classical mechanics - baez

8/2/2019 Classical Mechanics - Baez

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Lectures on

Classical Mechanics

by

John C. Baez

notes by Derek K. Wise

Department of Mathematics

University of California, Riverside

LaTeXed by Blair Smith

Department of Physics and Astronomy

Louisiana State University

2005


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c2005 John C. Baez & Derek K. Wise


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Preface

These are notes for a mathematics graduate course on classical mechanics at U.C.Riverside. Ive taught this course three times recently. Twice I focused on the Hamiltonianapproach. In 2005 I started with the Lagrangian approach, with a heavy emphasis onaction principles, and derived the Hamiltonian approach from that. This approach seemsmore coherent.

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

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

Later, Blair Smith from Louisiana State University miraculously appeared and volun-teered to turn the notes into LATEX . While not yet the book Id eventually like to write,the result may already be helpful for people interested in the mathematics of classicalmechanics.

The chapters in this LATEX version are in the same order as the weekly lectures, butIve merged weeks together, and sometimes split them over chapter, to obtain a moretextbook feel to these notes. For reference, the weekly lectures are outlined here.

Week 1: (Mar. 28, 30, Apr. 1)The Lagrangian approach to classical mechanics:deriving F = ma from the requirement that the particles path be a critical point of

the action. The prehistory of the Lagrangian approach: DAlemberts principle of leastenergy in statics, Fermats principle of least time in optics, and how DAlembertgeneralized his principle from statics to dynamics using the concept of inertia force.

Week 2: (Apr. 4, 6, 8)Deriving the Euler-Lagrange equations for a particle on anarbitrary manifold. Generalized momentum and force. Noethers theorem on conservedquantities coming from symmetries. Examples of conserved quantities: energy, momen-tum and angular momentum.

Week 3 (Apr. 11, 13, 15)Example problems: (1) The Atwood machine. (2) Africtionless mass on a table attached to a string threaded through a hole in the table, witha mass hanging on the string. (3) A special-relativistic free particle: two Lagrangians, one

with reparametrization invariance as a gauge symmetry. (4) A special-relativistic chargedparticle in an electromagnetic field.Week 4 (Apr. 18, 20, 22)More example problems: (4) A special-relativistic charged

particle in an electromagnetic field in special relativity, continued. (5) A general-relativisticfree particle.

Week 5 (Apr. 25, 27, 29)How Jacobi unified Fermats principle of least time andLagranges principle of least action by seeing the classical mechanics of a particle in apotential as a special case of optics with a position-dependent index of refraction. Theubiquity of geodesic motion. Kaluza-Klein theory. From Lagrangians to Hamiltonians.


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Week 6 (May 2, 4, 6)From Lagrangians to Hamiltonians, continued. Regular and

strongly regular Lagrangians. The cotangent bundle as phase space. Hamiltons equa-tions. Getting Hamiltons equations directly from a least action principle.

Week 7 (May 9, 11, 13)Waves versus particles: the Hamilton-Jacobi equation.Hamiltons principal function and extended phase space. How the Hamilton-Jacobi equa-tion foreshadows quantum mechanics.

Week 8 (May 16, 18, 20)Towards symplectic geometry. The canonical 1-form andthe symplectic 2-form on the cotangent bundle. Hamiltons equations on a symplecticmanifold. Darbouxs theorem.

Week 9 (May 23, 25, 27)Poisson brackets. The Schrodinger picture versus theHeisenberg picture in classical mechanics. The Hamiltonian version of Noethers theorem.Poisson algebras and Poisson manifolds. A Poisson manifold that is not symplectic.Liouvilles theorem. Weils formula.

Week 10 (June 1, 3, 5)A taste of geometric quantization. Kahler manifolds.If you find errors in these notes, please email me! I thank Sheeyun Park and Curtis

Vinson for catching lots of errors.


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Contents

1 From Newtons Laws to Langranges Equations 2

1.1 Lagrangian and Newtonian Approaches . . . . . . . . . . . . . . . . . . . . 21.1.1 Lagrangian versus Hamiltonian Approaches . . . . . . . . . . . . . 61.2 Prehistory of the Lagrangian Approach . . . . . . . . . . . . . . . . . . . . 6

1.2.1 The Principle of Minimum Energy . . . . . . . . . . . . . . . . . . 71.2.2 DAlemberts Principle and Lagranges Equations . . . . . . . . . . 81.2.3 The Principle of Least Time . . . . . . . . . . . . . . . . . . . . . . 101.2.4 How DAlembert and Others Got to the Truth . . . . . . . . . . . . 12

2 Equations of Motion 14

2.1 The Euler-Lagrange Equations . . . . . . . . . . . . . . . . . . . . . . . . . 142.1.1 Comments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

2.1.2 Lagrangian Dynamics . . . . . . . . . . . . . . . . . . . . . . . . . . 162.2 Interpretation of Terms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

3 Lagrangians and Noethers Theorem 20

3.1 Time Translation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203.1.1 Canonical and Generalized Coordinates . . . . . . . . . . . . . . . . 21

3.2 Symmetry and Noethers Theorem . . . . . . . . . . . . . . . . . . . . . . 223.2.1 Noethers Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

3.3 Conserved Quantities from Symmetries . . . . . . . . . . . . . . . . . . . . 243.3.1 Time Translation Symmetry . . . . . . . . . . . . . . . . . . . . . . 253.3.2 Space Translation Symmetry . . . . . . . . . . . . . . . . . . . . . . 253.3.3 Rotational Symmetry . . . . . . . . . . . . . . . . . . . . . . . . . . 26

3.4 Example Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 283.4.1 The Atwood Machine . . . . . . . . . . . . . . . . . . . . . . . . . . 283.4.2 Disk Pulled by Falling Mass . . . . . . . . . . . . . . . . . . . . . . 293.4.3 Free Particle in Special Relativity . . . . . . . . . . . . . . . . . . . 31

3.5 Electrodynamics and Relativistic Lagrangians . . . . . . . . . . . . . . . . 353.5.1 Gauge Symmetry and Relativistic Hamiltonian . . . . . . . . . . . . 353.5.2 Relativistic Hamiltonian . . . . . . . . . . . . . . . . . . . . . . . . 36

v


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CONTENTS 1

3.6 Relativistic Particle in an Electromagnetic Field . . . . . . . . . . . . . . . 37

3.7 Alternative Lagrangians . . . . . . . . . . . . . . . . . . . . . . . . . . . . 393.7.1 Lagrangian for a String . . . . . . . . . . . . . . . . . . . . . . . . . 393.7.2 Alternate Lagrangian for Relativistic Electrodynamics . . . . . . . . 41

3.8 The General Relativistic Particle . . . . . . . . . . . . . . . . . . . . . . . 433.8.1 Free Particle Lagrangian in GR . . . . . . . . . . . . . . . . . . . . 443.8.2 Charged particle in EM Field in GR . . . . . . . . . . . . . . . . . 45

3.9 The Principle of Least Action and Geodesics . . . . . . . . . . . . . . . . . 463.9.1 Jacobi and Least Time vs Least Action . . . . . . . . . . . . . . . . 463.9.2 The Ubiquity of Geodesic Motion . . . . . . . . . . . . . . . . . . . 48

4 From Lagrangians to Hamiltonians 51

4.1 The Hamiltonian Approach . . . . . . . . . . . . . . . . . . . . . . . . . . 514.2 Regular and Strongly Regular Lagrangians . . . . . . . . . . . . . . . . . . 54

4.2.1 Example: A Particle in a Riemannian Manifold with Potential V(q) 544.2.2 Example: General Relativistic Particle in an E-M Potential . . . . . 554.2.3 Example: Free General Relativistic Particle with Reparameteriza-

tion Invariance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 554.2.4 Example: A Regular but not Strongly Regular Lagrangian . . . . . 55

4.3 Hamiltons Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 564.3.1 Hamilton and Euler-Lagrange . . . . . . . . . . . . . . . . . . . . . 574.3.2 Hamiltons Equations from the Principle of Least Action . . . . . . 59

4.4 Waves versus ParticlesThe Hamilton-Jacobi Equations . . . . . . . . . . 614.4.1 Wave Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 614.4.2 The Hamilton-Jacobi Equations . . . . . . . . . . . . . . . . . . . . 63


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Chapter 1

From Newtons Laws to Langranges

Equations

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

Classical mechanics is a very peculiar branch of physics. It used to be considered thesum total of our theoretical knowledge of the physical universe (Laplaces daemon, theNewtonian clockwork), but now it is known as an idealization, a toy model if you will.The astounding thing is that probably all professional applied physicists still use classicalmechanics. So it is still an indispensable part of any physicists or engineers education.

