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Quick question about constant of motion

  • Thread starter Gregg
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  • #1
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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.
 

Answers and Replies

  • #2
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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]
 
  • #3
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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} ##
 
  • #4
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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.
 
  • #5
tiny-tim
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Hi Gregg! :smile:
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?
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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