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Lorentz force - conservative?

  1. Sep 18, 2006 #1
    [tex]\mathbf \nabla \times (\mathbf E + \mathbf v \times \mathbf B)[/tex]
    pluggin stuff from Maxwell equations
    [tex]= -\frac{\partial B}{\partial t} + \mathbf v (\mathbf \nabla \cdot B) - \mathbf B (\mathbf \nabla \cdot v)[/tex]
    [tex]\frac{\partial}{\partial t}(\mathbf \nabla \cdot \mathbf r) = 0[/tex]
    [tex]= -\frac{\partial B}{\partial t}[/tex]
    which is not zero in general. Or am I doing something wrong??

    Can this be the field is also holding energy?
    Last edited: Sep 18, 2006
  2. jcsd
  3. Sep 19, 2006 #2

    Meir Achuz

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    The Lorentz force is not conservative if dB/dt is not zero, as you have just shown. The dB/dt force is how the betatron accelerates particles.
    The circuit producing dB/dt puts energy into the system.
  4. Sep 19, 2006 #3
    Indeed. In fact, only conservative force fields can be written in terms of the gradient of a scalar field. Which is why

    [tex]\vec{E} = -\vec{\nabla}\phi[/tex]

    only applies when [itex]\partial B / \partial t = 0.[/itex]
  5. Sep 19, 2006 #4
    But... wait! Isn't electromagnetism (and all fundemental forces) supposed to be a conservative? What happens to [tex]\partial B / \partial t[/tex]?
  6. Sep 20, 2006 #5


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    Last edited: Sep 20, 2006
  7. Sep 21, 2006 #6
    Richard Feynman.
    I can't remember where exactly, probably in lectures on physics. It may take while for me to find where exactly.
  8. Sep 23, 2006 #7
    Found it:
    EM force is the last force I'd expect to be non-censervative since almost everything we observe is chemical which is by means of EM forces, yet conservative when observed carefully.

    From the "work-energy relation in EM" in Griffiths:
    OK, if the curl of force comes out zero, we say "this's conservative" right away. But what is the meaning of the term if it doesn't come out zero. Ie, I wonder the meaning of the Faraday term which came out.

    Please, can't anyone explain what's going on in this curl?!?
  9. Sep 23, 2006 #8
    It's usually because you are not considering the whole system:

    Poynting's theorem tells us how energy is transported by the electromagnetic field: this is done via EM radiation. Really, energy is conserved in EM, but only when one considers the whole universe as the system (which is rarely done). e.g. usually one doesn't include the battery which supplies power nor accounts for energy lost as radiation, all problems of non-conservation will disappear.
    Last edited: Sep 23, 2006
  10. Sep 24, 2006 #9
    So that extra term has something to do with energy flow to/from the field? Poynting's theorem seem to do the job, but it's a bit cumberstone.
    Can we show that Faraday term has to do with field?

    Sorry for such lame questions, I just haven't taken EMT course yet.
  11. Sep 24, 2006 #10


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    Last edited: Sep 24, 2006
  12. Sep 25, 2006 #11
    Forces are conservative when you 'fix' whatever object is supposedly creating them. If you imagine an infinite, constant electric field independent of the charges that produce it - that is, in a vacuum - it's conservative, because no matter what you do in the field, there's no way to change it and thus it'll be exactly the same when you go back in the opposite direction.

    In practice, any object that 'feels' an electric field also creates one of its own. This field affects the particles that generate the field you're considering, so the gross field itself is altered by your motion. It's no longer conservative.
  13. Sep 25, 2006 #12


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  14. Sep 26, 2006 #13
    Mhm, ok. What was wrong about what I said?
  15. Sep 28, 2006 #14
    Anyone? I'm curious now.
  16. Sep 28, 2006 #15
    The EM field is conservative, but you've only accounted for the energy stored in the particle. If you look at the total energy stored in an EM field, you get an energy conservation equation, that is unless you start pumping the system from the outside, but that's generally true.

    This is an interesting point, actually. If you start out with the lagrangian for a uniform magnetic field, you get that the total energy equals the hamiltonian, but there's no dependence on [tex]\vec{A}[/tex] when it all comes out. That's because you haven't written down the "whole hamiltonian" for the entire system, just the hamiltonian for how this charged particle interacts with an external field.
  17. Sep 28, 2006 #16


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    There is no violation of conservation of energy in EM.
    Frankly I do not see the meaning of taking the curl of the lorentz force. Taking the curl applies to vector fields (like the E and B fields). The lorentz force describes the force on a charged particle, it's not a field.

    Anyhoo, since the curl E = -dB/dt and curl B=uJ+(1/c^2)d(E/dt) they are not conservative in the sense that their curl is zero.
    However if you consider the entire system of charged particles and fields you do have conservation of energy. Whatever energy is lost in the fields is gained by the particles and vice versa.
  18. Sep 28, 2006 #17


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    Great post samalkhaiat. Keep spreading the knowledge. :smile:

    There's just one step I don't understand...

    Shouldn't the conclusion be simply "one can find a sclalar function
    [itex]U[/itex] such that

    [tex]m\frac{d}{dt}\vec{v} = -\frac{\partial}{\partial t} \vec{A} - -\vec{\nabla}U[/tex]"

    ?? What's the [itex](\vec{v}.\vec{\nabla}) \vec{A}[/itex] doing there? Thx?
  19. Sep 28, 2006 #18


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    Couldn't you see it as a force field changing its value everywhere in space as v changes?

    Poynting theorem. :cool:
  20. Sep 30, 2006 #19


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  21. Sep 30, 2006 #20


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