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,

Principle of Least Action with Derivation_1.gif

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

Principle of Least Action with Derivation_2.gif

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 Principle of Least Action with Derivation_3.gif of a variable x and its derivative Principle of Least Action with Derivation_4.gif.  We want to find an extremum of

Principle of Least Action with Derivation_5.gif

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, Principle of Least Action with Derivation_6.gif will be the same for any x we care about, as will Principle of Least Action with Derivation_7.gif.

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:

Principle of Least Action with Derivation_8.gif

The function A is totally arbitrary, except that we require it to vanish at the endpoints: Principle of Least Action with Derivation_9.gif.  The parameter α allows us to control how the variation A(t) enters into our path x(t;α).

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

Principle of Least Action with Derivation_10.gif

We know that α=0 corresponds to the extremum by definition of α.  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 Principle of Least Action with Derivation_11.gif since it’s an extremum.  Using this fact,

Principle of Least Action with Derivation_12.gif

So

Principle of Least Action with Derivation_13.gif

Integration by parts on the section term gives

Principle of Least Action with Derivation_14.gif

Since Principle of Least Action with Derivation_15.gif, the first term dies, and we get

Principle of Least Action with Derivation_16.gif

This must be zero.  Since A(t) is arbitrary except at the endpoints, we must have that the integrand is zero at all points:

Principle of Least Action with Derivation_17.gif

This is what was to be derived.

Least action: F=m a

Suppose we have the Newtonian kinetic energy, Principle of Least Action with Derivation_18.gif, and a potential that depends only on position, Principle of Least Action with Derivation_19.gif.  Then the Euler-Lagrange equations tell us the following:

Principle of Least Action with Derivation_20.gif

Principle of Least Action with Derivation_21.gif

Rearrangement gives

Principle of Least Action with Derivation_22.gif

Least action with no potential

Suppose we have no potential, U=0.  Then L=K, so the Euler-Lagrange equations become

Principle of Least Action with Derivation_23.gif

For Newtonian kinetic energy, Principle of Least Action with Derivation_24.gif, this is just

Principle of Least Action with Derivation_25.gif

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, Principle of Least Action with Derivation_26.gif.  Then the Euler-Lagrange equations become

Principle of Least Action with Derivation_27.gif

This is a parabola, as expected.

Constants of motion: Momenta

We may rearrange the Euler-Lagrange equations to obtain

Principle of Least Action with Derivation_28.gif

If it happens that Principle of Least Action with Derivation_29.gif, then Principle of Least Action with Derivation_30.gif is also zero.  This means that Principle of Least Action with Derivation_31.gif is a constant (with respect to time).  We call Principle of Least Action with Derivation_32.gif a (conserved) momentum of the system.

Linear Momentum

By noting that Newtonian kinetic energy, Principle of Least Action with Derivation_33.gif, is independent of the time derivatives of position, if potential energy depends only on position, we can infer that Principle of Least Action with Derivation_34.gif (and, similarly, Principle of Least Action with Derivation_35.gif and Principle of Least Action with Derivation_36.gif) are constant.  Then Principle of Least Action with Derivation_37.gif.  This is just standard linear momentum, m v.

Angular Momentum

Let us change to polar coordinates.

Principle of Least Action with Derivation_38.gif

Principle of Least Action with Derivation_39.gif

Using dot notation, this is

Principle of Least Action with Derivation_40.gif

Principle of Least Action with Derivation_41.gif

Note that θ does not appear in this expression.  If potential energy is not a function of θ (is only a function of r), then Principle of Least Action with Derivation_42.gif is constant.  This is standard angular momentum, Principle of Least Action with Derivation_43.gif.

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 Principle of Least Action with Derivation_44.gif.  What shape should the wire be in order to have the bead reach Principle of Least Action with Derivation_45.gif in as little time as possible.

Solution

Idea

Use the Euler equation to minimize the time it takes to get from Principle of Least Action with Derivation_46.gif to Principle of Least Action with Derivation_47.gif.

Implementation

Letting ds be the infinitesimal distance element and v be the travel speed,

Principle of Least Action with Derivation_48.gif

Now we apply the Euler equation to Principle of Least Action with Derivation_49.gif and change ty, Principle of Least Action with Derivation_50.gif.

