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Question about canonical transformations

  1. Sep 23, 2010 #1
    Hi everybody

    I've got a problem related to canonical transformations that I can`t solve:

    Given the expression of the canonical transformation

    [tex]
    Q=3q\cdot\big[ \exp\big((p+q)^5\big)+1\big] +3p\cdot \big[\exp((p+q)^5)+1\big]+p
    [/tex]
    [tex]
    P=p+q
    [/tex]
    I have to calculate an associated canonical transformation. Anybody can help me?

    Thanks
     
  2. jcsd
  3. Sep 23, 2010 #2
    Are you sure you have stated exactly your problem?
     
  4. Sep 23, 2010 #3
    Exact problems is (from exam):
    Given the transformation

    [tex]
    Q=3q\cdot\big[ \exp\big((p+q)^5\big)+1\big] +p\cdot \big[2\exp((p+q)^5)-1\big]
    [/tex]

    [tex]
    P=q+p
    [/tex]

    Modify it slightly in order to be canonical

    Answer. After imposing the condition for the Poisson bracket (equal to one) we get the result:
    [tex]
    \boxed{Q=3q\cdot\big[ \exp\big((p+q)^5\big)+1\big] +3p\cdot \big[\exp((p+q)^5)+1\big]+p}
    [/tex]

    In part two we're asked to obtain an associated canonical transformation. But after reading my books and papers about mechanics I still don't know what's an "associated trasformation mean". Is it the inverse transformation?
     
    Last edited: Sep 23, 2010
  5. Sep 24, 2010 #4
    OK. Now part of the problem is clear - supposing it is indeed a canonical transformation (I didn't check). But what the author of this exercise means by an "associated canonical transformation" - that I don't know.
     
  6. Sep 24, 2010 #5

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    Same here. I suppose the questioner might just be looking for an equivalent transformation, but written in a different functional form... something like

    [tex]Q=Q(q,P)=3P\left(e^{P^5}+1\right) +P-q[/tex] and [tex]P=q+p[/tex]
    instead of
    [tex]Q=Q(q,p)=3q\left(e^{(p+q)^5}+1\right)+3p\left(e^{(p+q)^5}+1\right)+p[/tex] and [tex]P=q+p[/tex]

    ...but that's just a guess
     
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