1. Aug 2, 2011

### demonelite123

Suppose i release a particle at (x=a,y=0) with (p_x = b, p_y = 0) and you release one in the transformed state (x=0, y=a) with (p_x = b, p_y = 0) where the transformation is that we rotate the coordinates but not the momenta. This is a non canonical transformation that leaves H invariant. Show that at later times the states of the two particles are not related by the same transformation.

i am not sure what to do on this problem. i can intuitively see that if the acceleration is 0, both particles move to the right at a constant velocity of p_x / m but i am not sure how mathematically describe the transformations of the two particles through time and show that they are different. can anyone give me some pointers? thanks.

2. Aug 2, 2011

### mathfeel

You are confusing canonical transformation with two different initial condition. You did not perform a canonical transformation. What you needs is 4 new variables (can be identical, of course if you wish) $\{Q_1, Q_2, P_1, P_2\}$ that is the function of the original variable $\{q_1, q_2, p_1, p_2\}$. That is a transformation.

What you probably have in mind is something like this: $\{Q_1 = q_2, Q_2 = q_1, P_1 = p_1, P_2 = p_2\}$. And you are right, it is not a canonical trasnformation.

3. Aug 2, 2011

### demonelite123

ok so i know the transformation equations are xnew= xcos(θ) - ysin(θ), ynew= xsin(θ) + ycos(θ), and px,new=px, py,new= py since the momentum coordinates are unchanged. i am still confused on what to do with the 2 sets of initial conditions given to me. using the second set of initial conditions for the transformed coordinates i see that xnew = -asin(θ), ynew = acos(θ), px,new = b, py,new = 0 which are the initial points for the transformed system. after this i need to find the transformations for both particles which describe their position and momentum coordinates at any time t. how would i do this?

4. Aug 2, 2011

### mathfeel

Suppose $x(t)$, $y(t)$, $p_x(t)$, $p_y(t)$ is a solution to the Hamilton's equations for a particular initial condition.

Then does $X(t)$, $Y(t)$ etc, constructed from the transformation still satisfies Hamilton's equations?

5. Aug 2, 2011

### demonelite123

hm it seems clear to me that if you write out Hamilton's equations for the untransformed state and solve for the x(t) and y(t), it won't have the same form as xnew(t) and ynew(t) since replacing the old coordinates with the new coordinates in the differential equations will obviously result in a different set of equations in this specific case of the rotation of just the coordinates. i imagine this sort of thing occurs for all transformations that are not canonical since the new coordinates do not satisfy the equations. is this the correct reasoning? thanks for your answers mathfeel.