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I Conservative Forces

  1. Aug 6, 2017 #1

    kmm

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    I actually have a few things I'm thinking about here. I'm curious as to whether a velocity dependent force field absolutely cannot be a conservative force field, in principle. I have at times come across statements in physics that I found out had mathematical exceptions for, but we don't actually run into in physics. For example, I think I read in Griffith's Quantum Mechanics that a good mathematician can give "pathological" examples of normalizable wave functions that don't actually go to zero at infinity, but we don't come across those in physics, so we ignore those and state that only wave functions that go to zero at infinity are normalizable. Now for my current concern, I'm aware that for a force to be conservative, it must be a function only of position, and the work done by that force on a particle as it moves between two points is the same for all paths. Now it's clear to me that velocity dependent forces we come across in physics, such as friction or air resistance are not conservative since the work required to move a particle between two points is path dependent. What isn't obvious to me is the claim that NO velocity dependent force field could do work that is path independent. Maybe I'm misunderstanding and this isn't actually the claim. If it is, while it seems correct intuitively, it doesn't seem obvious that one couldn't construct such a hypothetical force field. Maybe this would just be one of those "pathological" fields? In pondering this, it has also made me wonder when I'm studying physics, to what extent should I take mathematical statements in physics as universal, or just local to physics if it isn't explicitly stated?
     
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  3. Aug 7, 2017 #2

    andrewkirk

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    I think the answer lies in the fact that a force field is conservative iff it is the negative gradient of a scalar field, as proven here.

    In that case, the force applied to a body by a conservative field is entirely determined by the body's position. Hence the force cannot depend on velocity.
     
  4. Aug 7, 2017 #3

    kmm

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    OK, now thinking of it from the perspective of the derivative makes it clear. I was actually already aware of that, but for some reason, I was stuck trying to see it by trying to evaluate a path integral. Thanks for flipping my brain on this
     
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