It is so useful because the more accurate theories that we know of (general relativity

and quantum mechanics) make corrections to classical mechanics generally only in extremesituations (black holes, neutron stars, atomic structure, superconductivity, and so forth).Given that GR and QM are much harder theories to use and apply it is no wonder thatscientists will revert to classical mechanics whenever possible.

So, what is classical mechanics?

1.1 Lagrangian and Newtonian Approaches

We begin by comparing the Newtonian approach to mechanics to the subtler approach ofLagrangian mechanics. Recall Newtons law:

F = ma (1.1)

wherein we consider a particle moving in n. Its position, say q, depends on time t

,so it defines a function,

q :

n.

From this function we can define velocity,

v = q :

n

2


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1.1 Lagrangian and Newtonian Approaches 3

where q = dqdt

, and also acceleration,

a = q :

n.

Now let m > 0 be the mass of the particle, and let F be a vector field on n called the

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

m a(t) = F(q(t)) . (1.2)

This is a 2nd-order differential equation for q :

n which will have a unique solution

given some q(t0) and q(t0), provided the vector field F is nice by which we technicallymean smooth and bounded (i.e., |F(x)| < B for some B > 0, for all x

n).We can then define a quantity called kinetic energy:

K(t) :=1

2m v(t) v(t) (1.3)

This quantity is interesting because

d

dtK(t) = m v(t) a(t)

= F(q(t)) v(t)

So, kinetic energy goes up when you push an object in the direction of its velocity, andgoes down when you push it in the opposite direction. Moreover,

K(t1) K(t0) =

t1t0

F(q(t)) v(t) dt

=

t1t0

F(q(t)) q(t) dt

So, the change of kinetic energy is equal to the work done by the force, that is, the integralof F along the curve q : [t0, t1]

n. This implies (by Stokess theorem relating lineintegrals to surface integrals of the curl) that the change in kinetic energy K(t1) K(t0)is independent of the curve going from q(t0) = a to q(t1) = b iff

F = 0.

This in turn is true iffF = V (1.4)

for some function V : n

. This function is then unique up to an additive constant; wecall it the potential. A force with this property is called conservative. Why? Becausein this case we can define the total energy of the particle by

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


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4 From Newtons Laws to Langranges Equations

where V(t) := V(q(t)) is called the potential energy of the particle, and then we can

show that E is conserved: that is, constant as a function of time. To see this, note thatF = 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 of cool techniques. In the Lagrangianapproach we define a quantity

L := K(t) V(q(t)) (1.6)

called the Lagrangian, and for any curve q : [t0, t1]

n with q(t0) = a, q(t1) = b, wedefine the action to be

S(q) :=

t1t0

L(t) dt (1.7)

From here one can go in two directions. One is to claim that nature causes particlesto follow paths of least action, and derive Newtons equations from that principle. Theother is to start with Newtons principles and find out what conditions, if any, on S(q)follow from this. We will use the shortcut of hindsight, bypass the philosophy, and simplyuse the mathematics of variational calculus to show that particles follow paths that arecritical points of the action S(q) if and only if Newtons law F = ma holds. To do this,

n

q

qs + sq

t0 t1

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

let us look for curves (like the solid line in Fig. 1.1) that are critical points ofS, namely:

d

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


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1.1 Lagrangian and Newtonian Approaches 5

where

qs = q + sqfor all q : [t0, t1]

n with,q(t0) = q(t1) = 0.

To show that

F = ma d

dsS(qs)|t=0 = 0 for all q : [t0, t1]

n with q(t0) = q(t1) = 0 (1.9)

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

d

dsS(qs)

s=0

=d

ds

t1t0

1

2mqs(t) qs(t) V(qs(t)) dt

s=0

=

t1t0

d

ds

1

2mqs(t) qs(t) V(qs(t))

dt

s=0

=

t1t0

mqs

d

dsqs(t) V(qs(t))

d

dsqs(t)

dt

s=0

Note thatd

dsqs(t) =

d

ds

d

dtqs(t) =

d

dt

d

dsqs(t)

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

d

dtq(t)

So,

d

dsS(qs)

s=0

=

t1t0

mq

d

dtq(t) V(q(t)) q(t)

dt

Then we can integrate by parts, noting the boundary terms vanish because q = 0 at t1and t0:

d

dsS(qs)|s=0 =

t1t0

[mq(t) V(q(t))] q(t)dt

It follows that variation in the action is zero for all variations q iff the term in brackets

is identically zero, that is, mq(t) V(q(t)) = 0

So, the path q is a critical point of the action S iff

F = ma.

The above result applies only for conservative forces, i.e., forces that can be writtenas the minus the gradient of some potential. However, this seems to be true of the mostfundamental forces that we know of in our universe. It is a simplifying assumption thathas withstood the test of time and experiment.


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1.1.1 Lagrangian versus Hamiltonian Approaches

I am not sure where to mention this, but before launching into the history of the La-grangian approach may be as good a time as any. In later chapters we will describeanother approach to classical mechanics: the Hamiltonian approach. Why do we needtwo approaches, Lagrangian and Hamiltonian?

They both have their own advantages. In the simplest terms, the Hamiltonian ap-proach focuses on position and momentum, while the Lagrangian approach focuses onposition and velocity. The Hamiltonian approach focuses on energy, which is a functionof 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 veloc-ity. Our first task in understanding Lagrangian mechanics is to get a gut feeling for what

the Lagrangian means. The key is to understand the integral of the Lagrangian over time the action, S. We shall see that this describes the total amount that happened fromone moment to another as a particle traces out a path. And, peeking ahead to quantummechanics, the quantity exp(iS/), where is Plancks constant, will describe the changein phase of a quantum system as it traces out this path.

In short, while the Lagrangian approach takes a while to get used to, it providesinvaluable insights into classical mechanics and its relation to quantum mechanics. Weshall see this in more detail soon.

1.2 Prehistory of the Lagrangian Approach

Weve seen that a particle going from point a at time t0 to a point b at time t1 follows apath that is a critical point of the action,

S =

t1t0

K V dt

so that slight changes in its path do not change the action (to first order). Often, thoughnot always, the action is minimized, so this is called the Principle of Least Action.

Suppose we did not have the hindsight afforded by the Newtonian picture. Then wemight ask, Why does nature like to minimize the action? And why this action

KV dt?

Why not some other action?Why questions are always tough. Indeed, some people say that scientists shouldnever ask why. This seems too extreme: a more reasonable attitude is that we shouldonly ask a why question if we expect to learn something scientifically interesting in ourattempt to answer it.

There are certainly some interseting things to learn from the question why is actionminimized? First, note that total energy is conserved, so energy can slosh back andforth between kinetic and potential forms. The Lagrangian L = K V is big when mostof the energy is in kinetic form, and small when most of the energy is in potential form.


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1.2 Prehistory of the Lagrangian Approach 7

Kinetic energy measures how much is happening how much our system is moving

around. Potential energy measures how much could happen, but isnt yet thats whatthe word potential means. (Imagine a big rock sitting on top of a cliff, with the potentialto fall down.) So, the Lagrangian measures something we could vaguely refer to as theactivity or liveliness of a system: the higher the kinetic energy the more lively thesystem, the higher the potential energy the less lively. So, were being told that naturelikes to minimize 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 Earths gravitational field, as

in Fig. 1.2. The rock traces out a parabola, and we can think of it as doing this in order

K V small here...good!

Spend as much time as possible here

K V big...bad!

Get this over with quick!

Figure 1.2: A particles lazy motion, minimizes the action.

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

Here we are anthropomorphizing the rock by saying that it wants to minimize itsaction. This is okay if we dont take it too seriously. Indeed, one of the virtues of thePrinciple of Least Action is that it lets us put ourselves in the position of some physicalsystem and imagine what we would do to minimize the action.

There is another way to make progress on understanding why action is minimized:

history. Historically there were two principles that were fairly easy to deduce from ob-servations of nature: (i) the principle of minimum energy used in statics, and (ii) theprinciple of least time, used in optics. By putting these together, we can guess the prin-ciple of least action. So, let us recall these earlier minimum principles.

1.2.1 The Principle of Minimum Energy

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


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L1 L2

m1 m2

Figure 1.3: A principle of energy minimization determines a levers balance.

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

m1L1 = m2L2.

