Lectures on Classical Mechanics

John C. Baez Derek K. Wise

(version of January 15, 2018)

i

\begin{matrix}c \\ ◯ \end{matrix}2017 John C. Baez & Derek K. Wise ii

Preface

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

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

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

Later, Blair Smith from Louisiana State University miraculously appeared and volunteered to turn the notes into L^{A}TEX . While not yet the book I’d eventually like to write, the result may already be helpful for people interested in the mathematics of classical mechanics.

The chapters in this L^{A}TEX version are in the same order as the weekly lectures, but I’ve merged weeks together, and sometimes split them over chapter, to obtain a more textbook 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 particle’s path be a critical point of the action. The prehistory of the Lagrangian approach: D’Alembert’s “principle of least energy” in statics, Fermat’s “principle of least time” in optics, and how D’Alembert generalized 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 an arbitrary manifold. Generalized momentum and force. Noether’s theorem on conserved quantities coming from symmetries. Examples of conserved quantities: energy, momentum and angular momentum.

Week 3 (Apr. 11, 13, 15)—Example problems: (1) The Atwood machine. (2) A frictionless mass on a table attached to a string threaded through a hole in the table, with a 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 charged particle 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-relativistic free particle.

Week 5 (Apr. 25, 27, 29)—How Jacobi unified Fermat’s principle of least time and Lagrange’s principle of least action by seeing the classical mechanics of a particle in a potential as a special case of optics with a position-dependent index of refraction. The ubiquity of geodesic motion. Kaluza-Klein theory. From Lagrangians to Hamiltonians.

Week 6 (May 2, 4, 6)—From Lagrangians to Hamiltonians, continued. Regular and strongly regular Lagrangians. The cotangent bundle as phase space. Hamilton’s equations. Getting Hamilton’s equations directly from a least action principle.

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

Week 8 (May 16, 18, 20)—Towards symplectic geometry. The canonical 1-form and the symplectic 2-form on the cotangent bundle. Hamilton’s equations on a symplectic manifold. Darboux’s theorem.

Week 9 (May 23, 25, 27)—Poisson brackets. The Schrödinger picture versus the Heisenberg picture in classical mechanics. The Hamiltonian version of Noether’s theorem. Poisson algebras and Poisson manifolds. A Poisson manifold that is not symplectic. Liouville’s theorem. Weil’s formula.

Week 10 (June 1, 3, 5)—A taste of geometric quantization. Kähler manifolds.

If you find errors in these notes, please email me! I thank Sheeyun Park and Curtis Vinson for catching lots of errors.

John C. Baez

Contents

1 From Newtonian to Lagrangian Mechanics  1
1.1 Lagrangian and Newtonian Approaches . . . . . . . . . . . . . . . . . . . . 1
1.1.1 Lagrangian versus Hamiltonian Approaches . . . . . . . . . . . . . 5
1.2 Prehistory of the Lagrangian Approach . . . . . . . . . . . . . . . . . . . . 5
1.2.1 The Principle of Least Time . . . . . . . . . . . . . . . . . . . . . . 7
1.2.2 The Principle of Minimum Energy . . . . . . . . . . . . . . . . . . 9
1.2.3 Virtual Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
1.2.4 From Virtual Work to the Principle of Least Action . . . . . . . . . 11
2 Lagrangian Mechanics  15
2.1 The Euler–Lagrange Equations . . . . . . . . . . . . . . . . . . . . . . . . 15
2.2 Noether’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
2.2.1 Time Translation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
2.2.2 Symmetries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
2.2.3 Noether’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
2.2.4 Conservation of Energy . . . . . . . . . . . . . . . . . . . . . . . . . 22
2.3 Conserved Quantities from Symmetries . . . . . . . . . . . . . . . . . . . . 23
2.3.1 Time Translation Symmetry . . . . . . . . . . . . . . . . . . . . . . 24
2.3.2 Space Translation Symmetry . . . . . . . . . . . . . . . . . . . . . . 24
2.3.3 Rotational Symmetry . . . . . . . . . . . . . . . . . . . . . . . . . . 25
3 Examples  27
3.1 The Atwood Machine . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
3.2 Bead on a Rotating Rod . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
3.3 Disk Pulled by Falling Mass . . . . . . . . . . . . . . . . . . . . . . . . . . 30
3.4 Free Particle in Special Relativity . . . . . . . . . . . . . . . . . . . . . . . 32
3.4.1 Comments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34
3.5 Gauge Symmetries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36
3.5.1 Relativistic Hamiltonian . . . . . . . . . . . . . . . . . . . . . . . . 37
3.6 Relativistic Particle in an Electromagnetic Field . . . . . . . . . . . . . . . 38
3.7 Lagrangian for a String . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

3.8 Another Lagrangian for Relativistic Electrodynamics . . . . . . . . . . . . 41
3.9 The Free Particle in General Relativity . . . . . . . . . . . . . . . . . . . . 44
3.10 A Charged Particle on a Curved Spacetime . . . . . . . . . . . . . . . . . . 46
3.11 The Principle of Least Action and Geodesics . . . . . . . . . . . . . . . . . 46
3.11.1 Jacobi and Least Time vs Least Action . . . . . . . . . . . . . . . . 46
3.11.2 The Ubiquity of Geodesic Motion . . . . . . . . . . . . . . . . . . . 49
4 From Lagrangians to Hamiltonians  53
4.1 The Hamiltonian Approach . . . . . . . . . . . . . . . . . . . . . . . . . . 53
4.2 Regular and Strongly Regular Lagrangians . . . . . . . . . . . . . . . . . . 56
4.2.1 Example: A Particle in a Riemannian Manifold with Potential V (q) 56
4.2.2 Example: General Relativistic Particle in an E-M Potential . . . . . 56
4.2.3 Example: Free General Relativistic Particle with Reparameterization Invariance 57
4.2.4 Example: A Regular but not Strongly Regular Lagrangian . . . . . 57
4.3 Hamilton’s Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58
4.3.1 Hamilton and Euler–Lagrange . . . . . . . . . . . . . . . . . . . . . 59
4.3.2 Hamilton’s Equations from the Principle of Least Action . . . . . . 61
4.4 Waves versus Particles—The Hamilton-Jacobi Equations . . . . . . . . . . 63
4.4.1 Wave Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63
4.4.2 The Hamilton-Jacobi Equations . . . . . . . . . . . . . . . . . . . . 65

Chapter 1

From Newtonian to Lagrangian Mechanics

Classical mechanics is a peculiar branch of physics with a long history. It used to be considered the sum total of our theoretical knowledge of the physical universe (Laplace’s daemon, the Newtonian 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 classical mechanics. So it is still an indispensable part of any physicist’s or engineer’s 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 extreme situations (black holes, neutron stars, atomic structure, superconductivity, and so forth). Given that general relativity and quantum mechanics are much harder theories to apply, it is no wonder that scientists 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 of Lagrangian mechanics. Recall Newton’s law:

F = ma  (1.1)

wherein we consider a particle moving in ℝ^{n}. Its position, say q, depends on time t ∈ ℝ, so it defines a function,

q: ℝ ⟶ ℝ^{n}.

From this function we can define velocity,

v = q̇: ℝ ⟶ ℝ^{n}

where q̇ = \frac{dq}{dt}, and also acceleration,

a = q̈: ℝ ⟶ ℝ^{n}.

Now let m > 0 be the mass of the particle, and let F be a vector field on ℝ^{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: ℝ → ℝ^{n} which will have a unique solution given some q(t_{0}) and q̇(t_{0}), provided the vector field F is ‘nice’ — by which we technically mean smooth and bounded (i.e., |F (x)| < B for some B > 0, for all x ∈ ℝ^{n}).

We can then define a quantity called kinetic energy:

K(t) := \frac{1}{2}m v(t) · v(t).  (1.3)

This quantity is interesting because

\begin{aligned}\frac{d}{dt}K(t) & = m v(t) · a(t) \\ & = F (q(t)) · v(t). \end{aligned}

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

\begin{aligned}K(t_{1}) − K(t_{0}) & = ∫_{t_{0}}^{t_{1}} F (q(t)) · v(t) dt \\ & = ∫_{t_{0}}^{t_{1}} F (q(t)) · q̇(t) dt. \end{aligned}

So, the change of kinetic energy is equal to the work done by the force, that is, the integral of F along the curve q: [t_{0}, t_{1}] → ℝ^{n}. In 3 dimensions, Stokes’ theorem relating line integrals to surface integrals of the curl implies that the change in kinetic energy K(t_{1}) − K(t_{0}) is independent of the curve going from q(t_{0}) = a to q(t_{1}) = b iff

∇×F = 0.

This in turn is true iff

F = −∇V  (1.4)

for some function V : ℝ^{n} → ℝ.

In fact, this conclusion is true even when n ≠ 3, using a more general version of Stokes’ theorem: the integral of F along a curve in ℝ^{n} depends only on the endpoints of this curve iff F = −∇V for some function V. Moreover, this function is then unique up to an additive constant; we call this function the potential. A force with this property is called conservative. Why? Because in this case we can define 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 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 that F = ma implies

\begin{aligned}\frac{d}{dt} [K(t) + V (q(t))] & = F (q(t)) · v(t) + ∇V (q(t)) · v(t) \\ & = 0, (because F = −∇V). \end{aligned}

Conservative forces let us apply a bunch of cool techniques. In the Lagrangian approach we define a quantity

L := K(t) − V (q(t))  (1.6)

called the Lagrangian, and for any curve q: [t_{0}, t_{1}] → ℝ^{n} with q(t_{0}) = a, q(t_{1}) = b, we define the action to be

S(q) := ∫_{t_{0}}^{t_{1}} L(t) dt.  (1.7)

From here one can go in two directions. One is to claim that nature causes particles to follow paths of least action, and derive Newton’s equations from that principle. The other is to start with Newton’s 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 simply use the mathematics of variational calculus to show that particles follow paths that are ‘critical points’ of the action S(q) if and only if Newton’s law F = ma holds. To do this,

PSfrag replacements

ℝ^{n}
ℝ
q
q_{s}
t_{0}
t_{1}

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 of S, namely:

\frac{d}{ds}S(q_{s})|_{s=0} = 0  (1.8)

where

q_{s} = q + sδq

for all δq: [t_{0}, t_{1}] → ℝ^{n} with

δq(t_{0}) = δq(t_{1}) = 0.

To show that

F = ma ⇔ \frac{d}{ds}S(q_{s})|_{s=0} = 0 for all δq: [t_{0}, t_{1}] → ℝ^{n} with δq(t_{0}) = δq(t_{1}) = 0 (1.9)

we start by using the definition of the action and the chain rule:

\frac{d}{ds}S(q_{s})|_{s=0} = \frac{d}{ds} ∫_{t_{0}}^{t_{1}} \frac{1}{2}mq̇_{s}(t) · q̇_{s}(t) − V (q_{s}(t)) dt|_{s=0}

= ∫_{t_{0}}^{t_{1}} \frac{d}{ds} [\frac{1}{2}mq̇_{s}(t) · q̇_{s}(t) − V (q_{s}(t))] dt|_{s=0}

= ∫_{t_{0}}^{t_{1}} [mq̇_{s} · \frac{d}{ds}q̇_{s}(t) − ∇V (q_{s}(t)) · \frac{d}{ds}q_{s}(t)] dt|_{s=0}

Next note that

\frac{d}{ds}q_{s}(t) = δq(t)

so

\frac{d}{ds}q̇_{s}(t) = \frac{d}{ds} \frac{d}{dt}q_{s}(t) = \frac{d}{dt} \frac{d}{ds}q_{s}(t) = \frac{d}{dt}δq(t).

Thus we have

\frac{d}{ds}S(q_{s})|_{s=0} = ∫_{t_{0}}^{t_{1}} [mq̇ · \frac{d}{dt}δq(t) − ∇V (q(t)) · δq(t)] dt.

Next we can integrate by parts, noting the boundary terms vanish because δq = 0 at t_{1} and t_{0}:

\frac{d}{ds}S(q_{s})|_{s=0} = ∫_{t_{0}}^{t_{1}} [−mq̈(t) − ∇V (q(t))] · δq(t)dt .

It follows that variation in the action is zero for all variations δq iff the term in brackets is identically zero, that is,

−mq̈(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 written as minus the gradient of some potential. This is not true for all forces in nature: for example, the force on a charged particle in a magnetic field depends on its velocity as well as its position. However, when we develop the Lagrangian approach further we will see that it applies to this force as well!

1.1.1 Lagrangian versus Hamiltonian Approaches

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

They both have their own advantages. In the simplest terms, the Hamiltonian approach focuses on position and momentum, while the Lagrangian approach focuses on position and velocity. The Hamiltonian approach focuses on energy, which is a function of position and momentum — indeed, ‘Hamiltonian’ is just a fancy word for energy. The Lagrangian approach focuses on the Lagrangian, which is a function of position and velocity. 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’ from one moment to another as a particle traces out a path. And, peeking ahead to quantum mechanics, the quantity exp(iS/ℏ), where ℏ is Planck’s constant, will describe the ‘change in 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 provides invaluable insights into 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 time t_{0} to a point b at time t_{1} follows a path that is a critical point of the action,

S = ∫_{t_{0}}^{t_{1}} K − V dt

so that slight changes in its path do not change the action (to first order). Often, though not 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 we might ask, “Why does nature like to minimize the action? And why this action ∫ K−V dt? Why not some other action?”

