Lectures on Classical Mechanicsderek/classical-ebook.pdf · together, and sometimes split them over chapter, to obtain a more textbook feel to these notes. For reference, the weekly

Lectures onClassical Mechanics

John C. BaezDerek K. Wise

(version of February 3, 2013)

i

c©2013 John C. Baez & Derek K. Wise

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Preface

These are notes for a mathematics graduate courseon classical mechanics at U.C. Riverside. I’ve taughtthis course three times recently. Twice I focused onthe Hamiltonian approach. In 2005 I started with theLagrangian approach, with a heavy emphasis onaction principles, and derived the Hamiltonianapproach from that. This approach seems morecoherent.

Derek Wise took beautiful handwritten notes onthe 2005 course, which can be found on my website:

http://math.ucr.edu/home/baez/classical/

Later, Blair Smith from Louisiana State Universitymiraculously appeared and volunteered to turn thenotes into LATEX . While not yet the book I’deventually like to write, the result may already behelpful for people interested in the mathematics ofclassical mechanics.

The chapters in this LATEX version are in the sameorder as the weekly lectures, but I’ve merged weeks

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together, and sometimes split them over chapter, toobtain a more textbook feel to these notes. Forreference, the weekly lectures are outlined here.

Week 1: (Mar. 28, 30, Apr. 1)—The Lagrangianapproach to classical mechanics: deriving F = mafrom the requirement that the particle’s path be acritical point of the action. The prehistory of theLagrangian approach: D’Alembert’s “principle ofleast energy” in statics, Fermat’s “principle of leasttime” in optics, and how D’Alembert generalized hisprinciple from statics to dynamics using the conceptof “inertia force”.

Week 2: (Apr. 4, 6, 8)—Deriving theEuler-Lagrange equations for a particle on anarbitrary manifold. Generalized momentum andforce. Noether’s theorem on conserved quantitiescoming from symmetries. Examples of conservedquantities: energy, momentum and angularmomentum.

Week 3 (Apr. 11, 13, 15)—Example problems: (1)The Atwood machine. (2) A frictionless mass on atable attached to a string threaded through a hole in

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the table, with a mass hanging on the string. (3) Aspecial-relativistic free particle: two Lagrangians, onewith reparametrization invariance as a gaugesymmetry. (4) A special-relativistic charged particlein an electromagnetic field.

Week 4 (Apr. 18, 20, 22)—More exampleproblems: (4) A special-relativistic charged particlein an electromagnetic field in special relativity,continued. (5) A general-relativistic free particle.

Week 5 (Apr. 25, 27, 29)—How Jacobi unifiedFermat’s principle of least time and Lagrange’sprinciple of least action by seeing the classicalmechanics of a particle in a potential as a specialcase of optics with a position-dependent index ofrefraction. The ubiquity of geodesic motion.Kaluza-Klein theory. From Lagrangians toHamiltonians.

Week 6 (May 2, 4, 6)—From Lagrangians toHamiltonians, continued. Regular and stronglyregular Lagrangians. The cotangent bundle as phasespace. Hamilton’s equations. Getting Hamilton’sequations directly from a least action principle.

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Week 7 (May 9, 11, 13)—Waves versus particles:the Hamilton-Jacobi equation. Hamilton’s principalfunction and extended phase space. How theHamilton-Jacobi equation foreshadows quantummechanics.

Week 8 (May 16, 18, 20)—Towards symplecticgeometry. The canonical 1-form and the symplectic2-form on the cotangent bundle. Hamilton’sequations on a symplectic manifold. Darboux’stheorem.

Week 9 (May 23, 25, 27)—Poisson brackets. TheSchrodinger picture versus the Heisenberg picture inclassical mechanics. The Hamiltonian version ofNoether’s theorem. Poisson algebras and Poissonmanifolds. A Poisson manifold that is notsymplectic. Liouville’s theorem. Weil’s formula.

Week 10 (June 1, 3, 5)—A taste of geometricquantization. Kahler manifolds.

If you find errors in these notes, please email me! Ithank Sheeyun Park and Curtis Vinson for catchinglots of errors.

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John C. Baez

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Contents

1 From Newton’s Laws to Langrange’s Equa-tions 1

1.1 Lagrangian and Newtonian Approaches 2

1.1.1 Lagrangian versus HamiltonianApproaches . . . . . . . . . . . 9

1.2 Prehistory of the Lagrangian Approach 11

1.2.1 The Principle of Minimum Energy 15

1.2.2 D’Alembert’s Principle and La-grange’s Equations . . . . . . . 17

1.2.3 The Principle of Least Time . . 21

1.2.4 How D’Alembert and Others Gotto the Truth . . . . . . . . . . . 24

ix

x CONTENTS

2 Equations of Motion 29

2.1 The Euler-Lagrange Equations . . . . . 30

2.1.1 Comments . . . . . . . . . . . . 32

2.1.2 Lagrangian Dynamics . . . . . . 33

2.2 Interpretation of Terms . . . . . . . . . 37

3 Lagrangians and Noether’s Theorem 41

3.1 Time Translation . . . . . . . . . . . . 42

3.1.1 Canonical and Generalized Co-ordinates . . . . . . . . . . . . . 44

3.2 Symmetry and Noether’s Theorem . . 47

3.2.1 Noether’s Theorem . . . . . . . 49

3.3 Conserved Quantities from Symmetries 52

3.3.1 Time Translation Symmetry . . 53

3.3.2 Space Translation Symmetry . . 54

3.3.3 Rotational Symmetry . . . . . . 56

3.4 Example Problems . . . . . . . . . . . 59

3.4.1 The Atwood Machine . . . . . . 59

3.4.2 Disk Pulled by Falling Mass . . 61

3.4.3 Free Particle in Special Relativity 65

3.5 Electrodynamics and Relativistic Lagrangians 74

CONTENTS xi

3.5.1 Gauge Symmetry and Relativis-tic Hamiltonian . . . . . . . . . 75

3.5.2 Relativistic Hamiltonian . . . . 76

3.6 Relativistic Particle in an Electromag-netic Field . . . . . . . . . . . . . . . . 79

3.7 Alternative Lagrangians . . . . . . . . 84

3.7.1 Lagrangian for a String . . . . . 84

3.7.2 Alternate Lagrangian for Rela-tivistic Electrodynamics . . . . 87

3.8 The General Relativistic Particle . . . 92

3.8.1 Free Particle Lagrangian in GR 94

3.8.2 Charged particle in EM Field inGR . . . . . . . . . . . . . . . . 98

3.9 The Principle of Least Action and Geodesics 99

3.9.1 Jacobi and Least Time vs LeastAction . . . . . . . . . . . . . . 99

3.9.2 The Ubiquity of Geodesic Motion 104

4 From Lagrangians to Hamiltonians 111

4.1 The Hamiltonian Approach . . . . . . 112

4.2 Regular and Strongly Regular Lagrangians118

xii CONTENTS

4.2.1 Example: A Particle in a Rie-mannian Manifold with Poten-tial V (q) . . . . . . . . . . . . . 118

4.2.2 Example: General RelativisticParticle in an E-M Potential . . 119

4.2.3 Example: Free General Relativis-tic Particle with Reparameteri-zation Invariance . . . . . . . . 120

4.2.4 Example: A Regular but not StronglyRegular Lagrangian . . . . . . . 121

4.3 Hamilton’s Equations . . . . . . . . . . 1224.3.1 Hamilton and Euler-Lagrange . 1254.3.2 Hamilton’s Equations from the

Principle of Least Action . . . . 1294.4 Waves versus Particles—The Hamilton-

Jacobi Equations . . . . . . . . . . . . 1334.4.1 Wave Equations . . . . . . . . . 1354.4.2 The Hamilton-Jacobi Equations 139

Chapter 1

From Newton’s Lawsto Langrange’sEquations

(Week 1, March 28, 30, April 1.)

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2 From Newton’s Laws to Langrange’s Equations

1.1 Lagrangian and Newtonian

Approaches

We begin by comparing the Newtonian approach tomechanics to the subtler approach of Lagrangianmechanics. Recall Newton’s law:

F = ma (1.1)

wherein we consider a particle moving in Rn. Itsposition, say q, depends on time t ∈ R, so it defines afunction,

q : R −→ Rn.

From this function we can define velocity,

v = q : R −→ Rn

where q = dqdt

, and also acceleration,

a = q : R −→ Rn.

Now let m > 0 be the mass of the particle, and let Fbe a vector field on Rn called the force. Newton


claimed that the particle satisfies F = ma. That is:

ma(t) = F (q(t)) . (1.2)

This is a 2nd-order differential equation forq : R→ Rn which will have a unique solution givensome q(t0) and q(t0), provided the vector field F is‘nice’ — by which we technically mean smooth andbounded (i.e., |F (x)| < B for some B > 0, for allx ∈ Rn).

We can then define a quantity called kineticenergy:

K(t) :=1

2mv(t) · v(t) (1.3)

This quantity is interesting because

d

dtK(t) = mv(t) · a(t)

= F (q(t)) · v(t)

So, kinetic energy goes up when you push an objectin the direction of its velocity, and goes down when


you push it in the opposite direction. Moreover,

K(t1)−K(t0) =

∫ t1

t0

F (q(t)) · v(t) dt

=

∫ t1

t0

F (q(t)) · q(t) dt

So, the change of kinetic energy is equal to the workdone by the force, that is, the integral of F along thecurve q : [t0, t1]→ Rn. This implies (by Stokes’stheorem relating line integrals to surface integrals ofthe curl) that the change in kinetic energyK(t1)−K(t0) is independent of the curve going fromq(t0) = a to q(t0) = b if and only if

∇×F = 0.

This in turn is true if and only if

F = −∇V (1.4)

for some function V : Rn → R. This function is thenunique up to an additive constant; we call it thepotential. A force with this property is called


conservative. Why? Because in this case we candefine the total energy of the particle by

E(t) := K(t) + V (q(t)) (1.5)

where V (t) := V (q(t)) is called the potentialenergy of the particle, and then we can show that Eis conserved: that is, constant as a function of time.To see this, note that F = ma implies

d

dt[K(t) + V (q(t))] = F (q(t)) · v(t) +∇V (q(t)) · v(t)

= 0, (because F = −∇V ).

Conservative forces let us apply a whole bunch ofcool techniques. In the Lagrangian approach wedefine a quantity

L := K(t)− V (q(t)) (1.6)

called the Lagrangian, and for any curveq : [t0, t1]→ Rn with q(t0) = a, q(t1) = b, we definethe action to be

S(q) :=

∫ t1

t0

L(t) dt (1.7)


From here one can go in two directions. One is toclaim that nature causes particles to follow paths ofleast action, and derive Newton’s equations fromthat principle. The other is to start with Newton’sprinciples and find out what conditions, if any, onS(q) follow from this. We will use the shortcut ofhindsight, bypass the philosophy, and simply use themathematics of variational calculus to show thatparticles follow paths that are ‘critical points’ of theaction S(q) if and only if Newton’s law F = maholds. To do this, let us look for curves (like the solid

t0 t1

Rn

R

q

qs+sdq

Figure 1.1: A particle can sniff out the path of leastaction.


line in Fig. 1.1) that are critical points of S,namely:

d

dsS(qs)|s=0 = 0 (1.8)

whereqs = q + sδq

for all δq : [t0, t1]→ Rn with,

δq(t0) = δq(t1) = 0.

To show that

F = ma ⇔ d

dsS(qs)|t=0 = 0 for all δq : [t0, t1]→ Rn with δq(t0) = δq(t1) = 0

(1.9)we start by using integration by parts on thedefinition of the action, and first noting thatdqs/ds = δq(t) is the variation in the path:

d

dsS(qs)

∣∣∣∣s=0

=d

ds

∫ t1

t0

1

2mqs(t) · qs(t)− V (qs(t)) dt

∣∣∣∣s=0

=

∫ t1

t0

d

ds

[1

2mqs(t) · qs(t)− V (qs(t))

]dt

∣∣∣∣s=0

=

∫ t1

t0

[mqs ·

d

dsqs(t)−∇V (qs(t)) ·

d

dsqs(t)

]dt

∣∣∣∣s=0


Note that

d

dsqs(t) =

d

ds

d

dtqs(t) =

d

dt

d

dsqs(t)

and when we set s = 0 this quantity becomes just:

d

dtδq(t)

So,

d

dsS(qs)

∣∣∣∣s=0

=

∫ t1

t0

[mq · d

dtδq(t)−∇V (q(t)) · δq(t)

]dt

Then we can integrate by parts, noting the boundaryterms vanish because δq = 0 at t1 and t0:

d

dsS(qs)|s=0 =

∫ t1

t0

[−mq(t)−∇V (q(t))] · δq(t)dt

It follows that variation in the action is zero for allvariations δq if and only if the term in brackets isidentically zero, that is,

−mq(t)−∇V (q(t)) = 0


So, the path q is a critical point of the action S ifand only if

F = ma.

