Hamilton's Principle (2024)

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Michael Fowler, UVa

Beginnings of Dynamics

Galileo and Newton

Our text, Landau, begins (page 2!) by stating that the laws of dynamics come from the principleof least action, Hamilton’s principle. But where did Hamilton's principle come from? He was certainly very familiar with the laws of dynamics as developed by Galileo and Newton, and in fact his principle follows directly from them—but it gives a new and valuable perspective. To see how this comes about, we'll briefly follow the historical path.

To begin, then, with Galileo. His two major contributions todynamics were:

1. The realization, and experimental verification, thatfalling bodies have constant acceleration (provided air resistance can beignored) and all falling bodiesaccelerate at the same rate. (This was the first step to Newton's law of universal gravitation.)

2. Galilean relativity: As he put it himself, if you are in a closed room below decks in a ship movingwith steady velocity, no experiment on dropping or throwing something will lookany different because of the ship’s motion: you can’t detect the motion. As we would put it now, the laws of physicsare the same in all inertial frames.

Newton’s majorcontributions were his laws of motion,and his law of universal gravitationalattraction (which we'll discuss later).

His laws of motion are:

1. The law of inertia:A body moving at constant velocity will continue at that velocity unless actedon by a force. (Actually, Galileoessentially stated this law, but just for a ball rolling on a horizontal plane,with zero frictional drag.)

2. $\overrightarrow{F}=m\overrightarrow{a}.$

3. Action = reaction.

In terms of Newton’s laws, Galilean relativity is clear: ifthe ship is moving at steady velocity $\overrightarrow{v}$relative to the shore, than an object movingat $\overrightarrow{u}$relative to the ship is moving at $\overrightarrow{u}+\overrightarrow{v}$relative to the shore. If there is no forceacting on the object, it is moving at steady velocity in both frames: both areinertial frames, defined as frames in which Newton’s first law holds. And, since $\overrightarrow{v}$is constant, the acceleration is the same inboth frames, so if a force is introduced the second law is the same in the twoframes.

(Needless to say, all this is classical, meaningnonrelativistic and of course nonquantum, mechanics.)

Newton's Laws Explain Everything (in Principle...)

...in a classical mechanical system. Any such system can be analyzed as a (possiblyinfinite) collection of parts, or particles, having defined mutual interactions, so inprinciple Newton’s laws—just in Cartesian coordinates—can provide a description of the motion evolving fromany initial configuration of positions and velocities.

The problem is, though, that the equations are usuallyintractable—we can’tdo the math. In fact, the Cartesian coordinate positions and velocities might not be the bestchoice of parameters to specify the system’s configuration. For example, a simple pendulum is obviously more naturally described by the angle the string makes with the vertical, asopposed to the Cartesian coordinates of the bob. After Newton, a series of Frenchmathematicians reformulated his laws in terms of more useful coordinates—culminatingin Lagrange’s equations.

The Irish mathematician Hamilton then established thatthese improved dynamical equations could be derived using the calculus ofvariations to minimize an integral of a function, the Lagrangian, along a path in the system’s configuration space.

This integral is called the action, so the rule is that the systemfollows the path of least action from the initial to the final configuration.

Nitpicking footnote: strictly, we need the path to be a stationary point in the space of possible paths, usually it's the least-action path.

From Newton’s Laws in CartesianCo-ordinates to the Principle of Least Action

Calculus of Variations Done Backwards!

We’ve shown how (in lecture 2) given an integrand, we can finddifferential equations for the path in space time between two fixed points thatminimizes the corresponding path integral between those points.

Now we’ll do thereverse: we already know the differential equations in Cartesiancoordinates describing the path taken by a Newtonian particle in somepotential. We’ll show how to use that knowledge to construct the integrand suchthat the action integral is a minimum along that path. (This follows Jeffreysand Jeffreys, Mathematical Physics.)

We begin with the simplest nontrivial system, a particle ofmass $m$moving in one dimension from one point toanother in a specified time, we’ll assume it’s in a time-independent potential $U\left(x\right)$,so

$m\ddot{x}=-dU\left(x\right)/dx.$

Its path can be represented as a graph $x\left(t\right)$against time$—$forexample, for a ball thrown directly upwards in a constant gravitational fieldthis would be a parabola.

