Show that transformation is Canonical

[Gunthi]

Homework Statement

Given the transformation $$Q=qe^{\gamma t}$$ and $$P=pe^{-\gamma t}$$ with the Hamiltonean $$H=\frac{p^2e^{-2\gamma t}}{2m}+\frac{m\omega^2q^2e^{2\gamma t}}{2}$$ show that the transformation is Canonical

Homework Equations

I know that the condition for a transformation to be canonical is $$\sum_i p_i\dot{q_i}-P_i\dot{Q_i}=H-K+\frac{dF_2}{dt}$$ but don't know what to do with it...

I've managed to prove from $$\frac{dF_2}{dq_i}=p_i$$ and $$\frac{dF_2}{dP_i}=Q_i$$ that $$F_2=qPe^{\gamma t}$$.

 

[Gunthi]

I have another question related to this problem. Does having the generating function guarantee that the coordinate transformation associated with it is canonical? Or is it a necessary but not sufficient condition?

 

[fluidistic]

I know a way to check out whether the transformation is canonical but it doesn't involve the Hamiltonian.
If you know Poisson's brackets, if you can show that $[Q,P]_{q,p}=1$ then you'd be done.
To answer your question "Does having the generating function guarantee that the coordinate transformation associated with it is canonical? Or is it a necessary but not sufficient condition?", I'm not 100% sure but I'd say yes (99% sure).

 

[Gunthi]

I've been messing around with the coordinates and got this:

If we have a general invertible transformation of the type $$Q=Q(q,p,t)\Leftrightarrow q=q(Q,P,t)$$ and $$P=P(q,p,t)\Leftrightarrow p=p(P,Q,t)$$ then the following is true:

\begin{matrix}
\dot{q}=\frac{\partial q}{\partial Q}\dot{Q}+\frac{\partial q}{\partial P}\dot{P}\\
\dot{p}=\frac{\partial p}{\partial Q}\dot{Q}+\frac{\partial p}{\partial P}\dot{P}
\end{matrix}

From Hamilton's eqs we have (inserting the given coordinate transformations):
\begin{matrix}
\dot{q}=\frac{\partial H}{\partial p}=\frac{\partial H}{\partial P}e^{-\gamma t}=\frac{\partial q}{\partial Q}\dot{Q}\Rightarrow \frac{\partial H}{\partial P}=\dot{Q}\\
\dot{p}=-\frac{\partial H}{\partial q}= -\frac{\partial H}{\partial Q}e^{\gamma t}=\frac{\partial p}{\partial P}\dot{P}\Rightarrow -\frac{\partial H}{\partial Q}=\dot{P}
\end{matrix}

The form of Hamilton's equations is the same but from what I've seen in Goldstein, when one makes a canonical transformation the form is preserved not for the initial Hamiltonian but for the "Kamiltonian" $$K=H+\frac{dF_2}{dt}$$ i.e.

\begin{matrix}
\frac{\partial K}{\partial P}=\dot{Q}\\
-\frac{\partial K}{\partial Q}=\dot{P}
\end{matrix}

and because I've already proved that there exists a function $F_2$ I get the feeling that the proof is not correct. What am I doing wrong here?

I know a way to check out whether the transformation is canonical but it doesn't involve the Hamiltonian.
If you know Poisson's brackets, if you can show that $[Q,P]_{q,p}=1$ then you'd be done.
To answer your question "Does having the generating function guarantee that the coordinate transformation associated with it is canonical? Or is it a necessary but not sufficient condition?", I'm not 100% sure but I'd say yes (99% sure).

Thanks for the advice! Actually, calculating with the Poisson brackets is another part of this problem, but instead it is asked to evaluate $[Q,Q],[P,P],[Q,P],[q,H],[p,H]$ without explicit calculation.

For the first two the answer is obviously zero because of the crossed derivatives, the third one, being this a suposedly canonical transformation is 1 and for the last two I'm having a little bit of a problem because if I use $\frac{du}{dt}=[u,H]+\frac{\partial u}{\partial t}$ on q and p I am not sure if it is $\frac{dq}{dt}$ or $\frac{\partial q}{\partial t}$ (for p too) that is zero. I know, this is a dumb question, but when a function depends only on t does the partial derivative looses its meaning?

 

[fluidistic]

Ok, since you found out $F_2$ then I think you are done. Let me give you a useful link: http://books.google.com.ar/books?id...a transformation is canonical poisson&f=false.
About the Poisson's bracket question, you are right for the 3 first. As for the last 2 I must say I'm not sure; I'd consult Landau's book on classical mechanics first and then other books.
About your question about partial derivatives; when you have a function of only 1 variable a partial derivative is the same as a total derivative.
But I doubt q only depends on t.

 

[Gunthi]

Ok, since you found out $F_2$ then I think you are done. Let me give you a useful link: http://books.google.com.ar/books?id...a transformation is canonical poisson&f=false.
About the Poisson's bracket question, you are right for the 3 first. As for the last 2 I must say I'm not sure; I'd consult Landau's book on classical mechanics first and then other books.
About your question about partial derivatives; when you have a function of only 1 variable a partial derivative is the same as a total derivative.
But I doubt q only depends on t.

Thanks for the book! I tried to see if there was anything in Landau that could help but didn't find it yet. Anyhow, I found this http://solar.physics.montana.edu/dana/ph411/p_brack.pdf and I can't figure out why, in page 2 after equation (3) $\frac{\partial q_1}{\partial t}=0$. Isn't $q_1$ also a function of t? Why isn't $\frac{dq_1}{dt}$ also zero?

 

[fluidistic]

Thanks for the book! I tried to see if there was anything in Landau that could help but didn't find it yet. Anyhow, I found this http://solar.physics.montana.edu/dana/ph411/p_brack.pdf and I can't figure out why, in page 2 after equation (3) $\frac{\partial q_1}{\partial t}=0$. Isn't $q_1$ also a function of t? Why isn't $\frac{dq_1}{dt}$ also zero?

I'm in the same boat as you. I do not understand why it is so.
For sure, $q_1$ isn't a constant of motion, otherwise we'd have $\{ q_1 , H \}=0$. So there's indeed a dependency of $q_1$ with respect to time. Thus, I do not understand how it's possible that $\frac{\partial q _1 }{\partial t } =0$.
I wish someone could enlighten us!

 

[Gunthi]

I'm just posting this so that the thread doesn't seem closed.