‘Why’ questions are always tough. Indeed, some people say that scientists should never ask ‘why’. This seems too extreme: a more reasonable attitude is that we should only ask a ‘why’ question if we expect to learn something scientifically interesting in our attempt to answer it.

There are certainly some interesting things to learn from the question “why is action minimized?” First, note that total energy is conserved, so energy can slosh back and forth between kinetic and potential forms. The Lagrangian L = K − V is big when most of the energy is in kinetic form, and small when most of the energy is in potential form. Kinetic energy measures how much is ‘happening’ — how much our system is moving around. Potential energy measures how much could happen, but isn’t yet — that’s what the word ‘potential’ means. (Imagine a big rock sitting on top of a cliff, with the potential to fall down.) So, the Lagrangian measures something we could vaguely refer to as the ‘activity’ or ‘liveliness’ of a system: the higher the kinetic energy the more lively the system, the higher the potential energy the less lively. So, we’re being told that nature likes 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 Earth’s gravitational field, as in Fig. 1.2. The rock traces out a parabola, and we can think of it as doing this in order

PSfrag replacements 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 particle’s “lazy” motion minimizes the action.

to minimize its action. On the one hand, it wants to spend a lot much time near the top of its trajectory, since this is where the kinetic energy is least and the potential energy is 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 of action. The perfect compromise is a parabolic path!

Here we are anthropomorphizing the rock by saying that it ‘wants’ to minimize its action. This is okay if we don’t take it too seriously. Indeed, one of the virtues of the Principle of Least Action is that it lets us put ourselves in the position of some physical system 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 observations of nature: (i) the principle of least time, used in optics, and (ii) the principle of minimum energy, used in statics. By putting these together, we can guess the principle of least action. So, let us recall these earlier minimum principles.

1.2.1 The Principle of Least Time

In 1662, Pierre Fermat pointed out that light obeys the principle of least time: when a ray of light goes from one point to another, it chooses the path that takes the last time. It was known much earlier that moving freely through the air light moves in straight lines, which in Euclidean space are the shortest paths from one point to another. But more interesting than straight lines are piecewise straight paths and curves. Consider reflection of light from a mirror:

PSfrag replacements

A B C C′ θ_{1} θ_{2}

What path does the light take? The empirical answer was known at least since Euclid: it chooses B such that θ_{1} = θ_{2}, angle of incidence equals the angle of reflection. But Hero of Alexandria pointed out that this is precisely the path that minimizes the length of the trajectory subject to the condition that it must hit the mirror (at least at one point). In fact light traveling from A to B takes both the straight paths ABC and AC. Why is ABC the shortest path hitting the mirror? This follows from some basic Euclidean geometry:

B minimizes AB + BC ⇔ B minimizes AB + BC′
⇔ A, B, C′ lie on a line
⇔ θ_{1} = θ_{2}.

Note the introduction of the fictitious image C′ “behind” the mirror. A similar trick is now used in solving electrostatic problems: a conducting surface can be replaced by fictitious mirror image charges to satisfy the boundary conditions. (We also see this method in geophysics when one has a geological fault, and in hydrodynamics when there is a boundary between two media.)

However, the big clue came from refraction of light. Consider a ray of light passing from one medium to another: In 984 AD, the Persian scientist Ibn Sahl pointed out that

PSfrag replacements

medium 1 medium 2

θ_{1}

θ_{2}

each medium has some number n associated with it, called its index of refraction, such that

n_{1} sin θ_{1} = n_{2} sin θ_{2}.

This principle was rediscovered in the 1600s by the Dutch astronomer Willebrord Snellius, and is usually called Snell’s law. It is fundamental to the design of lenses.

In 1662, Pierre de Fermat pointed out in a letter to a friend that Snell’s law would follow if the speed of light were proportional to 1/n and light minimized the time it takes to get from A to C. Note: in this case it is the time that is important, not the length of the path. But the same is true for the law of reflection, since in that case the path of minimum length gives the same results as the path of minimum time.

So, not only is light the fastest thing around, it’s also always taking the quickest path from here to there!

In fact, this idea seems to go back at least to 1021, when Ibn al-Haytham, a scientist in Cairo, mentioned it in his Book of Optics. But the French physicists who formulated the principle of least action were much more likely to have been influenced by Fermat.

Fermat’s friend, Cureau de la Chambre, was unconvinced:

The principle which you take as the basis for your proof, namely that Nature always acts by using the simplest and shortest paths, is merely a moral, and not a physical one. It is not, and cannot be, the cause of any effect in Nature.

The same philosophical objection is often raised against the principle of least action. That is part of what makes the principle so interesting: how does nature “know” how to take the principle of least action? The best explanation so far involves quantum mechanics. But I am getting ahead of myself here.

1.2.2 The Principle of Minimum Energy

Another principle foreshadowing the principle of least action was the “principle of minimum energy”. Before physicists really got going in studying dynamics they thought a lot about statics. Dynamics is the study of moving objects, while statics is the study of objects at rest, or in equilibrium.

PSfrag replacements L_{1}

L_{2}
m_{1}
m_{2}

Figure 1.3: A principle of energy minimization determines a lever’s balance.

For example, Archimedes studied the laws of a see-saw or lever (Fig. 1.3), and he found that this would be in equilibrium if

m_{1}L_{1} = m_{2}L_{2}.

This can be understood using the “principle of virtual work”, which was formalized quite nicely by Johann Bernoulli in 1715. Consider moving the lever slightly, i.e., infinitesimally, In equilibrium, the infinitesimal work done by this motion is zero! The reason is that the

PSfrag replacements

dθ
dq_{1}
dq_{2}

Figure 1.4: A principle of energy minimization determines a lever’s balance.

work done on the i^{th} body is

dW_{i} = F_{i}dq_{i}

and gravity pulls down with a force m_{i}g, so

\begin{aligned}dW_{1} & = (0, 0, −m_{1}g) · (0, 0, −L_{1}dθ) \\ & = m_{1}gL_{1} dθ \end{aligned}

and similarly

dW_{2} = −m_{2}gL_{2} dθ

The total “virtual work” dW = dW_{1} + dW_{2} vanishes for all dθ (that is, for all possible infinitesimal motions) precisely when

m_{1}L_{1} − m_{2}L_{2} = 0

which is just as Archimedes wrote.

1.2.3 Virtual Work

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

The forces and constraints on a system may be time dependent. So equal small infinitesimal displacements of the system might result in the forces F_{i} acting on the system doing different amounts of work at different times. To displace a system by δr_{i} for each 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. It’s called ‘virtual’ because it cannot be realized: any actual displacement would occur over a finite time interval and possibly during which the forces and constraints might change. Now call the work done by this hypothetical virtual displacement, F_{i} · δr_{i}, the virtual work. Consider a system in the special state of being in equilibrium, i.e., when ∑ F_{i} = 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} F_{i} · δr_{i} = 0,  (when in equilibrium)  (1.10)

Note that in the above example we have two particles in ℝ^{3} subject to a constraint (they are pinned to the lever arm). However, a number n of particles in ℝ^{3} can be treated as a single quasi-particle in ℝ^{3n}, and if there are constraints it can move in some submanifold of ℝ^{3n}. So ultimately we need to study a particle on an arbitrary manifold. But, we’ll postpone such sophistication for a while.

For a particle in ℝ^{n}, the principle of virtual work says

q(t) = q_{0} satisfies F = ma,  (it’s in equilibrium)
⇕
dW = F · dq vanishes for all dq ∈ ℝ^{n},  (virtual work is zero for δq → 0)
⇕
F = 0,  (no force on it!)

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

∇V (q_{0}) = 0

that is, we have equilibrium at a critical point of the potential. The equilibrium will be stable if q_{0} is a local minimum of the potential V.

PSfrag replacements

V
ℝ^{n}

unstable equilibrium stable equilibrium

Figure 1.5: A principle of energy minimization determines a lever’s balance.

We can summarize all the above by proclaiming that we have a “principle of least energy” governing stable equilibria. We also have an analogy between statics and dynamics:

Statics  Dynamics
equilibrium, F = 0 F = ma
potential, V  action, S = ∫_{t_{0}}^{t_{1}} K − V dt
critical points of V critical points of S

1.2.4 From Virtual Work to the Principle of Least Action

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 in beauty). One way to make good guesses is to generalize.

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

F (q_{0}) · δq = 0, ∀δq ∈ ℝ^{n}

Around 1743, D’Alembert generalized this principle to dynamics in his Traite de Dynamique. He did it by inventing what he called the “inertia force”, −m a, and postulating that in dynamics equilibrium occurs when the so-called total force, F − ma, vanishes. Of course this is just a restatement of Newton’s law F = ma. But this allowed him to generalize the principle of virtual work from statics to dynamics. Namely, a particle will trace out a path q: [t_{0}, t_{1}] → ℝ^{n} obeying

[F (q(t)) − ma(t)] · δq(t) = 0  (1.11)

for all δq: [t_{0}, t_{1}] → ℝ^{n} with

δq(t_{0}) = δq(t_{1}) = 0.

Let us see how this principle implies the principle of least action. We create a family of paths parameterized by s in the usual way

q_{s}(t) = q(t) + s δq(t)

and define the variational derivative of any function f on the space of paths by

δf(q) = \frac{d}{ds}f(q_{s})|_{s=0}.  (1.12)

Then D’Alembert’s generalized principle of virtual work implies

∫_{t_{0}}^{t_{1}} [(F (q(t)) − mq̈(t)] · δq(t) dt = 0

for all δq, so if F = −∇V we have

0 = ∫_{t_{0}}^{t_{1}} [−∇V (q(t))) − mq̈(t)] · δq(t) dt
= ∫_{t_{0}}^{t_{1}} [−∇V (q(t)) · δq(t) + mq̇(t) · δq̇(t)] dt

where in the second step we did an integration by parts, which has no boundary terms since δq(t_{0}) = δq(t_{1}) = 0. Next, using

δV (q(t)) = \frac{d}{ds}V (q_{s}(t))|_{s=0} = ∇V (q) · \frac{dq_{s}(t)}{ds}|_{s=0} = ∇V (q(t)) · δq(t)

and

δ(q̇(t)^{2}) = 2q̇(t) · δq̇(t)

we obtain

0 = ∫_{t_{0}}^{t_{1}} [−∇V (q(t)) · δq(t) + mq̇(t) · δq̇(t)] dt

= ∫_{t_{0}}^{t_{1}} [−δV (q(t)) + \frac{m}{2}δ(q̇(t)^{2})] dt

= δ(∫_{t_{0}}^{t_{1}} [−V (q(t)) + \frac{m}{2}q̇(t)^{2}] dt)

= δ(∫_{t_{0}}^{t_{1}} [K(t) − V (q(t))] dt)

and thus

δS(q) = 0

where

S(q) = ∫_{t_{0}}^{t_{1}} [K(t) − V (q(t))] dt

is the action of the path q.

In fact, Joseph-Louis Lagrange presented a calculation of this general sort in 1768, and this idea underlies his classic text Mécanique Analitique, which appeared 20 years later. This is why K − V is called the Lagrangian.

I hope you now see that the principle of least action is a natural generalization of the principle of minimum energy from statics to dynamics. Still, there’s something unsatisfying about the treatment so far. I did not really explain why one must introduce the “inertia force”—except, of course, that we need it to obtain agreement with Newton’s F = ma.

We conclude with a few more words about this mystery. Recall from undergraduate physics 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 rotating reference frame. If you are inside the rotating system and you throw a ball straight ahead it will appear to curve away from your target, and if you did not know that you were rotating relative to the rest of the universe then you’d think there was a force on the ball equal to the centrifugal force. If you are inside a big rapidly rotating drum then you’ll also feel pinned to the walls. This is an example of an inertia force which comes from using a funny coordinate system. In fact, in general relativity one sees that—in a certain sense—gravity is an inertia force! But more about this later.

14  From Newtonian to Lagrangian Mechanics

Lagrangian Mechanics

In this chapter we’ll look at Lagrangian mechanics in more generality, and show the principle of least action is equivalent to some equations called the Euler–Lagrange equations.

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 space^{1}. So consider an arbitrary classical system as living in a space of points in some manifold Q. For example, the space for a spherical double pendulum would look like Fig. 2.1, where Q = S^{2} × S^{2}. So our system is “a particle

PSfrag replacements Q = S^{2} × S^{2}

Figure 2.1: Double pendulum configuration space.

in Q”, which means you have to disabuse yourself of the notion that we’re dealing with real particles: what we’re really dealing with is a single abstract particle in an abstract higher dimensional space. This single abstract particle represents two real particles if we are talking about the classical system in Fig. 2.1. Sometimes to make this clear we’ll talk about “the system taking a path”, instead of “the particle taking a path”. It is then clear that when we say, “the system follows a path q(t)” that we’re referring to the point q in configuration space Q that represents all of the particles in the real system.

1The tangent bundle T Q will be referred to as configuration space, later on when we get to the chapter on Hamiltonian mechanics we’ll find a use for the cotangent bundle T^{∗}Q, and normally we call this the phase space.

So as time passes, “the system” traces out a path

q: [t_{0}, t_{1}] ⟶ Q

and we define its velocity

q̇(t) ∈ T_{q(t)}Q

to be the tangent vector at q(t) given by the equivalence class [σ] of curves through q(t) with derivatives σ̇(t) = dq(s)/ds|_{s=t}. We’ll just write is as q̇(t).

