ω• = 0 (F.8.8) Having all the bits and pieces, we now assemble the three component equations of (F.8.1). The numbers show the Term above associated with each piece: Feff,1 = m1 a (F.8.1) ≈ - (GMEm1/r'13)r'1 - T r^1 – m1b•• S' – m1ω x (ω x r1) – 2m1 ω x v1 – m1ω• x r1 1 2 3 4 5 6 x^: m1 ax = - (GMEm1/r'13) x - (T/r 1)x 1 2 y^: m1 ay = - (GM Em1/r'13) y - (T/r 1)y + m1ω2y + 2m1ωvz 1 2 4 5 z^: m1 az = - (GM Em1/r'13) (b+z) - (T/r 1)z + m1ω2b + m1ω2z - 2m1ω vy (F.8.9) 1 2 3 4 5 Appendix F: The Dumbbell Satellite 266 We now rewrite the three equations dividing by m 1 and using (F.5.1) that GM E = ω2b3 . At the same time we replace velocity and acceleration components with dot notation components like x•• and y• : x•• = - [(ω2b3/r'13) + (T/m 1r1)]x y•• = - [(ω 2b3/r'13) + (T/m 1r1) - ω2]y + 2ωz• z•• = - [(ω 2b3/r'13) - ω2] (b+z) - (T/m 1r1)z - 2ω y• x2+y2+z2 = r12 ( F . 8 . 1 0 ) where r'12 = (r1+b)2 = r12 + b2 + 2 r1• b = r12 + b2 + 2 r1• [bz^] = r12 + b2 + 2bz . (F.8.11) Eq. (F.8.10) is a system of 4 equations in 4 unknow ns x,y,z,T. To simplify manipulations below, define A ≡ (ω2b3/r'13) B ≡ (T/m1r1) // rescaled tension so the system of equations becomes 1 x•• = - (A + B)x 2 y•• = - (A + B - ω2)y + 2ωz• 3 z•• = - (A - ω2) (b+z) - Bz - 2ω y• 4 x2+y2+z2 = r12 ( F . 8 . 1 2 ) We now wish to eliminate the rescaled tension B from the equation set. We first eliminate B between equations 1 and 2 : 1*y yx•• = - (A + B)xy 2*x xy•• = - (A + B - ω2)xy + 2ωxz• . Subtract so that the -(A + B)xy terms cancel, yx•• - xy•• = -ω 2xy - 2ωxz• . ( F . 8 . 1 3 ) Appendix F: The Dumbbell Satellite 267 Next, we eliminate B between equations 1 and 3: 1*z zx•• = - (A + B)xz 3*x x z•• = - (A - ω2)(b+z)x - Bxz - 2 ωxy• = -Abx -Azx + ω 2(b+z)x - Bxz - 2 ωxy• = - (A + B)xz - Abx + ω 2(b+z)x - 2ω xy• . Subtract so that the -(A + B)xy terms cancel, zx•• - x z•• = Abx - ω2(b+z)x + 2 ωxy• . ( F . 8 . 1 4 ) We now have a system of 3 equations in three unknowns x,y,z, where we now restore A = ( ω2b3/r'13) 1 yx•• - xy•• = - ω2xy - 2ωxz• 2 zx•• - x z•• = (ω 2b3/r'13)bx - ω2(b+z)x + 2 ωxy• where r' 12 = r12 + b2 + 2bz 3 x2+y2+z2 = r12 . ( F . 8 . 1 5 ) For convenience, we now reorder and rename e quations 1 and 2 of this set as follows: eq3 zx•• - x z•• = (ω 2b3/r'13)bx - ω2(b+z)x + 2 ωxy• eq2 yx•• - xy•• = - ω2xy - 2ωxz• . ( F . 8 . 1 6 ) Now apply the far approximation in equation eq3. From (F.6.24) we know that [(b/r' 1)3 - 1] ≈ - 3(r1/b)cosθ . (F.6.24) so (b/r'1)3 ≈ 1 - 3(z/b) // z = r 1cosθ and A ≡ (ω2b3/r'13) ≈ ω2[1 - 3(z/b) ] . (F.8.17) Equation eq3 above then becomes, zx•• - x z•• = (ω 2b3/r'13)bx - ω2(b+z)x + 2 ωxy• = ω 2[1 - 3(z/b) ]bx - ω2(b+z)x + 2 ωxy• = ω2bx - 3(z/b) ω2bx - ω2bx - ω2zx + 2ωxy• Appendix F: The Dumbbell Satellite 268 = - 3(z/b) ω2bx - ω2zx + 2ωxy• = - 3ω2xz - ω2zx + 2ωxy• = - 4 ω2xz + 2ωxy• . ( F . 8 . 1 8 ) This in the far approximation the equations of motion of mass m 1 of the dumbbell satellite are eq3 zx•• - x z•• = - 4ω2xz + 2ωxy• eq2 yx•• - xy•• = - ω 2xy - 2ωxz• x2+y2+z2 = r12 . ( F . 8 . 1 9 ) Once these three equations are solved for x(t), y(t) and z(t), we can find the tension T(t) from (say) the first equation of (F.8.12) : x••/x +A+B = 0 ⇒ B = - x••/x - A ⇒ (T/m 1r1) = - x••/x - (ω2b3/r'13) so T = - m 1r1[ x••/x + (ω2b3/r'13) ] so T = - m 1r1[ x••/x + ω2 ] . // far approximation (F.8.17) (F.8.20) F.9 Verification of the Cartesian equations of motion and stick tension In Section C.7 we showed that the x,y,z Foucault pe ndulum equations of motion were the same as the θ,φ ones. Here we repeat that task for the dumbbell satellite equations