The restricted canonical transformation group

  • #1
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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



(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?
 
Last edited:

Answers and Replies

  • #2
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Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
 

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