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Typed lecture notes titled Mechanics 2/23, teaching block 1, 2010/11, apparently from another author's course and kept in Phil's pendulum folder. Chapters cover the calculus of variations (Euler-Lagrange equation, brachistochrone, Fermat's principle), Lagrangian mechanics with constraints and conserved quantities (including the spherical pendulum and Noether's theorem), small oscillations and the double pendulum, rigid bodies, and Hamiltonian mechanics.

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Mechanics 2/23 Lecture notes, 2010/11, teaching block 1 Contents 0 Introduction 1 1 The calculus of variations 5 1.1 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 1.2 Generalisation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 1.3 Euler-Lagrange equation . . . . . . . . . . . . . . . . . . . . . . . . . 8 1.4 Solution of our problem . . . . . . . . . . . . . . . . . . . . . . . . . 10 1.5 Alternative version of the Euler-Lagrange equation . . . . . . . . . . 10 1.6 The brachistochrone . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 1.7 Functionals depending on several functions . . . . . . . . . . . . . . 16 1.8 Fermat’s principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 2 Lagrangian mechanics 21 2.1 Reminder: Newton . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 2.2 Lagrangian mechanics in Cartesian coordinates . . . . . . . . . . . . 22 2.3 Generalised coordinates and constraints . . . . . . . . . . . . . . . . 24 2.3.1 Gravitational ﬁeld . . . . . . . . . . . . . . . . . . . . . . . . 26 2.3.2 Pendulum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27 2.3.3 Inclined plane . . . . . . . . . . . . . . . . . . . . . . . . . . . 28 2.3.4 General properties of forces of constraint . . . . . . . . . . . 33 2.3.5 Derivation of Lagrange’s equations from Newton’s law in the general case . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35 2.4 Conserved quantities . . . . . . . . . . . . . . . . . . . . . . . . . . . 39 2.4.1 Energy conservation . . . . . . . . . . . . . . . . . . . . . . . 39 2.4.2 Conservation of generalised momenta . . . . . . . . . . . . . . 45 2.4.3 Spherical pendulum . . . . . . . . . . . . . . . . . . . . . . . 47 2.4.4 Noether’s theorem . . . . . . . . . . . . . . . . . . . . . . . . 55 3 Small oscillations 63 3.1 General theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63 3.2 Two springs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67 3.3 Small oscillations about equilibrium . . . . . . . . . . . . . . . . . . 70 3.4 The double pendulum . . . . . . . . . . . . . . . . . . . . . . . . . . 73 4 Rigid bodies 79 4.1 Angular velocity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80 4.2 Inertia tensor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81 4.3 Euler’s equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86 i ii CONTENTS 5 Hamiltonian mechanics 91 5.1 Hamilton’s equations . . . . . . . . . . . . . . . . . . . . . . . . . . . 91 5.2 Conservation laws and Poisson brackets . . . . . . . . . . . . . . . . 98 5.3 Canonical transformations . . . . . . . . . . . . . . . . . . . . . . . . 103 Chapter 0 Introduction In this course we are going to consider a formulation of mecha nics (variational mechanics) that is diﬀerent from the one in Mechanics 1 (Newt onian mechanics). I ﬁrst want to explain why such a formulation is called for, and discuss some of the main ideas. Newtonian mechanics The central result in Newtonian mechanics concerns the acce leration of a particle, i.e., the second derivative of its position rw.r.t. time. This acceleration is given by the ratio of the force Facting on the particle and its mass m, i.e., d2r dt2=F m. The force Ftypically depends on the particle position r; in addition it may depend on the positions of other particles and on the velocities of t he particles involved. Thus we obtain diﬀerential equations that contains both par ticle positions and their second (and sometimes its ﬁrst) derivatives, i.e., second-order ordinary diﬀer- ential equations . These diﬀerential equations provide a way of determining th e trajectories of par- ticles. However, their direct application can sometimes be come messy. Historically, the ﬁrst applications where this was felt in engineering and in astronomy. Variational mechanics Figure 1: Isaac Newton, Joseph Louis Lagrange, and William R owan Hamilton. To simplify the treatment of such complex problems, a variat ional formulation of mechanics was developped. There are two diﬀerent version s of variational me- 1 2 CHAPTER 0. INTRODUCTION chanics, respectively introduced by Joseph Louis Lagrange (1736-1813) and William Rowan Hamilton (1805-1865). To explain the key idea of these approaches, let us consider the simplest case possible: one particle on which n o forces are acting. We then haved2r dt2= 0, i.e., the particle travels with constant velocitydr dtalong a straight line. Now the important point is that the straight line is also the short- est connection between two points. Thus, instead of using a diﬀerential equ ation, we would have obtained a trajectory of the same form if we had p ostulated that without forces particles always follow the shortest line co nnecting two points. The main idea of variational mechanics is to generalise this simple observation to more general settings. Our goal is thus to formulate mecha nics through a varia- tional principle: Particles travel on trajectories that extremise (minimise , maximise) “something”. This “something” is not always the length, and one of our goals will be to ﬁnd what it is. In general it will be called the “action”. Advantages of variational mechanics The main advantages of the variational approach to mechanic s are: •In mechanics it is often helpful to work in coordinates adapted to the system we are interested in. For example, if we want to describe the m otion of a satellite in the gravitational ﬁeld of the Earth, it is helpf ul to use spherical coordinates with the centre of the Earth as the origin. More c omplicated systems require other complicated sets of coordinates. For example if we consider a particle moving in the gravitational ﬁeld of two m asses it would be better to use elliptic coordinates (see Fig. 2). In these coo rdinates two points are singled out, corresponding to the positions of the two ma sses. Figure 2: Elliptic coordinates. We shall see that in variational mechanics it becomes partic ularly simple to switch between coordinate systems. 3 •Many mechanical systems have constraints , i.e., conditions where particles or bodies are allowed to be and where not. A practical example wo uld be a train that is required to stay on the railroad track. In variationa l mechanics these constraints can easily be incorporated by choosing appropr iate coordinates, e.g., by taking the railroad track as a coordinate line. •Crucially, many areas of mathematical physics are formulated in the lan- guage developped in variational mechanics. For instance “H amiltonians” and “Lagrangians” play an important role in quantum mechanics o r chaos the- ory. Similarly the ideas underlying variational mechanics are important in the theory of diﬀerential equations . (Rough) outline of this course In this course, we will ﬁrst develop the mathematical tools n eeded for extremisation problems like the one sketched above ( variational calculus ). We will then consider both the Lagrangian and the Hamiltonian formulation of mechanics and several examples for their use. Reminder: Chain rule As a preparation for many calculations in this course I want t o remind how deriva- tives of expressions like f(u(t),v(t),t) are evaluated. According to the chain rule, we ﬁrst have to di ﬀerentiate w.r.t. the arguments of f, and then multiply with the derivatives of these arguments w .r.t.t. This yields d dtf(u(t),v(t),t) =∂f ∂u(u(t),v(t),t)du dt+∂f ∂v(u(t),v(t),t)dv dt+∂f ∂t(u(t),v(t),t). Dropping the arguments and setting ˙ u=du dt,˙v=dv dt, we can also write d f dt=∂f ∂u˙u+∂f ∂v˙v+∂f ∂t. It is important to distinguish between the partial derivative∂f ∂twith respect to the third argument of f, and the total derivative∂f ∂twhere also the t-dependence of the ﬁrst and the second argument are taken into account. An example would be frepresenting the termperature felt by a person, fde- noting the time, and u(t),v(t) the position of the person. Then the rate of change d f dtof the temperature felt will depend on the change of temperat ure with time (∂f ∂t) and on whether the person moves to a warmer or colder place (le ading to the terms ∂f ∂u˙uand∂f ∂v˙v). 4 CHAPTER 0. INTRODUCTION Chapter 1 The calculus of variations 1.1 Example Figure 1.1: Possible connections between ( x1,y1) and ( x2,y2). To introduce variational mechanics, we ﬁrst need to equip ou rselves with the tools needed to solve extremisation problems. We will start with the simplest ex- ample: Showing that the shortest connection between two points in R2is a straight line . Let us denote these two points by coordinates ( x1,y1) and ( x2,y2) in a Cartesian coordinate system, see Fig. 1.1. Curves connect ing these two points can then be described through functions y(x) that assign to each x-coordinate between x1andx2the corresponding y-coordinate. We now have to •determine the length lof the curve corresponding to each function y(x) •and ﬁnd a y(x) such that lbecomes minimal. To start with the ﬁrst task, we split the x-axis into small pieces of length dx. As seen in Fig. 1.2, this means that the curve is also split int o small pieces, whose length will be denoted by dl. We assume that the pieces are small enough such that inside each piece the curve can be considered as a straight li ne. This means that for each piece we can draw a triangle as in Fig. 1.2 with the wid thdx, the height 5 6 CHAPTER 1. THE CALCULUS OF VARIATIONS Figure 1.2: Determining the length of a curve y(x) inR2. dy, and the length dlas its sides. Since the slope of the curve must be given by the derivative y′(x),dydepends on dxas y′(x) =dy dx⇒dy=y′(x)dx . We can now determine the length of each piece using Pythagora s’ theorem, (dl)2= (dx)2+ (dy)2= (1 + y′(x)2)(dx)2 ⇒dl=/radicalbig 1 +y′(x)2dx . To obtain the overall length of the curve, we have to sum over t he lengths dlof all pieces. If we take the limit where the width dxof the pieces goes to zero, this sum can be replaced by the integral, and we obtain the Length of a curve in R2 l=/integraldisplayx2 x1/radicalbig 1 +y′(x)2dx . Our task is now to ﬁnd functions y(x)with y(x1) =y1andy(x2) =y2for which the length lbecomes minimal . 1.2 Generalisation The problem outlined above is the simplest example for a more general type of variational problems, where we maximise, minimise, or in ge neral look for stationary points of an integral of a function depending on y(x),y′(x) and x: 1.2. GENERALISATION 7 Stationarity problem Let K[y] =/integraldisplayx2 x1f(y(x),y′(x),x)dx . (1.1) Findy(x) satisfying the boundary conditions y(x1) =y1,y(x2) =y2for which K[y] becomes stationary. The stationarity problem considered here are is diﬀerent fr om those in Calculus, where the quantity to be minimised depended only on a single n umber or on a vector. In contrast K[y] depends on a function. The mapping from ytoK[y] is an example of a functional . A functional is a function that maps functions to real numbers. Not all functionals are of the from in Eq. (1.1) (e.g. we might also have integrals depending on y′′(x)), but those of the type in Eq. (1.1) play a particularly important role in mechanics. Another example for such a K[y] would beK[y] =/integraltextx2 x1(y(x)2+y′(x)2)e−xdx. To proceed, we have to understand better what it means for a fu nctional to be stationary . We will deﬁne stationarity of functionals in terms of a stat ionarity problem we are already familiar with: ﬁnding stationary poi nts of a function that only depends on one variable. As a preparation let us conside r a function y∗(x) satisfying our boundary conditions y∗(x1) =y1,y∗(x2) =y2and then look at functions of the type y∗(x) +ah(x). Hereais a real number, and h(x) is an abitrary but ﬁxed function that satisﬁes the boundary conditions h(x1) =h(x2) = 0. The functions y∗(x)+ah(x) then satisfy the same boundary conditions as y∗(x), i.e., y∗(x1)+ah(x1) =y1,y∗(x2)+ah(x2) =y2. We can evaluate the functional Kwith these functions as parameters, and get K[y∗+ah] =/integraldisplayx2 x1f(y∗(x) +ah(x),y′ ∗(x) +ah′(x),x)dx (1.2) which is just a real number. We can now consider the values of K[y∗+ah] we get for diﬀerent a, and look for which values of aourK[y∗+ah] becomes stationary. This is the type of stationarity problem we are know from calculus .K[y∗+ah] becomes stationary for all awith d daK[y∗+ah] = 0. (1.3) Now the point a= 0 corresponds to the function y∗, since for a= 0 we have y∗+ah=y∗. For Kto be stationary at y∗we thus have to demand that (1.3) holds for a= 0. Since hwas arbitrary, we have to demand this for all admissible functions h. We thus obtain the following deﬁnition of stationarity for functionals: Kisstationary at a function y∗if d daK[y∗+ah]|a=0= 0 (1.4) holds for all functions h(x) with h(x1) =h(x2) = 0. 8 CHAPTER 1. THE CALCULUS OF VARIATIONS 1.3 Euler-Lagrange equation We now want to derive a diﬀerential equation that can be used t o ﬁnd functions y∗(x) for which Kbecomes stationary. To do so, let us assume that the function y∗(x) is such a stationary point, and that it is also diﬀerentiabl e and satisﬁes our boundary conditions. First step: Use (1.4) If we use (1.2) and the chain rule, Eq. (1.4) turns into 0 =d daK[y∗+ah]|a=0 =/integraldisplayx2 x1/braceleftbiggd daf(y∗+ah,y′ ∗+ah′,x)/bracerightbigg a=0dx =/integraldisplayx2 x1/braceleftbigg∂f ∂y(y∗+ah,y′ ∗+ah′,x)h(x) +∂f ∂y′(y∗+ah,y′ ∗+ah′,x)h′(x)/bracerightbigg a=0dx. We thus obtain 0 =/integraldisplayx2 x1/braceleftbigg∂f ∂y(y∗(x),y′ ∗(x),x)h(x) +∂f ∂y′(y∗(x),y′ ∗(x),x)h′(x)/bracerightbigg dx (1.5) which needs to hold for all h(x) vanishing at x1andx2. (At this point, it is also convenient to drop the stars of y∗.) Second step: Integrate by parts We would like to simplify (1.5) such that both summands becom e proportional to h(x). This is easily achieved if we take the second summand and in tegrate by parts, /integraldisplayx2 x1∂f ∂y′ /bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright uh′ /bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright v′dx =∂f ∂y′h/vextendsingle/vextendsinglex2 x1−/integraldisplayx2 x1d dx/parenleftbigg∂f ∂y′/parenrightbigg h dx =−/integraldisplayx2 x1d dx/parenleftbigg∂f ∂y′/parenrightbigg h dx . Here the term∂f ∂y′h/vextendsingle/vextendsinglex2 x1could be dropped due to the boundary condition h(x1) = h(x2) = 0. Inserting this result into (1.5) we obtain /integraldisplayx2 x1/parenleftbigg∂f ∂y−d dx/parenleftbigg∂f ∂y′/parenrightbigg/parenrightbigg h(x) = 0 (1.6) for arbitrary h(x). Third step: Fundamental lemma of variational calculus Obviously, Eq. (1.6) is satisﬁed if the term in brackets vani shes. Indeed the funda- mental lemma of variational calculus guarantees that this i s the only way to satisfy the equation. 1.3. EULER-LAGRANGE EQUATION 9 Fundamental lemma of variational calculus Suppose that g(x) is continuous, and /integraldisplayx2 x1g(x)h(x) = 0 (1.7) holds for all h(x) satisfying h(x1) =h(x2) = 0. Then g(x) = 0 for all x∈(x1,x2). Proof: The proof proceeds by contradiction. Suppose that Eq. (1.7) holds for allh(x) and that we nevertheless have g(b)∝ne}ationslash= 0 for one b∈(x1,x2), say, g(b)>0 (see Fig. 1.3). Due to the continuity of g(x) this implies that we have g(x)>0 in an interval around b. We now pick h(x) such that h(x)>0 in this interval and h(x) = 0 outside. Then the integrand in Eq. (1.7) is positive in ou r interval and zero outside, and the integral must be positive as well. This cont radicts our assumption. Hence the theorem is proven. Figure 1.3: The continuous function g(x) is positive in an interval around b. The fundamental lemma of variational calculus implies that the term in brackets in Eq. (1.6) vanishes. We thus see that functions extremisin gK=/integraltextx2 x1f(y(x),y′(x),x)dx have to satisfy the Euler-Lagrange equation ∂f ∂y−d dx/parenleftbigg∂f ∂y′/parenrightbigg = 0 If we compare the Euler-Lagrange equation to the stationari ty condition for functions of a single variable, it appears natural to identi fy the term of the left- hand side with a kind of derivative. Indeed, δK δy=∂f ∂y−d dx∂f ∂y′ 10 CHAPTER 1. THE CALCULUS OF VARIATIONS is known as the functional derivative ofK, and one can build a whole theory of diﬀerentiation with respect to functions largely analogous to diﬀerentiation with respect to numbers or vectors.1This theory will not be considered further in this lecture, since the Euler-Lagrange equation is already suﬃcient for our purposes. 1.4 Solution of our problem The stationarity problem above was initially motivated by t he search for curves where the length l=/integraldisplayx2 x1(1 +y′(x)2)1/2dx becomes minimal. We are now ready to solve this problem. The E uler-Lagrange equation for f(y,y′,x) = (1 + y′2)1/2reads ∂(1 +y′2)1/2 ∂y−d dx∂(1 +y′2)1/2 ∂y′= 0. The summand on the left-hand side vanishes since (1 + y′2)1/2is independent of y. We thus ﬁnd d dx/bracketleftig (1 +y′2)−1/2y′/bracketrightig = 0 ⇒(1 +y′2)−1/2y′= const ⇒y′= const . The only curves with constant derivatives y′are straight lines y(x) =y′x+b . The constants y′,bare now determined by the boundary conditions y(x1) =y1,y(x2) = y2. To incorporate y(x1) =y1we set b=y1−y′·x1. This yields y(x) =y′·(x−x1) +y1 and indeed y(x1) =y1. The constant derivative y′must then coincide with the slope of the straight line leading from ( x1,y1) to (x2,y2) y′=y2−y1 x2−x1. 1.5 Alternative version of the Euler-Lagrange equation If the integrand finKdoes not depend explicitly on xthe Euler-Lagrange equation can be brought to a simpler form, also known as the Beltrami eq uation. 1See, e.g., Arfken, Mathematical Methods for Physicists or Gelfand and Fomin, Calculus of Variations . 1.6. THE BRACHISTOCHRONE 11 Alternative version of the Euler-Lagrange equation Iff=f(y,y′) the Euler-Lagrange equation turns into f−∂f ∂y′y′= const (where ”const” means that f−∂f ∂y′y′does not depend on x). Proof: Just use the chain rule to compute the total derivative of f−∂f ∂y′y′w.r.t. x: d dx/parenleftbigg f−∂f ∂y′y′/parenrightbigg =∂f ∂ydy dx+∂f ∂y′dy′ dx−/parenleftbiggd dx∂f ∂y′/parenrightbigg y′−∂f ∂y′dy′ dx Here the second and the fourth summand cancel. If we use the Eu ler-Lagrange equationd dx∂f ∂y′=∂f ∂ywe see that also the ﬁrst and third terms cancel, and we have d dx/parenleftbigg f−∂f ∂y′y′/parenrightbigg = 0, as desired. Note: In contrast to the original Euler-Lagrange equation, this i s a ﬁrst-order diﬀerential equation. It is typically much easier to use, si nce we can save the labour of taking derivatives w.r.t. yandx. If we apply this formula our problem of ﬁnding the curve y(x) with minimal length l=/integraltextx2 x1(1 +y′(x)2)1/2, we ﬁnd (1 +y′2)1/2−(1 +y′2)−1/2y′2= const ⇒y′= const as before, i.e., we again see that the shortest connection be tween two points is a straight line. 1.6 The brachistochrone Figure 1.4: Johann Bernoulli, Isaac Newton, Gottfried Leib niz, Guillaume de l’Hopital, and Jakob Bernoulli. A classical problem in the calculus of variations in the brac histochrone prob- lem (from Greek βρχσιζτoς χρoνoς , which means ”shortest time”). The brachis- tochrone problem was posed by Johann Bernoulli (1667-1746) inActa Eruditorum in 1696. He introduced the problem as follows:- 12 CHAPTER 1. THE CALCULUS OF VARIATIONS I, Johann Bernoulli, address the most brilliant mathematic ians in the world. Nothing is more attractive to intelligent people than an hon- est, challenging problem, whose possible solution will bes tow fame and remain as a lasting monument. Following the example set by Pa scal, Fermat, etc., I hope to gain the gratitude of the whole scient iﬁc commu- nity by placing before the ﬁnest mathematicians of our time a problem which will test their methods and the strength of their intel lect. If some- one communicates to me the solution of the proposed problem, I shall publicly declare him worthy of praise. Solution were given by Isaac Newton (1643-1727), Gottfried Leibniz (1646-1716), Guillaume de l’Hopital (1661-1704) and Jakob Bernoulli (16 54-1705, brother of the above). Figure 1.5: The brachistochrone problem. The problem is to ﬁnd the ideal form a a slide, such that a mass mthat is initially at rest at the origin (0 ,0) can be brought down to a point ( a,−b),b >0 in the shortest time possible, see Fig. 1.5. It is assumed that t he only force acting on the mass is gravity; there is no friction. The optimal form of the slide should then be represented as a function y(x) that assigns to each xcoordinate the corresponding height y. Among all possible paths between (0 ,0) and ( a,−b) the straight line would have the shortest length. On the other hand, the mass is the faster the lower it is since then more of its potential energy has been converted into kin etic energy. We thus expect that a slide where the mass goes down quickly (e.g. fro m (0,0) to (0 ,−b) and then to ( a,−b)) will be a good choice as well. One might guess that the optim al solution should lie somewhere in between these possibiliti es. Find time tfor given y(x) To solve the problem, we ﬁrst have to ﬁnd the time tthe mass needs to arrive at (a,−b) as a functional of the curve y(x). To get this time, we again split the curve into pieces. As we have seen in Section 2.1, the length of each piece is given by dl= (1 + y′(x)2)1/2dx. The time dtthe mass spends in each piece is then given by dt=dl vwhere vis the speed of the mass. This speed can be inferred from energ y 1.6. THE BRACHISTOCHRONE 13 conservation. At the starting point (0 ,0) the mass neither has potential nor kinetic energy. At a later point at height y <0 the potential energy is mgy, and the kinetic energy has increased tom 2v2. Energy conservation now implies that E= 0 = mgy+1 2mv2 ⇒v= (−2gy)1/2. Using this result for the speed and the length dlcalculated above we obtain the time spend in each piece of the curve as dt=dl v=/parenleftbigg1 +y′(x)2 −2gy(x)/parenrightbigg1/2 . If we integrate over dtwe see that the travel time of the mass is given by t=1 (2g)1/2/integraldisplaya 0/parenleftbigg1 +y′(x)2 −y(x)/parenrightbigg1/2 /bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright ≡fdx . Find minima We now have to ﬁnd the curve y(x) for which tbecomes minimal. To do so we use thealternative version of the Euler-Lagrange equation f−∂f ∂y′y′= const If we insert the fgiven above, we obtain /parenleftbigg1 +y′2 −y/parenrightbigg1/2 −1 2(1 +y′2)−1/22y′ (−y)1/2y′= const ⇒1 (−y)1/2(1 +y′2)1/2= const Thus the term ( −y)(1 + y′2) in the denominator must be constant. If we denote this constant by 2 Rwe obtain 1 +y′2=2R −y ⇒y′=∓/parenleftbigg2R+y −y/parenrightbigg1/2 (1.8) At least initially, the sign above must be negative since we w ould expect the mass to fall down. We have thus obtained a diﬀerential equation gi ving the derivative y′ as a function of y. Solving the diﬀerential equation We solve the diﬀerential equation by separation. We thus wri tey′=dy dx⇒dy y′=dx and then integrate on both sides. In our case the integration limits have to be 0 andaforxandy(0) = 0 and y(x) fory. We thus obtain /integraldisplayy(x) 0dy y′=/integraldisplayx 0d˜x=x 14 CHAPTER 1. THE CALCULUS OF VARIATIONS and, inserting the formula for y′, −/integraldisplayy(x) 0/parenleftbigg−y 2R+y/parenrightbigg1/2 dy=x . (1.9) To proceed we need to evaluate the integral over y. This can be accomplished with the trigonometric substitution y=−2Rsin2θ 2(1.10) Note that in the beginning of the curve we must have y= 0 and thus θ= 0. Diﬀerentiation now yields dy=−2Rsinθ 2cosθ 2dθ and the denominator in Eq. (1.9) simpliﬁes to (2R+y)1/2= (2R(1−sin2θ 2))1/2= (2R)1/2cosθ 2 If we substitute all these formulas into Eq. (1.9) we obtain x= 2R/integraldisplayθ(x) 0sin2θ 2dθ=R/integraldisplayθ(x) 0(1−cosθ)dθ=R(θ(x)−sinθ(x)).