temp3 v5 (reviewed)

Active and Passive Vectors, Version 5 PhL 7.4.12 This 26 page doc is the last in my "temp3 " series of versions, and it is the only one in that series I have reviewed (since it must be my best shot). The main idea was to invent a certain "bracket notation" in order to derive the G Rule, but I completely failed in this approach. I just did not grok that d/dt is frame specific. 1. The Passive View REVIEWED 2 2. The Active View REVIEWED 6 3. Computation of dR/dt and (dR/dt)V. REVIEWED 7 4. Computation of dV and dV/dt. REVIEWED 9 5. Passive view : suppose frame S' (body) is moving and S (space) is at rest . REVIEWED 14 6. Passive View: what Newtonian Mechanics looks like in a non-inertial frame REVIEWED 19 7. A Paradox and its Resolution REVIEWED 21 8. Comments on Goldstein's treatment of this subject. REVIEWED 22 9. Comparison with Goldstein notation (p 133) REVIEWED 23 10. Better interpretation of dΩ and ω REVIEWED 24 Overview (written at Review time 8.1.12) In Section 1 I consider "the passive view" of a vector V. I show two things. (1) that the components of V in a rotated frame S' are given by V' = RV. (2) that in this case V'n ≡ (V')n = (V)'n so parentheses are not really necessary. I then go on to draw vector V' in frame S and I was about to say this was "the active view", but then I suddenly wandered off by defining the bracket notation [a]S etc which just got me into trouble. In Section 2 (very short) I interpret V and V'= RV drawn in the same picture as "the active view" of a transformation acting on a vector. In Section 3 I show that (dR/dt)V = ω x (RV) for any vector V, and also (dR/dt)= [ -i ωJ ] R. When I was confused about dθ and dΩ in terms of "instantaneous rotation", I used these results. In Section 4 I discuss the idea that dV = dφ x V and dV/dt = ω x V with ω ≡ dφ/dt and I draw my nice cone picture. Basically this is a description of instantaneous circular motion. I then regard this as one term of the G Rule and I try to wedge in a second "intrinsic term" to obtain the G rule. But I fail to realize that the notion of (d/dt) is frame specific, so my result is fuzzily close but not really right. I always have a label on thing d/dt is acting on, instead of a label on operation d/dt itself. In Section 5 I take another flying leap at the G Rule with basically the same method, trying to clarify things a bit by using the bracket notation. But since I still miss the main point about (d/dt) and its frame dependence, the result is again mushy and I am stuck with an extra factor of R in my G Rule. In Section 6 still using the bracket notation and do the "fictitious forces analysis". Despite the bogus notation, this is the first time I ever derived the Coriolis force term. In Section 7 I note that some vectors don't transform simply as v' = Rv such as velocity between frames. This leads to some sort of "paradox" but it is all in my bracket notation so I ignore it now. In Section 8 I am trying to understand Goldstein's G Rule meaning in terms of my bracket notation, but no matter how hard I try, I cannot force agreement between him and me. In Section 9 (very brief) I basically try again to make sense of Goldstein's G Rule by comparing each of his equations to each of mine, it just won't fly. In Section 10 I comment on dΩ as instantaneous angle, but comments don't seem useful at all Review Comments: Here in red I ponder again my bracket notation but can find nothing useful to do with it. 1. The Passive View REVIEWED We have some real-world physical vector V which is being observed by two observers, one in frame S and the other in frame S'. The frames have a common origin. Here is a picture The vector V represents a certain specific aspect of some physical apparatus A. It could be some physical point in apparatus A, or it could be the velocity of some piece of apparatus A, etc. In this passive view, there is only one apparatus, but it is observed from two frames of reference. There is only one vector V and it is the same vector in both pictures. Define matrix R by e2 = R e'2 and more generally en = R e'n forn = 1,2...N Note that this sense of R is the same as now appears in the frames doc (that's good) In this 2D example, we have roughly R = Rz(10o) (right hand rule with = e3 out of plane of paper). The vector V has different components in the two frames. As just discussed, we can write V = (V)n en where (V)n = V en V = (V)'ne'n where (V)'n = V e'n One might wonder how the (V)'n might be expressed in terms of the (V)n and vice versa. Well, (V)'n = V e'n = [R V] [R e'n] // valid for ANY rotation R = [R V] en // now assuming rotation shown above = [R (V)m em] en // inserting first expansion above = (V)m (Rem)en = (V)m (Rem)k(en)k = (V)m Rki(em)i(en)k = (V)m Rkiδm,iδn,k = Rnm (V)m So we learn the that the components are related in this way (V)'n = Rnm (V)m (V)n = R-1nm (V)'m Suppose at this point we decide to define a new vector V' in this way V' ≡ RV. This notation means that (V')n = Rnm (V)m since the prime is part of the name of vector V'. On