Proof that canonical transformation implies symplectic condition

[Lagrange fanboy]

TL;DR Summary

Goldstein's classical mechanics shows the proof that if symplectic condition holds, then the transformation is canonical. The converse was claimed to be true, but I can't derive it.

Goldstein's Classical Mechanics makes the claim (pages 382 to 383) that given coordinates $q,p$, Hamiltonian $H$, and new coordinates $Q(q,p),P(q,p)$, there exists a transformed Hamiltonian $K$ such that $\dot Q_i = \frac{\partial K}{ \partial P_i}$ and $\dot P_i = -\frac{\partial K}{Q_i}$ if and only if $MJM^T= J$ where $M$ is the Jacobian of $Q,P$ with respect to $q,p$ and $J = \begin{bmatrix} O&I\\\\ -I&O \end{bmatrix}$. I understood the book's proof that $MJM^T= J$ implies the existence of such $K$. However, the proof of the converse was not given and I do not know how to derive it myself.

 

[wrobel]

A Hamiltonian function is not changed under the canonical transformation independent on time: K=H

 

[wrobel]

TL;DR Summary: Goldstein's classical mechanics shows the proof that if symplectic condition holds, then the transformation is canonical. The converse was claimed to be true, but I can't derive it.

Goldstein's Classical Mechanics makes the claim (pages 382 to 383) that given coordinates $q,p$, Hamiltonian $H$, and new coordinates $Q(q,p),P(q,p)$, there exists a transformed Hamiltonian $K$ such that $\dot Q_i = \frac{\partial K}{ \partial P_i}$ and $\dot P_i = -\frac{\partial K}{Q_i}$ if and only if $MJM^T= J$ where $M$ is the Jacobian of $Q,P$ with respect to $q,p$ and $J = \begin{bmatrix} O&I\\\\ -I&O \end{bmatrix}$. I understood the book's proof that $MJM^T= J$ implies the existence of such $K$. However, the proof of the converse was not given and I do not know how to derive it myself.

This proposition is actually wrong. Indeed, the transformation P=p, Q=2q does not satisfy $MJM^T= J$. But after this transformation a Hamiltonian system remains a Hamiltonian one with the new Hamiltonian K=2H.
The Hamiltonian formalism is a subtle enough mathematical topic to study it by physics textbooks

 

[vanhees71]

A Hamiltonian function is not changed under the canonical transformation independent on time: K=H

No, usually you have an additional term. The most convenient way is to use a generating function, e.g., $f(q,Q,t)$. Then the canonical transformation is given as
$$p=\partial_q f, \quad P=-\partial_Q f, \quad H'=H+\partial_t f.$$

 

[wrobel]

Please read the post you are quoting :)

 

[wrobel]

By the way: Most general definition of a canonical transformation of an extended phase space is as follows.
A transformation
$$(t,x,p)\mapsto(T,X,P)$$ is said to be canonical if
$$dp_i\wedge dx^i-dH\wedge dt=dP_i\wedge dX^i-dK\wedge dT,$$ where
$H=H(t,x,p),\quad K=K(T,X,P).$ Such transformations take a Hamiltonian system
$$\frac{dp}{dt}=-\frac{\partial H}{\partial x},\quad \frac{dx}{dt}=\frac{\partial H}{\partial p}$$
to a Hamiltonian one
$$\frac{dP}{dT}=-\frac{\partial K}{\partial X},\quad \frac{dX}{dT}=\frac{\partial K}{\partial P}.$$
But the most common type of canonical transformations is a special case of the above:
$$(t,x,p)\mapsto(T,X,P),\quad T=t.\qquad (*)$$
Introduce a notation
$$f=f(t,x,p),\quad \delta f=\frac{\partial f}{\partial x^i}dx^i+\frac{\partial f}{\partial p_i}dp_i.$$
Theorem. A transformation (*) is canonical iff $\delta P_i\wedge\delta X^i=\delta p_i\wedge\delta x^i.$

Here we assume that independent variables are $p,x,t$ that is
$$P=P(t,x,p),\quad X=X(t,x,p)\qquad (**)$$ and correspondingly $dx=\delta x,\quad dp=\delta p.$

If the transformation (**) is canonical and $X=X(x,p),\quad P=P(x,p)$ then
$$H(t,x,p)=K(t,X(x,p),P(x,p))$$

 

[PhDeezNutz]

Excuse the Juvenile comment but the first time I heard my classical mechanics teacher say "Kamiltonian" I struggled to withhold my laughter. I couldn't believe he said it with a straight face.

 

[wrobel]

Spoiler: offtop

Spoiler: offtop

Excuse the Juvenile comment but the first time I heard my classical mechanics teacher say "Kamiltonian" I struggled to withhold my laughter. I couldn't believe he said it with a straight face.

In Russian, the letter H in the word Hamiltonian is pronounced as g in the word gone. Oh, I see for English speakers this Kamiltonian sounds somehow like camel