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 f à L t f à K  U t
t1
t1
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

q
 L
 f0
 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 f  x s t. We want to find an extremum of
J Ã
t1
t0

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+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
J +/  Ã
t1
t0

f +x+t; /, x+t; /; t/  t
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
ˆ
 0
 0 since it’s an extremum. Using this fact,
J

Ã

 f x
 
t
 x    x 
 f x
t1
t0
+x+t/   A+t//  A+t/
  

x
 
A

+x+t/   A+t// 
    t
t
x


So
J

Ã
f
t1
x
t0
 f A
t

x t
A+t/ 
Integration by parts on the section term gives
Ã
t0
t1
 f A
f
 t  A+t/ 

x t
x
Since A+t0 /  A+t1 /  0, the first term dies, and we get
J

 Ã A+t/
t1
f
x
t0
 Ã A+t/
t1
t1
t0
t0

 f
 t
t x
 f
 t
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
x

 f
 0
t x
This is what was to be derived.
Least action: F  m a
Suppose we have the Newtonian kinetic energy, K 
Euler-Lagrange equations tell us the following:
1
2
¶
m v2 , and a potential that depends only on position, U  U +r/. Then the
Clear#U, m, r'
1
L
m r '#t'2 U#r#t'';
2
™r#t' L Dt#™r'#t' L, t, Constants ‘ m' m 0
U… #r#t'' m r…… #t' m 0
Rearrangement gives

U
¨
mr
r
F ma
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
q
For Newtonian kinetic energy, K 
1
2
 K
 f0
t q


m x 2 , this is just


mxf0
t

mxmv
x  x0  v t
This is a straight line, as expected.
Least action with gravitational potential
Suppose we have gravitational potential close to the surface of the earth, U  m g y, and Newtonian kinetic energy, K 
1
2

m y2 .
Then the Euler-Lagrange equations become
m g 

t
¨

m y  m g  m y f 0
¨
g  y
y  y0  a y t 
1
2
g t2
This is a parabola, as expected.
Constants of motion: Momenta
We may rearrange the Euler-Lagrange equations to obtain
L
q
If it happens that
L
q
 0, then
 L

t q

is also zero. This means that
 L

 t q
L

q
is a constant (with respect to time). We call
L

q
a
(conserved) momentum of the system.
 Linear Momentum
depends only on position, we can infer that
standard linear momentum, m v.
L

x
1
2
m v2 , is independent of the time derivatives of position, if potential energy


L
L
L
 1
(and, similarly, y and z ) are constant. Then x  x , 2 m x 2 0  m x. This is just
By noting that Newtonian kinetic energy, K 
4
Principle of Least Action.nb
 Angular Momentum
Let us change to polar coordinates.
x#t_' : r#t' Cos#T#t''
y#t_' : r#t' Sin#T#t''
K
1
2
Expand%FullSimplify%
m r +t/2 
1
2
1
2
m ,x '#t'2 y'#t'20)) ss TraditionalForm
m r+t/2  +t/2
Using dot notation, this is
K s. r_ '#t' ‘ OverDot#r' s. r_#t' ‘ r ss TraditionalForm

1 2
m r2
 m r2 
2
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



 m r2 
is constant. This is standard angular momentum, m r2   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 0. What shape should the
wire be in order to have the bead reach ,x f , y f 0 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 0.
 Implementation
Letting  s be the infinitesimal distance element and v be the travel speed,
T fÃ
tf
ti
s 
v
+ x/2  + y/2   y
2g y
s
v
t
1  +x '/2
x' 
x
y
(Assumption: bead starts at rest)
Principle of Least Action.nb
TÃ
1  +x '/2
yf
Now we apply the Euler equation to f 
y
2g y
0

and change t  y, x  x '.
1+x'/2
2g y
f
x
 f
 f0
 y x

f
x
1
f
 
x
 f
 f0 
 y x
0
x'
2gy
1  +x '/2
1
x'
1  +x '/
2gy
 Constant
2
Squaring both sides and making a special choice for the constant gives
+x '/2
2 g y,1  +x '/2 0

y
xÃ

2
x

0
yf
x
y
1

y s+2 A/
4g A
1  y s +2 A/
yÃ
0

y2
2 A y  y2
y
yf
y
2 A y  y2
To solve this, change variables:
y  A+1  cos+//,
 y  A sin+/  
FullSimplify$2 A y y2 s. y ‘ A +1 Cos#T'/(
A2 Sin#T'2
y
y
2A y y
2
A+1  cos+//
A2
sin +/
A sin+/    A+1  cos+//
2
x  Ã A+1  cos+//    A+  sin+//

