Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Uniqueness of Maxwell's equations

  1. Feb 19, 2014 #1
    Hi all,

    I'm trying to derive for myself the uniqueness proof for Maxwell's equations, but I'm a little stuck at the end. I've managed to prove the following:
    [tex]
    \dfrac{A^\mu}{\partial{t}}\nabla{A^\mu}|_S = \dfrac{A^\mu}{\partial{t}}|_{t_0} = \nabla{A^\mu}|_{t_0} =0 \Rightarrow \forall_{\vec{r}\in V, t > t_0} F^{\mu\nu} = 0
    [/tex]
    Where S is the surface bounding volume V, t0 is some initial time, and:
    [tex]
    A^\mu \equiv A_1^\mu - A_2^\mu,
    F^{\mu\nu} \equiv F_1^{\mu\nu} - F_2^{\mu\nu}
    [/tex]
    are the difference fields between the two solutions satisfying the same boundary conditions (at surface S, time t0, and with the same charge distributions)

    My problem is that I can't turn the left hand side of the first equation into any meaningful condition on the fields [itex]F^{\mu\nu}[/itex]. As far as I can tell you need to know the potentials at the boundaries in order for uniqueness to hold, rather than just the field values... What I'm looking for is some expression of [itex]F^{\mu\nu}[/itex] on the boundaries that implies the potential conditions on the left hand side of the equation
     
  2. jcsd
  3. Feb 27, 2014 #2
    Anyone? I've been googling this and nobody seems to provide a formal, complete proof for the uniqueness of solutions to Maxwell's equations. I'm trying to prove the uniqueness of an SU(N) gauge field theory, so the difficulty involved in the U(1) case is very discouraging...
     
  4. Feb 27, 2014 #3
    The solution is not unique. That's what's meant by gauge freedom. It is the freedom to chose different solutions for Aμ which correspond to the same physical situation
     
  5. Feb 27, 2014 #4
    Gauge freedom is on the potentials not the fields... As I showed in the OP the fields ARE unique as long as the derivatives of the potentials are known on the boundaries. My problem is that I can't turn the uniqueness conditions into expressions of the fields alone
     
  6. Feb 27, 2014 #5
    Do you have a formal, or clear statement of the uniqueness theorem?
     
  7. Feb 27, 2014 #6
    Yes, exactly what I stated in the OP... However the requirements for uniqueness involve knowledge of derivatives of the potentials, rather than of the field values and are not gauge-independent.

    Let there exist 2 solutions to Maxwell's equations for identical charge distributions, with identical boundary conditions for some (not yet defined) space-time boundaries (spatial boundaries or initial conditions). Label these solutions 1 and 2, and define the difference between them as:
    [tex]
    A^\mu \equiv A_1^\mu - A_2^\mu,
    F^{\mu\nu} \equiv F_1^{\mu\nu} - F_2^{\mu\nu}
    [/tex]

    These fields satisfy the empty space Maxwell's equations, and evaluate to 0 on any space-time boundary for which solutions 1 and 2 are constrained. The following statement can be derived for any volume V with boundary surface S, and some fixed time t0:
    [tex]
    \dfrac{\partial A^\mu}{\partial{t}}\nabla{A^\mu}|_S = \dfrac{\partial A^\mu}{\partial{t}}|_{t_0} = \nabla{A^\mu}|_{t_0} =0 \Rightarrow \forall_{\vec{r}\in V, t > t_0} F^{\mu\nu} = 0
    [/tex]

    Therefore, with sufficient boundary conditions solutions 1 and 2 are identical (the fields not the potentials, obviously). The condition:
    [tex]
    \dfrac{\partial A^\mu}{\partial{t}}\nabla{A^\mu}|_S = \dfrac{\partial A^\mu}{\partial{t}}|_{t_0} = \nabla{A^\mu}|_{t_0} =0
    [/tex]
    is not gauge-independent though, so I am looking for some expression of boundary conditions on the FIELDS that leads to this. The conditions are purely derivatives of the potentials, which is good, but I don't see an obvious transformation to the field variables

    *EDIT* sorry, I forgot to specify this but I used the Lorenz gauge in the derivation of the above equation so [itex]\partial_\mu A^\mu = 0[/itex]
     
    Last edited: Feb 27, 2014
  8. Feb 28, 2014 #7

    UltrafastPED

    User Avatar
    Science Advisor
    Gold Member

  9. Mar 1, 2014 #8
    No, I found this page while I was searching for an answer. The whole thing is pretty "hand-wavey" (e.g. 14 equations + 14 unknowns = unique solution), very informal, and under closer scrutiny suffers the exact same problems as me: the boundary conditions necessary for uniqueness are on the gauge-dependent potentials rather than the gauge independent fields.
    This link has plenty of uniqueness proofs, but I couldn't find a general one for an arbitrary system. It is very long though, so I may have just missed it... However, the ones I did see all simplified Maxwell's equations in some way, which makes the problem much simpler.
     
  10. Mar 2, 2014 #9
    I think it is possible to rewrite Maxwell's equation in a form (a symmetric, hyperbolic form) which lets you apply a general theorem on existence and uniqueness for such partial differential equations. I read about this in Geroch's paper "Partial Differential Equations of Physics", available here http://arxiv.org/abs/gr-qc/9602055.
     
  11. Mar 3, 2014 #10
    I'm sure you're right, but the thing is I've already proven uniqueness! (I don't really care about existence at the moment). The issue isn't whether or not the solutions are unique, its what conditions are necessary for uniqueness.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Uniqueness of Maxwell's equations
  1. Maxwell equations (Replies: 4)

  2. Maxwell equation (Replies: 1)

  3. Maxwells equation (Replies: 1)

Loading...