Initial and final positions are given: $x\left(t_1\right)=x_1$,$x\left(t_2\right)=x_2$,and the elapsed time is $t_2-t_1$.

Notice we have not specified the initial velocity$—$we don’thave that option. The differential equation is only second order, so itssolution is completely determined by the two (beginning and end) boundaryconditions.

We’re now ready to embark on the calculus of variations inreverse.

Trivially, multiplying both sides of the equation of motion byan arbitrary infinitesimal function the equality still holds:

$m\ddot{x}\delta x\left(t\right)=-\left(dU\left(x\right)/dx\right)\delta x\left(t\right)$

and in fact if this equationis true for arbitrary $\delta x\left(t\right)$,the original equation of motion holds throughout, because we can always choosea $\delta x\left(t\right)$nonzero only in the neighborhood of aparticular time $t$,from which the original equation must be true at that $t$.

By analogy with Fermat’s principle in the preceding section,we can picture this $\delta x\left(t\right)$as a slight variation in the path from theNewtonian trajectory, $x\left(t\right)\to x\left(t\right)+\delta x\left(t\right)$,and take the variation zero at the fixed ends, $\delta x\left(t_1\right)=\delta x\left(t_2\right)=0$.

In Fermat’s (light wave) case, the integrated time elapsed along the pathwas minimized—there waszero change to first order on going from the (physical) path to a neighboring path. Developing theanalogy, we’re looking for some dynamical quantity that has zero change tofirst order on going to a neighboring path having the same endpoints in spaceand time. We’ve fixed the time, what’sleft to integrate along the path?

For such a simple system, we don’t have many options! As we’ve discussed above, the equation ofmotion is equivalent to (putting in an overall minus sign that will proveconvenient)

${\displaystyle \underset{t_1}{\overset{t_2}{\int }}\left(-m\ddot{x}\left(t\right)-dU\left(x\left(t\right)\right)/dx\right)\delta x\left(t\right)}dt=0\text{toleadingorder,forallvariations}\delta x\left(t\right).$

Integrating the first term by parts (recalling $\delta \left(x\right)=0$at the endpoints):

$-{\displaystyle \underset{t_1}{\overset{t_2}{\int }}m\ddot{x}\left(t\right)\delta x\left(t\right)}dt={\displaystyle \underset{t_1}{\overset{t_2}{\int }}m\dot{x}\left(t\right)\delta \dot{x}\left(t\right)}dt={\displaystyle \underset{t_1}{\overset{t_2}{\int }}\delta \left(\frac{1}{2}m{\dot{x}}^2\left(t\right)\right)}dt={\displaystyle \underset{t_1}{\overset{t_2}{\int }}\delta T\left(x\left(t\right)\right)}dt$

using the standard notation $T$for kinetic energy.

The second term integrates trivially:

$-{\displaystyle \underset{t_1}{\overset{t_2}{\int }}\left(dU\left(x\right)/dx\right)\delta x\left(t\right)dt}=-{\displaystyle \underset{t_1}{\overset{t_2}{\int }}\delta U\left(x\right)dt}$

establishing that on making an infinitesimal variation from the physical path (the one that satisfies Newton's laws) there is zero first order change in the integral of kinetic energy minus potential energy.

The standard notation is

$$\delta S=\delta {\displaystyle \underset{t_1}{\overset{t_2}{\int }}\left(T-U\right)dt=\delta }{\displaystyle \underset{t_1}{\overset{t_2}{\int }}Ldt=0.}$$

The integral $S$is called the action integral, (also known as Hamilton’s Principal Function) andthe integrand $T-U=L$ is called the Lagrangian.

This equation isHamilton’s Principle of Least Action.

The derivation can be extended straightforwardly to aparticle in three dimensions, in fact to $n$interacting particles in three dimensions. We shall assume that the forces on particlescan be derived from potentials, including possibly time-dependent potentials,but we exclude frictional energy dissipation in this course. (It can be handled—see forexample Vujanovic and Jones, VariationalMethods in Nonconservative Phenomena,Academic press, 1989.)