of motion. We do these verifications to strengthen our confidence in all the equations since ther e are not many external sources for verification. A trusting reader can just skip this section. First, here are the satellite angular equations of motion from (F.5.7), eq1 θ•• + sinθcosθ(3ω 2 - φ•2) + ωsinθcosφ (ω cosθ cosφ - 2sinθ φ•) = 0 eq2 φ•• + 2cotθ θ• φ• – ωcosφ (ωsinφ -2θ•) = 0 . (F.5.7) (F.9.1) These were derived in (F.5.7) using the effective torque method and were then verified using the effective force method in (F.6.22) and (F.6.25). Meanwhile, here are the Cartesian equa tions of motion from Section F.8, eq3 zx•• - xx•• = - 4ω 2xz + 2ωxy• eq2 yx•• - xy•• = - ω2xy - 2ωxz• . (F.8.19) (F.9.2) Appendix F: The Dumbbell Satellite 269 Below we shall show that Task (a): eq2 of (F.9.2) ⇒ eq2 of (F.9.1) Task (b): [(sin θ)*eq3 - (cos θsinφ)*eq2] of (F.9.2) ⇒ eq1 of (F.9.1) That is to say, angular eq1 of (F.9.1) is a certain linear combination of eq3 and eq2 of (F.9.2). If we can show Task (a) and Task (b) above, then we have shown that (F.9.2) ⇔ (F.9.1), and this then serves as verification of (F.9.2). Maple must replace x,y,z and derivatives with r 1,θ,φ and derivatives. For coordina tes and first derivatives, (F.9.3) The second derivatives are messier, but Maple is happy to do the calculations, (F.9.4) Appendix F: The Dumbbell Satellite 270 Task (a): Show that eq2 of (F.9.2) ⇒ eq2 of (F.9.1) We enter eq2 of (F.9.2) and do some manipulations, suppressing the output except for the last step : (F.9.5) On the first red code line we enter eq2 of (F.9.2) and then divide the result by r 12. We then replace occurrences of cos2θ by 1-sin2θ. We use lhs = "left hand side" so we end up only with the left side of an equation which says stuff = 0. Symbol % refers to the last computed quantity. The blue result can be manually transcribed as 2sinθcosθ θ• φ• + sin2θ φ•• - ω2sin2θ cosφsinφ + 2ωsin2θ cosφ θ• = 0 . Now divide by sin2θ and reorder the four terms to get φ•• + 2cot(θ) θ• φ• - ω2cosφsinφ + 2ω cosφ θ• = 0 or φ•• + 2cot(θ) θ• φ• - ωcosφ(ωsinφ - 2θ•) = 0 (F.9.6) This is a match for eq2 of (F.9.1) so we have accomplished Task (a). Task (b): (sin θ)eq3 - (cos θsinφ)eq2 of (F.9.2) ⇒ eq1 of (F.9.1) The code continues from that shown above. Equation eq2 is already ente red, so we now enter eq3, form the linear combination for eq1, then process the results with a series of typical tortuous Maple steps, Appendix F: The Dumbbell Satellite 271 (F.9.7) We again manually transcribe the result -sinθcosθ φ•2 - 2ωsin2θcosφ φ• +4ω2sinθcosθ - ω2sinθcosθsin2φ + θ•• or -sinθcosθ φ• 2 - 2ωsin2θcosφ φ• +3ω2sinθcosθ +ω2sinθcosθ - ω2sinθcosθsin2φ + θ•• or θ•• + sinθcosθ 3ω 2 -sinθcosθ φ•2 + ω2sinθcosθ - ω2sinθcosθsin2φ - 2ωsin2θcosφ φ• or θ•• + sinθcosθ( 3ω 2 - φ•2) + ω2sinθcosθ(1-sin2φ) - 2ωsin2θcosφ φ• or θ•• + sinθcosθ( 3ω2 - φ•2) + ω2sinθcosθcos2φ - 2ωsin2θcosφ φ• or θ•• + sinθcosθ( 3ω2 - φ•2) + ωsinθcosφ (ω cosθ cosφ - 2sinθ φ•) (F.9.8) and after "pulling teeth" we do end up with a match for eq1 of (F.9.1), so Task (b) is accomplished. Tension equation verification Using the angular equations of motion (F.9.1), we no w show that the following two tension expressions are the same (the first is angular (F.6.26) while the second is Cartesian (F.8.20)) , T/(m 1r1) = ω2(3cos2θ - sin2θcos2φ) + θ•2 + φ•2 sin2θ - 2ω(θ• sinφ + φ• sinθcosθcosφ) (F.6.26) T/(m 1r1) = - [ x••/x + ω2 ] . (F.8.20) ( F . 9 . 9 ) Our task of showing (F.9.9) is the same as showing that Appendix F: The Dumbbell Satellite 272 x [ω2(3cos2θ - sin2θcos2φ) + θ•2 + φ•2 sin2θ - 2ω(θ• sinφ + φ• sinθcosθcosφ) ] = - x•• - ω2x or x [ω2(3cos2θ - sin2θcos2φ + 1) + θ•2 + φ•2 sin2θ - 2ω(θ• sinφ + φ• sinθcosθcosφ) ] = - x•• or L H S = R H S . ( F . 