Notes on Classical Mechanics - math.ucla.edulaurenst/Resources/classical_mechanics.pdf · Classical mechanics studies the motion of particles in Euclidean space Rn. A particle or

Notes on Classical Mechanics

Thierry Laurens

Last updated: October 4, 2020

Contents

I Newtonian Mechanics 5

1 Newton’s equations of motion 7

1.1 Empirical assumptions . . . . . . . . . . . . . . . . . . . . . . . . 7

1.2 Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

1.3 Conservative systems . . . . . . . . . . . . . . . . . . . . . . . . . 12

1.4 Nonconservative systems . . . . . . . . . . . . . . . . . . . . . . . 14

1.5 Reversible systems . . . . . . . . . . . . . . . . . . . . . . . . . . 16

1.6 Linear momentum . . . . . . . . . . . . . . . . . . . . . . . . . . 17

1.7 Angular momentum . . . . . . . . . . . . . . . . . . . . . . . . . 18

1.8 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

2 Examples 27

2.1 One-dimensional systems . . . . . . . . . . . . . . . . . . . . . . 27

2.2 Central fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

2.3 Closed bounded orbits . . . . . . . . . . . . . . . . . . . . . . . . 32

2.4 Kepler’s problem . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

2.5 Virial theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

2.6 Central field scattering . . . . . . . . . . . . . . . . . . . . . . . . 36

2.7 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

II Lagrangian Mechanics 45

3 Euler–Lagrange equations of motion 47

3.1 Principle of least action . . . . . . . . . . . . . . . . . . . . . . . 47

3.2 Conservative systems . . . . . . . . . . . . . . . . . . . . . . . . . 51

3.3 Equivalence to Newton’s equations . . . . . . . . . . . . . . . . . 54

3.4 Nonconservative systems . . . . . . . . . . . . . . . . . . . . . . . 56

3.5 Momentum and conservation . . . . . . . . . . . . . . . . . . . . 57

3.6 Noether’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . . 59

3.7 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61

2

CONTENTS 3

4 Constraints 694.1 The d’Alembert–Lagrange principle . . . . . . . . . . . . . . . . . 694.2 Gauss’ principle of least constraint . . . . . . . . . . . . . . . . . 714.3 Integrable constraints . . . . . . . . . . . . . . . . . . . . . . . . 744.4 Non-holonomic constraints . . . . . . . . . . . . . . . . . . . . . . 744.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76

5 Hamilton–Jacobi equation of motion 815.1 Hamilton–Jacobi equation . . . . . . . . . . . . . . . . . . . . . . 815.2 Separation of variables . . . . . . . . . . . . . . . . . . . . . . . . 845.3 Conditionally periodic motion . . . . . . . . . . . . . . . . . . . . 855.4 Geometric optics analogy . . . . . . . . . . . . . . . . . . . . . . 875.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89

III Hamiltonian Mechanics 93

6 Hamilton’s equations of motion 956.1 Hamilton’s equations . . . . . . . . . . . . . . . . . . . . . . . . . 956.2 Legendre transformation . . . . . . . . . . . . . . . . . . . . . . . 976.3 Liouville’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . . 1006.4 Poisson bracket . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1026.5 Canonical transformations . . . . . . . . . . . . . . . . . . . . . . 1056.6 Infinitesimal canonical transformations . . . . . . . . . . . . . . . 1086.7 Canonical variables . . . . . . . . . . . . . . . . . . . . . . . . . . 1106.8 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111

7 Symplectic geometry 1157.1 Symplectic structure . . . . . . . . . . . . . . . . . . . . . . . . . 1157.2 Hamiltonian vector fields . . . . . . . . . . . . . . . . . . . . . . 1167.3 Integral invariants . . . . . . . . . . . . . . . . . . . . . . . . . . 1187.4 Poisson bracket . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1207.5 Time-dependent systems . . . . . . . . . . . . . . . . . . . . . . . 1217.6 Locally Hamiltonian vector fields . . . . . . . . . . . . . . . . . . 1237.7 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124

8 Contact geometry 1278.1 Contact structure . . . . . . . . . . . . . . . . . . . . . . . . . . . 1278.2 Hamiltonian vector fields . . . . . . . . . . . . . . . . . . . . . . 1298.3 Dynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1308.4 Contact transformations . . . . . . . . . . . . . . . . . . . . . . . 1318.5 Time-dependent systems . . . . . . . . . . . . . . . . . . . . . . . 1348.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 136

Bibliography 139

4 CONTENTS

CONTENTS 5

Part I

Newtonian Mechanics

The Newtonian perspective is the most fundamental interpretation of themotion of a mechanical system. Newton’s principle of determinacy is an ex-perimental observation, the mathematical phrasing of which yields Newton’sequations. Together with Galileo’s principle of relativity, these form the ax-ioms of classical mechanics. We will also see how additional empirical assump-tions (conservative system, closed system, and the law of action and reaction)yield both physically observable effects (conservation laws and symmetries) andmathematical consequences (special properties for Newton’s equations and itssolutions). Although Newtonian mechanics are naturally phrased on Euclideanspace, we will observe the fundamental roles of configuration and phase spacein order to later generalize this perspective to include a broader class of systemsand coordinates.

6 CONTENTS

Chapter 1

Newton’s equations ofmotion

We build the theory starting from essential experimental observations and thendevelop some of its immediate consequences. This presentation is based on [Arn89,Ch. 1], [AKN06, Ch. 1], [Gol51, Ch. 1], and [Str15, Ch. 5–7].

1.1 Empirical assumptions

Classical mechanics studies the motion of particles in Euclidean space Rd. Aparticle or point mass is a model for a physical object consisting of two piecesof information, the body’s mass (a positive real number) and its position inEuclidean space Rd, and so the object’s spatial dimensions are ignored. We willconsider particle systems, which are collections of N such particles with positionsoften denoted xi ∈ Rd and masses mi. Altogether, the collection of all positionsx = (x1, . . . ,xN ) constitute the configuration space Rd×· · ·×Rd = RNd, whosedimension Nd is referred to as the degrees of freedom of the system.

The motion of a system of N particles is described by N maps xi : I → Rdfor I ⊂ R an interval, which are called the particle’s trajectories or the motionof the system. In order to describe how the particles trace their trajectorieswe need not only the particles’ velocities xi and accelerations xi, but also themomentum of the ith particle,

pi = mixi, i = 1, . . . , N. (1.1)

(We will denote time derivatives by f = df/dt.) The positions and momenta ofthe system span the Euclidean space RNd × RNd, which is called phase space,and its dimension is twice the number of degrees of freedom. We will often referto the trajectories of the system plotted in phase space as the system’s phaseportrait.

Newton’s principle of determinacy is the experimental observation that theinitial state of a mechanical system—the information of all the positions x(t0)

7

8 CHAPTER 1. NEWTON’S EQUATIONS OF MOTION

and velocities x(t0) at some moment in time t0—uniquely determines any sys-tem’s motion. Mathematically, this means for a particle system there exists afunction F : (RNd r ∆)× RNd × R→ RNd such that

p = F(x, x, t). (1.2)

Here, ∆ =⋃i<jxi = xj is the union of diagonals, so that we do not consider

two particles occupying the same position. eq. (1.2) is called Newton’s equation,or sometimes Newton’s second law. By default we will assume the particle massesmi are all constant, and so eq. (1.2) takes the form

mixi = Fi(x, x, t), i = 1, . . . , N. (1.3)

Historically, we would experimentally observe that a particle’s acceleration isinversely proportional to its mass in order to justify the definition eq. (1.1) of mo-mentum. Treating the velocity independently of position is a common practicein math to reduce a system of ODEs to first-order, so that instead of a systemof N second-order differential equations on configuration space we consider asystem of 2N first-order differential equations on phase space, equations (1.1)and (1.2).

In accordance with the uniqueness part of Newton’s principle and in order tofocus on physical techniques, we will always implicitly assume that F is smooth(infinitely differentiable) on (RNd r ∆) × RNd × R; the theorem of existenceand uniqueness for ODEs then tells us there always exists a unique solutionsto the system of differential equations eq. (1.2) for any initial state and thatthis solution is also smooth. Although the theorem only guarantees a solutionfor a finite interval of time, from experimental observation we expect that forany naturally occurring system that the the interval can be extended to all ofR, and so we will occasionally assume that the solutions to our mathematicalmodel eq. (1.3) we have the motion for all time.

Lastly, we will formulate one more mathematical assumption: that the phys-ical laws governing the system’s motion is independent of the choice of coor-dinates and origin we impose on Rd. A transformation Rd × R → Rd × R isGalilean provided that it is an affine transformation (a linear transformationand a translation) if it preserves time intervals and for any fixed t ∈ R is anisometry of configuration space Rd. So if g(x, t) = (x′, t′) is . The set of Galileantransformations form a group under function composition.

Example 1.1. It is straightforward to verify that the following are all Galileantransformations:

(1) Translations: g1(x, t) = (x + x0, t+ t0) for some fixed x0 ∈ Rd, t0 ∈ R

(2) Rotations and reflections: g2(x.t) = (Ax, t) for some fixed orthogonaltransformation A ∈ O(d)

(3) Uniform motion with constant velocity: g3(x, t) = (x + vt, t) for somefixed velocity v ∈ Rd.

1.2. ENERGY 9

In fact, these examples generate the entire Galilean group (see Exercise 1.3).Galileo’s principle of relativity is the experimental observation that for the spe-cial case of an isolated system invariant under any Galilean transformation, andthat a reference frame—a choice of origin and coordinate axes for Rd—can al-ways be chosen such that this condition is satisfied. Any such frame is calledinertial, and the principle also asserts that all coordinate systems in uniformrectilinear motion with respect to an inertial frame is also inertial. (This isobserved, for example, in a car traveling at a constant velocity and noting thatmotion inside the car is as if the car were at rest.) Physically, this is requiringthat space be homogeneous and isotropic and that time be homogeneous. Geo-metrically, this principle requires that if we graph the trajectories of a particlesystem in configuration space and apply a Galilean transformation to the entiregraph, then the resulting graph will still be trajectories.

If Newton’s equations eq. (1.3) with respect to such an inertial coordinatesystem, then they must be invariant with respect to the Galilean group. Letx(t) denote a solution in an inertial coordinate system. Applying the Galileangroup generators of Example 1.1, we find the following conditions on Fi:

(1) Translations: Fi(x, x, t) ≡ Fi(xj − xk, x).

(2) Rotations and reflections: Fi(Ax, Ax) = AFi(x, x) for A ∈ O(n).

(3) Uniform motion: Fi(xj − xk, x) ≡ Fi(xj − xk, xj − xk).

Note that the third type or transformations in Example 1.1 change neither x norxi−xj . In the case N = 1 of one particle, this requires that F is independent ofx, x, t and is rotationally invariant, and so F ≡ 0. Consequently, the particle’sacceleration in an inertial coordinate system is equal to zero; this is Newton’sfirst law.

1.2 Energy

The kinetic energy of the ith particle is

Ti =1

2mi|xi|2 =

1

2mi|pi|2 =

1

2xi · pi, (1.4)

and summing over i = 1, . . . , N gives us the total system kinetic energy T . Fromobservation, we know that the magnitude of the velocity and hence the kineticenergy can be increased and decreased by the force Fi acting on the ith particle,depending on the force’s magnitude and direction. This is measured throughthe work done by the force Fi on the ith particle from time t0 to t, defined bythe line integral

Wi =

∫ x(t)

x(t0)

Fi · dsi =

∫ t

t0

Fi(x(τ)) · x(τ) dτ. (1.5)

(We use dsi to denote the line element of the trajectory xi(t), and so the secondequality is just the definition of path integration.)


Proposition 1.2. The increase in total kinetic energy is equal to the total workon the system.

Proof. Summing over i = 1, . . . , N and differentiating the kinetic energy eq. (1.4)we obtain

T =

N∑i=1

mixi · xi =

N∑i=1

xi · Fi = x · F

by using Newton’s equations eq. (1.3). Integrating, we arrive at

T (t)− T (t0) =

∫ t

t0

T dt =

∫ t

t0

x · F dτ =

∫ x(t)

x(t0)

F · ds = W

where W is the total work.

Although work is measured in the physical space Rd, the total work is nat-urally defined on configuration space RNd. So the change in kinetic energy canalso be thought of as the work done by the tuple of forces F on the lifted pathx(t) in configuration space.

For some systems there is also a potential energy. Physically, a system ofparticles is called conservative if the force F depends only on the positions x(not on time or velocity x) and if the total work along any path connecting twopoints y, z in configuration space,

W =

∫ z

y

F(s) · ds, (1.6)

is independent of the path. The path here is arbitrary, and is not limitedto trajectories. This is equivalent to the work around any simple closed pathvanishing, since two paths with the same endpoints can be concatenated to forma closed path.

Example 1.3. If the interaction forces depend only on particle distances,

Fi =

N∑j=1

Fij , Fij = fij(rij)eij , rij = |xi − xj |, eij =xi − xjrij

,

then the system is conservative, with potential energy U =∑i<j Uij , Uij =∫

fij dr.

Let’s pause, fix N − 1 of the particles, and consider only wiggling the ithparticle. Let S be a smooth 2-dimensional submanifold of Rd with boundary.Using Stokes’ theorem, we can rewrite

0 =

∫∂S

Fi · dsi =

∫S

dFi (1.7)

1.2. ENERGY 11

where dFi is the exterior derivative of Fi when viewed as a 1-form in the ithcoordinate on Rd (for n = 3, dFi is the curl ∇i × Fi dotted with the unitnormal vector field). For this to hold for all such surfaces S, then we must havedFi = 0 (∇i × Fi = 0 for n = 3). So the physical definition of a conservativeforce is an integral formulation of Fi being closed when viewed as a 1-formon Rd, which avoids the issue of dFi not being defined at

⋃j 6=ixj. From

differential geometry, we expect that Fi being closed should imply that Fi isexact (Fi = −∇U(xi)) on Rd, although the 0-form antiderivative may not bedefined on

⋃j 6=ixj.

Theorem 1.4. A system is conservative if and only if there exists a potentialenergy, a smooth function U : RNd r ∆→ R such that F = −∇U .

Proof. First suppose that the work for all of the forces Fi does not depend onthe path. Then the line integral

U(x) = −∫ x

x0

F(s) · ds

is well defined as a function for a fixed choice of x0 ∈ RNd r ∆, and we haveF = −∇

∑i Ui(xi).

Conversely, if F is a conservative force with potential energy U , then thefundamental theorem of calculus tells us that∫ x

x0

F · ds = −U(x) + U(x0).

That is, the work is independent of the path.

Remark. In Theorem 1.4, it is assumed that F and U are defined on all ofRNd r ∆. The statement does not hold true in general on just an open subsetof configuration space RNd.

Example 1.5. The vector field

F(x, y) =

(y

x2 + y2,− x

x2 + y2

)is defined on R2r0, and on this set F may be written as the negative gradientof the planar polar coordinate angle. However, the work done on a particletraveling once clockwise about the unit circle is 2π, which corresponds to thefact that we cannot define an angular coordinate continuously on all of R2r0.Consequently, if we consider a particle subject to this force we have an exampleof a nonconservative force.

It is often helpful for physical intuition to picture a small ball rolling down thegraph of U(x). Suppose we have a solution x(t) to eq. (1.8) with E(x(t)) ≡ E0.Since kinetic energy is nonnegative, then a ball at position U(x(t)) is confined tothe region where U(x) ≤ E0, which is often called a potential well. The smaller


the potential energy the greater the kinetic energy and hence the greater thevelocity; this tells us that as the pebble gains velocity as it rolls into the well.This picture makes some facts very intuitive, like that local minima and maximaof U(x) are stable and unstable equilibria for the system, respectively. For abounded potential well in a conservative system, the ball rolls right through anyminima and up towards the boundary U−1(E0).

Now that we have two notions of mechanical energy—kinetic and potential—we may consider their sum E = T + U , called the total energy.

Proposition 1.6 (Conservation of energy). The total energy E = T + U of aconservative system is preserved under the motion: E(t) = E(t0).

Proof. As in Proposition 1.2, we differentiate:

d

dtE = T +∇U · x = x · F− F · x = 0.

1.3 Conservative systems

In section 1.2 we saw that for physically conservative systems the total mechan-ical energy is constant along trajectories. We will now give this phenomenona name, and examine the mathematical consequences of this seemingly simpleobservation. Exercise 1.6 provides an example where the conservative quantityis not an energy, for which the results of this section are more surprising.

Recall from section 1.1 that we can always reduce a system of ODEs to afirst-order system. Suppose we have a system of the form

x = f(x) (1.8)

which is conservative:

Definition 1.7. The system of ODEs eq. (1.8) is called conservative if there is asmooth function E(x) (that is nonconstant on open sets) such that d

dt (E(x(t))) =0 for any solution x(t) to eq. (1.8).

Note that geometrically this requires that trajectories x(t) lie in level sets ofE(x, x) in phase space.

Any point x∗ where f(x∗) is called a fixed point or equilibrium of eq. (1.8),since this implies that the constant solution x(t) ≡ x∗ is a solution. Aroundany point that is not an equilibrium, the implicit function theorem guarantees aneighborhood on which the level set through that point is the graph of a smoothfunction, and so non-equilibrium trajectories trace out smooth curves in phasespace. Moreover, if U(x) 6= E for all x then the level set cannot contain anequilibrium and must be smooth everywhere.

A fixed point is attracting if x(t) → x∗ as t → ∞ for all initial conditionssufficiently close to x∗.

Proposition 1.8. A conservative system cannot have any attracting (or repul-sive) fixed points.

1.3. CONSERVATIVE SYSTEMS 13

Proof. Suppose x∗ were an attracting fixed point. Then all points in an open ballaround x∗ would have to be at the same energy E(x∗) since energy is constanton trajectories and all trajectories in the ball flow to x∗. But then E(x) wouldbe constant on an open set, which contradicts our definition. Changing t 7→ −tyields the analogous argument for repulsive fixed points.

To examine the stability of a fixed point, we would like to Taylor expandabout x∗ to replace (1.8) with the linear system

d

dt(x− x∗) = x ≈ f(x∗) + f′(x∗) · (x− x∗) = f′(x∗) · (x− x∗), (1.9)

where f′(x∗) is the Jacobian matrix of f evaluated at x∗. From ODE theory, itturns out that this first term is sufficiently dominant to determine the stabilityof x∗ if f′(x∗) has at least one eigenvalue with nonzero real part. This fact issharp, as the following example illustrates:

Example 1.9. Consider the system

x = −y + ax(x2 + y2)y = x+ ay(x2 + y2)

⇐⇒ r = ar3

θ = 1

on R2, where a ∈ R is a constant. When a = 0 we obtain the linearized systemat (x∗, y∗) = (0, 0), which has eigenvalues ±i and predicts that the origin issurrounded by periodic solutions or closed orbits—such a fixed point is called acenter. However, when a < 0 (a > 0) we see that r(t) is decreasing (increasing)monotonically and so the origin becomes a stable (unstable) spiral.

Centers are delicate, in the sense that trajectories nearby need to perfectlymatch up after one revolution, and the neglected terms in the Taylor expansioneq. (1.9) can push them inwards or outwards. This example illustrates thatthe linearized system predicting a center is in general unilluminating as to thestability of the fixed point. However, a conserved quantity is enough to recoverthe prediction:

Theorem 1.10 (Nonlinear centers for conservative systems). Suppose x∗ is anisolated fixed point of eq. (1.8) and that E(x) is a conservative quantity. If x∗is a local minimum (or maximum) of E, then all trajectories sufficiently closeto x∗ are closed.

Sketch: Since E is constant on trajectories, each trajectory is containedin some contour of E. Near a local maximum or minimum the contours areclosed. The only remaining question is whether the trajectory actually goes allthe way around the contour or whether it stops at a fixed point on the contour.But because x∗ is an isolated fixed point, there cannot be any fixed points oncontours sufficiently close to x∗. Hence all trajectories in a sufficiently smallneighborhood of x∗ are closed orbits, and therefore x∗ is a center.

Remark. We need to assume that x∗ is isolated, otherwise there could actuallybe fixed points on the energy contour—see Exercise 1.7.


Example 1.11 (Harmonic oscillator). For the one-dimensional system mx =−kx, we have U(x) = 1

2kx2 and E(x, p) = 1

2 ( 1mp

2 + kx2). The trajectories areperiodic and trace out the elliptic level sets of E in phase space Rx × Rp. Thissystem has the explicit solution

x(t) = x0 cos

(√k

mt

)+

p0√km

sin

(√k

mt

),

p(t) = −√kmx0 sin

(√k

mt

)+ p0 cos

(√k

mt

),

which are parametric equations for an axes-parallel ellipse centered at the origin.

1.4 Nonconservative systems

From experience we know that in practice systems are rarely conservative—if wewere to test an ideally conservative system, dissipative forces like kinetic frictionwould dampen mechanical motion and prevent perpetual motion. In this casewe would now have d

dtE ≤ 0, which has new consequences for solutions.

Definition 1.12. Consider the two-dimensional ODE system (1.8) with a fixedpoint x∗. Suppose there exists a continuous function E(x) on a connectedneighborhood U of the fixed point that is smooth on U r x∗. x∗ is a strictglobal minimum with value zero, and for all c > 0 sufficiently small the levelsets E(x) = c within U form simple closed curves. If also:

(a) ddtE(x) ≤ 0 for all x in U r x∗ then E is a weak Liapunov function;

(b) ddtE(x) < 0 for all x in U r x∗ then E is a strong Liapunov function.

For our image of the ball rolling down the graph of the potential energy,the surface of the graph is now slightly sticky. The ball may still roll through aminimum, but does not have enough energy to approach the boundary U−1(E0)again and so the permitted region for the ball continually shrinks.

Example 1.13 (Damped harmonic oscillator). Consider the one dimensionalsystem mx = −bx − kx with b > 0 a positive damping constant. The totalenergy is still E = 1

2 (mx2 + kx2), but now

d

dtE = mxx+ kxx = −bx2 ≤ 0

and so total energy E is a weak (but not strong) Liapunov function. The originis globally attracting with three qualitatively different phase portraits:

• 0 < b < 2√km (under damped): the origin is a stable spiral and the system

oscillates infinitely many times with exponentially decaying amplitude;

1.4. NONCONSERVATIVE SYSTEMS 15

• b = 2√km (critically damped): the origin is a stable degenerate node and

the system moves toward the origin as quickly as possible; and

• b > 2√km (over damped): the origin is a stable node and the system

returns to the origin without oscillating.

Theorem 1.14. Consider the smooth two-dimensional ODE system (1.8) witha fixed point x∗.

(a) If there exists a weak Liapunov function near the fixed point x∗ then thefixed point is Liapunov stable: for all ε > 0 there exists δ > 0 such that‖x(t0)− x∗‖ < δ implies ‖x(t)− x∗‖ < ε for all t ≥ t0.

(b) If there exists a strong Liapunov function near the fixed point x∗ then x∗is also globally asymptotically stable: for all initial conditions the corre-sponding solutions x(t)→ x∗ as t→∞.

In particular, there can be no periodic solutions, unlike conservative systems.For a proof and more details, see [JS07, Ch. 10].

Now let us examine the behavior of solutions when the friction term dom-inates. For the damped harmonic oscillator in Example 1.13, this is the limitb mk (i.e. mk/b2 → 0)—see Exercise 1.8 for details. In this limit, the x termis negligible and we may be left with something of the form:

Definition 1.15. The ODE system (1.8) is a gradient system if it is of the formx = −∇U(x), for some smooth function U(x) (which for mechanical systemsmay not be the same as the potential energy).

Remark. Although conservative systems x = −∇U(x) and gradient systemsx = −∇U(x) look similar, they display entirely opposite behavior.

From the definition, we see that the vector field −∇U(x) in configurationspace which is tangent to trajectories always points in the direction of steepestdescent for U(x), and hence is orthogonal to level sets (rather than parallel aswith conservative systems). For our image of the ball rolling down the graph ofthe potential energy, the ball now slows and never reaches the first minimum itencounters, as if the potential energy graph was a water tank. Closed orbits areof course impossible again (see Exercise 1.9).

Example 1.16. In the over damping limit for the harmonic oscillator we areleft with x = −kx/b, and so U(x) = kx2/2b. This is now a first-order systemwith solution x(t) = x0e

−kt/b, which is the limiting (slow time scale) behaviorfor the over damped oscillator after the transient (fast time scale) behavior isnegligible. For the phase portrait we can take p = mx = −mkx/b, but we canno longer take a second arbitrary initial condition p(0).


1.5 Reversible systems

Other than conserved quantities, many mechanical systems obey another sym-metry: a symmetry in time. A mechanical system of the form

mx = F(x) (1.10)

that is independent of time and velocity (or if F is even in velocity in time) isinvariant under the change of variables t 7→ −t, since x picks up two factors of−1. Because p 7→ −p under this change of variables, phase space is reflectedabout the position axes (x,p) : p = 0 and the arrows on trajectories arereversed. The resulting trajectories are still trajectories for the original systemsince the equation is unchanged, and so the phase portrait must be symmetricabout the position axes with arrows reverse.

We will now give this phase space symmetry a name and study its mathe-matical consequences. Exercise 1.10 provides a two-dimensional example wherethe symmetry is not reflection across the x-axis.

Definition 1.17. The two-dimensional ODE system (1.8) is reversible if it isinvariant under the change of variables t 7→ −t, x 7→ R(x) for some smoothinvolution R (i.e. R(R(x)) = x).

Involutions include reflections across a hyperplane and rotations about an axisby π.

Recall from section 1.2 the difficulty of classifying an equilibrium as a center.

Theorem 1.18 (Nonlinear centers for reversible systems). Suppose that thesmooth two-dimensional ODE system (1.8) is reversible, and that x∗ is a fixedpoint both for the system (f(x∗) = 0) and for R (R(x∗) = x∗). If the linearizedsystem about x∗ has x∗ as a center, then all trajectories sufficiently close to x∗are indeed closed.

Sketch: After a change of variables, we may assume that x∗ = 0 and thatR is a linear reflection. Consider a trajectory that starts on the hyperplanefixed by R. Sufficiently near the origin, the flow swirls around the origin thanksto the dominant influence of the linear center, and so the trajectory eventuallyintersects the hyperplane again. By reversibility, we can reflect this trajectoryto obtain a twin trajectory with the same endpoints but with its arrow reversed.Together the two trajectories form a closed orbit, as desired. Hence all trajec-tories sufficiently close to the origin are closed.

This argument can also be used to show the existence of lone closed orhomoclinic orbits, once we know that the trajectory starts and returns to thehyperplane of symmetry.

1.6. LINEAR MOMENTUM 17

1.6 Linear momentum

Suppose the force on the ith particle can be decomposed into

Fi =

n∑j=1j 6=i

Fij + F′i, (1.11)

where Fij is the interaction force between the ith and jth particle, and F′i isthe external force on the ith particle. A system is called closed if there areno external forces. Further assume that the interaction forces obey the law ofaction and reaction or Newton’s third law—another experimental observationthat the force two particles exert on each other are equal and opposite: Fij =−Fji. Not all interaction forces obey this law, but we will always implicitlyassume that they do. If we were also working in an inertial frame then theinteraction forces would also be collinear: Fij = fijeij , where eij is the unitvector (xj − xi)/|xj − xi| from the ith to the jth particle.

Example 1.19. Any system of the form in Example 1.3, which includes themotion of gravitational bodies, obeys the law of action and reaction since eij =−eji. Interaction forces for non-mechanical systems may not obey this law. Forexample, a particle with electric charge q placed in an electromagnetic field isacted upon by the Lorentz force

F = q

[E +

1

c(v ×H)

], (1.12)

where E,H are the strengths of the electric and magnetic fields (they satisfythe Maxwell system of equations) and c is the speed of light.

The effect of all external forces together can be observed through the total(linear) momentum,

P =

N∑i=1

pi. (1.13)

Proposition 1.20 (Conservation of linear momentum). The change in the totalmomentum is equal to the total force

∑iFi, which for the decomposition (1.11)

is the total external force. In particular, for a closed system the total linearmomentum is conserved.

Proof.

P =

n∑i=1

pi =

n∑i=1

Fi =

n∑i,j=1i 6=j

Fij +

n∑i=1

F′i =

n∑i=1

F′i

In the last equality, we note that Fij = −Fji causes the double sum overinteraction forces to cancel pairwise.


By taking a dot product in the previous proposition, we see that sometimesthe individual components of the total momentum can still be conserved:

Corollary 1.21. If the total external force is perpendicular to an axis, then theprojection of the total momentum onto that axis is conserved.

The system center of mass (sometimes barycenter) is

X =

∑Ni=1mixi∑Ni=1mi

. (1.14)

This does not depend on the choice or origin, and it is special because the totalmomentum relative to this point vanishes (exercise Exercise 1.11). Moreover,the total momentum is mathematically the same as that of a particle of mass∑Ni=1mi lying at the center of mass:

P =

N∑i=1

pi = MX, M =

N∑i=1

mi. (1.15)

Proposition 1.22 (Newton’s first law). The center of mass moves as if allmasses were concentrated at it and all forces were applied to it. In particular,the center of mass of a closed system moves uniformly and rectilinearly.

Proof. Differentiating eq. (1.15) yields

MX = P =

n∑i=1

Fi.

The right-hand side vanishes for a closed system.

1.7 Angular momentum

Now that we have discussed linear momentum, we would like to consider angularmomentum, for which we will have to specialize to n = 3 for this section so thatwe may use the cross product. The angular momentum (about the origin) of aparticle is

Li = xi × pi = mixi × xi, (1.16)

and the torque (sometimes moment of force) is

Ni = xi × Fi. (1.17)

Since the force can be divided into interaction and external forces, we also havethe external torque N′i = xi×F′i. The relationship between angular momentumand external torque is analogous to before:

1.7. ANGULAR MOMENTUM 19

Proposition 1.23 (Conservation of angular momentum). The change in to-tal angular momentum is equal to the total torque, which for the decomposi-tion (1.11) is the total external torque. In particular, for a closed system thetotal angular momentum is conserved.

Proof. Using Newton’s equations (1.2) we have

Li =d

dt(xi × pi) = xi × pi + xi × pi = Ni + 0.

Note that the product xi×pi = 0 vanishes because xi is parallel to pi. Summingover i = 1, . . . , N yields L = N. As in the proof of Proposition 1.20 the totaltorque is equal to the total external torque,

L =

n∑i=1

xi × Fi =

n∑i,j=1i6=j

xi × Fij +

n∑i=1

xi × F′i =

n∑i=1

N′i.

This is because the pairwise vanishing guaranteed by the law of action andreaction,

xi × Fij + xj × Fji = (xi − xj)× Fij = 0.

In particular, for a closed system the external torque vanishes and we haveL = 0.

Taking a dot product we also obtain component-wise conservation:

Corollary 1.24. If the total external torque is perpendicular to an axis, thenthe projection of the total angular momentum onto that axis is conserved.

As with linear momentum, from outside the system the momentum evolvesas if it were all concentrated about the center of mass and all the externaltorques were applied to it.

Proposition 1.25. The total angular momentum and total torque about thecenter of mass is the sum of that of and about the center of mass.

Proof. We expand about the center of mass:

L =

N∑i=1

[X + (xi −X)]×mi

[X + (xi − X)

]= X×MX +

[N∑i=1

mi(xi −X)

]× X

+ X×

[N∑i=1

mi(xi − X)

]+

N∑i=1

(xi −X)×mi(xi − X)

= X×P + 0 + 0 +

N∑i=1

(xi −X)×mi(xi − X)

where we used eq. (1.15) for the first term and Exercise 1.11 for the vanishingof the square-bracketed terms. Since L = N, the second statement follows.


