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Uniqueness of solutions to maxwell eqns

  1. May 10, 2008 #1
    I am wondering if anyone knows of any conditions for uniqueness of solutions to maxwells equations. For electrostatics, I have seen uniqueness formulated in terms of the potential. I am asking here how this result generalizes to the non-electrostatic case.
  2. jcsd
  3. May 10, 2008 #2
    Gauge invariance of the 4-vector potential. If the potential [tex]A^\mu[/tex] satisfies Maxwell's equations then the potential [tex]A^\mu + \partial^\mu \lambda[/tex] also satisfies Maxwell's equations provided that [tex]\lambda[/tex] obeys the wave equation, i.e. [tex]\partial^\mu \partial_\mu \lambda = 0[/tex].

    The 4-vector potential has the electric potential as the time component, and the magnetic vector potential as the spatial components.
  4. May 11, 2008 #3
    additionally, to any solution, one can add source-free electromagnetic waves and the charge itself can be regauged. for a local regauging the source-free condition can be removed.
    Last edited: May 11, 2008
  5. May 12, 2008 #4
    To both posters: huh?

    I am sorry I don't understand. I am looking for conditions under which solutions to maxwells equations are unique. I don't see how your post relates to this.

    The reason I ask is because I recently solved a (magnetostatics) problem in which I was able to guess a solution that seemed to meet all appropriate boundary conditions. How could I be sure that this solution is correct? What is the precise meaning of "all appropriate boundary conditions"?
  6. May 12, 2008 #5
    For static solutions you don't have to worry about source-free solutions, since these are waves and not static. Thus, there is only one way in which
    [tex]\nabla^2 \phi[/tex] can be zero.
    On the other hand, if you allow the system to vary in time, then you have to deal with the d'Alembertian,
    [tex]\nabla^2 \phi - \frac{1}{c^2} \frac{\partial}{\partial t^2}\phi[/tex],
    and roughly, there are two ways this thing can be zero. Either each term is separately zero (thus, static solution which is a harmonic function), or the two terms add up to zero. The second case is the domain of electrodynamics.

    I realize I'm rambling here. Anyway, if you live in Minkowski space, then the Cauchy problem is well defined. Specify your initial conditions and "boundary conditions", i.e., location of any sources/conductors, etc., and the solution is unique. Since you can't actually reach infinity at finite time, you don't have to worry about waves "coming from infinity", which is a problem in some other spaces.

    I suggest you take a look at Jackson for a discussion.
  7. May 12, 2008 #6
    It was a direct reply to your question. Have you not seen gauge invariance before? There are no physical conditions for which solutions of Maxwell's equations are unique. It's all related to the issue of gauge invariance, which is arguably the most important topic in electrodynamics because the concept has far reaching consequences in more advanced theories found in modern physics.
  8. May 12, 2008 #7
    Gauge invariance has only to do with (classically) unobservable potentials, and so has nothing to do with the OP's question. [tex]E[/tex] and [tex]B[/tex] don't care about gauge invariance.
  9. May 12, 2008 #8
    Slick to put in classically since the Bohm-Aharanov effect has to do with the potentials and not the fields. Since potentials produce observable effects, I wouldn't dismiss them as saying only E and B matter. If the OP only wants to discuss E and B then he'll have to make that distinction himself.
  10. May 12, 2008 #9
    I can get even more slick =] The Aharanov-Bohm is also gauge invariant. The only reason the dependence on the gauge connection shows up is for topological reasons (the space isn't simply connected).
  11. May 12, 2008 #10
    yup my hats off to you, that is very slick. :-)
  12. May 12, 2008 #11
    Oddly enough, I'm having the same issues. Can you describle the problem?
    Last edited: May 12, 2008
  13. May 12, 2008 #12
    Unfortunately, I cannot get my hands on this book. Maybe google has something for me on the dAlembertian....

    I really don't care about uniqueness of potentials. After all, potentials are only unique up to some transformations. What I care about is uniqueness of the field. So in the electrostatics case, it seems that uniqueness comes from the fact that there exists a potential that satisfies Laplaces eqn (there is also a formulation of this uniqueness theorem in the presence of charge). In magneto-statics, I see now that the vector potential also satisfies a Laplace eqn, so the same uniqueness theorem holds. As a first step towards answering my question, I ask how would you formulate a uniqueness theorem for magnetostatics in the presence of current? Then, outside the boring worlds of statics, is there some way in which we still have uniqueness? (Ideally, if not, can you give an example?)

    It's kind of complicated. But basically there is a infinitely long wire carrying a current. "Above" the wire is air, below is a magnetic material. Find the B field everywhere. I bet there is an easy solution to this problem, but I guessed it on heuristics alone, then checked that Div B was 0, that Curl H was 0 and that the bound currents produced by this H field were consistent with the derived field.

    Edit: Also, thanks to everyone for their help!
  14. May 13, 2008 #13
    Basically you just add up the field contributions from the wire and from the magnetic material, to get your answer. I don't know if this is what you wanted to know. All along, throughout these various responses, is the assumption thay you wish to satisfy maxwells 4 equation, and not include the Lorentz force, by the way. So the linear sum idea assumes that the magnets don't effect the current and vice versa.
  15. May 13, 2008 #14
    I don't get it. The current necessarily affects the material (but we do assume that the material does not affect the current).

    Also, the field from the wire is easy to determine, but the field due to the material is not straightforward (I don't think) since that field is due to unknown bound currents. What I did was essentially assume that the net result is proportional to the field due to the wire. I don't think that need be the case in general. You get an answer that seems to make sense.
  16. May 13, 2008 #15
    I'm afraid we're still talking in circles. You you be more specific with your geometric arrangement?
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