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

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

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.