There is one last dynamical quantity of interest, the moment of inertia ofthe ith particle about the origin,

Ii = mi|xi|2. (1.18)

The moment of inertia plays the role for angular velocity,

ωi =xi × xi|xi|2

, (1.19)

analogous to mass for linear velocity. This is because

Li = xi × pi = xi × (miωi × xi) = mi [(xi · xi)ωi − (xi · ωi)xi]= Iiωi + 0, (1.20)

since xi = ωi×xi. As in the previous proposition, the formula L = Iω can alsobe broken into the sum of that of and about the center of mass.

Proposition 1.26. The total moment of inertia evolves according to I = 4T +2F ·x. For a conservative system with a homogeneous potential of degree k (i.e.U(x) = c|x|k) we have I = 4E − 2(k + 2)U .

Proof. The first statement is a straightforward calculation. For the second, wenote that for a homogeneous potential we have

F · x = −∇U · x = −kU, 4T = 4E − 4U.

1.8 Exercises

1.1 (Pendulum). Consider a mass m attached to the end of a rigid masslessrod of length ` with the other end suspended at a fixed point. We allow therod to rotate in a vertical plane, subject to a constant downward gravitationalacceleration g.

(a) Let x denote the angle from the vertical directly below the pivot, and showthat

x = −g`

sinx.

(We have not allowed for angular coordinates yet, so at the moment weare cheating and taking x ∈ R.)

(b) Sketch the potential energy and the phase portrait, and convince yourselfhow the trajectories correspond to a small ball rolling down the graph ofthe potential. Note that near the origin in the phase plane the diagramlooks like that of the harmonic oscillator (cf. Example 1.11), which is aconsequence of the small angle approximation sinx ≈ x. Identify the equi-libria (constant solutions) and the separatrix (the eye-shaped boundaryseparating different modes of behavior). How many trajectories make upthe separatrix from −π ≤ x ≤ π, and to what motion do they correspond?

1.8. EXERCISES 21

(c) Now add a damping term:

x = −bx− g

`sinx

where b > 0. Sketch the new phase portrait, and show that E decreasesmonotonically along all trajectories that are not equilibria.

1.2 (A way to compute π [Gal03]). In this example we will see a geometric aspectof phase space appear as a physically measurable quantity, which reinforcesthat phase space is inherent to a mechanical system and not merely abstract.Consider a frictionless horizontal ray with a vertical wall at the origin. Onesmall block of mass m is initially at rest on the surface, and a big block of massM m is pushed towards the small block so that the small block is sandwichedbetween the large block and the wall. We will count the number N of collisionsthe small block makes with the big block and the wall.

(a) Let v1 and v2 denote the velocities of the large and small blocks respec-tively, and consider the convenient rescaling y1 =

√Mv1, y2 =

√mv2.

Plot the initial energy level set in the (y1, y2)-plane, to which the motionis confined since the surface is frictionless.

(b) Initially we have v1 < 0, v2 = 0—plot this point in the same (y1, y2)-plane. Assume that the collision is purely elastic, so that when the blockscollide the total momentum is conserved. Plot the total momentum levelset which contains the initial point—the outcome velocities is determinedby the other intersection of the level sets. After the first collision the smallblock will hit the wall, and we will assume that this collision is elastic sothat v2 < 0 is replaced by −v2—plot this new point as well. Plot a fewmore iterates of this two-collision pattern in the (y1, y2)-plane.

(c) The pattern repeats until v1 > 0 and 0 < v2 < v1, so that the largeblock is moving away, the small block will not collide with the wall again,and the small block cannot catch up. Sketch the end zone region in the(y1, y2)-plane as well.

(d) Connect consecutive points occupied by the system in the (y1, y2)-plane.Since the lines are either vertical or parallel, the angle θ any any point be-tween any two consecutive lines is the same. Show that θ = arctan

√m/M .

(e) For any point, the angle θ at that point subtends an arc on the circleopposite that point. Show that the angle at the origin which subtends thesame arc is 2θ, and that the total number N of collisions that occur is themaximal integer such that Nθ < π.

(f) Take M = 100nm for n a positive integer. Show that 10−n · N → π asn → ∞, and so the number of collisions N begins to spell out the digitsof π for n large. In fact, N will be a number with n+ 1 digits whose firstn digits coincide with the first n decimal digits of the number π (startingwith 3).


1.3 (Galilean group generators). Show that every Galilean transformation g ofthe space Rd ×R can be written uniquely as the composition g1 g2 g3 of thethree types of Galilean transformations in Example 1.1. Start by writing g as ageneral affine transformation on Rd × R,

g(x, t) = (Ax + vt+ x0,b · x + kt+ t0),

and show that b = 0, k = 1, and A ∈ O(d). What is the dimension of theGalilean group for d = 3?

1.4. Suppose a mechanical system of N particles in Rd is in some inertial frameand all of the initial velocities are zero. Show that the particles always remainin the (d − 1)-dimensional linear subspace of Rd in which they were initiallycontained.

1.5 (Rotating reference frame). Suppose we have a system in an inertial co-ordinate system z ∈ R3 (like coordinates relative to a stationary sun), so thatNewton’s equations (1.3) obey the conditions (1)–(3) of section 1.1. Consider anon-inertial coordinates x (like coordinates relative to a fixed point on Earth’ssurface) defined by

t 7→ t, x = B(t)z + b(t),

where b(t) ∈ R3 is the new origin and B(t) ∈ O(3) for all t.

(a) Show that the equations of motion in the new frame are

mizi = Fi

(zk − zj , B

−1B(zk − zj) + (zk − zj))

+ Φi + Ψi,

where

Φi = −mi

(B−1Bzi +B−1b

), Ψi = −2miB

−1Bzi.

The new forces Φ and Ψ that appear on the right-hand side of the equa-tions of motion for z are called inertial or fictitious forces.

(b) Differentiate the definition of an orthogonal matrix to show that B−1B isskew-symmetric, and write B−1Bz = ω × z for some vector ω called theangular velocity of the moving frame. Now

Ψi = −2miω × z.

Ψi is called the Coriolis force, and depends on the velocity. In the northernhemisphere of the earth it deflects every body moving along the earth tothe right and every falling body eastward.

(c) Use the product rule for (B−1B)· and the skew-symmetry of B−1B toshow that

B−1B =(B−1B

)·+B−1BB−1B.

1.8. EXERCISES 23

Writing w = B−1b for the acceleration and α = ω for the angular accel-eration of the moving frame, we obtain

Φi = −mi [w + ω × (ω × zi) +α× zi] .

The second term is called the centrifugal force and the third term is theinertial rotation force or the Euler force, both of which only depend onposition. The former is always directed outward from the instantaneousaxis of rotation and acts even on a body at rest in the coordinate systemz. The latter is only present for nonuniform rotation.

1.6 (A non-mechanical conservative system). In dimensionless form, the Lotka–Volterra predator-prey model takes the form

x = x(1− y), y = µy(x− 1),

where µ > 0 is a constant.

(a) Use the chain rule to find a differential equation for dy/dx.

(b) Find a conserved quantity E(x, y) for this system by integrating this sep-arable differential equation.

(c) Show that all trajectories are periodic for initial conditions x0, y0 > 0.

1.7 (Minimum of a conservative system that is not a center). Consider thesystem

x = xy, y = −x2.

Show that E(x, y) = x2 +y2 is a conserved quantity and plot the phase portraitfor this system. Although the origin is a minimum for E, it is not an isolatedfixed point nor a center.

1.8 (Harmonic oscillator in the over damped limit). This is an example of the“over damped” limit yielding a gradient system, as advertised in section 1.4.The objective is to find what limit we need to take for the damped harmonicoscillator:

mx = −bx− kx, b, k > 0,

that justifies neglecting the x term.

(a) By equating the units of all terms, determine the dimensions of b and k.

(b) Define a new dimensionless variable τ via t = Tτ where T is a constantwith units time to be chosen. Find the new differential equation in termsof τ . Divide the equation by the coefficient of the x term, pick T to cancelthe coefficient of the x term, and check that T has units time.

(c) The coefficient of the x term should be mk/b2, and so the limit in whichthis term is negligible is ε := mk/b2 → 0. Find the general solutionx(t) = c1e

k1t + c2ek2t for the linear equation εx+ x+ x = 0.


(d) Recall that 1/|k| is called the characteristic time of ekt, since after a time1/|k| the function has decreased (since k < 0) by a factor of 1/e. Find theleading term in the Taylor expansion of 1/k1 and 1/k2 about ε = 0—theseare called the fast and slow time scales for the solution.

1.9. Show that nonconstant periodic solutions are impossible in a gradient sys-tem, by considering the change in U(x) around such an orbit. In particular,since we can always find an antiderivative in one dimension, this shows that anyone-dimensional first order ODE has no periodic solutions.

1.10 (A non-mechanical reversible system). Show that the system

x = −2 cosx− cos y, y = −2 cos y − cosx

is reversible, by sketching the phase portrait and finding a symmetry. Note thatthe presence of stable and unstable nodes guarantees that this system is notconservative.

1.11. (a) Show that the center of mass (1.14) does not depend on the choiceof origin.

(b) Show that both the “total position” and total momentum relative to thecenter of mass vanishes:

N∑i=1

mi(xi −X) = 0,

N∑i=1

mi(xi − X) = 0.

1.12 (Book stacking [Nah16]). We will stack books of mass 1 and length 1 ona table in an effort to produce the maximum amount of overhang.

(a) Place the first book with its left edge at x = 0 and its right edge linedup with the end of the table at x = 1. By considering the center of massof the book, determine the distance S(1) we can slide the book over theedge of the table before it falls.

(b) Starting with a stack of two books instead of one, we can reason as inpart (a) and slide the top book forward a distance of S(1) while keep-ing the bottom book stationary. By considering the center of mass ofthe two books, determine the distance S(2) we can slide this two-bookconfiguration before it falls.

(c) Now start with three books, slide the top one a distance of S(1) and thenthe top two books as in part (b) in order to produce an overhang S(2) fromthe edge of the bottom book. Determine the distance S(3) we can slidethe three-book configuration before it falls. Note that when multiplied by2, the sequence S(1), S(2), S(3), . . . begins to spell out a familiar series.

(d) Postulate a formula for S(n) and prove it by induction. Note that theoverhang S(n) tends to infinity as n→∞.

1.8. EXERCISES 25

1.13. Consider a conservative system in R3 with coordinates so that the centerof mass X = 0 is the origin and is at rest, X = 0.

(a) Show that K2 ≤ 2IT .

(b) Show that the trajectories in configuration space must be contained in the

intersection of the hyperplane x ∈ R3N :∑Ni=1mixi = 0 with the set

x ∈ R3N : U(x) + |L|2/(2I(x)) ≤ E, where L and E are constants ofthe motion.


Chapter 2

Examples

In this section we will examine some examples to see how these tools are appliedand to show their purpose and power. These represent only a small portion ofsolved problems and are meant to provide a baseline intuition; a thorough studyof classical mechanics should include many more, including a study of rigid bod-ies and spinning tops. The examples presented here are adapted from [Arn89,Ch. 2], [LL76, Ch. 3], and [Gol51, Ch. 3].

2.1 One-dimensional systems

Suppose we have a system with one degree of freedom of the form

mx = F (x). (2.1)

Any such system is conservative, since we can always find an antiderivative U(x)(unique up to an additive constant) such that F = −dU/dx = −U ′(x). Fromsection 1.2 we know that the total energy

E =1

2mx2 + U(x) (2.2)

is conserved for motions of the system. Since there is only one degree of freedom,this provides an implicit first-order equation (rather than second-order) for themotion of the system. In this section we will reap the consequences of thissimple observation.

Since the kinetic energy must be nonnegative then we have E ≥ U at everypoint in phase space. A point (x0, y0) in phase space will then be an equilibriumif y0 = 0 (so that x = 0) and if U ′(x0) = 0 (so that x = 0). Solving the energyequation (2.2) for x we obtain

dx

dt=

√2[E − U(x)]

m. (2.3)

27

28 CHAPTER 2. EXAMPLES

The velocity is zero for values of x with U(x) = E; such values are called turningpoints for the motion. Since the motion is confined to the region U(x) ≤ E (andmust be connected), these turning points provide limits for the motion of thesystem.

For naturally occurring systems we often have U(x)→ +∞ as x→ ±∞, andso there are often two turning points and the motion of the system is bounded;we will assume that this is the case. Such motion must be periodic since it isconfined to one component of the energy level set and the trajectory cannotdouble back. Treating eq. (2.3) as a separable differential equation, we arrive at

t(E) =

√m

2

∫dx√

E − U(x)+ t0

for t0 an integration constant. The period τ (not to be confused with the kineticenergy T ) of bounded motion between turning points x1(E) < x2(E) is thus theintegral of t(E) from x1 to x2 to x1. This second portion would be negative,and since time should be increasing we need to insert a factor of −1. Thereforewe arrive at two times the integral from x1 to x2:

τ(E) =√

2m

∫ x2(E)

x1(E)

dx√E − U(x)

. (2.4)

This is an explicit formula for the period in terms of the energy.The period of motion also has a geometric interpretation. Let S(E) denote

the area enclosed by a closed trajectory in the phase plane corresponding tothe energy E and bounded by the turning points (x1, 0) and (x2, 0). As insection 1.5, the time symmetry of eq. (2.1) tells us that the trajectory in phasemust be symmetric about the x-axis. Therefore, expressing the top half of thecurve as a function y(x) = x(x) we may write

S(E) = 2

∫ x2(E)

x1(E)

y(x) dx = 2

∫ x2(E)

x1(E)

√2[E − U(x)]

mdx

by eq. (2.3). We can differentiate this using the Leibniz integral rule to get

dS

dE= 2

∫ x2(E)

x1(E)

d

dE

√2[E − U(x)]

mdx

+ 2

√2[E − U(x2)]

m

dx2

dE− 2

√2[E − U(x1)]

m

dx1

dE

Since x1, x2 are turning points we know U(x1) = U(x2) = E, and so the secondand third terms vanish:

dS

dE= 2

∫ x2(E)

x1(E)

1

2

√2

m[E − U(x)]dx =

√2

m

∫ x2(E)

x1(E)

dx√E − U(x)

2.1. ONE-DIMENSIONAL SYSTEMS 29

Therefore, comparing this expression to what we found for the period in eq. (2.4),we conclude

τ = mdS

dE. (2.5)

That is, the period of motion for closed trajectories is proportional to the rateof change of the enclosed area with respect to energy.

Not only does the potential determine the period, but the converse is alsopartially true. Suppose now that the period of motion is experimentally knownin a region bounded by x1, x2 where U(x) has only one local minimum, andwe wish to find the potential energy U in terms of τ . For simplicity, redefinethe potential U and the coordinate x such that U obtains its minimum U = 0at x = 0. Away from this minimum, each value of U corresponds to exactlytwo values of x; this allows us to write x−(U), x+(U) in the two intervals(x1, 0), (0, x2) respectively and manipulate the period formula (2.4) by changingvariables:

τ(E) =√

2m

∫ 0

x1(E)

dx√[E − U(x)]

+√

2m

∫ x2(E)

0

dx√[E − U(x)]

=√

2m

∫ 0

E

dx−dU

dU√E − U

+√

2m

∫ E

0

dx+(U)

dU

dU√E − U

=√

2m

∫ E

0

(dx+

dU− dx−

dU

)dU√E − U

.

Dividing both sides of the equality by√W − E where W = U and integrating

with respect to E from 0 to W yields∫ W

0

τ(E) dE√W − E

=√

2m

∫ W

0

∫ E

0

(dx+

dU− dx−

dU

)dU dE√

(W − E)(E − U)

=√

2m

∫ W

0

(dx+

dU− dx−

dU

)[∫ W

U

dE√(W − E)(E − U)

]dU

=√

2m

∫ W

0

(dx+

dU− dx−

dU

)[tan−1

(W + U − 2E

2√

(W − E)(E − U)

)]E=W

E=U

dU

=√

2m

∫ W

0

(dx+

dU− dx−

dU

)π dU = π

√2m (x+(W )− x−(W ))

where in the last step we noted x+(0) = 0 = x−(0), and in the second equalitywe switched order of integration (we assumed U < W without loss of generality,since the integral over E is symmetric). Rearranging reveals the result:

x+(U)− x−(U) =1

π√

2m

∫ U

0

τ(E) dE√U − E

(2.6)

That is, the period determines the potential energy up to an additive constant.


2.2 Central fields

In this section we will consider the motion of a single particle of mass m in acentral field F. A vector field F in R3 is called central (about the origin) if allof the vectors are radial and that the magnitude of the force is only a functionof the radial coordinate r = |x|; in other words, F = F (r)r. (This definition ofcourse extends to Rd, but we will shortly need to restrict our attention to R3 inorder to discuss angular momentum.)

A central field must be conservative, and the corresponding potential energyU = U(r) depends only on the distance from the origin. This is because if wewrite F = F (r)r, then the work∫ x2

x1

F · ds =

∫ |x2|

|x1|F (r) dr

is path independent, and so from the conservative force Theorem 1.4 we knowthe force is conservative. Consequently, we can write F = −(dU/dr)r for aradial potential energy U(r).

The torque of the particle is

N = x× F = F (r)r(r × r) = 0,

and so by Proposition 1.23 we know the particle’s angular momentum L isconserved. Therefore L = x× p is constant as the system evolves, and since Lmust be perpendicular to x, then we know that (for L nonzero) the particle’smotion must be coplanar in the plane orthogonal to L. When L = 0, then x isparallel to the velocity x, and thus the particle’s motion must be collinear. Inboth cases, the motion is coplanar.

Using polar coordinates (r, φ) on this plane, the velocity is

x =d

dt

(r cosφr sinφ

)= r

(cosφsinφ

)+ rφ

(− sinφcosφ

)= rr + rφφ,

and so the angular momentum is

L = |x× p| = |(rr)× (mx)| = mr2φ. (2.7)

We can also observe this conservation geometrically. Write the area sector ∆Sswept by the radius vector x over an angle ∆φ to first order as

∆S =1

2x · x∆φ+O(∆t2) =

1

2r2φ∆t+O(∆t2). (2.8)

ThereforedS

dt=

1

2r2φ =

1

2mL (2.9)

This is known as Kepler’s second law : in equal times the particle’s radial vectorsweeps out equal areas.

2.2. CENTRAL FIELDS 31

We can now use the constant L to help us write the total energy as

E = T + U =1

2m(r2 + r2φ2) + U(r) =

1

2mr2 +

L2

2mr2+ U(r). (2.10)

If we define the effective potential to be

Ueff = U(r) +M2

2mr2, (2.11)

then (2.10) looks like a one-dimensional system for the coordinate r with thenew potential. The added term in the effective potential is called the centrifugalenergy. When the effective potential is equal to the total energy, then we haver = 0 which is a turning point for the particle. The particle is not at rest atsuch a point as was the case in the previous section however, unless the angularmomentum is zero. Rearranging the total energy (2.10) we get

dr

dt=

√2

m[E − Ueff(r)] (2.12)

This is a separable differential equation, with solution

t =

∫dr√

2m [E − Ueff(r)]

+ constant. (2.13)

We can also solve the separable differential equation (2.7) for φ to find

φ =

∫Mdr

r2√

2m [E − Ueff(r)]+ constant, (2.14)

where we plugged in dr/dt from eq. (2.12) and integrated. This yields an equa-tion for φ as a function of r. Note that the constancy of L requires that φ cannotchange sign.

As in the previous section, the motion is confined to the region Ueff ≤ E. Ifthe particle is unbounded for some finite energy E, then the potential energyat infinity U∞ = limr→∞ U(r) = limr→∞ Ueff must be finite—assuming it exists(as it often does for physical systems). If E > U∞, then we can define thevelocity at infinity v∞ via E = 1

2mv2∞+U∞. Conversely, in the region r → 0, in

order for the particle to reach the center U(r) must not outgrow the centrifugalenergy:

limr→0+

r2U(r) < − L2

2m. (2.15)

Suppose now that for some given Ueff we have motion bounded by two turn-ing points rmin and rmax, which confines the particle within an annulus definedby these two radii. Points where r = rmin are called pericenters and wherer = rmax are called apocenters. As before, the time symmetry for the one-dimensional system in r allow us to conclude that the particle’s motion is re-versible, and so the trajectory will be symmetric about any ray from the origin


through a pericenter or apocenter. According to our solution (2.14), the anglebetween successive pericenters (or apocenters) is

Φ = 2

∫ rmax

rmin

Ldr

r2√

2m[E − Ueff(r)](2.16)

by symmetry. In general, Φ 6= 2πm/n for some integers m,n, in which case theparticle’s orbit is not closed (and is in fact dense) in the annulus. There areonly two central fields in which all bounded orbits are closed: the first is theharmonic oscillator, U(r) = kr2 for k > 0 (cf. section 1.3), and the second iswhen U(r) = −k/r for k > 0 which will be covered in section 2.4.

2.3 Closed bounded orbits

There are two standard examples of central fields: the harmonic oscillator po-tential

U(r) = kr2, k > 0, (2.17)

and the gravitational potential

U(r) = −kr, k > 0. (2.18)

In this section, we will show that these are the only two central fields in whichall bounded orbits are closed.

Suppose U(r) is a central field in which all bounded orbits are closed. Thenfor any given orbit bounded by rmin ≤ r ≤ rmax, the angle between the pericen-ter and apocenter is

Φ =

∫ rmax

rmin

M

r2√

2m[E − U(r)]− L2/r2dr. (2.19)

If we make the substitution x = L/r, the angle

Φ =

∫ xmax

xmin

1√2m[E − U(L/x)− x2/2m]

dx (2.20)

looks like (2m)−1 times the period for oscillations in the one-dimensional systemW (x) = U(L/x) + x2/2m. Orbits near circular orbits have Φ ≈ Φcir which wecan calculate (cf. Exercise 2.2) to be

Φcir =πL

r2√U ′′(r)

= π

√U ′

3U ′ + rU ′′. (2.21)

For closed orbits we must have Φcir = 2πm/n for some integers m,n, andin particular must be independent of r. This yields a differential equation for

2.4. KEPLER’S PROBLEM 33

U(r), which requires U(r) = arα for α ≥ −2, α 6= 0, or U(r) = b log r. Pluggingthis back into Φcir yields

Φcir =π√α+ 2

, (2.22)

with the α = 0 case corresponding to U(r) = b log r. Now we split into cases.Right away, we see that if α = 0 then we have Φcir = π/

√2; this is not com-

mensurable with 2π and so small oscillations will not be closed.If α > 0, then U(r)→∞ as r →∞. Substituting x = xmaxy in the integral

for Φ yields

Φ =

∫ 1

ymin

1√2[W ∗(1)−W ∗(y)]

dy, W ∗(y) =y2

2+

1

x2max

U

(L

xmaxy

).

(2.23)As E →∞, note that xmax →∞ and the second term in W ∗ can be neglected.Moreover, in this limit we also have ymin → 0 and so we can evaluate the integralto be limE→∞ Φ(E,L) = π/2. But Φcir is a constant, and so this requires thatα = 2.

If α < 0, then U(r) = arα for −2 < α < 0. In the limit E → 0−, wecalculate

limE→0−

Φ(E,L) =

∫ 1

0

1√x−α − x2

dx =π

2 + α. (2.24)

Comparing this to eq. (2.22), we see that we must have α = −1.Therefore, the only two allowed cases are α = −1, 2 which correspond to

the gravitational and harmonic oscillator potentials. We have already seen thatall orbits for the harmonic oscillator are closed, and in the next section we willsolve the equations of motion for the gravitational potential and find that allbounded orbits are closed.

2.4 Kepler’s problem

Consider a particle whose motion is governed by the central field

U(r) = −kr, k > 0, (2.25)

which we might recognize as the gravitational field due to a fixed mass at theorigin, or the attractive electric force of a fixed charge at the origin. As discussedin section 2.2, this corresponds to the one-dimensional effective potential

Ueff(r) = −kr

+L2

2mr2. (2.26)

Note that limr→0+ Ueff(r) = +∞ and limr→∞ Ueff(r) = +∞. Since

dUeff

dr(r) =

k

r2− L2

mr3(2.27)


has only one root for r ∈ (0,∞) at r = L2/mk, then Ueff obtains a globalminimum with value

Ueff, min = Ueff

(L2

k

)= −mk

2

2L2. (2.28)

Since this must be the only extrema, we conclude that E ≥ 0 yields unboundmotion, and E < 0 requires that the particle be bound with E ≥ −mk2/2L2 ifL 6= 0.

Using the general solutions found in section 2.2, we find the equation ofmotion:

φ(r) = cos−1

Lr −

mkL√

2mE + m2k2

L2

(2.29)

where we have picked the origin such that the integration constant is zero.Rearranging, we get √

2mE +m2k2

L2cosφ =

L

r− mk

L

r

(mk

L+

√2mE +

m2k2

L2cosφ

)= L

r

(1 +

L

mk

√2mE +

m2k2

L2cosφ

)=

L2

mk

r =L2/mk

1 +√

1 + (2EL2/mk2) cosφ.

Define the quantities p = L2/mk and ε =√

1 + (2EL2/mk2), so that we maywrite

r(φ) =p

1 + ε cosφ. (2.30)

This is the parametric equation for a conic section with one focus at the origin,where ε ∈ [0,∞) is the eccentricity and 2p is the latus rectum. Note thatE ≥ 0 yields unbound orbits (ε = 1⇒ parabolas and ε > 1⇒ hyperbolas) and−mk2/2L2 < E < 0 yields bound orbits (ε = 0 ⇒ circles and 0 < ε < 1 ⇒ellipses). From geometry, the semi-major and -minor axes respectively are givenby

a =p

1− ε2=

k

2|E|, b =

p√1− ε2

=M√

2m|E|. (2.31)

For the planets in our solar system, the eccentricities are very small and thetrajectories are nearly circular. Kepler observed this, and in his first law statedthat the planetary orbits are circles with the Sun not at the center.

Now we determine the period τ of a bounded elliptic orbit. IntegratingKepler’s second law (2.9) over one orbit, we get

πab =

∫dS =

∫1

2mLdt =

1

2Lτ =⇒ τ =

2πmab

L= πk

√m

2|E|3. (2.32)

2.5. VIRIAL THEOREM 35

This is an explicit formula for the particle’s period in terms of its energy. Usingthe formula (2.31) for the semi-major axis a in terms of the energy E, we seethat

τ = πk

√8ma3

2k3= 2π

√ma3

k. (2.33)

This demonstrates Kepler’s third law : the square of the orbital period of aplanet is directly proportional to the cube of the semi-major axis of its orbit.

In practice, we know that a mass (or charge) sitting at the origin in order tocreate such a field is not unaffected by the particle’s presence—but this is easilyremedied. The two-body problem asks for the equations of motion for a closedsystem consisting of two interacting particles with position xi and mass mi, fori = 1, 2. The system is conservative with potential

U(x1,x2) ≡ U(|x1 − x2|) = − Gm1m2

|x1 − x2|(2.34)

where G is a constant called the gravitational constant. Define x = x1 − x2 tobe the distance between the particles and redefine the origin to be the center ofmass: m1x1 +m2x2 = 0. We can recover the positions form x via

x1 =m2x

m1 +m2, x2 = − m1x

m1 +m2. (2.35)

The total energy can be written solely in terms of x as

E =1

2m1|x1|2 +

1

2m2|x2|2 + U(x1,x2) =

1

2µ|x|2 + U(|x|), (2.36)

with the reduced mass µ ≡ m1m2/(m1 + m2). That is, the two-body systemis equivalent to a single-particle of mass µ moving in the external central fieldU(|x|). Solving for the motion of x in the reduced problem (2.36) yields thesolutions to the original problem (2.34) using the relations (2.35). In this way,any force that satisfies the law of action and reaction gives rise to a central field.In particular, for two gravitational bodies (or two attractive charges) this meansthe equations of motion for each will be conic sections with a shared focus atthe origin.

2.5 Virial theorem

The virial theorem is a formula for the time average of a system’s kinetic energy,which in the case of central field yields a remarkably simple relation. Considera mechanical system of N particles in Rd. It turns out the key quantity toexamine is the evolution of

∑i xi ·pi, which is not unreasonable since it is equal

to 2I where I is the moment of inertia. We calculate

d

dt

(∑i

xi · pi

)=∑i

xi · pi +∑i

xi · pi = 2T +∑i

Fi · xi.


Taking a time average of both sides, i.e.

f = limT→∞

1

T − t0

∫ T

t0

f(τ) dτ

(which is independent of choice of t0), we get

d

dt

(∑i

xi · pi

)= 2T +

∑i

Fi · xi. (2.37)

Note that for any bounded function f we have

df

dt= limT→∞

1

T

∫ T

0

df

dt(τ) dτ = lim

T→∞

f(T )− f(0)

T= 0.

Therefore, since all particle motion is bounded, then the quantity∑i xi · pi

must also be bounded, and so the left-hand side of eq. (2.37) vanishes:

T = −1

2

∑i

Fi · xi. (2.38)

This consequence is known as the virial theorem.If we return to a single particle moving in a central field as in the previous

section, then the force F = −dU/dr is conservative and is in the direction of r,and so

T =1

2U ′(r)r. (2.39)

Moreover, if we have a power-law force F (r) ∝ rn, then U(r) = arn+1 and so

U ′(r)r = (n+ 1)arn+1 = (n+ 1)U =⇒ T =n+ 1

2U. (2.40)

In particular, for gravitational force we have n = −2 and

T = −1

2U. (2.41)

2.6 Central field scattering

One-body scattering is a uniform beam of particles in R3, all of the same massand energy, approaching a central force which is assumed to decay far away fromthe origin, so that the particle motion begins and ends colinearly at infinity.Examples of such systems include a beam of asteroids passing by a star, or a(classical treatment of a) beam of charged particles passing near an attractive orrepulsive central charge. One of the beam characteristics is its intenstiy I: thenumber of particles crossing unit area normal to the initial direction of travelper unit time. For the solid angle Ω ⊂ S2 (where S2 has area 4π), the scattering

2.6. CENTRAL FIELD SCATTERING 37

cross section σ(Ω) is given by the number of particles scattered per unit solidangle per unit time divided by the incident intensity.

As established in section 2.2, we know the angular momentum L = |L| andenergy E for each particle is conserved. Define the impact parameter s for aparticle of mass m and initial velocity v0 via

L = mv0s = s√

2mE, (2.42)

because initially (and hence always since it is conserved) we have E = 12mv

20 .

Because the field is spherically symmetric, the scattering is rotationally symmet-ric about the incident beam’s axis, and so the quantities s and E are sufficientto determine the angle Θ between the incident and scattered beams. This sym-metry also tells us that the element of solid angle dΩ is related to the elementof angle dΘ by dΩ = 2π sin Θ dΘ. The particles between s, s + ds scatteredbetween Θ, Θ + dΘ is therefore

2πIs ds = −2πσ(Θ)I sin Θ dΘ. (2.43)

The minus sign arises for repulsive fields, which is what we will focus on due tophysical importance. Rearrangement yields

σ(Θ) = − s

sin Θ

ds

dΘ. (2.44)

An important example of central field scattering is for a beam of chargedparticles of charge −q incident towards a fixed charge −Q, for which the field isgiven by the Coulomb force

F(r) =qQ

r2r = ∇U, U(r) =

qQ

r. (2.45)

Although in the previous section we assumed that U = −k/r with k > 0 in orderto model celestial motion, nowhere did our calculations require that k = −qQbe positive, and so our solutions are still conic sections:

r(φ) =p

1 + ε cosφ= − L2

mqQ(1 + ε cosφ). (2.46)

The eccentricity is now given by

ε =

√1 +

2EL2

m(qQ)2=

√1 +

(2Es

qQ

)2

≥ 1. (2.47)

So the motion is necessarily hyperbolic for E, s > 0. Moreover, the radius r(φ)in eq. (2.46) must be positive, and so we also have the restriction

cosφ < −1

ε. (2.48)


Unlike in section 2.4 where the field center was the interior focus for hyperbolicmotion, the origin is now the hyperbola’s exterior focus. Since r → ∞ as φapproaches the bounds given by eq. (2.48), then the range Φ of φ and the totalangle of deflection Θ sum to π, and so

sin

(Θ

2

)= sin

(π

2− Φ

2

)= cos

(Φ

2

)=

1

ε

cot

(Θ

2

)=

√csc2

(Θ

2

)− 1 =

√ε2 − 1 =

2Es

qQ

This gives us the relation between s and Θ, and so using the cross-section (2.44)we find the Rutherford scattering cross section

σ(Θ) = −qQ2E cot

(Θ2

)sin Θ

d

dΘ

(qQ

2Ecot

(Θ

2

))

= −(qQ

2E

)2 cos(Θ/2)sin(Θ/2)

2 sin(Θ/2) cos(Θ/2)

(− 1

2 sin2(Θ/2)

)=

1

4

(qQ

2E

)21

sin4(Θ/2).

