Jacobian of a point transformation in phase space

Question

A common basic result of analytical mechanics is that point transformations $Q=Q(q)$ are canonical, i.e. they preserve the canonical structure of Hamilton equations. Also, a known theorem says that a necessary and sufficient condition for $(p,q) \to (P,Q)$ to be a canonical transformation is that its Jacobian $M$ has to be of the form $M J M^t=M^t J M = a J$, with $J$ the standard symplectic matrix and $a$ a constant different from zero.

As an exercise, I want to verify the theorem for a point transformation, but I am having trouble proving that $M^t J M = a J$ for dimension of $M$ higher than 2.

Being that $Q=Q(q)$ and $P=(N^t(q))^{-1}p$, where $N$ is the Jacobian of $Q(q)$ I write the Jacobian $M$ of the overall $(p,q) \to ((N^t)^{-1}p, Q(q))$ transformation as

$$\begin{bmatrix} \frac{\partial P}{\partial p} & \frac{\partial P}{\partial q}\\ \frac{\partial Q}{\partial p} & \frac{\partial Q}{\partial q} \end{bmatrix} = \begin{bmatrix} (N^t(q))^{-1} & R\\ 0 & N \end{bmatrix}$$

and I get $$ M^tJM= \begin{bmatrix} (N(q))^{-1} & 0\\ R^t & N^t \end{bmatrix} \begin{bmatrix} 0 & -I\\ I & 0 \end{bmatrix} \begin{bmatrix} (N^t(q))^{-1} & R\\ 0 & N \end{bmatrix} = \begin{bmatrix} (N(q))^{-1} & 0\\ R^t & N^t \end{bmatrix} \begin{bmatrix} 0 & -N\\ (N^t(q))^{-1} & R \end{bmatrix} = \begin{bmatrix} 0 & -I\\ I & -R^tN+N^tR \end{bmatrix} $$

where $R=[\frac{\partial (N^t(q))^{-1}p}{\partial q_1}...\frac{\partial (N^t(q))^{-1}p}{\partial q_n}]$

so that $$N^tR(i,j)=\begin{bmatrix} \frac{\partial Q_1}{\partial q_i} & ... &\frac{\partial Q_n}{\partial q_i} \end{bmatrix} \begin{bmatrix} \frac{\partial (N^t(q))^{-1}p}{\partial q_j} \end{bmatrix} $$ $$ R^tN(i,j)= \begin{bmatrix} \frac{\partial (N^t(q))^{-1}p}{\partial qi} \end{bmatrix}^t\begin{bmatrix} \frac{\partial Q_1}{\partial q_j} & ... &\frac{\partial Q_n}{\partial q_j} \end{bmatrix} $$

This last matrix is $J$ (so $a$ is 1) if $R^tN$ is symmetric, but I am not sure how to check it. Is the reasoning and the methodology correct?

Answer 1

Sketched calculation using the standard order $z=(q,p)$ and $Z=(Q,P)$:

1. Under a coordinate transformation of positions $$Q^i~=~f^i(q),\qquad N^i{}_j~:=~\frac{\partial Q^i}{\partial q^j}, $$ the momenta transform as a covector $$ P_i~=~p_k\frac{\partial q^k}{\partial Q^i}~=~p_kL^k{}_i,$$ $$ P~=~L^tp, \qquad L~:=~N^{-1}, \qquad L^i{}_j~=~\frac{\partial q^i}{\partial Q^j}. $$

2. Jacobian matrix in phase space: $$M^I{}_K ~=~\frac{\partial Z^I}{\partial z^K}~=~\begin{pmatrix} N & {\bf 0} \cr R & L^t\end{pmatrix}.$$

3. Condition for a symplectomorphism: $$ MJM^t~=~\begin{pmatrix} {\bf 0} & {\bf 1} \cr -{\bf 1} & RL\!-\!(RL)^t\end{pmatrix}~=~\begin{pmatrix} {\bf 0} & {\bf 1} \cr -{\bf 1} &{\bf 0}\end{pmatrix}~=:~J.$$

4. In the last equation was used that $$\begin{align} (RL)_{im} ~=~& R_{ij}L^j{}_m ~=~\frac{\partial P_i}{\partial q^j}L^j{}_m ~=~p_n\frac{\partial L^n{}_i}{\partial q^j}L^j{}_m\cr ~=~&-p_nL^n{}_{\ell}\frac{\partial N^{\ell}{}_k}{\partial q^j}L^k{}_iL^j{}_m\cr ~=~&-p_nL^n{}_{\ell}\frac{\partial^2 Q^{\ell}}{\partial q^j\partial q^k}L^k{}_iL^j{}_m ~=~(i\leftrightarrow m).\end{align}$$ $\Box$