Later DAlembert understood this using his principle of virtual work. He consideredmoving the lever slightly, i.e., infinitesimally, He claimed that in equilibrium the infinites-

ddq1dq2


imal work done by this motion is zero! He also claimed that the work done on the ith

body is, dWi = Fidqi

and gravity pulls down with a force mig so,

dWi = (0, 0, mg) (0, 0, L1d)

= m1gL1 d

and similarly,dW2 = m2gL2 d

Now DAlemberts principle says that equilibrium occurs when the virtual work dW =dW1 +dW2 vanishes for all d (that is, for all possible infinitesimal motions). This happens

whenm1L1 m2L2 = 0

which is just as Archimedes wrote.

1.2.2 DAlemberts Principle and Lagranges Equations

Lets go over the above analysis in more detail. Ill try to make it clear what we mean byvirtual work.


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The forces and constraints on a system may be time dependent. So equal small

infinitesimal displacements of the system might result in the forces Fi acting on thesystem doing different amounts of work at different times. To displace a system by ri foreach position coordinate, and yet remain consistent with all the constraints and forces at a

given instant of time t, without any time interval passing is called a virtual displacement.Its called virtual because it cannot be realized: any actual displacement would occurover a finite time interval and possibly during which the forces and constraints mightchange. Now call the work done by this hypothetical virtual displacement, Fi ri, thevirtual work. Consider a system in the special state of being in equilibrium, i.e.,. when

Fi = 0. Then because by definition the virtual displacements do not change the forces,we must deduce that the virtual work vanishes for a system in equilibrium,

i

Fi ri = 0, (when in equilibrium) (1.10)

Note that in the above example we have two particles in 3 subject to a constraint

(they are pinned to the lever arm). However, a number n of particles in 3 can be

treated as a single quasi-particle in 3n, and if there are constraints it can move in some

submanifold of 3n. So ultimately we need to study a particle on an arbitrary manifold.

But, well postpone such sophistication for a while.For a particle in

n, DAlemberts principle simply says,

q(t) = q0 satisfies F = ma, (its in equilibrium)

dW = F dq vanishes for all dq n, (virtual work is zero for q 0)

F = 0, (no force on it!)

If the force is conservative (F = V) then this is also equivalent to,

V(q0) = 0

that is, we have equilibrium at a critical point of the potential. The equilibrium will be

stable if q0 is a local minimum of the potential V.We can summarize all the above by proclaiming that we have a principle of least en-ergy governing stable equilibria. We also have an analogy between statics and dynamics,

Statics Dynamics

equilibrium, a = 0 F = ma

potential, V action, S =

t1t0

K V dt

critical points of V critical points of S


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V

n

unstable equilibrium

unstable equilibrium

stable equilibrium


d

dt

T

qi

T

qi= Qi. (1.11)

ddt

Lqi

L

qi= 0. (1.12)

DAlemberts principle is an expression for Newtons second law under conditionswhere the virtual 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 history surrounding the principles of Lagrangianmechanics. As well as hints from statics, there were also hints from the behavior of light,hints again that nature likes to minimize effort. In a vacuum light moves in straight lines,which in Euclidean space is the minimum distance. But more interesting than straightlines are piecewise straight paths and curves. Consider reflection of light from a mirror,

What path does the light take? The empirical answer was known since antiquity, itchooses B such that 1 = 2, so the angle of incidence equals the angle of reflection. Butthis is precisely the path that minimizes the distance of the trajectory subject to thecondition that it must hit the mirror (at least at one point). In fact light traveling fromA to B takes both the straight paths ABC and AC. Why is ABC the shortest path hitting


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A

B

C

C

1

2

2

the mirror? This follows from some basic Euclidean geometry:

B minimizes AB + BC B minimizes AB + BC

A,B,C lie on a line

1 = 2

Note the introduction of the fictitious image C behind the mirror, this is a trick oftenused in solving electrostatic problems (a conducting surface can be replaced by fictitiousmirror image charges to satisfy the boundary conditions), it is also used in geophysicswhen one has a geological fault, and in hydrodynamics when there is a boundary between

two media (using mirror image sources and sinks).The big clue leading to DAlemberts principle however came from refraction of light.

Snell (and predecessors) noted that each medium has some number n associated with it,

medium 1

medium 2

1

2

called the index of refraction, such that,

n1 sin 1 = n2 sin 2


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(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 to 1/n, then light willsatisfy Snells law if the light minimizes the time it takes to get from A to C. In the caseof refraction it is the time that is important, not just the path distance. But the sameis true for the law of reflection, since in that case the path of minimum length gives thesame results as the path of minimum time.

So, not only is light the fastest thing around, its also always taking the quickest pathfrom here to there!

1.2.4 How DAlembert and Others Got to the Truth

Sometimes laws of physics are just guessed using a bit of intuition and a gut feeling that

nature must be beautiful or elegantly simple (though occasionally awesomely complex inbeauty). One way to make good guesses is to generalize.

DAlemberts principle of virtual work for statics says that equilibrium occurs when

F(q0) q = 0, q n

DAlembert generalized this to dynamics by inventing what he called the inertia force=m a,and he postulated that in dynamics equilibrium occurs when the total force = F+inertiaforce, vanishes. Or symbolically when,

F(q(t)) ma(t) q(t) = 0 (1.13)

We then take a variational path parameterized by s,

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

whereq(t0) = q(t1) = 0

and with these paths, for any function f on the space of paths we can define the varia-tional derivative,

f :=d

dsf(qs)

s=0

(1.14)

Then DAlemberts principle of virtual work implies

t1

t0F(q) mq q dt = 0

for all q, so ifF = V, we get

0 =

t1t0

V(q) mq

qdt

=

t1t0

V(q) q + mqq

dt


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using

d

dsV(qs(t))

s=0

=dV

dq

dqs(t)

ds

s=0

and

(q2) =dq2s (t)

ds

then we have

0 =

t1

t0

dV

dq

dqs

dt +

m

2

dqs(t)

ds

dt

=d

ds

t1t0

d

dtV

qs(t)

+m

2qs(t)

s=0

dt

=

t1t0

d

dtV

qs(t)

+m

2qs(t)

2

dt

therefore

t1

t0

V(q) + K

dt

= 0

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

S(q) =

(K V) dt (1.15)

Weve described how DAlembert might have arrived at the principle of least actionby generalizing previously known energy minimization and least time principles. Still,theres something unsatisfying about the treatment so far. We do not really understandwhy one must introduce the inertia force. We only see that its necessary to obtainagreement with Newtonian mechanics (which is manifest in Eq.(1.13)).

We conclude with a few more words about this mystery. Recall from undergraduatephysics that in an accelerating coordinate system there is a fictional force = ma, which is

called the centrifugal force. We use it, for example, to analyze simple physics in a rotatingreference frame. If you are inside the rotating system and you throw a ball straight aheadit will appear to curve away from your target, and if you did not know that you wererotating relative to the rest of the universe then youd think there was a force on the ballequal to the centrifugal force. If you are inside a big rapidly rotating drum then youllalso feel pinned to the walls. This is an example of an inertia force which comes fromusing a funny coordinate system. In general relativity, one sees that in a certain sense gravity is an inertia force!


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Chapter 2

Equations of Motion

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

In this chapter well start to look at the Lagrangian equations of motion in moredepth. Well look at some specific examples of problem solving using the Euler-Lagrangeequations. First well show how the equations are derived.

2.1 The Euler-Lagrange Equations

We are going to start thinking of a general classical system as a set of points in an abstract

configuration space or phase space1. So consider an arbitrary classical system as livingin a space of points in some manifold Q. For example, the space for a spherical doublependulum would look like Fig. 2.1, where Q = S2 S2. So our system is a particle in

Q = S2 S2

Figure 2.1: Double pendulum configuration space.

Q, which means you have to disabuse yourself of the notion that were dealing with real

1The tangent bundle T Q will be referred to as configuration space, later on when we get to the chapteron Hamiltonian mechanics well find a use for the cotangent bundle TQ, and normally we call this thephase space.

14


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2.1 The Euler-Lagrange Equations 15

particles, we are not, we are dealing with a single quasi-particle in an abstract higher

dimensional space. The single quasi-particle represents two real particles if we are talkingabout the classical system in Fig. 2.1. Sometimes to make this clear well talk about thesystem taking a path, instead of the particle taking a path. It is then clear that whenwe say, the system follows a path q(t) that were referring to the point q in configurationspace Q that represents all of the particles in the real system.

So as time passes the system traces out a path

q : [t0, t1] Q

and we define its velocity,

q(t) Tq(t)Q

to be the tangent vector at q(t) given by the equivalence class [] of curves through q(t)with derivatives (t) = dq(s)/ds|s=t. Well just write is as q(t).