(1.11) Result One could attempt to solve (1.10) and (1.11) to get yin terms of xbut it is easier to represent the solution curve in parametrised form, leaving bothxandyas functions ofθ. Indeed (1.10) and (1.11) boil down to x(θ) = R(θ−sinθ) y(θ) = −2Rsin2θ 2=−R(1−cosθ) (1.12) A plot of the resulting x(θ),y(θ) is shown in ﬁgure 1.6. We see that for θ= 0,2π,4π,... the coordinate yreaches zero whereas xis equal to 0 ,2πR,4πR,... . At these points the curve has a cusp, and the slope becomes inﬁni te on both sides of the cusp. At θ=π,3π,5π,... there are minima with y=−2R. Strictly speaking our results only apply up to the ﬁrst minimum at θ=πsince we only considered the regime where ydecreases with increasing xand chose the sign in (1.8) accordingly. However by repeating our calculation with small modiﬁcatio ns for other values of θ, one can see that the solution obtained is actually valid for arbitary θ. It is instructive to to write (1.12) in vector notation, /parenleftbiggx(θ) y(θ)/parenrightbigg =/parenleftbigg0 −R/parenrightbigg +/parenleftbiggR 0/parenrightbigg θ+R/parenleftbigg−sinθ cosθ/parenrightbigg (1.13) Eq. (1.13) has the following interpretation: If we momentar ily forget about the third term on the right-hand side of (1.13), the vectors ( x,y) would start from (0 ,−R) at θ= 0; as θincreases they would then move on a straight line in positive x-direction. In contrast, the third term describes a motion around a circl e of radius R, i.e., a rotation. We thus see that as θincreases the points ( x,y) follow a superposition of a motion on a straight line and a rotation. 1.6. THE BRACHISTOCHRONE 15 Figure 1.6: θ,x, and yfor the brachistochrone problem. A related problem The same curve (1.13) but with opposite sign of yis obtained in a completely diﬀerent problem: the motion of a disk with radius Rrolling over the x-axis. This curve is called the cycloid curve. To also get the sign as in (1.13) we instead compare with the mo re artiﬁcial situation where a disk is “rolling” below thex-axis. If the disk is rolling into positive xdirection, its centre point moves on a straight line with inc reasing x. A point ( x,y) on the circumference of the disk follows that straight-line motion, but at the same time rotates around the centre. If one observes a rolling dis k, one can see that the point actually rotates to the left. This is in line with (1.13 ) since ( −sinθ,cosθ) =/parenleftbig cos/parenleftbigπ 2+θ/parenrightbig ,sin/parenleftbigπ 2+θ/parenrightbig/parenrightbig gives a rotation to the left if θis increased. During one revolution (i.e. an increase of θby 2π) the disk should move by a distance that coincides with the circumference 2 πR, i.e., xshould increase by 2 πRjust as in (1.13). Boundary conditions We still have to ﬁnd the right value for R. Moreover we must ﬁnd out which part of Fig. 1.6 should be taken as our ideal slide; this part must c ertainly start at the origin but we have to know at which value θendofθit has to end. Randθendcan be found by inserting the boundary conditions x(θend) =a,y(θend) =−binto Eq. (1.13). We then obtain the nonlinear system of equations a=R(θend−sinθend) −b=−R(1−cosθend) which will be considered further on the second problem sheet . Qualitatively the most interesting question is whether the slide will end to the left or to the right of the minimum at θ=π. In the ﬁrst case the slide will always go down (see Fig. 1.6a) whereas in the second case it will ﬁrst go down and then up again (see Fig. 1.6b). If we compare the curve in Fig. 1.6 to the line y=−2x π, we see that to the left of the minimum the curve always satisﬁe sy <−2x πwhereas to the right we have y >−2x πinserting x(θend) =a,y(θend) =−bwe thus see that the endpoint will be to the left if −b <−2a π⇔2a b< π . (1.14) 16 CHAPTER 1. THE CALCULUS OF VARIATIONS In this case the slide will always go down whereas otherwise i t ﬁrst goes down and then up. Figure 1.7: Solution of the brachistochrone problem: (a) fo r2a b< π, (b) for2a b> π. 1.7 Functionals depending on several functions So far we only considered variational problems that involve d one function y(x) map- pingRtoR. However one often encounters problems that involve severa l functions of this type. Let us thus consider ndiﬀerent functions y1(x),y2(x),... ,y n(x), each subject to boundary conditions at x1andx2, and then deﬁne a functional depending on all of them K[y1,... ,y n] =/integraldisplayx2 x1f(y1(x),... ,y n(x),y′ 1(x),... ,y′ n(x),x)dx . We want to ﬁnd y1(x),y2(x),... ,y n(x) such that Kbecomes stationary w.r.t. vari- ations of all nfunctions. Kwill be stationary w.r.t. variations of yj(x) if the corresponding Euler-Lagrange equation ∂f ∂yj−d dx∂f ∂y′ j= 0 (1.15) is satisﬁed. To make Kstationary w.r.t. variations of all function we thus have to demand that (1.15) holds for all jfrom 1 to n. Note: It is often convenient to collect all functions yj(x) into a vector-valued function y(x) = (y1(x),... ,y n(x)). 1.8 Fermat’s principle To illustrate the use of variational calculus in Rnwe consider an example from optics. (Geometric) optics can be based on a variational pri nciple formulated by Fermat: Fermat’s principle Light travels between two points on paths (rays) that take th e least (or stationary) time. These light rays are very similar to particle trajectories i n mechanics. To make use of Fermat’s principle we have to recall that the speed of ligh t is given byc nwhere 1.8. FERMAT’S PRINCIPLE 17 cis the speed of light in vacuum and the refraction index n >1 depends on the medium in which the light is propagating. In many applica tions the refraction index and thus the speed of light is constant. Then Fermat’s p rinciple implies that also the length of the rays is stationary. Light thus moves on straight lines . (It can also be reﬂected by a mirror; as seen on a problem sheet ray s where light is reﬂected from the mirror according to the reﬂection law ”ang le of incidence = angle of reﬂection” have stationary length.) We now want to consider the situation where the refraction index is not constant , i.e., we have n=n(x,y,z). Find tas functional of the path Figure 1.8: Propagation of a light ray in R3. We ﬁrst need to get the travel time tas a functional of the path. As shown in ﬁgure 1.8 the path can be described by functions y(x) and z(x) that assign to each value of xthe corresponding coordinates yandz. If we break the path into pieces, reasoning analogous to section 1.1 and 1.6 shows that the len gthdlof each piece is given by (dl)2= (dx)2+ (dy)2+ (dz)2 Using that dy=y′dx,dz =z′dxwe thus ﬁnd dl= (1 + y′2+z′2)1/2dx . The travel time corresponding to dlcan be obtained if we divide by the velocity of the pathc n. We then obtain dt=n cdl and integration yields the travel time of the whole ray t=/integraldisplayx2 x1n(x,y(x),z(x) c(1 +y′(x)2+z′(x)2)1/2dx 18 CHAPTER 1. THE CALCULUS OF VARIATIONS Find stationary points of t We now need to ﬁnd y(x),z(x) such that the travel time tbecomes extremal. We thus face an extremisation problem of the type discussed in s ection 1.7, with the identiﬁcations K=t (y1,y2) = ( y,z) f=n(x,y(x),z(x) c(1 +y′(x)2+z′(x)2)1/2 /bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright =u(x) To solve this extremisation problem we have to consider the Euler-Lagrange equations for both y(x) and z(x). For y(x) we obtain ∂f ∂y−d dx∂f ∂y′= 0 ⇒1 c∂n ∂yu−d dx/parenleftbiggn c1 2u−12y′/parenrightbigg = 0 ⇒d dx/parenleftbiggny′ u/parenrightbigg =∂n ∂yu (1.16) Forz(x) analogous reasoning yields ∂f ∂z−d dx∂f ∂z′= 0 ⇒d dx/parenleftbiggnz′ u/parenrightbigg =∂n ∂zu (1.17) These equations must now be solved for y(x),z(x) to get rays. Special case Figure 1.9: A ray of light in two dimensions. 1.8. FERMAT’S PRINCIPLE 19 For deﬁniteness, let us consider the special case where the refraction index depends only on x, i.e. n=n(x) and∂n ∂y=∂n ∂z= 0. Then the Euler-Lagrange equations (1.16) and (1.17) boil down to ny′ u= const nz′ u= const (1.18) Let us moreover assume that the z-coordinate is always zero, a condition that is certainly in line with (1.18). In this case the light rays onl y travel in the x−y-plane, see ﬁgure 1.9. If we insert the deﬁnition of uthe second equation in (1.18) then boils down to ny′ (1 +y′2)1/2= const (1.19) Eq. (1.19) can be simpliﬁed if we express the slope y′(x) of the curve in Fig. 1.9 through the angle θ(x) enclosed between the curve and the x-direction. We then write y′(x) = tan θ(x) and simplify the denominator in (1.19), (1 +y′(x)2)1/2= (1 + tan2θ(x))1/2=/parenleftbiggcos2θ(x) + sin2θ(x) cos2θ(x)/parenrightbigg1/2 =1 cosθ(x). Eq. (1.19) thus turns into n(x)sinθ(x) = const which is known as (the generalised version of) Snell’s law . For example, let us assume that the space x <0 is ﬁlled with a medium with constant refraction index n=n1(say, air) and the space x >0 is ﬁlled with a diﬀerent medium with n=n2(say, glass), see Fig. 1.10. Inside both media the rays travel on straight lines enclosing angles θ1andθ2with the x-direction. Given θ1the angle θ2is then determined by n1sinθ1=n2sinθ2 This equation is the original formulation of Snell’s law. 20 CHAPTER 1. THE CALCULUS OF VARIATIONS Figure 1.10: Snell’s law. Chapter 2 Lagrangian mechanics 2.1 Reminder: Newton We now want to formulate mechanics through a variational pri nciple. As a prepa- ration, we need to review some basic facts from Newtonian mec hanics. We consider systems of Nparticles with masses m1,m2,... ,m Nat positions r1,r2,...rN. Each of these positions is a vector ri= (xi,yi,zi)∈R3. The particle trajectories ri(t) are now determined by Newton’s second law mi¨ri(t) =Fi(r1(t),... ,rN(t),˙r1(t),... ,˙rN(t)) where Fiis theforce acting on the i-th particle. A particularly important type of forces are conservative forces: Forces are conservative if they can be written as derivatives of a potential energy U, Fi=−∂ ∂riU(r1,... ,rN). Here∂U ∂riis the gradient ∂U/∂x i ∂U/∂y i ∂U/∂z i . Example : In a uniform gravity ﬁeld (e.g. the gravity ﬁeld of the earth in the vicinity of the surface of the earth) a particle at r= (x,y,z) has the potential U=mgz , and thus feels the force F=− ∂U/∂x ∂U/∂y ∂U/∂z = 0 0 −mg  Finally, our system of Nparticles has the kinetic energy T=N/summationdisplay i=11 2mi˙r2 i. 21 22 CHAPTER 2. LAGRANGIAN MECHANICS 2.2 Lagrangian mechanics in Cartesian coordinates In Lagrangian mechanics, the laws of motion are formulated i n terms of variational calculus, i.e., by demanding that a certain functional shou ld become stationary. This approach has several advantages over direct use of Newt onian mechanics that will become clear over the course of this lecture. We will ﬁrs t develop the variational formulation for the simplest case, assuming that all forces are conservative and all particle coordinates are indicated in Cartesian coordinat es. The functions to be determined in mechanics are the particle trajectories ri(t). We thus need a variational principle that tells us how partic les move from given positions ri(t(1)) =r(1) iat time t(1)to given positions ri(t(2)) =r(2) iat time t(2). In our variational principle the role of the integrand fwill be played by the diﬀerence of the kinetic and potential energy, the so-calle dLagrangian L(r1,... ,rN,˙r1,... ,˙rN) =T(˙r1,... ,˙rN)−U(r1,... ,rN) The Lagrangian depends both on the particle positions (dete rmining U) and the ve- locities (determining T). We then deﬁne a functional called the action S, by taking the time integral of the Lagrangian from t(1)tot(2): S[r1,... ,rN] =/integraldisplayt(2) t(1)L(r1(t),... ,rN(t),˙r1(t),... ,˙rN(t))dt . Now the claim is that the particles travel on trajectories ri(t) for which Sbe- comes stationary. For historical reasons this famous princ iple (due to Hamilton and Maupertius) is known as the principle of least action (rathe r than stationary action, which would be the correct terminology). Principle of “least” action In a systen where all forces are conservative the trajectori es of particles moving from positions ri(t(1)) =r(1) ito positions ri(t(2)) =r(2) iare chosen such that the action becomes stationary w.r.t. variations that prese rve these boundary conditions, i.e., the Euler-Lagrange equations ∂L ∂ri=d dt∂L ∂˙ri(2.1) must be satisﬁed, for all i= 1... N. Here the vector equation (2.1) implies that the coordinates xiof the i-th particle must satisfy∂L ∂xi=d dt∂L ∂˙xi, and analogous equations hold for yiandzi. Altogether we thus obtain 3 Nequations for Nparticles described by 3 coordinates each. In the context of Lagrangian mechanics the Euler-Lagrange equati ons are usually called Lagrange’s equations. Note: When comparing to the section 1, we have to identify K=S,x=tand (y1(x),... ,y n(x)) = (r1(t),... ,rN(t)). Proof: The principle of least action is equivalent to Newton’s seco nd law. If all forces are conservative this can be shown in a surprising ly simple way. The 2.2. LAGRANGIAN MECHANICS IN CARTESIAN COORDINATES 23 derivative∂L ∂riin the Lagrange equation can be written as ∂L ∂ri=−∂U ∂ri=Fi where we used that the Lagrangian depends on the particle coo rdinates rionly through the potential and that derivatives of the potential yield forces. On the other hand, the Lagrangian depends on the velocities ˙rionly through the kinetic energy. We thus obtain ∂L ∂˙ri=∂T ∂˙ri=mi˙ri ⇒d dt∂L ∂˙ri=mi¨ri. By inserting these result into Eq. (2.1) we see that the Lagra nge equations are equivalent to Fi=mi¨ri, i.e., Newton’s second law. Examples: •If there is no potential and just a single particle we have L=T=1 2m˙r2and S=/integraltextt(2) t(1)1 2m˙r2(t)dtand the Lagrange equation reads ∂L ∂r=d dt∂L ∂˙r⇒0 =d dtm˙r⇒˙r= const , i.e., the particle is moving on a straight line with constant velocity. (Note that if we would just demand that the length becomes stationa ry, we would get slightly less: we would only see that there is a straight l ine, but not that this line is traversed with constant velocity.) •A particle at r= (x,y,z) in a uniform gravity ﬁeld has the Lagrangian L=T−U=1 2m( ˙x2+ ˙y2+ ˙z2)−mgz The Lagrange equations for x,yandzread ∂L ∂x=d dt∂L ∂˙x⇒ 0 =d dtm˙x⇒˙x= const ∂L ∂y=d dt∂L ∂˙y⇒ 0 =d dtm˙y⇒˙y= const ∂L ∂z=d dt∂L ∂˙z⇒ − mg=d dtm˙z⇒¨z=−g As expected, we get an acceleration gin negative x-direction. 24 CHAPTER 2. LAGRANGIAN MECHANICS Figure 2.1: Two-dimensional pendulum. 2.3 Generalised coordinates and constraints The great advantage of Lagrangian mechanics is that it can be elegantly generalised to arbitrary coordinate systems, and to systems where parti cles are not allowed to go into certain directions. Let us thus consider a system of several particles, whose pos itions are given in arbitrary coordinates (Cartesian, polar, spherical coord inates, diﬀerences between particle positions, whatever ...). These generalised coordinates are then denoted by q1,q2,... ,q d. If the particles are allowed to go anywhere , the number of coordinates needed to describe each particle is given by the number of dimension s of the system (i.e. usually three). Altogether the number of variables is thus d= #particles ·#dimensions . However there are also situations where particles are not al lowed to go in certain directions (i.e., there are constraints on the position of particles). In this case, we have to drop the coordinates corresponding to these directi ons and the number of variables is given by d= #particles ·#dimensions −#constraints . (2.2) Example: Consider a pendulum (see Fig. 2.1) where a mass m(the “bob”) is attached to some point through a rod of ﬁxed length. If we take the latter point as the origin, we can use polar coordinates. But only the angula r coordinate θchanges as the pendulum swings back and forth1, the distance ρfrom the origin stays ﬁxed and can therefore be dropped as a variable. We thus need only o ne variable instead of two. 1In contrast to usual polar coordinates, this coordinate is o ften taken as the angle enclosed with the negative y-direction, not the positive x-direction 2.3. GENERALISED COORDINATES AND CONSTRAINTS 25 We will see several more complicated examples of such constr aints later in the lecture. The space of all allowed positions of particles par ametrised by the co- ordinates q1,q2,... q dis called the conﬁguration space of the system. These coordinates are also called the degrees of freedom of the system. Relation between generalised and Cartesian coordinates The generalised coordinates determine the Cartesian parti cle positions ri, i.e., we can write rias a function of q1,... ,q d. Being slightly more general, we could could allow for rito depend on times as well, and write ri=ri(q1,... ,q d,t). (2.3) (This includes the case that e.g. the origin of our generalis ed coordinate system moves in time – a situation that won’t occur often in this cour se.) Given Eq. (2.3), the particle velocities ˙rican be determined using the chain rule, ˙ri=dri dt=d/summationdisplay α=1∂ri ∂qα˙qα+∂ri ∂t. (2.4) Here the right-hand side involves generalised coordinates , their derivatives and time. Hence we can express ˙rias a function of q1,... ,q d,˙q1,...˙qdandt. Lagrangian mechanics We can now use these relations to express all quantities rele vant for Lagrangian mechanics in terms of the new coordinates. Using (2.3) we can write the potential energy as a function U=U(q1,... ,q d,t). Expressing the velocities through (2.4) we can write the kin etic energy as a function T= (q1,... ,q d,˙q1,... ,˙qd,t). The Lagrangian thus turns into a function L(q1,... ,q d,˙q1,... ,˙qd,t) =T(q1,... ,q d,˙q1,... ,˙qd,t)−U(q1,... ,q d,t), and the action can be written as a functional depending on the functions q1(t),... ,q d(t), S[q1,... ,q d] =/integraldisplay(2) t(1)L(q1(t),... ,q d(t),˙q1(t),... ,˙qd(t),t)dt . The boundary conditions can be expressed as qα(t(1)) =q(1) α, q α(t(2)) =q(2) α. We now claim that a result analogous to Eq. (2.1) also holds fo rq1,... ,q d. This means that (i) the Lagrangian formulation of mechanics holds is valid for arbitrary coordinate systems and (ii) that it also holds for systems with constraints. The second point applies even though the forces that give ris e to these constraints (e.g. the tension force preventing us from increasing the le ngth of a pendulum) are typically non-conservative! 26 CHAPTER 2. LAGRANGIAN MECHANICS Principle of “least” action (general form) Consider systems where all forces are either conservative or give rise to con- straints . For such systems all qα(t) are chosen such that the action Sbecomes stationary w.r.t. variations of qα(t) that preserve the boundary conditions at t(1) andt(2). We thus have∂L ∂qα=d dt∂L ∂˙qα for all α= 1... d. Theproof of this statement will be given later, after discussing some examples. (The tricky bit will be the generalisation to systems with co nstraints. If not for the constraints, we could give a very short proof. Essentially, we could invoke the proof for Cartesian coordinates and then argue that an extremum of the action remains an extremum of the action regardless of which system of coord inates we are working in.) 2.3.1 Gravitational ﬁeld As a ﬁrst example for Lagrangian mechanics with generalised coordinates let us consider the trajectory r(t) of a mass m(e.g. the earth) in the gravitational ﬁeld of a mass M(e.g. the sun) at the origin. The corresponding gravitation al potential reads U=−GmM |r|and the kinetic energy is, of course, given by T=1 2m˙r2. The symmetry of the problem now suggests to work in spherical coordinates , where the vectors rare parametrised by their distance from the origin ρand two angles θandφ: r=ρ sinθcosφ sinθsinφ cosθ  In spherical coordinates the potential energy Uturns into U=−GmM ρ. To get the kinetic energy, we use that ˙r= ˙ρ sinθcosφ sinθsinφ cosθ +ρ˙θ cosθcosφ cosθsinφ −sinθ +ρ˙φsinθ −sinφ cosθ 0  where the three vectors multiplied with ˙ ρ,ρ˙θandρ˙φsinθare all normalised and perpendicular to each other. We thus get T=m 2˙r2=m 2( ˙ρ2+ρ2˙θ2+ρ2˙φ2sin2θ) Note that in contrast to the Cartesian case Tdoes not only depend on the derivatives ˙ρ,˙θ,˙φbut also on ρandθ. TheLagrangian can now be written as L(ρ,θ,φ, ˙ρ,˙θ,˙φ) =T−U=m 2( ˙ρ2+ρ2˙θ2+ρ2˙φ2sin2θ) +GmM ρ 2.3. GENERALISED COORDINATES AND CONSTRAINTS 27 and the action Scan be written as a functional of ρ(t),θ(t), and φ(t): S=/integraldisplayt(2) t(1)L(ρ(t),θ(t),φ(t),˙ρ(t),˙θ(t),˙φ(t))dt . (2.5) The principle of stationary action now gives rise to Lagrange equations forρ,θ andφ: ∂L ∂ρ=d dt∂L ∂˙ρ⇒mρ˙θ2+mρsin2θ˙φ2−GmM ρ2=m¨ρ ∂L ∂θ=d dt∂L ∂˙θ⇒mρ2sinθcosθ˙φ2=d dt(mρ2˙θ) ∂L ∂φ=d dt∂L ∂˙φ⇒0 =d dt(mρ2˙φsin2θ) We thus see that Lagrange’s formulation of mechanics makes i t simple to switch between coordinate systems: One only has to rewrite the Lagr angian in terms of the new coordinates and invoke Lagrange’s equations which h ave the same form in every coordinate system. 