The restricted canonical transformation group

[shinobi20]

Homework Statement

Show that the set of restricted canonical transformation forms a group. Verify this statement once using the invariance of Hamilton's principle under canonical transformation, and again using the symplectic condition.

Homework Equations

(Invariance of Hamilton's principle under canonical transformation)
$p_i\dot q_i - H = P_i\dot Q_i - K + \frac{dF}{dt}~~~~~$

(Symplectic Condition)
$MJM^T = J$ for some symplectic matrix $M$ and $J$ is such that the Hamilton's equations in symplectic notation can be written as $\dot η = J \frac{∂H}{∂η}$

The Attempt at a Solution

For the first part, suppose $~~p_2(p_1, q_1), q_2(p_1, q_1)~$ and $~p_3(p_2, q_2), q_3(p_2, q_2)$ are canonical transformations, then

$p_1\dot q_1 - H_1 = p_2\dot q_2 - K_1 + \frac{dF_1}{dt}~~$ and $~~p_2\dot q_2 - H_2 = p_3\dot q_3 - K_2 + \frac{dF_2}{dt}~~~$

By substituting $p_2\dot q_2$ from the right eq to the left, we have
$p_1\dot q_1 - H_1 = H_2 + p_3\dot q_3 - K_1 - K_2 + \frac{dF_1 + dF_2}{dt}~~$
$p_1\dot q_1 - (H_1 + H_2) = p_3\dot q_3 - (K_1 + K_2) + \frac{dF_1 + dF_2}{dt}~~$

Thus, $p_3(p_1, q_1), q_3(p_1, q_1)$ is a canonical transformation therefore belongs to the group.

For the second part, given the coordinate transformations $\dot ξ = M_1 \dot η~$ and $\dot χ = M_2 \dot ξ~$ →$~\dot χ = M_2M_1 \dot η$

$(M_2M_1)J(M_2M_1)^T = M_2M_1JM_1^TM_2^T = M_2JM_2^T = J$

Thus, it satisfies the symplectic condition.

Can anybody help me check if what I've done is correct?

 

[mark726]

Thank you.

Your solution looks correct to me. You have shown that the set of restricted canonical transformations forms a group by satisfying both the invariance of Hamilton's principle and the symplectic condition. Your steps and reasoning are clear and thorough. Great job!