Let be the space of smooth paths from a Q to b Q,

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

( is an infinite dimensional manifold, but we wont go into that for now.) Let the La-grangian=L for the system be any smooth function of position and velocity (not explicitlyof time, for simplicity),

L : T Q

and define the action, S:

S :

by

S(q) :=

t1t0

L(q, q) dt (2.1)

The path that the quasi particle will actually take is a critical point of S, in accord withDAlemberts principle of least action. In other words, a path q such that for anysmooth 1-parameter family of paths qs with q0 = q1, we have

d

dsS(qs)

s=0

= 0 (2.2)

We write,d

ds

s=0

as

so Eq.(2.2) can be rewritten

S = 0 (2.3)


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16 Equations of Motion

2.1.1 Comments

What is a 1-parameter family of paths? Well, a path is a curve, or a 1D manifold. Sothe 1-parameter family is nothing more nor less than a set of well-defined paths {qs}, eachone labeled by a parameter s. So a smooth 1-parameter family of paths will have q(s)everywhere infinitesimally close to q(s + ) for an infinitesimal hyperreal . So in Fig. 2.2

q0

qs

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

we can go from q0 to qs by smoothly varying s from s = 0 to s = sWhat does the condition S = 0 imply? Patience, we are just getting to that. We will

now start to explore what S = 0 means for our Lagrangian.

2.1.2 Lagrangian Dynamics

We were given that Q is a manifold, so it admits a covering of coordinate charts. Fornow, lets pick coordinates in a neighborhood U of some point q(t) Q. Next, consideronly variations qs such that qs = q outside U. A cartoon of this looks like Fig. 2.3 Then

Q

U

a

b

Figure 2.3: Local path variation.


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2.1 The Euler-Lagrange Equations 17

we restrict attention to a subinterval [t0, t

1] [t0, t1] such that qs(t) U for t

0 t t

1.

Lets just go ahead and rename t

0 and t

1 as t0 and t1 to drop the primes. We canuse the coordinate charts on U,

: U n

x (x) = (x1, x2, . . . , xn)

and we also have coordinates for the 1-forms,

d : T U T n =

n n

(x, y) d(x, y) = (x1, . . . , xn, y1, . . . , yn)

where y TxQ. We restrict L : TM

to T U TM, in our case the manifold isM = Q, and then we can describe it, L, using the coordinates xi, yi on T U. The i aregeneralized position coordinates, the yi are the associated generalized velocity coordinates.(Velocity and position are in the abstract configuration space Q). Using these coordinateswe get

S =

t1t0

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

=

t1t0

L(q, q) dt

=

t1

t0

Lxi

qi +L

y iqi

dt

where weve used the given smoothness of L and the Einstein summation convention forrepeated indices i. Note that we can write L as above using a local coordinate patchbecause the path variations q are entirely trivial outside the patch for U. Continuing,using the Leibniz rule

d

dt

Ly

q

=d

dt

L

yq +

L

yq

we have,

S =

t1t0

Lxi

d

dt

L

y i

qi(t) dt

= 0.

If this integral is to vanish as demanded by S = 0, then it must vanish for all pathvariations q, further, the boundary terms vanish because we deliberately chose q that


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18 Equations of Motion

vanish at the endpoints t0 and t1 inside U. That means the term in brackets must be

identically zero, ord

dt

L

y i

L

xi= 0 (2.4)

This is necessary to get S = 0, for all q, but in fact its also sufficient. Physicists alwaysgive the coordinates xi, yi on T U the symbols qi and qi, despite the fact that thesealso have another meaning, namely the xi and yi coordinates of the quantity,

q(t), q(t)

T U.

So in any case, physicists write,

d

dt

L

qi=

L

q i

and they call these the Euler-Lagrange equations.

2.2 Interpretation of Terms

The derivation of the Euler-Lagrange equations above was fairly abstract, the terms po-sition and velocity were used but were not assumed to be the usual kinematic notions

that we are used to in physics, indeed the only reason we used those terms was for theiranalogical appeal. Now well try to illuminate the E-L equations a bit by casting theminto the usual position and velocity terms.

So consider,

Q =

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

q i(V)i the force Fi

When we write V/qi = V were assuming the qi are Cartesian coordinates on Q.


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2.2 Interpretation of Terms 19

So translating our example into general terms, if we conjure up some abstract La-

grangian then we can think of the independent variables as generalized positions andvelocities, and then the Euler-Lagrange equations can be interpreted as equations relat-ing generalized concepts of momentum and force, and they say that

p = F (2.5)

So theres no surprise that in the mundane case of a single particle moving in 3 under

time t this just recovers Newton II. Of course we can do all of our classical mechanicswith Newtons laws, its just a pain in the neck to deal with the redundancies in F = mawhen we could use symmetry principles to vastly simplify many examples. It turns outthat the Euler-Lagrange equations are one of the reformulations of Newtonian physics

that make it highly convenient for introducing symmetries and consequent simplifications.Simplifications generally mean quicker, shorter solutions and more transparent analysisor at least more chance at insight into the characteristics of the system. The main thingis that when we use symmetry to simplify the equations we are reducing the numberof independent variables, so it gets closer to the fundamental degrees of freedom of thesystem and so we cut out a lot of the wheat and chaff (so to speak) with the full redundantNewton equations.

One can of course introduce simplifications when solving Newtons equations, its justthat its easier to do this when working with the Euler-Lagrange equations. Another goodreason to learn Lagrangian (or Hamiltonian) mechanics is that it translates better intoquantum mechanics.


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Chapter 3

Lagrangians and Noethers Theorem

If the form of a system of dynamical equations does not change under spatial translationsthen the momentum is a conserved quantity. When the form of the equations is similarlyinvariant under time translations then the total energy is a conserved quantity (a constantof the equations of motion). Time and space translations are examples of 1-parametergroups of transformations. Invariance under a group of transformations is precisely whatwe mean by a symmetry in group theory. So symmetries of a dynamical system giveconserved quantities or conservation laws. The rigorous statement of all this is the contentof Noethers theorem.

3.1 Time Translation

To handle time translations we need to replace our paths q : [t0, t1] Q by pathsq :

Q, and then define a new space of paths,

= {q :

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 a function of the space of paths.Nevertheless, if q = 0 outside of some finite interval, then the functional variation,

S :=

d

dsL

qs(t), qs(t)

s=0

dt

will converge, since the integral is smooth and vanishes outside this interval. Moreover,demanding that this S vanishes for all such variations q is enough to imply the Euler-

20


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3.1 Time Translation 21

Lagrange equations:

S =

d

dsL

qs(t), qs(t)

s=0

dt

=

L

qiqi +

L

qiqi

dt

=

L

qi

d

dt

L

qi

qi dt

where again the boundary terms have vanished since q = 0 near t = . To be explicit,the first term in

L

qi qi

=d

dtL

qi qi

d

dt

L

qi

q

vanishes when we integrate. Then the whole thing vanishes for all compactly supportedsmooth q iff

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 Hamiltons equationsif you change L to H you need to stickin a minus sign, and change variables from q to pi and eliminate pi.

3.1.1 Canonical and Generalized Coordinates

In light of this noted similarity with the Hamilton equations of motion, lets spend a fewmoments clearing up some terminology (I hate using jargon, but sometimes its unavoid-able, and sometimes it can be efficientprovided everyone is clued in).

Generalized Coordinates

For Lagrangian mechanics we have been using generalized coordinates, these are the{qi, qi}. The qi are generalized positions, and the qi are generalized velocities. The fullset of independent generalized coordinates represent the degrees of freedom of a particle,or system of particles. So if we have N particles then wed typically have 6N general-ized coordinates (the 6 is for 3 space dimensions, and at each point a position and amomentum). These can be in any reference frame or system of axes, so for example,


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22 Lagrangians and Noethers Theorem

in a Cartesian frame, with two particles, in 3D space wed have the 2 3 = 6 position

coordinates, 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 Cartesian velocity components. This makes12 = 6 2 = 6N coordinates, matching the total degrees of freedom as claimed. If weconstrain the particles to move in a plane (say place them on a table in a gravitationalfield) then we get 2N fewer degrees of freedom, and so 4N d.o.f. overall. By judiciouschoice of coordinate frame we can eliminate one velocity component and one positioncomponent for each particle.

It is also handy to respect other symmetries of a system, maybe the particles move on

a sphere for example, one can then define new positions and momenta with a consequentreduction in the number of these generalized coordinates needed to describe the system.

Canonical Coordinates

In Hamiltonian mechanics (which we have not yet fully introduced) we will find it moreuseful to transform from generalized coordinates to canonical coordinates. The canonicalcoordinates are a special set of coordinates on the cotangent bundle of the configurationspace manifold Q. They are usually written as a set of (qi, pj) or (x

i, pj) with the xs or qsdenoting the coordinates on the underlying manifold and the ps denoting the conjugatemomentum, which are 1-forms in the cotangent bundle at the point q in the manifold.