the other hand, we just showed above that (V)'n = Rnm (V)m Thus, in this situation we have shown that (V')n = (V)'n and then the simple notation V'n is unambiguous. This being the case, we might as well dispense with the parentheses and just write V'n ≡ (V')n = (V)'n We mention this because we will have situations in which (A')n ≠ (A)'n for certain vectors which will appear below. But for V we are happy with V'n ≡ (V')n = (V)'n . Here are some words to go with the equation V' = RV: " If in frame S we were to rotate vector V into a new vector V' also in frame S , that new vector V' would have components V'n which match those of vector V appearing in the S' picture above." So let's show this: So clearly, the components of vector V' in frame S are (V'1, V'2), where (a,b,c...) indicates a column vector to save space. Now, let's write our two expansions for this new vector V' V' = (V')nen where (V')n = V' en = [RV] [Re'n] = V e'n = (V)'n V' = (V')'ne'n where (V')'n = V' e'n = [RV] [e'n] = [R2V] [Re'n] = [R2V] en It is of course possible to draw this new vector V' in frame S' where it will have these new components which we call (V')'1, (V')'2 just to be able to distinguish them. Here is the above picture pair with this vector added: The notion of vector V' in frame S will have a use below in our "active view", whereas the vector V' in frame S' will not be useful at all. But we wanted to get it drawn lest it be "mysterious". Everything is OK to this point in this section 1. But what can we say about R2V which appears above without comment? We could draw it as a third vector on the left and call it V" and its components in Frame S would be the components of V' in Frame S' as shown. But I don't see this as buying anything although it would be true. At this point we might introduce the following notation: [V]F ≡ { V eFn} Then we have these four examples adopted later for components [V]S = (V1,V2) // first term is V e1 (V)i [V]S' = (V'1,V'2) // first term is V e'1 (V)'i [V']S = (V'1,V'2) // first term is V' e1 (V')i [V']S' = (V"1,V"2) // first term is V' e'1 (V')'i This concept is really about components, not bolded vectors, as the { V eFn} original form indicates! I am trying to have [V]F be an n-tuple of components and thus somehow itself be a bolded vector. I suppose there is nothing wrong with this, but not clear how you would use it. When we see the symbol V all by itself, we usually think of V = (V1,V2). That is what we meant by it in the notation V' = RV which appears above. And similarly, we thought of V' = (V'1,V'2) in this same equation. So by "default" we think of V as meaning [V]S and V' as meaning [V']S . V' = RV means for us above [V']S = R[V]S ok (V')i = Rij(V)j is what the above means However, since V' = RV is a relationship between two vectors, it should be a valid relationship when viewed from any observer frame. Thus we expect this to also be true [V']S' = R[V]S' or (V"1,V"2) = R (V'1,V'2) or V" = R V' . (V")i = Rij(V')j is what the above means Thus we get V" = R V' = R2V, but this is not new, we had it earlier. Suddenly we see yet another object V" which we could then add to our pictures above in an obvious way, but let's NOT add it. Think of V" as just a shorthand for (V"1,V"2). So here is how we can verify that the last line above is valid: V' = RV => RV' = R(RV) => V" = RV' QED So at this point I have said above that [V']S = R[V]S ↔ V' = RV = "default" [V']S' = R[V]S' ↔ V" = R V' = R2V I guess this is OK notation, but I don't see how it will be useful. The second line seems especially useless. Here are some other "default" meanings (but remember that these rotate oppositely to V ) en = [en]S e'n = [e'n]S e'n = R-1 en means [e'n]S = R-1[en]S and again the equation e'n = R-1 en must also be true in frame S' so that [e'n]S' = R-1[en]S'. I am not quite sure where I was going with this bracket idea. But I did later adopt a good "prime based notation" which I think is viable and lives now in frames doc. Comment: (1) Notice that we don't put brackets around the rotation matrix R to get [R]F for some frame F. This matrix R is a set of numbers which relates V' to V in the equation V' = RV. ok (2) What would tensor doc say? It would say V'a = RabVb (dev notation) where R is the linearized transformation matrix. Recall that R lives neither in x-space nor in x'-space, but straddles the two spaces. There is never an object R'ab. The equation V'a = RabVb is not a "covariant equation" since it is not in one space and since R is not a tensor. So tensor doc agrees with comment (1). review of the above section concluded. Bracket notation seems non useful. 