2.3.2 Pendulum To illustrate Lagrangian mechanics for systems with constr aints, let us consider the example of the pendulum. The generalised coordinate −π < θ < π depicted in Fig. 2.1 determines the x- and y-coordinates of the mass mas x=lsinθ y=−lcosθ . The derivatives of these coordinates read ˙x=lcosθ˙θ ˙y=lsinθ˙θ and determine the kinetic energy as T=1 2m( ˙x2+ ˙y2) =1 2ml2˙θ2 Thepotential energy is simply given by U=mgy=−mglcosθ TheLagrangian thus takes the form L=T−U=1 2ml2˙θ2+mglcosθ . TheLagrange equation for our generalised coordinate θnow reads ∂L ∂θ=d dt∂L ∂˙θ. 28 CHAPTER 2. LAGRANGIAN MECHANICS Figure 2.2: Inclined plane. If we do the derivatives ∂L ∂˙θ=ml2˙θ d dt∂L ∂˙θ=ml2¨θ ∂L ∂θ=−mglsinθ , this turns into −mglsinθ=ml2¨θ ⇒¨θ=−g lsinθ . (2.6) Now one can proceed as in Newtonian mechanics, i.e, approxim ate sin θbyθ. The resulting equation ¨θ=−g lθ has the solution θ(t) =Acos/parenleftbigg/radicalbiggg lt+φ/parenrightbigg where A and φare constants. 2.3.3 Inclined plane As a further example, we consider an inclined plane (see Fig. 2.2). This plane slides freely on a horizontal table, and a block slides freely on the plane. The mass of the plane is M, and the mass of the block is m. We thus have two constraints : The block must remain on the plane and the plane must remain on the table.2 Convenient generalised coordinates are the x-coordinate of the plane and the distance sof the block from the beginning of the plane, as sketched in Fi g. 2.2. These coordinates determine the position of the plane as /parenleftbiggx 0/parenrightbigg 2Note that we are not interested in the motion of the block once it has slid down the plane. 2.3. GENERALISED COORDINATES AND CONSTRAINTS 29 and the position of the block as r=/parenleftbiggx+scosα ssinα/parenrightbigg Thekinetic energy of the plane reads1 2M˙x2, whereas the block has the kinetic energy1 2m˙r2. To express1 2m˙r2in terms of our generalised coordinates we write ˙r=/parenleftbigg˙x+ ˙scosα ˙ssinα/parenrightbigg ˙r2= ( ˙x+ ˙scosα)2+ (˙ssinα)2= ˙x2+ ˙s2+ 2 ˙x˙scosα . The overall kinetic energy is thus obtained as T=1 2M˙x2+1 2m( ˙x2+ ˙s2+ 2 ˙x˙scosα) =1 2(M+m) ˙x2+1 2m˙s2+m˙x˙scosα . The only contribution to the potential energy Uis due to the height of the block, U=mgssinα . TheLagrangian thus reads L=T−U=1 2(M+m) ˙x2+1 2m˙s2+m˙x˙scosα−mgssinα . We now obtain two Lagrange equations , one for the coordinate xand one for s. Forxwe get ∂L ∂x−d dt∂L ∂˙x= 0. With the derivatives ∂L ∂˙x= (M+m) ˙x+mcosα˙s d dt∂L ∂˙x= (M+m)¨x+mcosα¨s ∂L ∂x= 0 this turns into −(M+m)¨x−mcosα¨s= 0. (2.7) The Lagrange equation for sreads ∂L ∂s−d dt∂L ∂˙s= 0. If we use the derivatives ∂L ∂˙s=m˙s+mcosα˙x d dt∂L ∂˙s=m¨s+mcosα¨x ∂L ∂s=−mgsinα 30 CHAPTER 2. LAGRANGIAN MECHANICS and cancel the factors −mthis simpliﬁes to ¨s+ cos α¨x+gsinα= 0. (2.8) We have thus obtained two coupled equations (2.7), (2.8) for the second derivatives ¨xand ¨s. To obtain separate equations for ¨ xand ¨s, we solve (2.7) for ¨ xand use it to eliminate ¨ xin (2.8). This yields ¨s−m M+mcos2α¨s+gsinα= 0 ⇒(M+m)¨s−mcos2α¨s+ (M+m)gsinα= 0 ⇒(M+msin2α)¨s+ (M+m)gsinα= 0 which ﬁnally leads to ¨s=−(M+m)gsinα M+msin2α= const . (2.9) If we substitute (2.9) back into (2.7) we get ¨x=mg M+msin2αsinαcosα= const . (2.10) Eq. (2.9) and (2.10) give the second derivatives of our gener alised coordinates as constants depending on the parameters of the problem. One ea sily checks that in special cases like α→0,α→π 2orm→0 these results agree with what we might expect. (E.g. for a perpendicular plane with α=π 2the block simply falls down with ¨s=−gwhereas the plane stays ﬁxed.) Assuming that the block and th e plane are initially at rest we can integrate (2.9) and (2.10), to ge t x(t) = x(0) +1 2¨xt2 s(t) = s(0) +1 2¨st2. For comparison: Inclined plane with Newton It is instructive to compare the above treatment of the incli ned plane with New- tonian mechanics. The main diﬀerence will be that in Newtoni an mechanics the constraints can no longer be built in by choosing appropriat e generalised coordi- nates. Instead one has to work in Cartesian coordinates and t ake into account all forces acting on the plane and the block; in particular this i ncludes forces that make sure that the constraints are satisﬁed. Forces acting on the inclined plane For the inclined plane Newton’s law implies FP=MaP where aP=/parenleftbigg¨x 0/parenrightbigg is the acceleration of the plane and FPis the sum of all forces acting on the plane (see Fig. 2.3): 2.3. GENERALISED COORDINATES AND CONSTRAINTS 31 Figure 2.3: Forces acting on the inclined plane. •This includes the gravitational force FgP=/parenleftbigg0 −Mg/parenrightbigg . •Moreover we have the constraint that the plane stays on the ta ble. Hence there must be a force coming from the table with keeps the plan e from “falling through” the table. This force is due to the rigidity of the ta ble (and can be felt by pressing on a table!) It must be oriented in a vertical direction, i.e., normal to the table, and thus be of the form N1=/parenleftbigg0 N1/parenrightbigg where N1is an undetermined constant. •In addition the block exerts a force on the plane. Due to the ge ometry of the system, this force should press the plane downwards, put als o push it to the right. The precise direction of this force should be normal t o the upper side of the inclined plane. This is easily understood if we realise t hat the force could never cause a tangential motion of the plane along the interf ace between the block and the plane. The normal force should thus be of the for m (compare Fig. 2.3) N2=/parenleftbiggN2sinα −N2cosα/parenrightbigg . The overall force acting on the plane is therefore given by FP=FgP+N1+N2=/parenleftbiggN2sinα −Mg+N1−N2cosα/parenrightbigg 32 CHAPTER 2. LAGRANGIAN MECHANICS Figure 2.4: Forces acting on the block. where N1andN2are undetermined constants. Now the two components of the vector equation FP=MaPread N2sinα=M¨x⇒N2=M¨x sinα(2.11) and −Mg+N1−N2cosα= 0⇒N1=Mg+N2cosα . (2.12) Inserting the former into the latter equation we obtain N1=M(g+ ¨xcotα). Forces acting on the block For the block Newton’s law assumes the form FB=maB with the acceleration aB=d2r dt2=d2 dt2/parenleftbiggx+scosα ssinα/parenrightbigg =/parenleftbigg¨x+ ¨scosα ¨ssinα/parenrightbigg . The force FBis a sum of •the gravitational force FgB=/parenleftbigg0 −mg/parenrightbigg •and a force N3exerted by the plane on the block. The latter originates from the rigidity of the plane and makes sure that the block does no t fall through the plane. Hence it is another force of constraint. Due to New ton’s third law it must be equal to the negative of the force that the block exe rts on the plane. We thus have (using (2.11)) N3=−N2=/parenleftbigg−N2sinα N2cosα/parenrightbigg =M¨x sinα/parenleftbigg−sinα cosα/parenrightbigg . 2.3. GENERALISED COORDINATES AND CONSTRAINTS 33 Summation yields the overall force FB=FgB+N3=/parenleftbigg−M¨x −mg+M¨xcotα/parenrightbigg . Now the two components of Newton’s law read m(¨x+ ¨scosα) =−M¨x ⇒¨x=−m m+M¨scosα (2.13) and m¨ssinα=−mg+M¨xcotα =−mg−mM m+M¨scos2α sinα(2.14) (where in the last step we used Eq. (2.13)). If we multiply Eq. (2.14) with sin αm+M m we obtain (m+M)¨ssin2α=−(m+M)gsinα−M¨scos2α ⇒(msin2α+M)¨s=−(m+M)gsinα ⇒¨s=−(M+m)gsinα M+msin2α= const . (2.15) Substitution into (2.13) then yields ¨x=mg M+msin2αsinαcosα= const . (2.16) Discussion We see that Lagrange’s equations and Newton’s law yield coin ciding results, given in Eqs. (2.9), (2.10) and in Eqs. (2.15), (2.16). The Lagrang ian treatment is considerably simpler since we can build in the constraints b y appropriately choosing coordinates. In the Newtonian approach we instead have to ta ke into account additional forces. These forces N1,N2,N3are forces of constraint. They keep the block on the plane and the plane on the table. These forces don’t appear in the Lagrangian treatment. However we can always determine them if we want (e.g., if we need the force exerted on the table to see if it breaks). We t hen have to solve Lagrange’s equations and determine determine the forces of constraint from m¨ri and the known conservative forces Fpotas in mi¨ri=Fpot i+Fconstraint i ⇒Fconstraint i =Fpot i−mi¨ri. 2.3.4 General properties of forces of constraint We will now discuss forces of constraint in more general term s. We want to speak of aforce of constraint when a force makes sure that the constraints are satisﬁed but has no further impact on the motion. Such forces must point in directions where the particles are forbidden from going and compensate all other forces that may point in these directions. They may have no componen ts in directions 34 CHAPTER 2. LAGRANGIAN MECHANICS where the particles are allowed to go. An example is the norma l force N1in the example of the inclined plane, see Fig. 2.5. This force point s in the direction normal to the table, and compensates the gravitational force and th e force from the block which push the inclined plane into a forbidden direction. Figure 2.5: Example for a force of constraint. Allowed and forbidden directions To formulate the above condition mathematically we have to l ook in more detail at the directions in which a system is, or is not, allowed to move . •First of all we note that all particle positions ( r1,... ,rN) form a 3 N-dimensional space (or a 2 N-dimensional space if the position vectors are in R2). •However not all positions are allowed. We parametrize the al lowed positions of particles by dgeneralised coordinates q1,... ,q dand (possibly) time as ri=ri(q1,... ,q d,t). This deﬁnes the d-dimensional conﬁguration space. •Starting from an allowed ( r1,... ,rN) we can go into all directions in the 3N-dimensional space that can be reached by changing the q1,... ,q d. For instance, if we change qαby an inﬁnitesimal amount δqα, each particle position richanges by δri=∂ri ∂qαδqα. This means that all inﬁnitesimal motions in the 3N-dimensional space that go in directions /parenleftbigg∂r1 ∂qα,... ,∂rN ∂qα/parenrightbigg (2.17) are allowed. The same applies to linear combinations of thes e directions. At every point ( r1,... ,rN) there are ddirections as in (2.17), one for each α= 1,... ,d . •This leaves 3 N−dlinearly independent forbidden directions (one for each constraint). They are perpendicular to the allowed directi ons. Now consider a set of forces ( Fc 1,Fc 2,... ,Fc N), where Fc 1acts on the ﬁrst particle, Fc 2acts on the second particle, etc. These forces are forces of c onstraint if the 3 N- dimensional vector ( Fc 1,Fc 2,... ,Fc N) points into a forbidden direction (such that other forces pointing into these directions are compensate d) and is perpendicular to 2.3. GENERALISED COORDINATES AND CONSTRAINTS 35 the allowed directions (such that the motion in allowed dire ctions is not aﬀected). This means that the scalar product of ( Fc 1,Fc 2,... ,Fc N) and any of the allowed directions must vanish. Forces of constraint can thus be deﬁ ned as follows: Deﬁnition (D’Alembert’s principle) A set of forces ( Fc 1,Fc 2,... ,Fc N) areforces of constraint if we have (Fc 1,Fc 2,... ,Fc N)·/parenleftbigg∂r1 ∂qα,... ,∂rN ∂qα/parenrightbigg =N/summationdisplay i=1Fc i·∂ri ∂qα= 0 (2.18) for all α= 1... d Note: The term “allowed direction” used here is not standard. Trad itionally, these allowed directions are rather referred to a “virtual displacements” . Fur- thermore note that the left-hand side of Eq. (2.18) has the di mension of work, since it involves a product of forces and changes of position s. The traditional way of stating D’Alembert’s principle is thus to say that virtua l displacements do no work. 2.3.5 Derivation of Lagrange’s equations from Newton’s law in the general case We have now learnt enough about constraints to give a proof fo r Lagrange’s equa- tions and the principle of least action for generalised coordinates andsystems with constraints . We will see that d’Alembert’s principle is crucial for show ing that forces of constraint don’t spoil the picture. We consider systems where all forces are either conservativ e forces or forces of constraint. For these systems Newton ’s second law reads: mi¨ri=Fi=−∂U ∂ri+Fc i (2.19) where Fiis the overall force acting on the i-th particle. It is written as a sum of the conservative force −∂U ∂riand the force of constraint Fc i. We now want to show that Lagrange ’s equations hold as well. I.e., if we parametrise the partic le positions by generalised coordinates where the constraint are automatically built in, ri=ri(q1,... ,q d,t), (2.20) then we have∂L ∂qα−d dt∂L ∂˙qα= 0 (2.21) for all α= 1... d. This means that Hamilton’s principle holds, i.e., that par ticles travel on trajectories for which the action S=/integraltext Ldtbecomes extremal. Preparation: Formulas for partial derivatives of ˙ri To prepare for a proof, we ﬁrst derive formulas for the deriva tives of the velocities ˙riw.r.t. the generalised coordinates qαand their derivatives ˙ qα. In Eq. (2.4) we 36 CHAPTER 2. LAGRANGIAN MECHANICS took the derivative of ri=ri(q1,... ,q d,t), (2.22) using the chain rule, and got ˙ri=dri dt=d/summationdisplay β=1∂ri ∂qβ(q1,... ,q d,t) ˙qβ+∂ri ∂t(q1,... ,q d,t). (2.23) We thus expressed the velocity as a function ˙ri=˙ri(q1,... ,q d,˙q1,... ,˙qd,t). We can now take derivatives of ˙riw.r.t. generalised coordinates and ˙ qβ’s. If we look at (2.23) we see that the derivative w.r.t. ˙ qβis just the term multiplying ˙ qβ. We thus have ∂˙ri ∂˙qβ=∂ri ∂qβ. (2.24) Since the two dots on the left-hand side have disappeared on t he right-hand side, this rule is also known as the “cancellation of dots” . For the derivatives w.r.t. generalised coordinates we will show that ∂˙ri ∂qα=d dt∂ri ∂qα. (2.25) This means that the total derivative w.r.t. t(which appears on the left-hand side as a dot) and the partial derivative w.r.t. qαcan be interchanged – which would be trivial if both derivatives were partial, but it is necessar y to give a proof because one of the derivatives is a total one. Proof: To evaluate the l.h.s., we use (2.23). The only qα-dependent terms in (2.23) are∂ri ∂qβand∂ri ∂t. Hence the chain rule yields ∂˙ri ∂qα=d/summationdisplay β=1∂2ri ∂qα∂qβ˙qβ+∂2ri ∂qα∂t. To compute the termd dt∂ri ∂qαon the r.h.s., we use∂ri ∂qαis a function of the generalised coordinates and time. Hence the chain rule brings about part ial derivatives with respect to these quantities, and the ﬁnal result d dt∂ri ∂qα=d/summationdisplay β=1∂2ri ∂qα∂qβ˙qβ+∂2ri ∂qα∂t agrees with the left-hand side. Eq. (2.25) is thus proven. Strategy We are now ready to prove Lagrange’s equation. We start from N ewton’s law, multiply both sides with∂ri ∂qα, and sum over i. This yields N/summationdisplay i=1mi¨ri∂ri ∂qα=N/summationdisplay i=1/parenleftbigg −∂U ∂ri+Fc i/parenrightbigg /bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright =Fi·∂ri ∂qα. (2.26) 2.3. GENERALISED COORDINATES AND CONSTRAINTS 37 A motivation for this strategy is that we would like to get an e xpression that involves partial derivatives w.r.t. qα– hence it is a good idea to multiply with∂ri ∂qα. We now have to simplify all three terms obtained to get Lagrange. Acceleration term For the ﬁrst term involving the acceleration ¨ri, we pull one of the two time deriv- atives in front such that is also acts on∂ri ∂qα, and then subtract the term where the time derivative acts on∂ri ∂qα, N/summationdisplay i=1mi¨ri·∂ri ∂qα=N/summationdisplay i=1mi/bracketleftbiggd dt/parenleftbigg ˙ri·∂ri ∂qα/parenrightbigg −˙ri·d dt∂ri ∂qα/bracketrightbigg . We then use Eq. (2.24) for the ﬁrst term, and Eq. (2.25) for the second term, N/summationdisplay i=1mi¨ri·∂ri ∂qα=N/summationdisplay i=1mi d dt/parenleftbigg ˙ri·∂˙ri ∂˙qα/parenrightbigg /bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright =1 2∂ ∂˙qα˙r2 i−˙ri·∂˙ri ∂qα/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright =1 2∂ ∂qα˙r2 i (2.27) As indicated in Eq. (2.27), the terms ˙ri·∂˙ri ∂˙qαand˙ri·∂˙ri ∂qαthus obtained can be written as derivatives of ˙r2 i. This is reassuring since ˙r2 ishows up in the kinetic energy Tincluded in L. We may thus hope to obtain derivatives of T. To do so we now pull all derivatives in front which leads to N/summationdisplay i=1mi¨ri·∂ri ∂qα=d dt∂ ∂˙qαN/summationdisplay i=11 2mi˙r2 i−∂ ∂qαN/summationdisplay i=11 2mi˙r2 i. We now recognise/summationtextN i=11 2mi˙r2 ias the kinetic energy and write N/summationdisplay i=1mi¨ri·∂ri ∂qα=d dt∂T ∂˙qα−∂T ∂qα. We have thus expressed the ﬁrst term in (2.26) through deriva tives of the kinetic energy, and obtained the intermediate result d dt∂T ∂˙qα−∂T ∂qα=N/summationdisplay i=1Fi·∂ri ∂qα. (2.28) In Eq. (2.28) we have not yet used our assumptions on the force s (i.e. that they are sums of conservative forces and forces of constraint). Eq. ( 2.28) can thus be seen as ageneralisation of Lagrange’s equation for arbitrary force s. Potential term The derivative of the potential in Eq. (2.26), multiplied wi th∂ri ∂qαand summed over, simply gives −N/summationdisplay i=1∂U ∂ri·∂ri ∂qα=−∂U ∂qα – again a term that we would like to see in Lagrange’s equation ! 38 CHAPTER 2. LAGRANGIAN MECHANICS Constraint term The ﬁnal term in (2.26) reads N/summationdisplay i=1Fc i·∂ri ∂qα This is exactly the scalar product of forces of constraint an d allowed directions (2.18) that vanishes due to d’Alembert’s principle. We thus have N/summationdisplay i=1Fc i·∂ri ∂qα= 0, and we see that due to d’Alembert’s principle the forces of co nstraint drop from our equations of motion. Result Summation of all three terms yields d dt∂T ∂˙qα−∂T ∂qα=−∂U ∂qα ⇒∂(T−U) ∂qα−d dt∂T ∂˙qα= 0 which is suspiciously close to Lagrange’s equations for L=T−U. The only thing that one might miss is a derivatived dt∂U ∂˙qα. But by deﬁnition, the potential is independent of ˙ qαand thusd dt∂U ∂˙qα= 0. If we now simply addd dt∂U ∂˙qαand replace T−UbyL, we obtain our desired result: Lagrange’s equation ∂L ∂qα−d dt∂L ∂˙qα= 0 for all α= 1,... ,d 2.4. CONSERVED QUANTITIES 39 2.4 Conserved quantities In mechanics it is often helpful to look for conserved quantities and for example check whether the energy, the momentum or the angular moment um of a particle remains ﬁxed. Def.: A quantity Ais aconserved if the total time derivativedA dtvanishes. We will show that Lagrangian mechanics provides an ideal fra mework to study these conserved quantities. In particular we will see that c onserved quantities always arise when the Lagrangian is independent of one of the argume ntsq1,... ,q d,t. 2.4.1 Energy conservation First of all we will show that if the Lagrangian does not depen d on time, the so-called generalised energy is conserved. (Usually this is just the e nergy itself.) Conservation of the generalised energy If the Lagrangian of a system is independent of the time t, i.e., L=L(q1,... ,q d/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright =q,˙q1,... ,˙qd/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright =˙q) then the generalised energy h≡d/summationdisplay α=1∂L ∂˙qα˙qα−L=∂L ∂˙q·˙q−L is a conserved quantity. Note that here we used the vector notation q= (q1,... ,q d) for the generalised coordinates. Proof •Remembering variational calculus we can use the alternative version of the Euler-Lagrange equation . We had seen that if a function f=f(y,y′) is independent of x, then the functional K=/integraltextx2 x1f(y(x),y′(x))dxbecomes stationary if ∂f ∂y′·y′−f= const . (Here the sign of the constant is ﬂipped compared to section 1 .) Thus the Euler-Lagrange equation was directly formulated in terms o f a conservation law. To apply this result to Lagrangian mechanics, we replac ex→t,y→q, f→LandK→S. Then we immediately obtain the statement above. •Since it was so simple, we can just redo the proof and take the t otal derivative 40 CHAPTER 2. LAGRANGIAN MECHANICS ofh: dh dt=d dt/parenleftbigg∂L ∂˙q·˙q/parenrightbigg −dL dt =/parenleftbiggd dt∂L ∂˙q/parenrightbigg ·˙q+∂L ∂˙q·¨q−∂L ∂q˙q−∂L ∂˙q·¨q If we use Lagrange’s equations to replaced dt∂L ∂˙qby∂L ∂qwe see that all terms above cancel and thusdh dt= 0. Example Let us consider the motion of a particle in one dimension. The corresponding Lagrangian L=1 2m˙x2−U(x) does not depend explicitly on time. Hence the generalised en ergy must be conserved; it reads h=∂L ∂˙x˙x−L =m˙x˙x−/parenleftbigg1 2m˙x2−U(x)/parenrightbigg =1 2m˙x2+U(x) (2.29) which is just the sum of kinetic and potential energy, i.e., t heenergy E=T+U itself. I now want to show that this result h=Eactually generalises to most practical cases (albeit there are exceptions). To check this we ﬁrst ne ed an explicit formula for the Lagrangian, and in particular for the kinetic energy of a mechanical system. We can then compute hand see whether it agrees with E. General formula for the kinetic energy If we use Cartesian coordinates for the particle positions ri, the kinetic energy of a mechanical system always has the form T=N/summationdisplay i=11 2mi˙r2 i. (2.30) We now want to see how this formula looks if we use generalised coordinates q1,... ,q d. The particle positions can then be parametrised by these ge neralised coordinates and (possibly) time as ri=ri(q1,... ,q d,t). According to the chain rule the velocity turns into ˙ri=d/summationdisplay α=1∂ri ∂qα˙qα+∂ri ∂t. (2.31) 2.4. CONSERVED QUANTITIES 41 If we insert this into the above formula for Twe get T=N/summationdisplay i=11 2mi d/summationdisplay α=1∂ri ∂qα˙qα·d/summationdisplay β=1∂ri ∂qβ˙qβ+ 2d/summationdisplay α=1∂ri ∂qα˙qα·∂ri ∂t+/parenleftbigg∂ri ∂t/parenrightbigg2 .