0
Full solution: The brachistochrone is described by
x  A+  sin+//
y  A+1  cos+//
There’s no analytic solution, but we can compute them.
5
6
Principle of Least Action.nb
Clear#x, y, A, T, soln, yf, xf, xmax, Tmax, Asol, f'; Manipulate#
Module#y Function#A, T, A +1 Cos#T'/', x Function#A, T, A +T Sin#T'/',
Module#soln FindRoot#x#A, T' m xf, y#A, T' m yf, A, 1, T, S', Module#
Asol A s. soln, Tmax T s. soln, ParametricPlot#x#Asol, T', y#Asol, T',
T, 0, Tmax, PlotRange ‘ 0, xmax, ymax, 0, PlotStyle ‘ Black'''',
xmax, 2 S, xmax , 0, 4 S, ymax, 2.5, ymax , 0, 20,
xf, 4, xf , 0, 10, yf, 2, yf , 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 0. 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
U f à  g y s
l
0
s 
+ x/2  + y/2   y
U Ã
yf
1  +x '/2
x' 
1  +x '/2  y
g 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
x


f
 y  x'
f
0
x
f
 x'
Since
f
x
 0,
f
x'
is constant, say a 
1 f
 g x'

y x'

f0
 g y x'
1  +x '/2
. Then
1+x'/2
x' 
x
a
y

y2  a2
Using the fact that
Ä
y
2
f cosh1
2
y a
y
a
integration of x ' gives
x+y/   a cosh1
where b is a constant of integration.
Plotting this for a  1, b  0 gives:
y
a
b
 b,
x
y
7
8
Principle of Least Action.nb
In[14]:=
t
t
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, ymin , 5, 5, ymax, 2, ymax , 5, 5)
a
b
ymin
ymax
2.0
Out[14]=
1.8
1.6
1.4
1.2
1.0
0.5
Problem: Bead on a Ring
From 8.033 Quiz #2
0.5
1.0
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    s 2?
 Solution
The general Lagrangian for the object in Cartesian coordinates is
Clear#x, y, z, t'; L
1
2
1
2
m ,x'#t'2 y'#t'2 z'#t'20 m g z#t' ss TraditionalForm
m ,x +t/2  y +t/2  z +t/2 0  g m z+t/
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  R cos+/
gives
10
Principle of Least Action.nb
Clear#r, T, I';
Defer#L' m +Lpolar Expand#FullSimplify#L s. x ‘ Function#t, R Cos#Z t' Sin#T#t''',
y ‘ Function#t, R Sin#Z t' Sin#T#t''', z ‘ Function#t,
ƒ
R R Cos#T#t'''''/ s. T '#t' ‘ T s. T#t' ‘ T ss TraditionalForm
0 m Defer#™T L Dt#"", t' ™ƒT L' m +EL Expand#FullSimplify#
™T#t' Lpolar ™t ™T'#t' Lpolar''/ s. T#t' ‘ T ss TraditionalForm
T ''#t' m +T ''#t' s. Solve#EL m 0, T ''#t''317/ ss TraditionalForm
L f g m R cos+/  g m R 
1
m R2 2 cos+2 / 
4
L
0f

 +t/ f

1 2
1
 m R2  m R2 2
2
4
 L
2 2
2 
 f g m R sin+/  m R  sin+/ cos+/  m R  +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 ''#t' s. Solve#EL m 0, T''#t''317/ m 0 ss TraditionalForm
Refine#Reduce#+T ''#t' s. Solve#EL m 0, T ''#t''317/ m 0, Cos#T#t''',
Sin#T#t'' œ 0 && R Cos#T#t'' œ 0 && g ! 0 && R Z œ 0' s. T#t' ‘ T ss TraditionalForm
R 2 sin++t// cos++t//  g sin++t//
R
f0
g
cos+/ f
R 2
In order for this to have a solution, we must have

If    s2, then cos+/  0, so   .
g
R
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 f  Ip

k:
The centrifugal effective force is
¨
g [ m sin+/  [ Fcent cos+/ f  I p
Fcent  m +[ sin+// 2
For small angles, sin+/ ! , cos+/ ! 1. Then equation (1) becomes
¨
g [ m   m [2  2 ! m [2 
¨
g
 
 2  ! 0
[
(1)
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
Clear#x, y, z, t'; L
1
2
1
2
m ,x'#t'2 y'#t'2 z'#t'20 m g z#t' ss TraditionalForm
m ,x +t/2  y +t/2  z +t/2 0  g m z+t/
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#[, T, I';
Defer#L' m +Lpolar Expand#FullSimplify#L s. x ‘ Function#t, [ Cos#: t' Sin#T#t''',
y ‘ Function#t, [ Sin#: t' Sin#T#t''', z ‘ Function#t, [ [ Cos#T#t'''''/ s.
ƒ
T '#t' ‘ T s. T#t' ‘ T ss TraditionalForm
0 m Defer#™T L Dt#"", t' ™ƒT L' m
+EL Expand#FullSimplify#™T#t' Lpolar ™t ™T'#t' Lpolar''/ s.
T#t' ‘ T ss TraditionalForm
T ''#t' m +T''#t' s. Solve#EL m 0, T''#t''317/ ss TraditionalForm
L f g m [ cos+/  g m [ 
1
m 2 [2 cos+2 / 
4
L
0f

 +t/ f

1 2
1
 m [2  m 2 [2
2
4
 L
2 
2 2
 f g m [ sin+/  m [  +t/  m  [ sin+/ cos+/
t 
2 [ sin++t// cos++t//  g sin++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
 +t/ f 
g
 2 +t/
[
This is the differential equation for a harmonic oscillator, with

g
[
 2
Principle of Least Action.nb
g
If 2  [ , the motion is no longer harmonic.
13