But Why Does It Follow This Path?

Fermat’s principle, that a light ray follows the path of least time, was easy to believe once it became clearthat light was a wave. The wave really propagates along all paths, and the phase changealong a particular path is simply the time taken to travel that path measuredin units of the light wave oscillation time. That means that if neighboringpaths have the same length to first order the light waves along them will addcoherently, otherwise they will interfere and essentially cancel. So the path of least time is heavily favored,and when we look on a scale much greater than the wavelength of the light, wedon’t even see the diffraction effects caused by imperfect cancellation, the lightrays might as well be streams of particles, mysteriously choosing the path ofleast time.

So what has this to do with Hamilton’s principle? Everything! A standard method in quantummechanic is the so-called sum over paths, for example to find theprobability amplitude for an electron to go from one point to another in agiven time under a given potential, you can sum over all possible paths itmight take, multiplying each path by a phase factor: and that phase factor (as established experimentally) isnone other than Hamilton’s action integral divided by Planck’s constant, $S/\hslash .$ So, since all systems are really quantum systems, the classical limit $S\gg \hslash$ is a short-wavelength limit, in which the path of a dynamical systemwill be that of least action, exactly analogous to Fermat's principle for light waves. This is covered in my notes on QuantumMechanics. The crucial insight was Dirac's, the method was developed by Feynman.

Bottom Line: All of classical mechanics follows from the Principle of Least Action. But that Principle itself follows from quantum mechanics!

Historical footnote: Lagrange presented these methods in a classicbook that Hamilton called a “scientific poem”.Lagrange thought mechanics properly belonged to pure mathematics, it wasa kind of geometry in four dimensions (space and time). Hamilton was the first to use the principleof least action to derive Lagrange’s equations in the present form. He built up the least action formalismdirectly from Fermat’s principle, considered in a medium where the velocity oflight varies with position and with direction of the ray. He saw mechanics as represented bygeometrical optics in an appropriate space of higher dimensions. But it didn’t apparently occur to him thatthis might be because it was really a wave theory! (See Arnold, Mathematical Methods of Classical Mechanics, for details.)

Deriving Lagrange’s Equations from the Least Action Principle

Chooosing the Best Coordinates

We started with Newton’s equations of motion, expressed inCartesian coordinates of particle positions. For many systems, these equationsare mathematically intractable. Running the calculus of variations argument inreverse, we established Hamilton’s principle: the system moves along the path throughconfiguration space for which the action integral, with integrand theLagrangian $L=T-U,$is a minimum.

We’re now free to beginfrom Hamilton’s principle, expressing the Lagrangian in variables that morenaturally describe the system, taking advantage of any symmetries (such asusing angle variables for rotationally invariant systems). Also, some forces donot need to be included in the description of the system: a simple pendulum isfully specified by its position and velocity, we do not need to know thetension in the string, although that wouldappear in a Newtonian analysis. The greater efficiency (and elegance) of theLagrangian method, for most problems, will become evident on working throughactual examples.

We’ll define a set of generalizedcoordinates $q=\left(q_1,\dots q_n\right)$by requiring that they give a completedescription of the configuration of the system (where everything is in space). Thestate of the system is specified by this set plus the corresponding velocities $\dot{q}=\left({\dot{q}}_1,\dots {\dot{q}}_n\right)$.

For example, the $x$-coordinate of a particular particle $a$is given by some function of the $q_i$’s, $x_a=f_{x_a}\left(q_1,\dots q_n\right),$and the corresponding velocity component ${\dot{x}}_a={\displaystyle \sum _k\frac{\partial f_{x_a}}{\partial q_k}{\dot{q}}_k}$.

The Lagrangian will depend on all these variables ingeneral, and also possibly on time explicitly, for example if there is atime-dependent external potential. (Butusually that isn’t the case.)