2. The Active View REVIEWED Instead of having two observers and one apparatus (which has a representative vector V in it), where the two observer frames are related by a rotation, now we have one observer frame S and we have two apparatuses which we call A and A'. This situation is basically that of the left picture above, which we repeat here: In the passive view, the frame S' was rotated by amount R-1 relative to frame S, since e'n = R-1 en. In the active view, the apparatus A' is rotated by amount R relative to apparatus A. above section is all OK. 3. Computation of dR/dt and (dR/dt)V. REVIEWED Neither of these concepts ever appears in frames doc, I wonder why? These results will be needed below, so we want to get them out of the way here so we don't clutter up issues which will be arising in the next section. Right here we make the change R → R-1 for all R which appear in Sections 1 and 2. This means that now we have V = RV' and e'n = R en . ok So the axes of frame S' are now rotated by R relative to the axes of frame S. We want to think now of these axes of frame S' as being in motion relative to those of frame S. We shall now derive our results two different ways. Method 1. We can parameterize R in this way, where θ = θ(t), R = R(θ) = exp(-iθJ) = exp(-iθkJk) where = i (Jk)ij = kij , (1) Here the Jk are the three standard rotation generators in 3 dimensional space and the θk are the corresponding rotation parameters. One can regard this as a rotation by amount θ about axis . Now consider frame S' to be tumbling around as seen from frame S. At time t, the unit vectors of frame S' have reached this position, e'n(t) = R(θ(t)) en . During the next interval dt of time, the following small tumble of these axes occurs e'n(t+dt) = R(dφ) e'n(t) . Therefore at time t the vector dφ describes the "instantaneous axis of rotation" of the e'n(t). We then have e'n(t+dt) = R(dφ) R(θ(t)) en ≡ R(t+dt) en , where we define R(t+dt) to get the total rotation as shown. Then consider dR = R(t+dt ) - R(t) = R(dφ)R(θ) - R(θ) = [ R(dφ) - 1] R(θ) ≈ [ -i dφJ ] R Then divide by dt to get dR/dt = [ -i ωJ ] R(θ) ω ≡ dφ/dt If we apply this to some arbitrary vector V, we get (dR/dt)V = [-i ω J ] RV = [-i ωkJk ] RV // on next line we will use i (Jk)ij = kij [(dR/dt)V]i = [-i ωk(Jk)ij ] [RV]j = - ωkkij [RV]j = εikj ωk[RV]j so (dR/dt)V = ω x (RV) (1) Method 2. Start with, V(t) = R(t)V'(t) . This equation tells us (using the default interpretation mentioned above) the S space components of a body frame vector V', and we allow that this body frame vector may be moving within S', hence V'(t) . Given some V'(t), we want to apply an extra small rotation R(dφ) so that we get V'(t+dt) = R(dφ) V'(t) = R(dφ) R(t) V ≡ R(t+dt) V . We argue that in going from V'(t) to V'(t+dt), we are rotating V'(t) about an "instantaneous axis of rotation" dφ. Then we find that dR = R(t+dt ) - R(t) = R(dφ)R(θ) - R(θ) = [ R(dφ) - 1] R(θ) ≈ [ -i dφJ ] R . This is the same dR as shown in Method 1, and we arrive by the same path at the same conclusion, (dR/dt)V = ω x (RV) . (1) Comment: The main idea here is this: R(t+dt) = R(dφ)R(θ(t)) where we then interpret dφ as the instantaneous axis of rotation at time t of e'n(t) or of V'(t). One might well wonder about this alternative approach, R1(t+dt ) = R(θ + dθ) . In this case one can consider the product R(θ + dθ)R-1(θ) . Since this is the product of two rotations, it must be a rotation. And since the product approaches 1 as dθ → 0, it must be a small rotation. Write it then as follows: R(θ + dθ)R-1(θ) = R(dΩ) . In working out the details, one will find that dΩ = f(,) dθ which says dΩ is proportional to dθ, but the direction of vector dΩ is in general some complicated function of the directions of θ and dθ (unless they happen to point in the same direction). Nevertheless, since the above says R(θ + dθ) = R(dΩ)R(θ) we can compute dR = R1(t+dt ) - R(t) = R(dΩ)R(θ) - R(θ) = [ R(dΩ) - 1] R(θ) ≈ [ -i dΩJ ] R . and then by the same path as above we would find that dR/dt = [ -i ω1J ] R(θ) ω1 ≡ dθ/dt (dR/dt)V = ω1 x (RV) Although these results are valid, the vectors dθ and ω1 do not represent the instantaneous axis of rotation in the sense described above, and for this reason these results are not very useful. The above section has been reviewed and is all OK. In retrospect, I originally used the above theorem (1) in the next section before I learned about the distinction between dθ and dφ . When I realized that dθ has nothing to do with the instantaneous rotation, I guess the need to differentiate the matrix R(θ) went away! I suppose the theorem itself is still OK. 4. Computation of dV and dV/dt. REVIEWED In the first line of Section 3 Method 2 above we had this statement, V(t) = R(t)V'(t) . (a) The case V'(t) = V'. If we momentarily regard vector V'(t) as stationary in frame S', so that V'(t) = V', then V(t) = R(t)V' V(t+dt) = R(t+dt)V' dV = V(t+dt) - V(t) = [ R(t+dt) - R(t)] V' = dR V' = [ -i dφJ ] R V' = dφ x [RV'] = dφ x V, where we are just following the steps used above with regard to i (Jk)ij = kij. That is to say, if V'(t) = V' is stationary in frame S', then we have dV = dφ x V (*) and dV/dt = ω x V ω ≡ dφ/dt . In this case, since frame S' is instantaneously rotating by dφ (or ω) relative to frame S, and since V' is motionless in frame S', the vector V seen in frame S is instantaneously traveling in a circle in frame S. Here is the usual picture associated with this idea , The tip of vector V is