(2.49)

In context, this formula is a classical approximation of a quantum system, andthus it has limitations. For example, if we were to integrate this expression overthe sphere we would obtain an infinite total scattering cross section, which is ofcourse physically impossible.

As with Kepler’s problem, in practice the central charge is not fixed butrather recoils as a result of the scattering. Here, we cannot get away with asimple substitution of the reduced mass since Θ measures the angle betweenthe initial and final vectors between the charges, and no longer the deflectionangle ϑ between the initial and final directions of the scattered particle. Assumethe “central” particle is initially stationary, and let the subscript 1 refer to thescattered particle, primed coordinates denote coordinates relative to the centerof mass (as opposed to the laboratory coordinates), and X represent the positionof the center of mass in laboratory coordinates. As before, we have x1 = X+x′1,v1 = x1 = X + v′1. In the final configuration, the center of mass velocity X stilllies along the initial direction by conservation of momentum (the two-particlesystem is conservative), and so geometrically we have

tanϑ =|v′1| sin Θ

|v′1| cos Θ + |X|. (2.50)

Using the coordinate relations (2.35) and the reduced mass µ of the two particles,we know

v′1 =µ

m1v, (2.51)

2.6. CENTRAL FIELD SCATTERING 39

where x ≡ x1−x2 as before. Moreover, energy must be conserved and so lettingv0 = X∞ denote the initial relative velocity, we get

|v′1| =µ

m1|v0| (2.52)

This, in tandem with the conservation of total linear momentum:

(m1 +m2)X = m1v0 ⇒ |X| = µ

m2|v0|, (2.53)

allows us to rewrite eq. (2.50) as

tanϑ =

µm1|v0| sin Θ

µm1|v0| cos Θ + µ

m2|v0|

=sin Θ

cos Θ + m1

m2

. (2.54)

As we would expect experimentally, when m2 m1 then the initially stationaryparticle does not experience significant recoil and ϑ ≈ Θ. We would now liketo determine the cross section σ′(ϑ) in the laboratory system. The number ofparticles scattered into a fixed element of solid angle must be the same in boththe laboratory and relative to the center of mass. Consequently,

2πIσ(Θ) sin(Θ) dΘ = −2πIs ds = 2πIσ′(ϑ) sin(ϑ) dϑ,

and so

σ′(ϑ) = σ(Θ)sin Θ

sinϑ

dΘ

dϑ= σ(Θ)

d(cos Θ)

d(cosϑ). (2.55)

Our model illustrates elastic scattering, in that the total kinetic energy is con-served. Since the second particle is initially at rest and is then set into motion,the velocity |v1| after scattering is less than |v0|. Starting with the law of cosineswe have

|v′1|2 = |v1|2 + |X|2 − 2|v1||X| cosϑ,

and so ∣∣∣∣v1

v0

∣∣∣∣2 − 2µ

m2

∣∣∣∣v1

v0

∣∣∣∣ cosϑ− m2 −m1

m2 +m1= 0, (2.56)

using the conserved quantities (2.52) and (2.53).Rutherford was interested in α-particle scattering, for which corrections due

to m1/m2 1 are small. When m1 = m2, the angle relation (2.54) becomes

tanϑ =sin Θ

cos Θ + 1= tan

(Θ

2

)⇒ ϑ =

Θ

2. (2.57)

So 0 ≤ ϑ ≤ π/2 and there can be no back scattering. The results (2.55)and (2.56) simplify to

σ′(ϑ) = 4 cosϑσ(2ϑ) ,

∣∣∣∣v1

v0

∣∣∣∣ = cosϑ. (2.58)


2.7 Exercises

2.1 (Pendulum period). Show that for the pendulum of Exercise 1.1,

x = −g`

sinx,

the motion with turning points ±θ0 has period

τ = 4

√l

gK

(sin

(θ0

2

)),

where

K(k) =

∫ π/2

0

dξ√1− k2 sin2 ξ

is the complete elliptic integral of the first kind. Taylor expanding about θ0 = 0,find the one-term expansion:

τ ≈ 2π

√l

g

(1 +

θ20

16+ . . .

).

Note that the zeroth order term is the constant period approximation obtainedwhen we take sinx ≈ x and replace the pendulum with a harmonic oscillator.

2.2 (Small one-dimensional oscillations). Let E0 be the value of the potentialenergy of a one-dimensional system at a minimum point ξ. Show that theperiod τ0 = limE→E0 τ(E) of small oscillations in a neighborhood of the pointξ is 2π/

√U ′′(ξ).

2.3. Show that for the central field with potential

U(r) = − k

r3,

the motion falls into the origin in finite time.

2.4 (Method of similarity). Suppose that the potential energy of a central fieldis a homogeneous function of degree ν:

U(αr) = ανU(r) for all α > 0.

Show that if a curve γ is the orbit of a motion, then the homothetic curve αγis also an orbit (under the appropriate initial conditions). Determine the ratioof the circulation times along these orbits. Conclude from this the isochronicityof the harmonic oscillator (ν = 2) and Kepler’s third law (ν = −1).

2.5 (Escape velocity). Let r0 denote the radius of the Earth, and g ≈ 9.8 m/ sec 2

the gravitational acceleration at the Earth’s surface. The gravitational potentialenergy of the Earth is

U(r) = −gr20

r.

Determine with what velocity a particle must be given on the surface of theEarth in order for it to travel infinitely far away.

2.7. EXERCISES 41

2.6 (Cosmic velocities). Consider the gravitational potential energy of the Earthas in the previous problem. The escape velocity v2 is sometimes called the secondcosmic velocity. The first cosmic velocity is the velocity of motion on a circularorbit of radius close to the radius of the earth. Find the magnitude of the firstcosmic velocity v1 and show that v2 =

√2v1.

2.7 (Geosynchronous orbit [Nah16]). It is useful for communication satellitesto be in geosynchronous orbit, so that their orbital period is that of the Earth’sand the satellite appears to hover in the sky. We will calculate the height ofthis orbit for Earth in two different ways.

(a) Let m be the satellite’s mass, M = 5.98 × 1024 kilograms be the Earth’smass, v be the satellite’s velocity, and Rs the radius of the satellite’scircular orbit. First equate the gravitational and centripetal accelerationsof a circular orbit, then substitute v = 2πRs/T where T = 86, 400 secondsis the period of the orbit, and solve for Rs.

(b) Use Kepler’s third law to calculate the same value for Rs.

2.8 (Satellite paradox [Nah16]). Satellites in low Earth orbit experience sig-nificant atmospheric drag, which counterintuitively increases the speed of thesatellite.

(a) For a circularly orbiting satellite in Earth’s gravitational potential, con-clude from the virial theorem that the satellite’s total energy is

E = −1

2mv2.

Alternatively, this relation can be obtained by equating the gravitationaland centripetal accelerations, solving for v, and substituting v into thekinetic energy.

(b) Differentiate the result from part (a) to determine dv/dt in terms of dE/dt.Assuming the energy loss rate (dissipated power) of the satellite is

dE

dt= −vfd, fd > 0,

show that dv/dt > 0.

2.9 (Solar and lunar tides [Nah16]). If we differentiate the gravitational poten-tial (2.34), we find the gravitational force between two bodies of masses m1 andm2 to have magnitude

F (r) = Gm1m2

r2


where r is the distance between the bodies’ centers and G is the universalgravitational constant. For the Earth, Sun, and Moon, we have

Ms = mass of the Sun = 2× 1030 kilograms,

Mm = mass of the Moon = 7.35× 1022 kilograms,

Rs = Earth–Sun separation = 93× 106 miles,

Rm = Earth–Moon separation = 2.39× 105 miles.

(a) Show that, even though the Sun is much farther from the Earth, that theSun’s gravitational force on the Earth is greater than the Moon’s, and findtheir ratio.

(b) Since the Earth is not a point mass, then the Sun’s gravitational force isstronger (weaker) on the side of the Earth closest (farthest) from the Sun.This causes water to bulge at the points closest an furthest from the Sun,which is called the solar high tides. Calculate the maximum difference ingravitational force in terms of Earth’s radius R for both the Sun and theMoon.

(c) Extract the leading term in the limit R/Rs 1 and R/Rm 1 for eachexpression in part (b). Calculate their ratio and conclude that, althoughthe Sun’s gravitational force is stronger, the lunar tides are more thantwice as large as the solar tides.

2.10 (Energy of the ocean tides [Nah16]). The high tides are not directly in linewith the centers of the Earth and Moon, but are rather carried ahead slightlyby the Earth’s rotation and frictional forces. This means that the Moon’s grav-itational pull on both tides produce torque: the Moon’s pull on the farther tideincreases the Earth’s rotational speed, but the stronger pull on the nearer tideis counter-rotational, and so the overall effect decreases the Earth’s rotationalspeed. Atomic clocks ave measured that the length of a day is increasing at therate of about 2 milliseconds per century.

(a) Let Ω denote the angular rotation rate of the Earth and let T denote thelength of a day in seconds, so that ΩT = 2π. By integrating the kineticenergy of an infinitesimal mass dm over the volume of the Earth, showthat the rotational energy E is given by

E =1

2Ω2I,

where

I =

∫∫∫r≤R

(x2 + y2) dm

is the moment of inertia.

2.7. EXERCISES 43

(b) Show that for a solid sphere of radius R and constant mass density ρ themoment of inertia is

I =8π

15R3ρ,

or in terms of the total mass M ,

I =2

5MR2.

The Earth is not a constant-density sphere, and so rather than 0.4 thecoefficient is approximately 0.3444.

(c) Write the rotational energy E as a function of the period T , and showthat

dE

dT= −4 · 0.3444Mπ2R2

T 3.

Taking T = 86, 400 seconds, the length of a day, and ∆T = 2 × 10−3

seconds, M = 5.98× 1024 kilograms, and R = 6.38× 106 meters, find thechange in the Earth’s rotational energy ∆E over a century. Dividing ∆Eby the number of seconds in a century, conclude that the power of theocean tides is 3, 260 gigawatts or 4.37 billion horsepower.

2.11 (Moon recession rate [Nah16]). As in Exercise 2.10, tidal friction decreasesthe Earth’s rotational angular momentum. Consequently, the Moon’s orbital an-gular momentum increases to conserve total angular momentum, which resultsin the Moon drifting away from the Earth. We will estimate this recession rate,assuming that all of the momentum is transferred to Moon’s orbit (and notrotation).

(a) Consider the Moon as a point mass m orbiting circularly about the Earthat a radius r, with speed v and angular speed ω radians per second. Showthat the Moon’s orbital angular momentum is Lm = mrv.

(b) The gravitational force on the Moon by the Earth has magnitude

F =GMm

r2,

where M is the mass of the Earth and G is the universal gravitationalconstant. Equating the gravitational and centripetal accelerations of theMoon, find v as a function of r and use this to find the angular momentumLm as a function of r.

(c) Using part (b) of Exercise 2.10, the spin angular momentum of the Earthis Le = 0.3444MR2Ω where Ω is Earth’s rotation rate. Expressing Ω interms of the day length T = 86, 400 seconds, find Le as a function of Tand calculate dLe/dT .


(d) Using the daily change ∆T = 2×10−5/365 seconds in the length of a day,approximate the daily change and then the yearly change in ∆Le.

(e) Equating change in the Moon’s orbital momentum ∆Lm with the changein Earth’s rotational momentum |∆Le|, find the yearly change in theMoon’s orbital radius. Using the values

M = mass of the Earth = 5.98× 1024 kilograms,

m = mass of the Moon = 7.35× 1022 kilograms,

r = radius of Moon’s orbit = 3.84× 108 meters,

R = Earth’s radius = 6.37× 106 meters,

G = gravitational constant = 6.67× 10−11 meters3

kilograms× seconds2 ,

find that the Moon is receding from the Earth at a rate of 3.75 centimetersof 1.48 inches per year. This value is in outstanding agreement withmeasurements made by a laser on Earth and corner cube reflectors placedon the Moon during the Apollo 11 mission.

2.7. EXERCISES 45

Part II

Lagrangian Mechanics

Newton’s equations are rephrased as the abstract and coordinate-free prin-ciple of least action, so that the configuration space can be endowed with anycoordinates or replaced by any manifold. The subsequent Euler–Lagrange equa-tions of motion are often the preferred perspective in physics, and consist of asecond-order ordinary differential equation for each degree of freedom. Theseequations are independent of coordinate choice, which is a powerful tool in ap-plications as we are free to choose the most convenient coordinates to describeour system. When the chosen coordinates align with symmetries of the system,the Euler–Lagrange equations are effective in identifying the corresponding con-served quantities, reducing the number of equations, and simplifying the sys-tem. Further study of the action—an initially abstract quantity introduced toformulate the principle of least action—leads to the Hamilton–Jacobi equationof motion: a single partial differential equation with two times the independentvariables as the degrees of freedom.


Chapter 3

Euler–Lagrange equationsof motion

We will reformulate the equations of motion as the coordinate-free principle ofleast action, the unpacking of which will demonstrate some of the fundamentaltechniques of variational calculus. This presentation of the calculus of variationsfollows [AKN06, Ch. 1], and in the rest of the chapter is based on [LL76, Ch. 1–2], [Arn89, Ch. 3], and [Gol51, Ch. 1–2].

3.1 Principle of least action

For a mechanical system with N particles, we will replace the configurationspace (Rd)N with a smooth manifold M . We will often denote by q any (local)coordinates on M (sometimes called generalized coordinates in physics), and thenumber of components n = dimM is called the number of degrees of freedom.

Definition 3.1. A Lagrangian system (M,L ) is a smooth n-dimensional man-ifold M called the configuration space together with a smooth time-dependentfunction L (q, v, t) : TM × I → R (where I ⊂ R is an interval) on the tangentbundle called the Lagrangian.

As we will see in section 3.2, the following abstract condition in fact encodesNewton’s equations.

Definition 3.2 (Hamilton’s principle of least action). A motion q(t) of theLagrangian system (M,L ) between any two times t0 < t1 is a smooth path onM that is a critical point of the action functional

S(q(t)) =

∫ t1

t0

L (q(t), q(t), t) dt. (3.1)

The principle is called “least” action because it turns out that the motion is oftena minimum, but we do not need this additional assumption. In this section wewill see how to extract equations of motion from this assumption.

47

48 CHAPTER 3. EULER–LAGRANGE EQUATIONS OF MOTION

The goal of variational calculus is to find the extrema of functionals. A pathfrom a1 ∈ M to a2 ∈ M (not necessarily distinct) starting at t1 and endingat t2 > t1 is a smooth map γ : [t1, t2] → M with γ(t1) = a1 and γ(t2) = a2.Let Ω denote the collection of all such paths, which can be thought of as an“infinite-dimensional manifold”. A functional Φ is a function from Ω into R.

Example 3.3. The arc length of the graph of a smooth function x(t) on some(t0, t1) into M = Rd is a functional. It takes as input a path γ on Rd and returns

Φ(γ) =

∫ t1

t0

√1 + x2 dt.

Intuitively, we expect the path of minimum length between two points to bethe line segment connecting those two points, and mathematically we see thatΦ(γ) ≥

∫ t1t0

√1 dt = t1 − t0 with equality if and only if x = 0. The more

systematic machinery we develop in this section should also return this answer.

We want to obtain a differential equation which a functional’s extrema mustsatisfy, arising from the vanishing of some sort of first derivative. For moreconcrete examples we do not need to assume arbitrary smoothness (cf. Exer-cise 3.2) nor even assume the existence of a minimizer (cf. Exercise 3.6), but herewe will assume existence (as before) and remain within the context of smoothmanifolds. A (fixed-endpoint) variation of a path γ ∈ Ω is a smooth mapH(s, t) = Hs(t) : (−ε, ε) × [t0, t1] → M for some ε > 0 such that H(0, ·) = γ,Hs = H(s, ·) ∈ Ω for all s ∈ (−ε, ε), and H(s, ti) = ai for i = 1, 2 for alls ∈∈ (−ε, ε). A variation can be thought of a path on Ω through γ at s = 0.

Just like with functions on Euclidean space, we would like to say that afunctional Φ is differentiable if

Φ(Hs) = Φ(γ)+sδΦ

(∂Hs

∂s

)+o(s) ⇐⇒ lim

s→0

Φ(Hs)− Φ(γ)

s= δΦ

(∂Hs

∂s

)for all variations H, where δΦ should be linear in ∂Hs/∂s. When this is thecase, δΦ is of course the derivative or first variation of Φ, and γ is a criticalpoint for Φ if δΦ ≡ 0. However, we first need to figure out what δΦ and ∂Hs/∂sare.

The derivative ∂Hs/∂s should be tangent to Ω at the path Hs. A tangentvector W to Ω at a path γ ∈ Ω is a function which associates to each t ∈ [t0, t1]a tangent vector Wt ∈ Tγ(t)M , and is smooth in the sense that for every smoothfunction f : M → R the function t 7→ df(Wt) ∈ R is smooth. The tangentspace TγΩ is the space of all tangent vectors W at γ such that Wt0 = 0 = Wt1 .Now we can define the tangent vector ∂Hs/∂s(0) to the variation Hs at s = 0to be

∂H

∂s(0, t) ∈ Tγ(t)M for t ∈ [t0, t1].

Since∂H

∂s(0, t0) = 0,

∂H

∂s(0, t1) = 0

3.1. PRINCIPLE OF LEAST ACTION 49

by the fixed endpoint requirement, then we indeed have ∂Hs/∂s(0) ∈ TγΩ. Nowwe can set

δΦ

(∂Hs

∂s

)=

d

dsΦ Hs

∣∣∣∣s=0

.

As we would hope, δΦ is well-defined and linear.For the rest of this section, we will restrict our attention to the action func-

tional S defined at the beginning of the section. If we take L (x,y, t) =√

1 + y2,we recover the functional of Example 3.3. The reason we insisted that L bedifferentiable is so that the action is differentiable.

Proposition 3.4 (Euler–Lagrange equations). The action functional is differ-entiable with derivative

δS

(∂Hs

∂s

)=

∫ t1

t0

(∂L

∂q− d

dt

∂L

∂q

)· ∂Hs

∂sdt. (3.2)

Consequently, a path q(t) is a critical point for the action (i.e. δS ≡ 0) if andonly if q(t) solves the differential equations

d

dt

[∂L

∂q

(q(t), q(t), t

)]− ∂L

∂q

(q(t), q(t), t

)= 0. (3.3)

The n-many second-order differential equations (3.3) are called the Euler–Lagrange equations for the functional S, or simply the Lagrange equations whenapplied to a mechanical system. Note that eq. (3.3) must hold for any choice ofcoordinates q, where ∂L /∂q is the gradient of L (q, v, t) in the q variables. Thederivative ∂L /∂q represents the derivative of the Lagrangian L (q, v, t) in thevelocity variables v, and should really be noted as (∂L /∂v)(q, q, t). Althougheq. (3.3) is often shortened to

d

dt

∂L

∂q− ∂L

∂q= 0,

we are meant to plug in q and q before taking the total time derivative d/dt,which will for example turn q into q.

Proof. Fix a set of coordinates q on M , let q(t) be a path, and let H be avariation of q(t). For the rest of this proof we will pull everything back to thedomain of q, the coordinate patch lying within Rn. Let x(t) denote the path inRn, L (x, x, t) denote the Lagrangian in Rn, and x(t) +h(t) the variation Hs(t)in Rn (we are suppressing the s-dependence of h). Note that the fixed endpointcondition requires that h(t0) = 0 = h(t1).

Now that we are in Euclidean space, we can proceed with the usual varia-tional argument. Since L is differentiable, we may Taylor expand to get

L (x + h, x + h, t) = L (x, x, t) +∂L

∂x· h +

∂L

∂x· h +O(h2).


Therefore,

S(γ + h)− S(γ) =

∫ t1

t0

[L (x + h, x + h, t)−L (x, x, t)

]dt

=

∫ t1

t0

(∂L

∂x· h +

∂L

∂x· h)

dt+O(h2).

We are able to pull the termO(h2) outside of the integral since L is continuouslydifferentiable. The integral in the rightmost expression will be the derivativeδS. Integrating by parts yields∫ t1

t0

∂L

∂x· h dt = −

∫ t1

t0

d

dt

(∂L

∂x

)· h dt+

(∂L

∂x· h)∣∣∣∣t1

t0

.

Because h(t0) = 0 = h(t1), we know that the boundary term must vanish:(∂L

∂x· h)∣∣∣∣t1

t0

= 0,

and plugging this back into δS produces eq. (3.2) as desired.

It is clear from the formula that if eq. (3.3) is satisfied then the functionalderivative (3.2) δS must be identically zero. Conversely, assuming that δS = 0for any variation, we know that∫ t1

t0

(∂L

∂x− d

dt

∂L

∂x

)· h dt = 0 (3.4)

for all smooth h : [t0, t1] → Rn with h(t0) = 0 = h(t1). We can conclude thatthe integrand without h identically vanishes (and so eq. (3.3) holds) using thefollowing basic analysis lemma.

Lemma 3.5. If a continuous function f : [t0, t1]→ R satisfies∫ t1t0f(t)h(t) dt =

0 for all smooth functions h : [t0, t1]→ R with h(t0) = 0 = h(t1), then f(t) = 0for all t0 ≤ t ≤ t1.

Remark. In the proof of Proposition 3.4, we only need that the coordinates arespanning in order to have enough directions h(t) to conclude that the Euler–Lagrange equation holds. Therefore, there is no issue in introducing more thann coordinates as long as they span the same space.

Corollary 3.6. Transforming the Lagrangian by a total time derivative:

L (q, q, t) 7→ L ′ = L (q, q, t) +d

dt[f(q, t)]

leaves the Euler–Lagrange equations (3.3) unchanged.


Proof. The new action is given by

S′ =

∫ t1

t0

L(q, q, t) +

∫ t1

t0

df

dtdt = S + f(q(2), t1)− f(q(1), t0)

and the addition of a constant to S does not affect the principle of least action.

Example 3.7. We will apply the Euler–Lagrange equations (3.3) to determinethe extrema of the arc length functional of Example 3.3. Using Cartesian coor-dinates x on Rd, the Euler–Lagrange equations are

0 =d

dt

∂L

∂x− ∂L

∂x=

d

dt

(x√

1 + x2

)− 0.

This meansx√

1 + x2= a

for some constant a ∈ Rn. Squaring and rearranging yields

x = b

for a new constant b ∈ Rn. Finally, integrating yields

x(t) = bt+ c

for b, c ∈ Rd. That is, the extrema of the arc length functional are straight lines.If we had instead chosen, say, polar coordinates, the Euler–Lagrange equationswould be different but the solutions would still describe straight lines.

3.2 Conservative systems

In the previous sections we saw how the principle of least action yields then = dimM second-order Euler–Lagrange equations

d

dt

∂L

∂q− ∂L

∂q= 0, (3.5)

which when applied to mechanical systems are simply referred to as Lagrange’sequations. In this section we will see how eq. (3.5) is a formulation of Newton’sequations, and so Newton’s principle of determinacy is contained in Hamilton’smore concise principle of least action. For this reason, many graduate physicstexts opt to begin with the principle of least action and view Newton’s equationsas a consequence.

First we will see how to apply the Euler–Lagrange equations in order toobtain the equations of motion. Let xi ∈ Rd be Cartesian coordinates on


Euclidean space. For a conservative system of N particles the Lagrangian isgiven by

L (x, x, t) = T (x)− U(x), (3.6)

where T and U are the kinetic and potential energies. The principle of leastaction implies that the motion x(t) = (x1(t), . . . ,xN (t)) of the system satisfiesthe Euler–Lagrange equations (3.5). For this Lagrangian we have

∂L

∂x=∂T

∂xi=

∂

∂x

(N∑i=1

12mix

2i

)= p,

∂L

∂x= −∂U

∂x= −∇U = F,

Therefore the Euler–Lagrange equations are Newton’s equations (1.2). Theadvantage now is that we are now longer restricted to Euclidean space.

Example 3.8. Consider a pendulum consisting of a mass m attached to the endof a rigid massless rod of length ` with the other end fixed, allowed to rotatein a vertical plane subject to a constant downward gravitational accelerationg. Let θ denote the angle from the vertical directly below the pivot, whichentirely describes the system. The configuration space is the circle S1, and theLagrangian is a (time-independent) function defined on (θ, θ) ∈ TS1 = S1 × R.The kinetic energy is

T = 12mv

2 = 12m`

2θ2,

and since the force acting on the mass is F = ma = −mgl sin θ (cf. Exercise 1.1)then the potential energy is

U = −mg` cos θ.

(We picked our integration constant so that U = 0 when the mass is at theheight of the pivot θ = π/2.) The Lagrangian is thus

L = T − U = 12m`

2θ2 +mg` cos θ,

and so the Euler–Lagrange equation is

0 =d

dt

∂L

∂θ− ∂L

∂θ= m`2θ +mg` sin θ =⇒ θ +

g

`sin θ = 0,

which agrees with what we found with Newton’s equation in Exercise 1.1.

In fact, the reason for the form of the Lagrangian (3.6) stems from Galileo’sprinciple of relativity, presented in section 1.1, now applied to the Lagrangian Lrather than the force. Consider a single particle in space whose potion is denotedby the Cartesian coordinates x ∈ Rd. In an inertial frame, the LagrangianL (x, x, t) of this particle cannot be explicitly dependent on position or timeby homogeneity: L ≡ L (x). Also L cannot be dependent on the direction of


v = x either since space is isotropic: L ≡ L (|v|2) (since L must be smooth).Since ∂L /∂x = 0, then the Euler–Lagrange equation for this particle is

d

dt

∂L

∂x= 0. (3.7)

The quantity ∂L /∂x is therefore constant in time, and since L ≡ L (v) then vmust be constant. This is the Lagrangian mechanics demonstration of Newton’sfirst law: in an inertial frame, any free motion takes place with constant velocity(both magnitude and direction).

Now we know that L ≡ L (|v|2), but we would like to conclude that for ourfree particle we must have L = T . Consider an inertial frame K moving withinfinitesimal velocity ε relative to an inertial frame K ′. Then v′ = v + ε and|v′|2 = |v|2 + 2v · ε+ |ε|2, and so

L (|v′|2) = L (|v|2) + 2∂L

∂|v|2v · ε+O

(|ε|2)

(3.8)

to first order in ε. However, since both frames are inertial the two Lagrangiansshould be equivalent for all ε, which means the rightmost term of eq. (3.8) mustbe a total time derivative (cf. Corollary 3.6). Since the Lagrangian can only bea function of |v|2, then this term could only be a total time derivative if it islinear in v. We therefore have that ∂L /∂|v|2 is independent of v, and henceL is proportional to |v|2. This allows us to write

L = 12m|v|

2 (3.9)

where m is the particle’s mass; we would experimentally observe that a particle’sacceleration is inversely proportional to its mass as in section 1.1. Note thatm cannot be negative since looking at the action (3.1) we see that Hamilton’sprinciple would yield maxima instead of minima. We did not use the thirdtype of Galilean transformations, but we can check that the expression (3.9) isinvariant with respect to rectilinear motion v′ = v + v0:

L ′(v′) = 12m|v

′|2 = 12m|v + v0|2 = 1

2m|v|2 +mv · v0 + 1

2mv20

= L (v) +d

dt

(mx · v0 + 1

2m|v0|2t),

and so the Euler–Lagrange equations are unaffected by Corollary 3.6.For a system of free noninteracting particles, the Lagrangian for each indi-

vidual particle cannot be dependent on the coordinates of any other, and so Lmust be additive:

L =

N∑i=1

12mi|vi|2 = T, (3.10)

which is the kinetic energy as claimed. Note that (at least symbolically)

|x|2 =

(ds

dt

)2

=(ds)2

(dt)2,


and so we only need to know the line element ds or metric ds2 in order totransform the kinetic energy into other coordinate systems. If we wish to expressthe Cartesian coordinates xi = fi(q1, . . . , qn) in terms of generalized coordinatesq = (q1, . . . , qn), then

T = 12

n∑i,j=1

aij(q)qiqj (3.11)

where aij are functions of the coordinates only. That is, the kinetic energy T ingeneralized coordinates is still a quadratic function of velocities, but may alsodepend on the other coordinates. Mathematically, a conservative Lagrangiansystem is determined by a Riemann manifold, with a metric that determinesthe kinetic energy, and a potential function.

For a general system of particles which may interact we have to insert afunction to the Lagrangian, which for a conservative system is the potentialenergy:

L (q, q) = T (q, q)− U(q). (3.12)

Now that we are no longer in Euclidean space, we require that for a conservativesystem on a manifold M the force (when viewed as a 1-form) be exact F =−∇U (and not merely closed). Now the time reversibility (cf. section 1.5) iseasily seen as the time independence of the Lagrangian—time reversal t 7→ −tin the quadratic kinetic energy (3.11) preserves each product qiqj . In otherwords, when the Lagrangian is time-independent the total energy is conserved(cf. section 1.5).

3.3 Equivalence to Newton’s equations

Now we will see how to obtain the Euler–Lagrange equations eq. (3.5) from New-tonian mechanics, so that the principle of least action is equivalent to Newton’sprinciple of determinacy and not stronger.

A mechanical system with a configuration manifold M can always be em-bedded —and in experiment is naturally—embedded in some Euclidean spacex ∈ RN . Within M , the motion of the system is dictated by some known forceF. The effect of constraining the motion to the manifold M can be thought ofas a force N orthogonal to M , called the constraint force. Newton’s equationsfor this system is

mixi = Fi + Ni. (3.13)

Rearranging, we see that mixi − Fi = Ni is orthogonal to M , and so

(mixi − Fi) · ξi = 0 (3.14)

for all vectors ξ = (ξ1, . . . , ξN ) tangent to M . This is Newton’s equation inthe tangent plane to the surface M . Summing over all particles, we get thedAlembert-Lagrange principle,

N∑i=1

(mixi − Fi) · ξi = 0, (3.15)

3.3. EQUIVALENCE TO NEWTON’S EQUATIONS 55

which dictates the motion of a system with constraints. Note that for a freesystem M = Rn we may take any vector ξ ∈ Rn, and so we recover Newton’sequations.

Let q = (q1, . . . , qn) be local coordinates on M . Then by the chain rule

xi =

n∑j=1

∂xi∂qj

qj ,

and so we may write the kinetic energy

T (q, q) =

N∑i=1

12mi|xi|2 =

n∑i,j=1

aij(q)qiqj

as a positive definite quadratic form on M .

Expressing the force in terms of the coordinates qj , the covectors Qj definedby the 1-form equation

N∑i=1

Fi dxi =

n∑j=1

Qj dqj , (3.16)

or equivalently

Qj =

N∑i=1

Fi∂xi∂qj

, (3.17)

are called the generalized forces. They dictate the evolution of the kinetic energyvia the following expression.

Proposition 3.9. The Newtonian motion q(t) of the system satisfies the equa-tion

d

dt

∂T

∂q− ∂T

∂q= Q. (3.18)

We will prove this in the next section.

Now that we are no longer in Euclidean space, we require that for a conser-vative system the 1-form Qdq is not merely closed but also exact and may bewritten as −dU for a potential energy U(q). For such a system we are naturallyled to define L = T − U so that

d

dt

∂L

∂q− ∂L

∂q= 0, (3.19)

and in section 3.1 we saw that this implies q(t) is a critical point for the actionfunctional. In this way, the principle of least action can be seen as a consequenceof Newton’s equations.


