1.8  Conserved Quantities

A function of the state of the system that is constant along a solution path is called a conserved quantity or a constant of motion. If C is a conserved quantity, then

for solution paths q. Following historical practice we also refer to constants of the motion as integrals of the motion.79 In this section, we will investigate systems with symmetry and find that symmetries are associated with conserved quantities. For instance, linear momentum is conserved in a system with translational symmetry, angular momentum is conserved if there is rotational symmetry, energy is conserved if the system does not depend on the origin of time. We first consider systems for which a coordinate system can be chosen that expresses the symmetry naturally, and later discuss systems for which no coordinate system can be chosen that simultaneously expresses all symmetries.

1.8.1  Conserved Momenta

If a Lagrangian L(t, q, v) does not depend on some particular coordinate qi, then

and the corresponding ith component of the Lagrange equations is

This is the same as80

So we see that

is a conserved quantity. The function is called the momentum state function. The value of the momentum state function is the generalized momentum. We refer to the ith component of the generalized momentum as the momentum conjugate to the ith coordinate.81 A generalized coordinate component that does not appear explicitly in the Lagrangian is called a cyclic coordinate. The generalized momentum component conjugate to any cyclic coordinate is a constant of the motion. Its value is constant along realizable paths; it may have different values on different paths. As we will see, momentum is an important quantity even when it is not conserved.

Given the coordinate path q and the Lagrangian L, the momentum path p is

with components

The momentum path is well defined for any path q. If the path is realizable and the Lagrangian does not depend on qi, then pi is a constant function

The constant value of pi may be different for different trajectories.

Examples of conserved momenta

The free-particle Lagrangian L(t, x, v) = (1/2) mv2 is independent of x. So the momentum state function, (t, q, v) = mv, is conserved along realizable paths. The momentum path p for the coordinate path q is p(t) = o [q](t) = m Dq(t). For a realizable path Dp(t) = 0. For the free particle the usual linear momentum is conserved for realizable paths.

For a particle in a central force field (section 1.6), the Lagrangian

depends on r but is independent of . The momentum state function is

It has two components. The first component, the ``radial momentum,'' is not conserved. The second component, the ``angular momentum,'' is conserved along any solution trajectory.

If the central-potential problem had been expressed in rectangular coordinates, then all of the coordinates would have appeared in the Lagrangian. In that case there would not be any obvious conserved quantities. Nevertheless, the motion of the system does not depend on the choice of coordinates, so the angular momentum is still conserved.

We see that there is great advantage in making a judicious choice for the coordinate system. If we can choose the coordinates so that a symmetry of the system is reflected in the Lagrangian by the absence of some coordinate component, then the existence of a corresponding conserved quantity will be automatic.82

1.8.2  Energy Conservation

Momenta are conserved by the motion if the Lagrangian does not depend on the corresponding coordinate. There is another constant of the motion, the energy, if the Lagrangian L(t, q, ) does not depend explicitly on the time: 0 L = 0.

Consider the time derivative of the Lagrangian along a solution path q:

Using Lagrange's equations to rewrite the second term yields

Isolating 0 L and combining the first two terms on the right side gives

where, as before, selects the velocity from the state. So we see that if 0 L = 0 then

is conserved along realizable paths. The function is called the energy state function.83 Let E = o [q] denote the energy function on the path q. The energy function has a constant value along any realizable trajectory if the Lagrangian has no explicit time dependence; the energy E may have a different value for different trajectories. A system that has no explicit time dependence is called autonomous.

Given a Lagrangian procedure L, we may compute the energy:

(define (Lagrangian->energy L)
  (let ((P ((partial 2) L)))
    (- (* P velocity) L)))

Energy in terms of kinetic and potential energies

In some cases the energy can be written as the sum of kinetic and potential energies. Suppose the system is composed of particles with rectangular coordinates x, the movement of which may be subject to constraints, and that these rectangular coordinates are some functions of the generalized coordinates q and possibly time t: x = f(t, q). We form the Lagrangian as L = T - V and compute the kinetic energy in terms of q by writing the rectangular velocities in terms of the generalized velocities:

The kinetic energy is

where v is the magnitude of v.

If the f functions do not depend explicitly on time (0 f = 0), then the rectangular velocities are homogeneous functions of the generalized velocities of degree 1, and T is a homogeneous function of the generalized velocities of degree 2, because it is formed by summing the square of homogeneous functions of degree 1. If T is a homogeneous function of degree 2 in the generalized velocities then

where the second equality follows from Euler's theorem on homogeneous functions.84 The energy state function is

So if f is independent of time, the energy function can be rewritten

Notice that if V depends on time the energy is still the sum of the kinetic energy and potential energy, but the energy is not conserved.

The energy state function is always a well defined function, whether or not it can be written in the form T + V, and whether or not it is conserved along realizable paths.