(2.32) This follows directly by inserting (2.31) into (2.30); the o nly nontrivial point is that when squaring/summationtextd α=1∂ri ∂qα˙qαI wrote the two factors with diﬀerent summation variables αandβto make clear that there are two diﬀerent sums, not just one. It would be nice if we could write Eq. (2.32) in a less clunky wa y. To do so we abbreviate the terms that are not derivatives of qby Mαβ(q1,... ,q d,t)≡N/summationdisplay i=1mi∂ri ∂qα·∂ri ∂qβ vα(q1,... ,q d,t)≡N/summationdisplay i=1mi∂ri ∂qα·∂ri ∂t c(q1,... ,q d,t)≡N/summationdisplay i=1mi/parenleftbigg∂ri ∂t/parenrightbigg2 . (2.33) The kinetic energy then turns into T=1 2N/summationdisplay α=1N/summationdisplay β=1Mαβ˙qα˙qβ+d/summationdisplay α=1vα˙qα+c 2. (2.34) To make this result even more compact we adopt a vector and mat rix notation. We thus collect all Mαβ’s into a matrix M≡ M11... M 1d ...... Md1... M dd  and all vα’s into a vector v= v1 ... vd . HereMis a symmetric matrix ( Mαβ=Mβα) because the matrix elements Mαβ deﬁned above remain the same if αandβare interchanged. With this Mandvthe kinetic energy assumes the form: General formula for the kinetic energy T=1 2˙q·M˙q+v·˙q+c 2(2.35) The term quadratic in ˙q,1 2˙q·M˙q, looks very much like the usual formula for the kinetic energy of a single particle in Cartesian coordin ates. The only diﬀerences are that ˙qis a vector containing derivatives of generalised coordina tes, and that the 42 CHAPTER 2. LAGRANGIAN MECHANICS mass is replaced by a matrix that may depend on q. This matrix is also called the mass matrix . The linear and constant terms are new. However they show up on ly if our transformation between Cartesian and generalised coordin ates involves time, i.e., ifri= (q1,... ,q d,t) with∂ri ∂t∝ne}ationslash= 0. In the usual situation that there is no time dependence, we have∂ri ∂t= 0 and the vαandcdeﬁned in (2.33) simply vanish. Then we only have the quadratic term. Formulas for gradients of linear and quadratic functions To get hwe must evaluate derivatives of L=T−Uw.r.t. qand˙q, i.e., we must learn how to deal with derivatives of scalar products like v·qand quadratic terms like˙q·M˙qwith respect to the vectors involved. We recall that the deri vative (gradient) w.r.t. a vector uis deﬁned as the vector containing partial derivatives w.r.t. the components of u, i.e., ∂f ∂u= ∂f ∂u1... ∂f ∂ud . We will show that derivatives of such terms are given by ∂(v·u) ∂u=v (2.36) ∂(u·Mu) ∂u= 2MuifMis symmetric (2.37) which are just the formulas we would get if v,uandMwere scalars. Proof of (2.36): We use that v·u=d/summationdisplay α=1vαuα. The partial derivatives are thus given by ∂(v·u) ∂uα=vα and collecting them into a vector yields vas in (2.36). Proof of (2.37): We use that u·Mu=d/summationdisplay α=1d/summationdisplay β=1Mαβuαuβ. The partial derivatives are thus given by ∂(u·Mu) ∂uγ=d/summationdisplay α=1d/summationdisplay β=1Mαβ/parenleftbigg∂uα ∂uγuβ+uα∂uβ ∂uγ/parenrightbigg . (2.38) 2.4. CONSERVED QUANTITIES 43 Here the derivative∂uα ∂uγreads ∂uα ∂uγ=δαγ≡/braceleftigg 1 ifα=γ 0 otherwise Therefore the ﬁrst sum in (2.38) only receives contribution s when α=γ. We thus drop∂uα ∂uγand the summation over αand replace the remaining αbyγ. The∂uβ ∂uγin the second term is handled in an analogous way. We then obtain ∂(u·Mu) ∂uγ=d/summationdisplay β=1Mγβuβ+d/summationdisplay α=1Mαγuα. If we now rename the summation variable αin the second sum into βand use that M is symmetric ( Mβγ=Mγβ) we get ∂(u·Mu) ∂uγ=d/summationdisplay β=1Mγβuβ+d/summationdisplay β=1Mβγ/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright =Mγβuβ= 2d/summationdisplay β=1Mγβuβ Collecting all partial derivatives∂(u·Mu) ∂uγinto a vector we thus get 2 Muas claimed. Generalised energy We are now prepared to evaluate the generalised energy. Due t o (2.35) the La- grangian is given by L=1 2˙q·M˙q+v·˙q+c 2−U . The generalised energy can now be computed from (2.29), h=∂L ∂˙q·˙q−L = (M˙q+v)·˙q−1 2˙q·M˙q−v·˙q−c 2+U =1 2˙q·M˙q−c 2+U For comparison, the energy is E=T+U=1 2˙q·M˙q+v·˙q+c 2+U . So the good news is: If the kinetic energy is quadratic in ˙q, i.e. v= 0andc= 0, then the energy coincides with the generalised energy This is the generic situation that we have if the transformat ion between Carte- sian and generalised coordinates is time independent and th us∂ri ∂t= 0. (Also recall thatUis independent of ˙q.) In a practical example we would ﬁrst check whether Ldepends on tor not. If it is independent, his conserved. If Tis quadratic in ˙qwe simply have h=E, otherwise we need to calculate hexplicitly. 44 CHAPTER 2. LAGRANGIAN MECHANICS Example where these conditions are satisﬁed: For the inclined plane we had L=T−U=/parenleftbigg1 2(˜M+m) ˙x2+1 2m˙s2+m˙x˙scosα/parenrightbigg −mgssinα . (Here I renamed the mass of the plane into ˜Mto avoid confusion with the matrix M).Lis independent of t, hence his conserved. All terms in Tinvolve products of two factors ˙ xor ˙s. Therefore the generalised energy coincides with the energ y and we have h=E=T+U=/parenleftbigg1 2(˜M+m) ˙x2+1 2m˙s2+m˙x˙scosα/parenrightbigg +mgssinα . Examples where these conditions are violated: •Consider a pendulum ﬁxed to a point (0,y(t))that is moving depend- ing on time with y(t) =y0sinωt. Using θ(see Fig. 2.6) as a generalised coordinate, we can write r(θ,t) =/parenleftbigg0 y(t)/parenrightbigg +l/parenleftbiggsinθ −cosθ/parenrightbigg , i.e. the potential is U=mg(y(t)−lcosθ). If we take the derivative ˙r=/parenleftbigg0 ˙y(t)/parenrightbigg +l˙θ/parenleftbiggcosθ sinθ/parenrightbigg we get the kinetic energy T=1 2m˙r2=1 2m(l2˙θ2+ 2lsinθ˙θ˙y(t) + ˙y(t)2) and the Lagrangian L(θ,˙θ,t) =T−U=1 2m(l2˙θ2+ 2lsinθ˙θ˙y(t) + ˙y(t)2)−mg(y(t)−lcosθ). Ldepends explicitly on t, sohis not conserved. Moreover the kinetic en- ergy contains terms linear in ˙θand independent of ˙θ. (Note that yis not a generalised coordinate so ˙θ˙yand ˙y2do not count as quadratic terms!) Hence our condition for h=E=T+Uis violated and we have to compute the generalised energy using h=∂L ∂˙θ˙θ−L=1 2m(l2˙θ2−˙y(t)2) +mg(y(t)−lcos(θ))∝ne}ationslash=T+U . •Another example (that will be on problem sheet 4) involves a p article in a magnetic ﬁeld . In this example a term linear in ˙qarises such that h∝ne}ationslash=E. However, Lis independent of t, sohis still conserved. 2.4. CONSERVED QUANTITIES 45 Figure 2.6: A pendulum ﬁxed to a point that is moving. 2.4.2 Conservation of generalised momenta If the Lagrangian is independent of one of the generalised co ordinates qα, this gives rise to another conservation law. Lagrange’s equations ∂L ∂qα=d dt∂L ∂˙qα imply: Thm: IfLisindependent of one of the generalised coordinates qα(∂L ∂qα= 0), then pα≡∂L ∂˙qα must be a conserved quantity (d dt∂L ∂˙qα= 0). Such coordinates not showing up in the Lagrangian are also ca lledignorable or cyclic coordinates. The quantity pαdeﬁned above is called the generalised mo- mentum associated to the generalised coordinate qα. Hence the statement could also be formulated as follows: If a coordinate is ignorable, then the corre- sponding generalised momentum is conserved . Examples a) Conservation of the linear momentum First of all, let us consider a single particle, described in Cartesian coordinates x,y,z. The Lagrangian of this particle can be written as L=T−U=1 2m( ˙x2+ ˙y2+ ˙z2)−U(x,y,z) 46 CHAPTER 2. LAGRANGIAN MECHANICS where U(x,y,z) is the potential. Now the generalised momenta px,pyandpz associated to x,yandzare obtained as px=∂L ∂˙x=m˙x py=∂L ∂˙y=m˙y pz=∂L ∂˙z=m˙z . We thus see that these generalised momenta just coincide wit h theusual (linear) momenta inx-,y- and z-direction, mass times component of the velocity. We now obtain the following conservation law: If the potential and thus Lagrangian are independent of x, ∂L ∂x=−∂U ∂x= 0, then the momentum in x-direction px=m˙xis conserved. Analogous results hold forpyandpz. Within Newtonian mechanics, we would have argued that due to∂U ∂x= 0 there is no force in x-direction and hence the corresponding component of the mom entum remains constant. b) Angular momentum conservation Let us now see what happens if we are in the x−y−plane and instead of Cartesian coordinates pick polar coordinates , r=ρ cosφ sinφ 0 . We then have ˙r= ˙ρ cosφ sinφ 0 +ρ˙φ −sinφ cosφ 0  and thus ˙r2= ˙ρ2+ρ2φ2. The Lagrangian reads L=T−U=1 2m( ˙ρ2+ρ2˙φ2)−U(ρ,φ) where the potential was written as a general function of ρandφ. So what are the generalised momenta associated to ρandφ? For ρwe get pρ=∂L ∂˙ρ=m˙ρ which is of the same type as the linear momenta, but now with th e velocity ˙ ρ. Hence pρcan be interpreted as the radial component of the linear momentum . The generalised momentum associated to φreads pφ=∂L ∂˙φ=mρ2˙φ . 2.4. CONSERVED QUANTITIES 47 pφhas the interpretation of an angular momentum . To check this, we use the deﬁnition of the angular momentum l=r×p=r×m˙rfrom Mechanics 1. Inserting our formula for rand˙rwe obtain l=ρ cosφ sinφ 0 × m˙ρ cosφ sinφ 0 +mρ˙φ −sinφ cosφ 0   = 0 0 mρ2˙φ . The only interesting component here is the z-component. (Any cross product of two vectors in the x−y−plane must point in z-direction.) As anticipated it coincides withpφ. Now which conservation law do we get? Since the kinetic energ y already depends onρ, the radial coordinate cannot be ignorable. But φcan be ignorable, if the potential Udepends only on ρand not on φ, ∂L ∂φ=−∂U ∂φ= 0. In this case the angular momentum is conserved . Such potentials independent ofφare also referred to as central ﬁelds . Central ﬁelds are rather common. If we have, e.g., a gravitational ﬁeld due to a mass at the origin, t he potential will only depend on the distance ρfrom the origin. c) Inclined plane For the inclined plane we have L=T−U=1 2(˜M+m) ˙x2+1 2m˙s2+m˙x˙scosα−mgssinα . and the generalised momenta are px=∂L ∂˙x= (˜M+m) ˙x+m˙scosα ps=∂L ∂˙s=m˙s+m˙xcosα . Lis independent of x, hence pxis conserved. But it depends on s, thus psis not conserved. 2.4.3 Spherical pendulum The spherical pendulum is a good example to illustrate the us e of conservation laws in Lagrangian mechanics. It is simply a pendulum that is allo wed to move in all three dimensions of space rather than in only two dimensions . To build a spherical pendulum, one simply takes a particle of mass mand pivots at the origin by a rigid rod of length l. Since the particle is allowed to move in three-dimensional space but its distance from the origin is ﬁxed to be l, the possible particle positions form a sphere of radius laround the origin . For simplicity, we will choose units in which m,gandlall become 1. 48 CHAPTER 2. LAGRANGIAN MECHANICS Figure 2.7: Spherical pendulum. Find Lagrangian To look for conservation laws, we ﬁrst have to ﬁnd suitable ge neralised coordinates and write down the Lagrangian. Since the possible particle p ositions are on a sphere, it is natural to use spherical coordinates , but with two twists: First, the radius is ﬁxed to be l= 1, so there are only two variables θandφ. Second, to be consistent with the treatment of the two-dimensional pendulum we deﬁne θto be the angle enclosed between the pendulum and the negative z-axis, not the positive one (see Fig. 2.7) which means that we replace θbyπ−θcompared to the usual deﬁnition of spherical coordinates. Our modiﬁed spherical coordinat es thus have the form r= sinθcosφ sinθsinφ −cosθ . (2.39) The velocity is now given by ˙r= cosθcosφ cosθsinφ −sinθ ˙θ+ −sinθsinφ sinθcosφ 0 ˙φ , and the kinetic energy reads T=1 2˙r2=1 2/parenleftig ˙θ2+ sin2θ˙φ2/parenrightig . The gravitational potential isU=mgz. Here zis−cosθ(see(2.39)) and we have setmandgto be equal to 1. Thus U=−cosθ . 2.4. CONSERVED QUANTITIES 49 TheLagrangian therefore has the form L=1 2˙θ2+1 2sin2θ˙φ2+ cos θ . Conserved quantities Now we want to look for conserved quantities, and thus for ign orable coordinates. The Lagrangian Ldoes not depend on φandt. This should give rise to two con- servation laws: •Since Lis independent of t, the generalised energy his conserved. Since the kinetic energy is quadratic in the derivatives ˙θand˙φthe generalised energy also coincides with the energy E=T+U. Thus we also have energy conservation : E=T+U=1 2˙θ2+1 2sin2θ˙φ2−cosθ= const . (2.40) •Since φis ignorable, the corresponding generalised momentum pφmust be a conserved quantity. We thus have pφ=∂L ∂˙φ= sin2θ˙φ= const . (2.41) pφcan be identiﬁed with the z-component of the angular momentum (corresponding to rotations about the z-axis, which is the symmetry axis of the spherical pendulum). We had already seen that this is the rig ht interpretation of the generalised momentum associated to φin polar coordinates. Similarly, if we use spherical coordinates the z-component of the angular momentum r×m˙r(where m= 1) is given by x˙y−y˙x = sin θcosφ(cosθsinφ˙θ+ sinθcosφ˙φ)−sinθsinφ(cosθcosφ˙θ−sinθsinφ˙φ) = sin2θ˙φ which indeed coincides with pφ. Conservation laws are so valuable for the description of mec hanical systems because they allow us to reduce the number of independent variables . In the present case, we can use (2.41) to get rid of ˙φin (2.40). If we solve (2.41) for ˙φwe get ˙φ=pφ sin2θ. (2.42) Inserting this into (2.40) yields E=1 2˙θ2+p2 φ 2sin2θ−cosθ= const (2.43) Now we have a diﬀerential equation for θonly. There is no φbecause TandUare independent of φ(which was the reason for angular momentum conservation in the ﬁrst place), and then ˙φcould be eliminated using (2.41). Instead we have two constants, Eandpφ. 50 CHAPTER 2. LAGRANGIAN MECHANICS Interpretation Since φis gone, Eq. (2.43) could be interpreted as describing a ﬁcti tiousparticle in one dimension , with the only coordinate θ. If we assume that this particle feels an eﬀective potential of the form Veﬀ(θ) =p2 φ 2sin2θ−cosθ (2.44) then Eq. (2.43) could be interpreted as an energy conservation law for our particle: E=1 2˙θ2+Veﬀ(θ) = const (2.45) Here the eﬀective potential contains both the original pote ntialU=−cosθand an extra termp2 φ 2sin2θthat originates from the ˙φ2-term in the kinetic energy; this term became part of the eﬀective potential due to the elimination of˙φ. We can also take the total time derivative, leading to 0 =dE dt=˙θ¨θ+V′ eﬀ(θ)˙θ . Division by ˙θthen gives ¨θ=−V′ eﬀ(θ). (2.46) This means that the acceleration of our ﬁctitious particle i s given by the negative derivative of the eﬀective potential, as one would expect. ( Recall that we have set the mass equal to one.) Method I: Integration There are two ways to proceed further. The ﬁrst one (which we w on’t carry to the end here) is to view (2.45) as a diﬀerential equation for θ, and try to integrate it. We thus take (2.45) and solve for ˙θ, ˙θ2= 2(E−Veﬀ(θ)) ⇒˙θ=/radicalbig 2(E−Veﬀ(θ)) We then use separation of variables. We write ˙θ=dθ dtas dt=dθ ˙θ and afterwards integrate on both sides. The integral over tgoes from 0 to t, whereas the integral over θgoes from θ(0) to θ(t). If we insert the explicit formula for ˙θthis yields t=/integraldisplayt 0dt=/integraldisplayθ(t) θ(0)dθ ˙θ=/integraldisplayθ(t) θ(0)dθ/radicalbig 2(E−Veﬀ(θ)) The solution of our problem is thus written as an integral. Th e evaluation of this integral is rather tricky, and outside the scope of this cour se. One would ﬁnally be lead to so-called elliptic functions, a kind of special func tions. One would then have to solve for θ(t), and afterwards determine φ(t). 2.4. CONSERVED QUANTITIES 51 Method II: Understand the behaviour of the solution An alternative approach is to ﬁrst understand the qualitative behaviour of the motion in the eﬀective potential Veﬀ. Then one can use the insight gained to say as much as possible about the solution without having to comput e the integral. This is the approach we are going to apply in the following. Case pφ= 0 First let us deal with the simple case that pφvanishes. According to (2.42) this implies ˙φ= 0, i.e., the angle φremains constant. This mains that the pendulum is only moving in a plane (enclosing an angle φwith the x-axis). The problem is thus reduced to the two-dimensional pendulum of Subsection 2.3. 1. Also the equation forθis the same. To see this explicitly, note that for pφ= 0 the eﬀective potential (2.44) simply reduces to Veﬀ(θ) =−cosθ , which was the gravitational potential. The equation ¨θ=−V′ eﬀ(θ) then boils down to ¨θ=−sinθ coinciding with Eq. (2.6). We thus get the same result as in Su bsection 2.3.1, with g,landmset equal to unity. Case pφ∝ne}ationslash= 0 Figure 2.8: Eﬀective potential of the spherical pendulum. Ifpφis nonzero, the general form of the eﬀective potential is given by Veﬀ(θ) =p2 φ 2sin2θ−cosθ . It is helpful to visualise the behaviour of this potential , see Fig. 2.8. Due to the ﬁrst summandp2 φ 2sin2θthe eﬀective potential diverges when sin θis zero, i.e., for θ→0 and for θ→π. In both limits, Veﬀtends to + ∞. Since in spherical coordinates θis restricted to be between 0 and π, it only makes sense to consider Veﬀ(θ) between these two values. So what is the behaviour in between? The sim plest possibility 52 CHAPTER 2. LAGRANGIAN MECHANICS would be that Veﬀjust has one minimum , and this is what indeed happens. For a proof, let us consider the equation determining the extrem a ofVeﬀ, 0 =V′ eﬀ(θ0) =−p2 φ sin3θ0cosθ0+ sinθ0. The eﬀective potential thus becomes extremal for angles θ0with sin4θ0=p2 φcosθ0. (2.47) We can show that this equation has only one solution. First le t us consider 0 ≤ θ0≤πand view the expressions sin4θ0on the l.h.s. and p2 φcosθ0on the r.h.s. as functions of θ0. It is easy to see that for θ= 0 the r.h.s. is larger and for θ=π 2 the l.h.s. is larger. Hence (2.47) must be satisﬁed somewher e in between. Since the l.h.s. sin4θ0increases monotonically with θ0and the r.h.s. p2 φcosθ0decreases monotonically, this can happen only once. So there is only on e minimum with θ0<π 2, see Fig. 2.8. For θ0>π 2there can be no solutions since the two sides have opposite signs. Figure 2.9: Minimal and maximal values of θfor the spherical pendulum. Now what will solutions look like? The eﬀective kinetic energy1 2˙θ2must always be positive (or zero). 1 2˙θ2=E−Veﬀ(θ)≥0. Hence the total energy must always be larger than the eﬀectiv e potential Veﬀ. This restricts the possible values of θ: For each Ewe may only have values of θwith Veﬀ(θ)≤E. Since the eﬀective potential becomes large for θgoing to 0 or π, this excludes values of θwhich are too close 0 or π, whereas values further away are permissible. The mimimal and maximal values of θfor a given value of Ewill be denoted by θminandθmax. They are determined by Veﬀ(θmin) =Veﬀ(θmin) =E . 2.4. CONSERVED QUANTITIES 53 As illustrated in Fig. 2.9, θminandθmaxcan be found graphically from the in- tersections of the graph of Veﬀ(θ) with a straight line corresponding to the energy E. Since the angle θis related to the height zviaz=−cosθ, the points θminand θmaxcorrespond to the lowest and highest points reached for the e nergy E. (θmin gives the lowest point and θmaxthe highest.) For a trajectory with given energy the angle and height will oscillate between these values. Th is is indicated by the thick line in Fig. 2.8. For θ=θminandθ=θmaxthe eﬀective kinetic energy1 2˙θ2 is zero the whole energy exists as eﬀective potential energy ; this includes both the gravitational potential and the kinetic energy due to the ch ange of φ. When moving between these values θchanges and thus part of the energy is put into1 2˙θ2. To visualise the trajectories of the pendulum, let us assume that we are looking at the pendulum from above and project the motion into the x-y-plane. The distance from the centre is then given by ρ=/radicalbig x2+y2= sinθ . Likeθandz, the distance ρoscillates between two values, ρmin= sin θminand ρmax= sin θmax. For each Ethe possible positions are thus enclosed between two circles with these radii. The trajectories must hit the inne r circle, then the outer circle, then the inner circle again, etc. Figure 2.10: Trajectories of the spherical pendulum in the x-y-plane. Now what about the angle φ? Due to angular momentum conservation, φ changes with a derivative ˙φ=pφ sin2θ. This leads to an eﬀective rotation about the centre, while ρoscillates between its minimal and maximal value. The results for ρandφtogether lead to trajectories as sketched in Fig. 2.10. Mass moving on a circle An important special case are orbits where θis constant and the mass just moves on a circle around the z-axis. In these cases the rod traces out a cone which is why these orbits are sometimes called conical orbits. Such solu tions are possible only forθ=θ0, i.e., if we are at the minimum of the eﬀective potential. Thi s is the 54 CHAPTER 2. LAGRANGIAN MECHANICS Figure 2.11: Trajectory of the sperical pendulum with θ(t) =θ0= const. only situation in which the acceleration ¨θ=−V′ eﬀ(θ) becomes zero and thus θcan remain constant. Let us now determine how fast the mass must be in this case. For the conical orbits, the angular velocity ˙φis given by (see (2.41)) ˙φ=pφ sin2θ0= const . If we insert Eq. (2.47) and thus pφ=sin2θ0 √cosθ0we obtain ˙φ=1√cosθ0. Mass moving close to a circle Now what happens if the energy is just slightly above the mini mum of the eﬀective energy? In this case the orbits should be close to the conical orbits derived above. But there will be a little bit of energy left that can be invest ed in going away from the minimum and thus increasing Veﬀ, and in changing θand thus having a nonvanishing ˙θ. We might thus expect that the solutions oscillate about the conical orbits considered above. To show this we use ¨θ=−V′ eﬀ(θ). If we know write θ(t) as a sum of the equilibrium value θ0and the deviation from the equilibrium δθ, θ(t) =θ0+δθ . the l.h.s. turns into δ¨θ. The r.h.s. can be approximated for small δθif we make a Taylor expansion V′ eﬀ(θ) =V′ eﬀ(θ0+δθ)≈V′ eﬀ(θ0)/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright =0+V′′ eﬀ(θ0)δθ . If we deﬁne Ω 0=/radicalbig V′′ eﬀ(θ0) we thus have δ¨θ=−Ω2 0δθ . 