9 . 1 0 ) We first get the complicated left hand side LHS entered: ( F . 9 . 1 1 ) We then compute RHS = - x•• as done earlier in this section, ( F . 9 . 1 2 ) Notice that RHS contains second derivatives θ•• and φ••. We shall eliminate these derivatives by manually solving the angular equations of motion (F.9.1) for Tdd = θ•• and Pdd = φ•• : To show that LHS = RHS, we define d = LHS-RHS and show that d = 0: ( F . 9 . 1 3 ) Appendix F: The Dumbbell Satellite 273 ( F . 9 . 1 4 ) Thus d = 0 and LHS = RHS and the two expressions for T in (F.9.9) are the same. F.10 Numerical solutions of the equations of motion (Cartesian Coordinates) We have done a lot of "work" in this Appendix F, and now it is time to "play", making use of our hard- won Cartesian equations of motion which don't ha ve the singularity problems had by the angular equations at θ = 0. Our task is to solve the set of equations (F.8.19) (eq1 now has a new meaning) : eq1 x 2+y2+z2 = r12 eq2 yx•• - xy•• = - ω2xy - 2ωxz• eq3 zx•• - x z•• = - 4ω2xz + 2ωxy• (F.8.19) (F.10.1) These equations describe the motion of mass m 1 of the dumbbell satellite in rotating Frame S as depicted in Fig (F.1.1). The position of mass m 1 is (x,y,z) where x2+y2+z2 = r12. The motion of mass m 2 is then determined by (D.2.8) m 1r1 = - m2r2 so (x2,y2,z2) = - (m 1/m2)(x,y,z). The length of the stick of the dumbbell satellite is s = r 1+ r2 = r1+ (m1/m2)r1 = [1 + (m 1/m2] r1. We enter eq2 and eq3 writing de rivatives for example as x•• = xdd (w = ω ) , Appendix F: The Dumbbell Satellite 274 ( F . 1 0 . 2 ) At this point zd = z• and zdd = z•• are unspecified. We use eq1 to compute z• and z•• in terms of x and y, using eq1 above: ( F . 1 0 . 3 ) When these expressions are installed, eq2 and e q3 becomes these formidable-looking equations which contain two unknown functions x(t) and y(t) and constants r 1 and ω : ( F . 1 0 . 4 ) The reader is reminded of the geometry of Fig (F.1.1) where z points up, away from Earth center, y points to the right and is in the plane of the satellite orbit, while x is perpendicular to the plane of the satellite. Appendix F: The Dumbbell Satellite 275 (F.10.5) Our plots below in the (x,y) plane are wh at a viewer would see looking at mass m 1 "from above", that is, from a point at perhaps z = b+2s on the z axis in the above figure. In-Plane Libration As our first test, we shall look for the "in-plane libra tion" behavior. We examined this behavior earlier below (F.7.3) in angular coordinates, and we now l ook in Cartesian coordinates. The initial conditions are: x(0) = 0 y(0) = 1 x•(0) = 0 y•(0) = 0 (F.10.6) With ω = 1, we expect to get a simple swinging back in the x=0 plane with period T = 3.63 sec as shown in (F.7.5). A half period is then 1.82 seconds. The Maple code to invoke a solution is as follows (fo r a numerical integration from t = 0 to t = 1.82 sec): ( F . 1 0 . 7 ) Appendix F: The Dumbbell Satellite 276 We show the result below on the left, and then from t = 0 to t = 0.91 (quarter period) on the right : (F.10.8) Thus both the "orbit" and the period for in-plane librati on are visually confirmed. If we run from t = 0 to t = 10, the graph is as on the left above since mass m 1 just swings back an forth in the same orbit, never leaving the y-axis. Out-of-Plane Libration We examined this behavior earlier below (F.7.6) in angular coordinates, and we now look in Cartesian coordinates. The initial conditions are now, x(0) = 