3.4 Nonconservative systems

Thus far we have considered Lagrangians of the form L = T−U , and the result-ing system is automatically conservative since it is derived from the potential.In this section we will extend Lagrangian mechanics to include nonconservativesystems. In the spirit of Proposition 3.9, we will see that the correct choice is

L = T +W, (3.20)

where W is the total work of the (generalized) forces (3.17):

W =

n∑i=1

Qiqi. (3.21)

As in the set up for the proof of Proposition 3.4, we can work in Euclideanspace. Consider a system whose motion is described by the n coordinates qj(t) ∈Rd from time t1 to t2, and write a fixed-endpoint variation of the motion asqj(t) + δqj(t) with δqj(t1) = δqj(t2) = 0. Repeating the integration by partsprocedure from Proposition 3.4, we get

δT =

∫ t2

t1

∑j

(∂T

∂qjδqj +

∂T

∂qjδqj

)dt

=

∑j

∂T

∂qjδqj

t2t1

+

∫ t2

t1

∑j

(∂T

∂qj− d

dt

∂T

∂qj

)δqj dt

=

∫ t2

t1

∑j

(∂T

∂qj− d

dt

∂T

∂qj

)δqj dt,

where in the last equality we noted that δqj(t1) = δqj(t2) = 0 for an endpoint-fixed variation.

The quantity δW represents the work done by the forces on the system intaking the actual to the varied path qj(t) + δqj(t). Adding it in, the principleof least action for the new Lagrangian (3.20) yields

0 = δ

∫ t2

t1

L(q, q, t) dt =

∫ t2

t1

(δT + δW ) dt

=

∫ t2

t1

∑j

[∂T

∂qj− d

dt

∂T

∂qj

]δqj +

∑j

Qjδqj

dt

=

∫ t2

t1

∑j

(∂T

∂qj− d

dt

∂T

∂qj+Qj

)δqj dt

Since this must be true for all δqj and each generalized coordinate qj can bevaried independently from the others, then we conclude

d

dt

∂T

∂qj− ∂T

∂qj= Qj , j = 1, . . . , d. (3.22)

3.5. MOMENTUM AND CONSERVATION 57

This is the extension of the Lagrange’s equations for nonconservative forces.Nowhere did we assume that the generalized forces are not derivable from

potentials, and thus as a sanity check that equations (3.22) reduce to the familiarformulation of Lagrange’s equations when the forces are conservative. In thiscase, we can use the same integration by parts procedure in reverse to obtain∫ t2

t1

δW dt =

∫ t2

t1

∑j

Qjδqj dt = −∫ t2

t1

∑j

[∂U

∂qj− d

dt

∂U

∂qj

]δqj dt

=

∑j

∂U

∂qjδqj

t2t1

−∫ t2

t1

∑j

(∂U

∂qjδqj +

∂U

∂qjδqj

)dt

= −∫ t2

t1

∑j

(∂U

∂qjδqj +

∂U

∂qjδqj

)dt = −

∫ t2

t1

δU dt.

In other words, δW = −δU , and so the new Lagrangian (3.20) generates thesame motion as the conservative Lagrangian L = T − U .

Even when the forces are nonconservative, all we needed for the previousparagraph is the ability to write the jth component of the generalized force as

Qj = −∂U∂qj

+d

dt

∂U

∂qj. (3.23)

In this case, we recover the familiar form of Lagrange’s equations (3.5) with noright-hand side, but now with a velocity-dependent potential.

So U = T −L is sometimes the potential energy even for nonconservativesystems, and the potential energy is generally time-dependent. A common caseis that the system A with coordinates qA is not closed, but it moves in anexternal field due to a system B with coordinates qB(t) independent of qA andthe entire system A+B is closed. This system has a Lagrangian of the form

L = TA(qA, qA)− U(qA, qB(t)). (3.24)

We may ignore TB since it depends only on time and is thus a full time deriva-tive. Equation (1.14) is a Lagrangian of the usual type, but with U being pos-sibly time-dependent. If system A is a single particle, then the Euler–Lagrangeequations yield

mq = −∂U∂q

(q, t) = F (q, t). (3.25)

As an example, if F ≡ F (t) is uniform (i.e. independent of position) thenU = F · x in Euclidean space.

3.5 Momentum and conservation

Mathematically there are two special cases when we can turn the second orderEuler–Lagrange equations into first order equations, which physically corre-spond to conservation laws.


For a Lagrangian system, the (generalized) momentum of a particle is definedto be

pi =∂L

∂qi. (3.26)

If qi = xi is a Cartesian coordinate, then the Lagrangian has a term 12mix

2i

and so pi = mixi is the linear momentum along the xi-axis. If qi = φi is theazimuthal angular coordinate in R3, then the Lagrangian has a term 1

2mir2i φ

2i

and so pi = mir2i φi is the angular momentum about the axis perpendicular to

the φ-plane.We can similarly define the (generalized) force as

Fi =∂L

∂qi=

d

dt

∂L

∂qi=

dpidt, (3.27)

so that Newton’s equation is still satisfied symbolically. This definition agreeswith the 1-form definition (3.16) for Qi, and so both forces Fi and momenta pishould be interpreted as covectors; conversely, the velocities qi are by definitiontangent vectors. This fact is built into the kinetic energy:

Ti = 12vi · pi = 1

2 〈vi,pi〉, (3.28)

and so it is indeed natural and not merely a convention.The Euler–Lagrange equations immediately imply that that when the La-

grangian is independent of a generalized coordinate qi then the force Fi ≡ 0vanishes and the momentum pi is conserved:

dpidt

=d

dt

∂L

∂qi=∂L

∂qi= 0. (3.29)

Mathematically, we can record this fact as saying:

Corollary 3.10. (Conservation of momentum) If the Lagrangian is indepen-dent of a variable qi, we can replace the second order Euler–Lagrange equationsby the first order equation

∂L

∂qi= constant.

Any such coordinate qi is called cyclic. This includes the conservation laws wesaw in sections 1.6 and 1.7. One of the strengths of Lagrangian mechanics is thatwhen there is a cyclic coordinate then there is one less Euler–Lagrange equa-tion to solve, which we already saw when conservation of angular momentumreduced a three-dimensional system to a two-dimensional system in section 2.2.Geometrically, this requires the trajectories q(t) in configuration manifold M tolie in the level sets of pi.

In section 3.2 we mentioned that when the cyclic coordinate is time then theconserved quantity is the total energy. We can extend our definition of the total

3.6. NOETHER’S THEOREM 59

energy (or Hamiltonian H) to any system as

E =

d∑i=1

qipi −L , (3.30)

since for a conservative system on Euclidean space we have

E =∑i

vi∂

∂vi

∑j

12mjv

2j − U(x)

−∑i

[12miv

2i − U(x)

]=∑i

miv2i −

∑i

12miv

2i + U = T + U.

More generally, we can show:

Proposition 3.11 (Conservation of energy). The solution q(t) to the Euler–Lagrange equation must satisfy the first order differential equation

H(q) := q · ∂L

∂q(q, q, t)−L (q, q, t) =

∫ t

t0

∂L

∂t(q(s), q(s), s) ds+ constant.

In particular, if the Lagrangian L is independent of t, then the antiderivativeon the right-hand side vanishes and the Hamiltonian H(q) is constant.

In the context of variational calculus this is sometimes called the second Euler–Lagrange equation. Notice again the natural paring of the velocity q with themomentum ∂L /∂q.

Proof. Using the chain rule and the Euler–Lagrange equations, we have

dL

dt=∂L

∂t+∂L

∂qi· qi +

∂L

∂qi· qi

=∂L

∂t+

(d

dt

∂L

∂qi

)· qi +

∂L

∂qi· qi

=∂L

∂t+

d

dt

(qi ·

∂L

∂qi

).

Integrating in time gives the desired equation for the Hamiltonian.

3.6 Noether’s theorem

The conservation of momentum (Corollary 3.10) demonstrates that a systemthat is continuously symmetric along the coordinate qi possesses the corre-sponding conserved quantity ∂L /∂qi. We will now demonstrate that everyone-parameter group of symmetries which preserves a Lagrangian system has aconserved quantity.


Consider a Lagrangian system consisting of a smooth n-dimensional manifoldM with a time-independent Lagrangian L (q, v, t) : TM → R. We say that adiffeomorphism h : M →M is a symmetry of the system if

L (h∗(q, v)) = L (q, v) for all (q, v) ∈ TM. (3.31)

Here, h∗(q, v) = dh(q, v) is the pushforward or differential of the point q ∈ Mand tangent vector v ∈ TqM , and is equal to the point h(q) ∈ M with thetangent vector dhq(v) ∈ Th(q)M .

Proposition 3.12 (Noether’s theorem). If the time-independent Lagrangiansystem (M,L ) has a differentiable one-parameter group of diffeomorphisms hs :M → M , s ∈ R, then the Lagrangian motion of the system has a conservedquantity (or “integral”) I : TM → R. In local coordinates,

I(q, v) =∂L

∂q(q, v) · d

dshs(q)

∣∣∣∣s=0

. (3.32)

Remark. The quantity I is independent of the choice of local coordinates q. Imeasures the rate of change of L (q, v) as v is varied in TqM in the direction ofthe tangent vector (d/ds)|s=0h

s(q), which is why the formula (3.32) looks likethe chain rule for d/ds|s=0L ((hs)∗(q, v)), and this does not intrinsically involvea choice of coordinates.

Proof. As in the set up for the proof of the Euler–Lagrange equations (Propo-sition 3.4), we may cover Md with coordinate patches and use a partition ofunity to reduce the statement to Euclidean space Rn. Let φ(t) := q(t) : R→Mdenote a solution to Lagrange’s equations. Since hs is a symmetry for eachs then hs φ will also be a solution, because (hs φ)(t) = (hs)∗(φ(t)) is thepushforward of the point φ(t), d

dt (hs φ)(t) = d(hs)φ (t)(φ(t)) = (hs)∗(φ(t))

is the pushforward of the tangent vector φ(t), and so by definition (3.31) theLagrangian evaluated at (hs φ)(t) is the same as the Lagrangian evaluated atφ(t).

Consider the map Φ : Rs ×Rt → Rn given by Φ(s, t) = (hs φ)(t). Since allof the symmetries hs preserve the Lagrangian L then

0 =∂L

∂s(Φ, Φ) =

∂L

∂q· ∂Φ

∂s+∂L

∂q· ∂Φ

∂s, (3.33)

where everything on the right-hand side is evaluated at (Φ(s, t), Φ(s, t)) ∈ TRn.As we have already argued, for fixed s the map Φ(s, ·) : R → Rn satisfiesLagrange’s equation:

∂

∂t

[∂L

∂q

(Φ(s, t), Φ(s, t)

)]=∂L

∂q

(Φ(s, t), Φ(s, t)

)

3.7. EXERCISES 61

for each s ∈ R. Plugging this into the right-hand side of eq. (3.33), we see that

0 =∂

∂t

[∂L

∂q

(Φ, Φ

)]· ∂Φ

∂s

(Φ, Φ

)+∂L

∂q

(Φ, Φ

)· ∂Φ

∂s

(Φ, Φ

)=

d

dt

[∂L

∂q· ∂Φ

∂s

] (Φ, Φ

)=

dI

dt

(Φ, Φ

)by the chain rule and using that ∂Φ/∂s = ∂2Φ/∂s∂t = d

dt (∂Φ/∂s).

Example 3.13 (Translational symmetry). Consider the conservativeN -particleLagrangian

L (x,v) =

N∑i=1

12mi|vi|2 − U(x), (3.34)

where X = (x1, . . . ,xN ), xi ∈ Rd and similarly for v. If the potential energyis invariant under translations along the first coordinate axis e1 ∈ Rn, then thesystem is symmetric with respect to the N translations

hsj : (Rd)N → (Rd)N , hsj(x1, . . . ,xN ) = (x1, . . . ,xj + se1, . . . ,xN )

for j = 1, . . . , N . Noether’s theorem yields N conserved quantities

Ij(x,v) =∂L

∂x· (0, . . . , 0, e1, 0, . . . , 0) = mjvj1,

which we recognize as the first component of the jth particle’s momentum pj =mjvj (cf. Corollary 1.21).

Example 3.14 (Rotational symmetry). Now suppose d = 3, and that theconservative N -particle Lagrangian (3.34) is invariant under rotations aboutthe first coordinate axis e1 ∈ R3. Then the system is symmetric with respectto the N rotations

hsj(x1, . . . ,xN ) = (x1, . . . , cos(s)xj + sin(s)e1 × xj + (1− cos s)xj1e1, . . . ,xN )

for j = 1, . . . , N . Noether’s theorem returns the N conserved quantities

Ij(x,v) =∂L

∂x· (0, . . . , 0, e1 × xj , 0, . . . , 0) = pj · (e1 × xj) = (xj × pj) · e1,

which we recognize as the first component of the jth particle’s angular momen-tum Lj = xj × pj (cf. Corollary 1.24).

3.7 Exercises

3.1 (Functional derivative is well-defined). Show that if Φ is a differentiablefunctional, then its differential is linear and is independent of the choice invariation of γ.


3.2. Prove the following version of Proposition 3.4 on Euclidean space withweaker assumptions: If L ∈ C1(Rn × Rn × [t0, t1];R) and q ∈ C1([t0, t1];Rd)is a (fixed-endpoint) critical point for the action functional defined in eq. (3.1),then ∂L /∂q(q, q, t) is a C1 function on [t0, t1] and q solves the Euler–Lagrangeequations (3.3).

3.3. (Geodesics on the sphere [Tro96, Ch. 1]) In Example 3.7, we saw thatthe geodesics—paths of shortest length between two given points—in Rd arestraight lines. We will repeat this procedure for the sphere S2.

(a) Using coordinates (φ, θ), we can parameterize a path x(t) in S2 ⊂ R3 as

x(t) = (cosφ(t) sin θ(t), sinφ(t) sin θ(t), cos θ(t)), t ∈ [0, 1]

(φ is the azimuth and θ is the zenith angle). Find the formula for the arclength functional Φ[x(t)].

(b) After rotating the sphere we may assume that θ(0) = 0, θ(1) = θ1, andφ(1) = 0. Find a simple lower bound for Φ[x(t)] using the φ′(t) term. Byconsidering when equality occurs, conclude that the geodesic connectingthe north pole x(0) to the point x(1) is the shorter arc of the great circle(an equator, or circle of maximum circumference on the sphere) connectingthe two points. This is another example where we can determine that thecritical path for the functional is a minimum.

3.4. (Brachistochrone [Tro96, Ch. 6]) The brachistochrone between two pointsin the plane is the curve on which a frictionless bead would traverse the quickestsubject to a downward gravitational acceleration. In 1696, Johann Bernoullichallenged mathematicians to find the shape of the brachistochrone, and it washis brother Jakob Bernoulli who provided a solution which was later refined intothe calculus of variations.

(a) After translating, we may assume that the initial point is the origin (0, 0)and the second point is given by some (x1, y1) with x1 > 0 and y1 < 0.Explain why it is reasonable to assume that the brachistochrone is thegraph of a function y(x), x ∈ [0, x1] as opposed to a general parametriccurve. Show that the time is takes the bead to traverse this curve is

Φ[y(x)] =

∫ x1

0

√1 + y′(x)2

v(x)dx

where v(x) is the bead’s speed.

(b) With constant downward acceleration g, show that v(x) =√

2gy(x).

(c) Using the conservation of total energy (3.30), find the first order differen-tial equation √

y

c2 − yy′ = 1.

3.7. EXERCISES 63

(d) Introducing a new dependent variable θ(x) so that

y = c2 sin2 θ

2=c2

2(1− cos θ), 0 ≤ θ < 2π,

show thatc2

2(1− cos θ) θ′ = 1.

(e) By integrating the two equations of the previous part, obtain the para-metric equations

x = c2(θ − sin θ) + c1,y = c2(1− cos θ),

0 ≤ θ ≤ θ1.

In order for x(0) = 0 = y(0), we must have c1 = 0. That is, the brachis-tochrone is a cycloid; these equations describe the path traced by a fixedpoint on a circle of radius c2 as it rolls along the x-axis in the lowerhalf-plane.

3.5. (Catenary [Tro96, Ch. 3]) The catenary is the shape taken by a cable hungby both ends under its own weight.

(a) Consider a cable of length L hung between two equal height supportsseparated by a distance H < L. Let y denote the vertical coordinatewith y = 0 at the point where the cable is fastened, and let y(s) denotethe shape of the cable where s is the arc length along the cable, so thaty(0) = 0 = y(L). If the weight per unit length is a constant W , explainwhy the cable shape y(s) minimizes the mass integral

F [y(s)] = W

∫ L

0

y(s) ds.

If we had chosen the horizontal coordinate x as the independent variable,the resulting functional is not convex.

(b) The functional with Lagrangian (y, y′, t) 7→ Wy is only convex and maynot have a unique minimizer. Show that in order to span the supports,the cable must satisfy the constraint∫ L

0

√1− y′(s)2 ds = H.

(Note that |y′(s)| = 1 requires x′(s) = 0 which would produce a cusp.)The new Lagrangian

L (y, y′, t) = Wy − λ√

1− y′2

is strictly convex for λ > 0.


(c) Apply the Euler–Lagrange equations and integrate once to obtain the firstorder differential equation

λy′(s)√1− y′2(s)

= s+ c

for the catenary, where c is a constant.

(d) Since we expect y(s) to be unique by convexity, then we may place addi-tional assumptions. Assume that the cable shape is symmetric about themidpoint ` = L/2, and conclude that y′(`) = 0 and c = −`. Solve for y′(s)and integrate on [0, s] to obtain

y(s) =√λ2 + (l − s)2 −

√λ2 + l2 on [0, `].

(e) Show that the constraint of part (b) yields∫ `

0

λ√λ2 + (`− s)2

ds =H

2.

Make the substitution (` − s) = λ sinh θ, evaluate the integral, and con-clude that

λ =`

sinh(H/L).

(f) Show that x(s) is given by

x(s) =

∫ s

0

√1− y′(t)2 dt =

H

2− λ sinh−1

(`− sλ

).

Together with y(s) from (d) we have a parametric equation for the cate-nary. Eliminate the variable s and conclude

y(x) = λ cosh

(x−H/2

λ

)−√λ2 + `2

for x ∈ [0, H]. That is, the catenary equation is hyperbolic cosine.

3.6. (Variational method for PDE [Eva10, Ch. 8]) In this exercise we will provethe existence of a solution to the elliptic divergence-form PDE

−∇ · (A(x)∇u(x)) = 0 for x ∈ Ω, u(x) = 0 for x ∈ ∂Ω,

for an open set Ω ⊂ Rn. Here, A(x) = (aij(x)) is a symmetric n × n matrixwith aij ∈ H2(Ω) (the Sobolev space), and we also assume that A is uniformlyelliptic:

λI ≤ A(x) ≤ ΛI for x ∈ Ω

(in the sense of positive definite matrices).

3.7. EXERCISES 65

(a) For u ∈ H10 (Ω) (the closure of C∞c (Ω) in H1(Ω)), show that the energy

functional

E[u] = 12

∫Ω

∇u(x) ·A(x)∇u(x) dx

is finite. Show that for φ ∈ C∞c (Ω) the first variation at u is

limε→0

E[u+ εφ]− E[u]

ε=

∫Ω

∇φ ·A∇u

Consequently, if u is a minimum of E, then the above expression vanishesfor all φ ∈ C∞c (Ω); such a function u ∈ H1

0 (Ω) is called a weak solution ofthe PDE and boundary condition, since one formal integration by partswould produce the PDE but we do not assume u is twice differentiable.

(b) Let uj be a sequence in H10 (Ω) such that

E[uj ]→ infH1

0 (Ω)E[u] as j →∞.

Using Poincare’s inequality, show that E[u] is bounded below and henceour sequence is bounded. Conclude that there exists a weakly convergentsubsequence ujk u in H1(Ω) using the Riesz representation theorem.

(c) Show that E is (sequentially) weakly lower semicontinuous:

E[u] ≤ lim infk→∞

E[ujk ].

Conclude that u ∈ H10 (Ω) is a minimum for E, and hence is a weak solution

to the PDE by part (a).

3.7 (Two pendulums connected by a spring). Consider two pendulums of unitlength and unit mass in a constant gravitational field g. Suppose they areconnected by a massless spring with spring constant k whose resting length isthe same as their distance of separation.

(a) Let θ1 and θ2 be the angles the pendulums make with the downwardvertical. Find the Lagrangian for the system for small angles, so thatsin θ ≈ θ.

(b) Define new variables

q1 =θ1 + θ2√

2, q2 =

θ1 − θ2√2

,

and show that Lagrangian separates into two harmonic oscillators:

L = 12 (q2

1 + q22) + 1

2 (ω21q

21 + ω2

2q22).

Find ω1 and ω2. When q2 = 0, we have θ1 = θ2 and so both pendulumsswing in phase with each other with frequency ω1. When q1 = 0, wehave θ2 = −θ1 and so the pendulums swing in exact opposite phase withfrequency ω2.


(c) For k 1 we will see that an exchange of energy occurs. Suppose thatat time t = 0 we have θ1 = 0 = θ2, θ2 = 0, and θ1 = v0. Using part (b),show that the motion is given by

θ1(t) =v0

2

(sin t+

1

ωsinωt

), θ2(t) =

v0

2

(sin t− 1

ωsinωt

)with ω = ω2. For k 1 we have 1/ω ≈ 1, and so

θ1(t) ≈ v0 cos εt sinω′t, θ2(t) ≈ −v0 cosω′t sin εt

for some ε and ω′. Show that to leading order as k → 0 we have

ε ≈ k

2, ω′ ≈ 1,

and so after a time T = π/2ε ≈ π/k the pendulums have switched rolesand now essentially only the second pendulum is oscillating.

3.8 (Charged particle in an electromagnetic field). The motion of a chargedparticle is an example of a nonconservative Lagrangian system. A particle withcharge q moving through the vector fields E,D,B,H on R3 with the scalarcharge density ρ and vector current density j obeys Maxwell’s equations (inGaussian units):

∇× E +1

c

∂B

∂t= 0, ∇ ·D = 4πρ,

∇×H − 1

c

∂D

∂t=

4π

cj, ∇ ·B = 0.

The force on the charge is given by the Lorentz law

F = q

[E +

1

c(v ×B)

].

(a) The fourth Maxwell equation requires that B is divergence-free, and so weintroduce a vector potential A for B so that B = ∇× A. Using the firstequation, introduce a scalar potential φ so that the electric field becomes

E = −∇φ− 1

c

∂A

∂t

and the Lorentz force is

F = q

[−∇φ− 1

c

∂A

∂t+

1

c(v × (∇×A))

].

(b) The Lorentz force is nonconservative, but if we can put it in the form (3.23)then we will have a Lagrangian for this system. The first term qφ is already

3.7. EXERCISES 67

of the desired form. Show that the x-component of the rightmost termv × (∇×A) may be rewritten as

(v × (∇×A))x =∂

∂x(v ·A)− dAx

dt+∂Ax∂t

.

By symmetry, we get the same relation for the other components with xreplaced by the respective variable.

(c) Show that the x-component of the Lorentz force can be written as

Fx = −∂U∂x

+d

dt

∂U

∂vx,

for the potential energy

U = qφ− q

cv ·A.

Symmetrically, the y- and z-components of the Lorentz force are also ofthis form if we replace x with the respective coordinates. Consequently,the Lagrangian for this system is

L = T − U = 12mv

2 − qφ+q

cv ·A

where m is the particle’s mass.

3.9 (Noether’s theorem with time dependence). Consider a time-dependentLagrangian system L : TM × R→ R.

(a) Prove Noether’s theorem for this system by applying Proposition 3.12 tothe extended configuration space M1 = M × R with Lagrangian

L1

(q, t,

dq

dτ,

dt

dτ

)= L

(q,

dq/dτ

dt/dτ, t

)dt

dτ

and the new “time” variable τ , to obtain a conserved quantity I(q, q, t) onM .

(b) Apply this to a time-independent Lagrangian L (q, v, t) ≡ L (q, v) to con-clude that the total energy is conserved (cf. Proposition 1.6).


Chapter 4

Constraints

In this chapter we focus on the theoretical formulations of constraints in La-grangian mechanics, without delving into the practical realizations of constraints;for an introduction to practical methods, see [AKN06, Sec. 1.6]. The treatmentof holonomic constraints is based on [AKN06, Ch. 1], and [Gol51, Ch. 1–2] fornon-holonomic constraints.

4.1 The d’Alembert–Lagrange principle

A holonomic constraint placed on a Lagrangian system requires that the sys-tem’s motion is confined to a submanifold S of the phase space TM that canbe locally expressed in the form

f1(q, q, t) = · · · = fk(q, q, t) = 0. (4.1)

A constraint that is not holonomic is called a non-holonomic constraint.

Example 4.1. Rigid-body motion in Euclidean space, in which the distancesbetween all particles are fixed, is a holonomic constraint. This constraint maybe expressed as

(xi − xj)2 − c2ij = 0 (4.2)

where cij are the inter-particle distances. Conversely, the jth particle of a gasinside of a rigid spherical container of radius a centered at the origin must obeythe equation

r2j − a2 ≤ 0,

which is non-holonomic. These two examples happen to be time-independent,but this need not be the case.

Tangent vectors ξ ∈ TqM which are tangent to the submanifold S satisfy

∂f1

∂q(q, q, t) · ξ = · · · = ∂fk

∂q(q, q, t) · ξ = 0 (4.3)

69

70 CHAPTER 4. CONSTRAINTS

are called virtual velocities of the constrained motion at the state (q, q) ∈ TMand time t. Analogous to the constraint forces of section 3.3, we can ensure thatthe motion q(t) is constrained to S by insisting that(

d

dt

∂L

∂q− ∂L

∂q

)· ξ = 0 (4.4)

for all virtual velocities ξ at the state (q(t), q(t)).

Definition 4.2 (d’Alembert–Lagrange principle). A motion of the Lagrangiansystem (M,L ) subject to the holonomic constraints (4.1) is a smooth trajectorysolving eq. (4.4) for all virtual velocities ξ.

By the definition (4.3) of the virtual velocities, this requires that the left-hand side of the Euler–Lagrange equations be within the span of the derivatives∂fi/∂q:

d

dt

∂L

∂q− ∂L

∂q=

k∑i=1

µi∂fi∂q

(4.5)

for some constants µi. These are called Lagrange’s equations with multipliers.Mathematically, we recognize the µi as playing the role of Lagrangian multipli-ers: in order for the action functional to obtain a minimum on the submanifoldS, the gradient (the left-hand side in eq. (4.5)) of the action must be orthogonalto the submanifold. Physically, if we imagine that the particle system is con-fined to the submanifold S via imaginary constraint forces, then eq. (4.4) statesthat the constraint forces do no work and the right-hand side of eq. (4.5) givesa formula for our constraint forces.

We can also observe the constraint effect via the principle of least action andthe space of paths. For a path γ ∈ Ω, let Γ be the subspace of the tangent spaceTγΩ consisting of vectors W ∈ TγΩ such that the vectors Wt are virtual veloc-ities for all t. Holder’s principle states that the effect of holonomic constraintson a Lagrangian system is given by a restricted principle of least action, and inthis way holonomic constraints effectively reduce a system’s degrees of freedom.

Corollary 4.3 (Holder’s principle). A path satisfying the d’Alembert–Lagrangecondition (4.4) is a motion of the Lagrangian system with constraints if andonly if the functional derivative δS of the action vanishes on the subspace Γ.

Proof. From Proposition 3.4 we recall that the first variation of the action S atthe path q(t) in the direction of the fixed-endpoint variation h(t) is given by

δS|q(∂Hs

∂s

)=

∫ t1

t0

(∂L

∂q− d

dt

∂L

∂q

)· ∂Hs

∂sdt.

Restricting δS to the subspace Γ is equivalent to requiring that ∂Hs∂s is a virtual

velocity. Since every virtual velocity could be attained this way, we see from theabove variation that δS|q vanishes on Γ if and only if the d’Alembert–Lagrangeprinciple (4.4) is satisfied.

4.2. GAUSS’ PRINCIPLE OF LEAST CONSTRAINT 71

Example 4.4. We can think of the pendulum of 3.8 as being a particle of massm in a vertical plane (x, y) subject to the downward gravitational force withpotential U(x, y) = mgy and subject to the holonomic constraints

x2 + y2 − `2 = 0, 2xx+ 2yy = 0

where ` is the length of the pendulum arm. Note that the second conditionis a time derivative of the first equation, but is necessary to specify the two-dimensional submanifold S in the four-dimensional tangent space TR2. If welet θ denote the angle from the downward vertical as before and r the distancefrom the pivot, then the Lagrangian is

L = 12m(r2θ2 + r2) +mgr cos θ

and we can rewrite the holonomic conditions as

f1(r, θ, r, θ) = r2 − `2 = 0, f2(r, θ, r, θ) = r = 0.

The first constraint does not place any restrictions on the virtual velocitiesξ = (ξr, ξθ) since ∂f1/∂(r, θ) = 0, but the second condition yields

0 =∂f2

∂(r, θ)· ξ = ξr.

The d’Alembert–Lagrange principle (4.4) then yields the condition

(mr −mg cos θ)ξr + (mr2θ + 2mrrθ +mgr sin θ)ξθ = 0.

Since ξr = 0 and ξθ is arbitrary, we conclude that the second parenthetical termabove vanishes. After using the constraints f1 and f2, we obtain the familiarequation of motion

θ +g

`sin θ = 0.

The holonomic constraints have effectively discarded the r equation and reducedthe degrees of freedom from two to one.

4.2 Gauss’ principle of least constraint

Unlike the principle of least action, the d’Alembert–Lagrange principle is not aminimum principle because the virtual velocities are not given by a functionalvariation. Gauss sought a rephrasing as a minimization problem, in which theactual constrained motion is the constrained motion which deviates the leastfrom the unconstrained motion.

The state of the system at time t0 is given by the position q0 = q(t0) andvelocity v0 = q(t0) at a given moment while the acceleration a(t0) = q(t0)is determined by the laws of motion and constraints. Consequently, we willconsider q0 and v0 as fixed while we vary a0. For a fixed state (q0, v0) ∈ TM


and time t0, we will refer to all paths q(t) allowed by the constraints withq(t0) = q0 and q(t0) = v0 as the conceivable motions; these lie in the constraintsubmanifold S but do not satisfy an equation of motion. A released motionsatisfies the unconstrained Euler–Lagrange equations (3.5) (and does not lie inS), and an actual motion is a conceivable motion satisfying the d’Alembert–Lagrange principle (4.4) (and hence lies in S).

By means of a partition of unity, we may reduce to a neighborhood aboutthe point q0 and work in Euclidean coordinates and use the vector notationq(t) = x(t). We will write an actual motion of the system xa(t) = xr(t) + δx(t)as a deviation from the released motion xr(t), and we assume that the initialpositions xa(t0) = xr(t0) = x0 and velocities xa(t0) = xr(t0) = v0 are fixed.Taylor expanding we have

xr(t) = x0 + v0(t− t0) + 12 xr(t0)(t− t0)2 +O

((t− t0)3

),

and if we replace xr(t) by xa(t) we only change x(t0) and hence

δx(t) = 12δx(t0)(t− t0)2 +O

((t− t0)3

). (4.6)

So far, our considerations have been independent of the constraints.In our local coordinates the d’Alembert–Lagrange principle requires(

d

dt

∂L

∂x− ∂L

∂x

)· ξ = 0 (4.7)

for all virtual displacements ξ. For simple physical systems ddt∂L∂x is the total

force mx and ∂L∂x is the system’s force −∇U without considering constraints,

and so ddt∂L∂x −

∂L∂x is the force mδx due to the constraints. We will now realize

this more generally by evaluating ddt∂L∂x −

∂L∂x at xa(t) = xr(t) + δx(t). Taylor

expanding about xr(t) we have(d

dt

∂L

∂x− ∂L

∂x

) ∣∣∣∣x=xa

=

(d

dt

∂L

∂x− ∂L

∂x

) ∣∣∣∣x=xr

+

[d

dt

(∂2L

∂x2 δx +∂2L

∂x∂xδx

)− ∂2L

∂x2δx− ∂2L

∂x∂xδv

]x=xr

+O(δx2).