Hamilton’s principle gives

$\delta S=\delta {\displaystyle \underset{t_1}{\overset{t_2}{\int }}L\left(q_i,{\dot{q}}_i,t\right)}dt=0$

that is,

${\displaystyle \underset{t_1}{\overset{t_2}{\int }}{\displaystyle \sum _i\left(\frac{\partial L}{\partial q_i}\delta q_i+\frac{\partial L}{\partial {\dot{q}}_i}\delta {\dot{q}}_i\right)}}dt=0.$

Integrating by parts,

$\delta S={\left[{\displaystyle \sum _i\frac{\partial L}{\partial {\dot{q}}_i}\delta q_i}\right]}_{t_1}^{t_2}+{\displaystyle \underset{t_1}{\overset{t_2}{\int }}{\displaystyle \sum _i\left(\frac{\partial L}{\partial q_i}-\frac{d}{dt}\left(\frac{\partial L}{\partial {\dot{q}}_i}\right)\right)}}\delta q_{i}dt=0.$

Requiring the path deviation to be zero at the endpointsgives Lagrange’s equations:

$$\frac{d}{dt}\left(\frac{\partial L}{\partial {\dot{q}}_i}\right)-\frac{\partial L}{\partial q_i}=0.$$

Non-uniqueness of the Lagrangian

The Lagrangian is not uniquely defined: two Lagrangians differing by the total derivative with respect to timeof some function will give the same identical equations on minimizing theaction,

$S^{\prime }={\displaystyle \underset{t_1}{\overset{t_2}{\int }}{L}^{\prime }\left(q,\dot{q},t\right)}dt={\displaystyle \underset{t_1}{\overset{t_2}{\int }}L\left(q,\dot{q},t\right)}dt+{\displaystyle \underset{t_1}{\overset{t_2}{\int }}\frac{df\left(q,t\right)}{dt}}dt=S+f\left(q\left(t_2\right),t_2\right)-f\left(q\left(t_1\right),t_1\right),$

and since $q\left(t_1\right),t_1,q\left(t_2\right),t_2$are all fixed, the integral over $df/dt$is trivially independent of path variations,and varying the path to minimize $S^{\prime }$gives the same result as minimizing $S$.This turns out to be important later$—$it givesus a useful new tool to change thevariables in the Lagrangian.

First Integral: Energy Conservationand the Hamiltonian

Since Lagrange’s equations are precisely a calculus of variationsresult, it follows from our earlier discussion that if the Lagrangian has no explicit time dependence then:

${\displaystyle \sum _i{\dot{q}}_i\frac{\partial L}{\partial {\dot{q}}_i}}-L=\text{constant}\text{.}$

(This is just the first integral $y^{\prime }\partial f/\partial y^{\prime }-f=\text{constant}$discussed earlier, now with $n$variables.)

This constant of motion is called the energy of the system, and denoted by $E$.We say the energy is conserved, even in the presence ofexternal potentials$—$providedthose potentials are time-independent.

(We’ll just mention that the function on the left-hand side,${\displaystyle \sum _i{\dot{q}}_i\partial L/\partial {\dot{q}}_i}-L,$is the Hamiltonian. We don’t discuss itfurther at this point because, as we’ll find out, it is more naturally treatedin other variables.)

We’ll now look at a couple of simple examples of theLagrangian approach.

Example 1: One Degree of Freedom:Atwood’s Machine

In 1784, the Rev. George Atwood, tutor at Trinity College, Cambridge,came up with a great demo for finding $g$.It’s still with us.

The traditional Newtonian solution of this problem is towrite $F=ma$for the two masses, then eliminate the tension$T$.(To keep things simple, we’ll neglect the rotational inertia of the toppulley.)

The Lagrangian approach is, of course, to write down theLagrangian, and derive the equation of motion.

Measuring gravitational potential energy from the top wheelaxle, the potential energy is

$U\left(x\right)=-m_{1}gx-m_{2}g\left(\ell -x\right)$

and the Lagrangian

$L=T-U=\frac{1}{2}\left(m_1+m_2\right){\dot{x}}^2+m_{1}gx+m_{2}g\left(\ell -x\right).$

Lagrange’s equation:

$$\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{x}}\right)-\frac{\partial L}{\partial x}=\left(m_1+m_2\right)\ddot{x}-\left(m_1-m_2\right)g=0$$

gives the equation of motion in just one step.

