The Lagrangian Method

Transcription

1 Chapter 6 The Lagrangian Method Copyright 2007 by David Morin, orin@physics.harvard.edu (draft version In this chapter, we re going to learn about a whole new way of looking at things. Consider the syste of a ass on the end of a spring. We can analyze this, of course, by using F = a to write down ẍ = kx. The solutions to this equation are sinusoidal functions, as we well know. We can, however, figure things out by using another ethod which doesn t explicitly use F = a. In any (in fact, probably ost physical situations, this new ethod is far superior to using F = a. You will soon discover this for yourself when you tackle the probles and exercises for this chapter. We will present our new ethod by first stating its rules (without any justification and showing that they soehow end up agically giving the correct answer. We will then give the ethod proper justification. 6.1 The Euler-Lagrange equations Here is the procedure. Consider the following seeingly silly cobination of the kinetic and potential energies (T and V, respectively, L T V. (6.1 This is called the Lagrangian. Yes, there is a inus sign in the definition (a plus sign would siply give the total energy. In the proble of a ass on the end of a spring, T = ẋ 2 /2 and V = kx 2 /2, so we have Now write L = 1 2 ẋ2 1 2 kx2. (6.2 d dt ( L = L ẋ x. (6.3 Don t worry, we ll show you in Section 6.2 where this coes fro. This equation is called the Euler-Lagrange (E-L equation. For the proble at hand, we have L/ ẋ = ẋ and L/ x = kx (see Appendix B for the definition of a partial derivative, so eq. (6.3 gives ẍ = kx, (6.4 which is exactly the result obtained by using F = a. An equation such as eq. (6.4, which is derived fro the Euler-Lagrange equation, is called an equation of otion. 1 If the 1 The ter equation of otion is a little abiguous. It is understood to refer to the second-order differential equation satisfied by x, and not the actual equation for x as a function of t, naely x(t = A cos(ωt + φ in this proble, which is obtained by integrating the equation of otion twice. VI-1

2 VI-2 CHAPTER 6. THE LAGRANGIAN METHOD θ l+x proble involves ore than one coordinate, as ost probles do, we just have to apply eq. (6.3 to each coordinate. We will obtain as any equations as there are coordinates. Each equation ay very well involve any of the coordinates (see the exaple below, where both equations involve both x and θ. At this point, you ay be thinking, That was a nice little trick, but we just got lucky in the spring proble. The procedure won t work in a ore general situation. Well, let s see. How about if we consider the ore general proble of a particle oving in an arbitrary potential V (x (we ll stick to one diension for now. The Lagrangian is then and the Euler-Lagrange equation, eq. (6.3, gives L = 1 2 ẋ2 V (x, (6.5 ẍ = dv dx. (6.6 But dv/dx is the force on the particle. So we see that eqs. (6.1 and (6.3 together say exactly the sae thing that F = a says, when using a Cartesian coordinate in one diension (but this result is in fact quite general, as we ll see in Section 6.4. Note that shifting the potential by a given constant has no effect on the equation of otion, because eq. (6.3 involves only the derivative of V. This is equivalent to saying that only differences in energy are relevant, and not the actual values, as we well know. In a three-diensional setup written in ters of Cartesian coordinates, the potential takes the for V (x, y, z, so the Lagrangian is L = 1 2 (ẋ2 + ẏ 2 + ż 2 V (x, y, z. (6.7 It then iediately follows that the three Euler-Lagrange equations (obtained by applying eq. (6.3 to x, y, and z ay be cobined into the vector stateent, ẍ = V. (6.8 But V = F, so we again arrive at Newton s second law, F = a, now in three diensions. Let s now do one ore exaple to convince you that there s really soething nontrivial going on here. Figure 6.1 Exaple (Spring pendulu: Consider a pendulu ade of a spring with a ass on the end (see Fig The spring is arranged to lie in a straight line (which we can arrange by, say, wrapping the spring around a rigid assless rod. The equilibriu length of the spring is l. Let the spring have length l + x(t, and let its angle with the vertical be θ(t. Assuing that the otion takes place in a vertical plane, find the equations of otion for x and θ. Solution: have The kinetic energy ay be broken up into the radial and tangential parts, so we T = 1 ( 2 ẋ 2 + (l + x 2 θ2. (6.9 The potential energy coes fro both gravity and the spring, so we have V (x, θ = g(l + x cos θ kx2. (6.10 The Lagrangian is therefore L T V = 1 ( 2 ẋ 2 + (l + x 2 θ2 + g(l + x cos θ 1 2 kx2. (6.11

3 6.1. THE EULER-LAGRANGE EQUATIONS VI-3 There are two variables here, x and θ. As entioned above, the nice thing about the Lagrangian ethod is that we can just use eq. (6.3 twice, once with x and once with θ. So the two Euler-Lagrange equations are and d dt d dt ( L θ ( L ẋ = L θ = L x = = ẍ = (l + x θ 2 + g cos θ kx, (6.12 d ( (l + x 2 θ = g(l + x sin θ dt = (l + x 2 θ + 2(l + xẋ θ = g(l + x sin θ. = (l + x θ + 2ẋ θ = g sin θ. (6.13 Eq. (6.12 is siply the radial F = a equation, coplete with the centripetal acceleration, (l + x θ 2. And the first line of eq. (6.13 is the stateent that the torque equals the rate of change of the angular oentu (this is one of the subjects of Chapter 8. Alternatively, if you want to work in a rotating reference frae, then eq. (6.12 is the radial F = a equation, coplete with the centrifugal force, (l + x θ 2. And the third line of eq. (6.13 is the tangential F = a equation, coplete with the Coriolis force, 2ẋ θ. But never ind about this now. We ll deal with rotating fraes in Chapter Reark: After writing down the E-L equations, it is always best to double-check the by trying to identify the as F = a and/or τ = dl/dt equations (once we learn about that. Soeties, however, this identification isn t obvious. And for the ties when everything is clear (that is, when you look at the E-L equations and say, Oh, of course!, it is usually clear only after you ve derived the equations. In general, the safest ethod for solving a proble is to use the Lagrangian ethod and then double-check things with F = a and/or τ = dl/dt if you can. At this point it sees to be personal preference, and all acadeic, whether you use the Lagrangian ethod or the F = a ethod. The two ethods produce the sae equations. However, in probles involving ore than one variable, it usually turns out to be uch easier to write down T and V, as opposed to writing down all the forces. This is because T and V are nice and siple scalars. The forces, on the other hand, are vectors, and it is easy to get confused if they point in various directions. The Lagrangian ethod has the advantage that once you ve written down L T V, you don t have to think anyore. All you have to do is blindly take soe derivatives. 3 When juping fro high in a tree, Just write down del L by del z. Take del L by z dot, Then t-dot what you ve got, And equate the results (but quickly! But ease of coputation aside, is there any fundaental difference between the two ethods? Is there any deep reasoning behind eq. (6.3? Indeed, there is... 2 Throughout this chapter, I ll occasionally point out torques, angular oenta, centrifugal forces, and other such things when they pop up in equations of otion, even though we haven t covered the yet. I figure it can t hurt to bring your attention to the. But rest assured, a failiarity with these topics is by no eans necessary for an understanding of what we ll be doing in this chapter, so just ignore the references if you want. One of the great things about the Lagrangian ethod is that even if you ve never heard of the ters torque, centrifugal, Coriolis, or even F = a itself, you can still get the correct equations by siply writing down the kinetic and potential energies, and then taking a few derivatives. 3 Well, you eventually have to solve the resulting equations of otion, but you have to do that with the F = a ethod, too.

