I Proof that canonical transformation implies symplectic condition

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Goldstein's Classical Mechanics asserts that a transformation of coordinates is canonical if and only if the symplectic condition, represented by the equation ##MJM^T= J##, holds true. The discussion reveals confusion regarding the proof of the converse, which is not provided in the text. An example is given where the transformation P=p and Q=2q does not satisfy the symplectic condition, yet the Hamiltonian system remains valid with a new Hamiltonian K=2H. The conversation also touches on the use of generating functions for canonical transformations and the general definition involving differential forms. The complexities of Hamiltonian formalism highlight the need for careful study beyond basic physics textbooks.
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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.
 
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A Hamiltonian function is not changed under the canonical transformation independent on time: K=H
 
Lagrange fanboy said:
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
 
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wrobel said:
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
$$p=\partial_q f, \quad P=-\partial_Q f, \quad H'=H+\partial_t f.$$
 
Please read the post you are quoting :)
 
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
$$H(t,x,p)=K(t,X(x,p),P(x,p))$$
 
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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.
 
PhDeezNutz said:
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
 
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