instantaneously rotating in circular motion where the circle has radius Vsinψ. In fact, the vector VT (Transverse) from the rotation axis out to the tip of vector V is this VT = V - Vcosψ . This vector VT is the radius vector of the circular motion. The motion is instantaneously circular at time t if the following is true dV VT = 0 . We can verify that motion (*) fulfills this requirement, dV VT = (dφ x V) (V - Vcosψ ) = 0 - 0 = 0 and this is why dV = dφ x V describes instantaneous circular motion. the above cone description is good and should be preserved somewhere as a stand-alone concept. Now dφ(t) and ω(t) may have some arbitrary time dependence, as determined by the mechanism that is driving the "platform" which is frame S'. These vectors may be changing in both magnitude and direction as time progresses. In this case, the actual trajectory of V(t) in frame S won't be a circular path, but will in general be some complicated three dimensional curve (all this time V' = constant in frame S' ). At any instant in time, however, a small piece of this curve near the point V(t) will be planar and will have some curvature in that plane, and V(t) will be doing instantaneous circular motion relative to dφ at that instant, This notion can be understood in terms of an iterative program of advancement where the first few steps of this program starting at time t would be V(t) dV(t) = dφ(t) x V(t) V(t+dt) = V(t) + dV(t) dV(t+dt) = dφ(t+dt) x V(t+dt) V(t+2dt) = V(t+dt) + dV(t+dt) etc. In this manner the trajectory V(t) is generated, and at any time one maintains the instantaneous circular motion indicated by dV(t) = dφ(t) x V(t) (b) The case V'(t) = V'(t) . We just redo the steps done above, adding changes as needed: V(t) = R(t)V'(t) V(t+dt) = R(t+dt)V'(t+dt) dV' = V'(t+dt) - V'(t) definition of dV' dV = V(t+dt) - V(t) = R(t+dt)V'(t+dt) - R(t)V'(t) = R(t+dt)[ V'(t) + dV'] - R(t)V'(t) = [R(t+dt) - R(t)] V'(t) + R(t+dt) dV' . In the last term we will use R(t+dt) = R(dφ)R(θ(t)) = [ 1 -i dφJ ] R R(t+dt) dV' = [1 -i dφJ ] R dV' = R dV' – i dφJ R dV' = RdV' + dφ x (RdV') For the first two terms we will use this fact shown earlier [R(t+dt) - R(t)] V'(t) = dφ x V Thus we end up with dV = dφ x V + RdV' + dφ x (RdV') we have two new pieces (last two terms) Here, the change dV' arises because V' is moving within frame S'. At this point, we need some way to identify the change of V' within frame S', so let's just define dV'body ≡ dV'(t) . Then we can define dVbody (no prime) to be this same change, but expressed in frame S components, dVbody ≡ R dV'body = R dV'(t) . Therefore we have shown that dV = dφ x V + dVbody + dφ x dVbody dV/dt = ω x V + dVbody/dt + ω x (dVbody/dt)dt and we then drop the last term in the limit dt → 0 and get these famous results dV = dφ x V + dVbody dV/dt = ω x V + dVbody/dt Had we not added the body label, the two dV objects in the first line above would be indistinguishable. How might we isolate the two terms in the last equation above? The first term is all there is if it happens that dVbody/dt = 0 which means dV'body/dt = 0 in terms of frame S' components. This in turn could happen if V'body = V', a constant, or it could happen just because we happen to have dV'body/dt = 0 at some particular time t. Note that we could have V'body = 0 while at the same time dV'body/dt ≠ 0. The objects dV'body/dt and dVbody/dt could in theory be functions of ω, since we allow an arbitrary body motion. For this reason, just setting ω = 0 to kill off the ω x V term won't necessarily expose all of the dVbody/dt term, since ω = 0 might kill off some part of dVbody/dt . Therefore we cannot say that the second term is all there is if we set ω = 0. It is of course all there is if we set ω x V = 0. When both terms are present, we loose the simple interpretation of vector V instantaneously rotating about the axis . Notice that we could define Vbody = R(t) V'(t), but this is in fact V(t). So we have this situation V = Vbody // = R(t)V'(t) but dV ≠ dVbody . In fact, we just showed that dV = dφ x V + dVbody . As defined above, dVbody is the "intrinsic" change in V within frame S' (but converted to S components) as if there were no frame rotation, whereas dV includes the effect of this rotation and gives the correct total change in V in frame S. It is convenient to make this notational definition, (dV/dt)body = (d/dt)bodyV ≡ (dVbody/dt) = (d/dt) Vbody because then we can write (*) as (d/dt) V = ω x V + (d/dt)bodyV and since this applies to any vector V, we can write it in the following operator notation, (d/dt) = (d/dt)body + ω x See for example equations 4-99 through 4-102 of Goldstein (page 133). Note that dΩGoldstein = dφ. We are now ready to see some Examples. Review comments: The above was my early attempt to understand/derive the G Rule, perhaps my first ever. When reading Goldstein I never really understood it and I should go review those notes now too. I just did not grasp the concept that d/dt has to have a frame index associated with the very operation itself when you are acting on vectors which are expanded on basis vectors that are frozen in one frame and move in the other. I did get something that looked pretty good at the time, but it was just not right. I ended up with something called (dVbody/dt) and then I fudged and just defined it to be (dV/dt)body . Perhaps my approach here could somehow be justified, but the G Rule derivation in frames doc is so much simpler and direct that it hardly seems worth while trying to repair/interpret the above discussion. 