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Goldstein Problem

  1. Sep 12, 2006 #1
    Can someone point me in the right direction with the derivation number 2 from Chapter 2 (3rd edition) of Goldstein?
     
  2. jcsd
  3. Sep 12, 2006 #2

    Kurdt

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    Not everybody has the book unfortunately, but if you tell us what derivation it is then perhaps we can help.
     
  4. Sep 12, 2006 #3
    The derivation is if the lagrangian contained velocity terms, derive what the conjugate momentum will be if the system is rotated by a angle in some direction.
     
  5. Sep 12, 2006 #4

    radou

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    I've got Goldstein, and I'm lost on that one too. (As on most of them, btw. :biggrin: )
     
  6. Sep 12, 2006 #5
    How would you define a potential with velocity dependent terms?
     
  7. Sep 13, 2006 #6

    Kurdt

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    Well I hope you know the lagrangian is given by:

    [tex] L=T-V = \frac{1}{2} m (\dot{x}+\dot{y}+\dot{z}) - V(x,y,z) [/tex]

    In cartesian coordinates. The conjugate momentum for a particular coordinate is given by:

    [tex] \frac{\partial L}{\partial \dot{x}}= p_x[/tex]

    Now if the system was rotated by a particular angle how would you then modify the original expression to deal with a rotation. If you consider the x,y,z coordinates to be a vector [tex]\mathbf{r}[/tex] rotated about any arbitrary unit vector [tex]\mathbf{\hat{a}}[/tex] by a small angle [tex]\theta[/tex].

    Now all you have to do is findout how the x,y,z coordinates are related to [tex]\mathbf{\hat{a}}[/tex] through vector [tex]\mathbf{r}[/tex] and plug them into the lagrangian and see what the turn up. In some books they set [tex]\mathbf{\hat{a}}[/tex] parallel to the z-axis for ease of computation and normally call it [tex]\mathbf{\hat{k}}[/tex] instead.

    Like I say I don't have the textbook but I assume this is something to do with conservation of angular momentum?

    Need any more pointers just post.
     
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