The first term on the right-hand side vanishes since the actual motion xa solvesLagrange’s equations. For the second term we use eq. (4.6) and take the limitt→ t0 to obtain (

d

dt

∂L

∂x− ∂L

∂x

) ∣∣∣∣x=xat=t0

=∂2L

∂x2

∣∣∣∣x=xrt=t0

δx(t0).

Therefore, we interpret eq. (4.7) as saying that

∂2L

∂q2

∣∣∣∣q=qrt=t0

(qa(t0)− qr(t0)) (4.8)

4.2. GAUSS’ PRINCIPLE OF LEAST CONSTRAINT 73

is orthogonal to the constraint submanifold S. On the other hand, eq. (4.8) isthe gradient of the functional

Z(q(t)) = 12 (q − qr) ·M(q − qr)

∣∣t=t0

(4.9)

for the Hessian matrix M = ∂2L∂q2 of the kinetic energy quadratic form at x =

xr and t = t0 (which for conservative systems is a positive definite matrixcontaining the particle masses). The quantity (4.9) is called Gauss’ compulsionand it measures how much the motion q(t) deviates from the released motion.

The fact that the gradient (4.8) of the compulsion functional (4.9) for theactual motion qa is orthogonal to the submanifold S is enough to conclude thatthe actual motion qa is a critical point for the compulsion functional. This is ageneral fact regarding Lagrange multipliers, but we will pause now to sketch whythis converse of the familiar Lagrange multiplier theorem is also true. Let f(q) bea functional on M with a critical point qa when restricted to the submanifold S.This happens if and only if D(f φ)|qa = 0 for all coordinates φ from Euclideanspace to a neighborhood of qa in S. Writing D(f φ) = ∇f ·Dφ and noting thatthe columns of the matrix Dφ span the tangent space TqS ⊂ TqM , we concludethat qa is a critical point on S if and only if the gradient ∇f(qa) ∈ TqM isorthogonal TqS.

Altogether, we have proven the following statement.

Theorem 4.5 (Gauss’ principle). Among the conceivable motions, the actualmotion q(t) is a critical point for the compulsion (4.9) with respect to the releasedmotion. In particular, if M is positive definite then the actual motion is theglobal minimum for the compulsion with respect to the released motion.

Gauss’ principle is analogous to the method of least squares in regressionanalysis, another result discovered by Gauss. In the method of least squares,there is an unknown data correlation function determined by n parameters anda larger number N > n of observations, the latter of which deviate slightlyfrom the desired function’s exact values due to observation error and hencethe overdetermined system for the function is inconsistent. The remedy is toconstruct the square sum error between the function and the data, and thenfind the desired function as a error minimizer in the n parameters. Here, thecompulsion is determined by the 2d initial conditions (q0, v0) ∈ TM where d =dimM is the number of degrees of freedom, and the system is overdetermineddue to the extra constraint conditions. Here, the actual motion q(t) plays therole of the function we seek determined by the smaller number of conditionsdimTS < 2d, the released motion is the measured data which over-determinesthe function with error, and the compulsion is the square sum error. Moreover,the matrix A of masses can be interpreted as weights in the method of leastsquares, which are inserted based on assumed reliability of the data accuracy.

See Exercise 4.1 for an example of Gauss’ principle.


4.3 Integrable constraints

Some authors further require that holonomic constraints be integrable (in thesense of Frobenius). To simplify the formulation of integrability, we will assumethe constraints that are time-independent and linear in q:

a1(q) · q = · · · = ak(q) · q = 0, (4.10)

where the covectors a1(q), . . . , ak(q) ∈ T ∗qM are linearly independent for allq ∈M .

Let ωi denote the 1-form ai(q) dq. We assume that the the constraints (4.10)are completely integrable: the 2-forms dωi for i = 1, . . . , k vanish on the spaceof velocities satisfying the d’Alembert–Lagrange condition (4.4). It then followsfrom Frobenius’ theorem that any path q(t) satisfying the d’Alembert–Lagrangecondition (4.4) must lie within a smooth integrable submanifold N of dimensiond− kCorollary 4.6. A path satisfying the d’Alembert–Lagrange condition (4.4) isa motion of the Lagrangian system with constraints if and only if the path re-stricted to N is a motion for the Lagrangian restricted to N .

Proof. From Holder’s principle (Corollary 4.3) we know that the d’Alembert–Lagrange condition is equivalent to insisting that the action variation δS van-ishes on a subspace Γ of conceivable variations. Here, Γ is a subspace of thetangent space Tq(t)ΩM to the paths ΩM on M . On the other hand, the motionof the Lagrangian system on N is given by taking the action S on the pathsΩN on N , and then insisting that its variation vanishes on the tangent spaceTq(t)ΩN . The key observation is that for the integrable submanifold N , the tan-gent space Tq(t)ΩN is equal to the subspace Γ ⊂ Tq(t)ΩM and so the conditionsbecome identical.

In particular, a holonomic system’s motion is determined by the restrictionof the Lagrangian to the constraint submanifold S. In this way, a system withholonomic constraints is like a new mechanical system with fewer degrees offreedom, which distinguishes holonomic from non-holonomic constraints.

4.4 Non-holonomic constraints

In the derivation of Lagrange’s equations (Proposition 3.4), the last step reliedupon the generalized coordinates qj being independent. As discussed in theprevious section, such qj can always be chosen for a system subject to holonomicconstraints. This is not always true in the non-holonomic case however, and sowe wish to develop a new tool for this situation.

We will consider the special case of m non-holonomic constraints that canbe expressed as the vanishing of a one-form:

n∑k=1

a`k dqk + a`t dt = 0, ` = 1, . . . ,m, (4.11)

4.4. NON-HOLONOMIC CONSTRAINTS 75

with coefficients alj dependent on q1, . . . , qn, t. Note that velocity-independentholonomic constraints also fit this requirement, since when the condition (4.1)is independent of q then taking the differential of both sides gives

n∑k=1

∂f

∂qkdqk +

∂f

∂tdt = 0. (4.12)

Consider endpoint-fixed variations of the system configurations at time t1and t2 as in the proof of Proposition 3.4. The displacement δq inside the time in-tegral happens for fixed time, and thus the constraint (4.11) for a path variationbecomes

n∑k=1

a`k dqk = 0, ` = 1, . . . ,m. (4.13)

If the system’s forces were otherwise conservative, the principle of least actionrequires ∫ t2

t1

n∑k=1

(∂L

∂qk− d

dt

∂L

∂qk

)δqk dt = 0. (4.14)

To make eq. (4.13) comparable to this, we multiply by the undetermined La-grange multiplier coefficients λ`, sum over `, and integrate from t1 to t2:∫ t2

t1

n∑k=1

m∑`=1

λà`kδqk dt = 0. (4.15)

Adding this to equation (3.15) we arrive at∫ t2

t1

n∑k=1

(∂L

∂qk− d

dt

∂L

∂qk+

m∑`=1

λà`k

)δqk dt = 0. (4.16)

As we have discussed, the generalized coordinates qk are no longer inde-pendent since they are related by the m constraints. However, the first d −mcoordinates can be chosen independently, and the remaining m coordinates aredetermined by the conditions (4.13). Pick the multipliers λ` such that

∂L

∂qk− d

dt

∂L

∂qk+

m∑`=1

λà`k = 0, k = d−m+ 1, . . . , d. (4.17)

This causes the last m terms of the summation in the variation (4.16) to vanish,leaving us with∫ t2

t1

n−m∑k=1

(∂L

∂qk− d

dt

∂L

∂qk+

m∑`=1

λà`k

)δqk dt = 0. (4.18)

Since the first d−m coordinates qk are independent, then we may conclude

∂L

∂qk− d

dt

∂L

∂qk+

m∑`=1

λà`k = 0, k = 1, . . . , n−m. (4.19)


Rearranging and including our requirements (4.17), we arrive at

d

dt

∂L

∂qk− ∂L

∂qk=

m∑`=1

λà`k, k = 1, . . . , d. (4.20)

This is the extension of Lagrange’s equations for non-holonomic constraints.Note that these are n equations in d+m unknowns—the n coordinates qk andthe m multipliers λ`—and so we must also consider the m equations of restraint

d∑k=1

a`kqk + a`t = 0, ` = 1, . . . ,m, (4.21)

obtained from (4.11) by dividing through by dt. Comparing these new equa-tions of motion (4.20) to the generalization for nonconservative forces (3.22),we observe that the quantities

∑` λà`k are a manifestation of the constraint

forces.

4.5 Exercises

4.1. Show explicitly that for the pendulum of Example 4.4 minimizing thecompulsion Z leads to the familiar equation of motion.

4.2 (Hoop rolling down an inclined plane). Consider a circular disk of mass Mand radius r rolling without slipping due to gravity down a stationary inclinedplane of fixed inclination φ.

(a) In a vertical plane, the disk requires three coordinates: for example, twoCartesian coordinates (x, y) for the center of mass and an angular coor-dinate to measure the disk’s rotation. If we pick the origin such that thesurface of the inclined plane is the line y = r − (tanφ)x, obtain a holo-nomic constraint of the form (4.1) for the center of mass corresponding tothe disk sitting on the plane.

(b) Consequently, we now can pick two generalized coordinates to describethe disk’s motion: let x denote the distance of the disk’s point of contactand the top of the inclined plane, and θ the angle through which the diskhas rotated from its initial state. By considering the arc length throughwhich the disk has rolled, show that rolling without slipping poses anotherholonomic constraint.

(c) In this case, it is easier to treat rolling without slipping as a non-holonomicconstraint of the type in section 4.4:

r dθ − dx = 0.

Show that the Lagrangian for this system is

L = 12M(x2 + r2θ2) +Mgx sinφ.

4.5. EXERCISES 77

(d) Apply Lagrange’s equations of the form (4.20) to determine the equationsof motion. Here, λ is the force of friction that causes the disk to rollwithout slipping. Substituting in the differential equation of restraint

rθ = x,

conclude that

x =g

2sinφ, θ =

g

2rsinφ, λ =

Mg

2sinφ.

4.3 (Solving Kepler’s problem using harmonic oscillators [KS65]). Consider thesystem of section 2.4 in which a particle x ∈ R3 of mass m under the influenceof a central potential

U(|x|) = −Mm

|x|.

The square function on C ' R2 given by

u := u1 + iu2 =

(u1

u2

)7→ x := u2 =

(u2

1 − u22

2u1u2

)(4.22)

is conformal except at the origin and maps conic sections centered at the originto conic sections with one focus at the origin. Since Kepler’s motion is given byconic sections, we are inspired to apply a transformation with similar propertiesto our system in order to turn the elliptic orbits into simple harmonic oscilla-tion. As a problem solving method this may seem idiosyncratic, but a similartransformation was used by Feynman to solve an open problem in physics.

Attached with the transformation (4.22) is the the linear transformation(dx1

dx2

)= 2

(u1 −u2

u2 u1

)(du1

du2

)of differentials. The matrix of this transformation has the following key prop-erties:

(a) the entries are linear homogeneous functions of the ui;(b) the matrix is orthogonal, in the sense that

• the dot product of any two different rows vanishes,• each row has the norm u2

1 + u22 + · · ·+ u2

n.

We would like to find such a transformation of differentials on Rn. It turns outsuch transformations can only exist for n = 1, 2, 4, or 8. Ultimately we wouldlike a transformation Rn → R3, and so we take n = 4 and choose such a matrix

A =

u1 −u2 −u3 u4

u2 u1 −u4 −u3

u3 u4 u1 u2

u4 −u3 u2 −u1


which satisfies these properties. (This matrix also can be witnessed from quater-nion multiplication.) Consequently, we set

dx1

dx2

dx3

0

= 2A

du1

du2

du3

du4

= 2

u1 du1 − u2 du2 − u3 du3 + u4 du4

u2 du1 + u1 du2 − u4 du3 − u3 du4

u3 du1 + u4 du2 + u1 du3 + u2 du4

u4 du1 − u3 du2 + u2 du3 − u1 du4

. (4.23)

Sadly, only the first three of these are complete differentials for three quantities:

x1 = u21 − u2

2 − u23 + u2

4

x2 = 2(u1u2 − u3u4)

x3 = 2(u1u3 + u2u4)

(4.24)

and the fourth line of eq. (4.23) yields the non-holonomic constraint:

u4 du1 − u3 du2 + u2 du3 − u1 du4 = 0. (4.25)

Explicit formulas exist for both the one-dimensional kernel and the inverse ofthe transformation (4.24).

(a) Let r = |x| denote the distance to the origin in R3. Show that

u21 + u2

2 + u23 + u2

4 = r.

Using the orthogonality of A, invert the transformation (4.23) of differ-entials and conclude that for a fixed point u ∈ R4 our transformationconformally maps the space orthogonal to the kernel at u onto R3.

(b) Let u, v ∈ R4 be two orthonormal vectors which satisfy the condition

u4v1 − u3v2 + u2v3 − u1v4 = 0

derived from the constraint (4.25). From part (a) we know that the planespanu, v is mapped conformally onto a plane of R3. Show that whenthis transformation is restricted to spanu, v onto its image, distancesfrom the origin are squared and angles at the origin are doubled. (Thecalculation of the image x of a point in spanu, v is rather lengthy, andto reach the conclusion it may help to compare the formula for x to theformula for Cayley–Klein parameters (see [Gol51, Sec. 4.5]).)

(c) In particular, it follows that a conic section centered at the origin in theplane spanu, v ⊂ R4 gets mapped to another conical section in R3 withone focus at the origin, the latter of which describes Kepler’s motion.Using part (b) and polar coordinates in the plane spanu, v, show thatellipses and hyperbolas in the plane spanu, v centered at the origin aremapped to ellipses and hyperbolas respectively, and from the limit caseconclude that a line in spanu, v is mapped to a parabola.

4.5. EXERCISES 79

(d) We want to take the motion x(t) ∈ R3 subject to the force P = (P1, P2, P3)and transpose it into motion u(t) in R4 subject to the force Q = (Q1, Q2,Q3, Q4) and the constraint (4.25). Use the transformation of differen-tials (4.23) to determine the kinetic energy T and the force Q in termsof the coordinates u and forces P , and use the formula for the force Q toshow that

u4Q1 − u3Q2 + u2Q3 − u1Q4 = 0. (4.26)

(e) Apply Lagrange’s equations (3.22) for nonconservative forces to obtain theequations of motion for u(t). These equations have Qi on the right-handside, so add them together according to the identity (4.26), simplify, andintegrate once to obtain

r(u4u1 − u3u2 + u2u3 − u1u4) = constant. (4.27)

The parenthetical term (i.e. u4u1−u3u2 +u2u3−u1u4) is exactly the con-straint (4.25) divided by dt, and so to ensure that the constraint is upheldwe pick the initial conditions for u and u so that the parenthetical termvanishes initially, and hence for all time by the conserved quantity (4.27).(Conversely, it can be shown that the equations of motion for u(t) and thecondition u4u1 − u3u2 + u2u3 − u1u4 = 0 yield mx = P .)

(f) The equations of motion for u have a singularity at r = 0. To deal withthis, substitute the regularizing time s

s =

∫ t

0

dt

r,

d

dt=

1

r

d

ds

for the time variable t to regularize the equations of motion for u. Plugin the Kepler forces Qi = −∂U(r)/∂ui and the (signed) semi-major axis

a0 =

(2

r− v2

M

)−1

to arrive at∂2ui∂s2

+M

4a0ui = 0.

That is, the preimage of bounded orbits (a0 > 0) under this transformationis simple harmonic motion in R4 with frequency ω =

√M/4a0. The

harmonic motion can be painstakingly transformed into a solution u(t) bycomputing and substituting the physical time t =

∫ s0r(s) ds.


Chapter 5

Hamilton–Jacobi equationof motion

Rather than n second-order ODEs, the Hamilton–Jacobi equation is one partialdifferential equation in 2d + 1 variables. This perspective is sometimes lesspowerful when solving for the motion, but it provides a physical interpretationof the previously abstract action functional. The material for this chapter isbased on [LL76, Ch. 7] and [Gol51, Ch. 9].

5.1 Hamilton–Jacobi equation

Let us return to the action functional. In section 3.1 we considered the variationof the action S(q) at the systems motion q(t) for paths with fixed endpoints. Asin the set up for the proof of the Euler–Lagrange equations (Proposition 3.4)we can reduce to an open subset of Euclidean space, where the variation of themotion q(t) between times t1 and t2 is given by q(t) + δq(t). Previously weassumed δq(t1) = δq(t2) = 0 and then prescribed the resulting variation δS tovanish in accordance with the principle of least action. Now we will considerthe action of the true progression of the system S[q(t)] as a function of thecoordinates at time t2 by allowing δq(t2) (and consequently δS) to vary.

Repeating the calculation of the action variation from Proposition 3.4, wenow have

δS(q; δq) =

[∂L

∂q· δq]t1t0

+

∫ t1

t0

(∂L

∂q− d

dt

∂L

∂q

)· δq dt =

∂L

∂q(t2) · q(t2).

The integral vanishes due to Lagrange’s equations of motion, but now only oneof the boundary terms vanish since δq(t1) = 0. If we let t := t2 vary now, thenwe conclude

∂S

∂qi=∂L

∂qi= pi (5.1)

81

82 CHAPTER 5. HAMILTON–JACOBI EQUATION OF MOTION

after recognizing ∂L /∂qi as the momentum pi.By definition of the action we have

dS

dt= L . (5.2)

On the other hand we know S = S[t, q(t)], and so the chain rule insists

dS

dt=∂S

∂t+∂S

∂q· q =

∂S

∂t+ p · q (5.3)

after using the derivative (5.1). Setting (5.2) and (5.2) equal yields

∂S

∂t= L − p · q = −H,

where H is the Hamiltonian (or total energy) (3.30). Using the derivative (5.1)we recognize this identity as a first order partial differential equation (PDE) forS(t, q):

0 =∂S

∂t+H

(t, q,

∂S

∂q

). (5.4)

This is the Hamilton–Jacobi equation, and for a system with n degrees of freedomthere are n+1 independent variables (t, q1, . . . , qn). It is often the case in practicethat the form of the action S(t, q) is unknown, and cannot be determined fromeq. (5.4) alone.

The solution, or complete integral, of this equation has n + 1 integrationconstants corresponding to the number of independent variables. Calling theseconstants α1, . . . , αn, and A (which play a physical role as we will see), we canwrite

S = f (t, q1, . . . , qn, α1, . . . , αn) +A;

we know one of these constants can be additive since the action appears in thePDE (5.4) only through its partial derivatives and hence is invariant under theaddition of a constant. Mathematically speaking, a solution S(t, q, α) to theHamilton–Jacobi equation (5.4) should be required to satisfy

det∂2S

∂q ∂α6= 0

to be a complete integral, in order to avoid incomplete solutions.The function f(t, q1, . . . , qn, α1, . . . , αn) (called a generating function) in-

duces a change of coordinates (a canonical transformation)—we will developthis idea more from the Hamiltonian perspective. Think of α1, . . . , αn as newmomenta, and let β1, . . . , βn denote new coordinates to be chosen. Note thatby the chain rule,

df

dt=∂f

∂t+∂f

∂q· q +

∂f

∂α· α. (5.5)

Set

βi := − ∂f

∂αi, (5.6)

5.1. HAMILTON–JACOBI EQUATION 83

and note that pi = ∂f/∂qi by eq. (5.1). Consider the new Hamiltonian

H ′(α, β, t) := H +∂f

∂t= H +

∂S

∂t= 0,

which has corresponding Lagrangian

L ′(β, α, t) := α · β −H ′ = α · β = −df

dt+∂f

∂t+∂f

∂q· q

= −df

dt−H + p · q = L − df

dt

by the chain rule (5.5). Since this new Lagrangian L ′ differs from the old L bya complete time derivative, then they generate the same motion by Corollary 3.6.

For any Lagrangian L , the definition of the Hamiltonian (3.30) requires

∂H

∂p=

∂

∂p(p · q −L (q, q, t)) = q,

∂H

∂q=

∂

∂q(p · q −L (q, q, t)) = −∂L

∂q= − d

dt

∂L

∂q= −p.

This calculation may appear questionable at first (is the velocity q really in-dependent of the position q and momentum p), but it is indeed correct; infact, we will later give it thorough justification as it is the basis of Hamiltonianmechanics. Applying this to our new Hamiltonian H ′(β, α, t), we observe that

α = 0, β = 0.

In other words, we have used the integration constants α to generate correspond-ing conserved quantities β. Recalling the definition (5.6) of the coordinates β,we conclude

∂S

∂αi= constant (5.7)

in the n coordinates and time. Note that even if we can obtain a partial solutionto the Hamilton–Jacobi equation involving only m constants αi, we still get thecorresponding m constants of motion.

Lastly, let us specialize to a conservative system. In this case, the Hamilto-nian H is the total energy E and is constant and the action integral from times0 to t becomes:

S[q(t)] =

∫ t

t0

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

∫ t

t0

(p · q −H) dt

=

∫ t

t0

p · q dt−∫ t

t0

E dt = S0 (q1, . . . , qn)− Et

for some S0(q) which is only a function of the coordinates. Since ∂S0/∂t = 0,the Hamilton–Jacobi equation then takes the special form

H

(q,∂S

∂q, t

)= E. (5.8)

In particular, we are no longer required to know the formula for the action.


5.2 Separation of variables

Sometimes we can reduce the Hamilton–Jacobi equation by one coordinate usingseparation of variables. Suppose that for some system with n degrees of freedom,we have a coordinate q1 with corresponding derivative ∂S/∂q1 that appear in theHamilton–Jacobi equation only in some particular combination φ(q1, ∂S/∂q1).That is, the Hamilton–Jacobi equation (5.4) takes the form

Φ

(t, q2, . . . , qn,

∂S

∂q2, . . . ,

∂S

∂qn, φ

(q1,

∂S

∂q1

))= 0 (5.9)

after rearranging the independent variables. We take the (additive) separationof variables ansatz

S = S′(q2, . . . , qn, t) + S1(q1)

for the solution. Plugging this back into our Hamilton–Jacobi equation (5.9) weget

Φ

(t, q2, . . . , qn,

∂S′

∂q2, . . . ,

∂S′

∂qn, φ

(q1,

dS1

dq1

))= 0. (5.10)

Note that q1 only influences φ and is entirely independent from the rest of theexpression. Since the variables are independent, we conclude that φ must beconstant:

φ

(q1,

dS1

dq1

)= α1, Φ

(t, q2, . . . , qn,

∂S′

∂q2, . . . ,

∂S′

∂qn, α1

)= 0. (5.11)

We now have a first order ODE and the Hamilton–Jacobi equation in terms ofthe remaining n− 1 coordinates. Since the resulting form of Φ is coherent withwhat we found in the previous section, then we conclude that our ansatz is cor-rect. The ability to remove a coordinate in the Hamilton–Jacobi considerationeven when it is not cyclic is a virtue of this approach.

If we do have some cyclic coordinate q1 for the system (i.e. S or L isindependent of q1), then the Hamilton–Jacobi equation (5.4) becomes

∂S

∂t+H

(q2, . . . , qn,

∂S

∂q1, . . . ,

∂S

∂qn, t

)= 0. (5.12)

Then q1 is of the type (5.9) for the function

φ

(q1,

∂S

∂q1

)=∂S

∂q1. (5.13)

Our result (5.11) yields

∂S

∂q1= α1, S = S′(q2, . . . , qn, t) + α1q1. (5.14)

Here, α1 = ∂S/∂q1 is just the conserved momentum corresponding to q1.

5.3. CONDITIONALLY PERIODIC MOTION 85

If this cyclic variable is time, then the system is conservative and we saw ineq. (5.8) that the action is given by

S = W (q1, . . . , qn)− Et. (5.15)

Here, −E is the conserved quantity associated with t, although it is not nec-essarily always the total energy. We have used W instead of S′ to denote thisspecial case, but they share the same purpose; W is known as the Hamilton’scharacteristic function. The Hamilton–Jacobi equation is now

H

(q1, . . . , qn,

∂W

∂q1, . . . ,

∂W

∂qn

)= −∂S

∂t= E, (5.16)

and E is just one integration constant αj of the motion of which S′ is inde-pendent. The corresponding conserved quantity (5.7) will give the coordinatesimplicitly as functions of the constants αi, βi, and time:

βi =∂S

∂αi=

∂W

∂αi− t for i = j,

∂W

∂αifor i 6= j.

(5.17)

Only the jth of these equations is time-dependent, and so one of the qi canbe chosen as an independent variable and the rest will be able to be written interms of this coordinate. Such solutions for the motion are called orbit equations.For the central forces of section 2.2, we were able to solve for the angle φ as afunction of the radius r.

5.3 Conditionally periodic motion

For this section, we will examine a system of n degrees of freedom with boundedmotion, such that every variable can be separated using the method of theprevious section. This means the action takes the form

S =

d∑i=1

Si(qi) + S′(t), (5.18)

where each coordinate Si is related to the corresponding momentum via

pi =∂Si∂qi

, Si =

∫pi dqi. (5.19)

The motion is bounded, and so this integral represents the area enclosed by aloop in the phase plane (qi, pi) (just as with the period integral (2.4)). Everytime qi returns to a value, Si has incremented by an amount 2πIi with

Ii =1

2π

∮pi dqi. (5.20)


These Ii are called the action variables.As in section 5.1, the generating function Si induces a canonical transforma-

tion of coordinates with the action variables Ii as the new momenta. We will seethis process more systematically from the Hamiltonian perspective. The newposition coordinates, called angle variables, are given by

wi =∂S

∂Ii=

n∑k=1

∂Sk(qk, I1, . . . , In)

∂Ii(5.21)

since this is the time integral of “∂H/∂p = q”. The generating functions Si aretime-independent, and so the new Hamiltonian H = E is just the old in termsof the new coordinates. The (Hamiltonian) equations of motion require

Ii = −∂H(I1, . . . , In)

∂wi= 0, wi =

∂H(I1, . . . , In)

∂Ii=∂E(I1, . . . , Is)

∂Ii, (5.22)

which can be immediately integrated to yield

Ii = constant, wi =∂E

∂Iit+ constant. (5.23)

As we have already observed, Si increments by 2πIi each time qi returns to itsoriginal value, and so the angle variables wi also increment by 2π. Consequentlythe derivative ∂E/∂Ii is the frequency of motion in the ith coordinate, whichwe were able to identify without solving the entire system.

Since the motion in these variables is periodic, any single-valued functionF (q, p) of the system coordinates and momenta will be periodic in the anglevariables with period 2π after being transformed to the canonical variables.Fourier expanding in each of the angle variables, we have

F =∑`∈Zn

Aèi`·w =

∑`1∈Z· · ·∑`n∈Z

A`1...`nei(`1w1+···+`nwn). (5.24)

Using eq. (5.23), we may write the angle variables as functions of time. Absorb-ing the integration constants of the wi into the coefficients A`, we get

F =∑`∈Zn

A` exp

(it` · ∂E

∂I

). (5.25)

Each term of this sum is periodic with frequency ` · (∂E/∂I). If the frequencies∂E/∂Ii are not commensurable however, then the total quantity F is not peri-odic. In particular, the coordinates q, p are may not be periodic, and the systemmay not return to any given state that it instantaneously occupies. However, ifwe wait long enough the system will come arbitrarily close to any given occupiedstate—this phenomenon is referred to as Poincare’s recurrence theorem. Suchmotion is called conditionally periodic.

Two frequencies ∂E/∂Ii that are commensurable (i.e. their ratio is a rationalnumber) are called a degeneracy of the system, and if all n are commensurable

5.4. GEOMETRIC OPTICS ANALOGY 87

the system is said to be completely degenerate. In the latter case, all motionis periodic, and so we must have a full set of 2n − 1 conserved quantities.Only n of these will be independent, and so they can be defined to be theaction variables I1, . . . , In. The remaining n− 1 constants may be chosen to bewi∂E/∂Ik − wk∂E/∂Ii for distinct i, k, since

d

dt

(wi∂E

∂Ik− wk

∂E

∂Ii

)= wi

∂E

∂Ik− wk

∂E

∂Ii=∂E

∂Ii

∂E

∂Ik− ∂E

∂Ik

∂E

∂Ii= 0.

Note, however, that since the angle variables are not single-valued, neither willbe the n− 1 constants of motion.

Consider a partial degeneracy, say, of frequencies 1 and 2. This means

k1∂E

∂I1= k2

∂E

∂I2(5.26)

for some k1, k2 ∈ Z. The quantity w1k2 − w2k1 will then be conserved, since

d

dt(w1k1 − w2k2) = w1k1 − w2k2 =

∂E

∂I1k1 −

∂E

∂I2k2 = 0. (5.27)

Note that this quantity is single-valued modulus 2π, and so a trigonometricfunction of it will be an actual conserved quantity.

In general, for a system with n degrees of freedom whose action is totallyseparable and has n single-valued integrals of motion, the system state movesdensely in a n-dimensional manifold in 2d-dimensional phase space. For degen-erate systems we have more than n integrals of motion, and consequently thesystem state is confined to a manifold of dimension less than n. When a systemhas less than n degeneracies, then there has fewer than n integrals of motionand the system state travels within a manifold of dimension greater than n.

5.4 Geometric optics analogy

In this section we will see that the level sets of the action propagate through con-figuration space mathematically similar to how light travels through a medium.This is not meant literally, but rather to bring a physical analogy to the formerlyabstract notion of the action.

Suppose we have a system for which the Hamiltonian is conserved and equalto the total energy. For simplicity, suppose the system describes just one particlemoving in three-dimensional Euclidean space using Cartesian coordinates q =x = (x, y, z) ∈ R3; much of the results we will show however hold without theseassumptions. Equation (5.15) tells us that the action is given by

S(q, t) = W (q)− Et. (5.28)

Since we have chosen Cartesian coordinates, rather than discussing the particle’smotion we may discuss the motion of the action level surfaces S(q, t) = b in


time within the same space. If we were to generalize this argument to multiple-particle systems, then instead of the particle’s motion in Cartesian space wemust consider the path that the system traces out in configuration space—thespace spanned by the coordinates q. At time t = 0, we have an equation forHamilton’s characteristic function W = b, and after a time step ∆t we thenhave W + ∆W = b+ E∆t+O(∆t2).

The propagation of this surface can be thought of as a wavefront. The changein the characteristic function W during the time interval dt is dW = E dt. Ifwe call the distance traveled normal to the wavefront ds, then we also have

∂W

∂s= |∇W |. (5.29)

The velocity u of the wavefront is then

u =ds

dt=dW/|∇W |dW/E

=E

|∇W |. (5.30)

As a conservative system, the Hamilton–Jacobi equation takes the form (5.16),which in Cartesian coordinates looks like

E = H

(q, p =

∂W

∂q

)= T

(p =

∂W

∂q

)+ U =

(∇W )2

2m+ U,

or after rearranging,(∇W )

2= 2m(E − U). (5.31)

Plugging this into the velocity (5.29), we have

u =E√

2m(E − U)=

E√2mT

=E

p. (5.32)

The faster the particle moves, the slower the action level sets propagate.The momentum is given by

p =∂W

∂q= ∇W. (5.33)

The gradient is of course normal to the level sets, and so this relation tells usthat the particle always moves normal to the level sets of the characteristicfunction W .