1 y(0) = 0 x•(0) = 0 y•(0) = 0 (F.10.9) With ω = 1, we expect to get a swinging back in the y=0 with period T = 3.14 sec as shown in (F.7.8). Here is what Maple has to say: Appendix F: The Dumbbell Satellite 277 (F.10.10) The period looks right since mass m 1 swings back close to its initia l position after 3.14 seconds, but one sees that the motion is not quite in the y=0 plane, so out-of-plane libration is an approximate concept as we noted earlier below (F.5.13). Note in the figure the fine scale of the vertical axis relative to horizontal. Here are orbits for a selection of final integration times: t = 4.1 t = 8.2 t = 15.9 t = 21 t = 41.6 (F.10.11) It does seem that the out-of-plane libration stays within a certain small band of de viation in the y direction which we shall leave to the reader to theoretically calculate. In each pl ot the time was selected to make the tail of the trace clearly visible. The author is reminded of a lecture given by Shelly Glashow on the question: What can be said about orbits on an arbi trarily-shaped-but-convex billiard table? Do they all eventually close on themselves, or might some never close? (Exercise for the reader). As regards the above plots, the fact that the ratio of the two libration frequencies is an irrational number 3 /2 might have some bearing on the closure of the orbits. Appendix F: The Dumbbell Satellite 278 In the examples above r 1 = 10 and we have used x(0) = 1 or y(0) = 1 to obtain "small oscillation". If in the in-plane libration case we use y(0) = 9, there is no change in the orbit, but the oscillation period is slightly altered. If in the out-of-plane libration case we use x(0) = 9, the orbit no longer maintains the narrow band as in the above examples. For example, going again to t = 64 seconds with x(0) = 9, (F.10.12) The Digits := 14 command tells Maple to compute the numerical integration with 14 decimal places of accuracy instead of the default 10 digits. Starting with a diagonal initial position x(0) = 1 y(0) = 1 x•(0) = 0 y•(0) = 0 t = 8 t = 64 ( F . 1 0 . 1 3 ) Appendix F: The Dumbbell Satellite 279 The t=64 result is reminiscent of a Lissajous pattern on an oscilloscope screen when the x and y axes are driven by different frequency sine ways. (See sine pl ots below.) In some sense these are the two libration frequencies. Attempting a Circular Orbit (Conical Solution) We have made many attempts to get a circular-like orbit by giving the mass m 1 an initial velocity kick in some useful direction, but this system does not want to cooperate. Here is an example : (F.10.14) What starts as a rough circle is soon distorted into a narrow orbit. In the case of the spherical pendulum we had a Conical Solution in (C.5.20) where θ = θ0 and φ• = constant. Assuming θ = θ0 in the satellite angular equations (F.5.7) gives sinθ 0cosθ0(3ω2 - φ•2) + ωsinθ0cosφ (ω cosθ0 cosφ - 2sinθ0 φ•) = 0 // φ^ φ•• – ω cosφ (ωsinφ) = 0 . / / θ^ Since this is two ODE's for the one function φ(t), it seems unlikely there is any general non-static solution. If φ• = 0 the equations become sinθ0cosθ0(3ω2) + ωsinθ0cosφ (ω cosθ0 cosφ) = 0 – ωcosφ (ωsinφ) = 0 . If φ = 0 the first equation requires that θ 0 = 0 or π /2 which are static vertical and horizontal positions. The same is true for φ = π/2 . Appendix F: The Dumbbell Satellite 280 Three-dimensional plots To make such plots, one must first extract the solu tion functions from the dsolve environment. For details on how this works and other information on dsolve (i ncluding a debugger's guide), see the author's Maple User Guide. Here we extract the functions calling them X,Y and Z , (F.10.15) The following code then creates a 3D orbit and superposes it on a contour sphere of radius r 1 , ( F . 