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

Γ = {q: [t_{0}, t_{1}] → Q|q(t_{0}) = a, q(t_{1}) = b}

(Γ is an infinite dimensional manifold, but we won’t go into that for now.) Let the Lagrangian for the system be any smooth function of position and velocity:

L : T Q ⟶ ℝ

and define the action

S: Γ ⟶ ℝ

by

S(q) = ∫_{t_{0}}^{t_{1}} L(q, q̇) dt  (2.1)

The path that our abstract particle will actually take is a critical point of S. In other words, it will choose a path q ∈ Γ such that for any smooth 1-parameter family of paths q_{s} ∈ Γ with q_{0} = q, we have

\frac{d}{ds}S(q_{s})|_{s=0} = 0.  (2.2)

For any function f on the space of paths we define its variational derivative by

δf(q) = \frac{d}{ds}f(q)|_{s=0}

so that Eq. (2.2) can be rewritten simply as

δS(q) = 0.  (2.3)

What is a “1-parameter family of paths”? It is nothing more nor less than a set of well-defined paths {q_{s}}, each one labeled by a parameter s. For a “smooth” 1-parameter family of paths, q_{s}(t) depends smoothly on both s and t. Thus, in Fig. 2.2 we can go from q_{0} to q_{s} by smoothly varying the parameter from 0 to a given value s.

PSfrag replacements q_{0}

q_{s}

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

PSfrag replacements

Q
U
a
b

Figure 2.3: Local path variation.

Since Q is a manifold, it admits a covering by coordinate charts. For now, let’s pick coordinates in a neighborhood U of some point q(t) ∈ Q. Next, consider only variations q_{s} such that q_{s} = q outside U. A cartoon of this looks like Fig. 2.3 Then we restrict attention to a subinterval [t′_{0}, t′_{1}] ⊆ [t_{0}, t_{1}] such that q_{s}(t) ∈ U for t′_{0} ≤ t ≤ t′_{1}.

Let’s just go ahead and rename t′_{0} and t′_{1} as “ t_{0} and t_{1}” to drop the primes. We can use the coordinate charts on U,

φ: U ⟶ ℝ^{n}
x ⟼ φ(x) = (x^{1}, x^{2}, …, x^{n})

and we also have coordinates for the tangent vectors

dφ: T U ⟶ T ℝ^{n} ≅ ℝ^{n} × ℝ^{n}
(x, y) ⟼ dφ(x, y) = (x^{1}, …, x^{n}, y^{1}, …, y^{n})

where y ∈ T_{x}Q. We can restrict L: T Q → ℝ to T U ⊆ T Q, and then we can describe L using the coordinates x^{i}, y^{i} on T U. The x^{i} are position coordinates, while the y^{i} are the associated velocity coordinates. Using these coordinates we have

δS = δ ∫_{t_{0}}^{t_{1}} L(q(t), q̇(t)) dt

= ∫_{t_{0}}^{t_{1}} δL(q, q̇) dt

= ∫_{t_{0}}^{t_{1}}(\frac{∂L}{∂x^{i}} δq^{i} + \frac{∂L}{∂y^{i}} δq̇^{i}) dt

where we’ve used the smoothness of L and the Einstein summation convention for repeated indices i. Note that we can write δL as above using a local coordinate patch because the path variations δq are entirely trivial outside the patch for U. Continuing, using the Leibniz rule

\frac{d}{dt}(\frac{∂L}{∂y}δq) = \frac{d}{dt} \frac{∂L}{∂y}δq + \frac{∂L}{∂y}δq̇

we have

δS = ∫_{t_{0}}^{t_{1}}(\frac{∂L}{∂x^{i}} − \frac{d}{dt} \frac{∂L}{∂y^{i}} )δq^{i}(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 we deliberately chose δq that vanish at the endpoints t_{0} and t_{1} inside U. That means the term in brackets must be identically zero, or

\frac{d}{dt} \frac{∂L}{∂y^{i}} − \frac{∂L}{∂x^{i}} = 0  (2.4)

This is necessary to get δS = 0, for all δq, but in fact it’s also sufficient. Physicists always give the coordinates x^{i}, y^{i} on T U the names “q^{i}” and “q̇^{i}”, despite the fact that these symbols also have another meaning, namely the x^{i} and y^{i} coordinates of the point

(q(t), q̇(t)) ∈ T U.

Thus, physicists write

\frac{d}{dt} \frac{∂L}{∂q̇^{i}} = \frac{∂L}{∂q^{i}}

and they call these the Euler–Lagrange equations.

Our derivation of these equations was fairly abstract: we used the terms “position” and “velocity”, but we did not assume these were the usual notions of position and velocity for a particle in ℝ^{3}, or even ℝ^{n}. So, to bring things down to earth, consider the good old familiar case where the configuration space Q is ℝ^{n} and the Lagrangian is

L(q, q̇) = \frac{1}{2}mq̇ · q̇ − V (q)

In this case

\frac{∂L}{∂q^{i}} = −\frac{∂V}{∂q^{i}} = F_{i}

are the components of the the force on the particle, while

\frac{∂L}{∂q̇^{i}} = mq̇^{i}

are the components of its mass times its velocity, also known as its momentum. In physics momentum is denoted by p for some obscure reason, so we say

\frac{∂L}{∂q̇^{i}} = p_{i}

and the Euler–Lagrange equations say simply

\frac{dp}{dt} = F.

The time derivative of momentum is force! Since dp/dt is also mass times acceleration, this is just another way of stating Newton’s law

F = ma.

Based on this example, we can make up nice names for the quantities in the Euler– Lagrange equation in general, for any Lagrangian L: T Q → ℝ. We define

F_{i} = \frac{∂L}{∂q^{i}}

and call this quantity the force, and we define

p_{i} = \frac{∂L}{∂q̇^{i}}

and call this quantity the momentum. The Euler–Lagrange equations then say

\frac{dp_{i}}{dt} = F_{i}.

Written this way, the general Euler–Lagrange equations are revealed to be a generalization of Newton’s law, with ma replaced by the time derivative of momentum.

Term Meaning for a particle in a potential Meaning in general
\frac{∂L}{∂q̇^{i}}  mv_{i}  momentum: p_{i}

\frac{∂L}{∂q^{i}}  −\frac{∂V}{∂q^{i}}  force: F_{i}

2.2 Noether’s Theorem

If the form of a system of dynamical equations does not change under spatial translations then the momentum is a conserved quantity. When the form of the equations is similarly invariant under time translations then the total energy is a conserved quantity (a constant of the equations of motion). Time and space translations are examples of 1-parameter groups of transformations. Invariance under a group of transformations is precisely what we mean by a symmetry in group theory. So symmetries of a dynamical system give conserved quantities or conservation laws. The rigorous statement of all this is the content of Noether’s theorem.

2.2.1 Time Translation

To handle time translations we need to replace our paths q: [t_{0}, t_{1}] → Q by paths q : ℝ → Q, and then define a new space of paths,

Γ = {q : ℝ → Q}.

The bad news is that the action

S(q) = ∫_{−∞}^{∞} L(q(t), q̇(t)) dt

typically will not converge, so S is then no longer a function of the space of paths. Nevertheless, if δq = 0 outside of some finite interval, then the functional variation,

δS := ∫_{−∞}^{∞} \frac{d}{ds}L(q_{s}(t), q̇_{s}(t))|_{s=0} dt

will converge, since the integral is smooth and vanishes outside this interval. Moreover, demanding that this δS vanishes for all such variations δq is enough to imply the Euler– Lagrange equations:

δS = ∫_{−∞}^{∞} \frac{d}{ds}L(q_{s}(t), q̇_{s}(t))|_{s=0} dt

= ∫_{−∞}^{∞} (\frac{∂L}{∂q_{i}}δq_{i} + \frac{∂L}{∂q̇_{i}}δq̇_{i}) dt

= ∫_{−∞}^{∞} (\frac{∂L}{∂q_{i}} − \frac{d}{dt} \frac{∂L}{∂q̇_{i}}) δq_{i} dt

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

\frac{∂L}{∂q̇^{i}} δq̇^{i} = \frac{d}{dt} (\frac{∂L}{∂q̇^{i}} δq^{i}) − (\frac{d}{dt} \frac{∂L}{∂q̇^{i}} ) δq

vanishes when we integrate. Then the whole thing vanishes for all compactly supported smooth δq iff

\frac{d}{dt} \frac{∂L}{∂q̇_{i}} = \frac{∂L}{∂q_{i}}.

So, we get Euler–Lagrange equations again.

Generalized Coordinates

2.2.2 Symmetries

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

Definition 2.1 (one-parameter family of symmetries). A 1-parameter family of symmetries of a Lagrangian system L : T Q → ℝ is a smooth map,

F : ℝ × Γ ⟶ Γ
(s, q) ⟼ q_{s},  with q_{0} = q

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

δL = \frac{dℓ}{dt}

for some ℓ : T Q → ℝ, that is,

\frac{d}{ds}L(q_{s}(t), q̇_{s}(t))|_{s=0} = \frac{d}{dt}ℓ(q_{s}(t), q̇_{s}(t))

for all paths q.

Remark: The simplest case is δL = 0, in which case we really have a way of moving paths around (q ↦ q_{s}) that doesn’t change the Lagrangian—i.e., a symmetry of L in the most obvious way. But δL = \frac{d}{dt}ℓ is a sneaky generalization whose usefulness will become clear.

2.2.3 Noether’s Theorem

Here’s a statement of the theorem. Note that ℓ in this theorem is the function associated with F in Definition 2.1.

Theorem 2.1 (Noether’s Theorem). Suppose F is a one-parameter family of symmetries of the Lagrangian system, L : T Q → ℝ. Then,

p_{i}δq^{i} − ℓ

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

\frac{d}{dt} [\frac{∂L}{∂y^{i}} (q(s)q̇(s)) \frac{d}{ds}q_{s}^{i}(t)|_{s=0} − ℓ(q(t), q̇(t))] = 0

Proof.

\frac{d}{dt}(p^{i}δq_{i} − ℓ) = ṗ_{i}δq^{i} + p_{i}δq̇^{i} − \frac{d}{dt}ℓ

= \frac{∂L}{∂q^{i}} δq^{i} + \frac{∂L}{∂q̇^{i}} δq̇^{i} − δL

= δL − δL = 0.

“Okay, big deal” you might say. Before this can be of any use we’d need to find a symmetry F. Then we’d need to find out what this p_{i}δq^{i} − ℓ business is that is conserved. So let’s look at some examples.

2.2.4 Conservation of Energy

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

All of our Lagrangian systems will have time translation invariance (because the laws 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

q_{s}(t) = q(t + s)

This indeed gives,

δL = L̇

for

\frac{d}{ds}L(q_{s})|_{s=0} = \frac{d}{dt}L = L̇

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

p_{i}δq^{i} − ℓ = p_{i}q̇^{i} − L

which we normally call the energy. For example, if Q = ℝ^{n}, and if

L = \frac{1}{2}mq̇^{2} − V (q)

then this quantity is

mq̇ · q̇ − (\frac{1}{2}mq̇ · q̇ − V ) = \frac{1}{2}mq̇^{2} + V (q).

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

Let’s repeat this example, this time with a specific Lagrangian. It doesn’t matter what the Lagrangian is, if it has 1-parameter families of symmetries then it’ll have conserved quantities, 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.)

2.3 Conserved Quantities from Symmetries

We’ve seen that any 1-parameter family

F_{s} : Γ ⟶ Γ
q ⟼ q_{s}

which satisfies

δL = ℓ̇

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

p_{i}δq^{i} − ℓ.

As usual we’ve defined

δL = \frac{d}{ds}L(q_{s}(t), q̇_{s}(t))|_{s=0} .

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

2.3.1 Time Translation Symmetry

For any Lagrangian system, L : T Q → ℝ, we have a 1-parameter family of symmetries

q_{s}(t) = q(t + s)

because

δL = L̇

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

H = p_{i}q̇^{i} − L  (2.5)

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

For example: a particle on ℝ^{n} in a potential V has Q = ℝ^{n}, L(q, q̇) = \frac{1}{2}mq̇^{2} − V (q). This system has

p_{i}q̇^{i} = \frac{∂L}{∂q̇^{i}} q̇^{i} = mq̇^{2} = 2K

so

H = p_{i}q̇^{i} − L = 2K − (K − V) = K + V

as you’d have hoped.

2.3.2 Space Translation Symmetry

For a free particle in ℝ^{n}, we have Q = ℝ^{n} and L = K = \frac{1}{2}mq̇^{2}. This has spatial translation symmetries, so that for any v ∈ ℝ^{n} we have the symmetry

q_{s}(t) = q(t) + s v

with

δL = 0

because δq̇ = 0 and L depends only on q̇ not on q in this particular case. (Since L does not depend upon q^{i} we’ll call q^{i} an ignorable coordinate; as above, these ignorables always give symmetries, hence conserved quantities. It is often useful therefore, to change coordinates so as to make some of them ignorable if possible!)

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

p_{i}δq^{i} = mq̇_{i}v^{i} = mq̇ · v.

Aside: Note the subtle difference between two uses of the term “momentum”; here it is a conserved quantity derived from space translation invariance, but earlier it was a different thing, namely the momentum ∂L/∂q̇^{i} = p_{i} conjugate to q^{i}. These two different “momentum’s” happen to be the same in this example!