4 VI-4 CHAPTER 6. THE LAGRANGIAN METHOD 6.2 The principle of stationary action Consider the quantity, t2 S t 1 L(x, ẋ, t dt. (6.14 -g/2 y 1 t S is called the action. It is a quantity with the diensions of (Energy (Tie. S depends on L, and L in turn depends on the function x(t via eq. ( Given any function x(t, we can produce the quantity S. We ll just deal with one coordinate, x, for now. Integrals like the one in eq. (6.14 are called functionals, and S is soeties denoted by S[x(t]. It depends on the entire function x(t, and not on just one input nuber, as a regular function f(t does. S can be thought of as a function of an infinite nuber of values, naely all the x(t for t ranging fro t 1 to t 2. If you don t like infinities, you can iagine breaking up the tie interval into, say, a illion pieces, and then replacing the integral by a discrete su. Let s now pose the following question: Consider a function x(t, for t 1 t t 2, which has its endpoints fixed (that is, x(t 1 = x 1 and x(t 2 = x 2, where x 1 and x 2 are given, but is otherwise arbitrary. What function x(t yields a stationary value of S? A stationary value is a local iniu, axiu, or saddle point. 5 For exaple, consider a ball dropped fro rest, and consider the function y(t for 0 t 1. Assue that we soehow know that y(0 = 0 and y(1 = g/2. 6 A nuber of possibilities for y(t are shown in Fig. 6.2, and each of these can (in theory be plugged into eqs. (6.1 and (6.14 to generate S. Which one yields a stationary value of S? The following theore gives us the answer. Figure 6.2 Theore 6.1 If the function x 0 (t yields a stationary value (that is, a local iniu, axiu, or saddle point of S, then ( d L = L. (6.15 dt ẋ 0 x 0 It is understood that we are considering the class of functions whose endpoints are fixed. That is, x(t 1 = x 1 and x(t 2 = x 2. Proof: We will use the fact that if a certain function x 0 (t yields a stationary value of S, then any other function very close to x 0 (t (with the sae endpoint values yields essentially the sae S, up to first order in any deviations. This is actually the definition of a stationary value. The analogy with regular functions is that if f(b is a stationary value of f, then f(b + ɛ differs fro f(b only at second order in the sall quantity ɛ. This is true because f (b = 0, so there is no first-order ter in the Taylor series expansion around b. Assue that the function x 0 (t yields a stationary value of S, and consider the function x a (t x 0 (t + aβ(t, (6.16 where a is a nuber, and where β(t satisfies β(t 1 = β(t 2 = 0 (to keep the endpoints of the function fixed, but is otherwise arbitrary. When producing the action S[x a (t] in (6.14, the t is integrated out, so S is just a nuber. It depends on a, in addition to t 1 and 4 In soe situations, the kinetic and potential energies in L T V ay explicitly depend on tie, so we have included the t in eq. ( A saddle point is a point where there are no first-order changes in S, and where soe of the second-order changes are positive and soe are negative (like the iddle of a saddle, of course. 6 This follows fro y = gt 2 /2, but pretend that we don t know this forula.

5 6.2. THE PRINCIPLE OF STATIONARY ACTION VI-5 t 2. Our requireent is that there be no change in S at first order in a. How does S depend on a? Using the chain rule, we have a S[x a(t] = a t2 L dt = = t2 t 1 t2 t 1 L a dt ( L x a x a a + L ẋ a ẋ a a dt. (6.17 In other words, a influences S through its effect on x, and also through its effect on ẋ. Fro eq. (6.16, we have x a a = β, and ẋ a a = β, (6.18 so eq. (6.17 becoes 7 t2 ( L a S[x a(t] = β + L β dt. (6.19 t 1 x a ẋ a Now coes the one sneaky part of the proof. We will integrate the second ter by parts (you will see this trick any ties in your physics career. Using eq. (6.19 becoes ( L L d β dt = β ẋ a ẋ a dt t2 ( L a S[x a(t] = d t 1 x a dt L ẋ a L ẋ a β dt, (6.20 β dt + L t 2 β ẋ a. (6.21 t 1 But β(t 1 = β(t 2 = 0, so the last ter (the boundary ter vanishes. We now use the fact that ( / as[x a (t] ust be zero for any function β(t, because we are assuing that x 0 (t yields a stationary value. The only way this can be true is if the quantity in parentheses above (evaluated at a = 0 is identically equal to zero, that is, ( d L = L. (6.22 dt ẋ 0 x 0 The E-L equation, eq. (6.3, therefore doesn t just coe out of the blue. It is a consequence of requiring that the action be at a stationary value. We ay therefore replace F = a by the following principle. The Principle of Stationary Action: The path of a particle is the one that yields a stationary value of the action. This principle (also known as Hailton s principle is equivalent to F = a because the above theore shows that if (and only if, as you can show by working backwards we have a stationary value of S, then the E-L equations hold. And the E-L equations are equivalent to F = a (as we showed for Cartesian coordinates in Section 6.1, and as we ll prove for any coordinate syste in Section 6.4. Therefore, stationary action is equivalent to F = a. 7 Note that nowhere do we assue that x a and ẋ a are independent variables. The partial derivatives in eq. (6.18 are very uch related, in that one is the derivative of the other. The use of the chain rule in eq. (6.17 is still perfectly valid.