It turns out that the qi together with the pj , form a coordinate system on the cotangentbundle TQ of the configuration space Q, hence these coordinates are called the canonicalcoordinates.

We will not discuss this here, but if you care to know, later on well see that therelation between the generalized coordinates and the canonical coordinates is given bythe Hamilton-Jacobi equations for a system.

3.2 Symmetry and Noethers Theorem

First, lets give a useful definition that will make it easy to refer to a type of dynamicalsystem symmetry. We want to refer to symmetry transformations (of the Lagrangian)governed by a single parameter.

Definition 3.1 (one-parameter family of symmetries). A 1-parameter family of symme-tries of a Lagrangian system L : T Q

is a smooth map,

F :

(s, q) qs, with q0 = q


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3.2 Symmetry and Noethers Theorem 23

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

L =d

dt

for some : T Q

, 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 L = 0, in which case we really have a way of moving

paths around (q qs) that doesnt change the Lagrangiani.e., a symmetry of L in themost obvious way. But L = ddt is a sneaky generalization whose usefulness will becomeclear.

3.2.1 Noethers Theorem

Heres a statement of the theorem. Note that in this theorem is the function associatedwith F in definition 3.1.

Theorem 3.1 (Noethers Theorem). Suppose F is a one-parameter family of symmetriesof the Lagrangian system, L : T Q

. Then,

piqi

is conserved, that is, its time derivative is zero for any path q satisfying the Euler-Lagrange equations. In other words, in boring detail,

d

dt

L

y i

q(s)q(s) d

dsqis(t)

s=0

q(t), q(t)

= 0

Proof.

d

dt

piq

i

= piqi +piq

i d

dt

= Lq i

qi + Lqi

qi L

= L L = 0.

Okay, big deal you might say. Before this can be of any use wed need to find asymmetry F. Then wed need to find out what this piqi business is that is conserved.So lets look at some examples.


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Example

1. Conservation of Energy. (The most important example!)

All of our Lagrangian systems will have time translation invariance (because thelaws of physics do not change with time, at least not to any extent that we can tell).So we have a one-parameter family of symmetries

qs(t) = q(t + s)

This indeed gives,

L = L

for

d

dsL(qs)

s=0

=d

dtL = L

so here we take = L simply! We then get the conserved quantity

piqi = piq

i L

which we normally call the energy. For example, ifQ = n, and if

L =

1

2 mq

2

V(q)

then this quantity is

mq q 1

2mq q V

=

1

2mq2 + V(q)

The term in parentheses is K V, and the left-hand side is K+ V.

Lets repeat this example, this time with a specific Lagrangian. It doesnt matter whatthe Lagrangian is, if it has 1-parameter families of symmetries then itll have conservedquantities, guaranteed. The trick in physics is to write down a correct Lagrangian in the

first place! (Something that will accurately describe the system of interest.)

3.3 Conserved Quantities from Symmetries

Weve seen that any 1-parameter family

Fs :

q qs


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3.3 Conserved Quantities from Symmetries 25

which satisfies

L =

for some function = (q, q) gives a conserved quantity

piqi

As usual weve defined

L :=d

dsL

qs(t), qs(t)

s=0

Lets see how we arrive at a conserved quantity from a symmetry.

3.3.1 Time Translation Symmetry

For any Lagrangian system, L : T Q

, we have a 1-parameter family of symmetries

qs(t) = q(t + s)

because

L = L

so we get a conserved quantity called the total energy or Hamiltonian,

H = piqi

L (3.1)

(You might prefer Hamiltonian to total energy because in general we are not in thesame configuration space as Newtonian mechanics, if you are doing Newtonian mechanicsthen total energy is appropriate.)

For example: a particle on n in a potential V has Q =

n, L(q, q) = 12 mq2 V(q).

This system has

piqi =

L

qiqi = mq2 = 2K

so

H = piqi L = 2K (K V) = K+ V

as youd have hoped.

3.3.2 Space Translation Symmetry

For a free particle in n, we have Q =

n and L = K = 12 mq2. This has spatial translation

symmetries, so that for any v n we have the symmetry

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


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3.3 Conserved Quantities from Symmetries 27

So,

L = mqiXi

j qj

= mq (Xq)

= 0

since X is skew symmetric as stated previously (X so(n)). So we get a conservedquantity, the angular momentum in the X direction.

(Note: this whole bunch of math above for L just says that the kinetic energy doesntchange when the velocity is rotated, without changing its magnitude.)

We write,

piq i = mqi (X q)i

(qi = Xq just as qi = Xq in our previous calculation), or if X has zero entries exceptin ij and ji positions, where its 1, then we get

m(qiqj qjq

i)

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

mqq

Note that above we have assumed one can construct a basis for so(n) using matricesof the form assumed for X, i.e., skew symmetric with 1 in the respectively ij and jielements, otherwise zero.

I mentioned earlier that we can do mechanics with any Lagrangian, but if we want tobe useful wed better pick a Lagrangian that actually describes a real system. But howdo we do that? All this theory is fine but is useless unless you know how to apply it.The above examples were for a particularly simple system, a free particle, for which theLagrangian is just the kinetic energy, since there is no potential energy variation for afree particle. Wed like to know how to solve more complicated dynamics.

The general idea is to guess the kinetic energy and potential energy of the particle (as

functions of your generalized positions and velocities) and then let,

L = K V

So we are not using Lagrangians directly to tell us what the fundamental physical lawsshould be, instead we plug in some assumed physics and use the Lagrangian approch tosolve the system of equations. If we like, we can then compare our answers with exper-iments, which indirectly tells us something about the physical lawsbut only providedthe Lagrangian formulation of mechanics is itself a valid procedure in the first place.


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3.4 Example Problems

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

To see how the formalisms in this chapter function in practise, lets do some problems.Its vastly superior to the simplistic F = ma formulation of mechanics. The Lagrangianformulation allows the configuration space to be any manifold, and allows us to easily useany coordinates we wish.

3.4.1 The Atwood Machine

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

x

xm1

m2

K =1

2(m1 + m2)(

d

dt( x))2 =

1

2(m1 + m2)x

2

V = m1gx m2g( x)

so

L = K V =1

2(m1 + m2)x

2 + m1gx + m2g( x)

The configuration space is Q = (0, ), and x (0, ) (we could use the owns symbol

here and write Q = (0, ) x ). Moreover T Q = (0, )

(x, x). As usual L : T Q

.Note that solutions of the Euler-Lagrange equations will only be defined for some timet

, as eventually the solutions reaches the edge of Q.The momentum is:

p =L

x= (m1 + m2)x

and the force is:

F =L

x= (m1 m2)g


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3.4 Example Problems 29

The Euler-Lagrange equations say

p = F

(m1 + m2)x = (m1 m2)g

x =m1 m2m1 + m2

g

So this is like a falling object in a downwards gravitational acceleration a =

m1m2m1+m2

g.

It is trivial to integrate the expression for x twice (feeding in some appropriate initialconditions) to obtain the complete solution to the motion x(t) and x(t). Note that x = 0when 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 falling mass. The disk is free to move ona frictionless surface, and it can thus whirl around the hole to which it is tethered to themass below.

r

r

m1

m2

no swinging allowed!

Here Q = open disk of radius , minus its center

= (0, ) S1 (r, )

T Q = (0, ) S1

(r,, r, )

K =1

2m1(r

2 + r22) +1

2m2(

d

dt( r))2

V = gm2(r )L =

1

2m1(r

2 + r22) +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

p =L

= m1r

2.


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Note that is an ignorable coordinateit doesnt appear in Lso theres a symmetry,

rotational symmetry, and p, the conjugate momentum, is conserved.The forces are,

Fr =L

r= m1r

2 gm2

F =L

= 0, ( is ignorable)

Note: in Fr the term m1r2 is recognizable as a centrifugal force, pushing m1 radially

out, while the term gm2 is gravity pulling m2 down and thus pulling m1 radially in.So, the Euler-Lagrange equations give,

pr = Fr, (m1 + m2)r = m1r2 m2g

p = 0, p = m1r2 = J = a constant.

