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  U
r
t; 2 5  5 - r
t L Dt
r'
t L, t, Constants m 0
U!
r
t
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.

x
t_  :
r
t Cos

t y
t_  :
r
t 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   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,

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
x
A,  xf, y
A,  yf , A, 1 , ,   , Module
Asol
A . soln, max
 . soln , ParametricPlot
x
Asol, , 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.

Plotting this for a  1, b  0 gives: 8 Principle of Least Action.nb

t t In[14]:= Clear
y; Manipulate ParametricPlot a ArcCosh    b, t , a ArcCosh    b, t , a a 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

ymin

ymax

2.0 Out[14]=

1.8

1.6

1.4

1.2

1.0 0.5 0.5 1.0

Problem: Bead on a Ring

From 8.033 Quiz #2 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 accelera- tion 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

1 2 2 2 Clear
x, y, z, t ; L
m x'
t  y'
t  z'
t   m g z
t  TraditionalForm 2 1 m x t 2  y t 2  z t 2  g m z t 2           Converting to polar coordinates, and using the constraints that    t and r  R, using the conversion

x  R sin  cos  t y  R sin  sin  t z  R Rcos     gives 10 Principle of Least Action.nb

Clear
r, , ; Defer
L Lpolar
Expand
FullSimplify
L . x - Function
t, R Cos
t Sin

t, y - Function t, R Sin t Sin  t , z - Function t,




R  R Cos

t   . '
t -  . 
t -   TraditionalForm 5  5
0 Defer
 L Dt
"", t   L EL Expand
FullSimplify
5  5 5  -  
t Lpolar t '
t Lpolar  .
t  TraditionalForm ''
t ''
t . Solve
EL 0, ''
t1  TraditionalForm

1 1  2 1 L  g m R cos   g m R  m R 2 2 cos 2    m R 2  m R 2 2   4   2 4 L   L   2 2 2  0   g m R sin   m R  sin  cos   m R  t  t          R 2 sin  t cos  t  g sin  t  t             R Finding the minimum value of  for the bead to be in equilibrium gives

''
t . Solve
EL 0, ''
t1 0  TraditionalForm Refine
Reduce
''
t . Solve
EL 0, ''
t1 0, Cos

t, Sin

t 8 0 && R Cos

t 8 0 && g 0 && R 8 0 . 
t -   TraditionalForm R 2 sin  t cos  t  g sin  t           0 R g cos     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 8.12

 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. Principle of Least Action.nb 11

 Solution by torque

(From the problem set solutions)

The torque about the pivot point is

p   I p

 ¨ k : g  m sin    Fcent cos    Ip (1)     The centrifugal effective force is

2 Fcent  m  sin      For small angles, sin 
, cos 
1. Then equation (1) becomes     ¨ g  m   m 2   2
m 2  ¨ g     2 
0  12 Principle of Least Action.nb

g     2 

g If 2  , the motion is no longer harmonic. 

 Solution by least action

The general Lagrangian for the object in Cartesian coordinates is

1 2 2 2 Clear
x, y, z, t ; L
m x'
t  y'
t  z'
t   m g z
t  TraditionalForm 2 1 m x t 2  y t 2  z t 2  g m z t 2           Converting to polar coordinates, and using the constraints that    t and r  , using the conversion

x   sin  cos  t y   sin  sin  t z   cos     gives
Clear
, , ;
Defer
L Lpolar
Expand
FullSimplify
L . x - Function
t, Cos
 t Sin

t, y - Function t,
Sin  t Sin  t , z - Function t,

Cos  t .






   '
t -  . 
t -   TraditionalForm 5  5 0 Defer
 L Dt
"", t   L
5  5 5 EL Expand
FullSimplify

t Lpolar t '
t Lpolar  . 
t -   TraditionalForm ''
t ''
t . Solve
EL 0, ''
t1  TraditionalForm

1 1  2 1 L  g m  cos   g m   m 2 2 cos 2    m 2  m 2 2   4   2 4 L   L    2  2 2 0   g m sin   m  t  m  sin  cos   t          2  sin  t cos  t  g sin  t  t              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 g  t     2  t      This is the differential equation for a harmonic oscillator, with

g     2  Principle of Least Action.nb 13

g If 2  , the motion is no longer harmonic.