Exercise 1.28.  
An analogous result holds when the f depend explicitly on time.

a.   Show that in this case the kinetic energy contains terms that are linear in the generalized velocities.

b.   Show that, by adding a total time derivative, the Lagrangian can be written in the form L = A - B, where A is a homogeneous quadratic form in the generalized velocities and B is independent of velocity.

c.   Show, using Euler's theorem, that the energy function is = A + B.

An example in which terms that were linear in the velocity were removed from the Lagrangian by adding a total time derivative has already been given: the driven pendulum.

Exercise 1.29.  
A particle of mass m slides off a horizontal cylinder of radius R in a uniform gravitational field with acceleration g. If the particle starts close to the top of the cylinder with zero initial speed, with what angular velocity does it leave the cylinder?

1.8.3  Central Forces in Three Dimensions

One important physical system is the motion of a particle in a central field in three dimensions, with an arbitrary potential energy V(r) depending only on the radius. We will describe this system in spherical coordinates r, , and , where is the colatitude and is the longitude. The kinetic energy has three terms:

As a procedure:

(define ((T3-spherical m) state)
  (let ((q (coordinate state))
        (qdot (velocity state)))
    (let ((r (ref q 0))
          (theta (ref q 1))
          (rdot (ref qdot 0))
          (thetadot (ref qdot 1))
          (phidot (ref qdot 2)))
      (* 1/2 m
         (+ (square rdot)
            (square (* r thetadot))
            (square (* r (sin theta) phidot)))))))

The Lagrangian is then formed by subtracting the potential energy:

(define (L3-central m Vr)
  (define (Vs state)
    (let ((r (ref (coordinate state) 0)))
      (Vr r)))
  (- (T3-spherical m) Vs))

Let's first look at the generalized forces (the derivatives of the Lagrangian with respect to the generalized coordinates). We compute these with a partial derivative with respect to the coordinate argument of the Lagrangian:

 (((partial 1) (L3-central 'm (literal-function 'V)))
  (up 't
      (up 'r 'theta 'phi)
      (up 'rdot 'thetadot 'phidot))))

The component of the force is zero because does not appear in the Lagrangian (it is a cyclic variable). The corresponding momentum component is conserved. Compute the momenta:

 (((partial 2) (L3-central 'm (literal-function 'V)))
  (up 't 
      (up 'r 'theta 'phi)
      (up 'rdot 'thetadot 'phidot))))

The momentum conjugate to is conserved. This is the z component of the angular momentum × (m ), for vector position and linear momentum m . We can show this by writing the z component of the angular momentum in spherical coordinates:

(define ((ang-mom-z m) state)
  (let ((q (coordinate state))
        (v (velocity state)))
     (ref (cross-product q (* m v)) 2)))
(define (s->r state)
  (let ((q (coordinate state)))
    (let ((r (ref q 0))
          (theta (ref q 1))
          (phi (ref q 2)))
      (let ((x (* r (sin theta) (cos phi)))
            (y (* r (sin theta) (sin phi)))
            (z (* r (cos theta))))
        (up x y z)))))
  ((compose (ang-mom-z 'm) (F->C s->r))
   (up 't 
       (up 'r 'theta 'phi)
       (up 'rdot 'thetadot 'phidot))))

The choice of the z-axis is arbitrary, so the conservation of any component of the angular momentum implies the conservation of all components. Thus the total angular momentum is conserved. We can choose the z-axis so all of the angular momentum is in the z component. The angular momentum must be perpendicular to both the radius vector and the linear momentum vector. Thus the motion is planar, = /2, and = 0. Planar motion in a central-force field was discussed in section 1.6.

We can also see that the energy state function computed from the Lagrangian for a central field is in fact T + V:

 ((Lagrangian->energy (L3-central 'm (literal-function 'V)))
  (up 't
      (up 'r 'theta 'phi)
      (up 'rdot 'thetadot 'phidot))))

The energy is conserved because the Lagrangian has no explicit time dependence.

Exercise 1.30.  Driven spherical pendulum
A spherical pendulum is a massive bob, subject to uniform gravity, that may swing in three dimensions, but remains at a given distance from the pivot. Formulate a Lagrangian for a spherical pendulum, driven by vertical motion of the pivot. What symmetry(ies) can you find? Find coordinates that express the symmetry. What is conserved? Give analytic expression(s) for the conserved quantity(ies).

1.8.4  Noether's Theorem

We have seen that if a system has a symmetry and if a coordinate system can be chosen so that the Lagrangian does not depend on the coordinate associated with the symmetry, then there is a conserved quantity associated with the symmetry. However, there are more general symmetries that no coordinate system can fully express. For example, motion in a central potential is spherically symmetric (the dynamical system is invariant under rotations about any axis), but the expression of the Lagrangian for the system in spherical coordinates exhibits symmetry around only one axis. More generally, a Lagrangian has a symmetry if there is a coordinate transformation that leaves the Lagrangian unchanged. A continuous symmetry is a parametric family of symmetries. Noether proved that for any continuous symmetry there is a conserved quantity.