2.4. CONSERVED QUANTITIES 55 This is the same diﬀerential equation as for a harmonic oscil lator. The solution involves sines or cosines with a frequency Ω 0. In other words: For orbits close to the conical orbits the angle θof the spherical pendulum oscillates with a frequency Ω 0=/radicalbig V′′ eﬀ(θ0). (Recall that we have set m= 1, otherwise there would be a mass here as well.) A simple calculation of the second derivative3yields Ω0=/radicalbigg 3cosθ0+1 cosθ0. Due to cos θ0>0 this is larger than ˙φ=1√cosθ0. Hence the radial oscillations compared to the conical orbits are faster than the oscillati on of the angle φdue to rotation. In a more general context the motion close to extrema of a pote ntial (equilibria ) will be discussed in Chapter 3. 2.4.4 Noether’s theorem So far we have seen that conserved quantities arise when one o f the the generalised coordinates qαor the time tdon’t show up in the Lagrangian. This is an example for a more general principle: In fact all symmetries of a system and its Lagrangian give rise to conservation laws. This principle will be very helpf ul to get counterparts for the conservation of linear and angular momenta (see Subsect ion 2.4.2) in a system with many particles. We will ﬁrst study a further example and then generalise. Translation symmetry Consider a system with Nparticles at r1,... ,rN. This system is called symmet- ric (or invariant) w.r.t. translations in direction dif the the Lagrangian remains the same when all particles positions are shifted by an arbitrary amount in direction d. This means that we have L(r1,... ,rN,˙r1,... ,˙rN,t) =L(r1+sd,... ,rN+sd,˙r1,... ,˙rN,t) for all real numbers s. Example: Consider the two-particle system with the Lagrangian L(r1,r2,˙r1,˙r2) =1 2m1˙r2+1 2m2˙r2 2+Gm1m2 |r1−r2|, (2.48) accounting for the kinetic energy of these particles and the potential due to their gravitational attraction. Both the derivatives ˙r1,˙r2and the diﬀerence |r1−r2| remain the same if all positions are increased by sd. Hence the Lagrangian is not changed if both particles are moved in direction d. Now our statement is: 3One has to show that V′′ eﬀ(θ) =3p2 φ sin4θcos2θ+p2 φ sin2θ+ cos θ. The result then follows if we insert θ=θ0andp2 φ=sin4θ0 cosθ0from Eq. (2.47). 56 CHAPTER 2. LAGRANGIAN MECHANICS Thm.: For s system that is invariant w.r.t. translations in direct iondthe com- ponent of the total linear momentum of all particles/summationtextN i=1piin direction dis a conserved quantity, i.e., N/summationdisplay i=1pi·d= const . Proof: For such a system L(r1+sd,... ,rN+sd,˙r1,... ,˙rN,t) does not depend ons, which means that the derivative of Lw.r.t. svanishes. This is true for all s, but it will be suﬃcient to consider the derivative at s= 0. We then get 0 =∂ ∂sL(r1+sd,... ,rN+sd,˙r1,... ,˙rN,t)/vextendsingle/vextendsingle s=0 =N/summationdisplay i=1∂L ∂ri·∂(ri+sd) ∂s/vextendsingle/vextendsingle s=0 =N/summationdisplay i=1 d dt∂L ∂˙ri/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright =pi ·d =d dt/parenleftiggN/summationdisplay i=1pi·d/parenrightigg . Here in the second line we used the chain rule. In the third lin e, we used Lagrange’s equations and that∂L ∂˙ri=pi; moreover the derivative of ri+sdwas evaluated. Finally the total time derivative was written in front. Sinc e this derivative vanishes we obtain the desired result N/summationdisplay i=1pi·d= const . Note: Many systems (e.g. (2.48) are invariant w.r.t. translation s inall directions d. Then all scalar products involving/summationtextN i=1pivanish, which is only possible if the total linear momentum/summationtextN i=1piis conserved . Noether’s theorem in its general form We now extend our results to systems with a rather general kin d of symmetry: We assume that the Lagrangian remains the same if the general ised coordinates (collected into the vector q= (q1,... ,q d)) are replaced by new values Q(q,s) = (Q1(q,s),... ,Q d(q,s)); these new values depend both on the old ones and on a parameter sthat tells us how much the coordinates have been changed. (In the above example swas the distance of a shift.) For s= 0 we assume that no change occurred, i.e., the coordinates are still the old ones. For s uch symmetries Emmy Noether (1882-1935) derived the following conservation la w: 2.4. CONSERVED QUANTITIES 57 Noether’s theorem Consider a system whose Lagrangian L(q,˙q,t) remains the same after replacing qbyQ(q,s) (where s∈RandQ(q,s) is a mapping with Q(q,0) =q), i.e., L(q,˙q,t) =L(Q,˙Q,t). (2.49) For such a system the quantity /summationdisplay α=1∂L ∂˙qα/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright pα∂Qα ∂s/vextendsingle/vextendsingle s=0(2.50) is conserved. Proof: According to (2.49) L(Q,˙Q,t) is independent of s, i.e., its derivative w.r.t. svanishes. Again we need this fact only for s= 0 (even though it holds true for all s). We get 0 =∂ ∂sL(Q,˙Q,t)/vextendsingle/vextendsingle s=0 =N/summationdisplay α=1/bracketleftigg ∂L ∂Qα∂Qα ∂s+∂L ∂˙Qα∂˙Qα ∂s/bracketrightigg /vextendsingle/vextendsingle s=0 =d/summationdisplay α=1/bracketleftigg ∂L ∂qα∂Qα ∂s/vextendsingle/vextendsingle s=0+∂L ∂˙qα∂˙Qα ∂s/vextendsingle/vextendsingle s=0/bracketrightigg =d/summationdisplay α=1/bracketleftbigg/parenleftbiggd dt∂L ∂˙qα/parenrightbigg∂Qα ∂s/vextendsingle/vextendsingle s=0+∂L ∂˙qα/parenleftbiggd dt∂Qα ∂s/vextendsingle/vextendsingle s=0/parenrightbigg/bracketrightbigg =d dtd/summationdisplay α=1∂L ∂˙qα∂Qα ∂s/vextendsingle/vextendsingle s=0 Here we ﬁrst evaluated the derivative w.r.t. susing the chain rule. Then s= 0 was invoked to replace Qα’s byqα’s. In the fourth line Lagrange’s equations were used and the derivatives w.r.t. tandsacting on Qαwere interchanged. Finally, the total time derivative was written in the beginning. Sinc e this derivative is zero we obtain the desired result d/summationdisplay α=1∂L ∂˙qα∂Qα ∂s/vextendsingle/vextendsingle s=0= const . Examples: Ignorable coordinates Let us ﬁrst check that Noether’s theorem contains the one we o btained in case of ignorable coordinates. If Lis independent of one variable qβ, Eq. (2.49) holds with Qβ=qβ+s,Qα=qαfor all α∝ne}ationslash=β. Then the constant above is d/summationdisplay α=1∂L ∂˙qα∂Qα ∂s/vextendsingle/vextendsingle s=0=d/summationdisplay α=1∂L ∂˙qαδαβ=∂L ∂˙qβ=pβ, 58 CHAPTER 2. LAGRANGIAN MECHANICS i.e. the generalised momentum associated to qβ. Translation symmetry Next we check that for the case of a system that is invariant w. r.t. translations in direction dthe conserved quantity (2.50) boils down to the correspondi ng compo- nent of the total linear momentum. Our coordinates qare now the positions of the Nparticles q= (r1,... ,rN) and the Lagrangian remains the same if these coordinates are replaced by Q(q,s) = (r1+sd,... ,rN+sd). The conserved quantity (2.50) thus takes the form N/summationdisplay i=1∂L ∂˙ri·∂(ri+sd) ∂s/vextendsingle/vextendsingle s=0=N/summationdisplay i=1pi·d which is indeed the d-component of the total linear momentum. Rotation symmetry Let us now apply Noether’s law to systems that can be rotated a bout an axis (for example the z-axis) without changing anything. Reminder: If we rotate a vector r= (x,y,z) about the z-axis by an angle swe obtain a new vector r′= (x′,y′,z′) with x′=xcoss−ysins y′=xsins+ycoss z′=z . The ﬁrst two lines are identical to what one gets when rotatin g about the origin in two dimensions. They were derived in Linear Algebra & Geom etry. The z- component just has to stay the same when rotating about the z-axis. In a matrix notation this result can be written as  x′ y′ z′ = coss−sins0 sinscoss0 0 0 1  /bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright R(s) x y z . The original and the rotated vectors have the same square r′2=x′2+y′2+z′2= (xcoss−ysins)2+(xsins+ycoss)2+z2=x2+y2+z2=r2 and thus the same norm |r′|=|R(s)r|=|r|. 2.4. CONSERVED QUANTITIES 59 Now assume that the Lagrangian of a many-particle system is invariant w.r.t. rotation of all particles about the z-axis, i.e., L(r1,... ,rN/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright =q,˙r1,... ,˙rN/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright =˙q,t) =L(R(s)r1,... ,R(s)rN/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright =Q,R(s)˙r1,... ,R(s)˙rN/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright =˙Q,t). Example: Just consider the two-particle system with gravity (2.48) a gain. Rota- tion by an angle sreplaces r1byRr1andr2byR(s)r2. Due to /parenleftbiggd dtR(s)r1/parenrightbigg2 = (R(s)˙r1)2=˙r2 1 /parenleftbiggd dtR(s)r2/parenrightbigg2 = (R(s)˙r2)2=˙r2 2 |R(s)r1− R(s)r2|=|R(s)(r1−r2)|=|r1−r2| the Lagrangian remains the same. Application of Noether’s theorem: For such systems Noether’s theorem gives the conserved quantity C=N/summationdisplay i=1∂L ∂˙ri·∂(R(s)ri) ∂s/vextendsingle/vextendsingle s=0=N/summationdisplay i=1pi·∂R(s) ∂s/vextendsingle/vextendsingle s=0ri To evaluate Cwe simply have to diﬀerentiate the rotation matrix w.r.t. s, to get ∂R(s) ∂s/vextendsingle/vextendsingle s=0= −sins−coss0 coss−sins0 0 0 0 /vextendsingle/vextendsingle s=0= 0−1 0 1 0 0 0 0 0 . If we write the vectors in components pi= (pi1,pi2,pi3),ri= (ri1,ri2,ri3) we obtain C=N/summationdisplay i=1 pi1 pi2 pi3 · 0−1 0 1 0 0 0 0 0  ri1 ri2 ri3  =N/summationdisplay i=1 pi1 pi2 pi3 · −ri2 ri1 0  =N/summationdisplay i=1(ri1pi2−ri2pi1). This is just the z-component of the sum of angular momenta/summationtext iri×pi, i.e. C=/parenleftiggN/summationdisplay i=1ri×pi/parenrightigg · 0 0 1  Thus our result is: For a system whose Lagrangian is invariant w.r.t. rotations about the z-axis the z-component of the total angular momentum/summationtextN i=1ri×piis conserved . 60 CHAPTER 2. LAGRANGIAN MECHANICS Analogous results hold for the y-axis, the z-axis and all other axes going through the origin. Therefore: For a system whose Lagrangian is invariant w.r.t. rotations aboutall axis going through the origin we have N/summationdisplay i=1ri×pi= const . Summary Symmetries of the Lagrangian lead to conservation laws: symmetry w.r.t. conserved quantity translations total momentum rotations total angular momentum Even energy conservation can be placed in this context: If th e Lagrangian does not depend on time this means that the system looks the same for al l times, i.e., it is invariant w.r.t. “translations in time”. The resulting con servation law is energy conservation: symmetry w.r.t. conserved quantity translations in time (generalised) energy These symmetries are satisﬁed by a large class of systems: Thm: If the Lagrangian of a system depends only •the squared velocities ˙r2 iand •the distances |ri−rj| of the particles, then all symmetries and conservation laws above are satisﬁed, i.e.: •translation symmetry ⇒total momentum conservation •rotation symmetry ⇒total angular momentum conservation •independence from t⇒(generalised) energy conservation Proof: (Just generalise what we said about example (2.48).) •Translation symmetry: The squared velocities ˙riare invariant under transla- tions as replacing ribyri+sddoesn’t change the derivative: /parenleftbiggd dt(ri+sd)/parenrightbigg2 =˙r2 i. The distances |ri−rj|are translation invariant as |(ri+sd)−(rj+sd)|=|ri−rj|. 2.4. CONSERVED QUANTITIES 61 •Rotation symmetry: The squared velocities are invariant under rotation, i.e. replacing ribyR(s)rias /parenleftbiggd dtR(s)ri/parenrightbigg2 = (R(s)˙ri)2=˙r2 i. Here we have used that rotation doesn’t change the norm or the square of a vector. The distances are invariant under rotation as |R(s)ri− R(s)rj|=|R(s)(ri−rj)|=|ri−rj| where we have used the same result about the norm or the square . •Independence from t : By assumption Ldoes not depend explicitly on time, only through ˙r2 iand|ri−rj|. Note: The total momentum and the total angular momentum each have t hree com- ponents. The generalised energy is just a real number. Hence there are altogether seven conserved numbers. Examples: •The kinetic energy/summationtext imi˙r2 idepends only on squared velocities. The same applies to many common potentials: e.g. the potential energ y of a spring depends only on the distance between the endpoints, the pote ntial due to gravitational attraction between two particles depends on ly the distance be- tween the particles. •Assume that our system is in the gravitational ﬁeld of the Earth . Then the potential will contain terms −GmimEarth |ri−rEarth| accounting for the gravitational attraction between each p article iand the Earth. These terms will satisfy the above conditions only if we take Earth as part of our system. Then we have a so-called isolated system where none of the particles interacts with anything outside the sy stem, and all our conservation laws hold. •Now let’s see which conservation laws if we don’t include Earth in our system , and use the approximation migzifor the gravitational potential of thei-th particle (while all other terms in the Lagrangian are inn ocent). –Translation symmetry: The potential remains invariant if we change all xiandyiby the same amount but not if we change the zi’s. Hence we only have translation invariance in x- and y-direction, and only the x andy-components of the total momentum are conserved. –Rotation symmetry: The gravitational potential remains invariant if we rotate all particles about the z-axis, as this does not change their z- coordinates. Hence the z-component of the total angular momentum is conserved. However rotations about the x- or the y-axis change the z- coordinates of the particles and thus the gravitational pot ential. Hence the corresponding components of the total angular momentum are not conserved. 62 CHAPTER 2. LAGRANGIAN MECHANICS –Independence from t: The gravitational potential does not depend ex- plicitly on time. Hence the generalised energy is conserved . We thus have only fourconserved numbers. Chapter 3 Small oscillations 3.1 General theory We will now study systems performing oscillations close to a minimum of a potential. The prototypical system displaying oscillations is the one -dimensional harmonic oscillator , with a potential energy that is quadratic in the coordinate , and a kinetic energy that is quadratic in the velocity. To generalise this example we investigate systems with an arbitrary number dof generalised coordinates (collected into a vector q= (q1,q2,... ,q d)) and a Lagrangian of the form: Generalised harmonic oscillator L=T−U=1 2˙q·M˙q−1 2q·Kq. (3.1) Here the potential energy is quadratic in qand kinetic energy is quadratic in ˙q.M andKare assumed to be constant, real, symmetric matrices of size d×d. Moreover, the matrix Mmust be positive deﬁnite , i.e., for all ˙q∝ne}ationslash= 0 we must have ˙q·M˙q>0. This condition is needed because the kinetic energy T=1 2˙q·M˙q, being the sum of the non-negative kinetic energies of the particles makin g up our system, must be non-negative as well (usually positive, unless all particl es are at rest and we thus have˙q= 0). Applications for Lagrangians of the above type are •systems composed of springs since the potential of a spring i s quadratic. •Much more generally, we will see that close to extrema of the potential (almost) any Lagrangian can be approximated by one of the for m (3.1). Lagrange equation The Lagrangian (3.1) has the derivatives ∂L ∂q=−Kq ∂L ∂˙q=M˙q. 63 64 CHAPTER 3. SMALL OSCILLATIONS Thus, Lagrange’s equation d dt∂L ∂˙q=∂L ∂q boils down to M¨q=−Kq (3.2) This is a linear diﬀerential equation, i.e., linear combina tions of solutions are also solutions. Normal modes As for any linear diﬀerential equation, it is a good idea to lo ok for particularly simple solutions, and then write general solutions as a line ar combination of these simple solutions. Motivated by the example of the harmonic o scillator we thus look for solutions of the form q(t) =ucos(ωt) (3.3) or q(t) =usin(ωt). (3.4) These solutions are called normal modes , and ω∈Cis called the normal fre- quency . If we insert (3.3) and its second derivative ¨q(t) =−uω2cos(ωt) into Lagrange’s equation (3.2) we obtain −Muω2cos(ωt) =−Kucos(ωt). The same result would be obtained for Eq. (3.3) with the cosin e replaced by a sine. To have these equations satisﬁed for all twe need (K−ω2M)u= 0. (3.5) This equation is similar to a familiar one: If Mwere replaced by the unit matrix 1, Eq. (3.5) would turn into the equation for eigenvalues and ei genvectors of a matrix K, (K−ω21)u= 0. Eq. (3.5) thus represents a generalised eigenvalue prob- lem. Accordingly the solutions for ω2are called the generalised eigenvalues , and the solutions for uare referred to as the generalised eigenvectors . Now how can we get uandω? •First, we have to realise that ( K−ω2M)u= 0 has nonzero solutions uonly if det(K−ω2M) = 0. (3.6) This is because multiplication of a matrix like K−ω2Mwith a nonzero vector can yield a zero result only if the determinant of the matrix i s equal to zero. Again (3.6) is analogous to the result known from Linear Alge bra for the case M= 1. Eq. (3.6) is called the characteristic (or secular) equation . det(K−ω2M) is called the characteristic (or secular) polynomial . To 3.1. GENERAL THEORY 65 show that it is a polynomial and determine its degree we use th at like any determinant of a d×d-matrix det(K−ω2M) = det K11−ω2M11K12−ω2M12... K21−ω2M21K22−ω2M22... .........  can be written as a sum over products of dmatrix elements. These matrix elements each involve a term proportional to ω2and a term independent of ω2. The product of dsuch matrix elements thus contains terms proportional to 1,ω2,ω4, etc., up to ω2d. Hence det( K−ω2M) is apolynomial of degree dinω2. The solutions for ω2in det( K−ω2M) = 0, i.e., the generalised eigenvalues, can now be obtained as roots of the characteris tic polynomial. As for any polynomial of degree dthere must be dcomplex roots. These roots will be denoted by ω2 1,ω2 2,... ω2 d. •The generalised eigenvectors u1,u2,... ,udcorresponding to ω2 1,ω2 2,... ω2 dare now obtained as solutions of the equation (K−ω2 jM)uj= 0. (3.7) We will focus on the case where all eigenvalues ω2 jare diﬀerent from each other. Then there is one generalised eigenvector ujfor each generalised eigenvalue (up to multiplication with a complex number).1 If a generalised eigenvalue ω2 jis zero, the corresponding normal modes ujcos(ωjt) anduisin(ωjt) have to be replaced by ujandujt. This can be seen as follows: q(t) =ujis a solution because insertion into (3.2) leads to Md2uj dt2=−Kuj⇐⇒0 =−Kuj which is satisﬁed due to ( K−ω2 1M)ujwithω2 j= 0. Similarly for q(t) =ujtwe get Md2uj dt2=−Kujt⇐⇒0 =−Kujt . General solution The general solution of our equations of motion (3.2) can be w ritten as a linear combination of the normal modes. If all eigenvalues are diﬀe rent from zero we thus obtain q(t) =d/summationdisplay j=1(ajujcos(ωjt) +bjujsin(ωjt)) (3.8) where the constants aj,bjare determined by the initial conditions. 1The characteristic equation may also have multiple roots. F or example, if det( K−ω2M) is proportional to ( ω2−1)3thenω2= 1 is a triple root, and three of the eigenvalues, say, ω2 1,ω2 2and ω2 3, coincide. One can show that for an n-fold root, the equation ( K−ω2 jM)u= 0 has nlinearly independent solutions (proof: see Anton, Elementary Linear Algebra orLang, Linear Algebra ). In the above example, we can thus pick linearly independent vec tors for u1,u2, and u3and then the remaining treatment is the same as in the case where all eigen values are diﬀerent. 66 CHAPTER 3. SMALL OSCILLATIONS Remarks: 1.Which values can ωjtake? First of all, we show that the generalised eigenvalues ω2 jare always real . Proof: Take the generalised eigenvalue equation (3.7) and multipl y with the complex conjugate of uj, i.e.,u∗ j. This gives u∗ j·(K−ω2 jM)uj= 0 and thus u∗ j·Kuj=ω2 ju∗ j·Muj⇒ω2 j=u∗ j·Kuj u∗ j·Muj. (3.9) Hereu∗ j·Kujsatisﬁes (u∗ j·Kuj)∗=uj·K∗u∗ j=uj·Ku∗ j=u∗ j·Kuj; the second equality sign follows because Kis real and the third one follows because Kis symmetric. u∗ j·Kujthus coincides with its complex conjugate, and is a real number. The same applies to u∗ j·Muj. Due to Eq. (3.9) this means that ω2 jis real. This leaves the following possibilities for ωj: •ωjcan be realand diﬀerent from zero. Then the normal modes ujcos(ωj) andujsin(ωjt) describe oscillations . Here the sign of ωjis not impor- tant. Changing the sign only ﬂips the sign of ujsin(ωjt) which can be compensated by also ﬂipping the sign of bjin our general solution (3.8). Hence in the present case ωjcan always be taken positive. •ωjcan be purely imaginary and diﬀerent from zero, i.e. ωj=iρjwhere ρjis real and diﬀerent from zero. The normal modes then take the form ujcos(ωjt) =ujcosh(ρjt) and ujsin(ωjt) =ujsinh(ρjt). Alternatively we can write the solutions as linear combinations of ujeρjtanduje−ρjt, i.e., our coordinates increase or decrease exponentially . Similarly as above ρjcan always be taken positive. •We have already shown that ωj= 0 leads to normal modes constant and linear in t. Due to ω2 jbeing real the generalised eigenvalue equation ( K−ω2 jM)uj= 0 is a real equation, and the eigenvectors ujcan be chosen real as well. 2.Are the eigenvectors orthogonal? ForM= 1, it was shown in Linear Algebra that the eigenvectors corr espond- ing to diﬀerent eigenvalues are orthogonal, i.e., uj·uk= 0 if ω2 j∝ne}ationslash=ω2 k. For general Mwe obtain a generalised orthogonality relation : For ω2 j∝ne}ationslash=ω2 k we have uj·Muk= 0. Proof: We write (ω2 j−ω2 k)uj·Muk=uk·ω2 jMuj−uj·ω2 kMuk =uk·Kuj−uj·Kuk = 0. 