We will now see how the level sets of the action propagate like waves. Forsome scalar-valued function φ, the wave equation of optics is

∇2φ− n2

c2∂2φ

∂t2= 0. (5.34)

If the refractive index n is constant, then there is a family of plane wave solu-tions:

φ(r, t) = φ0ei(k·x−ωt), (5.35)

5.5. EXERCISES 89

where the vector k ∈ R3 is the propagation direction and the magnitude ofk = 2π/λ = nω/c is the wave number.

In geometric optics, the refractive index n ≡ n(x) is not assumed to beconstant, but rather to be changing slowly on the scale of the wavelength. Con-sequentially, we now seek solutions to the wave equation (5.34) with the planewave ansatz (5.35):

φ(r, t) = eA(x)+ik0(L(x)−ct). (5.36)

where A(x) is related to the amplitude of the wave, k0 = ω/c is the wavenumber in vacuum (n ≡ 1), and L(x) is called the optical path length of thewave. Plugging in eq. (5.36), the wave equation (5.34) becomes

φ[∇2A+ (∇A)2 − k2

0(∇L)2 + k20n

2]

+ ik0

[∇2L+ 2∇A · ∇L

]= 0. (5.37)

In general, φ is nonzero and so the curly-bracketed expression must vanish.We want A and L to be real-valued by construction, and so we the real andimaginary parts in square brackets must also vanish:

∇2A+ (∇A)2 + k20

[n2 − (∇L)2

]= 0, ∇2L+ 2∇A · ∇L = 0. (5.38)

Now comes the geometric optics approximation: the wavelength λ = 2π/k issmall compared to the rate of change of the medium. In particular, the wavenumber in vacuum k0 = 2π/λ0 must be considerably large compared to thederivative terms in the first equation of (5.38), and thus we require

(∇L)2

= n2. (5.39)

This is called the eikonal equation.Returning to eq. (5.31), we see that the characteristic function W satis-

fies an eikonal equation for a “medium” of refractive index√

2m(E − U) = p(this equality holds for the single-particle case). This illustrates that the char-acteristic function is like a wavefront that propagates through the medium ofconfiguration space with refractive index p, in the geometric optics limit.

5.5 Exercises

5.1 (Harmonic oscillator). The harmonic oscillator Hamiltonian is given by

H =1

2mp2 +

mω2

2q2.

(a) Write down the Hamilton–Jacobi equation (5.4) for this system. Since thissystem is conservative, we expect a solution (up to an arbitrary additiveconstant) of the form

S(q, α, t) = W (q, α)− αt


where the constant α is the total energy. Plug this ansatz into theHamilton–Jacobi equation and conclude that

W = ±mω∫ √

2α

mω2− q2 dq.

This integral can be evaluated further, but it is not necessary for ourpurposes.

(b) The quantity β will implicitly give us the equation of motion q(α, β, t).Using the definition (5.6), show that

t+ β =1

ω

sin−1

(√m2αωq

)(plus sign)

cos−1(√

m2αωq

)(minus sign)

+ constant.

The integration constant may be absorbed into β which has yet to bedetermined, and since sin−1(x) = π/2− cos−1(x) we can take the secondcase and absorb the π/2 into β also. We should pick up a factor of −1on one side of the equation due to this last step, but this will not mattersince we have chosen cosine which is even. Altogether, we arrive at thefamiliar solution

q(t) =

√2α

mω2cos (ω(t+ β)) .

(c) To find the constants we must apply the initial conditions q(0) = q0,p(0) = p0. Determine α and β using p0 = (∂S/∂q)t=0 and q(0) = q0

respectively, and obtain the solution as a function of the initial values:

q(t) =

√q20 +

p20

m2ω2cos

ωt+ cos−1

q0√q20 +

p20m2ω2

.5.2 (Central field). Consider the motion of a particle in a central field as insection 2.2. In polar coordinates, we have

T =m

2(r2 + r2φ2), U = U(r),

pr =∂L

∂r= mr, pφ =

∂L

∂φ= mr2φ,

H =1

2m

(p2r +

p2φ

r2

)+ U(r).

(a) This Hamiltonian is both time-independent and cyclic in φ, and so weexpect an action of the form

S = W1(r) + αφφ− Et.

5.5. EXERCISES 91

Plug this into the Hamilton–Jacobi equation (5.16) for conservative sys-tems and integrate to arrive at

W = W1(r) + αφφ = ±∫ √

2m(E − U(r))−α2φ

r2dr + αφφ.

(b) Use eq. (5.17) to obtain the implicit equations of motion

β1 = ±∫ 1

22m dr√2m(E − U(r))− α2

φ

r2

− t,

β2 = ±∫ − 1

22αφr2 dr√

2m(E − U(r))− α2φ

r2

+ φ.

These match what we found in equations (2.13) and (2.14), where theconstant αφ = M is the angular momentum associated to the cyclic coor-dinate φ.

5.3 (Kepler’s problem). We will find the frequency of oscillations for Kepler’sproblem using action variables without solving the equations of motion. Con-sider a particle of massm in an inverse-square central force field, as in section 2.4:

T =m

2(r2 + r2θ2 + r2 sin2 θφ2), U = −k

r,

pr =∂L

∂r= mr , pθ =

∂L

∂θ= mr2θ, pφ =

∂L

∂φ= mr2 sin2 θφ,

H =1

2m

(p2r +

p2θ

r2+

p2φ

r2 sin2 θ

)− k

r.

The constant k is positive, since we must have an attractive force for boundedmotion.

(a) Write down the Hamilton–Jacobi equation (5.16) for conservative systems.Notice that all of the coordinates are separable, and so the characteristicfunction is of the form

W = Wr(r) +Wθ(θ) +Wφ(φ).

(b) The Hamiltonian is cyclic in φ, and so the derivative ∂W/∂φ = ∂Wφ/∂φ =αφ is the constant that is angular momentum about the z-axis. Plug thisin, group the terms involving only θ, and conclude that(

∂Wθ

∂θ

)2

+α2φ

sin2 θ= α2

θ,1

2m

[(∂Wr

∂r

)2

+α2θ

r2

]− k

r= E.


The equations for Wφ, Wθ, and Wr demonstrate the conservations of angu-lar momentum about the z-axis pθ, total angular momentum p, and totalenergy E, and from here they could be integrated to obtain the equationsof motion.

(c) Use the three differential equations of part (b) to obtain the action vari-ables:

Iφ = αφ = pφ,

Iθ =1

2π

∮ √α2θ −

α2φ

sin2 θdθ,

Ir =1

2π

∮ √2mE +

2mk

r−α2θ

r2dr.

(d) Let us look at the second action variable Iθ. We know from section 2.2that this motion is coplanar, so let ψ denote the angle in the plane of orbit.Set the momentum in the (r, θ, φ) variables and (r, ψ) variables equal, andconclude that pθ dθ = p dψ − pφ dφ. Conclude that

Iθ = p− pφ = αθ − αφ.

(e) Now for the third action variable Ir. The integral for Ir is evaluatedbetween two turning points r1, r2 for which the integrand pr = mr mustvanish. We can therefore integrate from r2 to r1 and back to r2, for whichthe integrand is first negative then positive, corresponding to the sign ofthe momentum pr = mr. In the complex plane, this integrand is analyticeverywhere but r = 0 and along the segment on the real axis connecting r1

and r2. Integrate around a counterclockwise simple path γ encompassingr1 and r2 to obtain

Ir = −α2θ +

mk√−2mE

= −(Iθ + Iφ) +mk√−2mE

.

The energy

E(I) = − mk2

2(Ir + Iθ + Iφ)2

is symmetric in the three action variables, and so the frequency of oscillationsin each coordinate r, θ, φ is the same:

∂E

∂Ir=∂E

∂Iθ=∂E

∂Iφ=

mk2

(Ir + Iθ + Iφ)3,

as is expected since the force is central.

5.5. EXERCISES 93

Part III

Hamiltonian Mechanics

Rather than working on the tangent bundle over configuration space, Hamil-tonian mechanics separates the position and momentum variables and is natu-rally phrased on the cotangent bundle. The equations of motion are a system offirst-order differential equations with two equations for each degree of freedom,and is often the preferred formulation in mathematics. The position and mo-mentum are of course not truly independent coordinates, and so an acceptablechange of variables must satisfy the criterion of a canonical transformation. Thisrelationship between the position and momenta induces a geometric structureon the tangent bundle, which is abstracted into symplectic geometry in the caseof conservative mechanics and contact geometry for dissipative mechanics.


Chapter 6

Hamilton’s equations ofmotion

We will develop the equations of motion and key tools of the Hamiltonian per-spective, with a primary focus on Euclidean space before exploring general ge-ometries. The material for this chapter is based on [Arn89, Ch. 3], [Gol51,Ch. 7–8], and [LL76, Ch. 7].

6.1 Hamilton’s equations

Recall that for a system with coordinates q = (q1, q2, . . . , qn) on a n-dimensionalmanifold M the Hamiltonian

H(q, p, t) := q · p−L (q, q, t) =

n∑i=1

qi∂L

∂qi−L (6.1)

is often the total energy of the system, and the components of p are the mo-menta pi := ∂L /∂qi. In general, the Hamiltonian is a smooth function functionH(q, p, t) : T ∗M × I → R where I ⊂ R is an interval, since the momentum isa covector and not a vector. In Proposition 3.11 we saw that the Hamiltoniandetermines the system’s motion, and that the quantity H is conserved wheneverthe Lagrangian L is time-independent.

When we defined the Hamiltonian H in section 3.5, we showed that it isequal to the total energy T +U of the system in Cartesian coordinates. We willnow show that the Hamiltonian H(q, p, t) is equal to the total energy T + Ufor any system with velocity-independent potential energy U ≡ U(q, t). As ineq. (3.11) the kinetic energy is a quadratic form in the velocity variables:

T = 12

∑i,j

aij(q, t)qiqj . (6.2)

95

96 CHAPTER 6. HAMILTON’S EQUATIONS OF MOTION

So T is a homogeneous function of order 2 in q: for any constant k we have

T (q, kq, t) = 12

∑i,j

aij(q, t)d

dt(kqi)

d

dt(kqj) = k2T (q, q, t). (6.3)

This determines the derivative of T , since

2kT =d

dk(k2T ) =

d

dkT (q, kq, t) =

(∂

∂(kq)T (q, kq, t)

)·(∂

∂kkq

)=∂T (kq)

∂(kq)· q,

and so taking k = 1, we conclude

∂T

∂q· q = 2T.

(This result is known as Euler’s homogeneous function theorem.) Since thepotential energy is velocity independent, then we have

H = p · q −L =∂L

∂q· q −L =

∂T

∂q· q − (T − U) = 2T − T + U = T + U

as desired.Treating the position q, velocity q, and momentum p as independent vari-

ables, the differential of the Hamiltonian must satisfy

∂H

∂pdp+

∂H

∂qdq +

∂H

∂tdt = dH = d(p · q −L )

= q dp+

(p− ∂L

∂q

)dq − ∂L

∂qdq − ∂L

∂tdt

= q dp+ 0 dq − p dq − ∂L

∂tdt

where in the last equality we used Lagrange’s equations. Of course the mo-mentum and velocity are not actually independent, but the coefficient of dq willappear with a factor in the term dp and the same cancellation occurs. Matchingterms we conclude

∂H

∂p= q,

∂H

∂q= −p. (6.4)

These 2n first-order differential equations are called Hamilton’s equations. Notethat equating the terms containing dt provides no information since this equalityfollows immediately from the definition of the Hamiltonian.

In Lagrangian mechanics we saw that a cyclic coordinate qk resulted in theconservation of pk = ∂L /∂qk, and thus pk entered the remaining n−1 equationsas a constant parameter. For Hamilton’s equations, a cyclic variable qk yields

pk =∂H

∂qk= 0, qk =

∂H

∂pk.

6.2. LEGENDRE TRANSFORMATION 97

The first of these equations expresses the conservation of pk, and the second isindependent of qk and is thus easily integrable. Consequently, we are left with2n− 2 equations in terms of the constant parameter pk.

To conclude this section, we will observe Hamilton’s equations as a conse-quence of the principle of least action. Although we have defined momentumand thus Hamilton’s equations themselves in terms of the Lagrangian and thegeneralized coordinates qi, we can also consider the momenta pi as variablessince we can still extract the Euler–Lagrange equations as long as we have atleast n independent coordinates (cf. the remark at the end of section 3.1). Forthe Lagrangian of a system with n degrees of freedom, we must specify 2n initialconditions—n for the coordinates qi and another n for the qi—and so derivingHamilton’s equations without the Lagrangian equations of motion allow us toregard the qi and pi both as independent coordinates linked only by the equa-tions of motion. To demonstrate this, suppose we have a system whose motionis described by the coordinates (q(t), p(t)) from time t0 to t1. Within a coordi-nate patch on Euclidean space, we may write any other time progression of thesystem as q(t) + δq(t) with δq(t0) = δq(t1) = 0. The principle of least actioninsists

0 = δS = δ

∫ t1

t0

L dt = δ

∫ t1

t0

(p · q−H(q,p, t)) dt

=

∫ t1

t0

∑i

(q · p + p · δq− ∂H

∂q· δq− ∂H

∂p· δp

)dt.

We may integrate the second term by parts:∫ t1

t0

piδqidt =

∫ t1

t0

pid

dt(δqi) dt = piδqi

∣∣∣∣t1t0

−∫ t1

t0

piδqi dt (6.5)

The boundary term vanishes by the fixed-endpoint condition δqi(t0) = δqi(t1) =0. Plugging this back in to the action variation yields

0 =

∫ t1

t0

[(q− ∂H

∂p

)· δp−

(p +

∂H

∂q

)· δq

]dt

Since this integral vanishes for any choice of δpi(t) and δqi(t), and we are alwaysable to pick paths such that these parameters are varied independently, then weconclude that the two parenthetical terms must vanish:

qi −∂H

∂pi= 0, pi +

∂H

∂qi= 0.

This agrees with Hamilton’s equations (6.4).

6.2 Legendre transformation

The Legendre transformation is an involution on the space of real-valued strictlyconvex functions, which naturally joins the Lagrangian and Hamiltonian per-


spectives. As is the case in physical application, the Lagrangian and Hamilto-nian are quadratic in q due to the kinetic energy, and are Legendre transformsof each other in the variable q.

Throughout this section, f : U → R will denote a smooth function on anopen convex set U ⊂ Rd that is (strictly) convex in the sense that the Hessianmatrix is positive definite, which is written as f ′′(x) > 0, and is defined by

f ′′(x)y · y =

(∂2f

∂xi∂xj(x)

)y · y > 0 for all 0 6= y ∈ Rd

at every point x ∈ U . We will now define the Legendre transform f∗(ξ) ofsuch a function f(x). Consider the distance F (x, ξ) = x · ξ − f(x) between thehyperplane x · ξ of “slope” ξ and the graph of the function f(x). Since f isconvex, then for fixed ξ the function F will be a concave function of x, and sothere is a unique point x∗(ξ) which maximizes F (x, ξ). We define the Legendretransform of f by f∗(ξ) = F (x∗(ξ), ξ), which mathematically is the function

f∗(ξ) = supx∈U

(x · ξ − f(x)) (6.6)

on the domain U∗ where this supremum is finite:

U∗ = sup

ξ ∈ Rd : sup

x∈U(x · ξ − f(x)) <∞

. (6.7)

Example 6.1. Consider the quadratic function

f(x) = 12Ax · x + b · x + c

on U = Rd, where A is a real symmetric positive definite matrix, b ∈ Rd, andc ∈ R. In order to maximize the distance F (x, ξ), we differentiate:

0 =∂F

∂x(x∗) =

∂

∂x

[x · ξ − 1

2Ax · x− b · x− c]x=x∗ = ξ −Ax∗ − b.

This equation has one critical point x∗ = A−1(ξ − b), which must be ourmaximum. Plugging this back in, we get

f∗(ξ) = F (x∗(ξ), ξ)

= A−1(ξ − b) · ξ − 12 (ξ − b) ·A−1(ξ − b)− b ·A−1(ξ − b)− c

= 12A−1(ξ − b) · (ξ − b)− c

on U∗ = Rd. In particular, we can take A = mI, b = 0, and c = 0 so thatf(x) = 1

2m|x|2 is the kinetic energy, and we get that f∗(ξ) = |ξ|2/2m is also

the kinetic energy if we recognize ξ as the momentum.

Next, we will demonstrate the transformation’s key properties.

Theorem 6.2 (Involutive property). The Legendre transform f∗ : U∗ → R off : U → R is a convex function, whose Legendre transform (f∗)∗ is f . In otherwords, the Legendre transform is an involution.

6.2. LEGENDRE TRANSFORMATION 99

Proof. First we check that the transform f∗ is also convex. As the uniquemaximizer of the distance F (x, ξ), we know that x∗ = x∗(ξ) is the uniquesolution to

0 =∂F

∂x(x∗) = ξ −Df(x∗) (6.8)

where Df(x∗) = ∇f(x∗) is the gradient. The function Df between open setsof Rd is invertible and so we can write x∗ = (Df)−1(ξ). By the inverse functiontheorem, the derivative of this function x∗(ξ) is

Dξx∗(ξ) = Dξ(Df)−1(ξ) = (D2f)−1(x∗(ξ)) (6.9)

where D2f(x) = f ′′(x) is the Hessian matrix whose entries are functions of x.The first derivative of f∗(ξ) is

Df∗(ξ) = Dξ

ξ · (Df)−1(ξ)− f

[(Df)−1(ξ)

]= (Df)−1(ξ) + ξ · (D2f)−1(x∗(ξ))−Df

[(Df)−1(ξ)

]· (D2f)−1(x∗(ξ))

= (Df)−1(ξ).

Therefore, the second derivative of our Legendre transform is

(f∗)′′(ξ) = D2f∗(ξ) = (D2f)−1(x∗(ξ)) > 0, (6.10)

which is positive definite since f ′′(x) is. That is, the Legendre transform f∗ isalso convex.

Now that we know

g(ξ) := f∗(ξ) = x∗(ξ) · ξ − f(x∗(ξ)) = Df−1(ξ) · ξ − f(Df−1(ξ)) (6.11)

is also convex, we may now consider its Legendre transform. Let ξ∗ be the pointwhich maximizes the distance function G(ξ,x) for g. This point is determinedby the condition

0 =∂G

∂ξ(ξ∗) =

∂

∂ξ[ξ · x− g(ξ)]ξ=ξ∗ = x−Dg(ξ∗), (6.12)

which we can solve asξ∗ = (Dg)−1(x) = Df(x) (6.13)

(using our previous calculation of Dg = Df∗). Consequently, the transform ofg(ξ) is given by

g∗(x) = G(ξ∗,x) = ξ∗ · x− g(ξ∗)

= Df(x) · x−Df−1(Df(x)) ·Df(x) + f(Df−1(Df(x))) = f(x)

as desired. For fixed x0, G(ξ,x0) geometrically represents the “y-coordinate” ofthe intersection of the line tangent to f(x) with slope ξ. Since f is convex, all ofthe tangent lines of f lie below the curve, and so the maximum of G(ξ,x) willbe at f(x0). Therefore, the transformation of an already transformed functiong(ξ) of some smooth f(x) must reproduce the original function f(x).


Returning to the context of mechanics, we make the following simple calcu-lation which is a consequence of the kinetic energy being quadratic.

Proposition 6.3. Let M be a real symmetric positive definite matrix. TheLegendre transformation of the conservative Lagrangian

L (q, q) = 12Mq · q − U(q)

in the velocity q variables with dual variable p yields the corresponding Hamil-tonian

H(q, p) = 12M

−1p · p+ U(q).

Since the velocity q is a tangent vector, then this duality and the correspondingnatural paring x · ξ = 〈q, p〉 shows us that momentum p is naturally a covector.

Proof. Take A = M , b = 0, and c = U(q)—which is constant with respect tothe velocity q—in Example 6.1.

6.3 Liouville’s theorem

Liouville’s theorem states that the density of trajectories in phase space sur-rounding a system configuration as it evolves in time is constant.

Suppose the Hamiltonian H(q, p) for a system with n degrees of freedomdoes not depend explicitly on time. We are interested in time progressions ofthe system in phase space, the 2n-dimensional space of the coordinates (q, p) =(q1, . . . , qn, p1, . . . , pn) ∈ R2n. Consider the vector field given by Hamilton’sequations: at each point (q, p) in phase space we know that the trajectory of thesystem at this configuration is tangent to the vector (−∂H/∂q, ∂H/∂p) whichis tangent to the trajectory passing through (q, p). We will (as always) supposethat the solution (q(t), p(t)) for any initial condition (q(0), p(0)) = (q0, p0) ofHamilton’s equations can be extended for all time. The phase flow is then theset of transformations φt : (q0, p0) 7→ (q(t), p(t)) with t ∈ R of phase space toitself. The phase flow forms a group under function composition parameterizedby the additive group t ∈ R and a group action on phase space R2d = Rdq ×Rdp:

(a) At t = 0 the phase flow is the identity map: φ0(q0, p0) = (q0, p0) for all(q0, p0) ∈ R2n.

(b) Composition corresponds to addition in t ∈ R: (φt0 φt1)(q0, p0) =φt0(q(t1), p(t1)) = (q(t0 + t1), p(t0 + t1)) = φt0+t1(q0, p0).

(c) Inversion corresponds to negation in t ∈ R: (φt φ−t)(q0, p0) = φt+(−t)

(q0, p0) = φ0(q0, p0) and vice versa.

(d) Function composition is associative and the group action is as well, becauseaddition is associative in R.

6.3. LIOUVILLE’S THEOREM 101

Theorem 6.4 (Liouville’s theorem). The density of system points in phasespace R2n = Rnp × Rnq around a phase curve remains constant along the curve.That is, for any measurable set D ⊂ R2n in phase space, the volume of φt(D)is equal to that of D.

Proof. First, let us simplify notation using the point x := (q, p) ∈ R2n and theHamiltonian vector field f : R2n → R2n, f(x) := (−∂H/∂q, ∂H/∂p), so that

x = f(x), φt(x) = x+ f(x)t+O(t2). (6.14)

Consider a region D(0) ⊂ R2n, and let D(t) = φt(D(0)) denote its flow and v(t)denote volume of D(t). We will prove the more general statement

if ∇ · f ≡ 0 then v(t) = v(0),

which is sufficient since for our vector field we have

∇ · f = − ∂2H

∂p∂q+∂2H

∂q∂p= 0

by the symmetry of second derivatives.The volume v(t) may be expressed as the integral

v(t) =

∫φt−t0 (D(t0))

dx =

∫D(t0)

det

[∂φt−t0i

∂xj

]dx (6.15)

by changing variables. Using the Taylor expansion (6.14), we can write[∂φt−t0i

∂xj

]= I + (t− t0)

[∂fi∂xj

]+O

((t− t0)2

).

It is true for any matrix A and scalar k that

det(I + hA) = 1 + h tr(A) +O(h2),

which is just a rewriting of Jacobi’s formula from matrix calculus. With this,the volume formula (6.15) becomes

v(t) =

∫D(t0)

[1 + (t− t0) tr

[∂fi∂xj

]+O

((t− t0)2

)]dx

=

∫D(t0)

[1 + (t− t0) (∇ · f) +O

((t− t0)2

)]dx.

Differentiating this, we conclude

dv

dt

∣∣∣∣t=t0

=

∫D(t0)

∇ · f dx (6.16)

The integrand is zero by premise, and so v(t) is constant in time: v(t) ≡ v(0).


Liouville’s theorem has many important consequences. For example, wenow know that for a Hamiltonian system there can be no asymptotically stableequilibrium points or asymptotically stable closed trajectories in phase space,since these would require that the density of phase curves to increase aroundsuch phenomena. We also have the following phenomenon.

Corollary 6.5. (Poincare’s recurrence theorem) Fix t ∈ R and D ⊂ R2n abounded region of phase space, and let φ := φt denote a phase flow group el-ement. Then for any positive measure set U ⊂ D there exists x0 ∈ U and apositive integer n such that φn(x0) ∈ U .

For bounded motion—as is the case for a conservative system with potentialenergy U(x) → +∞ as |x| → ∞, in which case a particle is confined to theregion D = (q, p) ∈ R2d : T + U ≤ E—this means that the system willreturn to an arbitrary vicinity of any given possible configuration (q, p) ∈ R2d

infinitely often, given enough time. If we were to open a connection between achamber of gas and a chamber of vacuum then the gas molecules will eventuallyall return to the initial chamber, seemingly in violation of the second law ofthermodynamics. Although it may appear that Poincare’s theorem contradictsLiouville’s, the time scales are often quite large—for a gas chamber it is longerthan the age of the universe—and so there is no conflict.

Proof. Since for a smooth Hamiltonian function Hamilton’s equations are asmooth system of ODEs, then the subsequent flow φ : R2n → R2n is injective byuniqueness of solutions and continuous by well-posedness. Liouville’s theoremtells us that the special form of this system that g is volume-preserving.

Consider the collection of sets U , φ(U), φ2(U), φ3(U), · · · ⊂ D. Since φ isvolume-preserving, all of these sets must have the same volume. D is boundedand thus has finite volume, and so it is impossible for all of these sets to bedisjoint. That is, there exists some distinct 0 < j < k such that φj(U)∩φk(U) 6=∅. Since φ is injective, this requires φk−j(U)∩U 6= ∅. Namely, we can pick somex0 ∈ U in this intersection, which gives φk−l(x0) ∈ U .

In fact, we can conclude that the set of points in U which do not return toU infinitely often has measure zero.

Together, a measure space X with a finite measure µ and a measurablefunction φ : X → X that is measure preserving (in the sense that µ(φ−1(A)) =µ(A) for all measurable A ⊂ X) constitute a measure preserving space, whichis the fundamental object of study in discrete dynamical systems.

6.4 Poisson bracket

For any two functions f and g of position q, momenta p, and time t we definethe Poisson bracket of f and g to be

f, g =

n∑k=1

(∂f

∂pk

∂g

∂qk− ∂f

∂qk

∂g

∂pk

). (6.17)

6.4. POISSON BRACKET 103

(Note that there is another popular convention which differs by a factor of −1.)It arises naturally as the time evolution of any function f(p, q, t); by the chainrule and Hamilton’s equations we have

df

dt=∂f

∂t+∑k

(∂f

∂qkqk +

∂f

∂pkpk

)=∂f

∂t+∑k

(∂f

∂qk

∂H

∂pk− ∂f

∂pk

∂H

∂qk

)=∂f

∂t+ H, f .

(6.18)

Consequently, a time-independent quantity f(q, p) is conserved exactly when itsbracket with the Hamiltonian H, f = 0 vanishes. Taking the quantity f to bea position or momentum coordinate, we see that Hamilton’s equations may bewritten as

qi =∂H

∂pi= H, qi , pi = −∂H

∂qi= H, pi . (6.19)

This perspective generalizes nicely to the Heisenberg formulation of quantummechanics, with the Poisson brackets replaced by operator commutators; anobservable without explicit time dependence form a Lax pair with the Hamil-tonian.

Next, we record some properties. It is easy to see that the Poisson bracketis bilinear (linear in both entries) and antisymmetric (f, g = −g, f) fromthe definition (6.17). The brackets also posses a Leibniz rule,

fg, h = f g, h+ g f, h , (6.20)

which can be seen by using the product rule for fg and expanding. They alsobehave nicely with the coordinates and momenta:

qi, qk = 0, pi, pk = 0, pi, qk = δik. (6.21)

Also, if f(p, q, t) is a function as before and g(p, q, t) ≡ G(f(q, p, t)), then theirbracket

f,G(f) =

d∑k=1

(∂f

∂pkG′(f)

∂f

∂qk− ∂f

∂qkG′(f)

∂f

∂pk

)= 0 (6.22)

vanishes by the chain rule.The Poisson bracket also satisfies the Jacobi identity :

f, g, h+ g, h, f+ h, f, g = 0, (6.23)

which we will now verify. Since f, g is a linear function where each termcontains a first derivative of f and g, then h, f, g (for example) is a linearfunction in which each term contains a second derivative. Let Dg(φ) = g, φand Dh(φ) = h, φ. Note that the first term of eq. (6.23) does not contain anysecond derivatives of f , and so all of the second derivatives of f are containedwithin

g, h, f+ h, f, g = g, h, f − h, g, f= Dg(Dh(f))−Dh(Dg(f)) = (DgDh −DhDg)f


To simplify notation and observe some cancellation, we write the operators Dg

and Dh as

Dg =∑j

(∂g

∂pj

∂

∂qj− ∂g

∂qj

∂

∂pj

)=∑k

ξk∂

∂xk

Dh =∑j

(∂h

∂pj

∂

∂qj− ∂h

∂qj

∂

∂pj

)=∑k

ηk∂

∂xk

where x = (q, p), ξ = (∂g∂p ,−∂g∂q ), and η = (∂h∂p ,−

∂h∂q ). Therefore

DgDh =∑k

ξk∂

∂xk

(∑l

ηl∂

∂xl

)=∑k,l

(ξk∂ηl∂xk

∂

∂xl+ ξkηl

∂2

∂xk∂xl

)

DhDg =∑k

ηk∂

∂xk

(∑l

ξl∂

∂xl

)=∑k,l

(ηk∂ξl∂xk

∂

∂xl+ ηkξl

∂2

∂xk∂xl

)In taking the difference of these, we see that the second terms in these lastequalities cancel and we are left with

DgDh −DhDg =∑k,l

(ξk∂ηl∂xk− ηk

∂ξl∂xk

)∂

∂xl

That is, all of the second derivatives of f cancel, leaving only first derivatives.Symmetrically, this must also be true for g and h, and since every term ofeq. (6.23) contains only terms containing exactly one second derivative, then allthe terms must cancel.

We will record one final property of the Poisson bracket.

Proposition 6.6. (Poisson’s theorem) If f(q, p) and g(q, p) are constants ofmotion for a Hamiltonian system, then their Poisson bracket f, g is also con-served.

Although powerful, it should be noted that the new quantity f, g may betrivial, especially considering that a system with n degrees of freedom can onlyhave up to 2n− 1 conserved quantities.

Proof. Since f is conserved and time-independent we have

0 =df

dt= H, f ,

and similarly for g. The Jacobi identity (6.23) yields

0 = H, f, g+ f, g,H+ g, H, f= H, f, g − f, H, g+ g, H, f= H, f, g .

6.5. CANONICAL TRANSFORMATIONS 105

Rearranging, we getd

dtf, g = H, f, g = 0,

and so f, g is an integral of the motion.

Example 6.7. If L1 = x2p3 − x3p2 and L2 = x3p1 − x1p3 are the first andsecond components of the angular momentum L = x × p of a particle x ∈ R3,then their bracket

L1, L2 = r2p3, r3p1+ r3p2, r1p3 = r2p1 − p2r1 = −L3

is the remaining component of the angular momentum.

6.5 Canonical transformations

For a system with n degrees of freedom, any choice of coordinates q = (q1, q2, . . . ,qn) is of course far from unique. In fact, we may choose any set of n independentquantities Qi(q, t), i = 1, . . . , n that describe the system’s state and Lagrange’sequations must still hold in terms of the new coordinates. For the Hamiltonianapproach, we would like to consider the more general transformation of (q, p) tothe 2n independent quantities Qi(q, p, t), Pi(q, p, t), and T (q, p, t) for i = 1, . . . , nso that on the extended phase space T ∗M × I, I ⊂ R the one-forms

p dq −H dt = P dQ−K dT + dS (6.24)

are equal, for some smooth functions K(Q,P, T ) and S(Q,P, T ). A transforma-tion satisfying eq. (6.24) is called canonical.

The definition (6.24) is made so that Hamilton’s equations

dQidT

=∂K

∂Pi,

dPidt

= − ∂K∂Qi

hold for the new variables and the new Hamiltonian K(Q,P, t) (sometimes re-ferred to as the Kamiltonian); in particular, if the canonical transformation istime-independent then we have K(Q,P ) = H(q, p). As remarked at the end ofsection 3.1, this is a consequence of the Lagrangians differing by a total timederivative:

p · q −H(q, p, t) = P · Q−K(Q,P, t) +d

dt[F (q(t), p(t), Q(t), P (t), t)] .