1 0 . 1 6 ) For a sample application, we start mass m 1 on the z axis and give it a good kick in the x and y directions with x•(0) = 10 and y•(0)= 10 to get an x,y plot : (F.10.17) Here then is the corresponding 3D plot Appendix F: The Dumbbell Satellite 281 (F.10.18) where the sphere of radius r 1 is gradually tipped down toward the viewer. Conventional plots In order to plot x(t), y(t) and so on, we first extrac t all functions from the dsolve system and then crudely add missing pieces like Xdd: ( F . 1 0 . 1 9 ) Here then is a plot of x(t),y(t),z(t) = red,black,blue for the above example: red = x(t) black = y(t) blue = z(t) (F.10.20) Appendix F: The Dumbbell Satellite 282 Here one sees red x and black y executing roughly sinus oidal motions. In a ballpark sense x and y are sinusoidal at their respective libration periods (3.14 and 3.63), and at least we see that black y has a longer period than red x (this black y period seems more like ~4). This is what creates the Lissajous pattern in our earlier figures. Meanwhile, blue z(t) is not co ming down much from its maximum value of z = 10. Here is a plot of x,x•,x•• = red.black,blue : red = x(t) black = x•(t) blue = x••(t) (F.10.21) Notice that red x and blue x•• always cross the axis at the same time, which allows x••/x to be finite at all values of t (see below). Tension in the the stick or tether The tension in the stick (tether) is stated in (F.8.20), T = - m 1r1[ x••/x +ω2 ] , (F.8.20) which we then plot with m 1 = 1 for the above example : (F.10.22) Appendix F: The Dumbbell Satellite 283 If we lower x•(0) = y•(0) = 10 to x•(0) = y•(0) = 1 to get a milder motion, the tension is less variable, (F.10.23) Here T is roughly equal to the "DC" value for the static dumbbell with θ = 0. From (F.6.30) that value is T ≈ 3m 1ω2r1 = 3 * 1 * 12 * 10 = 30 N . // tether tension (F.10.24) We have been using ω = 1, but for a low-Earth orbit satellite one has T orbit ≈ 88*60 seconds so ω = 2π/T orbit ≈ .0012 . (F.10.25) Then mass m 1 = 1 kg on a 20 meter tether with equal masses (r 1 = 10 m) would feel a tidal force of T ≈ 3m 1ω2r1 = 3 * 1 * (.0012)2 * 10 = .4320e-4 N = 43 μN (F.10.26) which is a very small tension. References 284 References Listed in alphabetical order by last name of first author. A quick web search on article title can usually locate documents with broken links. All li nks below were verified on 9 Feb 2017. M. Abramowitz and I.A. Segun, Handbook of Mathematical Functions, with Formulas, Graphs and Mathematical Tables (Dover, New York, 1964). Apart from the ma thematical tables of function values, this pre-computer reference has been replaced by NIST (2010), see below. G.B. Airy, "On the Vibration of a Free Pendulum in an Oval Differing little from a Straight Line", Memoirs of the Royal Astronomical Society, Vol. 20, p.121-130 (May 9, 1851). http://home.t01.itscom.net/allais/w hitep rior/airy/airyprecession.pdf (original) www.nawcc-index.net/Articles/ Airy -Vibration _of_Pendulum.pdf (typeset with annotation) The Airy precession formula appears in the middl e of page 8 in the form n/m = (3/8)bc/a2. E. I. Butikov, "A dynamical picture of the ocean tides" (2002), http://faculty.ifmo.ru/butikov/Oceanic_Tides.pdf . Y. Chen et al., "History of th e Tether Concept and Tether Missions: A Review", ISRN Astronomy and Astrophysics, Volume 2013 (2013), Article ID 502973, 7p https://www.hindawi.com/j ournals/isrn/ 2013/502973/ Gaspard-Gustave de