Consider a parametric coordinate transformation with parameter s:

To this parametric coordinate transformation there corresponds a parametric state transformation :

We require that the transformation (0) be the identity coordinate transformation x' = (0)(t, x'), and as a consequence (0) is the identity state transformation ( t, x', v' ) = (0)(t, x', v'). The Lagrangian L has a continuous symmetry corresponding to if it is invariant under the transformations

for any s. The Lagrangian L is the same function as the transformed Lagrangian (s).

That (s) = L for any s implies D(s) = 0. Explicitly, (s) is

where we have rewritten the velocity component of (s) in terms of the total time derivative. The derivative of is zero:

where we have used the fact that85

On a realizable path q we can use the Lagrange equations to rewrite the first term:

For s = 0 the paths q and q' are the same, because (0) is the identity, so [q] = [q'] and this equation becomes

Thus the state function ,

is conserved along solution trajectories. This is Noether's integral. The integral is the product of the momentum and a vector associated with the symmetry.

Illustration: motion in a central potential

For example, consider the central potential Lagrangian in rectangular coordinates:

and a parametric rotation Rz(s) about the z axis

The rotation is an orthogonal transformation so

Differentiating along a path, we get

so the velocities also transform by an orthogonal transformation, and vx2 + vy2 + vz2 = (vx')2 + (vy')2 + (vz')2 . Thus

and we see that L' is precisely the same function as L.

The momenta are


So the Noether integral is

which we recognize as minus the z component of the angular momentum: × (m ). Since the Lagrangian is preserved by any continuous rotational symmetry, all components of the vector angular momenta are conserved for the central-potential problem.

The procedure calls ((Rx angle-x) q), ((Ry angle-y) q), and ((Rz angle-z) q) rotate the rectangular tuple q about the indicated axis by the indicated angle.86 We use these to make a parametric coordinate transformation F-tilde:

(define (F-tilde angle-x angle-y angle-z)
  (compose (Rx angle-x) (Ry angle-y) (Rz angle-z) coordinate))

A Lagrangian for motion in a central potential is

(define ((L-central-rectangular m U) state)
  (let ((q (coordinate state))
        (v (velocity state)))
    (- (* 1/2 m (square v)) (U (sqrt (square q))))))

The Noether integral is then

(define Noether-integral
  (let ((L (L-central-rectangular 
            'm (literal-function 'U))))
    (* ((partial 2) L) ((D F-tilde) 0 0 0))))

    (up 't
        (up 'x 'y 'z)
        (up 'vx 'vy 'vz))))
(down (+ (* m vy z) (* -1 m vz y))
      (+ (* m vz x) (* -1 m vx z))
      (+ (* m vx y) (* -1 m vy x)))

We get all three components of the angular momentum.

79 In older literature conserved quantities are sometimes called first integrals.

80 The derivative of a component is equal to the component of the derivative.

81 Observe that we indicate a component of the generalized momentum with a subscript, and a component of the generalized coordinates with a superscript. These conventions are consistent with those commonly used in tensor algebra, which is sometimes helpful in working out complex problems.

82 In general, conserved quantities in a physical system are associated with continuous symmetries, whether or not one can find a coordinate system in which the symmetry is apparent. This powerful notion was formalized and a theorem linking conservation laws with symmetries was proved by Noether early in the 20th century. See section 1.8.4 on Noether's theorem.

83 The sign of the energy state function is a matter of convention.

84 A function f is homogenous of degree n if and only if f(ax) = anf(x). Euler's theorem says that if f is a homogeneous function of degree n, then Df(x)x = nf(x). The proof is as follows: Let gx(a) = f(ax). Then Dgx(a) = Df(ax)x. But gx(a) = anf(x) by the definition of homogeneity. Therefore Dgx(a) = nan-1f(x). Equating these, we find Df(ax)x = nan-1f(x). Specializing to a = 1 we obtain Df(x)x = nf(x) as required.

85 The total time derivative is like a derivative with respect to a real-number argument in that it does not generate structure, so it can commute with derivatives that generate structure. Be careful, though: it may not commute with some derivatives for other reasons. For example, Dt 1 ((s)) is the same as 1 Dt ((s)), but Dt 2 ((s)) is not the same as 2 Dt ((s)). The reason is that (s) does not depend on the velocity, but Dt ((s)) does.

86 The definition of the procedure Rx is

(define ((Rx angle) q)
  (let ((ca (cos angle)) (sa (sin angle)))
    (let ((x (ref q 0)) (y (ref q 1)) (z (ref q 2)))
      (up x
          (- (* ca y) (* sa z))
          (+ (* sa y) (* ca z))))))

The definitions of Ry and Rz are similar.