In physics the function F is called the generating function for the transforma-tion.

Remark. Some authors define canonical transformations as any diffeomor-phism which obeys Hamilton’s equations (6.5), and even falsely claim the equiv-alence of the two definitions (e.g. [LL76, Sec. 45]). The condition that a trans-formation preserves Hamilton’s equations (6.5) is much weaker—for exampleP = 2p, Q = q satisfies section 6.5 but not eq. (6.24)—but our definition hasdeeper geometric significance.


The first type of generating function we will consider is F ≡ F (q,Q, t).Applying the chain rule and rearranging eq. (6.24) yields

p · q − P · Q+K(Q,P, t)−H(q, p, t) =∂F

∂q· q +

∂F

∂Q· Q+

∂F

∂t.

Matching terms we find

pi =∂F

∂qi, Pi = − ∂F

∂Qi, K = H +

∂F

∂t. (6.25)

In this way, the generating function does indeed characterize the transformation.

Now consider a generating function of the form Φ ≡ Φ(q, P, t). Note that

d

dt

(d∑i=1

PiQi

)=

d∑i=1

PiQi +

d∑i=1

PiQi.

We add this to eq. (6.24) to solve for the remaining variables p and Q:

p · q −P · Q+ P ·Q+P · Q+K(Q,P, t)−H(q, p, t)

=d

dt[P ·Q+ F (q, P, t)] =

∂Φ

∂q· q +

∂Φ

∂P· P +

∂Φ

∂t

which requires that our generating function be of the form Φ(q, P, t) = P ·Q+F (q, P, t) and

pi =∂Φ

∂qi, Qi =

∂Φ

∂Pi, K = H +

∂Φ

∂t. (6.26)

Similar procedures yield the relations corresponding to generating functionsof p,Q or of p, P . Notice that in matching terms we need the 2n generatingfunction coordinates to be independent. For example, if the new coordinatesQi ≡ Qi(q, t) do not depend on the old momenta, then the coordinates Q andq are not independent, and so a generating function F (q,Q, t) would not work.Canonical transformations highlight the point that the Hamiltonian perspec-tive regards both q and p equally as independent coordinates. In particular,the transformation Qi = pi, Pi = −qi leaves the Hamiltonian formulation un-changed, which would not be legal in the Lagrangian framework.

Lastly, we can characterize time-independent canonical transformations withPoisson brackets. Statement (c) is often how a transformation is verified ascanonical in practice.

Proposition 6.8. For a time-independent transformation (q, p) → (Q,P ) thefollowing conditions are equivalent.

(a) The transformation is canonical, i.e. p dq = P dQ+ dS.

6.5. CANONICAL TRANSFORMATIONS 107

(b) The transformation preserves the Poisson bracket:

f, gp,q = f, gP,Q

for all smooth functions f and g on phase space, where the subscriptsdenote the respective variables of differentiation.

(c) The new variables satisfy the coordinate relations with respect to the oldPoisson brackets:

Qi, Qjp,q = 0, Pi, Pjp,q = 0, Pi, Qjp,q = δij .

Proof. (b) implies (c): This is immediate, since we can replace the differentiationvariables q, p with Q,P by assumption—for example,

Qi, Qjp,q = Qi, QjP,Q = 0.

(c) implies (b): By premise, we compute(∂P∂p

∂P∂q

∂Q∂p

∂Q∂q

)(0 −II 0

)(∂P∂p

∂Q∂p

∂P∂q

∂Q∂q

)=

(∂P∂p

∂P∂q

∂Q∂p

∂Q∂q

)(−∂P∂q −∂Q∂q∂P∂p

∂Q∂p

)

= −

[Pi, Pjp,q

]i,j

[Pi, Qjp,q

]i,j[

Qi, Pjp,q]i,j

[Qi, Qjp,q

]i,j

=

(0 −II 0

)

Plugging this in, we have

f, gp,q = −(∂f∂p

∂f∂q

)(0 −II 0

)(∂g∂p∂g∂q

)

= −(∂f∂P

∂f∂Q

)(∂P∂p

∂P∂q

∂Q∂p

∂Q∂q

)(0 −II 0

)(∂P∂p

∂Q∂p

∂P∂q

∂Q∂q

)(∂g∂P∂g∂Q

)

= −(∂f∂P

∂f∂Q

)(0 −II 0

)( ∂g∂P∂g∂Q

)= f, gP,Q

(a) implies (c): The time-independence requires that the new HamiltonianK(Q,P, t) = H(q, p, t) is the same as the old Hamiltonian. On the one hand,the chain rule and Hamilton’s equations insist

Qi =∂Qi∂q· q +

∂Qi∂p· p =

∂Qi∂q· ∂H∂p− ∂Qi

∂p· ∂H∂q

= H,Qip,q.

On the other hand, the new Hamilton’s equations (6.5) tell us that this shouldbe equal to

∂H

∂Pi=∂H

∂q· ∂q∂Pi

+∂H

∂p· ∂p∂Pi

.


Equating these, we have

∂Qi∂pj

= − ∂qj∂Pi

,∂Qi∂qj

=∂pj∂Pi

.

The analogous computation for Pi lead to the relations

∂Pi∂pj

=∂qj∂Qi

,∂Pi∂qj

= − ∂pj∂Qi

.

Together these are exactly the conditions of (c).

(c) implies (a): The same computation from the previous implication showsthat dp ∧ dq = dP ∧ dQ. Using Stokes’ theorem to pass to the boundary, wededuce that for any closed contour γ contractible to a point we have∮

γ

p dq =

∮γ

P dQ.

Therefore we conclude that the difference form p dq − P dQ is closed since itsantiderivative is well-defined.

6.6 Infinitesimal canonical transformations

A Hamiltonian flow can be thought of as a continuous canonical transformation.Consider a Hamiltonian system undergoing a canonical transformation

Qi = qi + δqi, Pi = pi + δpi, i = 1, . . . , d, (6.27)

which is infinitesimal in the sense that we will be sending δqi and δpi to zerolater to obtain a statement about derivatives. We expect the correspondinggenerating function to be a perturbation of that which generates the identitytransformation. It is the generating function

Φ0(q, P, t) =

n∑i=1

qiPi

which yields the identity transformation, since according to eq. (6.26) we have

pi =∂Φ0

∂qi= Pi, Qi =

∂Φ0

∂Pi= qi, K(Q,P ) = H(q, p).

Consequently, we seek a generating function for the transformation (6.27) of theform

Φ(q, P, t) =

n∑i=1

qiPi + εG(q, P ) (6.28)

6.6. INFINITESIMAL CANONICAL TRANSFORMATIONS 109

for some small parameter ε. We will be considering the first order behavior asε→ 0, and δqi and δpi will depend linearly on ε. By eq. (6.26), the function Φgenerates the transformation

pi =∂Φ

∂qi= Pi + ε

∂G

∂qi, Qi =

∂Φ

∂Pi= qi + ε

∂G

∂Pi= qi + ε

∂G

∂pi

∂pi∂Pi

.

The leading term of the derivative ∂pi∂Pi

= ∂∂Pi

(Pi − δpi) = 1 + O(δpi) is one,and with the factor of ε in front this derivative does not make a first ordercontribution. We made this change of variables so that we may take the functionG = G(q, p) to be dependent only on the old coordinates and momenta q, p.Solving for δqi and δpi in their definition (6.27), we get

δpi = Pi − pi = −ε∂G∂qi

+O(ε2), δqi = Qi − qi = ε∂G

∂pi+O(ε2). (6.29)

If we take the function G = H(q, p) to be the system’s Hamiltonian and let theparameter ε be an infinitesimal time interval δt, this becomes

δqi = δt∂H

∂pi+O(δt2), δpi = −δt∂H

∂qi+O(δt2). (6.30)

Taking δt→ 0 we recover Hamilton’s equations, which indicates that the Hamil-tonian is the instantaneous generator of the system’s motion with time.

We will now consider how the Hamiltonian is affected by this transforma-tion. Let u(q, p) be a smooth function, which is perturbed as the result of thetransformation (6.27) by

u(qi + δqi, pi + δpi) = u(qi, pi) + δu (6.31)

Taylor expanding, we have

δu =

d∑i=1

(∂u

∂qiδqi +

∂u

∂piδpi

)+O(δq2

i + δp2i )

=∑i

(∂u

∂qiε∂G

∂pi− ∂u

∂piε∂G

∂qi

)+O(ε2) = −ε u,G+O(ε2)

(6.32)

after plugging in for δqi and δpi using eq. (6.29). If we take the function u(q, p) =H(q, p) to be the Hamiltonian, then

dH

dε

∣∣∣∣ε=0

= −H,G . (6.33)

Since G is time-independent, then this quantity vanishes if and only if G isa constant of the motion, and so constants of the motion are instantaneousgenerators of those transformations which leave the Hamiltonian invariant. Thisis another appearance of Noether’s theorem (Proposition 3.12).


6.7 Canonical variables

For any system with bounded periodic motion there is always a change of vari-ables so that the motion is dictated by a linearly increasing angular variableand an energy parameter. In this section we will consider a conservative systemwith one degree of freedom and bounded motion. As in section 5.3 this analysisalso applies coordinate-wise to a system with periodic motion in each variable(but not necessarily jointly periodic motion).

Since the motion is bounded and necessarily periodic, we may consider theaction variable

I =1

2π

∮p dq (6.34)

which was first glimpsed in the period integral (2.4) and we later identified ineq. (5.20). The integral is evaluated over one period, around a loop in the phaseplane around which q returns to an extreme value once. We may also write

I =1

2π

∫∫dp dq (6.35)

where the double integral is evaluated over the region enclosed by the trajectory(q(t), p(t)) in phase space.

We would like to treat I as a momentum variable in order to obtain theHamiltonian and Hamilton’s equations in terms of this geometric quantity. Con-sider the generating function

Φ =

∫p(q, E) dq (6.36)

where E ≡ H(q(t), p(t)) is the constant energy along a trajectory and p(q, E)is implicitly determined. For our conservative system I ≡ I(E) is strictly afunction of the energy, and so the generating function Φ ≡ Φ(q, I) may beconsidered a function of I rather than E. The first equation of eq. (6.26) requiresthat

p =∂Φ

∂q(6.37)

which agrees with the fundamental theorem of calculus applied to the definition(6.36). The new coordinate—which we will denote by w—is dictated by thesecond equation of eq. (6.26):

w =∂Φ

∂I(6.38)

Together, w and I are called the canonical variables of the system, and individ-ually I is the action variable and w is the angle variable.

Since the generating function Φ(q, I, λ) does not explicitly depend on time,the last equation of eq. (6.26) tells us that the Hamiltonian is unaffected:

K = H +∂Φ

∂t= H, (6.39)

6.8. EXERCISES 111

and is just the old one in terms of the new variables. However, since the actionvariable I ≡ I(E) is determined by the energy then in the new variables H ≡E ≡ E(I). Hamilton’s equations are now

I = −∂H∂w

= 0, w =∂H

∂I=

dE

dI. (6.40)

The first equation is expected, since I is given by the system’senergy which isconstant for a conservative system. The latter can be easily integrated to obtain

w(t) =dE

dIt+ constant = ω(I)t+ constant, (6.41)

and so w is the phase of the oscillations.From their definitions (6.34) and (6.36) we can see that after each oscillation

the generating function Φ increments by 2πI. In turn, the equation of motion(6.41) tells us that the angle variable w increases by 2π with each oscillation,much like an actual angle coordinate. Conversely, the original variables q andp are unchanged after an integer number of periods.

6.8 Exercises

6.1 (Harmonic oscillator). The one-dimensional harmonic oscillator has thekinetic and potential energies

T = 12mv

2, U = 12kx

2.

Since this system is conservative, the Hamiltonian H = E is given by the totalenergy

H = T + U =1

2mp2 +

mω2

2x2

where ω2 = k/m.

(a) Directly apply Hamilton’s equations to obtain x + ω2x = 0, which hassolution

x(t) = x0 cos [ω(t− t0)]

as in Example 1.11. Hamilton’s equations may not make solving the mo-tion easier than using Lagrange’s or Newton’s equations, but rather pro-vide a new interpretation of the system’s motion.

(b) By considering the shape of trajectories in phase space, find the anglevariable I and show that the energy as a function of the angle variable Iis E(I) = ωI.

(c) Write down (but do not evaluate) the integral for the generating functionΦ, and show that the angle variable w is

w = tan−1

x√2Imω − x2

+ constant.


(d) By design, the angle variable w is cyclic for the new Hamiltonian E(I) =ωI. From the equality

w(t) =dE

dIt+ constant = ωt+ constant,

invert this equation to conclude

x(t) = x0 cos [ω(t− t0)]

as in part (a). This method is unwieldy in application, and is intended forthe purpose of physical understanding rather than solving for the equa-tions of motion.

6.2 (Charged particle in an electromagnetic field). In Exercise 3.8, we foundthe Lagrangian for a charged particle in R3 with charge q and mass m movingthrough an electromagnetic field to be

L = 12mv

2 − qφ+q

cv ·A

where φ and A are the scalar and vector potentials for the electric and magneticfields:

B = ∇×A, E = −∇φ− 1

c

∂A

∂t.

(a) Show that the Hamiltonian for this system is

H =1

2m

(p− q

cA)2

+ qφ.

(b) Although the electric and magnetic fields are uniquely expressed, thescalar and vector potentials that φ and A are not unique and they ap-pear explicitly in the Hamiltonian. In fact, substituting A′ = A+∇f forany smooth function f(x, y, z, t) leaves B unchanged since the curl of agradient is also zero. Show that for E to remain unchanged we need toalso substitute

φ′ = φ− 1

c

∂f

∂t.

Together, replacing (A, φ)→ (A′, φ′) is called a gauge transformation, anda specific pair (A, φ) is called a choice of gauge.

(c) Since the electric and magnetic fields are unaffected by the choice of gauge,then any physical laws in terms of the potentials should also be invariant.Show that under a gauge transformation the Hamiltonian becomes

H ′ = H − q

c

∂f

∂t,

and that Hamilton’s equations still hold in the new variables.

6.8. EXERCISES 113

6.3 (Young’s inequality). Show that for any Legendre transform pair f(x) andf∗(ξ) we have

x · ξ ≤ f(x) + f∗(ξ).

Apply this inequality to the function f(x) of Example 6.1.

6.4 (Example of Poisson’s theorem). Show that for the angular momentumL = x× p of a particle x ∈ R3 we have

L,L · n = L× n

for any unit vector n. Do so by picking a choice of Cartesian coordinates withrespect to n, writing down the infinitesimal canonical transformation (6.27)given by rotation about the n axis by an angle dθ, and apply the calculation(6.32) with u replaced by L. Conclude that Li, Lj = −Lk for i, j, k ∈ 1, 2, 3in cyclic order.


Chapter 7

Symplectic geometry

Symplectic geometry is the differential geometric generalization of the time-independent Hamiltonian structure of phase space to general manifolds, andthe physical results for conservative systems from the previous chapter can belifted to this perspective. The material for this section is based on [Arn89,Ch. 8] chapter 8 and [Lee13, Ch. 22], and we will try to meld the physics andmathematical notations.

7.1 Symplectic structure

Let M be a smooth even-dimensional manifold of dimension 2n. A symplecticform or symplectic structure on M is a two-form ω on M that is closed (dω =0) and nondegenerate: for each x ∈ M the map TxM → T ∗xM which takesξ 7→ ξyω = ω(ξ, ·) is invertible. The pair (M,ω) is called a symplectic manifold.There are other more easily verifiable criteria for a two-form ω (cf. Exercise 7.1),but conceptually this is the classification we will rely upon. Heuristically, thesymplectic form ω is an identification of the tangent and cotangent spaces ateach point in M , which bears resemblance to Riemannian structure but as weshall see exhibits drastically different behavior.

Example 7.1. Consider the cotangent bundle T ∗N of a n-dimensional manifoldN , or simply the Euclidean space T ∗Rn = Rnq × Rnp . Then the two-form

ω = dp ∧ dq =

n∑i=1

dpi ∧ dqi (7.1)

is a symplectic form on T ∗N , and is called the canonical symplectic form onthe cotangent bundle; we will later see how this form naturally encapsulatesHamilton’s equations. This form is closed, since it is the exterior derivative ofthe one-form

p dq =

n∑i=1

pi dqi, (7.2)

115

116 CHAPTER 7. SYMPLECTIC GEOMETRY

which is in turn called the tautological one-form on T ∗M ; we first saw thisform in the period integral (2.4) and then again later in defining the actionvariable (5.20). To see that ω is nondegenerate, fix (q, p) ∈ T ∗N and let(v1, . . . , vn, a1, . . . , an) ∈ T(q,p)(T

∗N) denote the dual basis to (dq,dp) so that

dqi(vj) = δij , dpi(aj) = δij , dqi(aj) = dpi(vj) = 0

for i, j = 1, . . . , n. Often in differential topology we write ∂qi for the dual vectorto dqi, but physically we like to think of the tangent vector to a point in phasespace as the velocity and (mass-times-)acceleration. The action of ω on thesebasis vectors is

ω(vi, vj) = ω(ai, aj) = 0, ω(vi, aj) = −ω(aj , vi) = δij

for i, j = 1, . . . , n. To check nondegeneracy, fix ξ ∈ T(q,p)(T∗N) and suppose we

have ω(ξ, η) = 0 for all η ∈ T(q,p)(T∗N). Expanding ξ =

∑ni=1(bivi + ciai) for

constants bi, ci ∈ R, we see that

0 = ω(ξ, vi) = −ci, 0 = ω(ξ, ai) = bi

and so ξ = 0 as desired.

In fact, the form of Example 7.1 is the fundamental example in the followingsense.

Theorem 7.2 (Darboux’s theorem). Let (M,ω) be a 2n-dimensional symplecticmanifold. For any ξ ∈M there exists local coordinates (q, p) centered at ξ withrespect to which ω has the representation (7.1).

For a proof see [Lee13, Th. 22.13].

Remark. Symplectic manifolds are automatically orientable because ωn is anonvanishing 2n form on M . Indeed, if we fix x ∈M and write ω =

∑i dpi∧dqi

at x, then the n-fold wedge product

ωn =∑I

dpi1 ∧ dqi1 ∧ · · · ∧ dpid ∧ dqid

where the sum ranges over all multi-indices I = (i1, . . . , in) of length n. Anyterm where I contains a repeated index is zero because dqi ∧ dqi = 0, and since2-forms commute under the wedge product then we can rewrite this as

ωd = d! dp1 ∧ dq1 ∧ · · · ∧ dpd ∧ dqd.

This is nonvanishing at x by definition of the nondegeneracy of ω.

7.2 Hamiltonian vector fields

Next let us see how to use the symplectic structure to obtain the familiar dynam-ics from a Hamiltonian. Let (M,ω) be a symplectic manifold and H : M → R

7.2. HAMILTONIAN VECTOR FIELDS 117

be a smooth function. Then dH is a differential one-form which associates acovector to each point in M . On the other hand, for Hamilton’s equations weneed to specify a vector field for the right-hand side of the differential equation.By definition, for each x ∈ M the map TxM → T ∗xM which takes ξ 7→ ξyω isinvertible, and so let J : T ∗xM → TxM denote its inverse. Then J dH is a vectorfield on M , called the Hamiltonian vector field associated to the HamiltonianH. The induced flow etJ dH : M →M , x0 7→ x(t) defined to solve the ODE

x(t) = J dH(x(t)), x(0) = x0 (7.3)

is called a Hamiltonian flow on M . In differential geometry the notation ω :TxM → T ∗xM is used for the map ξ 7→ ξyω, and so in place of J the notationsω−1 and

XH := ω−1(dH) ⇐⇒ XHyω = dH (7.4)

are often used.

Example 7.3. Consider the canonical symplectic form (7.1) on Euclidean phasespace T ∗Rn = Rnq × Rnp . On Euclidean space we already have a natural identi-fication of vectors and covectors via the dot product, and we can express J asa linear transformation in terms of this identification. Given a vector field

X =

n∑i=1

(bi∂

∂qi+ ci

∂

∂pi

)for some smooth coefficients bi and ci, we compute

Xyω =

n∑i=1

(ci dqi − bi dpi

)=

(c−b

)= J

(bc

)and hence

J =

(0 I−I 0

)(7.5)

for I the identity matrix. Writing

dH =∂H

∂qdq +

∂H

∂pdp =

n∑i=1

(∂H

∂qidqi +

∂H

∂pidpi)

=

(∂H∂q∂H∂p

),

we have (qp

)= J dH =

(∂H∂p

−∂H∂q

)which agrees with Hamilton’s equations (6.4).

Symplectic geometry is the generalization of time-independent Hamiltoniandynamics since the Hamiltonian function H is automatically conserved underits Hamiltonian flow.


Proposition 7.4 (Conservation of energy). Let (M,ω) be a symplectic manifoldand let H be a smooth function on M . Then H is constant along the integralcurves of J dH, and when dH 6= 0 the vector field dH is tangent to the levelsets of H.

Proof. Both are a consequence of

(J dH)(H) = dH(J dH) = ω(J dH,J dH) = 0,

which holds since ω is alternating.

7.3 Integral invariants

Consider a smooth diffeomorphism g : M → M . A k-form η is said to be anintegral invariant for g if ∫

g(N)

η =

∫N

η (7.6)

for all orientable k-dimensional submanifolds N with boundary.

Theorem 7.5. A Hamiltonian flow gt = etJ dH preserves the symplectic struc-ture: g∗t ω = ω. In other words, the symplectic form ω is an integral invariantof gt.

Proof. The map gt is smoothly homotopic to the identity via the family ofdiffeomorphisms gs, s ∈ [0, t], in the sense that at time s = 0 the map g0 :M → M is the identity, and at time s = t the map gt : M → M is whatwe are given. Fix N a smooth orientable 2-dimensional submanifold, and letΩN := gs(N) : s ∈ [0, t] denote the image of N under the homotopy. We canthink of ωN as an orientable 3-manifold in [0, t] ×M or as being immersed inM . With this choice of orientation we have

∂ΩN = gt(N) ∪ (−N) ∪ (−Ω∂N ). (7.7)

We claim that for any smooth curve γ in M we have

d

dt

∫Ωγ

ω =

∫gt(γ)

dH, (7.8)

where H is the Hamiltonian for the flow gt. Let φ : [a, b]→M be a parametriza-tion of γ. Then Ωγ is parameterized by Φ(s, x) := gs(φ(x)) and we have∫

Ωγ

ω =

∫ t

0

∫ b

a

ω

(∂Φ

∂x,∂Φ

∂s

)dxds

where ∂Φ∂s (s, x) is in TΦ(s,x)M . By definition of the Hamiltonian phase flow

gt = etJ dH the tangent vector ∂Φ∂s points in the direction of J dH, and so we

have ω(∂Φ∂x ,

∂Φ∂s ) = dH(∂Φ

∂x ). Therefore∫Ωγ

ω =

∫ t

0

∫ b

a

dH

(∂Φ

∂x

)dxds =

∫ t

0

(∫gs(γ)

dH

)ds,

7.3. INTEGRAL INVARIANTS 119

and the identity (7.8) follows from the fundamental theorem of calculus.For a closed curve like ∂N we note that∫

gt(∂N)

dH =

∫gt(∅)

H = 0,

and so the identity (7.8) implies ∫Ω∂N

ω = 0. (7.9)

Since ω is closed by definition, then by Stokes’ theorem we have

0 =

∫ΩN

dω =

∫∂ΩN

ω.

Decomposing the boundary ∂ΩN according to (7.7) we obtain

0 =

∫gt(N)

ω −∫N

ω −∫

Ω∂N

ω.

From (7.9) we know the last integral vanishes, and so we conclude that ω is anintegral invariant of gt.

In the previous section we saw that ωn defines a volume form on M and sowe immediately obtain an analog of Theorem 6.4.

Corollary 7.6 (Liouville’s theorem). Each of the forms ω2, ω4, . . . is preservedby a Hamiltonian flow. In particular, every Hamiltonian flow preserves thevolume form ωn.

Recall from section 6.5 that on a cotangent bundle M = T ∗Q with canonicalsymplectic form ω = dp ∧ dq, a (time-independent) canonical transformationg : T ∗Q→ T ∗Q satisfies

g∗(p dq) = p dq + dS. (7.10)

This definition does not make p dq an integral invariant for a canonical trans-formation, since the condition (7.6) only holds for closed curves N . Instead, weemploy the following useful observation.

Proposition 7.7. Let g : M → M be a smooth diffeomorphism. If η is a k-form such that (7.6) holds for only closed orientable k-submanifolds N , thendη is an integral invariant.

Proof. Let N be an orientable (k + 1)-submanifold. Then by Stokes’ theoremwe have ∫

N

dη =

∫∂N

η,

∫g(N)

dη =

∫∂g(N)

η =

∫g(∂N)

η.

Since ∂N is closed then these two right-hand sides are equal by premise. There-fore we conclude that (7.6) holds for the form dη, and hence dη is an integralinvariant.


Noting that d(pdq) = dp ∧ dq, we conclude:

Corollary 7.8. Canonical transformations preserve the symplectic form ω andthe volume form ωn.

Remark. In section 6.6 we saw that for Euclidean phase space T ∗Rn Hamil-tonian flows are canonical transformations. However, the converse of Proposi-tion 7.7 is not true in general, and so Theorem 7.5 does not imply the iden-tity (7.10). We must assume that M is simply connected in order for Hamilto-nian flows are canonical transformations, since then Theorem 7.5 implies (7.10).

7.4 Poisson bracket

In section 6.4 we saw how to phrase Hamiltonian dynamics in terms of thePoisson bracket, and in this section we will see how this notion manifests in termsof symplectic structure. For smooth functions f, g ∈ C∞(M) on a symplecticmanifold (M,ω) we define the Poisson bracket of f and g to be

f, g = ω(J df, J dg) = dg(J df) = (J df)(g). (7.11)

As in (6.18) the Poisson bracket H, f is also the evolution of the quantity funder the Hamiltonian flow H, since the Lie derivative of the function f alongJ dH is given by

LJ dHf =d

dt

∣∣∣∣t=0

f etJ dH = (J dH)(f) = H, f (7.12)

according to the definition (7.3) of the Hamiltonian flow. In particular, we againhave that f is conserved by the Hamiltonian flow of H if and only if H, f = 0.

Example 7.9. Let us check that the new definition (7.11) agrees with thephase space definition (6.17) on Euclidean space T ∗Rn = Rnq × Rnp . Using thecalculation of J from Example 7.3 we have

J df = J

(∂f

∂qdq +

∂f

∂pdp

)=

(0 I−I 0

)(∂f∂q

∂f∂p

)=∂f

∂p

∂

∂q− ∂f

∂q

∂

∂p.

Consequently, the definition (7.11) yields

f, g = (J df)(g) =∂f

∂p

∂g

∂q− ∂f

∂q

∂g

∂p,

which agrees with the first definition (6.17).

The properties listed in section 6.4 resemble those of the commutator becausethe Poisson bracket at the level of functions corresponds to the commutator ofthe respective Hamiltonian vector fields.

7.5. TIME-DEPENDENT SYSTEMS 121

Proposition 7.10. If (M,ω) is a symplectic manifold, then the Poisson bracketis the Hamiltonian of the commutator of the corresponding Hamiltonian vectorfields:

J df, g = [J df, J dg]. (7.13)

In particular, the vector space C∞(M) is a Lie algebra under the Poissonbracket: the Poisson is bilinear, antisymmetric, and satisfies the Jacobi identity.

Proof. In fact, it suffices to prove that the Poisson bracket satisfies the Jacobiidentity

f, g, h+ g, h, f+ h, f, g = 0 (7.14)

for arbitrary smooth functions f, g, h ∈ C∞(M). This is sufficient because wecan use the antisymmetry of ω to write

f, g, h = f, g, h − g, f, h,

and recognizing the Poisson bracket as a Lie derivative via (7.12) we conclude

LJ df,gh = LJ dfLJ dgh− LJ dgLJ dfh = L[J df,J dg]h.

Since h ∈ C∞(M) is arbitrary, the identity (7.13) must hold.It remains to demonstrate the Jacobi identity (7.14). The left-hand side

of (7.14) is a combination of terms which each have at least one second orderderivative. Isolating the terms containing second order derivatives of f andrecognize the Poisson bracket as a Lie derivative via (7.12), we have

f, g, h+ h, f, g = h, g, f − g, h, f= LJ dhLJ dgf − LJ dgLJ dhf.

However, we know LJ dhLJ dg−LJ dgLJ dh = L[J dh,J dg] is only a first order dif-ferential operator, and so we conclude that the terms with second order deriva-tives of f vanish. Lastly, since the left-hand side of (7.14) is symmetric in f ,g, and h, we conclude that there can be no second order derivatives and hencemust vanish.

7.5 Time-dependent systems

Recall that Hamilton’s equations (6.4) remain valid for time-dependent Hamil-tonian systems, while so far we have only allowed H to be a smooth function onphase space M . This section is dedicated to extending the symplectic geometrydeveloped thus far to time-dependent Hamiltonian systems.

Let (M,ω) be a 2n-dimensional symplectic manifold, and define the extendedphase space M×R. Given a possibly time-dependent Hamiltonian H, we definethe Poincare–Cartan one-form locally in terms of the canonical coordinates(q, p) on M guaranteed by Theorem 7.2 as

τ = p dq −H dt =

n∑i=1

pi dqi −H dt. (7.15)


Note that the first term above is the tautological one-form p dq on M , thedifferential of which yields the symplectic form ω. This is the form that weinsisted be preserved by a canonical transformation in eq. (6.24), and the samenotion of canonical transformations holds on a general extended phase spaceM × R in terms of the canonical coordinates (q, p, t).

Remark. The extended phase space M×R together with the Poincare–Cartanone-form τ do indeed define a contact manifold, but as we will see in the nextchapter it is not the natural contact extension of M since τ depends on thesystem’s Hamiltonian H.

On M × R we define the extended Hamiltonian vector field

YH = XH +∂

∂t, (7.16)

where XH(t) is the Hamiltonian vector field on M × t defined by (7.3). Inanalogy with the second condition of (7.4), the vector field (7.16) is the uniquesolution to

YHydτ = 0. (7.17)

The flow of YH is given by

qi =∂H

∂pi, pi = −∂H

∂qi, t = 1, (7.18)

which is just Hamilton’s equations (6.4) joined with the trivial equation t = 1. Itfollows that any smooth time-dependent function f on M ×R evolves accordingto

df

dt= H, fp,q +

∂f

∂t,

as was the case in section 6.4. In particular, a time-dependent Hamiltonian His no longer conserved under its own flow.

Example 7.11. [Cal41, Kan48] showed that in addition to naturally occur-ring time-dependent Hamiltonian systems, this framework can also be appliedto some systems which are time-independent and dissipative by introducing ar-tificial time dependence into the Hamiltonian. Consider the one-dimensionalHamiltonian system

H(Q,P, t) := e−γtP 2

2m+ eγtU(Q) (7.19)

on the extended phase space RQ ×RP ×Rt, where γ ≥ 0 is a constant and thecoordinatesQ,P are related to the physical coordinates q, p by the non-canonicaltransformation

P = eγtp, Q = q.

Then the equations of motion (7.18) yield

mq +mγq + U ′(q) = 0.

This represents a Newtonian system with a friction force that depends linearlyon the velocity, like the damped harmonic oscillator of Example 1.13.

7.6. LOCALLY HAMILTONIAN VECTOR FIELDS 123

7.6 Locally Hamiltonian vector fields

In section 7.1 we called a vector field V on a symplectic manifold (M,ω) Hamil-tonian if it is equal to J dH for some smooth function H ∈ C∞(M), and inTheorem 7.5 we saw that ω is necessarily invariant under the flow of J dH.In fact, any smooth vector field V on M whose flow leaves ω invariant (i.e.(etV )∗ω = ω) is called symplectic. However, a vector field being Hamiltonianis a global condition—in the sense that the corresponding Hamiltonian H mustextend smoothly to all of M—while being symplectic is a pointwise conditionand is hence only local.

