Find the error: If $L_x$ and $L_y$ are zero, then $L_z$ is conserved

Question

From Goldstein's Classical Mechanics (2nd ed.), problem 38 of chapter 9 basically says the following:

It's been shown that the Poisson bracket of two constants of the motion is also a constant of the motion (by the Jacobi identity). Applied the angular momentum of a particle/system, this says that if the components of angular momentum $L_x$ and $L_y$ are conserved, $L_z$ is also conserved because

$$\{L_x,L_y\}=L_z$$

This seems to imply that any system confined to move in a plane automatically has its angular momentum $L_z$ conserved, since $L_x$ and $L_y$ are identically zero. Immediately we can think of systems confined to a plane where $L_z$ isn't conserved (e.g. angular momentum of spring on a watch or that of a plane disk rolling down an incline). What objections can be made to this implication? Does the theorem above require any restrictions?

I can't find any solutions to this online, but here's my guess:

I think the problem is that this theorem relies on $L_x$ and $L_y$ being constants of the motion with respect to a system (certain phase-space with underlying hamiltonian) described by Hamilton's principle, and the cited examples where angular momentum wasn't conserved required constraint equations which would alter the form of the Lagrangian formalism (which the Hamiltonian formalism is based on). The theorem could then be formulated as follows: the Poisson bracket of two constants of the motion, with respect to a system described by the typical Hamilton's principle, is also a constant of the motion (by the Jacobi identity).

What do you think? Is my guess on the right track? If so, how could it be refined? If not, why is my guess invalid?

Answer 1

Refs. 1 and 2 only discuss a generic/bulk notion of a constant of motion in phase space; not any refinement thereof restricted to a subsurface of phase space. Let us here carefully explain the generic/bulk notion of a constant of motion used in Refs. 1 and 2. Both Refs. 1 and 2 mention Poisson's Theorem:

Poisson's Theorem: If $F$ and $G$ are constants of motion, then the Poisson bracket $\{F,G\}_{PB}$ is an constant of motion.

Example 1: If also $L_x$ and $L_y$ are constants of motion, then we can use Poisson's Theorem to deduce that $$L_z~=~\{L_x,L_y\}_{PB}\tag{1}$$ is a constant of motion as well.

Refs. 1 and 2 do not explicitly stress that the proof of Poisson's Theorem uses the notion of off-shell constants of motion. They are only equivalent to on-shell constants of motion, if they hold for all initial conditions.

The main point is that OP's examples are not on-shell constants of motion for all initial conditions.

Example 2: A non-relativistic point particle in a uniform gravitational field has Hamiltonian $$ H~=~\frac{{\bf p}^2}{2m}+mgy. \tag{2}$$ The angular momentum has the following Poisson brackets with the Hamiltonian: $$ \{L_x,H\}_{PB}~=~-mgz, \qquad \{L_y,H\}_{PB}~=~0, \qquad \{L_z,H\}_{PB}~=~mgx, \tag{3}$$ so only $L_y$ is a constant of motion.

If we restrict to the plane $\Sigma=\{z=0\}$, we see that $$\left. \{L_x,H\}_{PB}\right|_{\Sigma}~=~0 \tag{4}$$ does indeed commute, but $$\left. \{L_z,H\}_{PB}\right|_{\Sigma}~\neq~0, \tag{5}$$ so Poisson's Theorem does not apply to a subset/constrained surface of phase space without further conditions on the Hamiltonian flow. Constrained dynamics is the topic of Ref. 3.

References:

1. H. Goldstein, Classical Mechanics; Eq. (9.97).

2. L.D. Landau & E.M. Lifshitz, Mechanics, vol. 1, 1976; Eq. (42.3). [Note that Ref. 2 confusingly calls a constant of motion for an integral of motion. According to Wikipedia, an integral of motion is a constant of motion without explicit time dependence.]

3. M. Henneaux and C. Teitelboim, Quantization of Gauge Systems, 1994.

Answer 2

When you say that $L_x$ and $L_y$ vanish for a point confined to move in the plane $z=0$, you mean that the the solution $\vec{x}=\vec{x}(t)$, $\vec{p}=\vec{p}(t)$ describes a curve in the given plane with tangent vector parallel to that plane. So that, exactly along that curve, $$L_x(\vec{x}(t),\vec{p}(t))= L_y(\vec{x}(t),\vec{p}(t))=0\quad \forall t \in \mathbb R$$ When you instead compute $\{L_x,L_y\}$, you actually compute derivatives $x$ and $p$ of the involved functions regardless the particular curve you finally use: $$\{L_x,L_y\} = \sum_{i=1}^3 \frac{\partial L_x}{\partial x^i} \frac{\partial L_y}{\partial p_i}- \frac{\partial L_y}{\partial x^i} \frac{\partial L_x}{\partial p_i}\:.$$ You evaluate the derivatives on the given curve just at the end of the computation. The point is that these derivatives may not vanish on the given curve even if the function $L_x$ and $L_y$ vanish on it.

This fact is general. For instance $F(x,y)= x-y^2$ vanishes if evaluated on the curve $x=t^2$, $y=t$: $$F(x(t),y(t))= t^2-t^2=0 \quad \forall t \in \mathbb R\:.$$ However $\frac{\partial F}{\partial x}$ does not vanish on that curve: $$\frac{\partial F}{\partial x}(x(t),y(t))= 1-t^2\:.$$

Also the general statement that if $L_x$ and $L_y$ are conserved then $L_z$ is does not follow so straightforwardly from $$\{L_x,L_y\}=L_z\tag{1}$$ as it may seem at first glance. You need the so-called Jacobi identity of Poisson bracket and the fact that $$\frac{df(\vec{x}(t),\vec{p}(t))}{dt} = \{H,f\}(\vec{x}(t),\vec{p}(t))$$ to prove it.