Is a Rotated Frame a Canonical Transformation in Classical Mechanics?

[roeb]

Homework Statement

Verify that the change to a rotated frame is a canonical transformation:

[tex]\bar{x} = x cos\theta - y sin\theta[/tex]
[tex]\bar{y} = x sin \theta + y cos \theta[/tex]
[tex]\bar{p_x} = p_x cos \theta - p_y sin\theta[/tex]
[tex]\bar{p_y} = p_x sin \theta + p_y cos \theta[/tex]

Where [f,g] = poisson bracket

Homework Equations

The Attempt at a Solution

Hmm, well as far as I can tell in order to have a canonical transformation, we must say that:
[tex][q_j,q_k] = 0[/tex]
[tex][p_j, p_k] = 0[/tex]
[tex][q_j,p_k] = \delta_jk[/tex]

So here is what I've attempted:
I've got this funny feeling that
[tex] [\bar{x},\bar{p_y}] = \delta = 1 [/tex]
[tex] [\bar{y},\bar{p_x}] = \delta = 1 [/tex]
However that is just my intuition and I can't explain seem to find a mathematical way to explain why it isn't [tex]\bar{x}, \bar{p_x}[/tex] etc.

Nonetheless, I've done it both ways, and I seem to get the wrong answer either way:

[tex][\bar{x},\bar{p_y}] = \frac{\partial \bar{x}}{\partial x} \frac{\partial \bar{p_y}}{\partial p_x} + \frac{\partial \bar{x}}{\partial x} \frac{\partial \bar{p_y}}{\partial p_y} + \frac{ \partial \bar{x}}{\partial y} \frac{ \partial \bar{p_y}}{\partial p_x} + \frac{\partial \bar{x}}{\partial y} \frac{\partial \bar{p_y}}{\partial p_y} = cos\theta sin\theta + cos^2 \theta + -sin^2 \theta + - sin \theta cos \theta = cos^2 \theta - sin^2 \theta[/tex]

Likewise for [x,px] I get:
[tex]cos^2 \theta - 2 sin \theta cos \theta + sin^2 \theta[/tex]The math seems straightforward so I have a feeling I'm doing something incorrectly with respect to setting up the problem. Does anyone see where I've gone wrong?

 

[Phrak]

What canon is being invoked?

 

[roeb]

Thanks for your reply,

I'm afraid I'm not quite sure what you mean. This is stemming from an analysis of the Hamiltonian (Shankar QM Ch 2.7).

 

[Phrak]

sorry. Canon mean "by law". I'm not sure what law, or rule is being invoked.

It derives from "Papal Canon" as far as I can tell.

Whenever a professor uses these sort of words you should ask what they mean. "Manifestly" is another one. You shouldn't be afraid to ask. None of your peers know what they mean either. (And there's a good chance your professor doesn't know either.)

Honestly, I don't know what rule is to be taken as obvious. When a vector is rotated in a plane does it's length remain the same. This seems obvious, right? In other geometries it's not obvious, and is in fact false.

 

[gabbagabbahey]

Homework Statement

Verify that the change to a rotated frame is a canonical transformation:

[tex]\bar{x} = x cos\theta - y sin\theta[/tex]
[tex]\bar{y} = x sin \theta + y cos \theta[/tex]
[tex]\bar{p_x} = p_x cos \theta - p_y sin\theta[/tex]
[tex]\bar{p_y} = p_x sin \theta + p_y cos \theta[/tex]

Where [f,g] = poisson bracket

Homework Equations

The Attempt at a Solution

Hmm, well as far as I can tell in order to have a canonical transformation, we must say that:
[tex][q_j,q_k] = 0[/tex]
[tex][p_j, p_k] = 0[/tex]
[tex][q_j,p_k] = \delta_jk[/tex]

What you really want to show is that:

[tex][\bar{q}_j,\bar{q}_k] = 0[/tex]
[tex][\bar{p}_j, \bar{p}_k] = 0[/tex]
[tex][\bar{q}_j,\bar{p}_k] = \delta_{jk}[/tex]

Where [itex]\bar{q}_1=\bar{x}[/itex], [itex]\bar{q}_2=\bar{y}[/itex], [itex]\bar{p}_1=\bar{p}_x[/itex] and [itex]\bar{p}_2=\bar{p}_y[/itex]

So here is what I've attempted:
I've got this funny feeling that
[tex] [\bar{x},\bar{p_y}] = \delta = 1 [/tex]
[tex] [\bar{y},\bar{p_x}] = \delta = 1 [/tex]
However that is just my intuition and I can't explain seem to find a mathematical way to explain why it isn't [tex]\bar{x}, \bar{p_x}[/tex] etc.

This makes no sense, the kronecker delta without any indices is meaningless.

The equation [itex][\bar{q}_j,\bar{p}_k] = \delta_{jk}[/itex] implies that [itex][\bar{q}_1,\bar{p}_1]=[\bar{q}_2,\bar{p}_2] =1[/itex] and [itex][\bar{q}_1,\bar{p}_2]=[\bar{q}_2,\bar{p}_1] =0[/itex]...Or, in terms of [itex]\bar{x}[/itex] and [itex]\bar{y}[/itex], [itex][\bar{x},\bar{p}_x]=[\bar{y},\bar{p}_y] =1[/itex] and [itex][\bar{x},\bar{p}_y]=[\bar{y},\bar{p}_x] =0[/itex]...these are what you need to show are true (along with the other two conditions: [itex][\bar{q}_j,\bar{q}_k] = 0[/itex] and [itex][\bar{p}_j, \bar{p}_k] = 0[/itex])

Nonetheless, I've done it both ways, and I seem to get the wrong answer either way:

[tex][\bar{x},\bar{p_y}] = \frac{\partial \bar{x}}{\partial x} \frac{\partial \bar{p_y}}{\partial p_x} + \frac{\partial \bar{x}}{\partial x} \frac{\partial \bar{p_y}}{\partial p_y} + \frac{ \partial \bar{x}}{\partial y} \frac{ \partial \bar{p_y}}{\partial p_x} + \frac{\partial \bar{x}}{\partial y} \frac{\partial \bar{p_y}}{\partial p_y} = cos\theta sin\theta + cos^2 \theta + -sin^2 \theta + - sin \theta cos \theta = cos^2 \theta - sin^2 \theta[/tex]

No,

[tex][\bar{x},\bar{p}_y] =[\bar{q}_1,\bar{p}_2] =\sum_{i=1,2} \frac{\partial \bar{q}_1}{\partial q_i} \frac{\partial \bar{p}_2}{\partial p_i} - \frac{\partial \bar{q}_1}{\partial p_i} \frac{\partial \bar{p}_2}{\partial q_i}=\left(\frac{\partial \bar{q}_1}{\partial q_1} \frac{\partial \bar{p}_2}{\partial p_1} - \frac{\partial \bar{q}_1}{\partial p_1} \frac{\partial \bar{p}_2}{\partial q_1}\right)+\left( \frac{\partial \bar{q}_1}{\partial q_2} \frac{\partial \bar{p}_2}{\partial p_2} - \frac{\partial \bar{q}_1}{\partial p_2} \frac{\partial \bar{p}_2}{\partial q_2} \right)=\left(\frac{\partial \bar{x}}{\partial x} \frac{\partial \bar{p}_y}{\partial p_x} - \frac{\partial \bar{x}}{\partial p_x} \frac{\partial \bar{p}_y}{\partial x}\right)+\left( \frac{\partial \bar{x}}{\partial y} \frac{\partial \bar{p}_y}{\partial p_y} - \frac{\partial \bar{x}}{\partial p_y} \frac{\partial \bar{p}_y}{\partial y} \right)[/tex]