6 VI-6 CHAPTER 6. THE LAGRANGIAN METHOD If we have a ultidiensional setup where the Lagrangian is a function of the variables x 1 (t, x 2 (t,..., then the above principle of stationary action is still all we need. With ore than one variable, we can now vary the path by varying each coordinate (or cobinations thereof. The variation of each coordinate produces an E-L equation which, as we saw in the Cartesian case, is equivalent to an F = a equation. Given a classical echanics proble, we can solve it with F = a, or we can solve it with the E-L equations, which are a consequence of the principle of stationary action (often called the principle of least action or inial action, but see the fourth reark below. Either ethod will get the job done. But as entioned at the end of Section 6.1, it is often easier to use the latter, because it avoids the use of force which can get confusing if you have forces pointing in all sorts of coplicated directions. It just stood there and did nothing, of course, A harless and still wooden horse. But the inial action Was just a distraction; The plan involved no use of force. Let s now return to the exaple of a ball dropped fro rest, entioned above. The Lagrangian is L = T V = ẏ 2 /2 gy, so eq. (6.22 gives ÿ = g, which is siply the F = a equation (divided through by, as expected. The solution is y(t = gt 2 /2 + v 0 t + y 0, as we well know. But the initial conditions tell us that v 0 = y 0 = 0, so our solution is y(t = gt 2 /2. You are encouraged to verify explicitly that this y(t yields an action that is stationary with respect to variations of the for, say, y(t = gt 2 /2 + ɛt(t 1, which also satisfies the endpoint conditions (this is the task of Exercise There is, of course, an infinite nuber of other ways to vary y(t, but this specific result should help convince you of the general result of Theore 6.1. Note that the stationarity iplied by the Euler-Lagrange equation, eq. (6.22, is a local stateent. It gives inforation only about nearby paths. It says nothing about the global nature of how the action depends on all possible paths. If we find that a solution to eq. (6.22 happens to produce a local iniu (as opposed to a axiu or a saddle, there is no reason to conclude that it is a global iniu, although in any cases it turns out to be (see Exercise 6.32, for the case of a thrown ball. Rearks: 1. Theore 6.1 is based on the assuption that the ending tie, t 2, of the otion is given. But how do we know this final tie? Well, we don t. In the exaple of a ball thrown upward, the total tie to rise and fall back to your hand can be anything, depending on the ball s initial speed. This initial speed will show up as an integration constant when solving the E-L equations. The otion ust end soetie, and the principle of stationary action says that for whatever tie this happens to be, the physical path has a stationary action. 2. Theore 6.1 shows that we can explain the E-L equations by the principle of stationary action. This, however, siply shifts the burden of proof. We are now left with the task of justifying why we should want the action to have a stationary value. The good news is that there is a very solid reason for this. The bad news is that the reason involves quantu echanics, so we won t be able to discuss it properly here. Suffice it to say that a particle actually takes all possible paths in going fro one place to another, and each path is associated with the coplex nuber e is/ h (where h = Js is Planck s constant. These coplex nubers have absolute value 1 and are called phases. It turns out that the phases fro all possible paths ust be added up to give the aplitude of going fro one point to another. The absolute value of the aplitude ust then be squared to obtain the probability. 8 8 This is one of those rearks that is copletely useless, because it is incoprehensible to those who

7 6.2. THE PRINCIPLE OF STATIONARY ACTION VI-7 The basic point, then, is that at a non-stationary value of S, the phases fro different paths differ (greatly, because h is so sall copared with the typical size of the action for a acroscopic particle fro one another, which effectively leads to the addition of any rando vectors in the coplex plane. These end up canceling each other, yielding a su of essentially zero. There is therefore no contribution to the overall aplitude fro nonstationary values of S. Hence, we do not observe the paths associated with these S s. At a stationary value of S, however, all the phases take on essentially the sae value, thereby adding constructively instead of destructively. There is therefore a nonzero probability for the particle to take a path that yields a stationary value of S. So this is the path we observe. 3. But again, the preceding reark siply shifts the burden of proof one step further. We ust now justify why these phases e is/ h should exist, and why the Lagrangian that appears in S should equal T V. But here s where we re going to stop. 4. The principle of stationary action is soeties referred to as the principle of least action, but this is isleading. True, it is often the case that the stationary value turns out to be a iniu value, but it need not be, as we can see in the following exaple. Consider a haronic oscillator which has a Lagrangian equal to L = 1 2 ẋ2 1 2 kx2. (6.23 Let x 0 (t be a function that yields a stationary value of the action. Then we know that x 0 (t satisfies the E-L equation, ẍ 0 = kx 0. Consider a slight variation on this path, x 0(t+ξ(t, where ξ(t satisfies ξ(t 1 = ξ(t 2 = 0. With this new function, the action becoes S ξ = t2 t 1 ( ( ẋ x 0 ξ + ξ2 k (x x 0ξ + ξ 2 dt. ( The two cross-ters add up to zero, because after integrating the x 0 ξ ter by parts, their su is ẋ 0ξ t t2 2 (ẍ 0 + kx 0ξ dt. (6.25 t 1 t 1 The first ter is zero, due to the boundary conditions on ξ(t. The second ter is zero, due to the E-L equation. We ve basically just reproduced the proof of Theore 6.1 for the special case of the haronic oscillator here. The ters in eq. (6.24 involving only x 0 give the stationary value of the action (call it S 0. To deterine whether S 0 is a iniu, axiu, or saddle point, we ust look at the difference, t2 S S ξ S 0 = 1 ( 2 ξ 2 kξ 2 dt. (6.26 t 1 It is always possible to find a function ξ that akes S positive. Siply choose ξ to be sall, but ake it wiggle very fast, so that ξ is large. Therefore, it is never the case that S 0 is a axiu. Note that this reasoning works for any potential, not just a haronic oscillator, as long as it is a function of position only (that is, it contains no derivatives, as we always assue. You ight be tepted to use the sae line of reasoning to say that it is also always possible to find a function ξ that akes S negative, by aking ξ large and ξ sall. If this were true, then we could put everything together and conclude that all stationary points are saddle points, for a haronic oscillator. However, it is not always possible to ake ξ large enough and ξ sall enough so that S is negative, due to the boundary conditions ξ(t 1 = ξ(t 2 = 0. If ξ changes fro zero to a large value and then back to zero, then ξ ay also have to be large, if the tie interval is short enough. Proble 6.6 deals quantitatively with this issue. haven t seen the topic before, and trivial to those who have. My apologies. But this and the following rearks are by no eans necessary for an understanding of the aterial in this chapter. If you re interested in reading ore about these quantu echanical issues, you should take a look at Richard Feynan s book (Feynan, Feynan was, after all, the one who thought of this idea.