It’s usually pretty easy to figure out the kinetic energyand potential energy of a system, and thereby write down the Lagrangian. Thisis definitely less work than the Newtonian approach, which involves constraintforces, such as the tension in the string. This force doesn’t even appear in theLagrangian approach! Other constraintforces, such as the normal force for a bead on a wire, or the normal force fora particle moving on a surface, or the tension in the string of a pendulum$—$none ofthese forces appear in the Lagrangian. Notice, though, that these forces neverdo any work.

On the other hand, if you actually are interested in thetension in the string (will it break?) you use the Newtonian method, or maybework backwards from the Lagrangian solution.

Example 2: Lagrangian Formulation ofthe Central Force Problem

A simple example ofLagrangian mechanics is provided by the central force problem, a mass $m$acted on by a force $F_r=-dU\left(r\right)/dr$.

To contrast the Newtonian and Lagrangian approaches, we’llfirst look at the problem using just $\overrightarrow{F}=m\overrightarrow{a}$. To take advantage of the rotational symmetrywe’ll use $\left(r,\theta \right)$coordinates, and find the expression foracceleration by the standard trick of differentiating the complex number $z=r{e}^{i\theta }$twice, to get

$\begin{array}{c}m\left(\ddot{r}-r{\dot{\theta }}^2\right)=-dU\left(r\right)/dr\\ m\left(r\ddot{\theta }+2\dot{r}\dot{\theta }\right)=0.\end{array}$

The second equation integrates immediately to give

$m{r}^2\dot{\theta }=\ell ,$

a constant, the angular momentum. This can then be used to eliminate $\dot{\theta }$in the first equation, giving a second-order differentialequation for $r\left(t\right)$.

The Lagrangian approach, on the other hand, is first towrite

$L=T-U=\frac{1}{2}m\left({\dot{r}}^2+r^2{\dot{\theta }}^2\right)-U\left(r\right)$

and put it into the equations

$$\begin{array}{l}\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{r}}\right)-\left(\frac{\partial L}{\partial r}\right)=0,\\ \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{\theta }}\right)-\left(\frac{\partial L}{\partial \theta }\right)=0.\end{array}$$

Note now that since $L$doesn’t depend on $\theta$,the second equation gives immediately:

$$\frac{\partial L}{\partial \dot{\theta }}=\text{constant,}$$

and in fact $\partial L/\partial \dot{\theta }=m{r}^2\dot{\theta },$the angular momentum, we’ll call it $\ell$.

The first integral (see above) gives another constant:

$\dot{r}\frac{\partial L}{\partial \dot{r}}+\dot{\theta }\frac{\partial L}{\partial \dot{\theta }}-L=\text{constant}\text{.}$

This is just

$\frac{1}{2}m\left({\dot{r}}^2+r^2{\dot{\theta }}^2\right)+U\left(r\right)=E$

the energy.

Angular momentum conservation, $m{r}^2\dot{\theta }=\ell ,$then gives

$\frac{1}{2}m\left({\dot{r}}^2+\frac{{\ell }^2}{m^2{r}^2}\right)+U\left(r\right)=E$

giving a first-order differential equation for the radialmotion as a function of time. We’ll deal with this in more detail later. Note that it is equivalent to a particlemoving in one dimension in the original potential plus an effective potentialfrom the angular momentum term:

$E=\frac{1}{2}m{v}^2+U\left(r\right)+\frac{{\ell }^2}{m^2{r}^2}.$

This can beunderstood by realizing that for a fixed angular momentum, the closer theparticle approaches the center the greater its speed in the tangentialdirection must be, so, to conserve total energy, its speed in the radialdirection has to go down, unless it is in a verystrongly attractive potential (the usual gravitational or electrostaticpotential isn’t strong enough) so the radial motion is equivalent to that withthe existing potential plus the ${\ell }^2/m^2{r}^2$term, often termed the “centrifugal barrier”.

Exercise: Howstrong must the potential be to overcome the centrifugal barrier? (This canhappen in a black hole!)