Since this is conserved for all v we say that mq̇ ∈ ℝ^{n} is conserved. (In fact that whole Lie group G = ℝ^{n} is acting as a translation symmetry group, and we’re getting a 𝔮(= ℝ^{n})-valued conserved quantity!)

2.3.3 Rotational Symmetry

The free particle in ℝ^{n} also has rotation symmetry. Consider any X ∈ 𝔰𝔬(n) (that is a skew-symmetric n × n matrix), then for all s ∈ ℝ the matrix e^{sX} is in SO(n), that is, it describes a rotation. This gives a 1-parameter family of symmetries

q_{s}(t) = e^{sX}q(t)

which has

δL = \frac{∂L}{∂q^{i}} δq^{i} + \frac{∂L}{∂q̇^{i}} δq̇^{i} = mq̇_{i}δq̇^{i}.

Now q_{i} is ignorable, so ∂L/∂q^{i} = 0, and ∂L/∂q̇^{i} = p_{i}, and

δq̇^{i} = \frac{d}{ds}q̇_{s}^{i}|_{s=0}

= \frac{d}{ds} \frac{d}{dt}(e^{sX}q)|_{s=0}

= \frac{d}{dt}X q

= X q̇.

Thus,

δL = mq̇_{i}X_{j}^{i}q̇^{j}

= mq̇ · (X q̇) = 0

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

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

We write,

p_{i}δq^{i} = mq̇_{i} · (X q)^{i}

(δq^{i} = Xq just as δq̇^{i} = Xq̇ in our previous calculation), or if X has zero entries except in ij and ji positions, where it’s ±1, then we get

m(q̇_{i}q^{j} − q̇_{j}q^{i})

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

mq̇ × q.

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

I mentioned earlier that we can do mechanics with any Lagrangian, but if we want to be useful we’d better pick a Lagrangian that actually describes a real system. But how do 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 the Lagrangian is just the kinetic energy, since there is no potential energy variation for a free particle. We’d 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 laws should be, instead we plug in some assumed physics and use the Lagrangian approch to solve the system of equations. If we like, we can then compare our answers with experiments, which indirectly tells us something about the physical laws—but only provided the Lagrangian formulation of mechanics is itself a valid procedure in the first place.

Examples

To see how Lagrangian mechanics and Noether’s theorem works in practise, let’s do some problems. The Lagrangian approach is often vastly superior to the simplistic F = ma formulation of mechanics. The Lagrangian formulation allows the configuration space to be any manifold, and allows us to easily use any coordinates we wish.

3.1 The Atwood Machine

Consider a frictionless pulley with two masses, m_{1} and m_{2}, hanging from it:

PSfrag replacements

x
ℓ − x
m_{1}
m_{2}

We have

K = \frac{1}{2}(m_{1} + m_{2})(\frac{d}{dt}(ℓ − x))^{2} = \frac{1}{2}(m_{1} + m_{2})ẋ^{2}
V = −m_{1}gx − m_{2}g(ℓ − x)

so

L = K − V = \frac{1}{2}(m_{1} + m_{2})ẋ^{2} + m_{1}gx + m_{2}g(ℓ − x).

The configuration space is Q = (0, ℓ), and x ∈ (0, ℓ) (we could use the “owns” symbol ∋ here and write Q = (0, ℓ) ∋ x). Moreover T Q = (0, ℓ)×ℝ ∋ (x, ẋ). As usual L : T Q → ℝ. Note that solutions of the Euler–Lagrange equations will only be defined for some time t ∈ ℝ, as eventually the solutions reaches the “edge” of Q.

The momentum is:

p = \frac{∂L}{∂ẋ} = (m_{1} + m_{2})ẋ

and the force is:

F = \frac{∂L}{∂x} = (m_{1} − m_{2})g

The Euler–Lagrange equations say

ṗ = F
(m_{1} + m_{2})ẍ = (m_{1} − m_{2})g
ẍ = \frac{m_{1} − m_{2}}{m_{1} + m_{2}} g

So this is like a falling object in a downwards gravitational acceleration a = (\frac{m_{1}−m_{2}}{m_{1}+m_{2}} ) g.

It is trivial to integrate the expression for ẍ twice (feeding in some appropriate initial conditions) to obtain the complete solution to the motion x(t) and ẋ(t). Note that ẍ = 0 when m_{1} = m_{2}, and ẍ = g if m_{2} = 0.

3.2 Bead on a Rotating Rod

Now consider a bead of mass m sliding in a frictionless way on a rod rotating in a horizontal plane. The rod will go through the origin (0, 0) ∈ ℝ^{2} and rotate at a constant angular velocity ω, so if the angle of the rod is θ(t) at time t, we may as well assume

θ(t) = ωt.

The bead’s position on the rod will be given by a number q(t) ∈ ℝ depending on time. What will the bead do?

Since the bead lies on a line, namely the rod, its configuration space is Q = ℝ, and its Lagrangian is a function L: T Q → ℝ where T Q = ℝ × ℝ. Its position and velocity thus form a pair (q(t), q̇(t)) ∈ T Q.

Since the rod lies in a horizontal plane, its gravitational potential energy is constant, and doesn’t affect the Euler–Lagrange equations to assume this constant is zero. So, we take the bead’s potential energy to be

V = 0.

Its kinetic energy is

K = \frac{1}{2}mv · v

where v is its velocity. But what is its velocity? Its position in the plane at time t is

(x(t), y(t)) = (r(t) cos θ(t), r(t) sin θ(t))
= (|q(t)| cos(ωt), |q(t)| sin(ωt)).

The time derivative of |q(t)| is ±q(t), with the plus sign if q(t) > 0 and the minus sign if q(t) < 0. If q(t) = 0 we seem to be in trouble, because the absolute value is not differentiable at zero, but we’ll see a way around this later. Ignoring this case for now, the bead’s velocity in the plane is thus

v(t) = (ẋ(t), ẏ(t))
= \frac{d}{dt}(|q(t)| cos(ωt), |q(t)| sin(ωt))
= (±q̇(t) cos(ωt) − |q(t)|ω sin(ωt), ±q̇(t) sin(ωt) + |q(t)|ω cos(ωt)).

so we have

v(t) · v(t) = q̇^{2} cos^{2}(ωt) ∓ |q|q̇ sin(ωt) cos(ω(t)) + ω^{2}q^{2} sin^{2}(ωt)+
q̇^{2} sin^{2}(ωt) ± |q|q̇ sin(ωt) cos(ω(t)) + ω^{2}q^{2} cos^{2}(ωt)
= q̇^{2} + ω^{2}q^{2}.

Thus, the bead’s kinetic energy is

K = \frac{1}{2}mv · v = q̇^{2} + ω^{2}q^{2}.

This has a simple interpretation: the first term is the ‘radial’ part of the kinetic energy, while the second term is the ‘angular’ part.

The Lagrangian of the bead is

L(q, q̇) = K − V = q̇^{2} + ω^{2}q^{2}.

Note that this is perfectly well-defined and smooth at q = 0. Thus, our problem at that point is easily dealt with: simply define the Lagrangian as above. A more careful analysis shows this is reasonable.

The force on the bead is

F = \frac{∂L}{∂q} = mω^{2}q.

This is called centrifugal force, since it’s caused by the rotating rod and tends to pull the bead out. The bead’s momentum is

p = \frac{∂L}{∂q̇} = mq̇.

The Euler–Lagrange equation for the bead says

\frac{dp}{dt} = F

or

mq̈(t) = mω^{2}q(t)

or simply

q̈(t) = ω^{2}q(t).

The mass of the bead does not affect its equation of motion. This equation is easy to solve, too:

q(t) = Ae^{ωt} + Be^{−ωt}.

Thus, the bead is likely to shoot off to infinity as t → +∞, due to the centrifugal force. If ω > 0, the only exception is when A = 0: in this case the bead moves ever closer to the origin, or else just sits there if A = B = 0.

3.3 Disk Pulled by Falling Mass

Consider next a disk pulled across a table by a falling mass. The disk is free to move on a frictionless surface, and it can thus whirl around the hole to which it is tethered to the mass below.

PSfrag replacements

r ℓ − r m_{1} m_{2}

no swinging allowed!

Here Q = open disk of radius ℓ, minus its center

= (0, ℓ) × S^{1} ∋ (r, θ)

T Q = (0, ℓ) × S^{1} × ℝ × ℝ ∋ (r, θ, ṙ, θ̇)

K = \frac{1}{2}m_{1}(ṙ^{2} + r^{2}θ̇^{2}) + \frac{1}{2}m_{2}(\frac{d}{dt}(ℓ − r))^{2}
V = gm_{2}(r − ℓ)
L = \frac{1}{2}m_{1}(ṙ^{2} + r^{2}θ̇^{2}) + \frac{1}{2}m_{2}ṙ^{2} + gm_{2}(ℓ − r)

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

p_{r} = \frac{∂L}{∂ṙ} = (m_{1} + m_{2})ṙ
p_{θ} = \frac{∂L}{∂θ̇} = m_{1}r^{2}θ̇.

Note that θ is an “ignorable coordinate”—it doesn’t appear in L—so there’s a symmetry, rotational symmetry, and p_{θ}, the conjugate momentum, is conserved.

The forces are

F_{r} = \frac{∂L}{∂r} = m_{1}rθ̇^{2} − gm_{2}
F_{θ} = \frac{∂L}{∂θ} = 0 (θ is ignorable).

Note: in F_{r} the term m_{1}rθ̇^{2} is recognizable as a centrifugal force, pushing m_{1} radially out, while the term −gm_{2} is gravity pulling m_{2} down and thus pulling m_{1} radially in.

So, the Euler–Lagrange equations give

ṗ_{r} = F_{r}, (m_{1} + m_{2})r̈ = m_{1}rθ̇^{2} − m_{2}g
ṗ_{θ} = 0, p_{θ} = m_{1}r^{2}θ̇ = J = a constant.

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

θ̇ = \frac{J}{m_{1}r^{2}}

so

(m_{1} + m_{2})r̈ = \frac{J^{2}}{m_{1}r^{3}} − m_{2}g

Thus effectively we have a particle on (0, ℓ) of mass m = m_{1} + m_{2} feeling a force

F_{r} = \frac{J^{2}}{m_{1}r^{3}} − m_{2}g

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

V (r) = \frac{J^{2}}{2m_{1}r^{2}} + m_{2}gr

this is a sum of two terms that look like Fig. 3.1

If θ̇(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 the center of the disk that has been removed from Q, so then we’ve hit the boundary of Q and our solution is broken.

At r = r_{0}, the minimum of V (r), our disc mass m_{1} will be in a stable circular orbit of radius r_{0} (which depends upon J). Otherwise we get orbits like Fig. 3.2.

PSfrag replacements

r
r_{0}
V (r)

attractive gravitational potential repulsive centrifugal potential

no swinging allowed!

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

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

3.4 Free Particle in Special Relativity

In relativistic dynamics the parameter coordinate that parametrizes the particle’s path in Minkowski spacetime need not be the “time coordinate”, indeed in special relativity there are many allowed time coordinates.

Minkowski spacetime is,

ℝ^{n+1} ∋ (x^{0}, x^{1}, …, x^{n})

if space is n-dimensional. We normally take x^{0} as “time”, and (x^{1}, …, x^{n}) as “space”, but of course this is all relative to one’s reference frame. Someone else traveling at some high velocity relative to us will have to make a Lorentz transformation to translate from our coordinates to theirs.

This has a Lorentzian metric

g(v, w) = v^{0}w^{0} − v^{1}w^{1} − … − v^{n}w^{n}
= η_{μν}v^{μ}w^{ν}

where

η_{μν} = ( \begin{matrix}1 & 0 & 0 & … & 0 \\ 0 & −1 & 0 & … & 0 \\ 0 & 0 & −1 & & 0 \\ ⋮ & ⋮ & & ... & ⋮ \\ 0 & 0 & 0 & … & −1 \end{matrix} ) .

In special relativity we take spacetime to be the configuration space of a single point particle, so we let Q be Minkowski spacetime, i.e., ℝ^{n+1} ∋ (x^{0}, …, x^{n}) with the metric η_{μν} defined above. Then the path of the particle is,

q : ℝ(∋ t) ⟶ Q

where t is a completely arbitrary parameter for the path, not necessarily x^{0}, and not necessarily proper time either. We want some Lagrangian L : T Q → ℝ, i.e., L(q^{i}, q̇^{i}) such that the Euler–Lagrange equations will dictate how our free particle moves at a constant velocity. Many Lagrangians do this, but the “best” one should give an action that is independent of the parameterization of the path—since the parameterization is “unphysical”: it can’t be measured. So the action

S(q) = ∫_{t_{0}}^{t_{1}} L(q^{i}(t), q̇^{i}(t)) dt

for q: [t_{0}, t_{1}] → Q, should be independent of t. The obvious candidate for S is mass times arclength,

S = m ∫_{t_{0}}^{t_{1}} \sqrt{η_{ij}q̇^{i}(t)q̇^{j}(t)} dt

or rather the Minkowski analogue of arclength, called proper time, at least when q̇ is a timelike vector, i.e., η_{ij}q̇^{i}q̇^{j} > 0, which says q̇ points into the future (or past) lightcone and makes S real, in fact it’s then the time ticked off by a clock moving along the path q : [t_{0}, t_{1}] → Q. By “obvious candidate” we are appealing somewhat to physical intuition and

generalization. In Euclidean space, free particles follow straight paths, so the arclength or pathlength variation is an extremum, and we expect the same behavior in Minkowski spacetime. Also, the arclength does not depend upon the parameterization, and lastly, the mass m merely provides the correct units for ‘action’.