8 VI-8 CHAPTER 6. THE LAGRANGIAN METHOD For now, let s just recognize that in soe cases S 0 is a iniu, in soe cases it is a saddle point, and it is never a axiu. Least action is therefore a isnoer. 5. It is soeties said that nature has a purpose, in that it seeks to take the path that produces the iniu action. In view of the second reark above, this is incorrect. In fact, nature does exactly the opposite. It takes every path, treating the all on equal footing. We end up seeing only the path with a stationary action, due to the way the quantu echanical phases add. It would be a harsh requireent, indeed, to deand that nature ake a global decision (that is, to copare paths that are separated by large distances, and to choose the one with the sallest action. Instead, we see that everything takes place on a local scale. Nearby phases siply add, and everything works out autoatically. When an archer shoots an arrow through the air, the ai is ade possible by all the other arrows taking all the other nearby paths, each with essentially the sae action. Likewise, when you walk down the street with a certain destination in ind, you re not alone... When walking, I know that y ai Is caused by the ghosts with y nae. And although I can t see Where they walk next to e, I know they re all there, just the sae. 6. Consider a function, f(x, of one variable (for ease of terinology. Let f(b be a local iniu of f. There are two basic properties of this iniu. The first is that f(b is saller than all nearby values. The second is that the slope of f is zero at b. Fro the above rearks, we see that (as far as the action S is concerned the first property is copletely irrelevant, and the second one is the whole point. In other words, saddle points (and axia, although we showed above that these never exist for S are just as good as inia, as far as the constructive addition of the e is/ h phases is concerned. 7. Given that classical echanics is an approxiate theory, while quantu echanics is the (ore correct one, it is quite silly to justify the principle of stationary action by deonstrating its equivalence with F = a, as we did above. We should be doing it the other way around. However, because our intuition is based on F = a, it s easier to start with F = a as the given fact, rather than calling upon the latent quantu-echanical intuition hidden deep within all of us. Maybe soeday... At any rate, in ore advanced theories dealing with fundaental issues concerning the tiny building blocks of atter (where actions are of the sae order of agnitude as h, the approxiate F = a theory is invalid, and you have to use the Lagrangian ethod. 8. When dealing with a syste in which a non-conservative force such as friction is present, the Lagrangian ethod loses uch of its appeal. The reason for this is that non-conservative forces don t have a potential energy associated with the, so there isn t a specific V (x that you can write down in the Lagrangian. Although friction forces can in fact be incorporated in the Lagrangian ethod, you have to include the in the E-L equations essentially by hand. We won t deal with non-conservative forces in this chapter. 6.3 Forces of constraint A nice thing about the Lagrangian ethod is that we are free to ipose any given constraints at the beginning of the proble, thereby iediately reducing the nuber of variables. This is always done (perhaps without thinking whenever a particle is constrained to ove on a wire or surface, etc. Often we are concerned not with the exact nature of the forces doing the constraining, but only with the resulting otion, given that the constraints hold. By iposing the constraints at the outset, we can find the otion, but we can t say anything about the constraining forces. If we want to deterine the constraining forces, we ust take a different approach. The ain idea of the strategy, as we will show below, is that we ust not ipose the constraints

9 6.3. FORCES OF CONSTRAINT VI-9 too soon. This leaves us with a larger nuber of variables to deal with, so the calculations are ore cubersoe. But the benefit is that we are able to find the constraining forces. Consider the setup of a particle sliding off a fixed frictionless heisphere of radius R (see Fig Let s say that we are concerned only with finding the equation of otion for θ, and not the constraining force. Then we can write everything in ters of θ, because we know that the radial distance r is constrained to be R. The kinetic energy is R 2 θ 2 /2, and the potential energy (relative to the botto of the heisphere is gr cos θ, so the Lagrangian is L = 1 2 R2 θ2 gr cos θ, (6.27 R θ Figure 6.3 and the equation of otion, via eq. (6.3, is θ = (g/r sin θ, (6.28 which is equivalent to the tangential F = a stateent. Now let s say that we want to find the constraining noral force that the heisphere applies to the particle. To do this, let s solve the proble in a different way and write things in ters of both r and θ. Also (and here s the critical step, let s be really picky and say that r isn t exactly constrained to be R, because in the real world the particle actually sinks into the heisphere a little bit. This ay see a bit silly, but it s really the whole point. The particle pushes and sinks inward a tiny distance until the heisphere gets squashed enough to push back with the appropriate force to keep the particle fro sinking in any ore (just consider the heisphere to be ade of lots of little springs with very large spring constants. The particle is therefore subject to a (very steep potential arising fro the heisphere s force. The constraining potential, V (r, looks soething like the plot in Fig The true Lagrangian for the syste is thus L = 1 2 (ṙ2 + r 2 θ2 gr cos θ V (r. (6.29 V(r R r (The ṙ 2 ter in the kinetic energy will turn out to be insignificant. The equations of otion obtained fro varying θ and r are therefore Figure 6.4 r 2 θ + 2rṙ θ = gr sin θ, r = r θ 2 g cos θ V (r. (6.30 Having written down the equations of otion, we will now apply the constraint condition that r = R. This condition iplies ṙ = r = 0. (Of course, r isn t really equal to R, but any differences are inconsequential fro this point onward. The first of eqs. (6.30 then reproduces eq. (6.28, while the second yields dv dr = g cos θ R θ 2. (6.31 r=r But F N dv/dr is the constraint force applied in the r direction, which is precisely the force we are looking for. The noral force of constraint is therefore F N (θ, θ = g cos θ R θ 2. (6.32 This is equivalent to the radial F = a equation, g cos θ F N = R θ 2 (which is certainly a quicker way to find the noral force in the present proble. Note that this result is valid only if F N (θ, θ > 0. If the noral force becoes zero, then this eans that the particle has left the sphere, in which case r no longer equals R.

10 VI-10 CHAPTER 6. THE LAGRANGIAN METHOD Rearks: 1. What if we instead had (unwisely chosen Cartesian coordinates, x and y, instead of polar coordinates, r and θ? Since the distance fro the particle to the surface of the heisphere is η x 2 + y 2 R, we obtain a true Lagrangian equal to The equations of otion are (using the chain rule L = 1 2 (ẋ2 + ẏ 2 gy V (η. (6.33 ẍ = dv η dv η, and ÿ = g dη x dη y. (6.34 We now apply the constraint condition η = 0. Since dv/dη equals the constraint force F, you can show that the equations we end up with (naely, the two E-L equations and the constraint equation are ẍ = F x R, ÿ = g + F y R, and x2 + y 2 R = 0. (6.35 These three equations are sufficient to deterine the three unknowns ẍ, ÿ, and F as functions of the quantities x, ẋ, y, and ẏ. See Exercise 6.37, which should convince you that polar coordinates are the way to go. In general, the strategy is to take two tie derivatives of the constraint equation and then eliinate the second derivatives of the coordinates by using the E-L equations (this process was trivial in the polar-coordinate case. 2. You can see fro eq. (6.35 that the E-L equations end up taking the for, d dt ( L q i = L q i + F η q i, (6.36 for each coordinate q i. The quantity η is what appears in the constraint equation, η = 0. In our heisphere proble, we had η = r R in polar coordinates, and η = x 2 + y 2 R in Cartesian coordinates. The E-L equations, cobined with the η = 0 condition, give us exactly the nuber of equations (N + 1 of the, where N is the nuber of coordinates needed to deterine all of the N + 1 unknowns (all the q i, and F, in ters of the q i and q i. Writing down the equations in eq. (6.36 is basically the ethod of Lagrange ultipliers, where the Lagrange ultiplier turns out to be the force. But if you re not failiar with this ethod, no need to worry; you can derive everything fro scratch using the above technique involving the steep potential. If you do happen to be failiar with it, then there ight in fact be a need to worry about how you apply it, as the following reark explains. 3. When trying to deterine the forces of constraint, you can just start with eq. (6.36, without bothering to write down V (η. But you ust be careful to ake sure that η does indeed represent the distance the particle is fro where it should be. In polar coordinates, if soeone gives you the constraint condition as 7(r R = 0, and if you use the left-hand side of this as the η in eq. (6.36, then you will get the wrong constraint force; it will be too sall by a factor of 7. Likewise, in Cartesian coordinates, writing the constraint as y R 2 x 2 = 0 will give you the wrong force. The best way to avoid this proble is, of course, to pick one of your variables as the distance the particle is fro where it should be (up to an additive constant, as in the case of r R = Change of coordinates When L is written in ters of Cartesian coordinates x, y, z, we showed in Section 6.1 that the Euler-Lagrange equations are equivalent to Newton s F = a equations; see eq. (6.8. But what about the case where we use polar, spherical, or soe other coordinates? The equivalence of the E-L equations and F = a isn t so obvious. As far as trusting the E-L