Generalized Momenta and Forces

For the above orbital Lagrangian, $dL/d\dot{r}=m\dot{r}=p_r,$the momentum in the $r$-direction, and $dL/d\dot{\theta }=m{r}^2\dot{\theta }=p_{\theta },$the angular momentum associated with thevariable $\theta .$

The generalizedmomenta for a mechanical system are defined by

$$p_i=\frac{\partial L}{\partial {\dot{q}}_i}.$$

(Warning: thesegeneralized momenta are an essential part of the formalism, but do not alwaysdirectly correspond to the physicalmomentum of a particle, an example being a charged particle in a magneticfield, see my quantum notes.)

Less frequently used are the generalized forces, $F_i=\partial L/\partial q_i,$defined to make the Lagrange equations lookNewtonian, $F_i={\dot{p}}_i.$

Conservation Laws and Noether’s Theorem

Orbital Angular Momentum and Energy Conservation

The two integrals of motion for the orbital example above canbe stated as follows:

First: if the Lagrangian does not depend on thevariable $\theta ,\partial L/\partial \theta =0,$that is, it’s invariant under rotation, meaning it has circular symmetry, then

$$p_{\theta }=\frac{\partial L}{\partial \dot{\theta }}=\text{constant,}$$

angular momentum isconserved.

Second: As stated earlier, if the Lagrangian isindependent of time, that is, it’s invariantunder time translation, then energy is conserved. (This is nothingbut the first integral of the calculus of variations, recall that for anintegrand function $f\left(y,y^{\prime }\right)$not explicitly dependent on $x,$$y^{\prime }\partial f/\partial y^{\prime }-f$is constant.)

${\displaystyle \sum _i{\dot{q}}_i\partial L/\partial {\dot{q}}_i}-L=E,\text{aconstant}\text{.}$

Both these resultslink symmetries of the Lagrangian$—$invariance under rotation and timetranslation respectively$—$with conserved quantities.

This connection was first spelled out explicitly, and provedgenerally, by Emmy Noether, published in 1915.The essence of the theorem is that if the Lagrangian (which specifiesthe system completely) does not change when some continuous parameter isaltered, then some function of the $q_i,{\dot{q}}_i$stays the same$—$it iscalled a constant of the motion, an integral of the motion, or a conserved quantity.

To look further at this expression for energy, we take aclosed system of particles interacting with each other, but “closed” means nointeraction with the outside world (except possibly a time-independentpotential).

The Lagrangian for the particles is, in Cartesiancoordinates,

$L={\displaystyle \sum \frac{1}{2}{m}_i{v}_i^2-U\left({\overrightarrow{r}}_1,{\overrightarrow{r}}_2,\dots \right)}$.

A set of general coordinates $\left(q_1,\dots q_n\right)$,by definition, uniquely specifies the system configuration, so the coordinateand velocity of a particular particle $a$are given by

$x_a=f_{x_a}\left(q_1,\dots q_n\right),{\dot{x}}_a={\displaystyle \sum _k\frac{\partial f_{x_a}}{\partial q_k}{\dot{q}}_k}.$

From this it is clear that the kinetic energy term $T={\displaystyle \sum \frac{1}{2}{m}_i{v}_i^2}$is a hom*ogeneous quadratic function of the $\dot{q}$’s (meaning every term is of degree two), so

$L=\frac{1}{2}{\displaystyle \sum _{i,k}{a}_{ik}\left(q\right){\dot{q}}_i{\dot{q}}_k}-U\left(q\right).$

This being of degree two in the time derivatives means

${\displaystyle \sum _i{\dot{q}}_i\frac{\partial L}{\partial {\dot{q}}_i}=}{\displaystyle \sum _i{\dot{q}}_i\frac{\partial T}{\partial {\dot{q}}_i}=}2T.$

(If this isn’t obvious to you, check it out with a couple ofterms: ${\dot{q}}_1^2,{\dot{q}}_1{\dot{q}}_2.$)

Therefore for this system of interacting particles

$E={\displaystyle \sum _{i=1}^n{\dot{q}}_i\frac{\partial L}{\partial {\dot{q}}_i}}-L=2T-\left(T-U\right)=T+U.$

This expression for the energy is called the Hamiltonian:

$$H={\displaystyle \sum _{i=1}^n{p}_i{\dot{q}}_i}-L.$$

Momentum Conservation

Another conservation law follows if the Lagrangian isunchanged by displacing the whole system through a distance $\delta \overrightarrow{r}=\overrightarrow{\epsilon }.$This means, of course, that the system cannotbe in some spatially varying external field$—$it must bemechanically isolated.

