I must prove that the change of variables $\Psi: (p,q)\rightarrow(r,\alpha)$ such that $$q=\sqrt{2r}\cos(\alpha), \qquad p=\sqrt{2r}\sin(\alpha)$$ is symplectic.
What I know is: $$\dot{q}=p, \qquad \dot{p}=\sin(q)$$
And so I deduce: $\ddot{q}=\sin(q)$, and $H(p,q)=\frac{p^2}{2}+\cos(q)$ is the Hamiltonian.
How can I prove that $\Psi$ is symplectic?
Thank you!
The Jacobian matrix is $$M=\begin{bmatrix}\frac{\partial q}{\partial r}&\frac{\partial q}{\partial \alpha}\\\frac{\partial p}{\partial r}&\frac{\partial p}{\partial \alpha}\end{bmatrix}=\frac1{\sqrt{2r}}\begin{bmatrix}\cos\alpha&-2r\sin\alpha\\\sin\alpha&2r\cos\alpha\end{bmatrix}.$$ With $J=\begin{bmatrix}0&1\\-1&0\end{bmatrix}$, we have $$M^TJM=J.$$ So by definition $M$ is a symplectic matrix and as the Jacobian matrix is symplectic, the change of variable $\Psi$ is symplectic.