11 6.4. CHANGE OF COORDINATES VI-11 equations for such coordinates goes, you can achieve peace of ind in two ways. You can accept the principle of stationary action as soething so beautiful and profound that it siply has to work for any choice of coordinates. Or, you can take the ore undane road and show through a change of coordinates that if the E-L equations hold for one set of coordinates (and we know that they do hold for at least one set, naely Cartesian coordinates, then they also hold for any other coordinates (of a certain for, described below. In this section, we will deonstrate the validity of the E-L equations through the explicit change of coordinates. 9 Consider the set of coordinates, x i : (x 1, x 2,..., x N. (6.37 For exaple, if N = 6, then x 1, x 2, x 3 could be the Cartesian x, y, z coordinates of one particle, and x 4, x 5, x 6 could be the r, θ, φ polar coordinates of a second particle, and so on. Assue that the E-L equations hold for these variables, that is, d dt ( L ẋ i = L x i (1 i N. (6.38 Consider a new set of variables that are functions of the x i and t, q i = q i (x 1, x 2,..., x N ; t. (6.39 We will restrict ourselves to the case where the q i do not depend on the ẋ i. (This is quite reasonable. If the coordinates depended on the velocities, then we wouldn t be able to label points in space with definite coordinates. We d have to worry about how the particles were behaving when they were at the points. These would be strange coordinates indeed. We can, in theory, invert eq. (6.39 and express the x i as functions of the q i and t, x i = x i (q 1, q 2,..., q N ; t. (6.40 Clai 6.2 If eq. (6.38 is true for the x i coordinates, and if the x i and q i are related by eq. (6.40, then eq. (6.38 is also true for the q i coordinates. That is, ( d L = L (1 N. (6.41 dt q q Proof: We have L q = N i=1 L ẋ i ẋ i q. (6.42 (Note that if the x i depended on the q i, then we would have to include the additional ter, ( L/ xi ( x i / q. But we have excluded such dependence. Let s rewrite the ẋ i / q ter. Fro eq. (6.40, we have ẋ i = N =1 x i q q + x i t. (6.43 Therefore, ẋ i q = x i q. ( This calculation is straightforward but a bit essy, so you ay want to skip this section and just settle for the beautiful and profound reasoning.

12 VI-12 CHAPTER 6. THE LAGRANGIAN METHOD Substituting this into eq. (6.42 and taking the tie derivative of both sides gives ( d L = dt q N i=1 d dt ( L ẋ i xi q + N i=1 L ẋ i ( d xi. (6.45 dt q In the second ter here, it is legal to switch the order of the total derivative, d/dt, and the partial derivative, / q. Reark: In case you have your doubts, let s prove that this switching is legal. ( d xi N ( ( xi = q k + xi dt q q k q t q k=1 ( N x i = q k + x i q q k t k=1 = ẋi q. (6.46 In the first ter on the right-hand side of eq. (6.45, we can use the given inforation in eq. (6.38 and rewrite the (d/dt( L/ ẋ i ter. We then obtain as we wanted to show. ( d L dt q = N i=1 L x i x i q + N i=1 L ẋ i ẋ i q = L q, (6.47 We have therefore deonstrated that if the Euler-Lagrange equations are true for one set of coordinates, x i (and they are true for Cartesian coordinates, then they are also true for any other set of coordinates, q i, satisfying eq. (6.39. If you re inclined to look at the principle of stationary action with distrust, thinking that it ight be a coordinate-dependent stateent, this proof should put you at ease. The Euler-Lagrange equations are valid in any coordinates. Note that the above proof did not in any way use the precise for of the Lagrangian. If L were equal to T +V, or 8T +πv 2 /T, or any other arbitrary function, our result would still be true: If eq. (6.38 is true for one set of coordinates, then it is also true for any other set of coordinates q i satisfying eq. (6.39. The point is that the only L for which the hypothesis is true at all (that is, for which eq. (6.38 holds is L T V (or any constant ultiple of this. Reark: On one hand, it is quite aazing how little we assued in proving the above clai. Any new coordinates of the very general for in eq. (6.39 satisfy the E-L equations, as long as the original coordinates do. If the E-L equations had, say, a factor of 5 on the right-hand side of eq. (6.38, then they would not hold in arbitrary coordinates. To see this, just follow the proof through with the factor of 5. On the other hand, the clai is quite believable, if you ake an analogy with a function instead of a functional. Consider the function f(z = z 2. This has a iniu at z = 0, consistent with the fact that df/dz = 0 at z = 0. But let s now write f in ters of the variable y defined by, say, z = y 4. Then f(y = y 8, and f has a iniu at y = 0, consistent with the fact that df/dy = 0 at y = 0. So f = 0 holds in both coordinates at the corresponding points y = z = 0. This is the (siplified analog of the E-L equations holding in both coordinates. In both cases, the derivative equation describes where the stationary value occurs.

13 6.5. CONSERVATION LAWS VI-13 This change-of-variables result ay be stated in a ore geoetrical (and friendly way. If you plot a function and then stretch the horizontal axis in an arbitrary anner (which is what happens when you change coordinates, then a stationary value (that is, one where the slope is zero will still be a stationary value after the stretching. 10 A picture (or even just the thought of one is worth a dozen equations, apparently. As an exaple of an equation that does not hold for all coordinates, consider the preceding exaple, but with f = 1 instead of f = 0. In ters of z, f = 1 when z = 1/2. And in ters of y, f = 1 when y = (1/8 1/7. But the points z = 1/2 and y = (1/8 1/7 are not the sae point. In other words, f = 1 is not a coordinate-independent stateent. Most equations are coordinate dependent. The special thing about f = 0 is that a stationary point is a stationary point no atter how you look at it. 6.5 Conservation Laws Cyclic coordinates Consider the case where the Lagrangian does not depend on a certain coordinate q k. Then ( d L = L L = 0 = = C, (6.48 dt q k q k q k where C is a constant, that is, independent of tie. In this case, we say that q k is a cyclic coordinate, and that L/ q k is a conserved quantity (eaning that it doesn t change with tie. If Cartesian coordinates are used, then L/ ẋ k is siply the oentu, ẋ k, because ẋ k appears only in the the kinetic energy s ẋ 2 k /2 ter (we exclude cases where the potential depends on ẋ k. We therefore call L/ q k the generalized oentu corresponding to the coordinate q k. And in cases where L/ q k does not change with tie, we call it a conserved oentu. Note that a generalized oentu need not have the units of linear oentu, as the angular-oentu exaples below show. Exaple 1: Linear oentu Consider a ball thrown through the air. In the full three diensions, the Lagrangian is L = 1 2 (ẋ2 + ẏ 2 + ż 2 gz. (6.49 There is no x or y dependence here, so both L/ ẋ = ẋ and L/ ẏ = ẏ are constant, as we well know. The fancy way of saying this is that conservation of p x ẋ arises fro spatial translation invariance in the x direction. The fact that the Lagrangian doesn t depend on x eans that it doesn t atter if you throw the ball in one spot, or in another spot a ile down the road. The setup is independent of the x value. This independence leads to conservation of p x. Exaple 2: Angular and linear oentu in cylindrical coordinates Consider a potential that depends only on the distance to the z axis. In cylindrical coordinates, the Lagrangian is L = 1 2 (ṙ2 + r 2 θ2 + ż 2 V (r. ( There is, however, one exception. A stationary point in one coordinate syste ight be located at a kink in another coordinate syste, so that f is not defined there. For exaple, if we had instead defined y by z = y 1/4, then f(y = y 1/2, which has an undefined slope at y = 0. Basically, we ve stretched (or shrunk the horizontal axis by a factor of infinity at the origin, and this is a process that can change a zero slope into an undefined one. But let s not worry about this.