The above result applies only for conservativeforces, i.e., forces that can be written as the minusthe gradient of some potential. However, this seemsto be true of the most fundamental forces that weknow of in our universe. It is a simplifyingassumption that has withstood the test of time andexperiment.

1.1.1 Lagrangian versus HamiltonianApproaches

I am not sure where to mention this, but beforelaunching into the history of the Lagrangianapproach may be as good a time as any. In laterchapters we will describe another approach toclassical mechanics: the Hamiltonian approach. Whydo we need two approaches, Lagrangian andHamiltonian?


They both have their own advantages. In thesimplest terms, the Hamiltonian approach focuses onposition and momentum, while the Lagrangianapproach focuses on position and velocity. TheHamiltonian approach focuses on energy, which is afunction of position and momentum — indeed,‘Hamiltonian’ is just a fancy word for energy. TheLagrangian approach focuses on the Lagrangian,which is a function of position and velocity. Our firsttask in understanding Lagrangian mechanics is to geta gut feeling for what the Lagrangian means. Thekey is to understand the integral of the Lagrangianover time – the ‘action’, S. We shall see that thisdescribes the ‘total amount that happened’ from onemoment to another as a particle traces out a path.And, peeking ahead to quantum mechanics, thequantity exp(iS/~), where ~ is Planck’s constant,will describe the ‘change in phase’ of a quantumsystem as it traces out this path.

In short, while the Lagrangian approach takes awhile to get used to, it provides invaluable insightsinto classical mechanics and its relation to quantum


mechanics. We shall see this in more detail soon.

1.2 Prehistory of the

Lagrangian Approach

We’ve seen that a particle going from point a at timet0 to a point b at time t1 follows a path that is acritical point of the action,

S =

∫ t1

t0

K − V dt

so that slight changes in its path do not change theaction (to first order). Often, though not always, theaction is minimized, so this is called the Principleof Least Action.

Suppose we did not have the hindsight afforded bythe Newtonian picture. Then we might ask, “Whydoes nature like to minimize the action? And whythis action

∫K−V dt? Why not some other action?”

‘Why’ questions are always tough. Indeed, somepeople say that scientists should never ask ‘why’.


This seems too extreme: a more reasonable attitudeis that we should only ask a ‘why’ question if weexpect to learn something scientifically interesting inour attempt to answer it.

There are certainly some interseting things to learnfrom the question “why is action minimized?” First,note that total energy is conserved, so energy canslosh back and forth between kinetic and potentialforms. The Lagrangian L = K − V is big when mostof the energy is in kinetic form, and small when mostof the energy is in potential form. Kinetic energymeasures how much is ‘happening’ — how much oursystem is moving around. Potential energy measureshow much could happen, but isn’t yet — that’s whatthe word ‘potential’ means. (Imagine a big rocksitting on top of a cliff, with the potential to falldown.) So, the Lagrangian measures something wecould vaguely refer to as the ‘activity’ or ‘liveliness’of a system: the higher the kinetic energy the morelively the system, the higher the potential energy theless lively. So, we’re being told that nature likes tominimize the total of ‘liveliness’ over time: that is,


the total action.In other words, nature is as lazy as possible!For example, consider the path of a thrown rock in

the Earth’s gravitational field, as in Fig. 1.2. The

K V_small

K V_big

spend_time_here

get_done_quick

Figure 1.2: A particle’s “lazy” motion, minimizes theaction.

rock traces out a parabola, and we can think of it as


doing this in order to minimize its action. On the onehand, it wants to spend a lot much time near the topof its trajectory, since this is where the kinetic energyis least and the potential energy is greatest. On theother hand, if it spends too much time near the topof its trajectory, it will need to really rush to get upthere and get back down, and this will take a lot ofaction. The perfect compromise is a parabolic path!

There is another way to make progress onunderstanding ‘why’ action is minimized: history.Historically there were two principles that were fairlyeasy to deduce from observations of nature: (i) theprinciple of minimum energy used in statics, and(ii) the principle of least time, used in optics. Byputting these together, we can guess the principle ofleast action. So, let us recall these earlier minimumprinciples.


1.2.1 The Principle of MinimumEnergy

Before physicists really got going in studyingdynamical systems they used to study statics.Statics is the study of objects at rest, or inequilibrium. Archimedes studied the laws of a

m_1 m_2

L_1 L_2

Figure 1.3: A principle of energy minimization deter-mines a lever’s balance.

see-saw or lever (Fig. 1.3), and he found that thiswould be in equilibrium if

m1L1 = m2L2.

Later D’Alembert understood this using his“principle of virtual work”. He considered movingthe lever slightly, i.e., infinitesimally, He claimed thatin equilibrium the infinitesimal work done by this


dqdq_1

dq_2


motion is zero! He also claimed that the work doneon the ith body is,

dWi = Fidqi

and gravity pulls down with a force mig so,

dWi = (0, 0,−mg) · (0, 0,−L1dθ)

= m1gL1 dθ

and similarly,

dW2 = −m2gL2 dθ

Now D’Alembert’s principle says that equilibriumoccurs when the “virtual work” dW = dW1 + dW2

vanishes for all dθ (that is, for all possibleinfinitesimal motions). This happens when

m1L1 −m2L2 = 0


which is just as Archimedes wrote.

1.2.2 D’Alembert’s Principle andLagrange’s Equations

Let’s go over the above analysis in more detail. I’lltry to make it clear what we mean by virtual work.

The forces and constraints on a system may betime dependent. So equal small infinitesimaldisplacements of the system might result in theforces Fi acting on the system doing differentamounts of work at different times. To displace asystem by δri for each position coordinate, and yetremain consistent with all the constraints and forcesat a given instant of time t, without any timeinterval passing is called a virtual displacement. It’scalled ‘virtual’ because it cannot be realized: anyactual displacement would occur over a finite timeinterval and possibly during which the forces andconstraints might change. Now call the work done bythis hypothetical virtual displacement, Fi · δri, thevirtual work. Consider a system in the special state


of being in equilibrium, i.e.,. when∑

Fi = 0. Thenbecause by definition the virtual displacements donot change the forces, we must deduce that thevirtual work vanishes for a system in equilibrium,

∑i

Fi · δri = 0, (when in equilibrium) (1.10)

Note that in the above example we have twoparticles in R3 subject to a constraint (they arepinned to the lever arm). However, a number n ofparticles in R3 can be treated as a singlequasi-particle in R3n, and if there are constraints itcan move in some submanifold of R3n. So ultimatelywe need to study a particle on an arbitrary manifold.But, we’ll postpone such sophistication for a while.

For a particle in Rn, D’Alembert’s principle simply


says,

q(t) = q0 satisfies F = ma, (it’s in equilibrium)

m

dW = F · dq vanishes ∀dq ∈ Rn, (zero virtual work for

δq → 0)

mF = 0, (no force on it!)

If the force is conservative (F = −∇V ) then this isalso equivalent to,

∇V (q0) = 0

that is, we have equilibrium at a critical point of thepotential. The equilibrium will be stable if q0 is alocal minimum of the potential V .

We can summarize all the above by proclaimingthat we have a “principle of least energy” governingstable equilibria. We also have an analogy betweenstatics and dynamics,


stable_equilibrium

unstable_equilibrV

Rn

unstable_equilibrium


Statics Dynamics

equilibrium, a = 0 F = ma

potential, V action, S =

∫ t1

t0

K − V dt

critical points of V critical points of S

d

dt

(∂T

∂qi

)− ∂T

∂qi= Qi. (1.11)


d

dt

(∂L

∂qi

)− ∂L

∂qi= 0. (1.12)

D’Alembert’s principle is an expression forNewton’s second law under conditions where thevirtual work done by the forces of constraint is zero.

1.2.3 The Principle of Least Time

Time now to look at the second piece of historysurrounding the principles of Lagrangian mechanics.As well as hints from statics, there were also hintsfrom the behavior of light, hints again that naturelikes to minimize effort. In a vacuum light moves instraight lines, which in Euclidean space is theminimum distance. But more interesting thanstraight lines are piecewise straight paths and curves.Consider reflection of light from a mirror,

What path does the light take? The empiricalanswer was known since antiquity, it chooses B suchthat θ1 = θ2, so the angle of incidence equals theangle of reflection. But this is precisely the path that


q1 q2

A

B

C

C’

q2

minimizes the distance of the trajectory subject tothe condition that it must hit the mirror (at least atone point). In fact light traveling from A to B takesboth the straight paths ABC and AC. Why is ABC theshortest path hitting the mirror? This follows fromsome basic Euclidean geometry:

B minimizes AB +BC ⇔ B minimizes AB +BC ′

⇔ A,B,C ′ lie on a line

⇔ θ1 = θ2

Note the introduction of the fictitious image C′

“behind” the mirror, this is a trick often used insolving electrostatic problems (a conducting surfacecan be replaced by fictitious mirror image charges to


satisfy the boundary conditions), it is also used ingeophysics when one has a geological fault, and inhydrodynamics when there is a boundary betweentwo media (using mirror image sources and sinks).

The big clue leading to D’Alembert’s principlehowever came from refraction of light. Snell (and

q1

q2

medium 1

medium 2

predecessors) noted that each medium has somenumber n associated with it, called the index ofrefraction, such that,

n1 sin θ1 = n2 sin θ2

(having normalized n so that for a vacuum n = 1).Someone guessed the explanation, realizing that if


the speed of light in a medium is proportional to1/n, then light will satisfy Snell’s law if the lightminimizes the time it takes to get from A to C. Inthe case of refraction it is the time that is important,not just the path distance. But the same is true forthe law of reflection, since in that case the path ofminimum length gives the same results as the path ofminimum time.

So, not only is light the fastest thing around, it’salso always taking the quickest path from here tothere!

1.2.4 How D’Alembert and OthersGot to the Truth

D’Alembert’s principle of virtual work for staticssays that equilibrium occurs when

F (q0) · δq = 0, ∀δq ∈ Rn

D’Alembert generalized this to dynamics byinventing what he called the “inertia force”=−ma,and he postulated that in dynamics equilibrium


occurs when the total force = F+inertia force,vanishes. Or symbolically when,(

F (q(t))−ma(t))· δq(t) = 0 (1.13)

We then take a variational path parameterized by s,

qs(t) = q(t) + s δq(t)

where

δq(t0) = δq(t1) = 0

and with these paths, for any function f on the spaceof paths we can define the variational derivative,

δf :=d

dsf(qs)

∣∣∣s=0

(1.14)

Then D’Alembert’s principle of virtual work implies∫ t1

t0

(F (q)−mq

)·δq dt = 0


for all δq, so if F = −∇V , we get

0 =

∫ t1

t0

(−∇V (q)−mq

)· δq dt

=

∫ t1

t0

(−∇V (q) · δq +mqδq

)dt

using

d

dsV (qs(t))

∣∣∣∣s=0

=dV

dq

dqs(t)

ds

∣∣∣∣s=0

and

δ(q2) =dq2

s(t)

ds


then we have

0 =

∫ t1

t0

(−dVdq

dqsdt

+m

2

dqs(t)

ds

)dt

=d

ds

∫ t1

t0

(− d

dtV(qs(t)

)+m

2qs(t)

)∣∣∣∣s=0

dt

= δ

(∫ t1

t0

(− d

dtV(qs(t)

)+m

2qs(t)

2)dt

)therefore

δ(∫ t1

t0

(−V (q) +K

)dt)

= 0

so the path taken by the particle is a critical point ofthe action,

S(q) =

∫(K − V ) dt (1.15)

We’ve described how D’Alembert might havearrived at the principle of least action by generalizingpreviously known energy minimization and least timeprinciples.

(In general relativity, one sees that gravity is aninertia force!)


Chapter 2

Equations of Motion

(Week 2, April 4, 6, 8.)