3.2. TWO SPRINGS 67 In the ﬁrst line, the symmetry of Mwas used to replace one uj·Mukbyuk· Muj. In the second line, the eigenvector equation ω2 jMuj=Kuj(similarly fork) was invoked. The third line follows from the symmetry of K. If we normalise the eigenvectors according to uj·Muj= 1, and all generalised eigenvalues are diﬀerent, we even have uj·Muk=δjk. This generalises the relation uj·uk=δjkfor the eigenvector problem studied in Linear Algebra. 3.Are the eigenvectors linearly independent? Suppose that the generalised eigenvalues ω2 1,ω2 2,... ,ω2 dare distinct. Then the generalised eigenvectors u1,u2,... ,udare linearly independent . Proof: We have to show that 0 cannot be written as a nontrivial linear com- bination of the eigenvectors uj, i.e.,/summationtextd j=1cjuj= 0 implies that all coeﬃcients cjare 0. Let us thus assume that/summationtextd j=1cjuj= 0. Multiplication with ukand Mleads to d/summationdisplay j=1cjuk·Muj= 0. (3.10) According to 3. the product uk·Mujis equal to zero for all j∝ne}ationslash=k. Therefore the sum in (3.10) only receives a contribution from j=k, and we obtain ckuk·Muk= 0. (3.11) Since Mis positive deﬁnite we have uk·Muk∝ne}ationslash= 0 and thus (3.11) can only be satisﬁed if ck= 0 for all k, as claimed. 3.2 Two springs Let us apply the ideas above to a system of two springs, connec ting three freely moving blocks of mass m1,m2andm1. Both springs have the spring constant k and the natural length l. If the blocks are located at positions −l, 0 and las below the springs have their natural length and the potential is ze ro: Figure 3.1: A system of two springs connecting masses at −l, 0 and l. We now assume that the blocks are displaced from these positi ons by q1,q2and q3, and take q1,q2andq3as our generalised coordinates: The kinetic energy can 68 CHAPTER 3. SMALL OSCILLATIONS Figure 3.2: A system of two springs connecting masses at −l+q1,q2andl+q3. then be written as T=m1 2˙q2 1+m2 2˙q2 2+m1 2˙q2 3. In matrix notation Treads T=1 2˙q·M˙q with the mass matrix M= m10 0 0m20 0 0 m1 . Thepotential energy of each spring isk 2times the square of the displacement from the natural length. Since the length of the ﬁrst spring d iﬀers from the natural length by q2−q1, and the length of the second spring by q3−q2, we obtain U=k 2(q2−q1)2+k 2(q3−q2)2=k 2(q2 1+ 2q2 2+q2 3−2q1q2−2q2q3) In matrix notation we thus have U=1 2q·Kq where K= k−k0 −k2k−k 0−k k  (Note that when determining the entries of K, it is important that there is one (diagonal) term in Kcorresponding to for each of the squares q2 1,q2 2,q2 3, but two entries corresponding to each of the mixed terms q1q2,q2q3; hence the coeﬃcients ofq1q2andq2q3have to be divided by two.) With the KandMthus derived the secular equation reads 0 = det( K−ω2M) = det k−ω2m1 −k 0 −k 2k−ω2m2−k 0 −k k −ω2m1  = (k−ω2m1)2(2k−ω2m2)−2k2(k−ω2m1) = (k−ω2m1)(2k2−kω2m2−2kω2m1+ω4m1m2−2k2) = (k−ω2m1)ω2(m1m2ω2−km2−2km1) 3.2. TWO SPRINGS 69 and we obtain the following generalised eigenvalues ω2 1= 0 ω2 2=k m1 ω2 3=k m1+2k m2 For each ω2 jthe corresponding generalised eigenvector is obtained from 0 = (K−ω2 jM)uj= k−ω2m1 −k 0 −k 2k−ω2m2−k 0 −k k −ω2m1 uj We thus get: •forω2 1= 0 k−k0 −k2k−k 0−k k u1= 0 =⇒u1∝ 1 1 1  (here∝means “proportional to”) •forω2 2=k m1  0 −k 0 −k(2−m2 m1)k−k 0 −k 0 u2= 0 =⇒u2∝ 1 0 −1  •forω2 3=k m1+2k m2  −2m1 m2−k 0 −k−m1 m2k−k 0 −k−2m1 m2k u3= 0 =⇒u3∝ 1 −2m1 m2 1  Thegeneral solution is a linear combination of the normal modes (sketched in Fig. 3.3), i.e., q(t) = 1 1 1 (a1+b1t) + 1 0 −1 (a2cosω2t+b2sinω2t) + 1 −2m1 m2 1 (a3cosω3t+b3sinω3t) The form of the ﬁrst normal mode arise because no outside forc es act on the system (apart from gravity which is compensated by a force of constraint). Hence all particles may be translated by the same amount or move wit h the same constant velocity without changing anything. In the second mode the i nner mass remains ﬁxed while the two outer masses oscillate with opposite phas e. In the third mode all masses oscillate, the directions of oscillation of the o uter masse coincide and are opposite to the direction of oscillation of the inner mass. 70 CHAPTER 3. SMALL OSCILLATIONS Figure 3.3: Normal modes of a system with two springs, corres ponding to the generalised eigenvectors ω2 1,ω2 2andω2 3. 3.3 Small oscillations about equilibrium We now come to the main application of the generalised harmon ic oscillator (3.1): We will consider a rather large class of systems, namely thos e with an arbitrary potential and a kinetic energy that is quadratic in ˙q, L(q,˙q) =T−U=1 2˙q·M(q)˙q−U(q) =1 2d/summationdisplay α=1d/summationdisplay β=1˙qαMαβ(q) ˙qβ−U(q).(3.12) (Here M(q) is again a real symmetric, positive deﬁnite matrix.) We wil l show that close to stationary points of the potential the Lagrangian o f these systems can be approximated by a Lagrangian of the form (3.1). We will then a pply the theory of normal modes to the motion close to these points. Equilibria Points q∗where the potential is stationary are called equilibrium points . At these points the derivatives of Umust vanish, ∂U ∂q(q∗) = 0. It is important that particles can stay ﬁxed at equilibrium points . This is immediately clear in Newtonian mechanics: The (conserva tive) forces are given by derivatives of the potential. At equilibrium points thes e derivatives vanish and thus there are no forces. Hence particles may rest at these po ints. Theproof within Lagrangian mechanics looks as follows: With the Lagr angian in (3.12) Lagrange’s equations∂L ∂qγ=d dt∂L ∂˙qγtake the form 1 2d/summationdisplay α,β=1˙qα∂Mαβ ∂qγ˙qβ−∂U ∂qγ=d dtd/summationdisplay β=1Mγβ˙qβ=d/summationdisplay β=1/parenleftbiggdMγβ dt˙qβ+Mγβ¨qβ/parenrightbigg .(3.13) 3.3. SMALL OSCILLATIONS ABOUT EQUILIBRIUM 71 This equation is indeed satisﬁed for q(t) =q∗= const, since for constant qthe time derivatives vanish, and the term∂U ∂qγvanishes at stationary points of U. Approximation close to equilibrium To give an approximation for the Lagrangian Lthat is valid close to the equilibrium, we write q(t) =q∗+δq(t) which in particular implies ˙q=δ˙q. We then make a Taylor expansion up to second order in δq,δ˙q. For the potential this expansion reads U(q)≈U(q∗) +d/summationdisplay α=1∂U ∂qα(q∗) /bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright =0δqα+1 2d/summationdisplay α=1d/summationdisplay β=1∂2U ∂qα∂qβ(q∗) /bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright ≡Kαβδqαδqβ where ≈indicates dropping all terms of cubic or higher order. The li near term vanishes due todU dqα(q∗) = 0. The quadratic terms can be written in a simple form if we collect the second derivatives of the potential at the e quilibrium point into a matrix Kwith entries Kαβ=∂2U ∂qα∂qβ(q∗). We then obtain U(q)≈U(q∗) +1 2δq·Kδq (3.14) For the kinetic energy we write T=1 2q·M(q)q=1 2δ˙q·M(q)δ˙q ≈1 2δq·(M(q∗) + terms of order δqand higher) δq. Since we drop wall terms with more than two δq,δ˙qthis is approximated by T≈1 2δ˙q·M(q∗)δ˙q. In the vicinity of the equilibrium point q∗, the Lagrangian Lthus becomes L=T−U≈1 2δ˙q·M(q∗)δ˙q−U(q∗)−1 2δq·Kδq (3.15) This is just the form of the Lagrangian considered in Section (3.1), with δqandδ˙q taking the place of qand˙q. The only diﬀerence is the constant −U(q∗) which however does not aﬀect the equations of motion. Example: Imagine a linear molecule with three atoms of mass m1,m2and m1, and a potential that consists of a complicated term dependi ng on the distance between the ﬁrst and the second atom, and an analogous term de pending on the distance between the second and the third atom. Apart from th e form of the potential this situation is analogous to the system with two springs in Section 3.2 72 CHAPTER 3. SMALL OSCILLATIONS If we are close to equilibrium, a quadratic approximation of this potential gives rise to the same Lagrangian as for the two-spring system (with a sp ring constant that depends on the second derivatives of the complicated potent ial). Types of equilibria We are now prepared to study the motion close to equilibrium. We will see that there is a fundamental diﬀerence between equilibria corres ponding to minima and maxima of the potential. •If the potential becomes minimal atq∗, then U(q) =U(q∗)+δq·Kδq(with δq∝ne}ationslash= 0) must be larger than U(q∗). Hence Kispositive deﬁnite , δq·Kδq>0 for all δq∝ne}ationslash= 0. In this case all normal frequencies ωjare real , i.e., the particles oscillate around the equilibrium position. Such equilibria are calle dstable . Proof: Recall Eq. (3.9) and thus ω2 j=uj·Kuj uj·Muj. (3.16) If both MandKare positive deﬁnite, this implies ω2 j>0 and thus ωj∈R. Intuitive argument: Conservative forces always point in the direction where the potential decreases – e.g. the gravity points downwards , where the gravita- tional potential decreases. If the equilibrium correspond s to a minimum of the potential, the force pulls particles back equilibrium. Thi s leads to oscillations around the equilibrium conﬁguration. Note: The approximation of L, Eq. (3.15) is valid only in the vicinity of the equilibrium. Our theory thus describes correctly only the small oscillations about the minimum. Figure 3.4: Example of a stable equilibrium. •For amaximum of the potential the same argument as above implies that Kisnegative deﬁnite , i.e., we have δq·Kδq<0 for all δq∝ne}ationslash= 0 3.4. THE DOUBLE PENDULUM 73 In this case all normal frequencies ωjare purely imaginary (ωj=iρj, ρj∈R) and the normal modes are proportional to increasing and dec reasing exponentials eρjt,e−ρjt. Such equilibria are called unstable . Proof: We now have uj·Kuj<0 and uj·Muj>0. This means that ω2 j must be negative, and ωjmust be imaginary. Intuitive argument: The forces again point in the direction where the po- tential decreases – which this time means away from the equil ibrium. Thus particles are usually pushed away from the equilibrium. How ever if we ﬁnely tune the initial velocity a particle approaching the equili brium may just get slower and slower without ever changing direction or crossi ng the equilibrium. The latter situation corresponds to the normal modes uje−ρjt. Note: As above our solution is no longer applicable one we are far aw ay from the equilibrium, since then the quadratic approximation of the Lagrangian is no longer valid. Figure 3.5: Example of an unstable equilibrium. •IfKis neither positive or negative deﬁnite, there can be real, i maginary and zero normal frequencies. 3.4 The double pendulum As an example we consider the double pendulum depicted in Fig . 3.6. We assume that the two masses and the lengths of the two pendulums coinc ide. The motion of the double pendulum can be very complicated (“chaotic”) in g eneral, but we will see that it becomes simple close to equilibrium. The two mass es have positions r1=l/parenleftbiggsinθ1 −cosθ1/parenrightbigg ,r2=l/parenleftbiggsinθ1 −cosθ1/parenrightbigg +l/parenleftbiggsinθ2 −cosθ2/parenrightbigg and velocities ˙r1=l˙θ1/parenleftbiggcosθ1 sinθ1/parenrightbigg ,˙r2=l˙θ1/parenleftbiggcosθ1 sinθ1/parenrightbigg +l˙θ2/parenleftbiggcosθ2 sinθ2/parenrightbigg with ˙r2 1=l2˙θ2 1,˙r2 2=l2˙θ2 1+ 2l2cos(θ1−θ2)˙θ1˙θ2+l2˙θ2 2. 74 CHAPTER 3. SMALL OSCILLATIONS Figure 3.6: The double pendulum. The kinetic energy of the double pendulum is therefore obtai ned as T=1 2m(˙r2 1+˙r2 2) =1 2ml2/parenleftig 2˙θ2+ 2cos( θ1−θ2)˙θ1˙θ2+˙θ2 2/parenrightig . It is helpful to rewrite Tin matrix notation. We then get T=1 2˙q·M(q)˙q withq=/parenleftbiggθ1 θ2/parenrightbigg and the mass matrix M(q) =ml2/parenleftbigg2 cos( θ1−θ2) cos(θ1−θ2) 1/parenrightbigg . (3.17) Thepotential energy is given by U=−2mglcosθ1−mglcosθ2. Equilibria To ﬁnd equilibria , we have to set the derivatives of Uw.r.t. the generalised coor- dinates equal to zero. This leads to ∂U ∂θ1= 0⇒sinθ1= 0⇒θ1= 0 or θ1=π ∂U ∂θ2= 0⇒sinθ2= 0⇒θ2= 0 or θ2=π . The four equilibria found in this way are sketched in Fig. 3.7 . We would expect the equilibrium at θ1=θ2= 0 to be stable and the one at θ1=θ2=πto be unstable while the two others should be mixed (with one real and one pur ely imaginary normal frequency). We shall investigate in detail the motio n close to the stable and the unstable equilibrium. 3.4. THE DOUBLE PENDULUM 75 Figure 3.7: Equilibrium positions of the double pendulum. Stable equilibrium Since the ﬁrst equilibrium is at q∗=/parenleftbigg0 0/parenrightbigg (3.18) the deviations from equilibrium δqcoincide with the generalised coordinates q. The quadratic approximation of the potential is obtained ea sily if we use the Taylor expansion of the cosine cos θ= 1−1 2θ2+.... We then get U=−2mglcosθ1−mglcosθ2 ≈ −2mgl/parenleftbigg 1−θ2 1 2/parenrightbigg −mgl/parenleftbigg 1−θ2 2 2/parenrightbigg =1 2δq·Kδq+ const with K=mgl/parenleftbigg2 0 0 1/parenrightbigg . For the quadratic approximation of the kinetic energy , we just have to evaluate the mass matrix at the equilibrium position. We then get T≈1 2δ˙q·M(q∗)δ˙q M(q∗) =ml2/parenleftbigg2 1 1 1/parenrightbigg . 76 CHAPTER 3. SMALL OSCILLATIONS The generalised eigenvectors ω2and the normal frequencies ωcan now be obtained from the secular equation 0 = det( K−ω2M) = det ml2/parenleftbigg2g l−2ω2−ω2 −ω2 g l−ω2/parenrightbigg ⇒0 = 2/parenleftigg l−ω2/parenrightig2 −ω4 ⇒ω2= (2 ±√ 2)g l>0 As expected both normal frequencies are real. The correspon ding generalised eigen- vectors are easily obtained as u∝/parenleftbigg1 ∓√ 2/parenrightbigg . This leads to the two types of motion sketched in Fig. 3.8. Figure 3.8: Normal modes related to the stable equilibrium o f the double pendulum. Unstable equilibrium We now consider the equilibrium at q∗=/parenleftbiggπ π/parenrightbigg . We thus let q=/parenleftbiggθ1 θ2/parenrightbigg =q∗+δq=/parenleftbiggπ+δθ1 π+δθ2/parenrightbigg and approximate up to quadratic order in δqand its derivative. For the potential energy U=−2mglcosθ1−mglcosθ2 Taylor expansion of the cosine gives cos(θ1) = cos( π+δθ1) =−cos(θ1)≈ −1 +(δθ1)2 2 (similarly for θ2) and thus U=1 2δq·Kδq+ const 3.4. THE DOUBLE PENDULUM 77 with K=−mgl/parenleftbigg2 0 0 1/parenrightbigg . Thekinetic energy is approximated by T≈1 2δ˙q·M(q∗)δ˙q where the mass matrix at equilibrium again reads M(q∗) =ml2/parenleftbigg2 1 1 1/parenrightbigg . Compared to the stable equilibrium, only Khas changed sign. Since Kis pro- portional to gwe can thus copy the results from the stable equilibrium with the replacement g→ −g. This yields ω2=−(2±√ 2)g l<0 (i.e. we indeed have two imaginary frequencies) while the ge neralised eigenvectors u∝/parenleftbigg1 ∓√ 2/parenrightbigg remain the same. 78 CHAPTER 3. SMALL OSCILLATIONS Chapter 4 Rigid bodies An important application of mechanics is the motion of rigid bodies. Rigid bodies are formally deﬁned as follows: Arigid body is a system in which the distances between the particles do no t vary in time. Most solid objects around us are rigid bodies to a reasonable approximation, because the distances between the atoms do not vary in time. T he distances of the particles being ﬁxed, there are only few things that one can d o with a rigid body: translate it (i.e. move all particles by the same amount in the same dire ction) and rotate it about an arbitrary axis. In this lecture we will only be interested in rotations about axes through the origin . These rotations are important in the following two cases: •If the body is ﬁxed at the origin , we can no longer move all particles by the same amount in the same direction. And if we want to rotate the body we must make sure that the origin remains ﬁxed. This means tha t the axis of rotation must go through the origin. •Now let us consider a rigid body on which no outside forces are acting, e.g. a rigid body in space. In this case it is helpful to study the mo tion of the centre of mass, deﬁned by N/summationdisplay i=1mi Mri where M=/summationtextN i=1miis the overall mass of the system. As any isolated system, such a rigid body satisﬁes several conservation laws (see Se ction 2.4.4). In particular the overall momentum of all particles is conserv ed, N/summationdisplay i=1mi˙ri= const . This implies that d dtN/summationdisplay i=1mi Mri= const , i.e., the velocity of the centre of mass is constant. 79 80 CHAPTER 4. RIGID BODIES We now consider the case of zero velocity, and choose our coor dinate system such that the centre of mass is at the origin . Then again only rotations about axes through the origin are possible. Note that the body is still allowed to rotate about allaxes through the origin, and that the rotation axis can change in time. A main goal will be t o understand such changes of the rotation axis. 4.1 Angular velocity The angular velocity is an important quantity that characte rises both the axis and the speed of rotation. Def.: Assume that a rigid body is rotated about an axis that goes thr ough the origin and has the direction n(where nis a unit vector). If during the time dt the body is rotated by an angle dφ, itsangular velocity is deﬁned as ω=dφ dtn. If we know the angular velocity we can determine the velocity of every particle in the body. Fact: If a rigid body rotates with an angular velocity ω, the velocities of the particles are given by ˙ri=ω×ri. Proof: Figure 4.1: Rotation about an axis through the origin with di rection n. We show that both the norms and the directions of ˙riandω×ricoincide. We use that a particle rotating about an axis moves on a circle. A s seen in the picture the radius of this circle is |ri|sinθwhere |ri|is the distance from the origin and θ 4.2. INERTIA TENSOR 81 is the angle enclosed between the position vector and the dir ection of the axis. The product |ri|sinθcan also be written as |n×ri|(using that nis normalised). A particle moved by an angle dφnow travels a distance |n×ri|dφon the circle. The absolute value of its velocity is obtained by dividing out dt. We then get /vextendsingle/vextendsingle˙ri|=|ndφ dt/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright =ω×ri/vextendsingle/vextendsingle, (4.1) as desired. Now we consider the direction of the vectors. The circle that our particle moves on is in a plane perpendicular to the rotation axis, which mea ns that the velocity is perpendicular to n. In addition it is clear from the picture that the velocity is perpendicular to ri. The same holds true for the cross product n×ri. Hence also the directions coincide and the statement is proven. 4.2 Inertia tensor Analogies It would be nice if we could describe the rotation of rigid bod ies in a way analogous to to a motion that we already know very well: the translation al motion of a single particle. The most important properties of such a particle a re its momentum p, its velocity vand its mass m. These quantities are related by p=mv. For a rigid body, the analogue of the momentum is the total angular momentum, and the analogue of the velocity is the angular velocity. The questi on is therefore: Is there an analogue of the mass as well, and can we use this to write an e quation similar top=mv? We will see that an analogue of the mass indeed exists; howeve r it is not a really number but a matrix – the so-called inertia tensor . To ﬁnd it we write the total angular momentum as l=N/summationdisplay i=1ri×pi =N/summationdisplay i=1ri×mi˙ri =N/summationdisplay i=1ri×mi(ω×ri) =N/summationdisplay i=1miri×(ω×ri). (4.2) Reassuringly, Eq. (4.2) already expresses the total angula r momentum as a linear function of the angular velocity. To bring it closer to the fo rmp=mvwe use the following formula for combinations of two cross products: a×(b×c) =b(a·c)−c(a·b) 82 CHAPTER 4. RIGID BODIES (the so-called “back-cab formula”). We then obtain l=N/summationdisplay i=1mi{ω(ri·ri)−ri(ri·ω)}. It is now helpful to write out the ath component of the total angular momentum. Since the ri·riin the ﬁrst term and the ri·ωin the second term are just real numbers, for the ath component we have to replace the ωin the ﬁrst term by ωa a and the riin the second term by ria. Writing out the scalar products we thus obtain la=N/summationdisplay i=1mi/braceleftigg ωa3/summationdisplay c=1r2 ic−ria3/summationdisplay b=1ribωb/bracerightigg To let the formula look more symmetric, it would be helpful to introduce a sum- mation over bnot only in the second term, but also in the ﬁrst one. We thus write ωa=3/summationdisplay b=1δabωb with the Kronecker delta already used in Section 2.4.1. If we now pull the sum over bin front of all other terms and shift ωbto the end, we can write the ath component of the angular momentum as la=3/summationdisplay b=1N/summationdisplay i=1mi/braceleftigg δab3/summationdisplay c=1r2 ic−riarib/bracerightigg ωb. We thus obtain the following result: The total angular momentum land the angular velocity ωare related by la=3/summationdisplay b=1Iabωb where Iab=N/summationdisplay i=1mi/braceleftigg δab3/summationdisplay c=1r2 ic−riarib/bracerightigg . In matrix notation we have l=Iω where I= I11I12I13 I21I22I23 I31I32I33  is called the inertia tensor . Rotating coordinate system The inertia tensor Ideﬁned above depends on the coordinates riawhich themselves are functions of time. This can be avoided if we write our vect ors in a basis that depends on time: 4.2. INERTIA TENSOR 83 In the following, we will use basis vectors E1(t),E2(t),E3(t) that depend on time and rotate in the same way as the particle positions in the rig id body. This basis is chosen to be orthonormal, i.e., we have EA(t)·EB(t) =δAB. (4.3) Moreover we want to have a “right-handed” basis, which means that E1(t)×E2(t) =E3(t),E2(t)×E3(t) =E1(t),E3(t)×E1(t) =E2(t).