Relation between force in cartesian, polar.

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Homework Help Overview

The discussion revolves around the relationship between forces in Cartesian and polar coordinate systems, specifically in the context of generalized potentials and angular momentum. The original poster attempts to derive a relationship between the components of forces in these two systems, referencing a problem from Goldstein's textbook.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the derivation of force components and question the conditions under which certain terms vanish in the equations. There is a focus on the role of angular momentum in the generalized potential and its implications for the relationship between Cartesian and polar coordinates.

Discussion Status

Some participants have offered guidance on rewriting terms and explicitly expressing angular momentum components. There is an ongoing exploration of whether certain terms in the equations contribute to the overall relationship being investigated, with no explicit consensus reached on the resolution of the original poster's concerns.

Contextual Notes

Participants are considering the implications of angular momentum on the generalized potential and the specific forms of the equations involved. There is an acknowledgment of the complexity introduced by the transition between coordinate systems.

merrypark3
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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}}(<br /> - \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}} (<br /> 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:
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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?
 
chrisk said:
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
 
Explicitly write the angular momentum terms and see if any have r dot terms.
 
chrisk said:
Explicitly write the angular momentum terms and see if any have r dot terms.

Isn't x the component cartesian coordinate?
 
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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