In this chapter we’ll start to look at theLagrangian equations of motion in more depth. We’lllook at some specific examples of problem solvingusing the Euler-Lagrange equations. First we’ll showhow the equations are derived.

29

30 Equations of Motion

2.1 The Euler-Lagrange

Equations

Consider a classical system whose configurationspace is a manifold Q. For example, the space for aspherical double pendulum would look like Fig. 2.1,where Q = S2 × S2. So our system is “a particle in

Q=S2xS2

Figure 2.1: Double pendulum configuration space.

Q.”As time passes the system traces out a path

q : [t0, t1] −→ Q

and we define it’s velocity,

q(t) ∈ Tq(t)Q

2.1 The Euler-Lagrange Equations 31

to be the tangent vector at q(t) given by theequivalence class [σ] of curves through q(t) withderivatives σ(t) = dq(s)/ds|s=t. We’ll just write thisas q(t).

Let Γ be the space of smooth paths from a ∈ Q tob ∈ Q,

Γ = {q : [t0, t1]→ Q|q(t0) = a, q(t1) = b}

(Γ is an infinite dimensional manifold, but we won’tgo into that for now.) Let the Lagrangian=L for thesystem be any smooth function of position andvelocity (not explicitly of time, for simplicity),

L : TQ −→ R

and define the action, S:

S : Γ −→ R

by

S(q) :=

∫ t1

t0

L(q, q) dt (2.1)


The path that the quasi particle will actually take isa critical point of S, in accord with D’Alembert’sprinciple of least action. In other words, a path q ∈ Γsuch that for any smooth 1-parameter family ofpaths qs ∈ Γ with q0 = q1, we have

d

dsS(qs)

∣∣∣s=0

= 0 (2.2)

We write,d

ds

∣∣∣s=0

as “δ′′

so Eq.(2.2) can be rewritten

δS = 0 (2.3)

2.1.1 Comments

What is a “1-parameter family of paths”? Well, apath is a curve, or a 1D manifold. So the 1-parameterfamily is a set of well-defined paths {qs}, each onelabeled by a parameter s. So a smooth 1-parameterfamily of paths will have q(s) everywhereinfinitesimally close to q(s+ ε) for an infinitesimal


q0

q_s

Figure 2.2: Schematic of a 1-parameter family ofcurves.

hyperreal ε. So in Fig. 2.2 we can go from q0 to qs bysmoothly varying s from s = 0 to s = s

What does the condition δS = 0 imply? Patience,we are just getting to that. We will now start toexplore what δS = 0 means for our Lagrangian.

2.1.2 Lagrangian Dynamics

We were given that Q is a manifold, so it admits acovering of coordinate charts. For now, let’s pickcoordinates in a neighborhood U of some pointq(t) ∈ Q. Next, consider only variations qs such thatqs = q outside U . A cartoon of this looks like Fig. 2.3Then we restrict attention to a subinterval


a

bU

Q

Figure 2.3: Local path variation.

[t′0, t′1] ⊆ [t0, t1] such that qs(t) ∈ U for t′0 ≤ t ≤ t′1.

Let’s just go ahead and rename t′0 and t′1 as “ t0and t1” to drop the primes. We can use thecoordinate charts on U ,

ϕ : U −→ Rn

x 7−→ ϕ(x) = (x1, x2, . . . , xn)

and we also have coordinates for the 1-forms,

dϕ : TU −→ TRn ∼= Rn × Rn

(x, y) 7−→ dϕ(x, y) = (x1, . . . , xn, y1, . . . , yn)


where y ∈ TxQ. We restrict L : TM→ R toTU ⊆ TM, in our case the manifold is M = Q, andthen we can describe it, L, using the coordinatesxi, yi on TU . The i are generalized positioncoordinates, the yi are the associated generalizedvelocity coordinates. (Velocity and position are inthe abstract configuration space Q). Using thesecoordinates we get

δS = δ

∫ t1

t0

L(q(t), q(t)) dt

=

∫ t1

t0

δL(q, q) dt

=

∫ t1

t0

( ∂L∂xi

δqi +∂L

∂yiδqi)dt

where we’ve used the given smoothness of L and theEinstein summation convention for repeated indicesi. Note that we can write δL as above using a localcoordinate patch because the path variations δq areentirely trivial outside the patch for U . Continuing,


using the Leibniz rule

d

dt

(∂L∂yδq)

=d

dt

∂L

∂yδq +

∂L

∂yδq

we have,

δS =

∫ t1

t0

( ∂L∂xi− d

dt

∂L

∂yi

)δqi(t) dt

= 0.

If this integral is to vanish as demanded by δS = 0,then it must vanish for all path variations δq,further, the boundary terms vanish because wedeliberately chose δq that vanish at the endpoints t0and t1 inside U . That means the term in bracketsmust be identically zero, or

d

dt

∂L

∂yi− ∂L

∂xi= 0 (2.4)

This is necessary to get δS = 0, for all δq, but in factit’s also sufficient. Physicists always give thecoordinates xi, yi on TU the symbols “qi” and “qi”,

2.2 Interpretation of Terms 37

despite the fact that these also have anothermeaning, namely the xi and yi coordinates of thequantity, (

q(t), q(t))∈ TU.

So in any case, physicists write,

d

dt

∂L

∂qi=∂L

∂qi

and they call these the Euler-Lagrange equations.

2.2 Interpretation of Terms

The derivation of the Euler-Lagrange equationsabove was fairly abstract, the terms “position” and“velocity” were used but were not assumed to be theusual kinematic notions that we are used to inphysics, indeed the only reason we used those termswas for their analogical appeal. Now we’ll try toilluminate the E-L equations a bit by casting theminto the usual position and velocity terms.


So consider,

Q = R

L(q, q) =1

2mq · q − V (q)

=1

2mqiqi − V (q)

TERMS MEANING MEANING(in this example) (in general)

∂L

∂qimq the momentum pi

∂L

∂qi−(∇V )i the force Fi

When we write ∂V/∂qi = ∇V we’re assuming theqi are Cartesian coordinates on Q.

So translating our example into general terms, ifwe conjure up some abstract Lagrangian then we canthink of the independent variables as generalizedpositions and velocities, and then the Euler-Lagrangeequations can be interpreted as equations relating

2.2 Interpretation of Terms 39

generalized concepts of momentum and force, andthey say that

p = F (2.5)

So there’s no surprise that in the mundane case ofa single particle moving in R3 under time t this justrecovers Newton II. Of course we can do all of ourclassical mechanics with Newton’s laws, it’s just apain in the neck to deal with the redundancies inF = ma when we could use symmetry principles tovastly simplify many examples. It turns out that theEuler-Lagrange equations are one of thereformulations of Newtonian physics that make ithighly convenient for introducing symmetries andconsequent simplifications. Simplifications generallymean quicker, shorter solutions and more transparentanalysis or at least more chance at insight into thecharacteristics of the system. The main thing is thatwhen we use symmetry to simplify the equations weare reducing the number of independent variables, soit gets closer to the fundamental degrees of freedomof the system and so we cut out a lot of the wheatand chaff (so to speak) with the full redundant


Newton equations.One can of course introduce simplifications when

solving Newton’s equations, it’s just that it’s easierto do this when working with the Euler-Lagrangeequations. Another good reason to learn Lagrangian(or Hamiltonian) mechanics is that it translatesbetter into quantum mechanics.

Chapter 3

Lagrangians andNoether’s Theorem

If the form of a system of dynamical equations doesnot change under spatial translations then themomentum is a conserved quantity. When the formof the equations is similarly invariant under timetranslations then the total energy is a conservedquantity (a constant of the equations of motion).Time and space translations are examples of

41

42 Lagrangians and Noether’s Theorem

1-parameter groups of transformations. Invarianceunder a group of transformations is precisely what wemean by a symmetry in group theory. So symmetriesof a dynamical system give conserved quantities orconservation laws. The rigorous statement of all thisis the content of Noether’s theorem.

3.1 Time Translation

To handle time translations we need to replace ourpaths q : [t0, t1]→ Q by paths q : R→ Q, and thendefine a new space of paths,

Γ = {q : R→ Q}.

The bad news is that the action

S(q) =

∫ ∞−∞

L(q(t), q(t)

)dt

typically will not converge, so S is then no longer afunction of the space of paths. Nevertheless, if δq = 0

3.1 Time Translation 43

outside of some finite interval, then the functionalvariation,

δS :=

∫ ∞−∞

d

dsL(qs(t), qs(t)

)∣∣∣∣s=0

dt

will converge, since the integral is smooth andvanishes outside this interval. Moreover, demandingthat this δS vanishes for all such variations δq isenough to imply the Euler-Lagrange equations:

δS =

∫ ∞−∞

d

dsL(qs(t), qs(t)

)∣∣∣∣s=0

dt

=

∫ ∞−∞

(∂L

∂qiδqi +

∂L

∂qiδqi

)dt

=

∫ ∞−∞

(∂L

∂qi− d

dt

∂L

∂qi

)δqi dt

where again the boundary terms have vanished sinceδq = 0 near t = ±∞. To be explicit, the first term in

∂L

∂qiδqi =

d

dt

(∂L

∂qiδqi)−(d

dt

∂L

∂qi

)δq


vanishes when we integrate. Then the whole thingvanishes for all compactly supported smooth δq ifand only if

d

dt

∂L

∂qi=∂L

∂qi.

Recall that,

∂L

∂qi= pi, is the generalized momentum, by defn.

∂L

∂qi= pi, is the force, by the E-L eqns.

Note the similarity to Hamilton’s equations—if youchange L to H you need to stick in a minus sign, andchange variables from q to pi and eliminate pi.

3.1.1 Canonical and GeneralizedCoordinates

In light of this noted similarity with the Hamiltonequations of motion, let’s spend a few momentsclearing up some terminology (I hate using jargon,

3.1 Time Translation 45

but sometimes it’s unavoidable, and sometimes it canbe efficient—provided everyone is clued in).

Generalized Coordinates

For Lagrangian mechanics we have been usinggeneralized coordinates, these are the {qi, qi}. The qiare generalized positions, and the qi are generalizedvelocities. The full set of independent generalizedcoordinates represent the degrees of freedom of aparticle, or system of particles. So if we have Nparticles then we’d typically have 6N generalizedcoordinates (the “6” is for 3 space dimensions, and ateach point a position and a momentum). These canbe in any reference frame or system of axes, so forexample, in a Cartesian frame, with two particles, in3D space we’d have the 2× 3 = 6 positioncoordinates, and 2× 3 = 6 velocities,

{x1, y1, z1, x2, y2, z2}, {u1, v1, w1, u2, v2, w2}

where say u = vx, v = vy, w = vz are the Cartesianvelocity components. This makes 12 = 6× 2 = 6N


coordinates, matching the total degrees of freedom asclaimed. If we constrain the particles to move in aplane (say place them on a table in a gravitationalfield) then we get 2N fewer degrees of freedom, andso 4N d.o.f. overall. By judicious choice of coordinateframe we can eliminate one velocity component andone position component for each particle.

It is also handy to respect other symmetries of asystem, maybe the particles move on a sphere forexample, one can then define new positions andmomenta with a consequent reduction in the numberof these generalized coordinates needed to describethe system.

Canonical Coordinates

In Hamiltonian mechanics (which we have not yetfully introduced) we will find it more useful totransform from generalized coordinates to canonicalcoordinates. The canonical coordinates are a specialset of coordinates on the cotangent bundle of theconfiguration space manifold Q. They are usually

3.2 Symmetry and Noether’s Theorem 47

written as a set of (qi, pj) or (xi, pj) with the x’s orq’s denoting the coordinates on the underlyingmanifold and the p’s denoting the conjugatemomentum, which are 1-forms in the cotangentbundle at the point q in the manifold.

It turns out that the qi together with the pj, forma coordinate system on the cotangent bundle T ∗Q ofthe configuration space Q, hence these coordinatesare called the canonical coordinates.

We will not discuss this here, but if you care toknow, later on we’ll see that the relation between thegeneralized coordinates and the canonicalcoordinates is given by the Hamilton-Jacobiequations for a system.

3.2 Symmetry and Noether’s

Theorem

First, let’s give a useful definition that will make iteasy to refer to a type of dynamical systemsymmetry. We want to refer to symmetry


transformations (of the Lagrangian) governed by asingle parameter.

