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Given a canonical transformation, how does one find its type?

  1. Dec 17, 2013 #1
    I'm given the following transformation

    [tex]X=x \cos \alpha - \frac{p_y}{\beta} \sin \alpha[/tex]
    [tex]Y=y \cos \alpha - \frac{p_x}{\beta} \sin \alpha[/tex]
    [tex]P_X=\beta y \sin \alpha + p_x \cos \alpha[/tex]
    [tex]P_Y=\beta x \sin \alpha + p_y \cos \alpha[/tex]

    and I'm asked to find what type(s) of transformation it is. I'm not sure how to go about doing this without being given a generator. Do I basically try and find a generator? If so, how do I do this. Do I start from Hamilton's principle and try the definition of each type of generator?
  2. jcsd
  3. Dec 19, 2013 #2
    Not quite, that is a lot more work. You don't have to actually find the generator to determine what type it should be.

    Say, for the sake of argument, the generator [itex]F[/itex] is a function of [itex]p_x,y,X,Y[/itex] (it's not) so that [itex]F=F(p_x,y,X,Y)[/itex]. This means that you would be able to write the other coordinates (namely [itex]p_y,x,P_x,P_y[/itex]) in terms of these ones. I can see that this type of generating function wont work since when I try to write [itex]P_y[/itex] I find (you'll need to invert the transformation to see this):
    P_y = x\beta \sin\alpha + p_y\cos\alpha = \left(X\cos\alpha + \frac{\sin\alpha}{\beta}P_y\right) \beta\sin\alpha + p_y\cos\alpha
    and that simplifies to [itex]P_y=X\beta\tan\alpha + p_y\sec\alpha [/itex]. Therefore, I cannot write [itex]P_y[/itex] as a function of only those chosen coordinates. Try something else and see if you can find a coordinate set that works---then you will know what type of generating function you can have.
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