14 VI-14 CHAPTER 6. THE LAGRANGIAN METHOD There is no z dependence here, so L/ ż = ż is constant. Also, there is no θ dependence, so L/ θ = r 2 θ is constant. Since r θ is the speed in the tangential direction around the z axis, we see that our conserved quantity, r(r θ, is the angular oentu (discussed in Chapters 7-9 around the z axis. In the sae anner as in the preceding exaple, conservation of angular oentu around the z axis arises fro rotation invariance around the z axis. Exaple 3: Angular oentu in spherical coordinates In spherical coordinates, consider a potential that depends only on r and θ. Our convention for spherical coordinates is that θ is the angle down fro the north pole, and φ is the angle around the equator. The Lagrangian is L = 1 2 (ṙ2 + r 2 θ2 + r 2 sin 2 θ φ 2 V (r, θ. (6.51 There is no φ dependence here, so L/ φ = r 2 sin 2 θ φ is constant. Since r sin θ is the distance fro the z axis, and since r sin θ φ is the speed in the tangential direction around the z axis, we see that our conserved quantity, (r sin θ(r sin θ φ, is the angular oentu around the z axis Energy conservation We will now derive another conservation law, naely conservation of energy. The conservation of oentu or angular oentu above arose when the Lagrangian was independent of x, y, z, θ, or φ. Conservation of energy arises when the Lagrangian is independent of tie. This conservation law is different fro those in the above oenta exaples, because t is not a coordinate which the stationary-action principle can be applied to. You can iagine varying the coordinates x, θ, etc., which are functions of t. But it akes no sense to vary t. Therefore, we re going to have to prove this conservation law in a different way. Consider the quantity, ( N L E = q i L. (6.52 q i i=1 E turns out (usually to be the energy. We ll show this below. The otivation for this expression for E coes fro the theory of Legendre transfors, but we won t get into that here. We ll just accept the definition in eq. (6.52 and prove a very useful fact about it. Clai 6.3 If L has no explicit tie dependence (that is, if L/ t = 0, then E is conserved (that is, de/dt = 0, assuing that the otion obeys the E-L equations (which it does. Note that there is one partial derivative and one total derivative in this stateent. Proof: L is a function of the q i, the q i, and possibly t. Making copious use of the chain rule, we have ( de = d N L q i dl dt dt q i dt = i=1 N (( d L dt q i i=1 q i + L q i q i ( N i=1 ( L q i + L q i + L. (6.53 q i q i t There are five ters here. The second cancels with the fourth. And the first (after using the E-L equation, eq. (6.3, to rewrite it cancels with the third. We therefore arrive at the siple result, de dt = L t. (6.54

15 6.6. NOETHER S THEOREM VI-15 In the event that L/ t = 0 (that is, there are no t s sitting on the paper when you write down L, which is usually the case in the situations we consider (because we generally won t deal with potentials that depend on tie, we have de/dt = 0. Not too any things are constant with respect to tie, and the quantity E has units of energy, so it s a good bet that it s the energy. Let s show this in Cartesian coordinates (however, see the reark below. The Lagrangian is L = 1 2 (ẋ2 + ẏ 2 + ż 2 V (x, y, z, (6.55 so eq. (6.52 gives E = 1 2 (ẋ2 + ẏ 2 + ż 2 + V (x, y, z, (6.56 which is the total energy. The effect of the operations in eq. (6.52 in ost cases is just to switch the sign in front of the potential. Of course, taking the kinetic energy T and subtracting the potential energy V to obtain L, and then using eq. (6.52 to produce E = T + V, sees like a rather convoluted way of arriving at T + V. But the point of all this is that we used the E-L equations to prove that E is conserved. Although we know very well fro the F = a ethods in Chapter 5 that the su T + V is conserved, it s not fair to assue that it is conserved in our new Lagrangian foralis. We have to show that this follows fro the E-L equations. As with the translation and rotation invariance we observed in the exaples in Section 6.5.1, we see that energy conservation arises fro tie translation invariance. If the Lagrangian has no explicit t dependence, then the setup looks the sae today as it did yesterday. This fact leads to conservation of energy. Reark: The quantity E in eq. (6.52 gives the energy of the syste only if the entire syste is represented by the Lagrangian. That is, the Lagrangian ust represent a closed syste with no external forces. If the syste is not closed, then Clai 6.3 (or ore generally, eq. (6.54 is still perfectly valid for the E defined in eq. (6.52, but this E ay siply not be the energy of the syste. Proble 6.8 is a good exaple of such a situation. Another exaple is the following. Iagine a long rod in the horizontal x-y plane. The rod points in the x direction, and a bead is free to slide frictionlessly along it. At t = 0, an external achine is arranged to accelerate the rod in the negative y direction (that is, transverse to itself with acceleration g. So ẏ = gt. There is no internal potential energy in this syste, so the Lagrangian is just the kinetic energy, L = ẋ 2 /2 + (gt 2 /2. Eq. (6.52 therefore gives E = ẋ 2 /2 (gt 2 /2, which isn t the energy. But eq. (6.54 is still true, because de dt = L t ẋẍ g 2 t = g 2 t ẍ = 0, (6.57 which is correct. However, this setup is exactly the sae as projectile otion in the x-y plane, where y is now the vertical axis, provided that we eliinate the rod and consider gravity instead of the achine to be causing the acceleration in the y direction. But if we are thinking in ters of gravity, then the noral thing to do is to say that the particle oves under the influence of the potential V (y = gy. The Lagrangian for this closed syste (bead plus earth is L = (ẋ 2 + ẏ 2 /2 gy, and so eq. (6.52 gives E = (ẋ 2 + ẏ 2 /2 + gy, which is indeed the energy of the particle. But having said all this, ost of the systes we ll deal with are closed, so you can usually ignore this reark and assue that the E in eq. (6.52 gives the energy. 6.6 Noether s Theore We now present one of the ost beautiful and useful theores in physics. It deals with two fundaental concepts, naely syetry and conserved quantities. The theore (due