Definition 3.1 (one-parameter family ofsymmetries). A 1-parameter family of symmetries ofa Lagrangian system L : TQ→ R is a smooth map,

F : R× Γ −→ Γ

(s, q) 7−→ qs, with q0 = q

such that there exists a function `(q, q) for which

δL =d`

dt

for some ` : TQ→ R, that is,

d

dsL(qs(t), qs(t)

)∣∣∣∣s=0

=d

dt`(qs(t), qs(t)

)for all paths q.

Remark: The simplest case is δL = 0, in whichcase we really have a way of moving paths around


(q 7→ qs) that doesn’t change the Lagrangian—i.e., asymmetry of L in the most obvious way. ButδL = d

dt` is a sneaky generalization whose usefulness

will become clear.

3.2.1 Noether’s Theorem

Here’s a statement of the theorem. Note that ` inthis theorem is the function associated with F indefinition 3.1.

Theorem 3.1 (Noether’s Theorem). Suppose F is aone-parameter family of symmetries of theLagrangian system, L : TQ→ R. Then,

piδqi − `

is conserved, that is, it’s time derivative is zero forany path q ∈ Γ satisfying the Euler-Lagrangeequations. In other words, in boring detail,

d

dt

[∂L

∂yi(q(s)q(s)

) d

dsqis(t)

∣∣∣∣s=0

− `(q(t), q(t)

)]= 0


Proof.

d

dt

(piδq

i − `)

= piδqi + piδq

i − d

dt`

=∂L

∂qiδqi +

∂L

∂qiδqi − δL

= δL− δL = 0.

“Okay, big deal” you might say. Before this can beof any use we’d need to find a symmetry F . Thenwe’d need to find out what this piδqi − ` business isthat is conserved. So let’s look at some examples.

Example

1. Conservation of Energy. (The mostimportant example!)

All of our Lagrangian systems will have timetranslation invariance (because the laws ofphysics do not change with time, at least not to


any extent that we can tell). So we have aone-parameter family of symmetries

qs(t) = q(t+ s)

This indeed gives,

δL = L

for

d

dsL(qs)

∣∣∣∣s=0

=d

dtL = L

so here we take ` = L simply! We then get theconserved quantity

piδqi − ` = piq

i − L

which we normally call the energy. Forexample, if Q = Rn, and if

L =1

2mq2 − V (q)


then this quantity is

mq · q −(1

2mq · q − V

)=

1

2mq2 + V (q)

The term in parentheses is K − V , and theleft-hand side is K + V .

Let’s repeat this example, this time with a specificLagrangian. It doesn’t matter what the Lagrangianis, if it has 1-parameter families of symmetries thenit’ll have conserved quantities, guaranteed. The trickin physics is to write down a correct Lagrangian inthe first place! (Something that will accuratelydescribe the system of interest.)

3.3 Conserved Quantities from

Symmetries

We’ve seen that any 1-parameter family

Fs : Γ −→ Γ

q 7−→ qs


which satisfies

δL = ˙

for some function ` = `(q, q) gives a conservedquantity

piδqi − `

As usual we’ve defined

δL :=d

dsL(qs(t), qs(t)

)∣∣∣∣s=0

Let’s see how we arrive at a conserved quantity froma symmetry.

3.3.1 Time Translation Symmetry

For any Lagrangian system, L : TQ→ R, we have a1-parameter family of symmetries

qs(t) = q(t+ s)

because

δL = L


so we get a conserved quantity called the total energyor Hamiltonian,

H = piqi − L (3.1)

(You might prefer “Hamiltonian” to “total energy”because in general we are not in the sameconfiguration space as Newtonian mechanics, if youare doing Newtonian mechanics then “total energy”is appropriate.)

For example: a particle on Rn in a potential V hasQ = Rn, L(q, q) = 1

2mq2 − V (q). This system has

piqi =

∂L

∂qiqi = mq2 = 2K

so

H = piqi − L = 2K − (K − V ) = K + V

as you’d have hoped.

3.3.2 Space Translation Symmetry

For a free particle in Rn, we have Q = Rn andL = K = 1

2mq2. This has spatial translation


symmetries, so that for any v ∈ Rn we have thesymmetry

qs(t) = q(t) + s v

withδL = 0

because δq = 0 and L depends only on q not on q inthis particular case. (Since L does not depend uponqi we’ll call qi an ignorable coordinate; as above,these ignorables always give symmetries, henceconserved quantities. It is often useful therefore, tochange coordinates so as to make some of themignorable if possible!)

In this example we get a conserved quantity calledmomentum in the v direction:

piδqi = mqiv

i = mq · v

Aside: Note the subtle difference between two usesof the term “momentum”; here it is a conservedquantity derived from space translation invariance,but earlier it was a different thing, namely themomentum ∂L/∂qi = pi conjugate to qi. These two


different “momentum’s” happen to be the same inthis example!

Since this is conserved for all v we say thatmq ∈ Rn is conserved. (In fact that whole Lie groupG = Rn is acting as a translation symmetry group,and we’re getting a q(= Rn)-valued conservedquantity!)

3.3.3 Rotational Symmetry

The free particle in Rn also has rotation symmetry.Consider any X ∈ so(n) (that is a skew-symmetricn× n matrix), then for all s ∈ R the matrix esX is inSO(n), that is, it describes a rotation. This gives a1-parameter family of symmetries

qs(t) = esXq(t)

which has

δL =∂L

∂qiδqi +

∂L

∂qiδqi = mqiδq

i


now qi is ignorable and so ∂L/∂qi = 0, and∂L/∂qi = pi, and

δqi =d

dsqis

∣∣∣∣s=0

=d

ds

d

dt

(esXq

)∣∣∣∣s=0

=d

dtX q

= X q

So,

δL = mqiXij qj

= mq · (X q)

= 0

since X is skew symmetric as stated previously(X ∈ so(n)). So we get a conserved quantity, theangular momentum in the X direction.

(Note: this whole bunch of math above for δL justsays that the kinetic energy doesn’t change when thevelocity is rotated, without changing its magnitude.)


We write,piδq

i = mqi · (X q)i

(δqi = Xq just as δqi = Xq in our previouscalculation), or if X has zero entries except in ij andji positions, where it’s ±1, then we get

m(qiqj − qjqi)

the “ij component of angular momentum”. If n = 3we write these as,

mq×q

Note that above we have assumed one canconstruct a basis for so(n) using matrices of the formassumed for X, i.e., skew symmetric with ±1 in therespectively ij and ji elements, otherwise zero.

The above examples were for a particularly simplesystem, a free particle, for which the Lagrangian isjust the kinetic energy, since there is no potentialenergy variation for a free particle. We’d like toknow how to solve more complicated dynamics.

The general idea is to guess the kinetic energy andpotential energy of the particle (as functions of your

3.4 Example Problems 59

generalized positions and velocities) and then let

L = K − V

3.4 Example Problems

(Week 3, Apr. 11, 13, 15.)

3.4.1 The Atwood Machine

A frictionless pulley with two masses, m1 and m2,hanging from it. We have

m2

l x

x

m1


K =1

2(m1 +m2)(

d

dt(`− x))2 =

1

2(m1 +m2)x2

V = −m1gx−m2g(`− x)

so

L = K − V =1

2(m1 +m2)x2 +m1gx+m2g(`− x)

The configuration space is Q = (0, `), and x ∈ (0, `).Moreover TQ = (0, `)× R 3 (x, x). As usualL : TQ→ R. Note that solutions of theEuler-Lagrange equations will only be defined forsome time t ∈ R, as eventually the solutions reachesthe “edge” of Q.

The momentum is:

p =∂L

∂x= (m1 +m2)x

and the force is:

F =∂L

∂x= (m1 −m2)g


The Euler-Lagrange equations say

p = F

(m1 +m2)x = (m1 −m2)g

x =m1 −m2

m1 +m2

g

So this is like a falling object in a downwards

gravitational acceleration a =(m1−m2

m1+m2

)g.

We integrate the expression for x twice to obtainthe general solution to the motion x(t). Note thatx = 0 when m1 = m2, and x = g if m2 = 0.

3.4.2 Disk Pulled by Falling Mass

Consider next a disk pulled across a table by a fallingmass. The disk is free to move on a frictionlesssurface, and it can thus whirl around the hole towhich it is tethered to the mass below.

Here Q = open disk of radius `, minus it’s center

= (0, `)× S1 3 (r, θ)


r m1

m2

l rno swinging

TQ = (0, `)× S1 × R× R 3 (r, θ, r, θ)

K =1

2m1(r2 + r2θ2) +

1

2m2(

d

dt(`− r))2

V = gm2(r − `)

L =1

2m1(r2 + r2θ2) +

1

2m2r

2 + gm2(`− r)

having noted that ` is constant so d/dt(`− r) = −r.For the momenta we get,

pr =∂L

∂r= (m1 +m2)r

pθ =∂L

∂θ= m1r

2θ.

Note that θ is an “ignorable coordinate”—it doesn’tappear in L—so there’s a symmetry, rotational


symmetry, and pθ, the conjugate momentum, isconserved.

The forces are,

Fr =∂L

∂r= m1rθ

2 − gm2

Fθ =∂L

∂θ= 0, (θ is ignorable)

Note: in Fr the term m1rθ2 is recognizable as a

centrifugal force, pushing m1 radially out, while theterm −gm2 is gravity pulling m2 down and thuspulling m1 radially in.

So, the Euler-Lagrange equations give,

pr = Fr, (m1 +m2)r = m1rθ2 −m2g

pθ = 0, pθ = m1r2θ = J = a constant.

Let’s use our conservation law here to eliminate θfrom the first equation:

θ =J

m1r2


so

(m1 +m2)r =J2

m1r3−m2g

Thus effectively we have a particle on (0, `) of massm = m1 +m2 feeling a force

Fr =J2

m1r3−m2g

which could come from an “effective potential” V (r)such that dV/dr = −Fr. So integrate −Fr to findV (r):

V (r) =J2

2m1r2+m2gr

this is a sum of two terms that look like Fig. 3.1If θ(t = 0) = 0 then there is no centrifugal force

and the disk will be pulled into the hole. Thesolution is valid up until the time the disk reachesthe hole.

At r = r0, the minimum of V (r), our disc mass m1

will be in a stable circular orbit of radius r0 (whichdepends upon J). Otherwise we get orbits likeFig. 3.2.


r0

V(r)

attractive gra

repulsive centrif

r

Figure 3.1: Potential function for disk pulled by grav-itating mass.

3.4.3 Free Particle in SpecialRelativity

In relativistic dynamics the parameter coordinatethat parametrizes the particle’s path in Minkowskispacetime need not be the “time coordinate”, indeedin special relativity there are many allowed timecoordinates.

Minkowski spacetime is,

Rn+1 3 (x0, x1, . . . , xn)

if space is n-dimensional. We normally take x0 as“time”, and (x1, . . . , xn) as “space”, but of course


Figure 3.2: Orbits for the disc and gravitating masssystem.

this is all relative to one’s reference frame. Someoneelse travelling at some high velocity relative to uswill have to make a Lorentz transformation totranslate from our coordinates to theirs.

This has a Lorentzian metric

g(v, w) = v0w0 − v1w1 − . . .− vnwn

= ηµνvµwν


where

ηµν =

1 0 0 . . . 00 −1 0 . . . 00 0 −1 0...

.... . .

...0 0 0 . . . −1

In special relativity we take spacetime to be the

configuration space of a single point particle, so welet Q be Minkowski spacetime, i.e.,Rn+1 3 (x0, . . . , xn) with the metric ηµν definedabove. Then the path of the particle is,

q : R(3 t) −→ Q

where t is a completely arbitrary parameter for thepath, not necessarily x0, and not necessarily propertime either. We want some Lagrangian L : TQ→ R,i.e., L(qi, qi) such that the Euler-Lagrange equationswill dictate how our free particle moves at a constantvelocity. Many Lagrangians do this, but the “best”ones(s) give an action that is independent of the


parameterization of the path—since theparameterization is “unphysical” (it can’t bemeasured). So the action

S(q) =

∫ t1

t0

L(qi(t), qi(t)

)dt

for q : [t0, t1]→ Q, should be independent of t. Theobvious candidate for S is mass times arclength,

S = m

∫ t1

t0

√ηij qi(t)qj(t) dt

or rather the Minkowski analogue of arclength, calledproper time, at least when q is a timelike vector, i.e.,ηij q

iqj > 0, which says q points into the future (orpast) lightcone and makes S real, in fact it’s then thetime ticked off by a clock moving along the pathq : [t0, t1]→ Q. In Euclidean space, free particlesfollow straight paths, so the arclength or pathlengthvariation is an extremum, and we expect the samebehavior in Minkowski spacetime. Also, thearclength does not depend upon the


Lightlike

Timelike

Spacelik

parameterization, and lastly, the mass m merelyprovides the correct units for ‘action’.