Principle of Least Action with Derivation_51.gif

Squaring both sides and making a special choice for the constant gives

Principle of Least Action with Derivation_52.gif

To solve this, change variables:

Principle of Least Action with Derivation_53.gif

Principle of Least Action with Derivation_54.gif

Principle of Least Action with Derivation_55.gif

Principle of Least Action with Derivation_56.gif

Full solution: The brachistochrone is described by

Principle of Least Action with Derivation_57.gif

There’s no analytic solution, but we can compute them.

Principle of Least Action with Derivation_58.gif

Principle of Least Action with Derivation_59.gif

Classic Problem: Catenary

Problem

Suppose we have a rope of length l and linear mass density λ.  Suppose we fix its ends at points Principle of Least Action with Derivation_60.gif and Principle of Least Action with Derivation_61.gif.  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 with Derivation_62.gif

Note that if we choose to factor ds the other way (for y'), we get a mess.

Now we apply the Euler-Lagrange equation to Principle of Least Action with Derivation_63.gif and change ty, Principle of Least Action with Derivation_64.gif.

Principle of Least Action with Derivation_65.gif

Since Principle of Least Action with Derivation_66.gif, Principle of Least Action with Derivation_67.gif is constant, say Principle of Least Action with Derivation_68.gif.  Then

Principle of Least Action with Derivation_69.gif

Using the fact that

Principle of Least Action with Derivation_70.gif

integration of x' gives

Principle of Least Action with Derivation_71.gif

where b is a constant of integration.

Plotting this for a=1, b=0 gives:

Principle of Least Action with Derivation_72.gif

Principle of Least Action with Derivation_73.gif

Problem: Bead on a Ring

From 8.033 Quiz #2

Problem

Principle of Least Action with Derivation_74.gif

A bead of mass m slides without friction on a circular hoop of radius R. The angle θ is defined so that when the bead is at the bottom of the hoop, θ=0. The hoop is spun about its vertical axis with angular velocity ω. Gravity acts downward with acceleration g.
Find an equation describing how θ evolves with time.
Find the minimum value of ω for the bead to be in equilibrium at some value of θ other than zero.
(“equilibrium” means that Principle of Least Action with Derivation_75.gif and Principle of Least Action with Derivation_76.gif are both zero.) How large must ω be in order to make θ=π/2?

Solution

The general Lagrangian for the object in Cartesian coordinates is

Principle of Least Action with Derivation_77.gif

Principle of Least Action with Derivation_78.gif

Converting to polar coordinates, and using the constraints that φ=ω t and r=R, using the conversion

Principle of Least Action with Derivation_79.gif

gives

Principle of Least Action with Derivation_80.gif

Principle of Least Action with Derivation_81.gif

Principle of Least Action with Derivation_82.gif

Principle of Least Action with Derivation_83.gif

Finding the minimum value of ω for the bead to be in equilibrium gives

Principle of Least Action with Derivation_84.gif

Principle of Least Action with Derivation_85.gif

Principle of Least Action with Derivation_86.gif

In order for this to have a solution, we must have

Principle of Least Action with Derivation_87.gif

If θ=π/2, then cos(θ)=0, so ω=.

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 Ω. Find the pendulum’s frequency assuming the amplitude is small.

Principle of Least Action with Derivation_88.gif

Solution by torque

(From the problem set solutions)

Principle of Least Action with Derivation_89.gif

The torque about the pivot point is

Principle of Least Action with Derivation_90.gif

Principle of Least Action with Derivation_91.gif

The centrifugal effective force is

Principle of Least Action with Derivation_92.gif

For small angles, sin(θ)=θ, cos(θ)=1.  Then equation (1) becomes

Principle of Least Action with Derivation_93.gif

If Principle of Least Action with Derivation_94.gif, the motion is no longer harmonic.

Solution by least action

The general Lagrangian for the object in Cartesian coordinates is

Principle of Least Action with Derivation_95.gif

Principle of Least Action with Derivation_96.gif

Converting to polar coordinates, and using the constraints that φ=Ω t and r=, using the conversion

Principle of Least Action with Derivation_97.gif

gives

Principle of Least Action with Derivation_98.gif

Principle of Least Action with Derivation_99.gif

Principle of Least Action with Derivation_100.gif

Principle of Least Action with Derivation_101.gif

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 θ0 (Taylor expanding around θ=0 to the first order), we get

Principle of Least Action with Derivation_102.gif

This is the differential equation for a harmonic oscillator, with

Principle of Least Action with Derivation_103.gif

If Principle of Least Action with Derivation_104.gif, the motion is no longer harmonic.

Principle of Least Action with Derivation_105.gif

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