5. Passive view : suppose frame S' (body) is moving and S (space) is at rest . REVIEWED space frame body frame This section is another attempt at deriving the G rule , this time using the bracket notation. I seem to repeat the previous section just adding lots of brackets like [...]S. Most seems to be copy paste and edit of the previous sections. I think I was hoping that my bracket notation would clarify the G Rule, but instead I think it led to various paradoxes. See comments below. Right here we small make the convenience change R → R-1 for all R which appear in Sections 1 and 2 (but not Section 3). After this change, we have from Section 1 above, e'n= R(θ) en en = R-1(θ) e'n (1) where R(θ) describes the current position ( t = t, say) of the S' frame relative to the S frame. Now consider some vector V lying in the body S' frame. We know again from Section 1 (with change), V(t) = R(θ)V'(t) (2) which we can interpret in this hybrid manner (see end of Section 1 above for this notation), [V(t)]S = R(θ) [V(t)]S' . (3) The object [V(t)]S' is a vector V in the body frame S', and this vector may be moving in frame S'. Viewed from the space frame, what one sees is [V(t)]S. We now assume that frame S' is "instantaneously rotating" by some vector amount dθ in time dt. If we momentarily assume that [V(t)]S' is not moving within frame S', we would find that there is still a change in frame S, namely d[V(t)]S = R(dθ) [V(t)]S' – [V(t)]S' (4) Now a rotation can be written in terms of the rotation generator matrices in this manner R(θ) = exp(-i θ J) (5) where i (Jk)ij = kij . (6) Therefore, keeping only the first two terms in a Taylor expansion about 0, R(dθ) = exp(-i dθ J) ≈ 1 - i dθ J . (7) Using this in (4) gives d[V(t)]S = R(dθ) [V(t)]S' – [V(t)]S' = (1 - i dθ J) [V(t)]S' – [V(t)]S' = - i (dθ J) [V(t)]S' Writing this in components gives (d[V(t)]S )i = - i dθk (Jk)ij ([V(t)]S')j = -i εkij dθk ([V(t)]S')j = + i εikj dθk([V(t)]S')j which when written back in vector notation says d[V(t)]S = dθ x [V(t)]S' (11) Usually one simply starts with equation (11) with the following justification: If we rotate [V(t)]S' by amount dθ about an axis in the dθ direction, then d[V(t)]S is as shown in this picture, and then the radius out from the axis is the length of the vector [V(t)]S' times the sine of the cone half angle ψ, and the direction of d[V(t)]S is perpendicular to both dθ and [V(t)]S and that justifies (11) since the cross product involves sinψ. Our reason for not just starting with (11) will appear below. In the development above, we assumed that [V(t)]S' was not changing within frame S', but we now want to relax that constraint and allow it to move. From this movement alone we get d[V(t)]S = R(θ) d[V(t)]S' . (12) Notice the presence of matrix R(θ) which maps the change d[V(t)]S' expressed in frame S' coordinates to the coordinates of frame S. We can then add both the two motion effects just reviewed to get d[V(t)]S = R(θ) d[V(t)]S' + dθ x [V(t)]S' (13) If we now divide all three terms by dt, we get (d[V(t)]S/dt) = R(θ) (d[V(t)]S'/dt) + ω x [V(t)]S' where ω = dθ/dt (14) This was my new attempted G Rule where somehow the brackets make it look a little like Goldstein because something at least has a frame subscript. My problem of the moment was that R(θ) factor sitting on the right side, which did not appear in the official G Rule, so I had to assume the G Rule only applied when frame S and frame S' had their axes instantaneously lined up, which of course is totally wrong. If we now assume that at our time t of interest, frames S and S' are "lined up", then θ = 0 and R(θ) = 1 and we have (d[V(t)]S/dt) = (d[V(t)]S'/dt) + ω x [V(t)]S' where ω = dθ/dt ; lined up (15) But if the frames are lined up, then [V(t)]S' = [V(t)]S and we might just call it V(t). And we slightly change the location of the dt's in the derivatives, so (15) becomes, (d/dt [V(t)]S) = (d/dt [V(t)]S') + ω x V(t) where ω = dθ/dt ; lined up (16) Now we make the following definition where F is a frame (either S or S') (d/dt)F [V(t)] ≡ (d/dt [V(t)]F) Then (16) can be written as (d/dt)S [V(t)] = (d/dt)S' [V(t)] + ω x V(t) where ω = dθ/dt ; lined up (17) Since this is true for any vector V, we can write it as an operator equation (d/dt)S = (d/dt)S' + ω x where ω = dθ/dt ; lined up (18) Example #1. Suppose V = r, the position vector. Then [r]S' is the position of some point in frame S', and [r]S is the corresponding position in frame S, and we have [r]S = R(θ) [r]S' = [r]S' if "lined up" = r Then (16) reads (d/dt [r]S) = (d/dt [r]S') + ω x r where ω = dθ/dt ; lined up (19) We then make the obvious definitions [v]S = (d[r]S/dt) = velocity of point r in the space frame S (20a) [v]S' = (d[r]S'/dt) = velocity of point r in the body frame S ' (20b) and then (19) becomes [v]S = [v]S' + ω x r where ω = dθ/dt ; lined up (21) STOP for PARADOX. Above equation (16) we said that [V(t)]S' = [V(t)]S when the two frames are lined up, so applying that to V = v, we ought to have [v]S = [v]S', but this disagrees with (21). Example #2. Suppose V = v, the velocity vector. Then [v]S' is the velocity of some point in frame S', and [v]S is the corresponding velocity in frame S. Then (16) reads (d/dt [v]S) = (d/dt [v]S') + ω x v where ω = dθ/dt ; lined up (22) We then make the obvious definitions [a]S = (d[v]S/dt) = acceleration of point r in the space frame S (23a) [a]S' = (d[v]S'/dt) = acceleration of point r in the body frame S' (23b) and then (22) becomes [a]S = [a]S' + ω x v We see that this is the wrong result! The reason is this: Look back at (14) (d[V(t)]S/dt) = R(θ) (d[V(t)]S'/dt) + ω x [V(t)]S' where ω = dθ/dt and apply this with V = r and use the above velocity definitions to get [v(t)]S = R(θ) [v(t)]S' + ω x [r]S where ω = dθ/dt What we learn here is that velocity v is not really a "vector" because a vector must transform according to this equation, [V(t)]S = R(θ) [V(t)]S' . (3) Since v is not a true vector under the transformation from frame S' to frame S, we cannot make use of (22). Example #3. Suppose V = L = r x p = m r x v , the usual angular momentum vector. Then [L]S = m [r]S x [v]S I show in separate notes that [L]S = R[L]S' + m [r]S x (ω x [r]S) and therefore L is not a true vector and again (22) is not allowed. But both Goldstein and Marion say it is allowed! So my paradox remains. ________________________ We could have alternately described this same change in a different manner, [V(t+dt)]S = R(θ + dψ) [V(t)]S' That means that [V(t+dt)]S' = R(dθ) [V(t)]S' At this point, we wish to apply a small rotation R(dθ) to [V(t)]S' and see what change that causes in [V]S ________________________ Comment: Notice that, in reference to equation (2), we said that V was some "vector". By definition, a vector with respect to transformation R is something that "transforms as a vector according to (2)". When R is time-dependent, things that one normally thinks of as being "vectors" are in fact not vectors because their components do not satisfy (2). For example, the position vector r is a vector under R(t), but as we shall see, the velocity vector v is not a vector under R(t), though it is a vector if R is not time-dependent. The above concept is valid and made its way somewhere into frames doc. When a frame is rotating, things like r and v don't transform as vectors normally do, but I cannot find my frames doc ref right now. Now some time dt goes by and we get some changes: d[V]S = R d[V]S' + dR [V]S' . (4) The first term arises because there was a change within the body S' frame in our vector V, and that change then gets mapped to the space frame S by rotation R. The second term arises because even if [V]S' were not changing within the body frame, the body frame is moving, which causes a change of [V]S in the space frame. Now divide by dt to get d[V]S/dt = R (d[V]S'/dt) + (dR/dt) [V]S' . (5) Then use result 3.(4) for (dR/dt) acting on a vector to get (d[V]S/dt) = R (d[V]S'/dt) + ω x (R [V]S') . (6) Then use result (3) in the last term of (6) to get (d[V]S/dt) = R (d[V]S'/dt) + ω x [V]S . (7) We could apply d/dt to equation (7) and in so doing obtain four terms. Rather than do that here, we will do it in the special application of V = r of the next section. Review complete. See mid section red comment. The above section was a valiant "Plan B" effort to try to make something work but it was doomed to fail. The bracket notation is not "the solution" . Frames doc G Rule derivation has the solution. I just think of all the hours I spent on this stuff being endlessly confused. 6. Passive View: what Newtonian Mechanics looks like in a non-inertial frame REVIEWED Continuing with my bogus bracket notation and my bogus G Rule, I somehow end up with something that is "not to bad" for the fictional forces. This is a precursor to frames doc Sections on v and a. Somehow in my ghostly notation I did come up with the Coriolis force, this might be the first place I ever did that. The Goldstein derivation was totally hazy to me at this time and stayed hazy until very late in the game. So hazy that a whole Section of frames doc is devoted to Goldstein's section. I think my frames doc derivations are much better than Goldstein's because detail is provided! I see now why I was confused on my Goldstein reading a few years ago. So nothing in this section need be "preserved". Now as a particular application of 4.