So let’s takeL = m

√ηij qiqj (3.2)

and work out the Euler-Lagrange equations. We have

pi =∂L

∂qi= m

∂

∂qi

√ηij qiqj

= m2ηij q

j

2√ηij qiqj

= mηij q

j√ηij qiqj

=mqi‖q‖

(Note the numerator is “mass times 4-velocity”, atleast when n = 3 for a real single particle system, butwe’re actually in a more general n+ 1-dim spacetime,


so it’s more like the “mass times n+ 1-velocity”).Now note that this pi doesn’t change when wechange the parameter to accomplish q 7→ αq. TheEuler-Lagrange equations say,

pi = Fi =∂L

∂qi= 0

The meaning of this becomes clearer if we use“proper time” as our parameter (like parameterizinga curve by its arclength) so that∫ t1

t0

‖q‖dt = t1 − t0, ∀ t0, t1

which fixes the parametrization up to an additiveconstant. This implies ‖q‖ = 1, so that

pi = mqi‖q‖

= mqi

and the Euler-Lagrange equations say

pi = 0⇒ mqi = 0

so our (free) particle moves unaccelerated along astraight line, which is as we desired (expected).


Comments

This Lagrangian from Eq.(3.2) has lots of symmetriescoming from reparameterizing the path, so Noether’stheorem yields lots of conserved quantities for therelativistic free particle. This is in fact called “theproblem of time” in general relativity. Here we see itstarting to show up in special relativity.

These reparameterization symmetries work asfollows. Consider any (smooth) 1-parameter familyof reparameterizations, i.e., diffeomorphisms

fs : R −→ R

with f0 = 1R. These act on the space of pathsΓ = {q : R→ Q} as follows: given any q ∈ Γ we get

qs(t) = q(fs(t)

)where we should note that qs is physicallyindistinguishable from q. Let’s show that

δL = ˙, (when E-L eqns. hold)


so that Noether’s theorem gives a conserved quantity

piδqi − `

Here we go then:

δL =∂L

∂qiδqi +

∂L

∂qiδqi

= piδqi

=mqi‖q‖

d

dsqi(fs(t)

)∣∣∣∣s=0

=mqi‖q‖

d

dt

d

dsqi(fs(t)

)∣∣∣∣s=0

=mqi‖q‖

d

dtqi(fs(t)

)fs(t)ds

∣∣∣∣s=0

=mqi‖q‖

d

dt

(qiδfs

)=

d

dt

(piq

iδf)

where in the last step we used the E-L eqns., i.e.ddtpi = 0, so δL = ˙ with ` = piq

iδf .


So to recap a little: we saw the free relativisticparticle has

L = m‖q‖ = m√ηij qiqj

and we’ve considered reparameterization symmetries

qs(t) = q(fs(t)

), fs : R→ R

we’ve used the fact that

δqi :=d

dsqi(fs(t)

)∣∣∣∣s=0

= qiδf

so (repeating a bit of the above)

δL =∂L

∂qiδqi +

∂L

∂qiδqi

= piδqi, (since ∂L/∂qi = 0, and ∂L/∂qi = p)

= piδqi

= pid

dtδqi

= pid

dtqiδf

=d

dtpiq

iδf, and set piqiδf = `


so Noether’s theorem gives a conserved quantity

piδqi − ` = piq

iδf − piqiδf= 0

So these conserved quantities vanish! In short,we’re seeing an example of what physicists call gaugesymmetries. This is a good topic for starting a newsection.

3.5 Electrodynamics and

Relativistic Lagrangians

We will continue the story of symmetry andNoether’s theorem from the last section with a fewmore examples. We use principles of least action toconjure up Lagrangians for our systems, realizingthat a given system may not have a uniqueLagrangian but will often have an obvious naturalLagrangian. Given a Lagrangian we derive equationsof motion from the Euler-Lagrange equations.Symmetries of L guide us in finding conserved


quantities, in particular Hamiltonians from timetranslation invariance, via Noether’s theorem.

This section also introduces gauge symmetry, andthis is where we begin.

3.5.1 Gauge Symmetry andRelativistic Hamiltonian

What are gauge symmetries?

1. These are symmetries that permute differentmathematical descriptions of the same physicalsituation—in this case reparameterizations of apath.

2. These symmetries make it impossible tocompute q(t) given q(0) and q(0): since if q(t)is a solution so is q(f(t)) for anyreparameterization f : R→ R. We have a highdegree of non-uniqueness of solutions to theEuler-Lagrange equations.

3. These symmetries give conserved quantitiesthat work out to equal zero!


Note that (1) is a subjective criterion, (2) and (3)are objective, and (3) is easy to test, so we often use(3) to distinguish gauge symmetries from physicalsymmetries.

3.5.2 Relativistic Hamiltonian

What then is the Hamiltonian for special relativitytheory? We’re continuing here with the exampleproblem of §3.4.3. Well, the Hamiltonian comes fromNoether’s theorem from time translation symmetry,

qs(t) = q(t+ s)

and this is an example of a reparametrization (withδf = 1), so we see from the previous results that theHamiltonian is zero!

H = 0.

Explicitly, H = piδqi − ` where under q(t)→ q(t+ s)

we have δqi = qiδf , and so δL = d`/dt, which implies` = piδq

i. The result H = 0 follows.


Now you know why people talk about “theproblem of time” in general relativity theory, it’sglimmerings are seen in the flat Minkowski spacetimeof special relativity. You may think it’s nice andsimple to have H = 0, but in fact it means that thereis no temporal evolution possible! So we can’testablish a dynamical theory on this footing! That’sbad news. (Because it means you might have to solvethe static equations for the 4D universe as a whole,and that’s impossible!)

But there is another conserved quantity deservingthe title of “energy” which is not zero, and it comesfrom the symmetry,

qs(t) = q(t) + sw

where w ∈ Rn+1 and w points in some timelikedirection.


qs

q

w

In fact any vector w gives a conserved quantity,

δL =∂L

∂qiδqi +

∂L

∂qiδqi

= piδqi, (since ∂L/∂qi = 0 and ∂L/∂qi = pi)

= pi0 = 0

since δqi = wi, δqi = wi = 0. This is our ˙ fromNoether’s theorem with ` = 0, so Noether’s theoremsays that we get a conserved quantity

piδqi − ` = piw

namely, the momentum in the w direction. We knowp = 0 from the Euler-Lagrange equations, for our free

3.6 Relativistic Particle in an Electromagnetic Field 79

particle, but here we see it coming from spacetimetranslation symmetry;

p =(p0, p1, . . . , pn)

p0 is energy, (p1, . . . , pn) is spatial momentum.

We’ve just about exhausted all the basic stuff thatwe can learn from the free particle. So next we’ll addsome external force via an electromagnetic field.

3.6 Relativistic Particle in an

Electromagnetic Field

The electromagnetic field is described by a 1-form Aon spacetime, A is the vector potential, such that

dA = F (3.3)

is a 2-form containing the electric and magneticfields,

F ij =∂Aj∂xi− ∂Ai∂xj

(3.4)


We’d write (for Q having local charts to Rn+1),

A = A0dx0 + A1dx

1 + . . . Andxn

and then because d2 = 0

dA = dA0dx0 + dA1dx

1 + . . . dAndxn

and since the “Aj” are just functions,

dAj = ∂iAjdxi

using the summation convention and ∂i := ∂/∂xi.The student can easily check that the components forF = F01dx

0 ∧ dx1 + F02dx0 ∧ dx2 + . . ., agrees with

the matrix expression below (at least for 4D).So, for example, in 4D spacetime

F =

0 E1 E2 E3

−E1 0 B3 −B2

−E2 −B3 0 B1

−E3 B2 −B1 0

where E is the electric field and B is the magneticfield. The action for a particle of charge e is

S = m

∫ t1

t0

‖q‖ dt+ e

∫q

A


here

∫ t1

t0

‖q‖ dt = proper time,∫q

A = integral of A along the path q.

Note that since A is a 1-form it can be integrated (itis a linear combination of some basis 1-forms like the{dxi}).

(Week 4, April 18, 20, 22.)

Note that since A is a 1-form we can integrate itover an oriented manifold, but one can also write thepath integral using time t as a parameter, withAiq

i dt the differential, after dqi = qidt.

The Lagrangian in the above action, for a charge ewith mass m in an electromagnetic potential A is

L(q, q) = m‖q‖+ eAiqi (3.5)


so we can work out the Euler-Lagrange equations:

pi =∂L

∂qi= m

qi‖q‖

+ eAi

= mvi + eAi

where v ∈ Rn+1 is the velocity, normalized so that‖v‖ = 1. Note that now momentum is no longermass times velocity ! That’s because we’re in n+ 1-dspacetime, the momentum is an n+ 1-vector.Continuing the analysis, we find the force

Fi =∂L

∂qi=

∂

∂qi

(eAj q

j)

= e∂Aj∂qi

qj

So the Euler-Lagrange equations say (noting that


Ai = Aj

(q(t)

):

p = F

d

dt

(mvi + eAi

)= e

∂Aj∂qi

qj

mdvidt

= e∂Aj∂qi

qj − edAidt

mdvidt

= e∂Aj∂qi

qj − e∂Ai∂qj

qj

= e

(∂Aj∂qi− ∂Ai∂qj

)qj

the term in parentheses is F ij = the electromagneticfield, F = dA. So we get the following equations ofmotion

mdvidt

= eF ij qj, (Lorentz force law) (3.6)

(Usually called the “Lorenz” force law.)


3.7 Alternative Lagrangians

We’ll soon discuss a charged particle Lagrangian thatis free of the reparameterization symmetry. First aparagraph on objects other than point particles!

3.7.1 Lagrangian for a String

So we’ve looked at a point particle and tried

S = m · (arclength) +

∫A

or with ‘proper time’ instead of ‘arclength’, wherethe 1-from A can be integrated over a 1-dimensionalpath. A generalization (or specialization, dependingon how you look at it) would be to consider aLagrangian for an extended object.

In string theory we boost the dimension by +1 andconsider a string tracing out a 2D surface as timepasses (Fig. 3.3).

Can you infer an appropriate action for thissystem? Remember, the physical or

3.7 Alternative Lagrangians 85

becomes

Figure 3.3: Worldtube of a closed string.

physico-philosophical principle we’ve been using isthat the path followed by physical objects minimizesthe “activity” or “aliveness” of the system. Giventhat we presumably cannot tamper with the lengthof the closed string, then the worldtube quantityanalogous to arclength or proper time would be thearea of the worldtube (or worldsheet for an openstring). If the string is also assumed to be a source ofelectromagnetic field then we need a 2-form tointegrate over the 2D worldtube analogous to the1-form integrated over the pathline of the pointparticle. In string theory this is usually the“Kalb-Ramond field”, call it B. To recover


electrodynamic interactions it should beantisymmetric like A, but it’s tensor components willhave two indices since it’s a 2-form. The stringaction can then be written

S = α · (area) + e

∫B (3.7)

We’ve also replaced the point particle mass by thestring tension α [mass·length−1] to obtain the correctunits for the action (since replacing arclength by areameant we had to compensate for the extra lengthdimension in the first term of the above stringaction).

This may still seem like we’ve pulled a rabbit outof a hat. But we haven’t checked that this actionyields sensible dynamics yet! But supposing it does,then would it justify our guesswork and intuition inarriving at Eq.(3.7)? Well by now you’ve probablyrealized that one can have more than one form ofaction or Lagrangian that yields the same dynamics.So provided we supply reasonabe physically realisticheuristics then whatever Lagrangian or action thatwe come up with will stand a good chance of


describing some system with a healthy measure ofphysical verisimilitude.

That’s enough about string for now. The point wasto illustrate the type of reasoning that one can use inconjuring up a Lagrangian. It’s particularly usefulwhen Newtonian theory cannot give us a head start,i.e., in relativistic dynamics and in the physics ofextended particles.

3.7.2 Alternate Lagrangian forRelativistic Electrodynamics

In § 3.6, Eq.(3.5) we saw an example of a Lagrangianfor relativistic electrodynamics that had awkwardreparametrization symmetries, meaning that H = 0and there were non-unique solutions to theEuler-Lagrange equations arising from applyinggauge transformations. This freedom to change thegauge can be avoided.