Lets use our conservation law here to eliminate from the first equation:

=J

m1r2

so

(m1 + m2)r =J2

m1r3 m2g

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

Fr =J2

m1r3 m2g

which could come from an effective potential V(r) such that dV/dr = Fr. So integrateFr to find V(r):

V(r) =

J2

2m1r2 + m2gr

this is a sum of two terms that look like Fig. 3.1If (t = 0) = 0 then there is no centrifugal force and the disk will be pulled into the

hole until it gets stuck. At that time the disk reaches the hole, which is topologically thecenter of the disk that has been removed from Q, so then weve hit the boundary of Qand our solution is broken.

At r = r0, the minimum ofV(r), our disc mass m1 will be in a stable circular orbit ofradius r0 (which depends upon J). Otherwise we get orbits like Fig. 3.2.


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rr0

V(r)

attractive gravitational potential

repulsive centrifugal potential

Figure 3.1: Potential function for disk pulled by gravitating mass.

Figure 3.2: Orbits for the disc and gravitating mass system.

3.4.3 Free Particle in Special Relativity

In relativistic dynamics the parameter coordinate that parametrizes the particles pathin Minkowski spacetime need not be the time coordinate, indeed in special relativitythere are many allowed time coordinates.

Minkowski spacetime is, n+1 (x0, x1, . . . , xn)

if space is n-dimensional. We normally take x0 as time, and (x1, . . . , xn) as space,but of course this is all relative to ones reference frame. Someone else travelling at somehigh velocity relative to us will have to make a Lorentz transformation to translate fromour coordinates to theirs.

This has a Lorentzian metric

g(v, w) = v0w0 v1w1 . . . vnwn

= vw


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where

=

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 pointparticle, so we let Q be Minkowski spacetime, i.e.,

n+1 (x0, . . . , xn) with the metric defined above. Then the path of the particle is,

q :

( t) Q

where t is a completely arbitrary parameter for the path, not necessarily x0, and notnecessarily proper time either. We want some Lagrangian L : T Q

, i.e., L(qi, qi) suchthat the Euler-Lagrange equations will dictate how our free particle moves at a constantvelocity. Many Lagrangians do this, but the best ones(s) give an action that is inde-pendent of the parameterization of the pathsince the parameterization is unphysical(it cant be measured). So the action

S(q) =

t1t0

L

qi(t), qi(t)

dt

for q : [t0, t1] Q, should be independent oft. The obvious candidate for S is mass timesarclength,

S = m

t1t0

ij qi(t)qj(t) dt

or rather the Minkowski analogue of arclength, called proper time, at least when q is atimelike vector, i.e., ij q

iqj > 0, which says q points into the future (or past) lightconeand makes S real, in fact its then the time ticked off by a clock moving along the path q :[t0, t1] Q. By obvious candidate we are appealing somewhat to physical intuition and

Lightlike

Timelike

Spacelike

generalization. In Euclidean space, free particles follow straight paths, so the arclengthor pathlength variation is an extremum, and we expect the same behavior in Minkowski


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spacetime. Also, the arclength does not depend upon the parameterization, and lastly,

the mass m merely provides the correct units for action.So lets take

L = m

ij qiqj (3.2)

and work out the Euler-Lagrange equations. We have

pi =L

qi= m

qi

ij qiqj

= m2ij q

j

2

ij qiqj

= m

ij qj

ij qiqj =

mqi

q

(Note the numerator is mass times 4-velocity, at least when n = 3 for a real singleparticle system, but were actually in a more general n + 1-dim spacetime, so its morelike the mass times n + 1-velocity). Now note that this pi doesnt change when wechange the parameter to accomplish q q. The Euler-Lagrange equations say,

pi = Fi =L

q i= 0

The meaning of this becomes clearer if we use proper time as our parameter (like

parameterizing a curve by its arclength) so thatt1t0

qdt = t1 t0, t0, t1

which fixes the parametrization up to an additive constant. 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 a straight line, which is as we desired(expected).

Comments

This Lagrangian from Eq.(3.2) has lots of symmetries coming from reparameterizing thepath, so Noethers theorem yields lots of conserved quantities for the relativistic free


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particle. This is in fact called the problem of time in general relativity. Here we see it

starting to show up in special relativity.These reparameterization symmetries work as follows. Consider any (smooth) 1-

parameter family of reparameterizations, i.e., diffeomorphisms

fs :

with f0 = . These act on the space of paths = {q :

Q} as follows: given anyq we get

qs(t) = q

fs(t)

where we should note that qs is physically indistinguishable from q. Lets show that

L = , (when E-L eqns. hold)

so that Noethers theorem gives a conserved quantity

piqi

Here we go then:

L =L

q iqi +

L

qiqi

= piqi

= mqiq

dds

qi

fs(t)

s=0

=mqiq

d

dt

d

dsqi

fs(t)

s=0

=mqiq

d

dtqi

fs(t)fs(t)

ds

s=0

=mqiq

d

dt

qifs

=d

dt piqif

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

if.So to recap a little: we saw the free relativistic particle has

L = mq = m

ij qiqj

and weve considered reparameterization symmetries

qs(t) = q

fs(t)

, fs :


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3.5 Electrodynamics and Relativistic Lagrangians 35

weve used the fact that

qi := dds

qi

fs(t)

s=0= qif

so (repeating a bit of the above)

L =L

q iqi +

L

qiqi

= piqi, (since L/qi = 0, and L/qi = p)

= piqi

= pid

dtqi

= pi ddt

qif

=d

dtpiq

if, and set piqif =

so Noethers theorem gives a conserved quantity

piqi = piq

if piqif

= 0

So these conserved quantities vanish! In short, were seeing an example of what

physicists call gauge symmetries. This is a good topic for starting a new section.

3.5 Electrodynamics and Relativistic Lagrangians

We will continue the story of symmetry and Noethers theorem from the last section witha few more examples. We use principles of least action to conjure up Lagrangians forour systems, realizing that a given system may not have a unique Lagrangian but willoften have an obvious natural Lagrangian. Given a Lagrangian we derive equations ofmotion from the Euler-Lagrange equations. Symmetries ofL guide us in finding conservedquantities, in particular Hamiltonians from time translation invariance, via Noethers

theorem.This section also introduces gauge symmetry, and this is where we begin.

3.5.1 Gauge Symmetry and Relativistic Hamiltonian

What are gauge symmetries?

1. These are symmetries that permute different mathematical descriptions of the samephysical situationin this case reparameterizations of a path.


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2. These symmetries make it impossible to compute q(t) given q(0) and q(0): since if

q(t) is a solution so is q(f(t)) for any reparameterization f :

. We have a highdegree of non-uniqueness of solutions to the Euler-Lagrange equations.

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

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

3.5.2 Relativistic Hamiltonian

What then is the Hamiltonian for special relativity theory? Were continuing here with

the example problem of 3.4.3. Well, the Hamiltonian comes from Noethers theoremfrom time translation symmetry,

qs(t) = q(t + s)

and this is an example of a reparametrization (with f = 1), so we see from the previousresults that the Hamiltonian is zero!

H = 0.

Explicitly, H = piqi where under q(t) q(t + s) we have qi = qif, and so

L = d/dt, which implies = piqi. The result H = 0 follows.Now you know why people talk about the problem of time in general relativity

theory, its glimmerings are seen in the flat Minkowski spacetime of special relativity.You may think its nice and simple to have H = 0, but in fact it means that there is notemporal evolution possible! So we cant establish a dynamical theory on this footing!Thats bad news. (Because it means you might have to solve the static equations for the4D universe as a whole, and thats impossible!)

But there is another conserved quantity deserving the title of energy which is notzero, and it comes from the symmetry,

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

where w n+1 and w points in some timelike direction.

In fact any vector w gives a conserved quantity,

L =L

q iqi +

L

qiqi

= piqi, (since L/qi = 0 and L/qi = pi)

= pi0 = 0


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3.6 Relativistic Particle in an Electromagnetic Field 37

w

q

qs

since qi = wi, qi = wi = 0. This is our from Noethers theorem with = 0, soNoethers theorem says that we get a conserved quantity

piqi = piw

namely, the momentum in the w direction. We know p = 0 from the Euler-Lagrangeequations, for our free particle, but here we see it coming from spacetime translationsymmetry;

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

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

Weve just about exhausted all the basic stuff that we can learn from the free particle.So next well add some external force via an electromagnetic field.

