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When can I substitute transformations into the Hamiltonian?

  1. Mar 28, 2015 #1
    1. If I know that ##H(q_i,p_i,t)## is a valid Hamiltonian for which the hamilton equations hold. Now we are given that ##Q_j(q_i,p_i)## and ##P_j(q_i,p_i)## are canonical transformations. This means that there is a function ##K(Q_j,P_j)##, the new hamiltonian, for which the Hamilton equations hold in the new variables.

    QUESTION : What is this new function ##K##? Can I just substitute the transformations into the old ##H## to find ##K##? If not, when is this allowed?

    Reason for confusion: My notes from class tell me this is allowed. However any source on the net I find says that in general canonical transformations do not preserve the Hamiltonian and thus ##H## and ##K## are not always equal.
     
  2. jcsd
  3. Mar 29, 2015 #2
    If P and Q depend only p and q, and the there is no additional time dependence in the coordinate transformation, then
    $$K(Q, P) = H(q(Q,P), p(Q, P))$$

    Otherwise it can take forms such as
    $$K(Q, P) = H(q(Q,P), p(Q, P)) + \frac{\partial F_1(q(Q, P), Q, t)}{\partial t} $$
    Where for this type of transformation
    $$p = \frac{\partial F_1}{\partial q} $$
    $$P = -\frac{\partial F_1}{\partial Q} $$
    So in this case if ##F_1## has no time dependence then ##K ##is just ##H##at the corresponding coordinates.
     
  4. Mar 29, 2015 #3
    Thanks for the answer. This is exactly what I have thought up myself by now. One more question since you seem to know the subject if you don't mind?

    I'm a bit confused on the 4 ''types'' of generating functions and how their forms are found. For example for ##F(q,Q,p)## the form makes sense because it's just ##F=F(q,Q,p)##, but how does one knows to add the ##-QP## when dealing with the ##F(q,P,t)## type? It seems to work out alright but I seem to not grasp how one can make these 4 distinctions so clear and so strict? What if I take ##F=F(q,P,t)## as the generating function without the ##-QP## part?
     
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