# Show condition for canonical transformation

1. Oct 1, 2012

### aaaa202

1. The problem statement, all variables and given/known data
Consider the transformation from the variables (q,p) to (Q,P) by virtue of q = q(Q,P), p = p(Q,P) and H(q,p,t) = H(Q,P,t). Show that the equations of motion for Q,P are:
$\partial$H/$\partial$Q = -JDdP/dt
$\partial$H/$\partial$P = JDdQ/dt
where JD is the Jacobian determinant det($\partial$(q,p)/$\partial$(Q,P))
this shows the transformation is canonical only if JD=1.

2. Relevant equations

3. The attempt at a solution
I have tried to write some equations which might help me. They can be found on the attached picture. I would like to know which these can get me on track of the solution. Also I would like to know if my expression for the Jaciobian determinant is correct.
As a side question I would like to know why you can assume the two variables to have same hamiltonian. Is this because the transformation is not time dependent?

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2. Oct 1, 2012

### voko

I think you should start from the equations of motion in the original coordinates and, using the chain rule, express everything in the new coordinates. Then after some algebra you should get the result.

3. Oct 1, 2012

### aaaa202

I tried going that way but didn't really get anything pretty. What I got is attached - does it in any way resemble any of your steps?

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4. Oct 1, 2012

### voko

It is just a start. Now solve that for the time derivatives of Q and P. You will get at some stage expressions of the kind (dq/dQ dP/dq + dp/dQ dP/dp) (where "d" is the partial derivative symbol). Note that this is just dP/dQ = 0. Likewise, dq/dQ dQ/dq + dp/dQ dQ/dp = dQ/dQ = 1.

5. Oct 2, 2012

### aaaa202

This is where I could get to. But then the expressions get ugly and I get really see a way to get dH/dQ.

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6. Oct 2, 2012

### voko

Collect the terms at the time derivative of Q, and the partial derivatives of H, and simplify. That's what this is really all about.

7. Oct 2, 2012

### aaaa202

okay I think I got it now.

But another question. Since this approach is very messy:
Is it not possible to use the symplectic condition for a canonical transformation and prove it from there:

I.e. show that MJMT = J if and only if JD=0.
Here M is the jacobian matrix and J the sympletic matrix, which you probably know.

Last edited: Oct 2, 2012
8. Oct 2, 2012

### voko

By the way, you could simplify your life somewhat by introducing
A = dQ/dq, B = dQ/dp, C = dP/dq, D = dP/dp
a = dq/dQ, b = dq/dP, c = dp/dQ, d = dp/dP
X = dH/dQ Y = dH/dP

9. Oct 2, 2012

### aaaa202

I managed to show it. Although for some reason I get a 2 in front of my dH/dP such that:

2$\partial$H/$\partial$p =JDdQ/dt

did you get that too?

I have attached an explanation of where I get it (also note that I corrected the mistake I made on the last picture - if you noticed that)

Also: Would the symplectic approach described above work?

#### Attached Files:

• ###### hamiltonm.png
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Last edited: Oct 2, 2012
10. Oct 2, 2012

### voko

I don't understand that explanation.

I am not sure about he symplectic approach. If I remember correctly, the symplectic matrix is a basis transformation matrix in the cotangent space, so it should itself be the Jacobian in this case.

11. Oct 2, 2012

### aaaa202

Did you not have the same calculations? All my point was that I get a dH/dP term from two terms in the sum: namely dp/dP(dH/dP*dP/dp) + dq/dP(dH/dP*dP/dq) = 2dH/dP
did you not get these two?
note all d's are partials

12. Oct 2, 2012

### voko

Because they are partial. dp/dP * dP/dp do NOT cancel each other. However, as I remarked above, dp/dP * dP/dp + dq/dP *dP/dq = dP/dP = 1.

13. Oct 2, 2012

### aaaa202

ahh yes! I get it now. Thanks so much voko, you're always very helpful ;)