3.6 Relativistic Particle in an Electromagnetic Field

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

dA =F

(3.3)is a 2-form containing the electric and magnetic fields,

F =Aixj

Ajxi

(3.4)

Wed write (for Q having local charts to n+1),

A = A0dx0 + A1dx

1 + . . . Andxn


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3.7 Alternative Lagrangians 39

where v n+1 is the velocity, normalized so that v = 1. Note that now momentum is

no longer mass times velocity! Thats because were in n + 1-d spacetime, the momentumis an n + 1-vector. Continuing the analysis, we find the force

Fi =L

q i=

q i

e Ajq

j

= eAjq i

qj

So the Euler-Lagrange equations say (noting that Ai = Aj

q(t)

:

p = F

d

dt

mvi + eAi

= e

Ajq i

qj

mdvidt

= eAjq i

qj edAidt

mdvidt

= eAjq i

qj eAiqj

qj

= e

Ajq i

Aiqj

qj

the term in parentheses is Fij = the electromagnetic field, F = dA. So we get the followingequations of motion

mdvidt

= eFij qj, (Lorentz force law) (3.6)

(Usually called the Lorenz force law.)

3.7 Alternative Lagrangians

Well soon discuss a charged particle Lagrangian that is free of the reparameterizationsymmetry. First a paragraph on objects other than point particles!

3.7.1 Lagrangian for a String

So weve looked at a point particle and tried

S = m (arclength) +

A


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or with proper time instead of arclength, where the 1-from A can be integrated over

a 1-dimensional path. A generalization (or specialization, depending on how you look atit) would be to consider a Lagrangian for an extended object.

In string theory we boost the dimension by +1 and consider a string tracing out a 2Dsurface as time passes (Fig. 3.3).

becomes

Figure 3.3: Worldtube of a closed string.

Can you infer an appropriate action for this system? Remember, the physical orphysico-philosophical principle weve been using is that the path followed by physicalobjects minimizes the activity or aliveness of the system. Given that we presumablycannot tamper with the length of the closed string, then the worldtube quantity analogous

to arclength or proper time would be the area of the worldtube (or worldsheet for an openstring). If the string is also assumed to be a source of electromagnetic field then we needa 2-form to integrate over the 2D worldtube analogous to the 1-form integrated over thepathline of the point particle. In string theory this is usually the Kalb-Ramond field,call it B. To recover electrodynamic interactions it should be antisymmetric like A, butits tensor components will have two indices since its a 2-form. The string action canthen be written

S = (area) + e

B (3.7)

Weve also replaced the point particle mass by the string tension [masslength1] to

obtain the correct units for the action (since replacing arclength by area meant we had tocompensate for the extra length dimension in the first term of the above string action).This may still seem like weve pulled a rabbit out of a hat. But we havent checked that

this action yields sensible dynamics yet! But supposing it does, then would it justify ourguesswork and intuition in arriving at Eq.(3.7)? Well by now youve probably realizedthat one can have more than one form of action or Lagrangian that yields the samedynamics. So provided we supply reasonabe physically realistic heuristics then whateverLagrangian or action that we come up with will stand a good chance of describing somesystem with a healthy measure of physical verisimilitude.


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3.7 Alternative Lagrangians 41

Thats enough about string for now. The point was to illustrate the type of reasoning

that one can use in conjuring up a Lagrangian. Its particularly useful when Newtoniantheory cannot give us a head start, i.e., in relativistic dynamics and in the physics ofextended particles.

3.7.2 Alternate Lagrangian for Relativistic Electrodynamics

In 3.6, Eq.(3.5) we saw an example of a Lagrangian for relativistic electrodynamicsthat had awkward reparametrization symmetries, meaning that H = 0 and there werenon-unique solutions to the Euler-Lagrange equations arising from applying gauge trans-formations. This freedom to change the gauge can be avoided.

Recall Eq.(3.5), which was a Lagrangian for a charged particle with reparametrizationsymmetry

L = mq + eAiqi

just as for an uncharged relativistic particle. But theres another Lagrangian we can usethat doesnt have this gauge symmetry:

L =1

2mq q + eAiq

i (3.8)

This one even has some nice features.

It looks formally like 12 mv2, familiar from nonrelativistic mechanics.

Theres no ugly square root, so its everywhere differentiable, and theres no troublewith paths being timelike or spacelike in direction, they are handled the same.

What Euler-Lagrange equations does this Lagrangian yield?

pi =L

qi= mqi + eAi

Fi =L

q i= e

Ajq i

qj

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

d

dt

mqi + eAi

= e

Ajq i

qj

mqi = eFij qj

almost as before. (Ive taken to using F here for the electromagnetic field tensor to avoidclashing with F for the generalized force.) The only difference is that we have mqi insteadof mvi where vi = qi/q. So the old Euler-Lagrange equations of motion reduce to the


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new ones if we pick a parametrization with q = 1, which would be a parametrization

by proper time for example.Lets work out the Hamiltonian for this

L =1

2mq q + eAiq

i

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

L = m

qiqi + eAiqi

we got H = 0, time translation was a gauge symmetry. With the new Lagrangian itsnot! Indeed

H = piqi L

and now

pi =L

qi= mqi + eAi

so

H = (mqi + eAi)qi ( 12 mqiq

i + eAiqi)

= 12

mqiqi

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

H = piqi L = 2K (K V) = K+ V,but now the potential V = eAiq

i is linear in velocity, so now

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

As claimed H is not zero, and the fact that its conserved says q(t) is constant asa function of t, so the particles path is parameterized by proper time up to rescalingof t. That is, were getting conservation of speed rather than some more familiarconservation of energy. The reason is that this Hamiltonian comes from the symmetry

qs(t) = q(t + s)

instead of spacetime translation symmetry

qs(t) = q(t) + s w, w n+1

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

L(q, q) = 12

mq2 + Ai(q)qi

has time translation symmetry iff A is translation invariant (but its highly unlikely agiven system of interest will have A(q) = A(q + sw)). In general then theres no conservedenergy for our particle corresponding to translations in time.


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3.8 The General Relativistic Particle 43

01

2

3

4

12

3

4

5

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

w

Figure 3.4: Proper time rescaling vs spacetime translation.

3.8 The General Relativistic Particle

In GR spacetime, Q, is an (n + 1)-dimensional Lorentzian manifold, namely a smooth

(n + 1)-dimensional manifold with a Lorentzian metric g. We define the metric as follows.1. For each x Q, we have a bilinear map

g(x) : TxQ TxQ

(v, w) 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) = gijviwj in any basis, but for different bases gij will

have a different form.

3. g(x) varies smoothly with x.


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3.8.1 Free Particle Lagrangian in GR

The Lagrangian for a free point particle in the spacetime Q is

L(q, q) = m

g(q)(q, q)

= m

gij qiqj

just like in special relativity but with ij replaced by gij. Alternatively we could just aswell use

L(q, q) = 12

mg(q)(q, q)

= 12 mgij qiqj

The big difference between these two Lagrangians is that now spacetime translationsymmetry (and rotation, and boost symmetry) is gone! So there is no conserved energy-momentum (nor angular momentum, nor velocity of center of energy) anymore!

Lets find the equations of motion. Suppose then Q is a Lorentzian manifold withmetric g and L : T Q

is the Lagrangian of a free particle,

L(q, q) = 12

mgij qiqj

We find equations of motion from the Euler-Lagrange equations, which in this case startfrom

pi =L

qi= mgij q

j

The velocity q here is a tangent vector, the momentum p is a cotangent vector, and weneed the metric to relate them, via

g : TqM TqM

(v, w) g(v, w)

which gives

TqM T

q M

v g(v, ).

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

vector gijvj

. This is lurking behind the passage from qi

to the momentum mgijvj

.Getting back to the E-L equations,

pi =L

qi= mgij q

j

Fi =L

q i=

q i

12

mgjk (q)qj qk

= 12 migjk qj qk, (where i =

q i).


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3.8 The General Relativistic Particle 45

So the Euler-Lagrange equations say

d

dtmgij q

j = 12 migjk qj qk.

The mass factors away, so the motion is independent of the mass! Essentially we have ageodesic equation.

We can rewrite this geodesic equation as follows

d

dtgij q

j = 12

igjk qj qk

kgij qkqj + gij q

j = 12 igjk qj qk

gij qj =

12 igjk kgijq

j qk

= 12

igjk kgij j gki

qj qk

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

ijk =

igjk kgij jgki

the minus sign being just a convention (so that we agree with everyone else). This defineswhat we call the Christoffel symbols ijk . Then

qi = gij qj = ijk q

j qk

qi = ijk qj qk.

So we see that q can be computed in terms of q and the Christoffel symbols i

jk , whichis really a particular type of connection that a Lorentzian manifold has (the Levi-Civitaconnection), a connection is just the rule for parallel transporting tangent vectors aroundthe manifold.

Parallel transport is just the simplest way to compare vectors at different points inthe manifold. This allows us to define, among other things, a covariant derivative.

