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Quick question about constant of motion

  1. May 8, 2012 #1
    I have found that ##J = m r^2 \sin^2 \theta \dot{\phi}^2 ## is a constant of motion. How is this orbital angular momentum in the ##\theta = 0 ## direction? It is a particle moving in a central force field.
     
  2. jcsd
  3. May 8, 2012 #2
    Shouldn't that be [itex] J = m r^2 \sin^2(\theta) \dot{\phi} [/itex] ? I guess you should try to compare this to the "usual" definition, which is [itex] \mathbf{J} = \mathbf{r} \times \mathbf{p} [/itex]
     
  4. May 12, 2012 #3
    I can't work that out i can't do a cross product in polar coordinates.

    ## J = r p \sin (\theta) ##

    ## p = m r \dot{\phi} \sin (\theta) ## ?

    ## J = m r^2 \sin^2(\theta) \dot{\phi} ##
     
  5. May 14, 2012 #4
    The cross product works exactly like with the usual rectangular coordinates. If you want to make it more straightforward, take the absolute value so you don't have to worry about sign conventions. If you are really worried, you can also transform back to rectangular coordinates but that's a ton of unneeded work.
     
  6. May 14, 2012 #5

    tiny-tim

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    Hi Gregg! :smile:
    If you draw a "horizontal" circle, it has radius rsinθ, and the component of velocity parallel to the circle is φ'rsinθ :wink:
     
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