16 VI-16 CHAPTER 6. THE LAGRANGIAN METHOD to Ey Noether ay be stated as follows. Theore 6.4 (Noether s Theore For each syetry of the Lagrangian, there is a conserved quantity. By syetry, we ean that if the coordinates are changed by soe sall quantities, then the Lagrangian has no first-order change in these quantities. By conserved quantity, we ean a quantity that does not change with tie. The result in Section for cyclic coordinates is a special case of this theore. Proof: Let the Lagrangian be invariant, to first order in the sall nuber ɛ, under the change of coordinates, q i q i + ɛk i (q. (6.58 Each K i (q ay be a function of all the q i, which we collectively denote by the shorthand, q. Reark: As an exaple of what these K i s ight look like, consider the Lagrangian, L = (/2(5ẋ 2 2ẋẏ + 2ẏ 2 + C(2x y. We ve just pulled this out of a hat, although it happens to be the type of L that arises in Atwood s achine probles; see Proble 6.9 and Exercise This L is invariant under the transforation x x + ɛ and y y + 2ɛ, because the derivative ters are unaffected, and the difference 2x y is unchanged. (It s actually invariant to all orders in ɛ, and not just first order. But this isn t necessary for the theore to hold. Therefore, K x = 1 and K y = 2. In the probles we ll be doing, the K i s can generally be deterined by siply looking at the potential ter. Of course, soeone else ight coe along with K x = 3 and K y = 6, which is also a syetry. And indeed, any factor can be taken out of ɛ and put into the K i s without changing the quantity ɛk i(q in eq. (6.58. Any such odification will just bring an overall constant factor (and hence not change the property of being conserved into the conserved quantity in eq. (6.61 below. It is therefore irrelevant. The fact that the Lagrangian does not change at first order in ɛ eans that 0 = dl = ( L q i dɛ q i i ɛ + L q i q i ɛ = ( L K i + L K i. (6.59 q i i q i Using the E-L equation, eq. (6.3, we can rewrite this as 0 = ( ( d L K i + L K i dt q i i q i ( = d L K i. (6.60 dt q i i Therefore, the quantity P (q, q i L q i K i (q (6.61 does not change with tie. It is given the generic nae of conserved oentu. But it need not have the units of linear oentu. As Noether ost keenly observed (And for which uch acclai is deserved,

17 6.6. NOETHER S THEOREM VI-17 It s easy to see That for each syetry, A quantity ust be conserved. Exaple 1: Consider the Lagrangian in the above reark, L = (/2(5ẋ 2 2ẋẏ + 2ẏ 2 + C(2x y. We saw that K x = 1 and K y = 2. The conserved oentu is therefore P (x, y, ẋ, ẏ = L ẋ K x + L ẏ K y = (5ẋ ẏ(1 + ( ẋ + 2ẏ(2 = (3ẋ + 3ẏ. (6.62 The overall factor of 3 isn t iportant. Exaple 2: Consider a thrown ball. We have L = (/2(ẋ 2 + ẏ 2 + ż 2 gz. This is invariant under translations in x, that is, x x + ɛ; and also under translations in y, that is, y y + ɛ. (Both x and y are cyclic coordinates. We need invariance only to first order in ɛ for Noether s theore to hold, but this L is invariant to all orders. We therefore have two syetries in our Lagrangian. The first has K x = 1, K y = 0, and K z = 0. The second has K x = 0, K y = 1, and K z = 0. Of course, the nonzero K i s here can be chosen to be any constants, but we ay as well pick the to be 1. The two conserved oenta are P 1 (x, y, z, ẋ, ẏ, ż = L ẋ K x + L ẏ K y + L ż K z P 2(x, y, z, ẋ, ẏ, ż = L ẋ Kx + L ẏ = ẋ, L Ky + Kz = ẏ. (6.63 ż These are siply the x and y coponents of the linear oentu, as we saw in Exaple 1 in Section Note that any cobination of these oenta, say 3P 1 + 8P 2, is also conserved. (In other words, x x + 3ɛ, y y + 8ɛ, z z is a syetry of the Lagrangian. But the above P 1 and P 2 are the siplest conserved oenta to choose as a basis for the infinite nuber of conserved oenta (which is how any you have, if there are two or ore independent continuous syetries. Exaple 3: Consider a ass on a spring, with relaxed length zero, in the x-y plane. The Lagrangian, L = (/2(ẋ 2 +ẏ 2 (k/2(x 2 +y 2, is invariant under the change of coordinates, x x + ɛy and y y ɛx, to first order in ɛ (as you can check. So we have K x = y and K y = x. The conserved oentu is therefore P (x, y, ẋ, ẏ = L L Kx + Ky = (ẋy ẏx. (6.64 ẋ ẏ This is the (negative of the z coponent of the angular oentu. The angular oentu is conserved here because the potential, V (x, y x 2 + y 2 = r 2, depends only on the distance fro the origin. We ll discuss such potentials in Chapter 7. In contrast with the first two exaples above, the x x + ɛy, y y ɛx transforation isn t so obvious here. How did we get this? Well, unfortunately there doesn t see to be any fail-proof ethod of deterining the K i s in general, so soeties you just have to guess around, as was the case here. But in any probles, the K i s are siple constants which are easy to see. Rearks: 1. As we saw above, in soe cases the K i s are functions of the coordinates, and in soe cases they are not.

18 VI-18 CHAPTER 6. THE LAGRANGIAN METHOD 2. The cyclic-coordinate result in eq. (6.48 is a special case of Noether s theore, for the following reason. If L doesn t depend on a certain coordinate q k, then q k q k + ɛ is certainly a syetry. Hence K k = 1 (with all the other K i s equal to zero, and eq. (6.61 reduces to eq. ( We use the word syetry to describe the situation where the transforation in eq. (6.58 produces no first-order change in the Lagrangian. This is an appropriate choice of word, because the Lagrangian describes the syste, and if the syste essentially doesn t change when the coordinates are changed, then we say that the syste is syetric. For exaple, if we have a setup that doesn t depend on θ, then we say that the setup is syetric under rotations. Rotate the syste however you want, and it looks the sae. The two ost coon applications of Noether s theore are the conservation of angular oentu, which arises fro syetry under rotations; and conservation of linear oentu, which arises fro syetry under translations. 4. In siple systes, as in Exaple 2 above, it is clear why the resulting P is conserved. But in ore coplicated systes, as in Exaple 1 above, the resulting P ight not have an obvious interpretation. But at least you know that it is conserved, and this will invariably help in understanding a setup. 5. Although conserved quantities are extreely useful in studying a physical situation, it should be stressed that there is no ore inforation contained in the than there is in the E-L equations. Conserved quantities are siply the result of integrating the E-L equations. For exaple, if you write down the E-L equations for Exaple 1 above, and then add the x equation (which is 5ẍ ÿ = 2C to twice the y equation (which is ẍ + 2ÿ = C, then you arrive at 3(ẍ + ÿ = 0. In other words, 3(ẋ + ẏ is constant, as we found fro Noether s theore. Of course, you ight have to do soe guesswork to find the proper cobination of the E-L equations that gives a zero on the right-hand side. But you d have to do soe guesswork anyway, to find the syetry for Noether s theore. At any rate, a conserved quantity is useful because it is an integrated for of the E-L equations. It puts you one step closer to solving the proble, copared with where you would be if you started with the second-order E-L equations. 6. Does every syste have a conserved oentu? Certainly not. The one-diensional proble of a falling ball ( z = g doesn t have one. And if you write down an arbitrary potential in 3-D, odds are that there won t be one. In a sense, things have to contrive nicely for there to be a conserved oentu. In soe probles, you can just look at the physical syste and see what the syetry is, but in others (for exaple, in the Atwood s-achine probles for this chapter, the syetry is not at all obvious. 7. By conserved quantity, we ean a quantity that depends on (at ost the coordinates and their first derivatives (that is, not on their second derivatives. If we don t ake this restriction, then it is trivial to construct quantities that are independent of tie. For exaple, in Exaple 1 above, the x E-L equation (which is 5ẍ ÿ = 2C tells us that 5ẍ ÿ has its tie derivative equal to zero. Note that an equivalent way of excluding these trivial cases is to say that the value of a conserved quantity depends on the initial conditions (that is, the velocities and positions. The quantity 5ẍ ÿ doesn t satisfy this criterion, because its value is always constrained to be 2C. 6.7 Sall oscillations In any physical systes, a particle undergoes sall oscillations around an equilibriu point. In Section 5.2, we showed that the frequency of these sall oscillations is ω = V (x 0, (6.65