Recall Eq.(3.5), which was a Lagrangian for acharged particle with reparametrization symmetry

L = m‖q‖+ eAiqi


just as for an uncharged relativistic particle. Butthere’s another Lagrangian we can use that doesn’thave this gauge symmetry:

L =1

2mq · q + eAiq

i (3.8)

This one even has some nice features.

• It looks formally like “12mv2”, familiar from

nonrelativistic mechanics.

• There’s no ugly square root, so it’s everywheredifferentiable, and there’s no trouble with pathsbeing timelike or spacelike in direction, theyare handled the same.

What Euler-Lagrange equations does thisLagrangian yield?

pi =∂L

∂qi= mqi + eAi

Fi =∂L

∂qi= e

∂Aj∂qi

qj


Very similar to before! The E-L eqns. then say

d

dt

(mqi + eAi

)= e

∂Aj∂qi

qj

mqi = eF ij qj

almost as before. (I’ve taken to using F here for theelectromagnetic field tensor to avoid clashing with Ffor the generalized force.) The only difference is thatwe have mqi instead of mvi where vi = qi/‖q‖. So theold Euler-Lagrange equations of motion reduce to thenew ones if we pick a parametrization with ‖q‖ = 1,which would be a parametrization by proper time forexample.

Let’s work out the Hamiltonian for this

L =1

2mq · q + eAiq

i

for the relativistic charged particle in anelectromagnetic field. Recall that for ourreparametrization-invariant Lagrangian

L = m√qiqi + eAiq

i


we got H = 0, time translation was a gaugesymmetry. With the new Lagrangian it’s not! Indeed

H = piqi − L

and now

pi =∂L

∂qi= mqi + eAi

so

H = (mqi + eAi)qi − (1

2mqiq

i + eAiqi)

= 12mqiq

i

Comments. This is vaguely like how anonrelativistic particle in a potential V has

H = piqi − L = 2K − (K − V ) = K + V,

but now the “potential’ V = eAiqi is linear in

velocity, so now

H = piqi − L = (2K − V )− (K − V ) = K.


As claimed H is not zero, and the fact that it’sconserved says ‖q(t)‖ is constant as a function of t,so the particle’s path is parameterized by propertime up to rescaling of t. That is, we’re getting“conservation of speed” rather than some morefamiliar “conservation of energy”. The reason is thatthis Hamiltonian comes from the symmetry

qs(t) = q(t+ s)

instead of spacetime translation symmetry

qs(t) = q(t) + sw, w ∈ Rn+1

the difference is illustrated schematically in Fig. 3.4.Our Lagrangian

L(q, q) = 12m‖q‖2 + Ai(q)q

i

has time translation symmetry if and only if A istranslation invariant (but it’s highly unlikely a givensystem of interest will have A(q) = A(q + sw)). Ingeneral then there’s no conserved “energy” for ourparticle corresponding to translations in time.


w

qs=q(t+s) qs=q+sw

01234

12345

Figure 3.4: Proper time rescaling vs spacetime trans-lation.

3.8 The General Relativistic

Particle

In GR spacetime, Q, is an (n+ 1)-dimensionalLorentzian manifold, namely a smooth(n+ 1)-dimensional manifold with a Lorentzianmetric g. We define the metric as follows.

3.8 The General Relativistic Particle 93

1. For each x ∈ Q, we have a bilinear map

g(x) : TxQ× TxQ −→ R(v, w) 7−→ g(x)(v, w)

or we could write g(v, w) for short.

2. With respect to some basis of TxQ we have

g(v, w) = gijviwj

gij =

1 0 . . . 00 −1 0...

. . ....

0 0 . . . −1

Of course we can write g(v, w) = gijv

iwj in anybasis, but for different bases gij will have adifferent form.

3. g(x) varies smoothly with x.


3.8.1 Free Particle Lagrangian in GR

The Lagrangian for a free point particle in thespacetime Q is

L(q, q) = m√g(q)(q, q)

= m√gij qiqj

just like in special relativity but with ηij replaced bygij. Alternatively we could just as well use

L(q, q) = 12mg(q)(q, q)

= 12mgij q

iqj

The big difference between these two Lagrangiansis that now spacetime translation symmetry (androtation, and boost symmetry) is gone! So there isno conserved energy-momentum (nor angularmomentum, nor velocity of center of energy)anymore!

Let’s find the equations of motion. Suppose then Qis a Lorentzian manifold with metric g andL : TQ→ R is the Lagrangian of a free particle,

L(q, q) = 12mgij q

iqj


We find equations of motion from the Euler-Lagrangeequations, which in this case start from

pi =∂L

∂qi= mgij q

j

The velocity q here is a tangent vector, themomentum p is a cotangent vector, and we need themetric to relate them, via

g : TqM× TqM−→ R(v, w) 7−→ g(v, w)

which gives

TqM−→ T ∗qMv 7−→ g(v,−).

In coordinates this would say that the tangent vectorvi gets mapped to the cotangent vector gijv

j. This islurking behind the passage from qi to the momentummgijv

j.


Getting back to the E-L equations,

pi =∂L

∂qi= mgij q

j

Fi =∂L

∂qi=

∂

∂qi

(12mgjk(q)q

j qk)

= 12m∂igjkq

j qk, (where ∂i =∂

∂qi).

So the Euler-Lagrange equations say

d

dtmgij q

j = 12m∂igjkq

j qk.

The mass factors away, so the motion is independentof the mass! Essentially we have a geodesic equation.

We can rewrite this geodesic equation as follows

d

dtgij q

j = 12∂igjkq

j qk

∴ ∂kgij qkqj + gij q

j = 12∂igjkq

j qk

∴ gij qj =

(12∂igjk − ∂kgij

)qj qk

= 12

(∂igjk − ∂kgij − ∂jgki

)qj qk


where the last line follows by symmetry of themetric, gik = gki. Now let,

Γijk = −(∂igjk − ∂kgij − ∂jgki

)the minus sign being just a convention (so that weagree with everyone else). This defines what we callthe Christoffel symbols Γijk. Then

qi = gij qj = −Γijkq

j qk

∴ qi = −Γijkqj qk.

So we see that q can be computed in terms of q andthe Christoffel symbols Γijk, which is really aparticular type of connection that a Lorentzianmanifold has (the Levi-Civita connection), aconnection is just the rule for parallel transportingtangent vectors around the manifold.

Parallel transport is just the simplest way tocompare vectors at different points in the manifold.This allows us to define, among other things, acovariant derivative.


3.8.2 Charged particle in EM Fieldin GR

We can now apply what we’ve learned inconsideration of a charged particle, of charge e, in anelectromagnetic field with potential A, in ourLorentzian manifold. The Lagrangian would be

L = 12mgjkq

j qk + eAiqi

which again was conjured up be replacing the flatspace metric ηij by the metric for GR gij. Notsurprisingly, the Euler-Lagrange equations then yieldthe following equations of motion,

mqi = −mΓijkqj qk + eF ij q

j.

If you want to know more about Lagrangians forgeneral relativity we recommend the paper byPeldan [?], and also the book by Misner, Thorne &Wheeler [?].


3.9 The Principle of Least

Action and Geodesics

(Week 4, April 18, 20, 22.)

3.9.1 Jacobi and Least Time vs LeastAction

We’ve mentioned that Fermat’s principle of leasttime in optics is analogous to the principle of leastaction in particle mechanics. This analogy is strange,since in the principle of least action we fix the timeinterval q : [0, 1]→ Q. Also, if one imagines a forceon a particle resulting from a potential gradient atan interface as analogous to light refraction then youalso get a screw-up in the analogy (Fig. 3.5).

Nevertheless, Jacobi was able to reinterpret themechanics of a particle as an optics problem andhence “unify” the two minimization principles. First,let’s consider light in a medium with a varying indexof refraction n (recall 1/n ∝ speed of light). Suppose


light

light

faster

slower

particle

particle

slower

faster

n high

n low

V high

V low

Figure 3.5: Least time versus least action.

it’s in Rn with its usual Euclidean metric. If the lights trying to minimize the time, its trying to minimizethe arclength of its path in the metric

gij = n2δij

that is, the index of refraction n : Rn → (0,∞), timesthe usual Euclidean metric

δij =

1 0. . .

0 1

This is just like the free particle in general

relativity (minimizing it’s proper time) except that


now gij is a Riemannian metric

g(v, w) = gijviwj

where g(v, v) ≥ 0

So we’ll use the same Lagrangian:

L(q, q) =√gij(q)qiqj

and get the same Euler-Lagrange equations:

d2qi

dt2+ Γijkq

j qk = 0 (3.9)

if q is parameterized by arclength or more generally

‖q‖ =√gij(q)qiqj = constant.

As before the Christoffel symbols Γ are built fromthe derivatives of the metric g.

Now, what Jacobi did is show how the motion of aparticle in a potential could be viewed as a specialcase of this. Consider a particle of mass m in


Euclidean Rn with potential V : Rn → R. It satisfiesF = ma, i.e.,

md2qi

dt2= −∂iV (3.10)

How did Jacobi see (3.10) as a special case of (3.9)?He considered a particle of energy E and he chosethe index of refraction to be

n(q) =

√2

m

(E − V (q)

)which is just the speed of a particle of energy E whenthe potential energy is V (q), since√

2

m(E − V ) =

√2

m

1

2m‖q‖2 = ‖q‖.

Note: this is precisely backwards compared to optics,where n(q) is proportional to the reciprocal of thespeed of light!! But let’s see that it works.

L =√gij(q)qiqj

=√n2(q)qiqj

=√

2/m(E − V (q))q2


where q2 = q · q is just the usual Euclidean dotproduct, v · w = δijv

iwj. We get the Euler-Lagrangeequations,

pi =∂L

∂qi=

√2

m(E − V ) · q

‖q‖

Fi =∂L

∂qi= ∂i

√2

m(E − V (q)) · ‖q‖

=1

2

−2/m∂iV√2/m(E − V q)

· ‖q‖

Then p = F says,

d

dt

√2/m(E − V (q)) · qi

‖q‖= − 1

m∂iV

‖q‖√2/m(E − V )

Jacobi noticed that this is just F = ma, ormqi = −∂iV , that is, provided we reparameterize q sothat,

‖q‖ =√

2/m(E − V (q)).

Recall that our Lagrangian gives reparameterizationinvariant Euler-Lagrange equations! This is theunification between least time (from optics) and leastaction (from mechanics) that we sought.


3.9.2 The Ubiquity of GeodesicMotion

We’ve seen that many classical systems trace outpaths that are geodesics, i.e., paths q : [t0, t1]→ Qthat are critical points of

S(q) =

∫ t1

t0

√gij qiqj dt

which is proper time when (Q, g) is a Lorentzianmanifold, or arclength when (Q, g) is a Riemannianmanifold. We have

1. The metric at q ∈ Q is,

g(q) : TqQ× TqQ→ R(v, w) 7→ g(v, w)

and it is bilinear.

2. w.r.t a basis of TqQ

g(v, w) = δijviwj


3. g(q) varies smoothly with q ∈ Q.

An important distinction to keep in mind is thatLorentzian manifolds represent spacetimes, whereasRiemannian manifolds represent that we’d normallyconsider as just space.

We’ve seen at least three important things.

(1) In the geometric optics approximation, light inQ = Rn acts like particles tracing out geodesicsin the metric

gij = n(q)2δij

where n : Q→ (0,∞) is the index of refractionfunction.

(2) Jacobi saw that a particle in Q = Rn in somepotential V : Q→ R traces out geodesics in themetric

gij =2

m(E − V )δij

if the particle has energy E (where1 V < E).

1The case V > E, if they exist, would be classically forbiddenregions.


(3) A free particle in general relativity traces out ageodesic on a Lorentzian manifold (Q, q).

In fact all three of these results can be generalized tocover every problem that we’ve discussed!

(1′) Light on any Riemannian manifold (Q, q) withindex of refraction n : Q→ (0,∞) traces outgeodesics in the metric h = n2g.

(2′) A particle on a Riemannian manifold (Q, q)with potential V : Q→ R traces out geodesicsw.r.t the metric

h =2

m(E − V )g

if it has energy E. Lots of physical systemscan be described this way, e.g., the Atwoodmachine, a rigid rotating body (Q = SO(3)),spinning tops, and others. All of these systemshave a Lagrangian which is a quadratic functionof position, so they all fit into this framework.