So let’s take

L = m\sqrt{η_{ij}q̇^{i}q̇^{j}}  (3.1)

and work out the Euler–Lagrange equations. We have

p_{i} = \frac{∂L}{∂q̇^{i}} = m\frac{∂}{∂q̇^{i}} \sqrt{η_{ij}q̇^{i}q̇^{j}}

= m\frac{2η_{ij}q̇^{j}}{2\sqrt{η_{ij}q̇^{i}q̇^{j}}}

= m\frac{η_{ij}q̇^{j}}{\sqrt{η_{ij}q̇^{i}q̇^{j}}} = \frac{mq̇_{i}}{∥q̇∥} .

Now note that this p_{i} doesn’t change when we change the parameter to accomplish q̇ ↦ αq̇. The Euler–Lagrange equations say

ṗ_{i} = F_{i} = \frac{∂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 that

∫_{t_{0}}^{t_{1}}∥q̇∥dt = t_{1} − t_{0}, ∀ t_{0}, t_{1}

which fixes the parametrization up to an additive constant. This implies ∥q̇∥ = 1, so that

p_{i} = m\frac{q̇_{i}}{∥q̇∥} = mq̇_{i}

and the Euler–Lagrange equations say

ṗ_{i} = 0 ⇒ mq̈_{i} = 0

so our free particle moves unaccelerated along a straight line, as one would expect.

3.4.1 Comments

This Lagrangian from Eq.(3.1) has lots of symmetries coming from reparameterizing the path, so Noether’s theorem yields lots of conserved quantities for the relativistic free 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) 1parameter family of reparameterizations, i.e., diffeomorphisms

f_{s}: ℝ ⟶ ℝ

with f_{0} = 1_{ℝ}. These act on the space of paths Γ = {q : ℝ → Q} as follows: given any q ∈ Γ we get

q_{s}(t) = q(f_{s}(t))

where we should note that q_{s} is physically indistinguishable from q. Let’s show that

δL = ℓ̇, (when Euler–Lagrange eqns. hold)

so that Noether’s theorem gives a conserved quantity

p_{i}δq^{i} − ℓ

Here we go then:

δL = \frac{∂L}{∂q^{i}} δq^{i} + \frac{∂L}{∂q̇^{i}} δq̇^{i}

= p_{i}δq̇^{i}

= \frac{mq̇_{i}}{∥q̇∥} \frac{d}{ds}q̇^{i}(f_{s}(t))|_{s=0}

= \frac{mq̇_{i}}{∥q̇∥} \frac{d}{dt} \frac{d}{ds}q^{i}(f_{s}(t))|_{s=0}

= \frac{mq̇_{i}}{∥q̇∥} \frac{d}{dt}q̇^{i}(f_{s}(t))\frac{f_{s}(t)}{ds}|_{s=0}

= \frac{mq̇_{i}}{∥q̇∥} \frac{d}{dt} (q̇^{i}δf_{s})

= \frac{d}{dt} (p_{i}q̇^{i}δf)

where in the last step we used the Euler–Lagrange equations, i.e. \frac{d}{dt}p_{i} = 0, so δL = ℓ̇ with ℓ = p_{i}q̇^{i}δf.

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

L = m∥q̇∥ = m\sqrt{η_{ij}q̇^{i}q̇^{j}}

and we’ve considered reparameterization symmetries

q_{s}(t) = q(f_{s}(t)), f_{s} : ℝ → ℝ

we’ve used the fact that

δq^{i} = \frac{d}{ds}q^{i}(f_{s}(t))|_{s=0} = q̇^{i}δf

so (repeating a bit of the above)

δL = \frac{∂L}{∂q^{i}} δq^{i} + \frac{∂L}{∂q̇^{i}} δq̇^{i}

= p_{i}δq̇^{i}, (since ∂L/∂q^{i} = 0, and ∂L/∂q̇^{i} = p) = p_{i}δq̇^{i}

= p_{i} \frac{d}{dt}δq^{i}

= p_{i} \frac{d}{dt}q̇^{i}δf

= \frac{d}{dt}p_{i}q̇^{i}δf, and set p_{i}q̇^{i}δf = ℓ

so Noether’s theorem gives a conserved quantity

p_{i}δq^{i} − ℓ = p_{i}q̇^{i}δf − p_{i}q̇^{i}δf

= 0

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

3.5 Gauge Symmetries

What are gauge symmetries?

1. These are symmetries that permute different mathematical descriptions of the same physical situation—in this case reparameterizations of a path.

2. These symmetries make it impossible to compute q(t) given q(0) and q̇(0): since if q(t) is a solution so is q(f(t)) for any reparameterization f : ℝ → ℝ. We have a high degree 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 to test, so we often use (3) to distinguish gauge symmetries from physical symmetries.

3.5.1 Relativistic Hamiltonian

What then is the Hamiltonian for special relativity theory? We’re continuing here with the example problem of §3.4. Well, the Hamiltonian comes from Noether’s theorem from time translation symmetry,

q_{s}(t) = q(t + s)

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

H = 0.

Explicitly, H = p_{i}δq̇^{i} − ℓ where under q(t) → q(t + s) we have δq̇^{i} = q̇^{i}δf, and so δL = dℓ/dt, which implies ℓ = p_{i}δq^{i}. 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 it’s nice and simple to have H = 0, but in fact it means that there is no temporal evolution possible! So we can’t establish a dynamical theory on this footing! That’s bad news. (Because it means you might have to solve the static equations for the 4D universe as a whole, and that’s impossible!)

But there is another conserved quantity deserving the title of “energy” which is not zero, and it comes from the symmetry,

q_{s}(t) = q(t) + s w

where w ∈ ℝ^{n+1} and w points in some timelike direction.

PSfrag replacements

w
q
q_{s}

In fact any vector w gives a conserved quantity,

δL = \frac{∂L}{∂q^{i}} δq^{i} + \frac{∂L}{∂q̇^{i}} δq̇^{i}
= p_{i}δq̇^{i}, (since ∂L/∂q^{i} = 0 and ∂L/∂q̇^{i} = p_{i})
= p_{i}0 = 0

since δq^{i} = w^{i}, δq̇^{i} = ẇ^{i} = 0. This is our ℓ̇ from Noether’s theorem with ℓ = 0, so Noether’s theorem says that we get a conserved quantity

p_{i}δq^{i} − ℓ = p_{i}w^{i}

namely, the momentum in the w direction. We know ṗ = 0 from the Euler–Lagrange equations, for our free particle, but here we see it coming from spacetime translation symmetry:

p =(p_{0}, p_{1}, …, p_{n})
p_{0} is energy, (p_{1}, …, p_{n}) is spatial momentum.

We’ve just about exhausted all the basic stuff that we can learn from the free particle. So next we’ll 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  (3.2)

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

F_{ij} = \frac{∂A_{j}}{∂x^{i}} − \frac{∂A_{i}}{∂x^{j}}  (3.3)

We can write (for Q having local charts to ℝ^{n+1}),

A = A_{0}dx^{0} + A_{1}dx^{1} + … A_{n}dx^{n}

and then because d^{2} = 0

dA = dA_{0}dx^{0} + dA_{1}dx^{1} + … dA_{n}dx^{n}

and since the A_{j} are just functions,

dA_{j} = ∂_{i}A_{j}dx^{i}

using the summation convention and ∂_{i} := ∂/∂x^{i}. The reader can easily check that the components for F = F_{01}dx^{0} ∧ dx^{1} + F_{02}dx^{0} ∧ dx^{2} + …, agrees with the matrix expression below (at least in 4 dimensions).

So, for example, in 4-dimensional spacetime

F = ( \begin{matrix}0 & E_{1} & E_{2} & E_{3} \\ −E_{1} & 0 & B_{3} & −B_{2} \\ −E_{2} & −B_{3} & 0 & B_{1} \\ −E_{3} & B_{2} & −B_{1} & 0 \end{matrix} )

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

S = m ∫_{t_{0}}^{t_{1}}∥q̇∥ dt + e ∫_{q} A

here

∫_{t_{0}}^{t_{1}}∥q̇∥ dt = proper time,

∫_{q} A = integral of A along the path q.

Note that since A is a 1-form we can integrate it over an oriented manifold, but one can also write the path integral using time t as a parameter, with A_{i}q̇^{i} dt the differential, after dq^{i} = q̇^{i}dt.

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

L(q, q̇) = m∥q̇∥ + eA_{i}q̇^{i}  (3.4)

so we can work out the Euler–Lagrange equations:

p_{i} = \frac{∂L}{∂q̇^{i}} = m\frac{q̇_{i}}{∥q̇∥} + eA_{i}
= mv_{i} + e A_{i}

where v ∈ ℝ^{n+1} is the velocity, normalized so that ∥v∥ = 1. Note that now momentum is no longer mass times velocity! That’s because we’re in n + 1-d spacetime, the momentum is an n + 1-vector. Continuing the analysis, we find the force

F_{i} = \frac{∂L}{∂q^{i}} = \frac{∂}{∂q^{i}} (e A_{j}q̇^{j})

= e\frac{∂A_{j}}{∂q^{i}} q̇^{j}

So the Euler–Lagrange equations say (noting that A_{i} = A_{j}(q(t)):

ṗ = F
\frac{d}{dt}(mv_{i} + eA_{i}) = e\frac{∂A_{j}}{∂q^{i}} q̇^{j}
m\frac{dv_{i}}{dt} = e\frac{∂A_{j}}{∂q^{i}} q̇^{j} − e\frac{dA_{i}}{dt}
m\frac{dv_{i}}{dt} = e\frac{∂A_{j}}{∂q^{i}} q̇^{j} − e\frac{∂A_{i}}{∂q^{j}} q̇^{j}

= e (\frac{∂A_{j}}{∂q^{i}} − \frac{∂A_{i}}{∂q^{j}} ) q̇^{j}

the term in parentheses is F_{ij} = the electromagnetic field, F = dA. So we get the following equations of motion

m\frac{dv_{i}}{dt} = eF_{ij}q̇^{j}, (Lorentz force law)

(3.5)

(Usually called the “Lorenz” force law.)

3.7 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’, where the 1-from A can be integrated over a 1-dimensional path. A generalization (or specialization, depending on how you look at it) 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 2D surface as time passes (Fig. 3.3).

Figure 3.3: Worldtube of a closed string.

Can you infer an appropriate action for this system? Remember, the physical or physico-philosophical principle we’ve been using is that the path followed by physical objects minimizes the “activity” or “aliveness” of the system. Given that we presumably cannot 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 open string). If the string is also assumed to be a source of electromagnetic field then we need a 2-form to integrate over the 2D worldtube analogous to the 1-form integrated over the pathline 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, but its tensor components will have two indices since it’s a 2-form. The string action can then be written

S = α · (area) + e ∫ B  (3.6)

We’ve also replaced the point particle mass by the string tension α [mass·length^{−1}] to obtain the correct units for the action (since replacing arclength by area meant we had to compensate for the extra length dimension in the first term of the above string action).

This may still seem like we’ve pulled a rabbit out of a hat. But we haven’t checked that this action yields sensible dynamics yet! But supposing it does, then would it justify our guesswork and intuition in arriving at Eq.(3.6)? Well by now you’ve probably realized that one can have more than one form of action or Lagrangian that yields the same dynamics. So provided we supply reasonabe physically realistic heuristics then whatever Lagrangian or action that we come up with will stand a good chance of describing some system with a healthy measure of physical verisimilitude.

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

3.8 Another Lagrangian for Relativistic Electrodynamics

In § 3.6, Eq.(3.4) we saw an example of a Lagrangian for relativistic electrodynamics that had awkward reparametrization symmetries, meaning that H = 0 and there were non-unique solutions to the Euler–Lagrange equations arising from applying gauge transformations. This freedom to change the gauge can be avoided.

Recall Eq.(3.4), which was a Lagrangian for a charged particle with reparametrization symmetry

L = m∥q̇∥ + eA_{i}q̇^{i}

just as for an uncharged relativistic particle. But there’s another Lagrangian we can use that doesn’t have this gauge symmetry:

L = \frac{1}{2}mq̇ · q̇ + eA_{i}q̇^{i}  (3.7)

This one even has some nice features.

• It looks formally like “\frac{1}{2}mv^{2}”, familiar from nonrelativistic mechanics.

• There’s no ugly square root, so it’s everywhere differentiable, and there’s no trouble with paths being timelike or spacelike in direction, they are handled the same.

What Euler–Lagrange equations does this Lagrangian yield?

p_{i} = \frac{∂L}{∂q̇^{i}} = mq̇_{i} + eA_{i}
F_{i} = \frac{∂L}{∂q^{i}} = e\frac{∂A_{j}}{∂q^{i}} q̇^{j}

Very similar to before! The Euler–Lagrange equations then say

\frac{d}{dt}(mq̇_{i} + eA_{i}) = e\frac{∂A_{j}}{∂q^{i}} q̇^{j}
mq̈_{i} = eF_{ij}q̇^{j}

almost as before. (I’ve taken to using F here for the electromagnetic field tensor to avoid clashing with F for the generalized force.) The only difference is that we have mq̈_{i} instead of mv̇_{i} where v_{i} = q̇_{i}/∥q̇∥. So the old Euler–Lagrange equations of motion reduce to the new ones if we pick a parametrization with ∥q̇∥ = 1, which would be a parametrization by proper time for example.

