Canonical Transformation and Symplectic Conditions

I have one question regarding Canonical transformation and symplectic matrix. I have read some notions from the following note:

For me it is not clear how it is obtained the following relation:

$$\dot y = MJM^t \frac{\partial H}{\partial y}.$$

Can you explain me, please.

Update1

What I have tried.

I'll consider the case when $n=2$.

We try to make a transformation $y \to y(x)$.

relation 1(a): $$\dot y = \frac{\partial y}{\partial x} \dot x $$ relation 2(a): $$\dot x = J \frac{\partial H}{\partial x}$$BUT

relation 3(a): $$\frac{\partial H}{\partial x}=\frac{\partial H}{\partial y}\frac{\partial y}{\partial x}$$

So, we have:

$$\dot y= \frac{\partial y}{\partial x} \dot x = \frac{\partial y}{\partial x} J \frac{\partial H}{\partial x}=\frac{\partial y}{\partial x} J\frac{\partial H}{\partial y}\frac{\partial y}{\partial x}.$$

From here I get stuck.

Looks similar but I don't know how to continue to obtain same form.

I don't know why there appear the transpose of that matrix.

Can you help me to write in coordinates $\displaystyle \frac{\partial y}{\partial x}$.

UPDATE2

from relation 1(a):

$$\dot y = \frac{\partial y }{\partial x} \dot x $$ In matrix notation, for $n=2$, I'll have: relation (1b)

$$\begin{pmatrix} \dot y_{1}\\ \dot y_{2} \end{pmatrix}=\begin{pmatrix} \frac{\partial y_{1}}{\partial x_{1}}& \frac{\partial y_{1}}{\partial x_{2}}\\ \frac{\partial y_{2}}{\partial x_{1}}&\frac{\partial y_{2}}{\partial x_{2}} \end{pmatrix} \begin{pmatrix} \dot x_1\\\dot x_2 \end{pmatrix}.$$

Now, From relation 3(a) I want to obtain in a matrix form relation 3(b):

$$\begin{pmatrix} \frac{\partial H}{\partial x_1}\\ \frac{\partial H}{\partial x_2} \end{pmatrix}=\begin{pmatrix} \frac{\partial H}{\partial y_1}\\ \frac{\partial H}{\partial y_2} \end{pmatrix} \begin{pmatrix} \frac{\partial y_1}{\partial x_1} & \frac{\partial y_1}{\partial x_2}\\ \frac{\partial y_2}{\partial x_1} & \frac{\partial y_2}{\partial x_2} \end{pmatrix}.$$

So,the relation from UPDATE1

$$\dot y=\frac{\partial y}{\partial x} J\frac{\partial H}{\partial y}\frac{\partial y}{\partial x}.$$

will become: $$\begin{pmatrix} \dot y_{1}\\ \dot y_{2} \end{pmatrix}=\begin{pmatrix} \frac{\partial y_{1}}{\partial x_{1}}& \frac{\partial y_{1}}{\partial x_{2}}\\ \frac{\partial y_{2}}{\partial x_{1}}&\frac{\partial y_{2}}{\partial x_{2}} \end{pmatrix} \begin{pmatrix} 0 & 1\\ -1 & 0 \end{pmatrix}\begin{pmatrix} \frac{\partial H}{\partial y_1}\\ \frac{\partial H}{\partial y_2} \end{pmatrix} \begin{pmatrix} \frac{\partial y_1}{\partial x_1} & \frac{\partial y_1}{\partial x_2}\\ \frac{\partial y_2}{\partial x_1} & \frac{\partial y_2}{\partial x_2} \end{pmatrix} $$

And from here, again, it seems impossible for me... more and more, the matrix multiplication is not possible...

Thanks!

1 Answer

Don't know if you're still interested in this, but here goes.

Take the coordinate transformation $\vec{x}\to\vec{y}(\vec{x})$, i.e. we consider $\vec{y}$ to be a function of $\vec{x}$. Let the Hamiltonian function in terms of the new coordinates be denoted $\tilde{H}$, that is, $\tilde{H}(\vec{y}(\vec{x}))=H(\vec{x})$.

Rather than doing the calculation in terms of matrices, it is easier to do it in terms of components (tensor notation). Then you don't have to worry about tracking rows/columns, and also the notation is a lot more compact. Let $y_i$ denote the $i$'th component of $\vec{y}$, etc. Also, $M_{ij}$ denotes the $ij$ component of the $M$ matrix, which is $\frac{\partial y_i}{\partial x_j}$ (there is a typo in the definition of $M$ in your link). Our goal is to show$$\dot{y}_i=\left(M J M^T \frac{\partial H}{\partial \vec{y}}\right)_i.$$

Consider the LHS above, $\frac{d}{dt}\left(y_i (\vec{x}(t))\right)$. By the chain rule, this is equal to $$\frac{\partial y_i}{\partial x_j}\frac{dx_j}{dt}$$ with implicit summation over $j$. But $\frac{\partial y_i}{\partial x_j}=M_{ij}$, so$$\dot{y}_i=\frac{\partial y_i}{\partial x_j}\frac{dx_j}{dt}=M_{ij}\dot{x}_j=(M\dot{\vec{x}})_i$$$$\implies \dot{\vec{y}}=M\dot{\vec{x}}=MJ\frac{\partial H}{\partial \vec{x}}.$$

Now recalling $H(\vec{x})=\tilde{H}(\vec{y}(\vec{x}))$, apply the chain rule to get$$\frac{\partial H}{\partial x_i}=\frac{\partial \tilde{H}(\vec{y})}{\partial y_j}\frac{\partial y_j}{\partial x_i}=M_{ji}\frac{\partial \tilde{H}}{\partial y_j}=(M^T)_{ij}\frac{\partial \tilde{H}}{\partial y_j}=\left(M^T \frac{\partial \tilde{H}}{\partial \vec{y}}\right)_i$$$$\frac{\partial H}{\partial \vec{x}}=M^T \frac{\partial \tilde{H}}{\partial \vec{y}}.$$

Substitute this into the above gives$$\dot{\vec{y}}=MJM^T\frac{\partial \tilde{H}}{\partial\vec{y}}$$as required.