(3′) Kaluza-Klein Theory. A particle withcharge e on a Lorentzian manifold (Q, q) in anelectromagnetic vector potential follows a pathwith

qi = −Γijkqj qk +

e

mF ij q

j

where

F ij = ∂iAj − ∂jAi

but this is actually geodesic motion on themanifold Q× U(1) where U(1) = {eiθ : θ ∈ R}is a circle.

Let’s examine this last result a bit further. To getthe desired equations for motion on Q× U(1) weneed to given Q× U(1) a cleverly designed metricbuilt from g and A where the amount of“spiralling”—the velocity in the U(1) direction ise/m. The metric h on Q× U(1) is built from g andA in a very simple way. Let’s pick coordinates xi onQ where i ∈ {0, . . . , n} since we’re inn+ 1-dimensional spacetime, and θ is our local


QxU(1)

a geodesic

Q

the apparent

coordinate on S1. The components of h are

hij = gij + AiAj

hθi = hθi = −Aihθθ = 1

Working out the equations for a geodesic in thismetric we get

qi = −Γijkqj qk +

e

mF ij q

j

qθ = 0,

if qθ = e/m

since F ij is part of the Christoffel symbols for h.To summarize this section on least time versus

least action we can say that every problem that


we’ve discussed in classical mechanics can beregarded as geodesic motion!


Chapter 4

From Lagrangians toHamiltonians

In the Lagrangian approach we focus on the positionand velocity of a particle, and compute what theparticle does starting from the Lagrangian L(q, q),which is a function

L : TQ −→ R

where the tangent bundle is the space ofposition-velocity pairs. But we’re led to consider

111

112 From Lagrangians to Hamiltonians

momentum

pi =∂L

∂qi

since the equations of motion tell us how it changes

dpidt

=∂L

∂qi.

4.1 The Hamiltonian

Approach

In the Hamiltonian approach we focus on positionand momentum, and compute what the particle doesstarting from the energy

H = piqi − L(q, q)

reinterpreted as a function of position andmomentum, called the Hamiltonian

H : T ∗Q −→ R

where the cotangent bundle is the space ofposition-momentum pairs. In this approach, position

4.1 The Hamiltonian Approach 113

and momentum will satisfy Hamilton’s equations :

dqi

dt=∂H

∂pi,

dpi

dt= −∂H

∂qi

where the latter is the Euler-Lagrange equation

dpi

dt=∂L

∂qi

in disguise (it has a minus sign since H = pq − L).To obtain this Hamiltonian description of

mechanics rigorously we need to study this map

λ : TQ −→ T ∗Q

(q, q) 7−→ (q, p)

where q ∈ Q, and q is any tangent vector in TqQ (notthe time derivative of something), and p is acotangent vector in T ∗qQ := (TqQ)∗, given by

qλ−→ pi =

∂L

∂qi

So λ is defined using L : TQ→ R. Despiteappearances, λ can be defined in a coordinate-free


TqQ

TQ

T(q,qdot)TqQ

QTqQ

q

Figure 4.1:

way, as follows (referring to Fig. 4.1). We want todefine “ ∂L

∂qi” in a coordinate-free way; it’s the

“differential of L in the vertical direction”—i.e., theqi directions. We have

π : TQ −→ Q

(q, q) 7−→ q

and

dπ : T (TQ) −→ TQ


has kernel1 consisting of vertical vectors :

V TQ = ker dπ ⊆ TTQ

The differential of L at some point (q, q) ∈ TQ is amap from TTQ to R, so we have

(dL)(q,q) ∈ T ∗(q,q)TQ

that is,

dL(q,q) : T(q,q)TQ −→ R.

We can restrict this to V TQ ⊆ TTQ, getting

f : V(q,q)TQ −→ R.

But note

V(q,q)TQ = T (TqQ)

1The kernel of a map is the set of all elements in the domainthat map to the null element of the range, so ker dπ = {v ∈TTQ : dπ(v) = 0 ∈ TQ}.


TqQ

T(q,qdot)TqQ

Qq

V(q,qdot)TQ

and since TqQ is a vector space,

T(q,q)TqQ ∼= TqQ

in a canonical way2. So f gives a linear map

p : TqQ −→ R

that is,

p ∈ T ∗qQ2The fiber TvV at v ∈ V of vector manifold V has the same

dimension as V .


this is the momentum!

(Week 6, May 2, 4, 6.)

Given L : TQ→ T ∗Q, we now know acoordinate-free way of describing the map

λ : TQ −→ T ∗Q

(q, q) 7−→ (q, p)

given in local coordinates by

pi =∂L

∂qi.

We say L is regular if λ is a diffeomorphism from TQto some open subset X ⊆ T ∗Q. In this case we candescribe what our system is doing equally well byspecifying position and velocity,

(q, q) ∈ TQor position and momentum

(q, p) = λ(q, q) ∈ X.We call X the phase space of the system. In practiceoften X = T ∗Q, then L is said to be strongly regular.


4.2 Regular and Strongly

Regular Lagrangians

This section discusses some examples of the abovetheory.

4.2.1 Example: A Particle in aRiemannian Manifold withPotential V (q)

For a particle in a Riemannian manifold (Q, q) in apotential V : Q→ R has Lagrangian

L(q, q) =1

2mgij q

iqj − V (Q)

Here

pi =∂L

∂qi= mgij q

j

so

λ(q, q) =(q,mg(q,−)

)

4.2 Regular and Strongly Regular Lagrangians 119

so3 L is strongly regular in this case because

TqQ −→ T ∗qQ

v −→ g(v,−)

is 1-1 and onto, i.e., the metric is nondegenerate.Thus λ is a diffeomorphism, which in this caseextends to all of T ∗Q.

4.2.2 Example: General RelativisticParticle in an E-M Potential

For a general relativistic particle with charge e in anelectromagnetic vector potential A the Lagrangian is

L(q, q) =1

2mgij q

iqj − eAiqi

and thus

pi =∂L

∂qi= mgij q

iqj + eAi.

3The missing object there “−” is of course any tangent vec-tor, not inserted since λ itself is an operator on tangent vectors,not the result of the operation.


This L is still strongly regular, but now each map

λ |TqQ : TqQ −→ T ∗qQ

q 7−→ mg(q,−) + eA(q)

is affine rather than linear4.

4.2.3 Example: Free GeneralRelativistic Particle withReparameterization Invariance

The free general relativistic particle withreparameterization invariant Lagrangian has,

L(q, q) = m√gij qiqj

This is terrible from the perspective of regularityproperties—it’s not differentiable when gij q

iqj

4All linear transforms are affine, but affine transformationsinclude translations, which are nonlinear. In affine geometrythere is no defined origin. For the example the translation isthe “+eA(q)” part.

4.2 Regular and Strongly Regular Lagrangians 121

vanishes, and undefined when the same is negative.Where it is defined

pi =∂L

∂qi=mgij q

j

‖q‖

(where q is timelike), we can ask about regularity.Alas, the map λ is not 1-1 where defined sincemultiplying q by some number has no effect on p!(This is related to the reparameterizationinvariance—this always happens withreparameterization-invariant Lagrangians.)

4.2.4 Example: A Regular but notStrongly Regular Lagrangian

Here’s a Lagrangian that’s regular but not stronglyregular. Let Q = R and

L(q, q) = f(q)

so that

p =∂L

∂q= f ′(q)


This will be regular but not strongly so if f ′ : R→ Ris a diffeomorphism from R to some proper subsetU ⊂ R. For example, take f(q) = eq sof ′ : R ∼→ (0,∞) ⊂ R. So

L(q, q) = eq //

positive slope

or

L(q, q) =√

1 + q2 //

slope between−1 and 1

and so forth.

4.3 Hamilton’s Equations

Now let’s assume L is regular, so

λ : TQ∼−→ X ⊆ T ∗Q

(q, q) 7−→ (q, p)

4.3 Hamilton’s Equations 123

This lets us have the best of both worlds: we canidentify TQ with X using λ. This lets us treat qi, pi,L, H, etc., all as functions on X (or TQ), thuswriting

qi (function on TQ)

for the function

qi ◦ λ−1 (function on X)

In particular

pi :=∂L

∂qi(Euler-Lagrange eqn.)

which is really a function on TQ, will be treated as afunction on X. Now let’s calculate:

dL =∂L

∂qidqi +

∂L

∂qidqi

= pidqi + pidq

i


while

dH = d(piqi − L)

= qidpi + pidqi − dL

= qidpi + pidqi − (pidq

i + pidqi)

= qidpi − pidqi

so

dH = qidpi − pidqi.

Assume the Lagrangian L : TQ→ R is regular, so

λ : TQ∼−→ X ⊆ T ∗Q

(q, q) 7−→ (q, p)

is a diffeomorphism. This lets us regard both L andthe Hamiltonian H = piq

i − L as functions on thephase space X, and use (qi, qi) as local coordinateson X. As we’ve seen, this gives us

dL = pidqi + pidq

i

dH = qidpi − pidqi.


But we can also work out dH directly, this timeusing local coordinates (qi, pi), to get

dH =∂H

∂pidpi +

∂H

∂qidqi.

Since dpi, dqi form a basis of 1-forms, we conclude:

qi =∂H

∂pi, pi = −∂H

∂qi

These are Hamilton’s Equations.

4.3.1 Hamilton and Euler-Lagrange

Though qi and pi are just functions of X, when theEuler-Lagrange equations hold for some pathq : [t0, t1]→ Q, they will be the time derivatives of qi

and pi. So when the Euler-Lagrange equations hold,Hamilton’s equations describe the motion of a pointx(t) =

(q(t), p(t)

)∈ X. In fact, in this context,

Hamilton’s equations are just the Euler-Lagrange


equations in disguise. The equation

qi =∂H

∂pi

really just lets us recover the velocity q as a functionof q and p, inverting the formula

pi =∂L

∂qi

which gave p as a function of q and q. So we get aformula for the map

λ−1 : X −→ TQ

(q, p) 7−→ (q, q).

Given this, the other Hamilton equation

pi = −∂H∂qi

is secretly the Euler-Lagrange equation

d

dt

∂L

∂qi=∂L

∂qi, or p =

∂L

∂qi


These are the same because

∂H

∂qi=

∂

∂qi(piq

i − L)

= −∂L∂qi

.

Example: Particle in a Potential V (q)

For a particle in Q = Rn in a potential V : Rn → Rthe system has Lagrangian

L(q, q) =m

2‖q‖2 − V (q)

which gives

p = mq

q =p

m, (though really that’s q =

gijpjm

)

and Hamiltonian

H(q, p) = piqi − L =

1

m‖p‖2 −

(‖p‖2

2m− V

)=

1

2m‖p‖2 + V (q).


So Hamilton’s equations say

qi =∂H

∂pi⇒ q =

p

m

pi = −∂H∂qi

⇒ p = −∇V

The first just recovers q as a function of p; the secondis F = ma.

Note on Symplectic Structure

Hamilton’s equations push us toward the viewpointwhere p and q have equal status as coordinates onthe phase space X. Soon, we’ll drop the requirementthat X ⊆ T ∗Q where Q is a configuration space. Xwill just be a manifold equipped with enoughstructure to write down Hamilton’s equationsstarting from any H : X → R.

The coordinate-free description of this structure isthe major 20th century contribution to mechanics: asymplectic structure.

This is important. You might have some particlesmoving on a manifold like S3, which is not


symplectic. So the Hamiltonian mechanics point ofview says that the abstract manifold that you arereally interested in is something different: it must bea symplectic manifold. That’s the phase space X.We’ll introduce symplectic geometry morecompletely in later chapters.

4.3.2 Hamilton’s Equations from thePrinciple of Least Action

Before, we obtained the Euler-Lagrange equations byassociating an “action” S with any q : [t0, t1]→ Qand setting δS = 0. Now let’s get Hamilton’sequations directly by assigning an action S to anypath x : [t0, t1]→ X and setting δS = 0. Note: wedon’t impose any relation between p and q, q! Therelation will follow from δS = 0.

Let P be the space of paths in the phase space Xand define the action

S : P −→ R


by

S(x) =

∫ t1

t0

(piqi −H)dt

where piqi −H = L. More precisely, write our path x

as x(t) =(q(t), p(t)

)and let

S(x) =

∫ t1

t0

[pi(t)

d

dtqi(t)−H

(q(t), p(t)

)]dt

we write ddtqi instead of qi to emphasize that we

mean the time derivative rather than a coordinate inphase space.

