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How can a potential depending on velocities give equal and opposite forces?

 
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Jul26-09, 06:45 AM   #1
 

How can a potential depending on velocities give equal and opposite forces?


On page 10 of Goldstein's classical mechanics, it was said:

"If [tex]V_{jj}[/tex] were also a function of the difference of some other pair of vectors associated with the particles, such as their velocities or (to step into the domain of modern physics) their intrinsic "spin" angular momenta, then the forces would still be equal and opposite, but would not necessarily lie along the direction between the particles."

What does this mean? IMO if a potential is dependent on velocities, the forces derived typically will not respect any aspects of the law of action and reaction. Why would they still be equal and opposite?
 
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Jul28-09, 05:28 AM   #2
 
Quote by sadness View Post
On page 10 of Goldstein's classical mechanics, it was said:

"If [tex]V_{jj}[/tex] were also a function of the difference of some other pair of vectors associated with the particles, such as their velocities or (to step into the domain of modern physics) their intrinsic "spin" angular momenta, then the forces would still be equal and opposite, but would not necessarily lie along the direction between the particles."

What does this mean? IMO if a potential is dependent on velocities, the forces derived typically will not respect any aspects of the law of action and reaction. Why would they still be equal and opposite?
The point is that the book precises [tex]V_{jj}[/tex] being function of the difference of some vector associated to the particles, that is not generally of velocities for instance as you say, but for differences of them in such a way that (1.33) still holds but not (1.34), when now the dependence is not just of the relative position and the forces not central, as mentioned in page 7. Then this weak action-reaction law is satisfied, but not the "strong".
 
Jul28-09, 05:50 AM   #3
 
Quote by Rebel View Post
The point is that the book precises [tex]V_{jj}[/tex] being function of the difference of some vector associated to the particles, that is not generally of velocities for instance as you say, but for differences of them in such a way that (1.33) still holds but not (1.34), when now the dependence is not just of the relative position and the forces not central, as mentioned in page 7. Then this weak action-reaction law is satisfied, but not the "strong".
Thanks. I understood what the authors meant now.
 
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