3.8.2 Charged particle in EM Field in GR

We can now apply what weve learned in consideration of a charged particle, of charge e,in an electromagnetic field with potential A, in our Lorentzian manifold. The Lagrangian

would beL = 12 mgjk q

j qk + eAiqi

which again was conjured up be replacing the flat space metric ij by the metric for GRgij . Not surprisingly, the Euler-Lagrange equations then yield the following equations ofmotion,

mqi = mijk qj qk + eFij q

j .

If you want to know more about Lagrangians for general relativity we recommend thepaper by Peldan [Pel94], and also the black book of Misner, Thorne & Wheeler [WTM71].


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3.9 The Principle of Least Action and Geodesics

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

3.9.1 Jacobi and Least Time vs Least Action

Weve mentioned that Fermats principle of least time in optics is analogous to the prin-ciple of least action in particle mechanics. This analogy is strange, since in the principleof least action we fix the time interval q : [0, 1] Q. Also, if one imagines a force on aparticle resulting from a potential gradient at an interface as analogous to light refractionthen you also get a screw-up in the analogy (Fig. 3.5).

light

light

particle

particle

faster

faster slower

slower

V high

V low

n high

n low

Figure 3.5: Least time versus least action.

Nevertheless, Jacobi was able to reinterpret the mechanics of a particle as an opticsproblem and hence unify the two minimization principles. First, lets consider light ina medium with a varying index of refraction n (recall 1/n speed of light). Suppose itsin

n with its usual Euclidean metric. If the light s trying to minimize the time, its tryingto minimize the arclength of its path in the metric

gij = n2ij

that is, the index of refraction n : n (0, ), times the usual Euclidean metric

ij =

10

. . .

0 1


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3.9 The Principle of Least Action and Geodesics 47

This is just like the free particle in general relativity (minimizing its proper time)

except that now gij is a Riemannian metric

g(v, w) = gij viwj

where g(v, v) 0

So well use the same Lagrangian:

L(q, q) =

gij (q)qiqj

and get the same Euler-Lagrange equations:

d2qi

dt2 + ijk qj qk = 0 (3.9)

if q is parameterized by arclength or more generally

q =

gij(q)qiqj = constant.

As before the Christoffel symbols are built from the derivatives of the metric g.Now, what Jacobi did is show how the motion of a particle in a potential could be

viewed as a special case of this. Consider a particle of mass m in Euclidean n with

potential V : n

. It satisfies F = ma, i.e.,

md2

qi

dt2= iV (3.10)

How did Jacobi see (3.10) as a special case of (3.9)? He considered a particle of energyE and he chose the index of refraction to be

n(q) =

2

m

E V(q)

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

2

m

(E V) = 2

m

1

2

mq2 = q.

Note: this is precisely backwards compared to optics, where n(q) is proportional to thereciprocal of the speed of light!! But lets see that it works.

L =

gij(q)qiqj

=

n2(q)qiqj

=

2/m(E V(q))q2


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3.9 The Principle of Least Action and Geodesics 49

3. g(q) varies smoothly with q Q.

An important distinction to keep in mind is that Lorentzian manifolds represent space-times, whereas Riemannian manifolds represent that wed normally consider as just space.

Weve seen at least three important things.

(1) In the geometric optics approximation, light in Q = n acts like particles tracing

out geodesics in the metricgij = n(q)

2ij

where n : Q (0, ) is the index of refraction function.

(2) Jacobi saw that a particle in Q = n in some potential V : Q

traces outgeodesics in the metric

gij = 2m

(E V)ij

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

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

In fact all three of these results can be generalized to cover every problem that wevediscussed!

(1) Light on any Riemannian manifold (Q, q) with index of refraction n : Q (0, )traces out geodesics in the metric h = n2g.

(2) A particle on a Riemannian manifold (Q, q) with potential V : Q

traces outgeodesics w.r.t the metric

h =2

m(E V)g

if it has energy E. Lots of physical systems can be described this way, e.g., theAtwood machine, a rigid rotating body (Q = SO(3)), spinning tops, and others.All of these systems have a Lagrangian which is a quadratic function of position, sothey all fit into this framework.

(3) Kaluza-Klein Theory. A particle with charge e on a Lorentzian manifold (Q, q)in an electromagnetic vector potential follows a path with

qi = ijk qj qk +

e

mFij q

j

whereFij = iAj j Ai

but this is actually geodesic motion on the manifold Q U(1) where U(1) = {ei :

} is a circle.

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


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Lets examine this last result a bit further. To get the desired equations for motion on

Q U(1) we need to given Q U(1) a cleverly designed metric built from g and A wherethe amount of spirallingthe velocity in the U(1) direction is e/m. The metric h on

Q U(1)

a geodesic

Q

the apparent path

Q U(1) is built from g and A in a very simple way. Lets pick coordinates xi on Q wherei {0, . . . , n} since were in n + 1-dimensional spacetime, and is our local coordinateon S1. The components ofh are

hij = gij + AiAj

hi = hi = Aih = 1

Working out the equations for a geodesic in this metric we get

qi = ijk qj qk +

em

Fij qj

q = 0,

if q = e/m

since Fij 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 weve discussed in classical mechanics can be regarded as geodesic motion!


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Chapter 4

From Lagrangians to Hamiltonians

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

L : T Q

where the tangent bundle is the space of position-velocity pairs. But were led to considermomentum

pi =L

qi

since the equations of motion tell us how it changes

dpidt

=L

q i.

4.1 The Hamiltonian Approach

In the Hamiltonian approach we focus on position and momentum, and compute whatthe particle does starting from the energy

H = piqi L(q, q)

reinterpreted as a function of position and momentum, called the Hamiltonian

H : TQ

where the cotangent bundle is the space of position-momentum pairs. In this approach,position and momentum will satisfy Hamiltons equations :

dqi

dt=

H

pi,

dpi

dt=

H

qi

51


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52 From Lagrangians to Hamiltonians

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

: T Q TQ

(q, q) (q, p)

where q Q, and q is any tangent vector in TqQ (not the time derivative of something),

and p is a cotangent vector in Tq Q := (TqQ), given by

q

pi =L

qi

So is defined using L : T Q

. Despite appearances, can be defined in a coordinate-free way, as follows (referring to Fig. 4.1). We want to define Lqi in a coordinate-free

TqQ

TqQ

T QT(q,q)TqQ

qQ

Figure 4.1:

way; its the differential of L in the vertical directioni.e., the qi directions. We have

: T Q Q

(q, q) q

and

d : T(T Q) T Q


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4.1 The Hamiltonian Approach 53

has kernel1 consisting of vertical vectors :

V T Q = ker d T T Q

The differential of L at some point (q, q) T Q is a map from T T Q to

, so we have

(dL)(q,q) T

(q,q)T Q

that is,

dL(q,q) : T(q,q)T Q

.

We can restrict this to V T Q T T Q, getting

f : V(q,q)T Q

.

But noteV(q,q)T Q = T(TqQ)

TqQ

V(q,q)T Q

T(q,q)TqQ

q

and since TqQ is a vector space,

T(q,q)TqQ = TqQ

in a canonical way2. So f gives a linear map

p : TqQ

that is,

p Tq Q

this is the momentum!

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

Given L : T Q TQ, we now know a coordinate-free way of describing the map

: T Q T

Q(q, q) (q, p)

given in local coordinates by

pi =L

qi.

1The kernel of a map is the set of all elements in the domain that map to the null element of therange, so ker d = {v T T Q : d(v) = 0 T Q}.

2The fiber TvV at v V of vector manifold V has the same dimension as V.


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54 From Lagrangians to Hamiltonians

We say L is regular if is a diffeomorphism from T Q to some open subset X TQ.

In this case we can describe what our system is doing equally well by specifying positionand velocity,

(q, q) T Q

or position and momentum

(q, p) = (q, q) X.

We call X the phase space of the system. In practice often X = TQ, then L is said tobe strongly regular.

4.2 Regular and Strongly Regular Lagrangians

This section discusses some examples of the above theory.

4.2.1 Example: A Particle in a Riemannian Manifold with Po-

tential V(q)

For a particle in a Riemannian manifold (Q, q) in a potential V : Q

has Lagrangian

L(q, q) = 12

mgij qiqj V(Q)

Here

pi =L

qi= mgij q

j

so

(q, q) =

q,mg(q, )

so3 L is strongly regular in this case because

TqQ T

q Q

v g(v, )

is 1-1 and onto, i.e., the metric is nondegenerate. Thus is a diffeomorphism, which inthis case extends to all of TQ.

3The missing object there is of course any tangent vector, not inserted since itself is an operatoron tangent vectors, not the result of the operation.

8/2/2019

classical mechanics - baez

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