Consequently, a smooth vector field V on M is called locally Hamiltonian (asopposed to globally Hamiltonian) if for each point p there exists a neighborhoodon which V is Hamiltonian. As an extension of Theorem 7.5, locally Hamiltonianvector fields are exactly those which are symplectic.

Proposition 7.12. Let (M,ω) be a symplectic manifold. A smooth vector fieldV on M is symplectic if and only if it is locally Hamiltonian. If M is also simplyconnected, then every locally Hamiltonian vector field is globally Hamiltonian.

Proof. Differentiating the condition (etV )∗ω = ω with respect to t at t = 0, wesee that a vector field V is symplectic if and only if LV ω = 0. From Cartan’smagic formula we have

LV ω = d(V yω) + V ydω = d(V yω)

since ω is closed. Therefore V is symplectic if and only if V yω is closed.If V is locally Hamiltonian, then in a neighborhood of any point there exists

a function f so that V = J df , and hence V yω = df which is certainly closed.Conversely, if V is a symplectic vector field, then V yω is closed and hence exactin a neighborhood of any point, and writing V yω = df we deduce that V = J dfon a neighborhood as desired.

Now suppose M is also simply connected. Then every closed one form isexact, and so V yω is closed if and only if V yω = df and V = J df for a smoothfunction f defined on all of M .

A smooth vector field V is called an infinitesimal symmetry of the Hamil-tonian H if both ω and H are invariant under the flow etV of V . The secondcondition (etV )∗H = H can be recast as V (H) = 0, as can be seen by differen-tiating H etV = H with respect to t at t = 0. With this notion, we can provethe following analogue of Proposition 3.12.

Proposition 7.13 (Noether’s theorem). Let (M,ω) be a symplectic manifoldand H a fixed Hamiltonian. If f is a conserved quantity, then the Hamiltonianvector field J df is an infinitesimal symmetry. Conversely, if M is also simplyconnected then each infinitesimal symmetry is the Hamiltonian vector field of aconserved quantity, and the quantity is unique up to the addition of a functionthat is constant on each component of M .


Proof. First suppose f is a conserved quantity. Then from the identity (7.12)we know that H, f = 0. However, since the Poisson bracket is antisymmetric,we deduce that 0 = f,H = (J df)(H) as well. This demonstrates that H isconserved by J df and from Theorem 7.5 we know that ω is also conserved byJ df , and hence J df is an infinitesimal symmetry.

Now suppose that M is simply connected and V is an infinitesimal symmetry.Then V is symplectic by definition, and by Proposition 7.12 we know V isglobally Hamiltonian. Writing V = J df , since H is conserved by f then we have0 = (J df)(H) = f,H = −H, f, and therefore f is a conserved quantity. Ifg is another function with J dg = V = J df , then by definition of J we haved(f − g) = (J df − J dg)yω = 0 and hence f − g is constant on the componentsof M .

7.7 Exercises

7.1. Show that the following criteria for a nondegenerate two-form ω are equiv-alent.

(a) For each x ∈ M the map TxM → T ∗xM which takes ξ 7→ ω(ξ, ·) is invert-ible.

(b) For each x ∈ M and ξ ∈ TxM nonzero there exists η ∈ TxM such thatω(ξ, η) 6= 0.

(c) The local matrix representation of ω in terms of some (hence every) basisis invertible.

7.2. (a) Show that rotation on the two-dimensional sphere S2 is a Hamilto-nian flow.

(b) Show that translations gt(q, p) = (q+ t, p) on the torus R2/Z2 is a locallyHamiltonian flow, but is not globally Hamiltonian.

7.3. Let (M,ω) be a 2n-dimensional compact symplectic manifold.

(a) Show that the n-fold wedge product ωn is not exact.

(b) Show that the de-Rham cohomology groups H2kdR(M) are nontrivial for

k = 1, . . . , n.

(c) Conclude that the two-dimensional sphere S2 is the only sphere that ad-mits a symplectic structure.

7.4. (Real symplectic matrices) A linear transformation R2n → R2n is calledsymplectic if it preserves the canonical symplectic form (7.1). The set of all sym-plectic matrices forms a group with matrix multiplication and is often denotedby Sp(2n,R).

7.7. EXERCISES 125

(a) Prove that the determinant of any symplectic matrix is equal to 1.

(b) Show that a matrix A is symplectic if and only if ATJA = J for the matrixJ defined in (7.5).

(c) Show that the Jacobian of a canonical transformation (7.10) is symplec-tic. (Note that this calculation implicitly appears in the proof of Propo-sition 6.8.)

7.5. (Symplectic and complex structure) Identify Euclidean phase space R2n =Rnq × Rnp with the complex space Cn via zj := qj + ipj .

(a) Show that matrix multiplication by matrix J (7.5) on R2n correspondsto multiplication by −i on Cn. This is why the matrix (7.5) is called J ,and in fact some authors (e.g. [Arn89]) choose the opposite sign for thecanonical symplectic form (7.1) and use the notation I in place of J .

(b) Show that the Hermitian inner product (z, w) 7→∑zjwj on Cn corre-

sponds to (ξ, η) 7→ ξ · η + i ξ · Jη on R2n. In other words, the scalarproduct and symplectic product are the real and imaginary parts of theHermitian inner product, respectively.

7.6. (Hamiltonian PDE [Gar71]) The Hamiltonian mechanics we have devel-oped thus far originates from the time derivative of trajectories in phase spaceand hence provides a class of ODEs on a finite dimensional manifold; however,this can be extended to PDE when we consider the time derivative of trajec-tories in an infinite dimensional function space. Given a smooth functional Fof, say, smooth real-valued periodic functions C∞(R/Z) we denote the Frechetderivative kernel by δF/δq(x), so that

dF∣∣q(f) =

d

ds

∣∣∣∣s=0

F (q + sf) =

∫δF

δq(x)f(x) dx.

The Frechet space C∞(R/Z;R) will now take the role of the manifold M . Specif-ically, for two smooth functionals F (q) and G(q) on C∞(R/Z) we define thePoisson bracket

F,G =

∫ (δF

δq

)′(x)

δG

δq(x) dx.

In other words, we have replaced the dot product on the Euclidean phase spaceR2n and the matrix J (7.5) with the L2 real inner product and the skew-symmetric operator J = ∂

∂x . Given a smooth functional H(q) on C∞(R/Z),we define the associated Hamiltonian PDE

∂q

∂t=

∂

∂x

(∂H

∂q

)in the spirit of (7.3) and (7.12).

It turns out that many of the results of finite dimensional Hamiltonian me-chanics have analogs in this infinite dimensional setting. For example, the global


minimum of a classical Hamiltonian is a (Liapunov) stable equilibrium, and theglobal minimum of a Hamiltonian functional is orbitally stable (i.e. concentra-tion compactness).

(a) Check that this Poisson bracket is bilinear, antisymmetric, and satisfiesthe Jacobi identity.

(b) Given a Hamiltonian functional H(q), show that a smooth functional F (q)is constant for solutions q to the PDE associated to H if and only ifH,F = 0 as a functional on C∞(R/Z).

(c) Show that for any Hamiltonian functional H(q) both the Hamiltonian Hand the mass functional

M(q) :=

∫ 1

0

q(x) dx

are automatically conserved for solutions q to the PDE associated to H.

(d) Show that the momentum functional

P (q) := 12

∫ 1

0

q(x)2 dx

does indeed generate translations, in the sense that the solution to theassociated PDE with initial data q(0, x) = f(x) is q(t, x) = f(x+ t).

(e) The Korteweg–de Vries (KdV) equation is the PDE associated to theHamiltonian

HKdV(q) :=

∫ 1

0

(12q′(x)2 − q(x)3

)dx,

and arises as the long-wavelength and shallow-water limit for unidirec-tional water waves of height q from the undisturbed water level q = 0.Show that the mass M(q), momentum P (q), and energy HKdV(q) are allconstant for solutions q to KdV. In fact, these are the first three of aninfinite hierarchy of conserved quantities for KdV.

Chapter 8

Contact geometry

Just as symplectic geometry extends the structure of conservative Hamiltoniandynamics on phase space, contact geometry is the natural generalization ofnonconservative dynamics on the product of phase space with the time axis.The material for this chapter is based on [BCT17] and [Lee13, Ch. 22].

8.1 Contact structure

A contact manifold is a smooth (2n+ 1)-dimensional manifold M paired with acontact form η. A contact form η is a one-form required to satisfy the followingnondegeneracy condition: for each x ∈M the restriction of dηx to the subspaceker(ηx) ⊂ TxM is nondegenerate (i.e. dηx is a symplectic tensor for all x ∈M).The rank-2n distribution N ⊂ TM satisfying Nx = ker(ηx) for each x ∈ M iscalled a contact structure on M ; it plays a fundamental role and is sometimestaken in the literature to be the defining geometric concept instead of η.

Similar to how the symplectic form nondegeneracy condition is equivalent tothe nonvanishing of the n-fold wedge product (ω)n, this nondegeneracy conditionturns out to be equivalent to

η ∧ (dη)n 6= 0; (8.1)

the proof of this equivalence is Exercise 8.1. Consequently, η ∧ (dη)n defines avolume form on M , and so in particular M must be orientable. Although thecondition (8.1) is sometimes easily verifiable in practice, we will conceptuallybe relying on the first condition.

Example 8.1. On the Euclidean contact space RS × T ∗Rn = RS × Rnq × Rnpwe have the canonical contact form

η = dS − p dq = dS −n∑i=1

pi dqi. (8.2)

127

128 CHAPTER 8. CONTACT GEOMETRY

Note that this is the combination of dS and the tautological one-form (7.2) onT ∗Rn. A straightforward computation shows that η ∧ (dη)n = dS ∧ dq ∧ dp isthe Euclidean volume form on Rt × T ∗Rn, but we can also check that dη is asymplectic tensor. Note that dη = −dp ∧ dq, and so the rank-2n distributionN ⊂ TR2n+1 annihilated by η is spanned by the vector fields

Xi =∂

∂qi+ pi

∂

∂S, Yi =

∂

∂pi

for i = 1, . . . , n. Moreover, we have

dη(Xi, Xj) = 0, dη(Yi, Yj) = 0, dη(Xi, Yj) = δij

for i, j = 1, . . . , n, and it follows that dη|N is nondegenerate as in Example 7.1.

This example is again fundamental in the following sense.

Theorem 8.2 (Contact Darboux theorem). If (M,η) is a (2n+1)-dimensionalcontact manifold, then for each x ∈ M there exist local coordinates (q, p, S)centered at x with respect to which η has the representation (8.2).

For a proof see [Lee13, Th. 22.31].The contact structure automatically induces an associated vector field called

the Reeb field, which heuristically points orthogonally to the distribution N andplays the role of S-axis.

Proposition 8.3 (The Reeb field). If (M,η) is a contact manifold, then thereexists a unique smooth vector field ξ on M called the Reeb field satisfying

ξydη = 0, η(ξ) = 1. (8.3)

Proof. The map Φ which takes X 7→ Xydη defines a smooth bundle homo-morphism Φ : TM → T ∗M , and for each x ∈ M it reduces to a linear mapΦx : TxM → T ∗xM . Since dηx restricted to the subspace Nx is nondegenerateby definition, then Φx|Nx is injective and hence Φx has rank at least 2n. On theother hand, we know that Φx cannot have rank 2n+ 1 because then dηx wouldbe nondegenerate and contradict that TxM is odd-dimensional. Therefore, weconclude that ker Φx is one-dimensional. Moreover, since ker(Φx) is not con-tained in Nx = ker(ηx) by definition, we know there exists a unique ξ ∈ ker(Φx)with ηx(ξx) = 1; these correspond to the two conditions (8.3) respectively.

The smoothness of ξ follows from the smoothness of η. Note that ker Φ ⊂TM is a smooth rank-one subbundle, and so around any x ∈ M we can pick asmooth nonvanishing section X of ker Φ near x. Since η(X) 6= 0, then we canwrite ξ = η(X)−1X as a composition of smooth maps near x.

Example 8.4. For the Euclidean contact space of Example 8.1, we see thatReeb field is

ξ =∂

∂S

as the two conditions (8.3) are easily verified.

8.2. HAMILTONIAN VECTOR FIELDS 129

8.2 Hamiltonian vector fields

Given a smooth function H on a contact manifold (M,η), the associated contactHamiltonian vector field XH is uniquely determined by the two conditions

η(XH) = H, (XHydη)|N = −dH|N . (8.4)

Since dη|N is nondegenerate by definition, there is a unique vector field Y onN satisfying the second condition of (8.4). The vector field XH := Y + Hξ isthen the unique solution of the conditions (8.4).

In comparison to symplectic Hamiltonian vector fields, the first conditionof (8.4) looks like the primitive of ω(XH) = dH (which we wrote as XH =J dH) and the second condition is like XHyω = dH; in the symplectic casethese conditions were redundant (cf. (7.4)), but now we need both in order todetermine XH on and off of the kernel Nx of ηx.

Example 8.5. Let us see what the contact Hamiltonian vector field XH lookslike for the Euclidean contact space of Example 8.1 (and hence also the ex-pression of XH in the local coordinates guaranteed by Theorem 8.2). Given asmooth function H(s, q, p) it is easily verified that

XH =

n∑i=1

(pi∂H

∂pi−H

)∂

∂S+∂H

∂pi∂

∂qi−(∂H

∂qi+ pi

∂H

∂S

)∂

∂pi(8.5)

satisfies the two conditions (8.4), from which we obtain the differential equationsystem

dS

dt=

n∑i=1

pi∂H

∂pi−H, dqi

dt=∂H

∂pi,

dpi

dt= −∂H

∂qi− pi ∂H

∂S. (8.6)

When H is independent of s, the second two sets of equations reduce to Hamil-ton’s equations of motion (6.4). We also recognize the quantity S as the actionS(q, t); indeed, if we use the formula (5.1) for the momentum and use theHamilton–Jacobi equation (5.4) we see that the time derivative of the action is

dS

dt=

n∑i=1

∂S

∂qidqi

dt+∂S

∂t=

n∑i=1

pidqi

dt−H.

To conclude the section, let us consider a one-dimensional physical examplefor to illustrate how contact geometry encapsulates time-independent noncon-servative dynamics.

Example 8.6. Take n = 1 in the Euclidean contact space of Example 8.1, andconsider the Hamiltonian

H(q, p, S) =p2

2m+ U(q) + γS (8.7)


where γ is a real constant. The contact Hamilton’s equations (8.6) read

q =p

m, p = −U ′(q)− γp, S =

p2

2m− U(q)− γS.

This represents a Newtonian system with a friction force that depends linearlyon the velocity, like the damped harmonic oscillator of Example 1.13. Note thatas opposed to Example 7.11, in this approach the momentum coordinate stillcoincides with the physical momentum defined via the velocity.

8.3 Dynamics

Using the expression (8.5) of a contact Hamiltonian vector field XH in termsof the local coordinates (q, p, S) of Theorem 8.2, we see that a smooth functionF (q, p, S) on a contact manifold (M,η) evolves according to

dF

dt= XH(F )

= −H∂F

∂S+

n∑i=1

pi[∂H

∂pi∂F

∂S− ∂H

∂S

∂F

∂pi

]+

n∑i=1

[∂H

∂pi∂F

∂qi− ∂H

∂qi∂F

∂pi

]

= −H∂F

∂S+

n∑i=1

piH,Fpi,S + H,Fp,q.

(8.8)

The last term above we recognize from (6.18) and the first two terms are cor-rections introduced by the contact structure. The notation H,Fpi,S shouldbe interpreted as a convenient shorthand defined via the above formula with nodeeper meaning.

Taking F = H(q, p, S) to be the Hamiltonian for the vector field XH in theformula (8.8) above, we see that the Hamiltonian evolves according to

dH

dt= −H∂H

∂S. (8.9)

From this we see that H is a conserved quantity if and only if H is independentof S or H = 0. In particular, for a conservative Hamiltonian H ≡ H(q, p) werecover the conservation of the Hamiltonian. However, in general the rate ofdecrease of the Hamiltonian is proportional to the system’s energy H and itsdissipation ∂H/∂S.

Specifically, let us consider a Hamiltonian of the form

H(q, p, S) = H0(q, p) + f(S) (8.10)

where H0(q, p) is the mechanical energy of the system (e.g. H0(q, p) = p2/2m+U(q)), then according to the formula (8.8) the mechanical energy obeys

dH0

dt= −

n∑i=1

pi∂H0

∂pif ′(S). (8.11)

8.4. CONTACT TRANSFORMATIONS 131

We interpret this as saying that f(S) is the potential for the system’s dissipativeforce. Moreover, the evolution (8.9) of the Hamiltonian can be integrated toobtain

H(t) = H(0) exp

[−∫ t

0

f ′(S)

].

Plugging in the Hamiltonian (8.10), we obtain an implicit equation which inprinciple determines the action S = S(q, p, t); thus the equations of motion (8.6)reduce to the 2n equations for the positions qi and momenta pi.

Example 8.7. If we take d = 1 and the dissipative potential f(S) = γS to belinear as in Example 8.6 then equation (8.11) becomes

H0 = −mγq2

which agrees with what we found for the damped harmonic oscillator of Ex-ample 1.13. The energy of this system decays exponentially to zero, accordingto

H(t) = H(0)e−γt.

Solving for the action S we obtain

S(q, p, t) =1

γ

[H(0)e−γt − p2

2m− U(q)

].

8.4 Contact transformations

A contact transformation is a smooth transformation which leaves the contactstructure invariant. As opposed to canonical transformations, we allow thecontact form fη in the new coordinates to differ by a smooth nonvanishingfactor f ∈ C∞(M) since we are interested in the contact structure N . Interms of the canonical coordinates (q, p, S) guaranteed by Theorem 8.2, the newcoordinates (q, p, S) must satisfy

η = dS − p dq = f(dS − p dq) = fη. (8.12)

Writing (S, q, p) as functions of (q, p, S), this is equivalent to the conditions

f =∂S

∂S−

n∑j=1

pj∂qj

∂S, −fpi =

∂S

∂qi−

n∑j=1

pj∂qj

∂qi, 0 =

∂S

∂pi−

n∑j=1

pj∂qj

∂pi

for i = 1, . . . , d. Note that canonical transformations are independent of S, Sand are defined by the condition (8.12) with f ≡ 1.

Since we allow for a conformal factor f in the definition (8.12), the volumeform is also rescaled. Indeed, if η = fη then dη = df ∧ η + f dη, and so

η ∧ (dη)n = fd+1η ∧ (dη)n


is the new volume form. In the case of canonical transformations we have f ≡ 1,and hence we recover volume preservation.

As with canonical transformations in section 6.5, we may consider a contacttransformation as being generated by a generating function. For an example,assume that the coordinates (q, q, S) are independent. Then the differential ofthe generating function S = S(q, q, S) may be written

dS =∂S

∂SdS +

n∑i=1

∂S

∂qidqi +

n∑i=1

∂S

∂qidqi.

Plugging this into (8.12), we see that the remaining coordinates are determinedin terms of S by

f =∂S

∂S, fpi = − ∂S

∂qi, pi =

∂S

∂qi.

Taking f ≡ 1 and comparing to eq. (6.25), we see that S is related to thecanonical transformation generating function F (q, q) via

S = S − F (q, q).

In particular, we conclude that all physicist’s time-independent canonical trans-formations (cf. the remark of section 6.5) are a special case of contact transfor-mations.

Recall that for symplectic structures the Hamiltonian dynamics generateinstantaneous canonical transformations; this was first seen in section 6.6 com-putationally, and then abstractly as an equivalence in Proposition 7.12. Inanalogy with symplectic vector fields, we will call a smooth vector field V on Mif a contact vector field if its flow etV preserves the contact structure N , in that

d(etV)x

(Nx) = NetV x (8.13)

for all t ∈ R and x ∈M in the domain of definition of the flow etV .

Proposition 8.8. Let (M,η) be a contact manifold. A smooth vector field Von M is a contact vector field if and only if it is a contact Hamiltonian vectorfield.

Proof. From the condition (8.13) and the definition of the Lie derivative, we seethat V is a contact vector field if and only if LV η = 0 on N .

First assume that V is a contact Hamiltonian vector field, and write V =XH for a Hamiltonian H. Then from Cartan’s magic formula and the firstcondition (8.4) defining XH we have

LXHη = d [η(XH)] +XHydη = dH +XHydη.

From the second condition of (8.4) we know that XHydη is equal to −dH onN . By the definition (8.3) of the Reeb field it then follows that

LXHη = −ξ(H)η = −∂H∂S

η, (8.14)

8.4. CONTACT TRANSFORMATIONS 133

where the last equality is merely the expression in terms of the local canonicalcoordinates. Comparing this to the definition (8.12) of a contact transformation,we see that the flow is a contact transformation with f = −ξ(H) = −∂H/∂S.In particular, if we restrict to the contact structure N , we have η = 0 and henceLXHη = 0 as desired.

Conversely, assume that V is contact vector field. Then Cartan’s magicformula reads

0 = (LV η)|N = dη(XH)|N + (V ydη)|N .

Consider the smooth function H = η(V ), defined so that the first condition ofthe definition (8.4) holds. Then we obtain the second condition of (8.4) fromthe above equality, and so we conclude that V = XH is the contact Hamiltonianvector field for H.

Heuristically, we do not expect the volume form to be preserved by a generalcontact Hamiltonian flow since contact dynamics includes dissipative systems.Using eq. (8.14) we see that the volume form evolves according to

LXH [η ∧ (dη)n] = (LXHη) ∧ (dη)n +

n−1∑i=0

η ∧ (dη)i ∧ [d(LXHη)] ∧ (dη)n−1−i

= −(n+ 1)∂H

∂Sη ∧ (dη)n.

This illustrates the connection between the Hamiltonian’s S-dependence to thesystem’s dissipation, and consequently systems for which ∂H

∂S is nonvanishingare called dissipative.

Instead, we have a variant of Liouville’s theorem due to [BT], in which arescaled volume form is preserved away from the zero set H−1(0).

Proposition 8.9 (Canonical measure for dissipative contact systems). Let(M,η) be a (2n + 1)-dimensional contact manifold and H a smooth functionon M . Then the volume form

|H|−(n+1) η ∧ (dη)n

is an invariant measure for the contact Hamiltonian flow for H along orbitsoutside of H−1(0). Moreover, up to scalar multiplication it is the unique suchmeasure whose density with respect to the standard volume form depends onlyon H.

Proof. For a smooth function ρ on M , a computation using (8.14) shows that

LXH [ρ η ∧ (dη)n] = (LXHρ) η ∧ (dη)n − (n+ 1)∂H

∂Sρ η ∧ (dη)n

=

[XH(ρ)− (n+ 1)

∂H

∂Sρ

]η ∧ (dη)n.


If we assume ρ = ρ(H) then

XH(ρ) = −Hρ′(H)∂H

∂S,

and so the vanishing of LXH [ρ η ∧ (dη)n] occurs exactly when ρ solves

ρ′(H) = −(n+ 1)H−1ρ.

This equation has the solution ρ(H) = |H|−(n+1) and it is unique up to scalarmultiplication.

8.5 Time-dependent systems

Thus far we have allowed H to be a function on the contact manifold M , andhence have only considered time-independent dissipative systems. In this sec-tion, we present the extension introduced in [BCT17] of contact Hamiltoniansystems to include time-dependence.

For (M,η) a (2n+ 1)-dimensional contact manifold, we define the extendedmanifold M × R. In analogy with the Poincare–Cartan one-form (7.15), givena possibly time-dependent Hamiltonian H we extend the contact form to

θ = dS − p dq +H dt = dS −n∑i=1

pi dqi +H dt (8.15)

in terms of the canonical coordinates (q, p, S) on M guaranteed by Theorem 8.2.On M × R we define the extended contact Hamiltonian vector field

YH = XH +∂

∂t.

In place of the conditions (8.4), it can be checked that this vector field is uniquelydetermined by

θ(YH) = 0, YHydθ = −∂H∂S

θ. (8.16)

Here, the first condition is analogous to how (7.17) replaced (7.4) for time-dependent symplectic systems, and the second condition is the analog of (8.14)(which serves as a rephrasing for the second condition of (8.4) that does notinvolve N).

The flow of YH is given by

dS

dt=

n∑i=1

pi∂H

∂pi−H, dqi

dt=∂H

∂pi,

dpi

dt= −∂H

∂qi− pi ∂H

∂S, t = 1,

which are the old equations of motion (8.6) joined with the trivial equationt = 1. It follows that any smooth time-dependent function F on M ×R evolvesaccording to

dF

dt= −H∂F

∂S+

n∑i=1

piH,Fpi,S + H,Fp,q +∂F

∂t, (8.17)

8.5. TIME-DEPENDENT SYSTEMS 135

using the notation of eq. (8.8). In particular, we see that under its own flowthe Hamiltonian now changes according to both its dissipation ∂H/∂S and itstime-dependence.

Lastly, let us extend the notion of contact transformations to our extendedmanifold M × R. In terms of canonical coordinates, a time-dependent contacttransformation (q, p, S, t) 7→ (q, p, S, t) must satisfy

θ = dS − p dq +K dt = f(dS − pdq +H dt) = fθ (8.18)

for a smooth nonvanishing factor f ∈ C∞(M × R) and a new HamiltonianK ∈ C∞(M × R). Expanding dq and dS, we see that the new Hamiltonianmust satisfy

fH =∂S

∂t−

n∑i=1

pi∂qi

∂t+K.

As before, we may consider a contact transformation as being generatedby a generating function. For example, let us assume that the coordinates(q, q, S, t) are independent. After substituting the differential of the generatingfunction S = S(q, q, S, t) into (8.18), we see that the remaining coordinates aredetermined in terms of S by

f =∂S

∂S, fpi = − ∂S

∂qi, pi =

∂S

∂qi, fH =

∂S

∂t+K. (8.19)

The first three conditions are unchanged and the last condition defines the newHamiltonian K = K(q, q, S, t). Taking f ≡ 1 and comparing to eq. (6.25), we seethat S is related to the canonical transformation generating function F (q, q, t)via

S = S − F (q, q, t).

However, now there is an additional constraint on S imposed by the in-variance of the second condition of (8.16). After the transformation we musthave

YHydθ = −∂K∂S

θ.

Using θ = fθ, YH = YH , and the extended contact Hamiltonian vector fieldconditions (8.16), this yields

f∂K

∂S= f

∂H

∂S+ df(YH).

In the special case f ≡ 1 we note that if H is independent of S then K = 0is a solution, in which case the last condition of (8.19) becomes the familiarHamilton–Jacobi equation (5.4). However, in general f may be S-dependent andso the notion of contact transformations is strictly more general than even thephysicist’s notion of canonical transformations (cf. the remark of section 6.5).


Example 8.10. In Example 7.11 we introduced a non-canonical coordinatetransformation (q, p) 7→ (q, eγt) to describe a dissipative system using time-dependent Hamiltonian dynamics. However, it is easily checked that the trans-formation

q = q, p = eγtp, S = eγtS, t = t

satisfies eq. (8.15) and is hence a time-dependent contact transformation. Here,the conformal factor is f = eγt, and the Hamiltonians are expressed in theirrespective coordinates: H is given by eq. (8.7) and K is given by eq. (7.19).

8.6 Exercises

8.1. Show that a smooth one-form η on a (2n + 1)-dimensional manifold Msatisfies the nondegeneracy condition of a contact form if and only if it satisfiesthe nonvanishing top form condition (8.1).

8.2 (Contact structure on S2n+1). On the Euclidean space R2n+2 consider thecoordinates (x1, . . . , xn+1, y1, . . . , yn+1) and define the one-form

θ :=

n+1∑i=1

(xi dyi − yi dxi

).

The standard contact form on the sphere S2n+1 is η := ι∗θ, where ι :=: S2n+1 →R2n+2 is the inclusion map.

(a) Show that the vector fields

V =

n+1∑i=1

(xi

∂

∂xi+ yi

∂

∂yi

), W =

n+1∑i=1

(xi

∂

∂yi− yi ∂

∂xi

)satisfy V ydθ = 2θ and Wydθ = −d

(x2 + y2

).

(b) Let S ⊂ T (R2n+1 r 0) denote the subbundle spanned by V and W , andlet

S⊥ =⋃

p∈S2n+1

X ∈ TpR2n+2 : dθp(Vp, Xp) = dθp(Wp, Xp) = 0

denote its symplectic complement. Show that θ is indeed a contact formwith respect to the contact structure S⊥.

(c) Show that the corresponding Reeb field is given by W restricted S2n+1.

8.3. (Solving the damped parametric oscillator via expanding coordinates) Con-sider the one-dimensional damped parametric oscillator

H =p2

2m+m

2ω2(t)q2 + γS

8.6. EXERCISES 137

with time-dependent frequency ω(t) and damping parameter γ. Show that thenew expanding coordinates

q = eγt/2q, p = eγt/2(p+ 12mγq), S = eγt(S + 1

4mγq2), t = t

define a contact transformation, with respect to which the Hamiltonian takesthe form

K =p2

2m+m

2

(ω2(t)− 1

4γ2)q2.

This new Hamiltonian K now corresponds to an undamped parametric oscillator

with the new frequency√ω2(t)− 1

4γ2. The undamped oscillator has been ex-

tensively studied and solutions for the equations of motion can be obtained. Inthe harmonic oscillator case ω(t) ≡ ω0 the Hamiltonian K is a conserved quan-tity, and the coordinates expand exponentially in time so that the trajectoriesform closed orbits at a slower frequency.

8.4. (Solving the damped parametric oscillator via conserved quantities) Con-sider the one-dimensional damped parametric oscillator

H =p2

2m+m

2ω2(t)q2 + γS

with time-dependent frequency ω(t) and damping parameter γ.

(a) We seek a solution F to eq. (8.17) with vanishing left-hand side. Substi-tuting the quadratic ansatz

F (q, p, S, t) := β(t)p2 − 2ξ(t)qp+ η(t)q2 + ζ(t)S,

obtain a system of first-order equations for β, η, ξ, and ζ, where the ζ hassolution ζ(t) = ζ0e

γt.

(b) Using the substitution β(t) := 12me

γtα2(t), show that the β equation issolved if and only if α satisfies the Ermakov equation

α+ [ω2(t)− 14γ

2]α = α−3,

and the remaining equations become

η(t) = 12me

γt[α− 12γα]2 + α−2, ξ(t) = 1

2eγt[α− 1

2γα] + 14 .

(c) Conclude that the quantity F (q, p, S, t) = I(q, p, t) + ζ0G(q, p, S, t) is con-served, with

I =m

2eγt[ p

mα−

(α− γ

2α)q]2

+( qα

)2, G = eγt(S − 1

2qp),

and α(t) solves the Ermakov equation. Moreover, since F is invariant forall initial conditions and ζ0 is determined solely by initial conditions, thenI and G must be separately conserved.


(d) Show that the new coordinates

q = arctan

[(α− 1

2γα)α− α2 p

mq

], p = I(q, p, t), S = G(q, p, S, t),

(and t = t) define a contact transformation, with respect to which the newHamiltonian is simply K = Iα−2. Solve the new equations of motion forq, p, S and obtain the solution

q(t) =

√2I

meγtα(t) cosφ(t), S(t) = Ge−γt + 1

2q(t)p(t),

p(t) =√

2mIeγt[(α− γ

2α)

cosφ(t)− 1

αsinφ(t)

], φ(t) =

∫ t

t0

dτ

α2(τ).

Here, α(t) solves the Ermakov equation and the conserved quantities Iand G are determined by the initial conditions.

Bibliography

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Notes on Classical Mechanics - math.ucla.edulaurenst/Resources/classical_mechanics.pdf · Classical mechanics studies the motion of particles in Euclidean space Rn. A particle or