(4.4) Example: If the body rotates about the z-axis with angular velocity ω= 0 0 ω we can use E1(t) = cosωt sinωt 0 ,E2(t) = −sinωt cosωt 0 ,E3(t) = 0 0 1 . Expansion in the new basis: All vectors v(t) can now be expanded in the basisE1(t),E2(t),E3(t) and the coeﬃcients will be denoted by capital letters, v(t) =3/summationdisplay A=1VA(t)EA(t) For instance we will have ri(t) =3/summationdisplay A=1RiAEA(t) ω(t) =3/summationdisplay A=1ΩA(t)EA(t) l(t) =3/summationdisplay A=1LA(t)EA(t). Note that here the new particle coordinates RiAdon’t depend on time, because the basis vectors are rotated in the same way as the particles and the whole time dependence of ri(t) is now in the rotation of the basis vectors EA(t). That’s the reason why we introduced these basis vectors in the ﬁrst plac e. We don’t know yet how the total angular momentum and the angular velocity are g oing to depend on time, so in contrast to RiAthe new coordinates of these quantities still have to be written as functions of t. Due to Eq. (4.3) all scalar products of vectors can be written as sums over products of their components in the new basis: v(t)·w(t) =/summationdisplay AVA(t)EA(t)·/summationdisplay BWB(t)EB(t) =/summationdisplay A/summationdisplay BVA(t)WB(t)δAB =/summationdisplay AVA(t)WA(t). 84 CHAPTER 4. RIGID BODIES For scalar products with basis vectors this imples v(t)·EA(t) =VA(t). To apply these ideas to Eq. (4.2) we multiply with EA(t). Repeated application of the formula for the scalar product and manipulations simi lar to the previous calculation give LA(t) = l(t)·EA(t) =N/summationdisplay i=1mi/bracketleftbig (ω(t)·EA(t))(ri(t)·ri(t))−(ri(t)·EA(t))(ri(t)·ω(t))/bracketrightbig =N/summationdisplay i=1mi/bracketleftbig ΩA(t)/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright = /C8 BδABΩB(t)/summationdisplay CR2 iC−RiA/summationdisplay BRiBΩB/bracketrightbig =/summationdisplay B/bracketleftiggN/summationdisplay i=1mi/parenleftigg δAB/summationdisplay CR2 iC−RiARiB/parenrightigg/bracketrightigg /bracehtipupleft /bracehtipdownright/bracehtipdownleft /bracehtipupright ≡IABΩB(t) =/summationdisplay BIABΩB(t). HereIABis independent of time, like the mass in p=m˙r! In many cases IABcan be made diagonal by choosing a convenient set of rotating vec tors. The diagonal elements obtained in this way (i.e. the eigenvalues of Iare also called moments of inertia . Continuously distributed mass Typically a rigid body contains lots of particles (atoms). I n this case a further improvement is very helpful. Instead of summing over all pos itions Riof atoms, we assume that the mass is distributed continuously over the ri gid body. It is helpful to describe this distribution in terms of the mass density: Def.: Themass density ρ(R1,R2,R3) of a rigid body is deﬁned such that the mass in a volume from R1toR1+dR1, from R2toR2+dR2, and form R3to R3+dR3is given by ρ(R1,R2,R3)dR1dR2dR3. Typically, the mass is just distributed uniformly over the r igid body. Then the mass density is simply the overall mass of the body divided by the overall volume. The elements of the inertia tensor can now be determined as by IAB=/integraldisplay /integraldisplay /integraldisplay ρ(R1,R2,R3)/parenleftigg δAB/summationdisplay CR2 C−RARB/parenrightigg dR1dR2dR3 (4.5) Proof: The contribution of the volume from R1toR1+dR1, from R2to R2+dR2, and from R3toR3+dR3toIABis given by ρ(R1,R2,R3)dR1dR2dR3/parenleftigg δAB/summationdisplay CR2 C−RARB/parenrightigg . Integration over the whole body gives Eq. (4.5). 4.2. INERTIA TENSOR 85 Example An example for a rigid body is a box with coordinates −a 2< R1<a 2,−b 2< R2<b 2 and−c 2< R3<c 2. and a constant density. If we denote the mass of the box by M and use that the volume is abcwe thus have ρ=M abc= const. Application of Eq. (4.5) then gives I=/integraldisplayc/2 −c/2/integraldisplayb/2 −b/2/integraldisplaya/2 −a/2ρ R2 2+R2 3−R1R2−R1R3 −R2R1R2 1+R2 3−R2R3 −R3R1−R3R2R2 1+R2 2 dR1dR2dR3 =ρabc/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright =M b2+c2 120 0 0a2+c2 120 0 0a2+b2 12  where diﬀerent rows and columns now correspond to diﬀerent d irections in the rotating coordinate system ﬁxed to the rigid body. The oﬀ-di agonal elements vanish due to/integraltexta/2 −a/2R1dR1= 0, etc. We see that Iis constant in time and even diagonal. Figure 4.2: A box centred at the origin. Example (optional) Consider a sphere with radius R, volume V=4 3πR3and constant density ρ=M V. We use spherical coordinates R1= rsinθcosφ,R2=rsinθsinφ,R3=rcosθ. The inertia tensor is I= /CIR 0 /CIπ 0 /CI2π 0ρ /BCR2 2+R2 3−R1R2 −R1R3 −R2R1 R2 1+R2 3−R2R3 −R3R1 −R3R2 R2 1+R2 2 /BD/BTr2sinθdφdθdr = /CIR 0 /CIπ 0 /CI2π 0ρr2 /BCsin2θsin2φ+ cos2θ −sin2θcosφsinφ −sinθcosφcosθ −sin2θsinφcosφsin2θcos2φ+ cos2θ −sinθsinφcosθ −cosθsinθcosφ −cosθsinθsinφ sin2θ /BD/BTr2sinθdφdθdr =M 4 3πR3/DG /DF/DE /DH =ρ·1 5R5·2π·4 3 /BC1 0 0 0 1 0 0 0 1 /BD/BT=2 5MR2 /BC1 0 0 0 1 0 0 0 1 /BD/BT Here the oﬀ-diagonal elements vanish due to /CA2π 0cosφsinφdφ= /CA2π 01 2sin2φ= 0, /CA2π 0cosφdφ= 0, /CA2π 0sinφdφ= 0. For the diagonal elements we need the φ-integrals /CA2π 0cos2φdφ= /CA2π 0sin2φdφ=π, the θ-integrals /CAπ 0cos2sinθ=−1 3cos2θ|π 0= 2 3, /CAπ 0sin3θdθ= /CAπ 0sindθ− /CAπ 0cos2sinθ= 2 −2 3=4 3, and the r-integral /CAR 0r4dr=1 5R5. 86 CHAPTER 4. RIGID BODIES 4.3 Euler’s equations We now want to ﬁnd out how for a rigid body the angular velocity components Ω1(t),Ω2(t),Ω3(t) (and thus the axis and speed of rotation) change in time. For simplicity we only consider rigid bodies on which no externa l forces are acting. Then we use that for such isolated systems the total angular momen tumlis a conserved quantity. We thus obtain 0 =d dtl =d dt/summationdisplay ALA(t)EA(t) =˙L1(t)E1(t) +˙L2(t)E2(t) +˙L3(t)E3(t) +L1(t)˙E1(t) +L2(t)˙E2(t) +L3(t)˙E3(t). (4.6) (Note that the conservation of ldoes not imply conservation of L1(t),L2(t) and L3(t)!) We now need the derivatives of the basis vectors E1(t),E2(t),E3(t). Since these basis vectors are rotated in the same way as the particl e positions their time derivatives are also obtained by taking a cross product with ω(t). This gives ˙E1(t) = ω(t)×E1(t) = (Ω 1E1(t) + Ω 2(t)E2(t) + Ω 3(t)E3(t))×E1(t) =−Ω2(t)E3(t) + Ω 3(t)E2(t) where in the second line ω(t) was expanded in terms of the rotating basis vectors and in the third line I used that we have a right-handed coordi nate system (Eq. (4.4)) and that the cross product E1(t)×E1(t) is zero. Analogous reasoning for the other basis vectors gives ˙E2(t) = Ω 1(t)E3(t)−Ω3(t)E1(t) ˙E3(t) = −Ω1(t)E2(t) + Ω 2(t)E1(t). If we sustitute all this into Eq. (4.6) and collect the terms p roportional to each of the basis vectors we get ˙L1(t) + Ω 2(t)L3(t)−Ω3(t)L2(t) = 0 ˙L2(t) + Ω 3(t)L1(t)−Ω1(t)L3(t) = 0 ˙L3(t) + Ω 1(t)L2(t)−Ω2(t)L1(t) = 0 . (4.7) Now we can eliminate the LA(t)’s using that LA(t) =/summationtext BIABΩB(t). If we choose our rotating basis in such a way that Ibecomes diagonal (with diagonal elements I1,I2andI3), this relation turns into LA(t) =IAΩA(t). This implies ˙LA(t) =IA˙ΩA(t). Eq. (4.7) thus turns into 4.3. EULER’S EQUATIONS 87 Euler’s equations I1˙Ω1= (I2− I3)Ω2Ω3 I2˙Ω2= (I3− I1)Ω3Ω1 I3˙Ω3= (I1− I2)Ω1Ω2 (4.8) where the Ω A’s depend on time whereas the I’s do not. Note: Alternatively we could have used Ω A=LA IAto eliminate the Ω A’s, yielding ˙L1=I2− I3 I2I3L2L3 ˙L2=I3− I1 I3I1L3L2 ˙L3=I1− I2 I1I2L1L3. Euler’s equations (4.8) determine the time evolution of the ΩA’s and thus of the rotation axis. Unfortunately, they are nonlinear (the Ω A’s and their derivatives appear both linearly and quadratically) and thus diﬃcult to solve. However, there is a special case (the symmetric top) in which they eﬀectivel y become linear and can be solved easily. Example: Symmetric top Figure 4.3: The symmetric top. An example for a symmetric top is shown in Fig. 4.3. We see that the top is symmetric w.r.t. rotations around the R3-axis. Hence the moment of inertia I1 88 CHAPTER 4. RIGID BODIES corresponding to the R1-axis and the moment of inertia I2corresponding to the R2-axis should coincide, but diﬀer from the moment of inertia I3corresponding to rotations around the R3-axis. Denoting the two diﬀerent moments of inertia by I⊥ andI/bardbl, we thus have I1=I2=I⊥∝ne}ationslash=I3=I/bardbl. If we insert these momenta into Euler’s equations (4.8), we ﬁ nd I⊥˙Ω1= (I⊥− I/bardbl)Ω2Ω3 I⊥˙Ω2= (I/bardbl− I⊥)Ω3Ω1 I/bardbl˙Ω3= 0 (4.9) Crucially, the third line implies that the time derivative o f Ω3vanishes. Hence Ω 3 (the component of the angular velocity corresponding to rot ations around the Z- axis) is a constant. The only thing we still have to do is to sol ve for Ω 1and Ω 2. – This is now a simple task since the ﬁrst two equations are line ar in Ω 1and Ω 2. To proceed we collect the constants I/bardbl,I⊥and Ω 3into one, denoted by ν=I⊥− I/bardbl I⊥Ω3= const . The ﬁrst two equations in (4.9) thus turn into ˙Ω1=−νΩ2 (4.10) ˙Ω2=νΩ1 (4.11) These equations can now easily be solved. If we take the deriv ative of (4.10) and insert (4.11) we get ¨Ω1=−ν˙Ω2=−ν2Ω1. This is the well-known equation of a harmonic oscillator. Th e solution reads Ω1=Ccos(νt+φ), where Candφare constants that have to be chosen to satisfy the initial co nditions. Ω2can now be determined from Eq. (4.10) as Ω2=−1 ν˙Ω1=Csin(νt+φ) These equations just indicate a rotation about the R3-axis with frequency ν. We thus see that the rotation axis of a symmetric top is not ﬁxed ( even in a coordinate system following the rigid body), but precedes about the R3-axis. Example: Rotation axis of the Earth Another example for a rigid body is the Earth. Of course, the E arth has an almost spherical shape. But it is not quite a sphere – the radius of th e Equator is a little bit larger than the distance between the Equator plane and th e North pole. Hence, just like for the symmetric top the moments of inertia corres ponding to the R1-and R2-direction coincide with each other, but diﬀer slightly fro m the moment of inertia 4.3. EULER’S EQUATIONS 89 corresponding to the R3-axis. Therefore, Leonhard Euler courageously applied to the Earth the same results as for the symmetric top. For the Ea rth we have I/bardbl− I⊥ I⊥≈1 305 and (since it rotates around itself once in a day) Ω3=2π 1 day. One would thus conclude that the Earth’s rotation axis rotat es (precedes) relative to the surface of the Earth with a frequency ν=2π 305 days, i.e., with a period of 305 days. Measurements by Set Carlo Cha ndler (1846-1913) conﬁrmed the existence of this precession (which is now also called the Chandler wobble ), but he got a period of 433 days! This discrepancy was explai ned by Simon Newcomb (1835-1909) as being due to the slight non-rigidity of the Earth. Actually the Earth is quite rigid, slightly more rigid than steel – but the diﬀerence from a perfect rigid body is comparable to the diﬀerence of the shap e of the Earth from a perfect sphere, which was causing the eﬀect in the ﬁrst place . It is not possible to keep into account one of these small eﬀects and neglect the ot her. 90 CHAPTER 4. RIGID BODIES Chapter 5 Hamiltonian mechanics 5.1 Hamilton’s equations Hamiltonian mechanics is a formulation of mechanics simila r to the one by Lagrange, but with important advantages. To motivate it, let us recall some results from the previous chapters. Reminder We started from Newtonian mechanics , in particular from Newton’s second law F=ma. We then derived Lagrange ’s formulation of mechanics. The main advan- tage of Lagrange’s formalism is that it looks the same for all coordinate systems, and one can easily account for constraints. The central object i n Lagrangian mechanics is theLagrangian L(q,˙q,t) =T−U, depending on the generalised coordinates q, their derivatives ˙qand time t. The trajectories of particles are then obtained from Lagrange’s equations. To solve Lagrange’s equations it is v ery helpful to identify conserved quantities of the system. These conserved quantities can be read oﬀ from the Lagrangian: •If the Lagrangian is independent of time the generalised energy h(q,˙q,t) =∂L ∂˙q·˙q−L(q,˙q,t) is conserved. •If the Lagrangian is independent of one of the coordinates qα, the correspond- inggeneralised momentum pα=∂L ∂˙qα(5.1) is conserved. Hamilton Even though conservation laws can be used in Lagrangian mech anics, the fundamen- tal quantities in Lagrangian mechanics are still the Lagran gianL, the coordinates qand their derivatives ˙q, not the (potentially) conserved quantities handp. The treatment of conservation laws would be much cleaner if inst ead we could build 91 92 CHAPTER 5. HAMILTONIAN MECHANICS the whole theory around handp. We would thus replace the Lagrangian as the main quantity of interest by the generalised energy, and use the momenta p instead of the derivatives ˙q. This is the idea of Hamiltonian mechanics. To replace ˙qone can use that pis determined by q,˙qandtthrough (5.1) p=∂L ∂˙q=p(q,˙q,t). (5.2) We now assume that this equation can be solved uniquely for ˙q. – This uniqueness is a condition for the applicability of Hamiltonian mechani cs. Then ˙qcan be written as a function of q,pandt, ˙q=˙q(q,p,t). (5.3) We now want to build a theory based on the generalised energy h. But instead of writing has a function of q,˙qandtwe want to replace ˙qas an argument by the momentum p. This can be done if we substitute all occurrences of ˙qby˙q(q,p,t) and replace∂L ∂˙qbyp. The resulting function is called the Hamiltonian. Hamiltonian H(q,p,t) =h(q,˙q(q,p,t),t) =p·˙q(q,p,t)−L(q,˙q(q,p,t),t) The Hamiltonian can be used to get equations of motion simila r to Lagrange’s equations. These equations have the following form: Hamilton’s equations ˙qα=∂H ∂pα ˙pα=−∂H ∂qα(5.4) Proof: To prove Hamilton’s equations we compute the partial deriva tives of the Hamiltonian. There are three terms in Hthat depend on H: the initial pand two functions ˙q(q,p,t). Taking into account all these terms we obtain the derivati ve ∂H ∂pαas ∂H ∂pα= ˙qα+p·∂˙q ∂pα−∂L ∂˙q/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright =p·∂˙q ∂pα= ˙qα Here the second and third term cancel due to p=∂L ∂˙q. The ﬁrst equation in (5.4) is thus proven. To compute∂H ∂qαwe take into account the q-dependence of the two ˙q(q,p,t)’s and of the explicit dependence of Lonq. We thus get ∂H ∂qα=p·∂˙q ∂qα−∂L ∂qα−∂L ∂˙q/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright =p·∂˙q ∂qα=−∂L ∂qα 5.1. HAMILTON’S EQUATIONS 93 where again two terms have cancelled. If we now use Lagrange’ s equation∂L ∂qα= d dt∂L ∂˙qαand the deﬁnition of the momentum we can write the result as ∂H ∂qα=−d dt∂L ∂˙qα/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright =pα=−˙pα, proving second equation of (5.4). A related result concerns the time derivative of the Hamiltonian: ∂H ∂t=−∂L ∂t Proof: Diﬀerentiating w.r.t. all three occurrences of tand again using the deﬁnition of pwe get ∂H ∂t=p·∂˙q ∂t−∂L ∂˙q/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright =p·∂˙q ∂t−∂L ∂t=−∂L ∂t Examples: a) Particle in 1d A particle of mass min moving in one dimension has the Lagrangian L=1 2m˙q2−U(q) where U(q) is the potential. The generalised momentum is obtained as p=∂L ∂˙q=m˙q , (5.5) i.e., it is simply the linear momentum (mass times velocity) . Eq. (5.5) can be solved for ˙q, ˙q=p m. (5.6) TheHamiltonian is now obtained as H(q,p,t) =p˙q−L where all ˙ q’s have to be replaced byp m. We thus get H(q,p,t) =p2 m−/parenleftbiggp2 2m−U(q)/parenrightbigg =p2 2m+U(q), andHamilton’s equations read ˙q=∂H ∂p=p m(5.7) ˙p=−∂H ∂q=−∂U ∂q. (5.8) 94 CHAPTER 5. HAMILTONIAN MECHANICS Here (5.7) is equivalent to the deﬁnition of p(see Eq. (5.6)); hence this equation merely provides a consistency check, whereas (5.8) gives so mething new. We also note that Eqs. (5.7) and (5.8) could be combined into ¨q=˙p m=−1 m∂U ∂q. (5.9) Since−∂U ∂qis the conservative force corresponding to the potential U(q), Eq. (5.9) coincides with Newton’s second law. b) Particle in 2d, polar coordinates The Lagrangian of a particle in two dimensions can be written in polar coordinates ρ,φas L=1 2m˙ρ2+1 2mρ2˙φ2−U(ρ,φ). We can now determine the momenta associated to ρandφand solve for ˙ ρand˙φ, pρ=∂L ∂˙ρ=m˙ρ=⇒˙ρ=pρ m pφ=∂L ∂˙φ=mρ2˙φ=⇒˙φ=pφ mρ2. TheHamiltonian is thus given by H(ρ,φ,p ρ,pφ,t) = pρ˙ρ+pφ˙φ−L =p2 ρ m+p2 φ mρ2−/parenleftigg p2 ρ 2m+p2 φ 2mρ2−U(ρ,φ)/parenrightigg =p2 ρ 2m+p2 φ 2mρ2+U(ρ,φ), andHamilton’s equations read ˙ρ=∂H ∂pρ=pρ m ˙φ=∂H ∂pφ=pφ mρ2 ˙pρ=−∂H ∂ρ=p2 φ mρ3−∂U ∂ρ ˙pφ=−∂H ∂φ=−∂U ∂φ. Here the ﬁrst two equations again provide a consistency chec k with the deﬁnitions of pρandpφ. The third and fourth equation give the time derivative of th e momenta. Again these time derivatives involve partial derivatives o f the potential. Finally we mention that for a central ﬁeld (a potential Uthat depends only on ρand thus satisﬁes∂U ∂φ= 0) the angular momentum pφis a conserved quantity (i.e. ˙φ= 0). In the context of Lagrangian mechanics this was already seen in Section 2.4.2. 5.1. HAMILTON’S EQUATIONS 95 c) Quadratic kinetic energy In the two previous examples the Hamiltonian coincided with the energy E=T+U, written as a function of q,pandt. This agrees with our previous observation for the generalised energy h, the only diﬀerence being that the Hamiltonian is written as a function of q,pandtrather than q,˙qandt. We thus expect that the same result holds for the whole class of systems considered previously, satisfying the following two conditions: •The kinetic energy Tis quadratic in ˙q, T=1 2˙q·M(q,t)˙q; hereM(q,t) is a matrix (the mass matrix) that must be real, symmetric an d positive deﬁnite. Positive deﬁniteness makes sure that in c ase of non-vanishing velocities Tis positive. •Following our deﬁnition of Uwe assume that the potential is independent of ˙q, i.e., U=U(q,t). The Lagrangian is thus of the form L=1 2˙q·M(q,t)˙q−U(q,t). Themomentum can now be obtained as p=∂L ∂˙q=M(q,t)˙q (5.10) where we used that M(q,t) is symmetric and applied the rule for gradients of quadratic functions from Section 3.5.2. Eq. (5.10) can be so lved for ˙q, to get ˙q=M(q,t)−1p. Here we have used that Mis an invertible matrix. As shown in Linear Algebra, this follows from our conditions on M. (Idea of proof: Since Mis real symmetric it can be diagonalised, i.e., it has eigenvalues and eigenvect ors. To get the inverse we leave the eigenvectors as they are and invert the eigenvalue s. This won’t work if the eigenvalues are zero. However due to the positive deﬁnitene ss ofMthe eigenvalues are positive and there are no diﬃculties.) If we use the deﬁni tion of Hand replace all˙qbyM(q,t)−1ptheHamiltonian is obtained as H=p·˙q−L =p·M(q,t)−1p− 1 2(M(q,t)−1p)·M(q,t)(M(q,t)−1p)/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright =p−U(q,t)  Interchanging the two factors in the scalar product ( M(q,t)−1p)·p, we see that this product cancels half of the initial term p·M(q,t)−1p. We thus obtain H=1 2p·M(q,t)−1p+U(q,t) 96 CHAPTER 5. HAMILTONIAN MECHANICS where1 2p·M(q,t)−1pis the kinetic energy. We have thus veriﬁed our expectation: For systems of the type deﬁned above the Hamiltonian is equal toE=T+U, written as a function of q,pandt. Let us now determine Hamilton’s equations . For ˙qwe obtain ˙q=∂H ∂p=M(q,t)−1p, again in line with our formula for p. The formula for ˙pis easier when written in components; it reads ˙pα=−∂H ∂qα=−1 2p·∂M(q,t)−1 ∂qαp−∂U ∂qα. Phase space The essential diﬀerence between Lagrangian and Hamiltonia n mechanics is that in Hamiltonian mechanics coordinates and momenta are treated on equal foot- ing. Therefore, while Lagrange’s equations were in conﬁgurati on space, Hamilton’s equation are equations in phase space . Phase space is the set {(q,p)}of all allowed generalised coordinates and momenta. Each point in phase sp ace is characterised by one qand one p, which are sometimes referred to as canonical coordinates . Since there is one generalised momentum for each coordinate , the dimension of phase space is 2 d, where dis the number of coordinates (or degrees of freedom). Going from from conﬁguration space to phase space has the fol lowing advan- tages: •(Potentially) conserved quantities like momenta and the Ha miltonian play a much more central role than in Lagrangian mechanics. Hence conservation lawscan be treated in a more systematic way, and can be used more eﬃ ciently to ﬁnd solutions. •An important advantage of Lagrangian mechanics was the free dom to choose any coordinates in phase space that we like. If one works in ph ase space, this freedom in picking coordinates is increased further. Instead of just going from Cartesian coordinates in conﬁguration space to, say, p olar coordinates we can make variable transformations that mix coordinates and mo- menta . This helps us to ﬁnd variables in which a given problem becom es particularly simple. •For many systems it is not possible to solve the equations of m otion exactly. A famous example is three-body problem in which one consider s the gravita- tional attraction between three bodies. If one wants to get t rajectories of these systems one has to resort to approximation methods (“perturbation the- ory”). It turns out that these approximation methods can als o be formulated more elegantly in phase space, using Hamiltonian mechanics . •Phase space has interesting geometrical properties, with connections to pure mathematics (symplectic geometry, diﬀerential forms). Outside classical mechanics, the Hamiltonian formulation also has close connec- tions to quantum theory . 