The Principle of Least Action

Transcription

1 The Principle of Least Action Jason Gross, December 7, 2010 Introduction Recall that we defined the Lagrangian to be the kinetic energy less potential energy, LK U, at a point. The action is then defined to be the integral of the Lagrangian along the path, S t0 t 1Lt t0 t 1K U t It is (remarkably!) true that, in any physical system, the path an object actually takes minimizes the action. It can be shown that the extrema of action occur at This is called the Euler equation, or the Euler-Lagrange Equation. L q L t q 0 Derivation Courtesy of Scott Hughes s Lecture notes for (Most of this is copied almost verbatim from that.) Suppose we have a function fx,x ;t of a variable x and its derivative x xt. We want to find an extremum of J t0 t 1fxt,x t;tt Our goal is to compute xt such that J is at an extremum. We consider the limits of integration to be fixed. That is, xt 1 will be the same for any x we care about, as will xt 2. Imagine we have some xt for which J is at an extremum, and imagine that we have a function which parametrizes how far our current path is from our choice of x: xt;xtat The function A is totally arbitrary, except that we require it to vanish at the endpoints: At 0 At 1 0. The parameter allows us to control how the variation At enters into our path xt;. The correct path xt is unknown; our goal is to figure out how to construct it, or to figure out how f behaves when we are on it. Our basic idea is to ask how does the integral J behave when we are in the vicinity of the extremum. We know that ordinary functions are flat --- have zero first derivative --- when we are at an extremum. So let us put J t0 t 1fxt;,x t;;tt

2 2 Principle of Least Action.nb We know that 0 corresponds to the extremum by definition of. However, this doesn t teach us anything useful, sine we don t know the path xt that corresponds to the extremum. But we also know We know that J 0 since it s an extremum. Using this fact, 0 J t 1 x t0 x x x t So x x xtatat A xtat t t J t0 t 1 x A At x t t Integration by parts on the section term gives Since At 0 At 1 0, the first term dies, and we get t 1 t A 1 t 1At t0 x tat t x t 0 t 0 t x t J t 1At t 0 x t x t This must be zero. Since At is arbitrary except at the endpoints, we must have that the integrand is zero at all points: This is what was to be derived. x t x 0 Least action: F ma Suppose we have the Newtonian kinetic energy, K 1 2 mv2, and a potential that depends only on position, U Ur. Then the Euler-Lagrange equations tell us the following: ClearU,m,r L 1 2 mr't2 Urt; rt LDt r't L,t,Constants m 0 U rtmr t 0 Rearrangement gives U r mr F ma

3 Principle of Least Action.nb 3 Least action with no potential Suppose we have no potential, U 0. Then LK, so the Euler-Lagrange equations become For Newtonian kinetic energy, K 1 2 mx 2, this is just This is a straight line, as expected. K q K t q 0 t mx 0 mx mv xx 0 vt Least action with gravitational potential Suppose we have gravitational potential close to the surface of the earth, U mgy, and Newtonian kinetic energy, K 1 my 2. 2 Then the Euler-Lagrange equations become mg t my mg mÿ0 gÿ yy 0 a y t 1 2 gt2 This is a parabola, as expected. Constants of motion: Momenta We may rearrange the Euler-Lagrange equations to obtain L q t If it happens that L 0, then L q t q is also zero. This means that L q is a constant (with respect to time). We call L q a (conserved) momentum of the system. L q Linear Momentum By noting that Newtonian kinetic energy, K 1 2 mv2, is independent of the time derivatives of position, if potential energy depends only on position, we can infer that L L (and, similarly, x y standard linear momentum, mv. L and z L ) are constant. Then x x 1 mx 2 mx. This is just 2

4 4 Principle of Least Action.nb Angular Momentum Let us change to polar coordinates. xt_ : rtcost yt_ : rtsint K ExpandFullSimplify 1 2 mx't2 y't 2 TraditionalForm 1 2 mr t mrt2 t 2 Using dot notation, this is K.r_'t OverDotr. r_t r TraditionalForm mr 2 mr 2 Note that does not appear in this expression. If potential energy is not a function of (is only a function of r), then L mr2 is constant. This is standard angular momentum, mr 2 rmrrmv. Classic Problem: Brachistochrone ( shortest time ) Problem A bead starts at x0, y0, and slides down a wire without friction, reaching a lower point x f, y f. What shape should the wire be in order to have the bead reach x f, y f in as little time as possible. Solution Idea Use the Euler equation to minimize the time it takes to get from x i, y i to x f, y f. Implementation Letting s be the infinitesimal distance element and v be the travel speed, t f s T ti v t s x 2 y 2 y 1x ' 2 x' x y v 2gy (Assumption: bead starts at rest)

5 Principle of Least Action.nb 5 T 0 y f 1x' 2 2gy y Now we apply the Euler equation to f 1x' 2 2gy and change ty, x x'. y x x y x 0 1 2gy x 0 1 2gy x 0 x' 1x' 2 x' 1x ' 2 Constant Squaring both sides and making a special choice for the constant gives To solve this, change variables: x' 2 2gy1x' 2 x y x 0 y f x 1 4gA 2 y2a 1y2A y y 0 y f y2 2Ayy 2 y 2Ayy 2 y ya1cos, ya sin FullSimplify2Ayy 2. y A 1Cos A 2 Sin 2 y 2Ayy 2 Full solution: The brachistochrone is described by There s no analytic solution, but we can compute them. A1cos y A sina1cos A 2 sin 2 x A1cosAsin 0 xasin ya1cos