(7) above, let V = r, the position vector : (d[r]S/dt) = R (d[r]S'/dt) + ω x [r]S (1) Note that I still have this annoying R factor sitting in my totally bogus G Rule. We next identify [v]S = (d[r]S/dt) = velocity of point r in the space frame S (2a) [v]S' = (d[r]S'/dt) = velocity of point r in the body frame S ' (2b) so (1) can be rewritten as [v]S = R [v]S' + ω x [r]S . (3) Next, apply d/dt to equation (3) to get (d[v]S/dt) = (dR/dt) [v]S' + R (d[v]S'/dt) + (dω/dt) x [r]S + ω x (d[r]S/dt) . (4) We now identify these accelerations, [a]S = (d[v]S/dt) = acceleration of point r in the space frame S (5a) [a]S' = (d[v]S'/dt) = acceleration of point r in the body frame S' (5b) so that, along with (2a) in the last term, we get [a]S = (dR/dt) [v]S' + R [a]S' + (dω/dt) x [r]S + ω x [v]S . (6) Then use 3.(4) for (dR/dt) acting on a vector to get [ here dR/dt used first time in this whole doc! ] [a]S = ω x (R[v]S') + R [a]S' + (dω/dt) x [r]S + ω x [v]S . (7) Finally, use (3) to replace [v]S in the last term : [a]S = ω x (R[v]S') + R [a]S' + (dω/dt) x [r]S + ω x (R[v]S') + ω x (ω x [r]S) . (8) We see that two ω x terms are the same, so we add them, and then put R [a]S' first to get [a]S = R [a]S' + 2 ω x (R[v]S') + (dω/dt) x [r]S + ω x (ω x [r]S) (9) This then is the final and very general result relating the acceleration in the space frame S to that in the body frame S'. Obviously, if ω is constant in time (such as the rotation of the earth) the above simplifies to [a]S = R [a]S' + 2 ω x (R[v]S') + ω x (ω x [r]S) (10) If in addition it happens that the space and body frames are exactly aligned at the instant t which is of interest to us, we may replace R = 1 in both places in (10). Then of course [r]S = [r]S' and we might as well just call it r. Then we get at this instant in time, [a]S = [a]S' + 2 ω x [v]S' + ω x (ω x r) . (11) In the space frame S (inertial since at rest) we have Newton's law being valid, [F]S = m [a]S . (12) Using (11) we can say that [Feff]S' ≡ m [a]S' = m [a]S – 2 ω x [v]S' – ω x (ω x r) = [F]S – 2 ω x [v]S' – ω x (ω x r) . (13) This says that we can pretend there is a Newton's law in the (non-inertial) rotating frame, [Feff]S' ≡ m [a]S' as long as we are willing to include the two "fictitious forces" shown in (13). The second fictitious force is just the usual centrifugal force, pointing exactly away from the rotation axis. The first term is the new object of interest : the Coriolis force– 2 ω x [v]S' . Review of the above completed, so comment at the start. All dust in the wind. 7. A Paradox and its Resolution REVIEWED This small section hopefully illuminates the Comment made above equation 4.(4). Note that several equations in this section are wrong so please don't use them! Equation 4.(7) is supposedly valid for any vector V. What happens if we take V = v, the velocity of a particle. It then says, (d[V]S/dt) = R (d[V]S'/dt) + ω x [V]S 4.(7) (d[v]S/dt) = R (d[v]S'/dt) + ω x [v]S 4.(7) with V = v We identify accelerations as in 5.(5) and this then says [a]S = R [a]S' + ω x [v]S . Comparison with 5.(6) shows that this last equation is wrong because we are missing two terms! [ well, this is really a long story, all cleared up in frames doc]. So what is going on here?? If we trace backwards from 4.(7) we get to 4.(3) which says, for V = v, v = R v' which we could be written, according to the end of Section 1, [v]S = R [v]S' . But this is not true for V = v ! In fact, what is true is 5.(3) [v]S = R [v]S' + ω x [r]S What this means is that, although v "looks like" a vector, it does not transform as a vector since we have this extra ω x [r]S term (when R is time dependent). We are only allowed to apply 4.(7) to objects that really transform as a vector under our transformation between frames S and S'. V = r really does transform as a vector. 8. Comments on Goldstein's treatment of this subject. REVIEWED The theory of transforming between rotating frames is discussed on pages 132-135 of Herbert Goldstein's famous book "Classical Mechanics" (1950, 7th printing 1965), with some support material on earlier pages and some implications on later pages. These pages form Sections 4-8 and the start of 4-9 of that book. Goldstein never draws a picture of the two frames in question, which we have called S (space) and S' (body). He seems to make an assumption (which I cannot find stated anywhere) that, at the moment of interest t, the two frames S and S' line up with each other, which means we can set R = 1 in any of our final expressions above [ I am still worrying about this issue!] . With this assumption, we may quote some of our results from above, replacing our generic vector V by generic vector G, (d[G]S/dt) = R (d[G]S'/dt) + ω x [G]S . 4.(7) (d[G]S/dt) = (d[G]S'/dt) + ω x [G]S . 4.(7) with R=1 Goldstein writes the last equation above in the following manner, (dG/dt)space = (dG/dt)body + ω x G Goldstein (4-100) First of all, if R = 1, then since G is a "vector" we know from 4.(3) that [G]S = R [G]S' = [G]S'. Since we then have [G]S = [G]S' , we might as well just call it G, as he does above. As an example, consider our equation 5.(3) from above, but in which we set R = 1 and so [r]S = [r]S' = r , [v]S = [v]S' + ω x r . 5.