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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, S (^) t 0
t 1 L t (^) t 0
t 1 K 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 L q ^
t
L
q^ ^ ^0 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
J (^) t 0
t 1 f x t , x^ ^ t ; t t
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 t 1 will be the same for any x we care about, as will x t 2 . 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 t 0 A t 1 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 J (^) t 0
t 1 f x t ; , x^ ^ t ; ; t t
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 0 since it’s an extremum. Using this fact,
J
(^) t 0
t 1 f x
x
f x^
x^
t x ^
x t ^ ^ A t ^ ^ A t x^
t
x t A t ^ A t So
J
(^) t 0
t 1 f x
A t
f x^
A
t
t
Integration by parts on the section term gives
t 0
t 1 f x^
A
t ^ t^ ^ A t ^
f x^ ^ t 0
t 1 (^) t 0
t 1 A t
t
f x^ ^ ^ t
Since A t 0 A t 1 0, the first term dies, and we get
J
t 0
t 1 A t
f x
t
f x^ ^
t
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 x
t
f x^ ^
This is what was to be derived.
Least action: F m a
Suppose we have the Newtonian kinetic energy, K 12 m v^2 , and a potential that depends only on position, U U r^ . Then the Euler-Lagrange equations tell us the following:
ClearU, m, r L 1 2
m r 't^2 Urt; rt L Dtr't L, t, Constants m 0 Urt m rt 0
Rearrangement gives
U
r
m r^ ¨ F m a
Angular Momentum
Let us change to polar coordinates. xt_ : rt Cost yt_ : rt Sint K ExpandFullSimplify 12 m x 't^2 y't^2 TraditionalForm 1 2
m r t ^2
m r t ^2 t ^2
Using dot notation, this is K . r_ 't OverDotr . r_t r TraditionalForm 1 2
^2 m r^2 m r
Note that does not appear in this expression. If potential energy is not a function of (is only a function of r ), then L m r^2 is constant. This is standard angular momentum, m r^2 r m r r m 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,
T (^) t i
tf s v ^ t s x ^2 y ^2 y 1 x '^2 x ' ^ x y v 2 g y (Assumption: bead starts at rest)
T 0
y (^) f 1 x '^2 2 g y
y
Now we apply the Euler equation to f 1 2 g yx '^2 and change t y , x^ ^ x '.
f x
y
f x^ ^
f x
f x^ ^
2 g y
x ' 1 x '^2 y
f x^ ^
2 g y
x ' 1 x '^2
Constant
Squaring both sides and making a special choice for the constant gives
x '^2 2 g y 1 x '^2
4 g A
x y
2
y 2 A 1 y 2 A ^
y^2 2 A y y^2
x (^) 0
y (^) f x y ^ y^ ^ 0
y (^) f y 2 A y y^2
y
To solve this, change variables:
y A 1 cos, y A sin FullSimplify2 A y y^2 . y A 1 Cos A^2 Sin^2 y 2 A y y^2
y
A 1 cos A^2 sin^2
A sin A 1 cos
x (^) 0
A 1 cos A sin
Full solution: The brachistochrone is described by
x A sin y A 1 cos
There’s no analytic solution, but we can compute them.
U
0
l g y s
s x ^2 y ^2 y 1 x '^2 x '
x y U (^) y 0
y (^) f g y 1 x '^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 g y 1 x '^2 and change t y , x^ ^ x '.
f x
y
f x '
f x
f x '
g y x ' 1 x '^2
Since ^ fx 0, ^ xf ' is constant, say a (^) ^1 g^ ^ fx ' y x ' 1 x '^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
x y a cosh^1
y a
b
where b is a constant of integration.
Plotting this for a 1, b 0 gives:
In[14]:= Cleary; ManipulateParametricPlot a ArcCosh t a
b, t, a ArcCosh t a
b, t, t, ymin, ymax , PlotStyle Black, a, 1 , 5, 5 , b, 0 , 5, 5 , ymin, 0, ymin , 5, 5 , ymax, 2, ymax , 5, 5
Out[14]=
a b ymin ymax
1.0 0.5 0.5 1.
Problem: Bead on a Ring
From 8.033 Quiz #
Clearr, , ; DeferL Lpolar ExpandFullSimplifyL . x Functiont, R Cos t Sint, y Functiont, R Sin t Sint, z Functiont, R R Cost . 't . t TraditionalForm 0 Defer L Dt"", t L EL ExpandFullSimplify t Lpolar t 't Lpolar . t TraditionalForm ''t ''t . SolveEL 0, ''t 1 TraditionalForm
L g m R cos g m R
m R^2 ^2 cos 2
^2 m R^2
m R^2 ^2
0
L
t
L
^
g m R sin m R^2 ^2 sin cos m R^2 t
t
R ^2 sin t cos t g sin t R Finding the minimum value of for the bead to be in equilibrium gives ''t . SolveEL 0, ''t 1 0 TraditionalForm RefineReduce ''t . SolveEL 0, ''t 1 0, Cost, Sint 0 && R Cost 0 && g 0 && R 0 . t TraditionalForm R ^2 sin t cos t g sin t R ^0 cos
g R ^2 In order for this to have a solution, we must have
g R If 2, then cos 0, so .
Problem 11.8: K & K 8.
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.
Solution by torque
(From the problem set solutions)
The torque about the pivot point is p Ip
k^ ^ : g m sin F cent cos ^ ¨ Ip^ (1) The centrifugal effective force is F cent m sin ^2 For small angles, sin , cos 1. Then equation (1) becomes
g m m ^2 ^2 m ^2
¨
¨ g
^2 0
If ^2 g , the motion is no longer harmonic.