Let’s work out the Hamiltonian for this

L = \frac{1}{2}mq̇ · q̇ + eA_{i}q̇^{i}

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

L = m\sqrt{q̇_{i}q̇^{i}} + eA_{i}q^{i}

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

H = p_{i}q̇^{i} − L

and now

p_{i} = \frac{∂L}{∂q̇^{i}} = mq̇^{i} + eA_{i}

so

H = (mq̇_{i} + eA_{i})q̇^{i} − (\frac{1}{2}mq̇_{i}q̇^{i} + eA_{i}q̇^{i})
= \frac{1}{2}mq̇_{i}q̇^{i}

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

H = p_{i}q̇^{i} − L = 2K − (K − V) = K + V,

but now the “potential’ V = eA_{i}q̇^{i} is linear in velocity, so now

H = p_{i}q̇^{i} − L = (2K − V) − (K − V) = K.

As claimed H is not zero, and the fact that it’s conserved says ∥q̇(t)∥ is constant as a function of t, so the particle’s path is parameterized by proper time up to rescaling of t. That is, we’re getting “conservation of speed” rather than some more familiar “conservation of energy”. The reason is that this Hamiltonian comes from the symmetry

q_{s}(t) = q(t + s)

instead of spacetime translation symmetry

q_{s}(t) = q(t) + s w, w ∈ ℝ^{n+1}

the difference is illustrated schematically in Fig. 3.4.

PSfrag replacements q_{s}(t) = q(t + s) q_{s}(t) = q(t) + sw

w

Figure 3.4: Proper time rescaling vs spacetime translation.

Our Lagrangian

L(q, q̇) = \frac{1}{2}m∥q̇∥^{2} + A_{i}(q)q̇^{i}

has time translation symmetry iff A is translation invariant (but it’s highly unlikely a given system of interest will have A(q) = A(q +sw)). In general then there’s no conserved “energy” for our particle corresponding to translations in time.

3.9 The Free Particle in General Relativity

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

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

g(x) : T_{x}Q × T_{x}Q ⟶ ℝ
(v, w) ⟼ g(x)(v, w)

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

2. With respect to some basis of T_{x}Q we have

g(v, w) = g_{ij}v^{i}w^{j}

g_{ij} = ( \begin{matrix}1 & 0 & … & 0 \\ 0 & −1 & & 0 \\ ⋮ & & ... & ⋮ \\ 0 & 0 & … & −1 \end{matrix} )

Of course we can write g(v, w) = g_{ij}v^{i}w^{j} in any basis, but for different bases g_{ij} will have a different form.

3. g(x) varies smoothly with x.

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

L(q, q̇) = m\sqrt{g(q)(q̇, q̇)}
= m\sqrt{g_{ij}q̇^{i}q̇^{j}}

just like in special relativity but with η_{ij} replaced by g_{ij}. Alternatively we could just as well use

L(q, q̇) = \frac{1}{2}mg(q)(q̇, q̇)
= \frac{1}{2}mg_{ij}q̇^{i}q̇^{j}

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

Let’s find the equations of motion. Suppose then Q is a Lorentzian manifold with metric g and L : T Q → ℝ is the Lagrangian of a free particle,

L(q, q̇) = \frac{1}{2}mg_{ij}q̇^{i}q̇^{j}

We find equations of motion from the Euler–Lagrange equations, which in this case start from

p_{i} = \frac{∂L}{∂q̇^{i}} = mg_{ij}q̇^{j}

The velocity q̇ here is a tangent vector, the momentum p is a cotangent vector, and we need the metric to relate them, via

g : T_{q}M × T_{q}M ⟶ ℝ
(v, w) ⟼ g(v, w)

which gives

T_{q}M ⟶ T_{q}^{∗}M
v ⟼ g(v, −).

In coordinates this would say that the tangent vector v^{i} gets mapped to the cotangent vector g_{ij}v^{j}. This is lurking behind the passage from q̇^{i} to the momentum mg_{ij}v^{j}.

Getting back to the Euler–Lagrange equations,

p_{i} = \frac{∂L}{∂q̇^{i}} = mg_{ij}q̇^{j}
F_{i} = \frac{∂L}{∂q^{i}} = \frac{∂}{∂q^{i}} (\frac{1}{2}mg_{jk}(q)q̇^{j}q̇^{k})

= \frac{1}{2}m∂_{i}g_{jk}q̇^{j}q̇^{k}, (where ∂_{i} = \frac{∂}{∂q^{i}}).

So the Euler–Lagrange equations say

\frac{d}{dt}mg_{ij}q̇^{j} = \frac{1}{2}m∂_{i}g_{jk}q̇^{j}q̇^{k}.

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

We can rewrite this geodesic equation as follows

\frac{d}{dt}g_{ij}q̇^{j} = \frac{1}{2}∂_{i}g_{jk}q̇^{j}q̇^{k}
∴ ∂_{k}g_{ij}q̇^{k}q̇^{j} + g_{ij}q̈^{j} = \frac{1}{2}∂_{i}g_{jk}q̇^{j}q̇^{k}
∴ g_{ij}q̈^{j} = (\frac{1}{2}∂_{i}g_{jk} − ∂_{k}g_{ij})q̇^{j}q̇^{k}
= \frac{1}{2}(∂_{i}g_{jk} − ∂_{k}g_{ij} − ∂_{j}g_{ki})q̇^{j}q̇^{k}

where the last line follows by symmetry of the metric, g_{ik} = g_{ki}. Now let,

Γ_{ijk} = −(∂_{i}g_{jk} − ∂_{k}g_{ij} − ∂_{j}g_{ki})

the minus sign being just a convention (so that we agree with everyone else). This defines what we call the Christoffel symbols Γ_{jk}^{i}. Then

q̈_{i} = g_{ij}q̈^{j} = −Γ_{ijk}q̇^{j}q̇^{k}
∴ q̈^{i} = −Γ_{jk}^{i}q̇^{j}q̇^{k}.

So we see that q̈ can be computed in terms of q̇ and the Christoffel symbols Γ_{jk}^{i}, which is really a particular type of connection that a Lorentzian manifold has (the Levi-Civita connection), a “connection” is just the rule for parallel transporting tangent vectors around the manifold.

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

3.10 A Charged Particle on a Curved Spacetime

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

L = \frac{1}{2}mg_{jk}q̇^{j}q̇^{k} + eA_{i}q̇^{i}

which again was conjured up be replacing the flat space metric η_{ij} by the metric for GR g_{ij}. Not surprisingly, the Euler–Lagrange equations then yield the following equations of motion,

mq̈_{i} = −mΓ_{ijk}q̇^{j}q̇^{k} + eF_{ij}q̇^{j}.

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

3.11 The Principle of Least Action and Geodesics

3.11.1 Jacobi and Least Time vs Least Action

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

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

light particle

faster slower V high V low n high n low

Figure 3.5: Least time versus least action.

it’s in ℝ^{n} with its usual Euclidean metric. If the light is trying to minimize the time, its trying to minimize the arclength of its path in the metric

g_{ij} = n^{2}δ_{ij}

that is, the index of refraction n : ℝ^{n} → (0, ∞), times the usual Euclidean metric

δ_{ij} = ( \begin{matrix}1 & & & & 0 \\ & . & . & . \\ 0 & & & & 1 \end{matrix} )

This is just like the free particle in general relativity (minimizing its proper time) except that now g_{ij} is a Riemannian metric

g(v, w) = g_{ij}v^{i}w^{j}
where g(v, v) ≥ 0

So we’ll use the same Lagrangian:

L(q, q̇) = \sqrt{g_{ij}(q)q̇^{i}q̇^{j}}

and get the same Euler–Lagrange equations:

\frac{d^{2}q^{i}}{dt^{2}} + Γ_{jk}^{i}q̇^{j}q̇^{k} = 0  (3.8)

if q is parameterized by arclength or more generally

∥q̇∥ = \sqrt{g_{ij}(q)q̇^{i}q̇^{j}} = constant.

As before the Christoffel symbols Γ 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 ℝ^{n} with potential V : ℝ^{n} → ℝ. It satisfies F = ma, i.e.,

m\frac{d^{2}q^{i}}{dt^{2}} = −∂_{i}V  (3.9)

How did Jacobi see (3.9) as a special case of (3.8)? He considered a particle of energy E and he chose the index of refraction to be

n(q) = \sqrt{\frac{2}{m}(E − V (q))}

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

\sqrt{\frac{2}{m}(E − V)} = \sqrt{\frac{2}{m} \frac{1}{2}m∥q̇∥^{2}} = ∥q̇∥.

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

L = \sqrt{g_{ij}(q)q̇^{i}q̇^{j}}
= \sqrt{n^{2}(q)q̇^{i}q̇^{j}}
= \sqrt{2/m(E − V (q))q̇^{2}}

where q̇^{2} = q̇ · q̇ is just the usual Euclidean dot product, v · w = δ_{ij}v^{i}w^{j}. We get the Euler–Lagrange equations,

p_{i} = \frac{∂L}{∂q̇^{i}} = \sqrt{\frac{2}{m}(E − V)} · \frac{q̇}{∥q̇∥}
F_{i} = \frac{∂L}{∂q^{i}} = ∂_{i}\sqrt{\frac{2}{m}(E − V (q))} · ∥q̇∥

= \frac{1}{2} \frac{−2/m∂_{i}V}{\sqrt{2/m(E − V q)}} · ∥q̇∥

Then ṗ = F says,

\frac{d}{dt}\sqrt{2/m(E − V (q))} · \frac{q̇_{i}}{∥q̇∥} = −\frac{1}{m}∂_{i}V \frac{∥q̇∥}{\sqrt{2/m(E − V)}}

Jacobi noticed that this is just F = ma, or mq̈_{i} = −∂_{i}V, that is, provided we reparameterize q so that,

∥q̇∥ = \sqrt{2/m(E − V (q))}.

Recall that our Lagrangian gives reparameterization invariant Euler–Lagrange equations! This is the unification between least time (from optics) and least action (from mechanics) that we sought.

3.11.2 The Ubiquity of Geodesic Motion

We’ve seen that many classical systems trace out paths that are geodesics, i.e., paths q : [t_{0}, t_{1}] → Q that are critical points of

S(q) = ∫_{t_{0}}^{t_{1}} \sqrt{g_{ij}q̇^{i}q̇^{j}} dt

which is proper time when (Q, g) is a Lorentzian manifold, or arclength when (Q, g) is a Riemannian manifold. We have

1. The metric at q ∈ Q is,

g(q) : T_{q}Q × T_{q}Q → ℝ
(v, w) ↦ g(v, w)

and it is bilinear.

2. w.r.t a basis of T_{q}Q

g(v, w) = δ_{ij}v^{i}w^{j}

3. g(q) varies smoothly with q ∈ Q.

An important distinction to keep in mind is that Lorentzian manifolds represent spacetimes, whereas Riemannian manifolds represent that we’d normally consider as just space.

We’ve seen at least three important things.

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

out geodesics in the metric

g_{ij} = n(q)^{2}δ_{ij}

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

(2) Jacobi saw that a particle in Q = ℝ^{n} in some potential V : Q → ℝ traces out geodesics in the metric

g_{ij} = \frac{2}{m}(E − V)δ_{ij}

if the particle has energy E (where^{1} 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 we’ve discussed!

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

(1′) Light on any Riemannian manifold (Q, q) with index of refraction n : Q → (0, ∞) traces out geodesics in the metric h = n^{2}g.

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

h = \frac{2}{m}(E − V)g

if it has energy E. Lots of physical systems can be described this way, e.g., the Atwood 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, so they 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

q̈_{i} = −Γ_{ijk}q̇^{j}q̇^{k} + \frac{e}{m}F_{ij}q̇^{j}

where

F_{ij} = ∂_{i}A_{j} − ∂_{j}A_{i}

but this is actually geodesic motion on the manifold Q × U(1) where U(1) = {e^{iθ} : θ ∈ ℝ} is a circle.

Let’s 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 where the amount of “spiralling”—the velocity in the U(1) direction is e/m. The metric h on

PSfrag replacements Q × U(1) a geodesic

Q

the apparent path

Q×U(1) is built from g and A in a very simple way. Let’s pick coordinates x^{i} on Q where i ∈ {0, …, n} since we’re in n + 1-dimensional spacetime, and θ is our local coordinate on S^{1}. The components of h are

h_{ij} = g_{ij} + A_{i}A_{j}
h_{θi} = h_{θi} = −A_{i}
h_{θθ} = 1

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

q̈_{i} = −Γ_{ijk}q̇^{j}q̇^{k} + \frac{e}{m}F_{ij}q̇^{j}
q̈_{θ} = 0,

if q̇_{θ} = 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 be regarded as geodesic motion!

52  Examples

From Lagrangians to Hamiltonians

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

L : T Q ⟶ ℝ

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

p_{i} = \frac{∂L}{∂q̇^{i}}

since the equations of motion tell us how it changes:

\frac{dp_{i}}{dt} = \frac{∂L}{∂q^{i}} .

