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, L K U, at a point. The action is then defined to be the integral of the Lagrangian along the path, t1 t1 S L t K U t _t0 _t0 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 L L 0 q t q This is called the Euler equation, or the Euler-Lagrange Equation. Derivation Courtesy of Scott Hughes’s Lecture notes for 8.033. (Most of this is copied almost verbatim from that.) Suppose we have a function f x, x; t of a variable x and its derivative x x t. We want to find an extremum of t1 J f x t , x t ; t t _t0 Our goal is to compute x t such that J is at an extremum. We consider the limits of integration to be fixed. That is, x t1 will be the same for any x we care about, as will x t2 . Imagine we have some x t 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: x t; x t A t The function A is totally arbitrary, except that we require it to vanish at the endpoints: A t0 A t1 0. The parameter allows us to control how the variation A t enters into our path x t; . The “correct” path x t 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 t1 J f x t; , x t; ; t t _t0 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 x t that corresponds to the extremum. But we also know We know that J 0 since it’s an extremum. Using this fact, $ 0 J t1 f x f x t _t0 x x x x t A t A t x A x t A t t t So J t1 f f A A t t _t0 x x t Integration by parts on the section term gives t t1 f A f 1 t1 f t A t A t t t t _ 0 x t x t0 _ 0 t x Since A t0 A t1 0, the first term dies, and we get J t1 f f A t t _t0 x t x This must be zero. Since A t is arbitrary except at the endpoints, we must have that the integrand is zero at all points: f f 0 x t x This is what was to be derived. Least action: F m a 1 2 Suppose we have the Newtonian kinetic energy, K m v , and a potential that depends only on position, U U Rr . Then the 2 Euler-Lagrange equations tell us the following: Clear U, m, r 1 2 L m r' t Urt; 2 5 5 - rt L Dt r' t L, t, Constants m 0 U!rt m r !! t 0 Rearrangement gives U ¨ m r r F m a Principle of Least Action.nb 3 Least action with no potential Suppose we have no potential, U 0. Then L K, so the Euler-Lagrange equations become K K 0 q t q 1 For Newtonian kinetic energy, K m x 2, this is just 2 m x 0 t m x m v x x0 v t This is a straight line, as expected. Least action with gravitational potential 1 Suppose we have gravitational potential close to the surface of the earth, U m g y , and Newtonian kinetic energy, K m y 2. 2 Then the Euler-Lagrange equations become ¨ m g m y m g m y 0 t ¨ g y 1 2 y y0 ay t g t 2 This is a parabola, as expected. Constants of motion: Momenta We may rearrange the Euler-Lagrange equations to obtain L L q t q L L L L If it happens that 0, then is also zero. This means that is a constant (with respect to time). We call a q t q q q (conserved) momentum of the system. Linear Momentum By noting that Newtonian kinetic energy, K 1 m v 2, is independent of the time derivatives of position, if potential energy 2 L L L L 1 2 depends only on position, we can infer that (and, similarly, and ) are constant. Then m x m x . This is just x y z x x 2 standard linear momentum, m v . 4 Principle of Least Action.nb Angular Momentum Let us change to polar coordinates. xt_ : rt Cos t yt_ : rt Sin t 1 2 2 K Expand FullSimplify m x' t y' t TraditionalForm 2 1 1 m r t 2 m r t 2 t 2 2 2 Using dot notation, this is K . r_ ' t - OverDot r . r_ t - r TraditionalForm 2 1 2 m r m r 2 2 2 L 2 Note that does not appear in this expression. If potential energy is not a function of (is only a function of r), then m r is constant. This is standard angular momentum, m r 2 r m r rm v . Classic Problem: Brachistochrone (“shortest time”) Problem A bead starts at x 0, y 0, 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 xi, yi to x f , y f . Implementation Letting s be the infinitesimal distance element and v be the travel speed, tf s T t _ti v x s x 2 y 2 y 1 x ' 2 x' y v 2 g y (Assumption: bead starts at rest) Principle of Least Action.nb 5 2 y f 1 x' T y _0 2 g y 1 x' 2 Now we apply the Euler equation to f and change t y, x x'. 2 g y f f 0 x y x f 0 x f 1 x' x 2 g y 1 x' 2 f 1 x' 0 Constant y x 2 g y 1 x ' 2 Squaring both sides and making a special choice for the constant gives x' 2 1 2 g y 1 x' 2 4 g A x 2 y 2 A y2 y 1 y 2 A 2 A y y2 y f x y f y x y y 0 y 0 _ _ 2 A y y2 To solve this, change variables: y A 1 cos , y A sin 2 FullSimplify 2 A y y . y - A 1 Cos A2 Sin 2 y A 1 cos y A sin A 1 cos 2 A y y2 A2 sin 2 x A 1 cos A sin _0 Full solution: The brachistochrone is described by x A sin y A1 cos There’s no analytic solution, but we can compute them. 6 Principle of Least Action.nb Clear x, y, A, , soln, yf, xf, xmax, max, Asol, f ; Manipulate Module y Function A, , A 1 Cos , x Function A, , A Sin , Module soln FindRoot xA, xf, y A, yf , A, 1 , , , Module Asol A . soln, max . soln , ParametricPlot xAsol, , y Asol, , , 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 xmax ymax xf yf 0.0 1 2 3 4 5 6 0.5 1.0 1.5 2.0 2.5 Classic Problem: Catenary Problem Suppose we have a rope of length l and linear mass density . Suppose we fix its ends at points x0, y0 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, y x , and minimize this quantity using the Euler-Lagrange equations. Implementation Suppose we have curve parameterized by t, x t , y t . The potential energy associated with this curve is Principle of Least Action.nb 7 l U g y s _0 x s x 2 y 2 y 1 x ' 2 x' y y f U g y 1 x' 2 y _y0 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 g y 1 x' 2 and change t y, x x '. f f 0 x y x' f 0 x f g y x ' x' 1 x' 2 f f f y x ' Since 0, is constant, say a 1 . Then x x' g x' 1 x' 2 x a x' y y2 a2 Using the fact that y y cosh 1 b, a ` y2 a2 integration of x' gives y x y a cosh 1 b a where b is a constant of integration.