5.1. HAMILTON’S EQUATIONS 97 Orders of equations Lagrangian and Hamiltonian mechanics diﬀer in the orders of the equations in- volved. Lagrange ’s equations contain the conﬁguration-space coordinates qand their ﬁrst and second derivatives ˙qand¨q. The q’s and ˙q’s appear as arguments of the Lagrangian; the ¨q’s come into play because we take an additional time derivati ve ind dt∂L ∂˙q. Hence Lagrange’s equations are second order diﬀerential equations in conﬁguration space . In contrast Hamilton’s equations are ﬁrst-order equations in phase space : They specify ˙qand˙pas functions of qandp, and there are no second derivatives. Going from Lagrange to Hamilton could thus be seen as a trade: reducing the order of equations to the ﬁrst order while at the same time dou bling the number of variables needed. This trick is often used in the theory of di ﬀerential equations. The ﬁrst-order nature of Hamilton’s equations has an import ant consequence: If we know all phase-space variables, i.e., the coordinates q(t)and the momenta p(t)of a mechanical system at a time t, this determines the trajectories of the particles for all times . Note that this is not true in con- ﬁguration space: Given initial conditions only for the posi tionsqwe would not be able to determine the motion of the system at later times. To explain this statement, let us assume that we know q(t) and p(t). Let us then try to ﬁnd the coordinates and momenta after a short time inte rvalǫ, i.e.q(t+ǫ) andp(t+ǫ). Since ǫis small, we can approximate q(t+ǫ) by a Taylor expansion: q(t+ǫ)≈q(t) +˙q(t)ǫ . Since Hamilton’s equations are ﬁrst order, we know ˙q(t). We can thus write q(t+ǫ)≈q(t) +∂H ∂p(q(t),p(t),t)ǫ . The same can be done for the momentum p(t+ǫ): p(t+ǫ)≈p(t) +˙p(t)ǫ=p(t)−∂H ∂q(q(t),p(t),t)ǫ . Now this procedure can be carried further and further. Once w e have q(t+ ǫ),p(t+ǫ), we can the get coordinates and momenta q(t+ 2ǫ),p(t+ 2ǫ) after a further time step ǫby making another Taylor expansion. The changes of the coordinates and momenta are then given by ˙q(t+ǫ)ǫ=∂H ∂p(q(t+ǫ),p(t+ǫ),t)ǫ and˙p(t+ǫ)ǫ=−∂H ∂q(q(t+ǫ),p(t+ǫ),t)ǫ. We can continue like this, adding more and more time steps, until we have the trajectories for all ti mes. A sketch of this procedure for systems with one-dimensional qandpis shown in Fig. 5.1. For a ﬁxed ǫ∝ne}ationslash= 0, our procedure gives a good approximation of qandpat later times. It can be used for example if we want to numerically determine the trajectory. However there is a small error since we dropped a ll higher-order terms in the Taylor expansion. If we want to prove that qandpat a given time determine the trajectories for all time, we therefore have to take the l imitǫ→0. Our observation has an important consequence: Through each point in phase space there is only one unique trajectory . Hence, if we take a picture of phase space like Fig. 5.1 and draw all possible trajectori es, there will be no intersections. 98 CHAPTER 5. HAMILTONIAN MECHANICS Figure 5.1: Stepwise solution of Hamilton’s equations for o ne-dimensional qandp. 5.2 Conservation laws and Poisson brackets In Lagrangian mechanics, we had seen that conservation laws could be read oﬀ from the Lagrangian: If Lis independent of time, the generalised energy is conserved , if it is independent of one coordinate, the corresponding mome ntum is conserved. In Hamiltonian mechanics, these conservation laws are replac ed by the following: •If the Hamiltonian does not depend explicitly on time (i.e.∂H ∂t= 0) it is conserved (i.e.dH dt= 0). Proof: Using the chain rule and Hamilton’s equations, we get d dtH(q,p,t) =∂H ∂q/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright =−˙p·˙q+∂H ∂p/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright =˙q·˙p+∂H ∂t =−˙p·˙q+˙q·˙p+∂H ∂t=∂H ∂t which is zero if Hdoes not depend explicitly on t. •If the Hamiltonian does not depend on a coordinate qα(i.e.∂H ∂qα= 0) the corresponding momentum pαis conserved (i.e.dpα dt= ˙pα= 0). Proof: This follows immediately from Hamilton’s equation ˙ pα=−∂H ∂qα. •If the Hamiltonian does not depend on a momentum pα(i.e.∂H ∂pα= 0) the corresponding coordinate qαis conserved (i.e.dqα dt= ˙qα= 0). Proof: This follows immediately from Hamilton’s equation ˙ qα=∂H ∂pα. 5.2. CONSERVATION LAWS AND POISSON BRACKETS 99 Compared to Lagrangian mechanics we see that the ﬁrst conser vation law has a cleaner form (only Hshows up, not Landh) and that there is a new conservation law for coordinates; the latter appears because coordinate s and momenta are treated on equal footing. How do arbitrary functions F(q,p,t)change in time? A conserved quantity need not be a coordinate or a momentum. H ence we should also ﬁnd out when an arbitrary function F(q,p,t) is conserved and how it changes in time if it is not conserved. To do so we compute the total tim e derivative dF dt=∂F ∂q·˙q+∂F ∂p·˙p+∂F ∂t. If we now use Hamilton’s equations we obtain dF dt=∂F ∂q·∂H ∂p−∂F ∂p·∂H ∂p+∂F ∂t. (5.11) Here∂F ∂q·∂H ∂p−∂F ∂p·∂H ∂pis called the Poisson bracket of FandH. The general deﬁnition of the Poisson bracket is: Def.: ThePoisson bracket of two functions F(q,p,t) and G(q,p,t) is given by {F,G}=∂F ∂q·∂G ∂p−∂F ∂p·∂G ∂q=d/summationdisplay γ=1/parenleftbigg∂F ∂qγ∂G ∂pγ−∂F ∂pγ∂G ∂qγ/parenrightbigg . With the Poisson bracket we can now restate Eq. (5.11) as foll ows: The time derivative of a arbitrary function F(q,p,t) is given by dF dt={F,H}+∂F ∂t. This leads to the following conservation law: IfFis independent of t, i.e. F=F(q,p), then it is a conserved quantity if its Poisson bracket with the Hamiltonian vanishes, i.e. if {F,H}= 0. Poisson brackets allow to check easily whether a given quant ity is conserved. Examples a) Fundamental Poisson brackets The simplest Poisson brackets are those between coordinate s and momenta. They have the following form: {qα,qβ}= 0 {pα,pβ}= 0 {qα,pβ}=δαβ. We see that all Poisson brackets between coordinates and mom enta vanish except the one involving a coordinate and the corresponding moment um. 100 CHAPTER 5. HAMILTONIAN MECHANICS Proof: The ﬁrst two formulas are trivial: When the functions FandGare both coordinates, the derivatives∂F ∂pand∂G ∂pin the deﬁnition of {F,G}vanish and the result is zero. Similarly when the functions FandGare both momenta, the derivatives∂F ∂qand∂G ∂qvanish and again the Poisson bracket is zero. The Poisson bracket involving coordinates and momenta {qα,pβ}=d/summationdisplay γ=1/parenleftbigg∂qα ∂qγ∂pβ ∂pγ−∂qα ∂pγ∂pβ ∂qγ/parenrightbigg . is more complicated. To evaluate it we note that the derivati ve∂qα ∂qγis 1 if α=γand 0 otherwise. Hence it can be written as the Kronecker delta δαβ. For the momenta we similarly have∂pβ ∂pγ=δβγ. In contrast the derivatives∂qα ∂pγand∂pβ ∂qγare always zero. We thus obtain {qα,pβ}=d/summationdisplay γ=1δαγδβγ. Here the summand is 1 if all coeﬃcients coincide, i.e., if α=β=γ. Otherwise it is zero. Hence for α∝ne}ationslash=βall summands are zero, while for α=βone summand is 1. The sum thus yields {qα,pβ}=δαβ as required. b) Components of the angular momentum We now evaluate the Poisson brackets involving the componen ts of the angular momentum l=r×p= x y z × px py pz , i.e., lx=ypz−zpy ly=zpx−xpz lz=xpy−ypx. The Poisson bracket of the ﬁrst two components is {lx,ly}=∂lx ∂x∂ly ∂px+∂lx ∂y∂ly ∂py+∂lx ∂z∂ly ∂pz−(lx↔ly) where ( lx↔ly) indicates the last three summands of the Poisson bracket fo r which the roles of lxandlyare interchanged. Now we can use that∂lx ∂x= 0,∂ly ∂y= 0, ∂lx ∂px= 0 and∂ly ∂py= 0. This leaves only two summands, and we get {lx,ly}=∂lx ∂z∂ly ∂pz−∂ly ∂z∂lx ∂pz= (−py)(−x)−pxy=lz. We thus see that the Poisson bracket of the ﬁrst two component s of the angular momentum yields the third component. 5.2. CONSERVATION LAWS AND POISSON BRACKETS 101 Properties of Poisson brackets We will now derive a few rules that simplify the computation o f more complicated Poisson brackets. Here F,F1,F2,G,H are functions of q,p(and possibly time), anda1anda2are numbers. 1. Linearity: {a1F1+a2F2,G}=a1{F1,G}+a2{F2,G} 2. Antisymmetry: {F,G}=−{G,F} 3. Product rule: {FG,H }=F{G,H}+G{F,H} 4. Jacobi identity: {{F,G},H}+{{G,H},F}+{{H,F},G}= 0 These rules imply that we should think of {F,G}as a kind of cross product of phase-space functions. It should be a product since the deﬁn ition involves products of derivatives of FandG; also the linearity property is expected to hold for prod- ucts. However like the cross product, this product is an anti symmetric one, since it changes sign if the two factors are interchanged. Let us now prove these rules: 1. Linearity can be checked trivially if we insert a1F1+a2F2into the deﬁnition of the Poisson bracket. 2. To prove antisymmetry we rearrange the terms in {F,G}and see that the result coincides with −{G,F}: {F,G}=∂F ∂q·∂G ∂p−∂F ∂p·∂G ∂q =−/parenleftbigg∂G ∂q·∂F ∂p−∂G ∂p·∂F ∂q/parenrightbigg =−{G,F}. 3. To prove the product rule for the Poisson bracket, we use th e product rule for derivatives: {FG,H }=∂(FG) ∂q·∂H ∂p−∂(FG) ∂p·∂H ∂q =F∂G ∂q·∂H ∂p+G∂F ∂q·∂H ∂p−F∂G ∂p·∂H ∂q−G∂F ∂p·∂H ∂q =F{G,H}+G{F,H} 4. The proof of Jacobi’s identity is a bit messy. It is listed b elow for complete- ness: For simplicity we’ll consider the case of one degree of freed om. The calculation in higher dimensions is essentially the same; the only complic ation is keeping track of indices. 102 CHAPTER 5. HAMILTONIAN MECHANICS LetF=F(q, p),G=G(q, p) and H=H(q, p) be functions on phase space. We wish to show that {{F, G}, H}+{{G, H}, F}+{{H, F}, G}= 0. (5.12) Let’s evaluate the ﬁrst term explicitly. We have that {F, G}=FqGp−FpGq, where Fq=∂F/∂q , and similarly Fp,GqandGp. Then {{F, G}, H}=∂{F, G} ∂qHp−∂{F, G} ∂pHq =∂(FqGp−FpGq) ∂qHp−∂(FqGp−FpGq) ∂pHq, or {{F, G}, H}= (FqqGp+FqGpq−FpqGq−FpGqq)Hp−(FqpGp+FqGpp−FppGq−FpGqp)Hq, (5.13) where Fqq=∂2F/∂q∂q , and similarly for Fqp,Fpp,Gqq, etc. Note that Fqp=Fpq, i.e., the ordering of mixed partials doesn’t matter; similarly fo rGqpandHqp. Similarly, for the other two terms in (5.12), {{G, H}, F}= (GqqHp+GqHpq−GpqHq−GpHqq)Fp−(GqpHp+GqHpp−GppHq−GpHqp)Fq, (5.14) and {{H, F}, G}= (HqqFp+HqFpq−HpqFq−HpFqq)Gp−(HqpFp+HqFpp−HppFq−HpFqp)Gq. (5.15) Now it is simply a matter of adding together terms from (5.13) , (5.14) and (5.15) to see that everything cancels. For example, consider the term FqqGpHp. This appears in both (5.13) and (5.15), but with opposite signs. Likewise , you can check that every other term appears in just two of the expressions (5.13), (5. 14) and (5.15) but with opposite signs. In the rules 1, 3 and 4 above always the ﬁrst argument of the Poi sson bracket was something interesting (a sum, a product or another Poisson b racket). Completely analogous rules hold if the second argument is a sum, product or Poisson bracket . They can all be proven from the rules already derived, if we u se the antisymmetry rule to interchange the two arguments. Example To illustrate the above rules let us evaluate the Poisson bra cket{py,lz}: {py,lz}={py,xpy−ypx} ={py,xpy} − {py,ypx} ={py,x}py+{py,py}x− {py,y}px− {py,py}y =−{py,y}px ={y,py}px =px Here we used linearity to get from the ﬁrst to the second line, and the product rule to get from the second to the third line. The third line th en contains only Poisson brackets of coordinates and momenta, which vanish a part from the term −{py,y}. Finally we used antisymmetry and the fundamental Poisson b rackets to get−{py,y}={y,py}= 1. 5.3. CANONICAL TRANSFORMATIONS 103 Example (optional) Consider a phase space with one coordinate qand one momentum p. Let us evaluate the Poisson bracket of qαcos(βp) and qαsin(βp): {qαcosβ(βp), qαsin(βp)}= qα{cos(βp), qαsin(βp)}+ cos( βp){qα, qαsin(βp)} = qαqα{cos(βp),sin(βp)}+qαsin(βp){cos(βp), qα}+ cos( βp)qα{qα,sin(βp)}+ cos( βp)sin( βp){qα, qα} = 0 + qαsin(βp) /AW −∂cos(βp) ∂p∂qα ∂q /AX + cos( βp)qα /AW ∂qα ∂q∂sin(βp) ∂p /AX + 0 = αβsin2(βp)q2α−1+αβcos2(βp)q2α−1 = αβq2α−1 In the ﬁrst two lines we used the product rule. Then the deriva tives were evaluated. Applications As mentioned above, Poisson brackets can be used to check whether a quantity is conserved . Moreover Poisson brackets can be used to generate new conser- vation laws . This is due to Poisson’s theorem: Poisson’s theorem : If two phase-space functions A(q,p) and B(q,p) are con- served then their Poisson bracket {A,B}is conserved as well. Proof: We assume that AandBare conserved, i.e., that their Poisson bracket with the Hamiltonian Hvanishes. We then use the Jacobi identity to evaluate the Poisson bracket of {A,B}withH. The result is zero: {{A,B},H}=−{{B,H}/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright =0,A} − {{ H,A}/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright =0,B}= 0. Hence the Poisson bracket of {A,B}withHvanishes and {A,B}is a conserved quantity. We can thus get additional conserved quantities of a system b y computing the Poisson brackets of the conserved quantities we already kno w. For example, if we know that lxandlyare conserved then their Poisson bracket {lx,ly}=lzmust be conserved as well, and if pyandlzare conserved the same must hold for {py,lz}= px. However computing Poisson brackets need not always give so mething new; the Poisson bracket {A,B}may also be linearly dependent of AandBor trivial (say, zero). 5.3 Canonical transformations A strong advantage of Lagrangian mechanics was that it the eq uations of motion have the same form in all systems of coordinate. This freedom of picking coordinates becomes even larger in Hamiltonian mechanics: Now we can per form coordinate transformations in phase space, including transformations that mix generalised coordinates and generalised momenta . Not all of these transformations will be useful, but only those in which the basic ingredients of Ha miltonian mechanics (such as the Poisson brackets) remain the same. These transf ormation are called canonical transformations and deﬁned as follows: 104 CHAPTER 5. HAMILTONIAN MECHANICS Def.: Consider a transformation of coordinates in phase space Qα=Qα(q,p) Pα=Pα(q,p) and the Poisson bracket deﬁned by derivatives w.r.t. QandP {F,G}Q,P≡∂F ∂Q·∂G ∂P−∂F ∂P·∂G ∂Q. The transformation is called canonical if the Poisson bracket {F,G}Q,Pcoincides with{F,G}=∂F ∂q·∂G ∂p−∂F ∂p·∂G ∂qif we insert for FandGthe coordinates and momenta, {qα,qβ}Q,P={qα,qβ}= 0 {pα,pβ}Q,P={pα,pβ}= 0 {qα,pβ}Q,P={qα,pβ}=δαβ. (5.16) In other words, canonical transformations preserve the fun damental Poisson brack- ets. Example: Consider a system with just one coordinate qand one momenum p. We can now deﬁne a transformation that just interchanges the coordinate and the momentum and ﬂips a sign: Q=−p P=q This transformation is canonical because {q,q}Q,P= 0 {p,p}Q,P= 0 {q,p}Q,P=∂q ∂Q∂p ∂P−∂q ∂P∂p ∂Q=∂P ∂Q∂(−Q) ∂P−∂P ∂P∂(−Q) ∂Q= 1. Here the ﬁrst two lines are trivial as the Poisson bracket of a ny quantity with itself is zero. Thm.: If the transformation leading from q,ptoQ,Pis canonical we have {F,G}Q,P={F,G}for all phase-space functions FandG. In other words, canonical transformations preserve all Poisson brackets. Proof: We assume that (5.16) holds and try to show that {F,G}Q,P={F,G}. 5.3. CANONICAL TRANSFORMATIONS 105 We get {F,G}Q,P=∂F ∂Q·∂G ∂P−∂F ∂P·∂G ∂Q =d/summationdisplay α=1/parenleftbigg∂F ∂qα∂qα ∂Q+∂F ∂pα∂pα ∂Q/parenrightbigg ·d/summationdisplay β=1/parenleftbigg∂G ∂qβ∂qβ ∂P+∂G ∂pβ∂pβ ∂P/parenrightbigg −/bracketleftbigg∂ ∂Q↔∂ ∂P/bracketrightbigg =d/summationdisplay α=1d/summationdisplay β=1 ∂F ∂qα∂G ∂qβ/parenleftbigg∂qα ∂Q·∂qβ ∂P−∂qα ∂P·∂qβ ∂Q/parenrightbigg /bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright ={qα,qβ}Q,P=0 +∂F ∂qα∂G ∂pβ/parenleftbigg∂qα ∂Q·∂pβ ∂P−∂qα ∂P·∂pβ ∂Q/parenrightbigg /bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright ={qα,pβ}Q,P=δαβ +∂F ∂pα∂G ∂qβ/parenleftbigg∂pα ∂Q·∂qβ ∂P−∂pα ∂P·∂qβ ∂Q/parenrightbigg /bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright ={pα,qβ}Q,P=−{qβ,pα}Q,P=−δαβ +∂F ∂pα∂G ∂pβ/parenleftbigg∂pα ∂Q·∂pβ ∂P−∂pα ∂P·∂pβ ∂Q/parenrightbigg /bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright ={pα,pβ}Q,P=0  =d/summationdisplay α=1/parenleftbigg∂F ∂qα∂G ∂pα−∂F ∂pα∂G ∂qα/parenrightbigg ={F,G}. In the ﬁrst line, I simply wrote down the deﬁnition of the Pois son bracket. In the second line, the ﬁrst two derivatives were evaluated usi ng the chain rule (i.e., ∂F ∂Q=/summationtextd α=1/parenleftig ∂F ∂qα∂qα ∂Q+∂F ∂pα∂pα ∂Q/parenrightig , and it was noted that∂F ∂P·∂G ∂Qcan be obtained from∂F ∂Q·∂G ∂Pby interchanging the derivatives w.r.t. QandP. Then the derivatives were grouped together to obtain the fundamental Poisson bra ckets. Hamilton’s equations Canonical transformation also preserve Hamilton’s equati ons. Thm.: If the transformation q,p→Q,Pis time independent and canonical, we have ˙Qα=∂H ∂Pα ˙Pα=−∂H ∂Qα, i.e., if we express the Hamiltonian in terms of Q,Pand take derivatives w.r.t. these quantities we get Hamilton’s equations just as for q,p. 106 CHAPTER 5. HAMILTONIAN MECHANICS Proof: We have ˙Qα={Qα,H}+∂Qα ∂t/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright =0 ={Qα,H}Q,P =d/summationdisplay β=1/parenleftbig∂Qα ∂Qβ/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright =δαβ∂H ∂Pβ−∂Qα ∂Pβ/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright =0∂H ∂Qβ/parenrightbig =∂H ∂Pα Here I used that the total time derivative of every phase spac e function can be written as a Poisson bracket with the Hamiltonian plus a part ial time derivative. In our case the partial time derivative vanishes because the Qα’s do not depend explicitly on time. Then the Poisson bracket was written in t erms of Q,P, and in this form it could be evaluated easily. Similarly we obtain ˙Pα={Pα,H}+∂Pα ∂t/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright =0 ={Pα,H}Q,P =d/summationdisplay β=1/parenleftbig∂Pα ∂Qβ/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright =0∂H ∂Pβ−∂Pα ∂Pβ/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright =δαβ∂H ∂Qβ/parenrightbig =−∂H ∂Qα. Example To illustrate canonical transformations, let us consider t he one-dimensional har- monic oscillator. If we set the mass equal to 1, the harmonic o scillator has the Hamiltonian H=1 2p2+1 2ω2q2. Since Hdoes not depend explicitly on time, it is a conserved quantit y. This means that in phase space the particles can only move on trajectori es for which H= 1 2p2+1 2ω2q2= const, i.e., on ellipses (or circles if ω= 1), see Fig. 5.2. But we still have to determine how fast the particles go along these ellip ses. Theusual way to solve this problem would be to use Hamilton’s equation s with variables qandp. One thus gets ˙q=∂H ∂p=p ˙p=−∂H ∂q=−ω2q . This yields ¨ q= ˙p=−ω2qand thus q=Asin(ωt+φ) where Aandφare constants. Forpwe obtain p= ˙q=Aωcos(ωt+φ). 5.3. CANONICAL TRANSFORMATIONS 107 Figure 5.2: The harmonic oscillator in phase space. Alternatively we can make a coordinate transformation inspired by the form of the curves. If the phase-space curves were circles, it wou ld be convenient to use polar coordinates in phase space . For ellipses one takes p=√ 2Iωcosθ q=/radicalbigg 2I ωsinθ . (5.17) (One easily sees that for ω= 1, this boils down to polar coordinates with a radius r=√ 2I, and we would have H=1 2p2+1 2q2=1 2r2=I.) The coordinate Ican assume arbitrary positive values. Since q,premain the same if the θis increased by 2πthe deﬁnition range of θshould be taken as [0 ,2π). Now one of these variables should be interpreted as a generalised coordinate and the ot her one as a generalised momentum. It turns out that the appropriate choice is to take θ(the“angle” ) as the coordinate and I(the“action” ) as the momentum. We have to show that this transformation is canonical . This can be done by checking that the fundamental Poisson bracket {q,p}θ,Iis 1 (note that {q,q}θ,I= 0 and{p,p}θ,I= 0 are trivial), {q,p}θ,I=∂q ∂θ∂p ∂I−∂q ∂I∂p ∂θ =/radicalbigg 2I ωcosθ/radicalbiggω 2Icosθ−/radicalbigg 1 2Iωsinθ/parenleftig −√ 2Iω/parenrightig sinθ = 1. Expressed in terms of θandItheHamiltonian now becomes H=1 2p2+1 2ω2q2 =Iωcos2θ+Iωsin2θ =Iω , 108 CHAPTER 5. HAMILTONIAN MECHANICS andHamilton’s equations read ˙θ=∂H ∂I=ω ˙I=−∂H ∂θ= 0 We have thus managed to express the Hamiltonian in terms of on ly one variable, I. Since the coordinate θdoes no show up, the corresponding momentum Ihas become a conserved quantity – indicating that the particle s tays on an ellipse in phase space. Also we have seen that θincreases with constant speed ω. However onceθreaches 2 π,pandqbecome the same as for θ= 0, i.e. θis reset to zero. If we draw the phase-space trajectories in a coordinate system parametrised by θand I, the trajectories thus look like in Fig. 5.3: The ellipses fr om Fig. 5.2 have been turned into straight lines. Figure 5.3: The harmonic oscillator in action-angle variab les. Integrable systems There are many more systems for which one can ﬁnd a canonical t ransformation such that the motion in phase space becomes trivial (going al ong straight lines): Def.: A system is called integrable if there is a canonical transformation q,p→θ,I (with q,p,θ,Idenoting d-dimensional vectors) such that the Hamiltonian can be written as a function of Ionly: H=H(I). As for the harmonic oscillator the components of θare called angle variables and the components of Iare called action variables . For integrable systems Hamilton’s equations read ˙θα=∂H ∂Iα ˙Iα=−∂H ∂θα= 0 The second equation means that all Iα’s are conserved quantities. Since in the ﬁrst equation ˙θ=∂H ∂Iαdepends only on these Iα’s it must be constant as well and θα 5.3. CANONICAL TRANSFORMATIONS 109 increases linearly. Thus, expressed in terms of θ,Ithe phase-space motion indeed goes along straight lines. Now which systems are integrable? It is important that there must be ddiﬀer- ent conserved quantities Iα. For a one-dimensional system one conservation law (for the energy) is enough, in higher-dimensional systems we nee d more. Also it must be possible to pick these quantities as momenta after a suita ble canonical transfor- mation. Hence the must be mutually independent and their Poi sson brackets Iα,Iβ must vanish. Roughly speaking, integrable systems are “nice” systems wh ich satisfy enough conservation laws to allow for an analytical solution. Most systems considered in this lecture are of this type. For example the spherical pend ulum is decribed by two generalised coordinates and has two conserved quantiti es (the energy and the angular momentum), which we had already used to obtain an ana lytical solution.