4.1 The Hamiltonian Approach

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

H = p_{i}q̇^{i} − L(q, q̇)

reinterpreted as a function of position and momentum We call this the Hamiltonian

H: T^{∗}Q ⟶ ℝ

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

\frac{dq^{i}}{dt} = \frac{∂H}{∂p_{i}} , \frac{dp^{i}}{dt} = −\frac{∂H}{∂q_{i}}

where the latter is the Euler–Lagrange equation

\frac{dp^{i}}{dt} = \frac{∂L}{∂q_{i}}

in disguise (it has a minus sign since H = pq̇ − L).

To obtain this Hamiltonian description of mechanics rigorously we need to study this map

λ : T Q ⟶ T^{∗}Q
(q, q̇) ⟼ (q, p)

where q ∈ Q, and q̇ is any tangent vector in T_{q}Q (not the time derivative of something), and p is a cotangent vector in T_{q}^{∗}Q := (T_{q}Q)^{∗}, given by

q̇ ⟶^{λ} p_{i} = \frac{∂L}{∂q̇^{i}} .

So λ is defined using L : T Q → ℝ. Despite appearances, λ can be defined in a coordinatefree way, as follows (referring to Fig. 4.1). To understand “\frac{∂L}{∂q̇^{i}} ” in a coordinate-free way,

PSfrag replacements T_{q}Q

T Q
T_{(q,q̇)}T_{q}Q
q
Q

Figure 4.1:

we treat it as the “differential of L in the vertical direction”—i.e., the q̇^{i} directions. To do this, note that the differential of

\begin{matrix}π: T Q & ⟶ & Q \\ (q, q̇) & ⟼ & q \end{matrix}

is a map

dπ: T T Q ⟶ T Q.

We define the bundle of vertical vectors to be

V T Q = {v ∈ T T Q : dπ(v) = 0} ⊆ T T Q.

That is, it consists of tangent vectors to T Q that are sent to zero by dπ. Intuitively, they point ‘straight up’. The differential of L at some point (q, q̇) ∈ T Q is a map from T T Q to ℝ, so we have

(dL)_{(q,q̇)} ∈ T_{(q,q̇)}^{∗}T Q

that is,

dL_{(q,q̇)} : T_{(q,q̇)}T Q ⟶ ℝ.

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

f : V_{(q,q̇)}T Q ⟶ ℝ.

But note

V_{(q,q̇)}T Q ≅ T_{q,q̇}(T_{q}Q)

\underline{PSfrag replacements} T_{q}Q V_{(q,q̇)}T Q T_{(q,q̇)}T_{q}Q q Q and since T_{q}Q is a vector space,

T_{(q,q̇)}T_{q}Q ≅ T_{q}Q

in a canonical way^{1}. So f gives a linear map

p : T_{q}Q ⟶ ℝ

that is,

p ∈ T_{q}^{∗}Q

this is the momentum!

Given L: T Q → T^{∗}Q, we now know a coordinate-free way of describing the map

λ: T Q ⟶ T^{∗}Q
(q, q̇) ⟼ (q, p)

given in local coordinates by

p_{i} = \frac{∂L}{∂q̇^{i}} .

1Each tangent space T_{v}V of a vector space V is isomorphic to V in a canonical way.

We say L is regular if λ is a diffeomorphism from T Q to some open subset X ⊆ T^{∗}Q. In this case we can describe what our system is doing equally well by specifying position and velocity,

(q, q̇) ∈ T Q

or position and momentum

(q, p) = λ(q, q̇) ∈ X.

We call X the phase space of the system. In practice often 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 above theory.

4.2.1 Example: A Particle in a Riemannian Manifold with Potential V (q)

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

L(q, q̇) = \frac{1}{2}mg_{ij}q̇^{i}q̇^{j} − V (q)

Here

p_{i} = \frac{∂L}{∂q̇^{i}} = mg_{ij}q̇^{j}

so

λ(q, q̇) = (q, mg(q̇, −))

L is strongly regular in this case because

T_{q}Q ⟶ T_{q}^{∗}Q
v ⟶ g(v, −)

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

4.2.2 Example: General Relativistic Particle in an E-M Potential

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

L(q, q̇) = \frac{1}{2}mg_{ij}q̇^{i}q^{j} − eA_{i}q̇^{i}

and thus

p_{i} = \frac{∂L}{∂q̇^{i}} = mg_{ij}q̇^{i}q^{j} + eA_{i}.

This L is still strongly regular, but now each map

λ |_{T_{q}Q} : T_{q}Q ⟶ T_{q}^{∗}Q
q̇ ⟼ m g(q̇, −) + eA(q)

is affine rather than linear^{2}.

4.2.3 Example: Free General Relativistic Particle with Reparameterization Invariance

The free general relativistic particle with reparameterization invariant Lagrangian has

L(q, q̇) = m\sqrt{g_{ij}q̇^{i}q̇^{j}}.

This is terrible from the perspective of regularity—it’s not differentiable when g_{ij}q̇^{i}q̇^{j} vanishes, and undefined when the same is negative. Where it is defined

p_{i} = \frac{∂L}{∂q̇^{i}} = \frac{mg_{ij}q̇^{j}}{∥q̇∥}

(where q̇ is timelike), we can ask about regularity. Alas, the map λ is not 1-1 where defined since multiplying q̇ by some number has no effect on p! (This is related to the reparameterization invariance—this always happens with reparameterization-invariant Lagrangians.)

4.2.4 Example: A Regular but not Strongly Regular Lagrangian

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

L(q, q̇) = f(q̇)

so that

p = \frac{∂L}{∂q̇} = f′(q̇)

This will be regular but not strongly so if f′: ℝ → ℝ is a diffeomorphism from ℝ to some proper subset U ⊂ ℝ. For example, take f(q̇) = eq̇ so f′: ℝ →^{∼} (0, ∞) ⊂ ℝ. So

L(q, q̇) = eq̇ [figure: positive slope] or

2All linear transforms are affine, but affine transformations include translations, which are nonlinear.

In affine geometry there is no defined origin. For the example the translation is the “+eA(q)” part.

L(q, q̇) = \sqrt{1 + q̇^{2}} [figure: slope between −1 and 1]

and so forth.

4.3 Hamilton’s Equations

Now let’s assume L is regular, so

λ: T Q ⟶ X ⊆ T^{∗}Q
(q, q̇) ⟼ (q, p)

is a diffeomorphism. This lets us have the best of both worlds: we can identify T Q with X using λ. This lets us treat q^{i}, p^{i}, L, H, etc., all as functions on X (or T Q), thus writing

q̇^{i} (function on T Q)

for the function

q̇^{i} ∘ λ^{−1} (function on X)

In particular

ṗ_{i} := \frac{∂L}{∂q^{i}} (Euler–Lagrange equation)

which is really a function on T Q, will be treated as a function on X. Now let’s calculate:

dL = \frac{∂L}{∂q^{i}} dq^{i} + \frac{∂L}{∂q̇^{i}} dq̇^{i}
= ṗ_{i}dq^{i} + p_{i}dq̇^{i}

while

dH = d(p_{i}q̇^{i} − L)
= q̇^{i}dp_{i} + p_{i}dq̇^{i} − dL
= q̇^{i}dp_{i} + p_{i}dq̇^{i} − (ṗ_{i}dq^{i} + p_{i}dq̇^{i})
= q̇^{i}dp_{i} − ṗ_{i}dq̇^{i}

so

dH = q̇^{i}dp_{i} − ṗ^{i}dq_{i}.

Assume the Lagrangian L: T Q → ℝ is regular, so

λ: T Q ⟶^{∼} X ⊆ T^{∗}Q
(q, q̇) ⟼ (q, p)

is a diffeomorphism. This lets us regard both L and the Hamiltonian H = p_{i}q̇^{i} − L as functions on the phase space X, and use (q^{i}, q̇^{i}) as local coordinates on X. As we’ve seen, this gives us

dL = ṗ_{i}dq^{i} + p_{i}dq̇^{i}
dH = q̇^{i}dp_{i} − ṗ_{i}dq^{i}.

But we can also work out dH directly, this time using local coordinates (q^{i}, p_{i}), to get

dH = \frac{∂H}{∂p^{i}} dp_{i} + \frac{∂H}{∂q^{i}} dq^{i}.

Since dp_{i}, dq^{i} form a basis of 1-forms, we conclude:

q̇^{i} = \frac{∂H}{∂p_{i}} ,  ṗ_{i} = −\frac{∂H}{∂q_{i}}

These are Hamilton’s Equations.

4.3.1 Hamilton and Euler–Lagrange

Though q̇^{i} and ṗ_{i} are just functions of X, when the Euler–Lagrange equations hold for some path q: [t_{0}, t_{1}] → Q, they will be the time derivatives of q^{i} and p_{i}. So when the Euler–Lagrange equations hold, Hamilton’s equations describe the motion of a point x(t) = (q(t), p(t)) ∈ X. In fact, in this context, Hamilton’s equations are just the Euler– Lagrange equations in disguise. The equation

q̇^{i} = \frac{∂H}{∂p_{i}}

really just lets us recover the velocity q̇ as a function of q and p, inverting the formula

p_{i} = \frac{∂L}{∂q̇^{i}}

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

λ^{−1} : X ⟶ T Q
(q, p) ⟼ (q, q̇).

Given this, the other Hamilton equation

ṗ_{i} = −\frac{∂H}{∂q^{i}}

is secretly the Euler–Lagrange equation

\frac{d}{dt} \frac{∂L}{∂q̇^{i}} = \frac{∂L}{∂q^{i}} ,

or ṗ_{i} = \frac{∂L}{∂q^{i}} for short. These are the same because

\frac{∂H}{∂q^{i}} = \frac{∂}{∂q^{i}} (p_{i}q̇^{i} − L) = −\frac{∂L}{∂q^{i}} .

Example: Particle in a Potential V (q) For a particle in Q = ℝ^{n} in a potential V : ℝ^{n} → ℝ the system has Lagrangian

L(q, q̇) = \frac{m}{2}∥q̇∥^{2} − V (q)

which gives

p = mq̇
q̇ = \frac{p}{m},  (though really that’s q̇ = \frac{g^{ij}p_{j}}{m})

and Hamiltonian

H(q, p) = p_{i}q̇^{i} − L = \frac{1}{m}∥p∥^{2} − (\frac{∥p∥^{2}}{2m} − V )

= \frac{1}{2m}∥p∥^{2} + V (q).

So Hamilton’s equations say

q̇^{i} = \frac{∂H}{∂p_{i}} ⇒ q̇ = \frac{p}{m}
ṗ_{i} = −\frac{∂H}{∂q^{i}} ⇒ ṗ = −∇V

The first just recovers q̇ as a function of p; the second is F = ma.

Note on Symplectic Structure

Hamilton’s equations push us toward the viewpoint where p and q have equal status as coordinates on the phase space X. Soon, we’ll drop the requirement that X ⊆ T^{∗}Q where Q is a configuration space. X will just be a manifold equipped with enough structure to write down Hamilton’s equations starting from any H: X → ℝ.

The coordinate-free description of this structure is the major 20th century contribution to mechanics: a symplectic structure.

This is important. You might have some particles moving on a manifold like S^{3}, which is not symplectic. So the Hamiltonian mechanics point of view says that the abstract manifold that you are really interested in is something different: it must be a symplectic manifold. That’s the phase space X. We’ll introduce symplectic geometry more completely in later chapters.

4.3.2 Hamilton’s Equations from the Principle of Least Action

Before, we obtained the Euler–Lagrange equations by associating an “action” S with any q: [t_{0}, t_{1}] → Q and setting δS = 0. Now let’s get Hamilton’s equations directly by assigning an action S to any path x: [t_{0}, t_{1}] → X and setting δS = 0. Note: we don’t impose any relation between p and q, q̇! The relation will follow from δS = 0.

Let P be the space of paths in the phase space X and define the action

S: P ⟶ ℝ

by

S(x) = ∫_{t_{0}}^{t_{1}}(p_{i}q̇^{i} − H)dt

since p_{i}q̇^{i} − H is another way of writing the Lagrangian. More precisely, write our path x as x(t) = (q(t), p(t)) and let

S(x) = ∫_{t_{0}}^{t_{1}} [p_{i}(t)\frac{d}{dt}q^{i}(t) − H(q(t), p(t))] dt

we write \frac{d}{dt}q^{i} instead of q̇^{i} to emphasize that we mean the time derivative rather than a coordinate in phase space.

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

δS = δ ∫ (p_{i}q̇^{i} − H)dt
= ∫ (δp_{i}q̇^{i} + p_{i}δq̇^{i} − δH)dt

then integrating by parts,

= ∫ (δp_{i}q̇^{i} − ṗ_{i}δq^{i} − δH)dt

= ∫ (δp_{i}q̇^{i} − ṗ_{i}δq^{i} − \frac{∂H}{∂q^{i}} δq^{i} − \frac{∂H}{∂p_{i}} δp_{i})dt

= ∫ (δp_{i}(q̇^{i} − \frac{∂H}{∂p_{i}} ) + δq^{i}(−ṗ_{i} − \frac{∂H}{∂q^{i}} )) dt

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

q̇^{i} = \frac{∂H}{∂p_{i}} , p_{i} = −\frac{∂H}{∂q^{i}}

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 a third version based on paths in position-velocity space T Q. But when our Lagrangian is regular we have a diffeomorphism λ: T Q →^{∼} X, so this third principle of least action is just a reformulation of principle 2. However, the really interesting principle of least action involves paths in the extended phase space where we have an additional coordinate for time: X × ℝ.