Coriolis, "S ur les équations du mouvement relatif des systèmes de corps". J. De l'Ecole royale polytechnique 15: 144–154 (1835). This paper (in French) begins at page 37 of the following pdf file: http://www.bibnum.education.fr/sites/de fault/f iles/coriolis-force-texte.pdf M.L. Cosmo and E.C. Lorenzini, Tethers in Space Handbook, 3rd Ed . (Smithsonian, 1997). Available at https://ntrs.nasa.gov/archive/nasa /casi.ntrs.nasa. gov/19980018321. pdf www.tethers.com/papers/TethersInSpace.pdf (cleaner photos) M.J. Crowe, A History of Vector Analysis ( Dover Books, Mineola, NY, 2011) M.J. Crowe, talk (17p, 2002) related to his book A History of Vector Analysis . www.worrydream.com/refs/Crowe-HistoryOfVectorAnalysis.pdf [Bateman] A. Erdelyi et al., Higher Transcendental Functions , Volumes 1,2, and 3 (McGraw-Hill, New York, 1953); Tables of Integral Transforms , Volumes 1 and 2 (McGraw-Hill, New York, 1954). This is the 5-volume Bateman Manuscript Project. H. Goldstein, Classical Mechanics (Addison-Wesley, Boston, 1950). H. Goldstein, C. Poole, J. Safko [GPS], Classical Mechanics, 3rd Ed. (Addison-Wesley, Boston, 2001). Herbert Goldstein (1922-2005) was the sole author of the first two editi ons 1950 and 1980. References 285 [GR7] I.S. Gradshteyn and I.M. Ryzhik, Table of Integrals, Series, and Products, 7th Ed. (Academic Press, New York, 2007). The 8th Edition [GR8] appeared 2 Oct 2014 (no CD). Dan Zwillinger is collecting errata here for the 9th Edition : http://www.mathtable.com/gr/ W.M. Lai, D. Rubin and E. Krempl, Introduction to Continuum Mechanics, 4th Ed. (Butterworth- Heinemann/Elsevier, Amsterdam, 2010). This book has had the same three authors since its first edition in 1974. P. Lucht, A Maple User's Guide (http://user.xmission.com/~rimrock/ , 2016). P. Lucht, Tensor Analysis and Curvilinear Coordinates (http://user.xmission.com/~rimrock/ , 2016). P. Lucht, Tensor Products, Wedge Prod ucts and Differential Forms (http://user.xmission.com/~rimrock/ , 2016). J.B. Marion, Classical Dynamics of Particles and Systems, 2nd Ed. (Academic Press, New York, 1970). [NIST] F.W.J. Olver, D.W. Lozier, R.F. Boisvert and C.W. Clark, NIST Handbook of Mathematical Functions (Cambridge University Press, 2010). NIST is the U.S. National Institute of Standards and Technology (formerly National Bureau of Standards). This work is an update of the well-known 1964 version with editors Abramowitz and Segun, known affec tionately as "A&S". Th e greatly expanded 2010 edition (968 p) can be freely accessed online at dlmf.nist.gov which also has errata. The book ($30-$60) comes with a CD containing a bookmarked PDF file which of course has been bootlegged onto the web. Olver died in 2013. R.A. Schumacher and B. Tarbet, "A Short Fou cault Pendulum Free of Ellipsoidal Precession", https://arxiv.org/abs/0902.1829 K.R. Symon, Mechanics, 2nd Ed. (Addison-Wesley, Reading MA, 1960). J.L. Synge and B.A. Griffith, Principles of Mechanics, 2nd Ed . (McGraw-Hill, New York, 1949). This book has a lot to say about tops and pendulums. Th e Airy precession formula appears on page 381. J.R. Taylor, Classical Mechanics (University Science Books, Mill Valley, CA, 2005). S.T. Thornton and J.B. Marion [T&M], Classical Dynamics of Particles and Systems, 5th Ed ., (Thomson- Brooks/Cole, Belmont, CA, 2003). This textbook carries on the legacy of the founding author of the first two editions 1965 and 1970 (Jerry B. Marion 1929-1981). E.B. Wilson (notes of J.W. Gibbs), Vector Analysis (Dover, New York, 1960). https://www.forgottenbooks.com/en/download_pdf/Vector_Analy sis_1000079578.pdf