Poisson brackets of the Kepler Problem

Question

For the hamiltonian of a particle of unit mass in a kepler potential:

$$H = \frac{1}{2}\mathbf{p} \cdot \mathbf{p} - \frac{\mu}{r}$$

The angular momentum vector is given by: $\mathbf{L} = \mathbf{r} \times \mathbf{p}$

I know and can show that the poisson brackets of $\mathbf{r} \cdot \mathbf{r}$, $\mathbf{r} \cdot \mathbf{p}$ and $\mathbf{p} \cdot \mathbf{p}$ with any component of the angular momentum vector vanish algebraically, but what is the geometric reasoning behind this? I am trying to develop a better intuition about this. Could someone explain? Thanks!

Answer 1

The classical poisson bracket with the generator of any symmetry gives the infinitesimal evolution with respect to that symmetry. The most familiar statement of this is that the time-evolution of any observable $f$ on the phase space is given by

$$ \partial_t f = \{H,f\}$$

Similarily, for a rotation around the $i$-th axis with angle parameter $\phi$, the behaviour under this rotation is governed by

$$ \partial_\phi f = \{L_i,f\}$$

Therefore, if the Poisson bracket $\{L_i ,f\}$ vanishes for all $i$, $f$ is invariant under rotation, i.e. a scalar.

Answer 2

Let me give a further comment (not exactly an answer)

The quantity $\mathbf{r} \cdot \mathbf{r}$ represents the magnitude of the radius as such it does not change under rotation (poisson commutator with ang. momenutm $L$)

The quantity $\mathbf{p} \cdot \mathbf{p}$ represents the magnitude of (linear) momentum, as such it also does not change under rotation

The quantity $\mathbf{r} \cdot \mathbf{p}$ represents the magnitude of action, as such it also does not change under rotation