It is natural to work in Cartesian coordinates to analyzethis, each particle is moved the same distance ${\overrightarrow{r}}_i\to {\overrightarrow{r}}_i+\delta {\overrightarrow{r}}_i={\overrightarrow{r}}_i+\overrightarrow{\epsilon }$,so

$\delta L={\displaystyle \sum _i\frac{\partial L}{\partial {\overrightarrow{r}}_i}\cdot \delta {\overrightarrow{r}}_i=\overrightarrow{\epsilon }\cdot }{\displaystyle \sum _i\frac{\partial L}{\partial {\overrightarrow{r}}_i},}$

where the “differentiation by a vector” notation meansdifferentiating with respect to each component, then adding the three terms. (I’mnot crazy about this notation, but it’s Landau’s, so get used to it.)

For an isolated system, we must have $\delta L=0$on displacement, moving the whole thingthrough empty space in any direction $\overrightarrow{\epsilon }$changes nothing, so it must be that the vectorsum ${\displaystyle \sum _i\partial L/\partial {\overrightarrow{r}}_i=0}$,so from the Cartesian Euler-Lagrange equations, writing$\dot{\overrightarrow{r}}=\overrightarrow{v},$

$$0={\displaystyle \sum _i\partial L/\partial {\overrightarrow{r}}_i=}{\displaystyle \sum \frac{d}{dt}}\left(\frac{\partial L}{\partial {\dot{\overrightarrow{r}}}_i}\right)={\displaystyle \sum \frac{d}{dt}}\left(\frac{\partial L}{\partial {\overrightarrow{v}}_i}\right)=\frac{d}{dt}{\displaystyle \sum \frac{\partial L}{\partial {\overrightarrow{v}}_i}},$$

and taking the system to be composed of particles of mass $m_i$and velocity ${\overrightarrow{v}}_i$,

${\displaystyle \sum _i\frac{\partial L}{\partial {\overrightarrow{v}}_i}}={\displaystyle \sum _i{m}_i{\overrightarrow{v}}_i=\overrightarrow{P}=\text{constant,}}$

the momentum ofthe system.

This vector conservation law is of course three separate directionalconservation laws, so even if there is anexternal field, if it doesn’t vary in a particular direction, the component oftotal momentum in that direction will be conserved.

In the Newtonian picture, conservation of momentum in aclosed system follows from Newton’s third law.In fact, the above Lagrangian analysis is really Newton’s third law indisguise. Since we’re working in Cartesian coordinates, $\partial L/\partial {\overrightarrow{r}}_i=-\partial V/\partial {\overrightarrow{r}}_i={\overrightarrow{F}}_i$,the force on the $i$th particle, and if there are no externalfields, ${\displaystyle \sum _i\partial L/\partial {\overrightarrow{r}}_i=0}$just means that if you add all the forces onall the particles, the sum is zero. For the Lagrangian of atwo particle system to be invariant under translation through space, thepotential must have the form $V\left({\overrightarrow{r}}_1-{\overrightarrow{r}}_2\right)$, from which automatically ${\overrightarrow{F}}_{12}=-{\overrightarrow{F}}_{21}$.

Center of Mass

If an inertial frame of reference $K^{\prime }$is moving at constant velocity $\overrightarrow{V}$relative to inertial frame $K,$the velocities of individual particles in theframes are related by ${\overrightarrow{v}}_i={{\overrightarrow{v}}^{\prime }}_i+\overrightarrow{V},$so the total momenta are related by

$\overrightarrow{P}={\displaystyle \sum _i{m}_i{\overrightarrow{v}}_i=}{\displaystyle \sum _i{m}_i{{\overrightarrow{v}}^{\prime }}_i}+\overrightarrow{V}{\displaystyle \sum _i{m}_i}={\overrightarrow{P}}^{\prime }+M\overrightarrow{V},M={\displaystyle \sum _i{m}_i.}$

If we choose $\overrightarrow{V}=\overrightarrow{P}/M,$then ${\overrightarrow{P}}^{\prime }={\displaystyle \sum _i{m}_i{{\overrightarrow{v}}^{\prime }}_i}=0,$the system is “at rest” in the frame $K^{\prime }.$Of course, the individual particles might bemoving, what is at rest in $K^{\prime }$is the centerof mass defined by

$M{\overrightarrow{R}}_{\text{cm}}={\displaystyle \sum _i{m}_i{\overrightarrow{r}}_i.}$

(Check this by differentiating both sides with respect totime.)

