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## 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

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**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?

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