Recall the action

S(x) = ∫ (p_{i}q̇^{i} − H) dt

= ∫ p_{i} \frac{dq^{i}}{dt} dt − H dt

= ∫ p_{i}dq^{i} − H dt

We can interpet the integrand as a 1-form

β = p_{i}dq^{i} − H dt

on X × ℝ, which has coordinates {p_{i}, q^{i}, t}. So any path

x: [t_{0}, t_{1}] ⟶ X

gives a path

σ: [t_{0}, t_{1}] ⟶ X × ℝ
t ⟼ (x(t), t)

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

S(x) = ∫ p_{i} dq^{i} − H dt = ∫_{σ} β

4.4 Waves versus Particles—The Hamilton-Jacobi Equations

In quantum mechanics we discover that every particle—electrons, photons, neutrinos, etc.—is a wave, and vice versa. Interestingly Newton already had a particle theory of light (his “corpuscules”) and various physicists argued against it by pointing out that diffraction is best explained by a wave theory. We’ve talked about geometrized optics, an approximation in which light consists of particles moving along geodesics. Here we start with a Riemannian manifold (Q, g) as space, but we use the new metric

h_{ij} = n^{2}g_{ij}

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

4.4.1 Wave Equations

Huygens considered this same setup (in simpler language) and considered the motion of a wavefront: and saw that the wavefront is the envelope of a bunch of little wavelets centered at points along the big wavefront:

[figure]

[figure] [figure] [figure]

[figure] [figure]

balls of radius ϵ centered at points of the old wavefront

[figure]

In short, the wavefront moves at unit speed in the normal 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(q_{0}, q_{1}) = inf_{Υ}(arclength)

where Υ = {paths from q_{0} to q_{1}}. (Secretly this d(q_{0}, q_{1}) is the least action—the infimum of action over all paths from q_{0} to q_{1}.) Using this we get the wavefronts centered at q_{0} ∈ Q as the level sets

{q: d(q_{0}, q) = c}

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

PSfrag replacements q_{0}

Figure 4.2:

to be smooth—we say a catastrophe occurs—and then the wavefronts are no longer the level sets. This sort of situation can happen for topological reasons (as when the waves smash into each other in the back of Fig. 4.2) and it can also happen for geometrical

Figure 4.3:

reasons (Fig. 4.3). Assuming no such catastrophes occur, we can approximate the waves of light by a wavefunction:

ψ(q) = A(q)e^{ik d(q,q_{0})}

where k is the wavenumber of the light (i.e., its color) and A: Q → ℝ describes the amplitude of the wave, which drops off far from q_{0}. This becomes the eikonal approximation in optics^{3} once we figure out what A should be.

Hamilton and Jacobi focused on distance d: Q × Q → [0, ∞) as a function of two variables. This is now called Hamilton’s principal function, and we will denote it as W. They noticed that

\frac{∂}{∂q_{1}^{i}} W (q_{0}, q_{1}) = (p_{1})_{i}

[figure: q_{0}; •; p_{1}; •; q_{1}]

where p_{1} is a cotangent vector pointing normal to the wavefronts.

4.4.2 The Hamilton-Jacobi Equations

We’ve seen that in optics, particles of light move along geodesics, but wavefronts are level sets of the distance functions:

[figure]

3Eikonal comes form the Greek word for ‘image’ or ‘likeness’, in optics the eikonal approximation is the basis for ray tracing methods.

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

ψ: Q ⟶ ℂ
ψ(q_{1}) = A(q_{1})e^{ik W (q_{0},q_{1})}

where (Q, h) is a Riemannian manifold, h is the optical metric, q_{0} ∈ Q is the light source, k is the frequency and

W: Q × Q ⟶ [0, ∞)

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

W (q_{0}, q_{1}) = inf_{q∈Υ} S(q)

where Υ is the space of paths from q_{0} to q and S(q) is the action of the path q, i.e., its arclength. This is begging to be generalized to other Lagrangian systems! (At least it is retrospectively, with the advantage of our historical perspective.) We also saw that

\frac{∂}{∂q_{1}^{i}} W (q_{0}, q_{1}) = (p_{1})_{i},

[figure: q_{0}; •; p_{1}; •; q_{1}]

“points normal to the wavefront”—really the tangent vector

p_{1}^{i} = h^{ij}(p_{1})_{j}

points in this direction. In fact kp_{1}_{i} is the momentum of the light passing through q_{1}. This foreshadows quantum mechanics! After all, in quantum mechanics, the momentum is an operator that acts to differentiating the wavefunction.

Jacobi generalized this to the motion of point particles in a potential V : Q → ℝ, using the fact that a particle of energy E traces out geodesics in the metric

h_{ij} = \frac{2(E − V)}{m} g_{ij}.

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

Suppose Q is any manifold and L: T Q → ℝ is any function (Lagrangian). Define Hamilton’s principal function

W: Q × ℝ × Q × ℝ ⟶ ℝ

by

W (q_{0}, t_{0}; q_{1}, t_{1}) = inf_{q∈Υ} S(q)

where

Υ = {q: [t_{0}, t_{1}] → Q, q(t_{0}) = q_{0}, & q(t_{1}) = q_{1}}

and

S(q) = ∫_{t_{0}}^{t_{1}} L(q(t), q̇(t)) dt

Now W is just the least action for a path from (q_{0}, t_{0}) to (q_{1}, t_{1}); it’ll be smooth if (q_{0}, t_{0}) and (q_{1}, t_{1}) are close enough—so let’s assume that is true. In fact, we have

\frac{∂}{∂q_{1}^{i}} W (q_{0}, q_{1}) = (p_{1})_{i},

[figure: (q_{0}, t_{0}); •; p_{1}; •; (q_{1}, t_{1})]

where p_{1} is the momentum of the particle going from q_{0} to q_{1}, at time t_{1}, and

\frac{∂W}{∂q_{0}^{i}} = −(p_{0})_{i},  (-momentum at time t_{0})
\frac{∂W}{∂t_{1}} = −H_{1},  (-energy at time t_{1})
\frac{∂W}{∂t_{0}} = H_{0},  (+energy at time t_{0})

(H_{1} = H_{0} as energy is conserved). These last four equations are the Hamilton–Jacobi equations. The mysterious minus sign in front of energy was seen before in the 1-form,

β = p_{i}dq^{i} − H dt

on the extended phase space X × ℝ. Maybe the best way to get the Hamilton–Jacobi equations is from this extended phase space formulation. But for now let’s see how Hamilton’s principal function W and variational principles involving least action also yield the Hamilton–Jacobi equations.

Given (q_{0}, t_{0}), (q_{1}, t_{1}), let

q: [t_{0}, t_{1}] ⟶ Q

be the action-minimizing path from q_{0} to q_{1}. Then

W (q_{0}, t_{0}; q_{1}, t_{1}) = S(q)

Now consider varying q_{0} and q_{1} a bit

t_{0}  t_{1} q

and thus vary the action-minimizing path, getting a variation δq which does not vanish at t_{0} and t_{1}. We get

δW = δS

= δ ∫_{t_{0}}^{t_{1}} L(q, q̇) dt

= ∫_{t_{0}}^{t_{1}}(\frac{∂L}{∂q^{i}} δq^{i} + \frac{∂L}{∂q̇^{i}} δq̇^{i}) dt

= ∫_{t_{0}}^{t_{1}}(\frac{∂L}{∂q^{i}} δq^{i} − ṗ_{i}δq^{i}) dt + p_{i}δq^{i}|_{t_{0}}^{t_{1}}

= ∫_{t_{0}}^{t_{1}}(\frac{∂L}{∂q^{i}} − ṗ_{i})δq^{i} dt

the term in parentheses is zero because q minimizes the action and the Euler–Lagrange equations hold. So we δq^{i} have

δW = p_{1}_{i}δq_{1}^{i} − p_{0}_{i}δq_{0}^{i}

and so

\frac{∂W}{∂q_{1}^{i}} = p_{1i}, and \frac{∂W}{∂q_{0}^{i}} = −p_{0}_{i}

These are two of the four Hamilton–Jacobi equations! To get the other two, we need to vary t_{0} and t_{1}:

t_{0}

t_{1} + Δt_{1} t_{0} + Δt_{0}

t_{1}

• •

• •

Now change in W will involve Δt_{0} and Δt_{1}

(you can imagine Δt_{0} < 0 in this figure if you like).

We want to derive the Hamilton–Jacobi equations describing the derivatives of Hamilton’s principal function

W (q_{0}, t_{0}; q_{1}, t_{1}) = inf_{q∈Υ} S(q)

where Υ is the space of paths q: [t_{0}, t_{1}] → Q with q(t_{0}) = q, q(t_{1}) = q_{1} and

S(q) = ∫_{t_{0}}^{t_{1}} L(q, q̇) dt

where the Lagrangian L: T Q → ℝ will now be assumed regular, so that

λT Q: ⟶ X ⊆ T^{∗}Q
(q, q̇) ⟼ (q, p)

is a diffeomorphism. We need to ensure that (q_{0}, t_{0}) is close enough to (q_{i}, t_{1}) that there is a unique q ∈ Υ that minimizes the action S, and assume that this q depends smoothly on U = (q_{0}, t_{0}; q_{1}, t_{1}) ∈ (Q × ℝ)^{2}. We’ll think of q as a function of U:

(t_{0}, q_{0})

(t_{1}, q_{1}) q •

•

(Q × ℝ)^{2} → Υ
u ↦ q
defined only when (q_{0}, t_{0}) and
(q_{1}, t_{1}) are sufficiently close.

Then Hamilton’s principal function is

W (u) := W (q_{0}, t_{0}; q_{1}, t_{1}) = S(q)
= ∫_{t_{0}}^{t_{1}} L(q, q̇) dt
= ∫_{t_{0}}^{t_{1}}(pq̇ − H(q, p)) dt
= ∫_{t_{0}}^{t_{1}} p dq − H dt
= ∫_{C} β

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

C(t) = (q(t), p(t), t) ∈ X × ℝ.

Note that C depends on the curve q ∈ Υ, which in turn depends upon u = (q_{0}, t_{0}; q_{1}, t_{1}) ∈ (Q×ℝ)^{2}. We are 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–Jacobi equations from β. Let u_{s} be a 1-parameter family of points in (Q × ℝ)^{2} and work out

\frac{d}{ds}W (u_{s}) = \frac{d}{ds} ∫_{C_{s}} β

where C_{s} depends on u_{s} as above

[figure: C_{s}; •; •; C_{0}; •; •; X × ℝ; A_{s}; B_{s}; .................................; .................................; _{.................................}]

Let’s compare

∫_{C_{s}} β and ∫_{A_{s}+C_{s}+B_{s}} = ∫_{A_{s}} β + ∫_{C_{s}} β + ∫_{B_{s}} β

Since C_{0} minimizes the action among paths with the given end-points, and the curve A_{s} + C_{s} + B_{s} has the same end-points, we get

\frac{d}{ds} ∫_{A_{s}+C_{s}+B_{s}} β = 0

(although A_{s} + C_{s} + B_{s} is not smooth, we can approximate it by a path that is smooth). So

\frac{d}{ds} ∫_{C_{s}} β = \frac{d}{ds} ∫_{B_{s}} β − \frac{d}{ds} ∫_{A_{s}} β at s = 0.

Note

\frac{d}{ds} ∫_{A_{s}} β = \frac{d}{ds} ∫ β(A′_{r}) dr
= β(A′_{0})

where A′_{0} = v is the tangent vector of A_{s} at s = 0. Similarly,

\frac{d}{ds} ∫_{B_{s}} β = β(w)

where w = B′_{0}. So,

\frac{d}{ds}W (u_{s}) = β(w) − β(v)

where w keeps track of the change of (q_{1}, p_{1}, t_{1}) as we move C_{s} and v keeps track of (q_{0}, p_{0}, t_{0}). Now since β = p^{i}dq_{i} − Hdt, we get

\frac{∂W}{∂q_{1}^{i}} = p_{1}^{i}
\frac{∂W}{∂t_{1}} = −H

and similarly

\frac{∂W}{∂q_{0}^{i}} = −p_{0}^{i}
\frac{∂W}{∂t_{0}} = H

So, if we define a wavefunction:

ψ(q_{0}, t_{0}; q_{1}, t_{1}) = e^{iW (q_{0},t_{0};q_{1},t_{1})/ℏ}

then we get

\frac{∂ψ}{∂t_{1}} = −\frac{i}{ℏ}H_{1}ψ
\frac{∂ψ}{∂q_{1}^{i}} = \frac{i}{ℏ}p_{1}ψ

At the time of Hamilton and Jacobi’s research this would have been new... but nowadays it is thoroughly familiar from quantum mechanics!

72  From Lagrangians to Hamiltonians

Bibliography

[Pel94] Peter Peldan. Actions for gravity, with generalizations: A review. Classical and Quantum Gravity, 11:1087, 1994.

[WTM71] J. A. Wheeler, K. S. Thorne, and C. W. Misner. Gravitation. W. H. Freeman, New York, 1971.