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Symmetries and Maxwells equations

  1. Dec 3, 2011 #1
    Hi

    Maxwells Equations for a time-invariant system are separable, hence we can write a solution as E(r, t) = E(r)E(t). They also mention that if the system is radially invariant, then that implies that the solution splits into a product of radial and angular functions (with 2π periodic angular functions).

    Is it a general rule that when the system described by Maxwells equations has a symmetry, then the solutions become separable? If yes, does this go beyond Maxwells Equations?

    Best,
    Niles.
     
  2. jcsd
  3. Dec 3, 2011 #2

    Vanadium 50

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    No. It's only true for 1/r and r2 potentials.
     
  4. Dec 3, 2011 #3
    Thanks. Where does that spatial dependence come into play when looking at Maxwells Equations? Is in through ε(r)=n(r)2?

    Do you have a suggestion for a reference that explains this in more details?

    Best,
    Niles.
     
  5. Dec 3, 2011 #4

    Vanadium 50

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    It's described in Volume 1 of Landau's Mechanics. Mathematically, it's because there is a hidden SO(4) symmetry in the equations describing the 1/r potential, and this symmetry (the same one that gives the n-l degeneracy in the hydrogen atom) ensures that the angular piece is separated out.
     
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