Relation between force in cartesian, polar.

  • Thread starter merrypark3
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  • #1
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Goldstein(3rd) 1.15

Generalized potential, U as follows.

[tex] U( \stackrel{\rightarrow}{r} ,\stackrel{\rightarrow}{v})=V(r)+\sigma\cdot L[/tex]

L is angular momentum and [tex]\sigma[/tex] is a fixed vector.


(b) show thate the component of the forces in the two coordinate systems(cartesin, spherical polar) are related to each other as
[tex]Q_{j}=F_{i}\cdot \frac{\partial r_{i}} {\partial q_{j}} \cdots (a) [/tex]

So I did,
[tex] Q_{j}= - \frac{\partial U}{\partial q_{j}} + \frac{d}{dt} (\frac{\partial U}{\partial \dot q_{j}})[/tex]

[tex]
=- \frac{\partial x_{k}}{\partial q_{j}} \frac{\partial U}{\partial x_{k}} + \frac{d}{dt}(\frac{\partial \dot x_{k}}{\partial \dot q_{j}} \frac{\partial U}{\partial \dot x_{k}})
[/tex]

[tex]
=- \frac{\partial x_{k}}{\partial q_{j}} \frac{\partial U}{\partial x_{k}} + \frac{d}{dt}(\frac{\partial x_{k}}{\partial q_{j}} \frac{\partial U}{\partial \dot x_{k}})
[/tex]

[tex]
=- \frac{\partial x_{k}}{\partial q_{j}} \frac{\partial U}{\partial x_{k}} + \frac{d}{dt}(\frac{\partial x_{k}}{\partial q_{j}}) \frac{\partial U}{\partial \dot x_{k}}+ \frac{\partial x_{k}}{\partial q_{j}} \frac{d}{dt}(\frac{\partial U}{\partial \dot x_{k}})
[/tex]

[tex]=\frac{\partial x_{k}}{\partial q_{j}}(
- \frac{\partial U}{\partial x_{k}} + \frac{d}{dt} (\frac{\partial U}{\partial \dot x_{k}}))+ \frac{d}{dt}(\frac{\partial x_{k}}{\partial q_{j}}) \frac{\partial U}{\partial \dot x_{k}}
[/tex]

[tex]=\frac{\partial x_{k}}{\partial q_{j}} (
F_{k})+ \frac{d}{dt}(\frac{\partial x_{k}}{\partial q_{j}}) \frac{\partial U}{\partial \dot x_{k}}
[/tex]



If the last term in the last line vanishes, [tex] Q_{j} [/tex] and [tex] F_{k} [/tex] satisfies the relation (a), but it DOESN't vanish. What's my problem??
I've evaluated the last term in this condition, but It doesn't....
 
Last edited:

Answers and Replies

  • #2
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Replace x with r:

[tex]\frac{d}{dt}(\frac{\partial r_{k}}{\partial q_{j}}) \frac{\partial U}{\partial \dot r_{k}}[/tex]

Does U change with respect to r dot?
 
  • #3
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Replace x with r:

[tex]\frac{d}{dt}(\frac{\partial r_{k}}{\partial q_{j}}) \frac{\partial U}{\partial \dot r_{k}}[/tex]

Does U change with respect to r dot?
Yes, There's L(angular momentum) in the U
 
  • #4
288
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Explicitly write the angular momentum terms and see if any have r dot terms.
 
  • #5
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Explicitly write the angular momentum terms and see if any have r dot terms.
Isn't x the component cartesian coordinate?
 
  • #6
288
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You could have started the problem as this:

[tex] Q_{r}= - \frac{\partial U}{\partial r} + \frac{d}{dt} (\frac{\partial U}{\partial \dot r})[/tex]

and expressed the generalized potential as this:

[tex] U( \stackrel{\rightarrow}{r} ,\stackrel{\rightarrow}{v})=V(r)+\sigma_{\theta}\L_{\theta}+\sigma_{\phi}\L_{\phi}[/tex]

Express the L terms explicitly in terms of m, r, theta, phi, and sigma and the second term on the right side of the generalized force equation will equal zero.
 
Last edited:

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