6 6 Principle of Least Action.nb Clearx, y, A,,soln, yf,xf, xmax, max, Asol, f; Manipulate Moduley FunctionA,, A 1 Cos, x FunctionA,, A Sin, Modulesoln FindRootxA, xf, ya, yf, A, 1,,, Module Asol A.soln, max.soln,parametricplotxasol,, yasol,,,0, max,plotrange 0, xmax, ymax,0, PlotStyle Black, xmax,2,x max, 0, 4, ymax, 2.5, y max,0, 20, xf,4,x f, 0, 10, yf, 2, y f,0, 5 x max y max x f y f Classic Problem: Catenary Problem Suppose we have a rope of length l and linear mass density. Suppose we fix its ends at points x 0, y 0 and x f, y f. What shape does the rope make, hanging under the influence of gravity? Solution Idea Calculate the potential energy of the rope as a function of the curve, yx, and minimize this quantity using the Euler-Lagrange equations. Implementation Suppose we have curve parameterized by t, xt, yt. The potential energy associated with this curve is

7 Principle of Least Action.nb 7 U 0 lgys s x 2 y 2 y 1x ' 2 x' x y U y0 y fgy 1x' 2 y Note that if we choose to factor s the other way (for y'), we get a mess. Now we apply the Euler-Lagrange equation to f gy 1x' 2 and change ty, x x '. x y x 0 x' 0 x' gyx' 1x' 2 Since 0, is constant, say a 1 x x' g x' yx' 1x' 2. Then x' x y a y 2 a 2 Using the fact that y y 2 a 2 cosh 1 y a b, integration of x' gives xya cosh 1 y a b where b is a constant of integration. Plotting this for a1, b0 gives:

8 8 Principle of Least Action.nb In[14]:= Cleary; ManipulateParametricPlotaArcCosh t a b,t, aarccosht a b,t, t, ymin, ymax,plotstyle Black, a,1, 5,5, b,0, 5,5, ymin, 0, y min, 5,5, ymax, 2, y max, 5, 5 a b y min y max Out[14]= Problem: Bead on a Ring From Quiz #2

9 Principle of Least Action.nb 9 Problem A bead of mass m slides without friction on a circular hoop of radius R. The angle is defined so that when the bead is at the bottom of the hoop, 0. The hoop is spun about its vertical axis with angular velocity. Gravity acts downward with acceleration g. Find an equation describing how evolves with time. Find the minimum value of for the bead to be in equilibrium at some value of other than zero. ( equilibrium means that and are both zero.) How large must be in order to make 2? Solution The general Lagrangian for the object in Cartesian coordinates is Clearx, y,z, t; L 1 2 m x't2 y't 2 z't 2 mgzt TraditionalForm 1 2 mx t 2 y t 2 z t 2 gmzt Converting to polar coordinates, and using the constraints that t and rr, using the conversion gives xr sin cost yr sin sint zrrcos

10 10 Principle of Least Action.nb Clearr,, ; DeferL Lpolar ExpandFullSimplifyL. x Functiont, RCostSint, y Functiont,RSintSint,zFunctiont, RRCost. 't. t TraditionalForm 0 Defer LDt"",t L EL ExpandFullSimplify t Lpolar t 't Lpolar. t TraditionalForm ''t ''t. SolveEL 0, ''t1 TraditionalForm L gmrcosgmr 1 4 mr2 2 cos mr mr2 2 0 L t L gmrsinmr 2 2 sin cosmr 2 t t R2 sint costg sint R Finding the minimum value of for the bead to be in equilibrium gives ''t.solveel 0, ''t1 0 TraditionalForm RefineReduce''t. SolveEL 0, ''t1 0, Cost, Sint 0&&RCost 0&&g 0&&R 0. t TraditionalForm R 2 sint costg sint 0 R cos g R 2 In order for this to have a solution, we must have g R If 2, then cos0, so. Problem 11.8: K & K 8.12 Problem A pendulum is rigidly fixed to an axle held by two supports so that it can only swing in a plane perpendicular to the axle. The pendulum consists of a mass m attached to a massless rod of length l. The supports are mounted on a platform which rotates with constant angular velocity. Find the pendulum s frequency assuming the amplitude is small.

11 Principle of Least Action.nb 11 Solution by torque (From the problem set solutions) The torque about the pivot point is k : p I p g msinf cent cos I p (1) The centrifugal effective force is F cent m sin 2 For small angles, sin, cos 1. Then equation (1) becomes g mm 2 2 m 2 g 2 0

12 12 Principle of Least Action.nb g 2 If 2 g, the motion is no longer harmonic. Solution by least action The general Lagrangian for the object in Cartesian coordinates is Clearx, y,z, t; L 1 2 m x't2 y't 2 z't 2 mgzt TraditionalForm 1 2 mx t 2 y t 2 z t 2 gmzt Converting to polar coordinates, and using the constraints that t and r, using the conversion gives xsin cost ysin sint z cos Clear,, ; DeferL Lpolar ExpandFullSimplifyL. x Functiont, CostSint, y Functiont, SintSint,zFunctiont, Cost. 't. t TraditionalForm 0 Defer LDt"",t L EL ExpandFullSimplify t Lpolar t 't Lpolar. t TraditionalForm ''t ''t. SolveEL 0, ''t1 TraditionalForm L gm cosgm 1 4 m2 2 cos m m2 2 0 L L t gm sinm 2 tm 2 2 sin cos t 2 sint costg sint Note that this is, after minor changes of variable, the exact same equation that we found in the previous problem. We should( ve) expect(ed) this. Making the first order approximation that 0 (Taylor expanding around 0 to the first order), we get t g 2 t This is the differential equation for a harmonic oscillator, with g 2

13 If 2 g, the motion is no longer harmonic. Principle of Least Action.nb 13