The energy of a mechanical system in its rest frame is oftencalled its internal energy, we’lldenote it by $E_{\mathrm{int}}.$(This includes kinetic and potentialenergies.) The total energy of a movingsystem is then

$E=\frac{1}{2}M{\overrightarrow{V}}^2+E_{\mathrm{int}}.$

(Exercise: Verify this.)

Angular Momentum Conservation

Conservation of momentum followed from the invariance of theLagrangian on being displaced in arbitrary directions in space, the hom*ogeneityof space, angular momentum conservation is the consequence of the isotropy of space$—$there isno preferred direction.

So angular momentum of an isolated body in space isinvariant even if the body is not symmetric itself.

The strategy is just as before, except now instead of aninfinitesimal displacement we make an infinitesimal rotation,

$\delta \overrightarrow{r}=\delta \overrightarrow{\varphi }\times \overrightarrow{r}$

and of course the velocities will also be rotated:

$\delta \overrightarrow{v}=\delta \overrightarrow{\varphi }\times \overrightarrow{v}.$

We must have

$\delta L={\displaystyle \sum _i\left(\frac{\partial L}{\partial {\overrightarrow{r}}_i}\cdot \delta {\overrightarrow{r}}_i+\frac{\partial L}{\partial {\overrightarrow{v}}_i}\cdot \delta {\overrightarrow{v}}_i\right)}=0.$

Now $\partial L/\partial {\overrightarrow{v}}_i=\partial L/\partial {\dot{\overrightarrow{r}}}_i={\overrightarrow{p}}_i$by definition, and from Lagrange’s equations

$\partial L/\partial {\overrightarrow{r}}_i=\left(d/dt\right)\left(\partial L/\partial {\dot{\overrightarrow{r}}}_i\right)={\dot{\overrightarrow{p}}}_i,$

so the isotropy of space implies that

${\displaystyle \sum _i\left({\dot{\overrightarrow{p}}}_i\cdot \delta \overrightarrow{\varphi }\times {\overrightarrow{r}}_i+{\overrightarrow{p}}_i\cdot \delta \overrightarrow{\varphi }\times {\overrightarrow{v}}_i\right)=0.}$

Notice the second term is identically zero anyway, since twoof the three vectors in the triple product are parallel: $\left(d{\overrightarrow{r}}_i/dt\right)\times {\overrightarrow{p}}_i={\overrightarrow{v}}_i\times m{\overrightarrow{v}}_i=0$

That leaves the first term.The equation can be written:

$\delta \overrightarrow{\varphi }\cdot \frac{d}{dt}{\displaystyle \sum _i{\overrightarrow{r}}_i\times {\overrightarrow{p}}_i}=0,$

Integrating, we find that

${\displaystyle \sum _i{\overrightarrow{r}}_i\times {\overrightarrow{p}}_i}=\overrightarrow{L}$

is a constant of motion, the angular momentum.

The angular momentum of a system is different aboutdifferent origins. (Think of a single moving particle.) The angular momentum in the rest frame isoften called the intrinsic angular momentum, the angular momentum in a frame inwhich the center of mass is at position $\overrightarrow{R}$and moving with velocity $\overrightarrow{V}$is

$\overrightarrow{L}={\overrightarrow{L}}_{\text{cmframe}}+\overrightarrow{R}\times \overrightarrow{P}.$

(Exercise: Check this.)

For a system of particles in a fixed external central field $V\left(r\right),$the system is invariant with respect torotations about that point, soangular momentum about that point is conserved.For a field “cylindrically” invariant for rotations about an axis,angular momentum about that axis is conserved.

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