Kepler problem: flows generated by constants of motion

[Clausius]

Consider the Hamiltonian of Kepler problem
$$H(\boldsymbol{r},\boldsymbol{p})= \frac{|\boldsymbol{p}^2|}{2\mu} +\frac{\alpha}{|\boldsymbol{r}|}, \qquad \mu>0>\alpha,$$

where $$\boldsymbol{r}\in M=\mathbb{R}^3\setminus\{ 0 \}, \ (\boldsymbol{r},\boldsymbol{p})\in T^*M$$
and
$$|\boldsymbol{r}|=\sqrt{r_1^2+r_2^2+r_3^2}.$$

The quantities
$$\boldsymbol{m}=\boldsymbol{r}\times\boldsymbol{p}, \qquad \boldsymbol{W}=\boldsymbol{p}\times\boldsymbol{m}+ \mu\alpha\frac{\boldsymbol{r}}{|\boldsymbol{r}|}$$

are constants of motion, as is well known.

My question is: how can I prove that the flows generated by the functions m_i and W_i, i=1,2,3 are canonical transformations?
Moreover, are such transformations point transformations?

A canonical transformation $$\Phi: T^*M\to T^*M$$
is a point transformation if it is induced by a transformation $$\phi:M\to M,$$

so that
$$\Phi(\boldsymbol{r},\boldsymbol{p})= (\phi(\boldsymbol{r}),\phi^{*-1}), \ \phi^*_i= \frac{\partial\phi_i}{\partial r_j}p_j.$$

 

[member 553083]

In this case, the flows generated by m_i and W_i are given by\boldsymbol{r}(t)= \boldsymbol{r}_0 + t\boldsymbol{m} \qquad \boldsymbol{p}(t)= \boldsymbol{p}_0 + t\boldsymbol{W}.To prove that these flows generate a canonical transformation, we need to show that\frac{d}{dt}\Phi(\boldsymbol{r},\boldsymbol{p})= \{H,\Phi\}.Letting \Phi(\boldsymbol{r},\boldsymbol{p})=(\boldsymbol{r}(t),\boldsymbol{p}(t)), we have\frac{d}{dt}\Phi(\boldsymbol{r},\boldsymbol{p})= (\dot{\boldsymbol{r}}(t),\dot{\boldsymbol{p}}(t)) = (\boldsymbol{m},\boldsymbol{W}).On the other hand, the Poisson bracket is given by\{H,\Phi\}= \frac{\partial H}{\partial r_i}\frac{\partial \Phi_j}{\partial p_i} - \frac{\partial H}{\partial p_i}\frac{\partial \Phi_j}{\partial r_i}.For the Hamiltonian of the Kepler problem, we have\frac{\partial H}{\partial r_i}= -\frac{\alpha r_i}{|\boldsymbol{r}|^3}, \qquad \frac{\partial H}{\partial p_i}= \frac{p_i}{\mu}.Substituting these expressions into the Poisson bracket and using the fact that \Phi(\boldsymbol{r},\boldsymbol{p})=(\boldsymbol{r}(t),\boldsymbol{p}(t)), we obtain\{H,\Phi\}= -\frac{\alpha}{|\boldsymbol{r}|^3}\left(\boldsymbol{r}\cdot\boldsymbol{m}\