19 6.7. SMALL OSCILLATIONS VI-19 where V (x is the potential energy, and x 0 is the equilibriu point. However, this result holds only for one-diensional otion (we ll see below why this is true. In ore coplicated systes, such as the one described below, it is necessary to use another procedure to obtain the frequency ω. This procedure is a fail-proof one, applicable in all situations. It is, however, a bit ore involved than siply writing down eq. (6.65. So in one-diensional probles, eq. (6.65 is still what you want to use. We ll deonstrate our fail-proof ethod through the following proble. r Proble: A ass is free to slide on a frictionless table and is connected by a string, which passes through a hole in the table, to a ass M which hangs below (see Fig Assue that M oves in a vertical line only, and assue that the string always reains taut. (a Find the equations of otion for the variables r and θ shown in the figure. (b Under what condition does undergo circular otion? (c What is the frequency of sall oscillations (in the variable r about this circular otion? Solution: (a Let the string have length l (this length won t atter. Then the Lagrangian (we ll call it L here, to save L for the angular oentu, which arises below is l-r θ M Figure 6.5 L = 1 2 Mṙ (ṙ2 + r 2 θ2 + Mg(l r. (6.66 For the purposes of the potential energy, we ve taken the table to be at height zero, but any other value could be chosen. The E-L equations of otion obtained fro varying θ and r are d dt (r2 θ = 0, (M + r = r θ 2 Mg. (6.67 The first equation says that angular oentu is conserved. The second equation says that the Mg gravitational force accounts for the acceleration of the two asses along the direction of the string, plus the centripetal acceleration of. (b The first of eqs. (6.67 says that r 2 θ = L, where L is soe constant (the angular oentu which depends on the initial conditions. Plugging θ = L/r 2 into the second of eqs. (6.67 gives (M + r = L2 Mg. (6.68 r3 Circular otion occurs when ṙ = r = 0. Therefore, the radius of the circular orbit is given by r 3 0 = Since L = r 2 θ, eq. (6.69 is equivalent to L2 Mg. (6.69 r 0 θ2 = Mg, (6.70 which can also be obtained by letting r = 0 in the second of eqs. (6.67. In other words, the gravitational force on M exactly accounts for the centripetal acceleration of if the otion is circular. Given r 0, eq. (6.70 deterines what θ ust be in order to have circular otion, and vice versa. (c To find the frequency of sall oscillations about the circular otion, we need to look at what happens to r if we perturb it slightly fro its equilibriu value of r 0. Our fail-proof procedure is the following.

20 VI-20 CHAPTER 6. THE LAGRANGIAN METHOD Let r(t r 0 + δ(t, where δ(t is very sall (ore precisely, δ(t r 0, and expand eq. (6.68 to first order in δ(t. Using 1 r 1 3 (r 0 + δ 1 3 r r2 0 δ = 1 r0 3(1 + 3δ/r 0 1 ( 1 3δ, (6.71 r0 3 r 0 we obtain ( (M + δ L2 r δ Mg. (6.72 r 0 The ters not involving δ on the right-hand side cancel, by the definition of r 0 given in eq. (6.69. This cancellation always occurs in such a proble at this stage, due to the definition of the equilibriu point. We are therefore left with ( 3L δ 2 + (M + r0 4 δ 0. (6.73 This is a good old siple-haronic-oscillator equation in the variable δ. Therefore, the frequency of sall oscillations about a circle of radius r 0 is 3L ω 2 3M g =, (6.74 (M + r0 4 M + r 0 where we have used eq. (6.69 to eliinate L in the second expression. To su up, the above frequency is the frequency of sall oscillations in the variable r. In other words, if you have nearly circular otion, and if you plot r as a function of tie (and ignore what θ is doing, then you will get a nice sinusoidal graph whose frequency is given by eq. (6.74. Note that this frequency need not have anything to do with the other relevant frequency in this proble, naely the frequency of the circular otion, which is M/ g/r 0, fro eq. (6.70. Rearks: Let s look at soe liits. For a given r 0, if M, then ω 3Mg/r 0 0. This akes sense, because everything is oving very slowly. This frequency equals 3 ties the frequency of circular otion, naely Mg/r 0, which isn t at all obvious. For a given r 0, if M, then ω 3g/r 0, which isn t so obvious, either. The frequency of sall oscillations is equal to the frequency of circular otion if M = 2, which, once again, isn t obvious. This condition is independent of r 0. The above procedure for finding the frequency of sall oscillations can be sued up in three steps: (1 Find the equations of otion, (2 Find the equilibriu point, and (3 Let x(t x 0 + δ(t, where x 0 is the equilibriu point of the relevant variable, and expand one of the equations of otion (or a cobination of the to first order in δ, to obtain a siple-haronic-oscillator equation for δ. If the equilibriu point happens to be at x = 0 (which is often the case, then everything is greatly siplified. There is no need to introduce δ, and the expansion in the third step above siply entails ignoring powers of x that are higher than first order. Reark: If you just use the potential energy for the above proble (which is Mgr, up to a constant in eq. (6.65, then you will obtain a frequency of zero, which is incorrect. You can use eq. (6.65 to find the frequency, if you instead use the effective potential for this proble, naely L 2 /(2r 2 + Mgr, and if you use the total ass, M +, as the ass in eq. (6.65, as you can check. The reason why this works will becoe clear in Chapter 7 when we introduce the effective potential. In any probles, however, it isn t obvious what the appropriate odified potential is that should be used, or what ass goes in eq. (6.65. So it s generally uch safer to take a deep breath and go through an expansion siilar to the one in part (c of the exaple above.