(3) with R=1 which we might also write as [v]space = [v]body + ω x r This then is an example of Goldstein (4-100). Since (4-100) is valid for any (true!) vector G, we can abstract out the vector G and treat it as an operator equation of sorts (which can only act on true vectors) (d/dt)space = (d/dt)body + ω x Goldstein (4-102) In our notation, this could be written several ways, removing the G vector acted upon, (d[ ]S/dt) = (d[ ]S'/dt) + ω x or (d/dt)[ ]S = (d/dt)[ ]S' + ω x and if we define operator (d/dt)S[ ] ≡ (d/dt)[ ]S [ I am defining things again!] , then finally (d/dt)S[ ] = (d/dt)S'[ ] + ω x or (d/dt)S = (d/dt)S' + ω x or (d/dt)space = (d/dt)body + ω x and this last form then aligns Goldstein (4-102). The operator notation is nice, but it can be a little confusing what it really means, so we feel it is best to keep the vector G present and, assuming R = 1, use this form: (d[G]S/dt) = (d[G]S'/dt) + ω x G . our 4.(7) with R=1 (dG/dt)space = (dG/dt)body + ω x G Goldstein (4-100) Review complete. I keep trying to "force it " to work, but it does not work. More dust in the wind. –––––––––––––––––––––––––––––– 9. Comparison with Goldstein notation (p 133) REVIEWED Goldstein uses vector G instead of V for his generic vector. So I will now translate some equations above into G form, and then do side-by side comparison with Goldstein: [dG]S = [dG]S(body) + [dΩ]S x [G]S (23a) (dG)space = (dG)(body) + dΩ x G p 133 (4-99) [dG/dt]S = [dG/dt]S(body) + [ω]S x [G]S where ω = dΩ/dt (23b) (dG/dt)space = (dG/dt)(body) + ω x G p 133 (4-100) (d/dt) = (d/dt)(body) + ω x (21) (d/dt)space = (d/dt)(body) + ω x p 133 (4-102) Several comments are in order: (1) All of Goldstein's equations shown above have all quantities evaluated in frame S (space). He uses what I call the "default meanings" such that [G]S = G, [ω]S = ω . [ note word evaluated! ] (2) His body notation is ambiguous I feel because it suggests that the term is evaluated in the body frame when in fact it is a body change which is evaluated in the space frame. His notation does not get that point across. [ that is because I have it wrong! ] 10. Better interpretation of dΩ and ω REVIEWED I think this is a correct distinction. We showed in the last section that dR(t) = [-i dΩ J ] R dR(t)/dt ≈ [-i ω J ] R ω = dΩ/dt Now consider the basis vector equation (after the change noted at the start of Section 3) e'n(t) = R(t) en de'n(t)/dt = dR(t)/dt en = [-i ω J ] R en = [-i ω J ] e'n(t) = [-i ωkJk ] e'n(t) = -iωk[Jk e'n(t)] and taking a component (using i (Jk)ij = kij) [de'n(t)/dt]i = -iωk[Jk e'n(t)]i = -iωk(Jk)ij [e'n(t)]j = - ωk kij [e'n(t)]j = εikj ωk [e'n(t)]j so that de'n(t)/dt = ω x e'n(t) Therefore ω is the amount and direction of the instantaneous rate of rotation of any axis of the body frame. The entire body frame is rotating according to this vector ω . The above can be regarded as a generic equation which we would normally evaluate with all terms of the type [..]S. Perhaps here is another derivation of our basic result V = V'ne'n dV = dV'n e'n + V'n de'n dV/dt = (dV'n/dt) e'n + V'n de'n/dt dV/dt = (dV/dt)(body) + V'n ω x e'n(t) dV/dt = (dV/dt)(body) + ω x V (6.1) The key idea here is that (dV'n/dt) e'n = rate of change of vector V seen in body frame = (dV/dt)(body) which I think agrees with the pictures and discussion above. We can regard (6.1) as being an equation which we can evaluate in any frame, and we usually evaluate it in frame S in which case it becomes [dV/dt]S = [(dV/dt)(body)]S + [ω]S x [V]S (18b) Comments on the bracket notation. There is a distinction between "a bolded vector" and "an n-tuple which contains a number of components of that bolded vector in some basis". a the bolded vector [a]S ≡ {a1, a2} n-tuple of components in basis ei [a]S' ≡ {(a)'1, (a)'2} n-tuple of components in basis e'i You can dot a with another vector, but you cannot dot [a]S with another vector. [a]S e1 = {a1, a2} e1 = ? = {a1, a2} {1, 0} = a1 [a]S e'1 = {a1, a2} e'1 = ? = {a1, a2} {( e'1)1, (e'1)2} = a1 ( e'1)1 + a2 (e'1)2 = ??? Maybe this is what I was trying to say a = {a1, a2}S = {(a)'1, (a)'2}S' where {x,y}S ≠ {x,y}S' Then a = ([a]S)S = ([a]S')S' I suppose. Consider a vector equation a = Qb, where Q is a matrix. We can write ai = Qijbj where ai is in the n-tuple [a]S and bi is in the n-tuple [b]S You might crudely think of this as saying [a]S = Q[b]S where you imagine Q as its matrix self. Now if Q is a rank 2 tensor and a,b are rank 1 tensors, then in frame S' we have (a)'i = (Q)'ij(b)'j Then using the above crude notation you might say [a]S' = Q'[b]S' where probably Q' = R-1QR or something like that where Q' is not the same matrix as Q. ( It just happens that Q' = Q in the case that Q = R. ) So in general, if you have a vector equation involving [..]S vectors, you cannot just "evaluate" the equation in frame S' by replacing all those vectors with [..]S' and leaving matrices as they were. I conclude that this bracket notation is totally useless and I don't know why I got so involved with it thinking it was going to "do something for me". It was just "an idea" and it cost me lots of hours. Sometimes ideas do that.