Let’s show δS = 0⇔Hamilton’s equations.

δS = δ

∫(piq

i −H)dt

=

∫ (δpiq

i + piδqi − δH

)dt


then integrating by parts,

=

∫ (δpiq

i − piδqi − δH)dt

=

∫ (δpiq

i − piδqi −∂H

∂qiδqi − ∂H

∂piδpi

)dt

=

∫ (δpi

(qi − ∂H

∂pi

)+ δqi

(−pi −

∂H

∂qi

))dt

This vanishes ∀δx = (δq, δp) if and only if Hamilton’sequations

qi =∂H

∂pi, pi = −∂H

∂qi

hold. Just as we hoped.We’ve seen two principles of “least action”:

1. For paths in configuration space Q,δS = 0⇔Euler-Lagrange equations.

2. For paths in phase space X,δS = 0⇔Hamilton’s equations.

Additionally, since X ⊆ T ∗Q, we might consider athird version based on paths in position-velocity


space TQ. But when our Lagrangian is regular wehave a diffeomorphism λ : TQ

∼→ X, so this thirdprinciple of least action is just a reformulation ofprinciple 2. However, the really interesting principleof least action involves paths in the extended phasespace where we have an additional coordinate fortime: X × R.

Recall the action

S(x) =

∫(piq

i −H) dt

=

∫pidqi

dtdt−H dt

=

∫pidq

i −H dt

We can interpet the integrand as a 1-form

β = pidqi −H dt

on X × R, which has coordinates {pi, qi, t}. So anypath

x : [t0, t1] −→ X

4.4 Waves versus Particles—The Hamilton-Jacobi Equations 133

gives a path

σ : [t0, t1] −→ X × Rt 7−→ (x(t), t)

and the action becomes the integral of a 1-form overa curve:

S(x) =

∫pi dq

i −H dt =

∫σ

β

4.4 Waves versus

Particles—The

Hamilton-Jacobi

Equations

(Week 7, May 9, 11, 13.)

In quantum mechanics we discover that everyparticle—electrons, photons, neutrinos, etc.—is awave, and vice versa. Interestingly Newton already


had a particle theory of light (his “corpuscules”) andvarious physicists argued against it by pointing outthat diffraction is best explained by a wave theory.We’ve talked about geometrized optics, anapproximation in which light consists of particlesmoving along geodesics. Here we start with aRiemannian manifold (Q, g) as space, but we use thenew metric

hij = n2gij

where n : Q→ (0,∞) is the index of refractionthroughout space (generally not a constant).


4.4.1 Wave Equations

Huygens considered this same setup (in simplerlanguage) and considered the motion of a wavefront:

//

and saw that the wavefront is the envelope of abunch of little wavelets centered at points along the


big wavefront:

��

��

��

��

balls of radius εcentered at points

of the old wavefront��

In short, the wavefront moves at unit speed in thenormal direction with respect to the “optical metric”h. We can think about the distance function

d : Q×Q −→ [0,∞)

on the Riemannian manifold (Q, h), where

d(q0, q1) = infΥ

(arclength)

where Υ = {paths from q0 to q1}. (Secretly thisd(q0, q1) is the least action—the infimum of action


over all paths from q0 to q1.) Using this we get thewavefronts centered at q0 ∈ Q as the level sets

{q : d(q0, q) = c}

or at least for small c > 0, as depicted in Fig. 4.2.For larger c the level sets can cease to be smooth—we

q0

Figure 4.2:

say a catastrophe occurs—and then the wavefrontsare no longer the level sets. This sort of situation canhappen for topological reasons (as when the wavessmash into each other in the back of Fig. 4.2) and itcan also happen for geometrical reasons (Fig. 4.3).Assuming no such catastrophes occur, we canapproximate the waves of light by a wavefunction:

ψ(q) = A(q)eik d(q,q0)


Figure 4.3:

where k is the wavenumber of the light (i.e., itscolor) and A : Q→ R describes the amplitude of thewave, which drops off far from q0. This becomes theeikonal approximation in optics5 once we figure outwhat A should be.

Hamilton and Jacobi focused on distanced : Q×Q→ [0,∞) as a function of two variables andcalled it W=Hamilton’s principal function. Theynoticed that:

∂

∂qi1W (q0, q1) = (p1)i

5Eikonal comes form the Greek word for ‘image’ or ‘likeness’,in optics the eikonal approximation is the basis for ray tracingmethods.


q0•

p1• //

q1

where p1 is a cotangent vector pointing normal to thewavefronts.

4.4.2 The Hamilton-Jacobi Equations

We’ve seen that in optics, particles of light movealong geodesics, but wavefronts are level sets of thedistance functions:

_k{��% 1 DS_k{��

%2DS

_ekr{�� % + 3:DLSY _ ekrz��%+2:DLSY

_cgkouz��! % ) -27=DJNSW[ _ cgkoty��!%)-27=DJOSW[

qqqqqqqqq88


at least while the level sets remain smooth. In theeikonal approximation, light is described by waves

ψ : Q −→ Cψ(q1) = A(q1)eikW (q0,q1)

where (Q, h) is a Riemannian manifold, h is theoptical metric, q0 ∈ Q is the light source, k is thefrequency and

W : Q×Q −→ [0,∞)

is the distance function on Q, or Hamilton’s principalfunction:

W (q0, q1) = infq∈Υ

S(q)

where Υ is the space of paths from q0 to q and S(q)is the action of the path q, i.e., its arclength. This isbegging to be generalized to other Lagrangiansystems! (At least it is retrospectively, with theadvantage of our historical perspective.) We also sawthat

∂

∂qi1W (q0, q1) = (p1)i,


q0•

p1• //

q1

“points normal to the wavefront”—really the tangentvector

pi1 = hij(p1)j

points in this direction. In fact kp1i is themomentum of the light passing through q1. Thisforeshadows quantum mechanics! After all, inquantum mechanics, the momentum is an operatorthat acts to differentiating the wavefunction.

Jacobi generalized this to the motion of pointparticles in a potential V : Q→ R, using the factthat a particle of energy E traces out geodesics inthe metric

hij =2(E − V )

mgij.

We’ve seen this reduces point particle mechanics tooptics—but only for particles of fixed energy E.Hamilton went further, and we now can go furtherstill.


Suppose Q is any manifold and L : TQ→ R is anyfunction (Lagrangian). Define Hamilton’s principalfunction

W : Q× R×Q× R −→ R

by

W (q0, t0; q1, t1) = infq∈Υ

S(q)

where

Υ ={q : [t0, t1]→ Q, q(t0) = q0, & q(t1) = q1

}and

S(q) =

∫ t1

t0

L(q(t), q(t)

)dt

Now W is just the least action for a path from(q0, t0) to (q1, t1); it’ll be smooth if (q0, t0) and (q1, t1)are close enough—so let’s assume that is true. Infact, we have

∂

∂qi1W (q0, q1) = (p1)i,


(q0, t0)•

p1• //

(q1, t1)

where p1 is the momentum of the particle going fromq0 to q1, at time t1, and

∂W

∂qi0= −(p0)i, (-momentum at time t0)

∂W

∂t1= −H1, (-energy at time t1)

∂W

∂t0= H0, (+energy at time t0)

(H1 = H0 as energy is conserved). These last fourequations are the Hamilton-Jacobi equations. Themysterious minus sign in front of energy was seenbefore in the 1-form,

β = pidqi −H dt

on the extended phase space X × R. Maybe the bestway to get the Hamilton-Jacobi equations is from


this extended phase space formulation. But for nowlet’s see how Hamilton’s principal function W andvariational principles involving least action also yieldthe Hamilton-Jacobi equations.

Given (q0, t0), (q1, t1), let

q : [t0, t1] −→ Q

be the action-minimizing path from q0 to q1. Then

W (q0, t0; q1, t1) = S(q)

Now consider varying q0 and q1 a bit

t0 t1

q

and thus vary the action-minimizing path, getting avariation δq which does not vanish at t0 and t1. We


get

δW = δS

= δ

∫ t1

t0

L(q, q) dt

=

∫ t1

t0

(∂L

∂qiδqi +

∂L

∂qiδqi)dt

=

∫ t1

t0

(∂L

∂qiδqi − piδqi

)dt+ piδq

i

∣∣∣∣t1t0

=

∫ t1

t0

(∂L

∂qi− pi

)δqi dt

the term in parentheses is zero because q minimizesthe action and the Euler-Lagrange equations hold.So we δqi have

δW = p1iδqi1 − p0iδq

i0

and so∂W

∂qi1= p1i, and

∂W

∂qi0= −p0i


These are two of the four Hamilton-Jacobi equations!To get the other two, we need to vary t0 and t1:

t0t1 + ∆t1t0 + ∆t0t1

•••

•

Now change inW will involve ∆t0and ∆t1

(you can imagine ∆t0 < 0 in this figure if you like).

We want to derive the Hamilton-Jacobi equationsdescribing the derivatives of Hamilton’s principalfunction

W (q0, t0; q1, t1) = infq∈Υ

S(q)

where Υ is the space of paths q : [t0, t1]→ Q withq(t0) = q, q(t1) = q1 and

S(q) =

∫ t1

t0

L(q, q) dt


where the Lagrangian L : TQ→ R will now beassumed regular, so that

λTQ −→ X ⊆ T ∗Q

(q, q) 7−→ (q, p)

is a diffeomorphism. We need to ensure that (q0, t0)is close enough to (qi, t1) that there is a unique q ∈ Υthat minimizes the action S, and assume that this qdepends smoothly on U = (q0, t0; q1, t1) ∈ (Q× R)2.We’ll think of q as a function of U :

(t0, q0)

(t1, q1)q •

•

(Q× R)2 → Υu 7→ q

defined only when (q0, t0) and(q1, t1) are sufficiently close.


Then Hamilton’s principal function is

W (u) := W (q0, t0; q1, t1) = S(q)

=

∫ t1

t0

L(q, q) dt

=

∫ t1

t0

(pq −H(q, p)

)dt

=

∫ t1

t0

p dq −H dt

=

∫C

β

where β = pdq −H(q, p)dt is a 1-form on theextended phase space X × R, and C is a curve in theextended phase space:

C(t) =(q(t), p(t), t

)∈ X × R.

Note that C depends on the curve q ∈ Υ, which inturn depends upon u = (q0, t0; q1, t1) ∈ (Q× R)2. Weare after the derivatives of W that appear in the


Hamilton-Jacobi relations, so let’s differentiate

W (u) =

∫C

β

with respect to u and get the Hamilton-Jacobiequations from β. Let us be a 1-parameter family ofpoints in (Q× R)2 and work out

d

dsW (us) =

d

ds

∫Cs

β

where Cs depends on us as above

Cs** ••

C0

,,•

•

X × RAs

VV BsWW................................

................................

................................

...


Let’s compare∫Cs

β and

∫As+Cs+Bs

=

∫As

β +

∫Cs

β +

∫Bs

β

Since C0 minimizes the action among paths with thegiven end-points, and the curve As + Cs +Bs has thesame end-points, we get

d

ds

∫As+Cs+Bs

β = 0

(although As + Cs +Bs is not smooth, we canapproximate it by a path that is smooth). So

d

ds

∫Cs

β =d

ds

∫Bs

β − d

ds

∫As

β at s = 0.

Note

d

ds

∫As

β =d

ds

∫β(A′r) dr

= β(A′0)


where A′0 = v is the tangent vector of As at s = 0.Similarly,

d

ds

∫Bs

β = β(w)

where w = B′0. So,

d

dsW (us) = β(w)− β(v)

where w keeps track of the change of (q1, p1, t1) as wemove Cs and v keeps track of (q0, p0, t0). Now sinceβ = pidqi −Hdt, we get

∂W

∂qi1= pi1

∂W

∂t1= −H

and similarly

∂W

∂qi0= −pi0

∂W

∂t0= H


So, if we define a wavefunction:

ψ(q0, t0; q1, t1) = eiW (q0,t0;q1,t1)/~

then we get

∂ψ

∂t1= − i

~H1ψ

∂ψ

∂qi1=i

~p1ψ

At the time of Hamilton and Jacobi’s research thiswould have been new... but nowadays it isthoroughly familiar from quantum mechanics !

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