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Maxwell eqs: 8 eqs for 6 unknowns - too many eqs?

  1. Jun 13, 2010 #1
    I know I really need to understand partial diff eqs better for this, but I don't know what subtopic to look for in a pde text.

    Maxwell eqs are 2 scalar eqs plus 2 vector eqs (3 components) giving eight total equations, coupled in the 6 components of E(x) and B(x). Why not just 6 equations?

    This is easier to see in the covariant form

    [tex]\partial_{\mu}F^{\mu\nu}=j^\nu[/tex] , [tex]\nu=0,1,2,3[/tex]
    [tex]\epsilon^{\alpha\beta\gamma\delta}\partial_\gamma F_{\alpha\beta} =0[/tex] , [tex]\delta=0,1,2,3[/tex]

    F is a 4x4 anti-symmetric tensor, so it is made of six independent functions.

    The equations are first order with respect to space and time derivatives. Contrast for example the Dirac equation which is also first order in space and time derivatives but is really only 4 equations for 4 unknown functions.

    Do the Maxwell equations contain a hidden redundancy?
  2. jcsd
  3. Jun 14, 2010 #2


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  4. Jun 14, 2010 #3
    Thanks. But how does this relate? Doesn't gauge freedom only have to do with the vector potential? I am concerned that the equations in terms of the fields E and B (which have no gauge freedom, correct?) seem more than necessary.
  5. Jun 14, 2010 #4
    The current has four components, so you have 10 quantities and 8 equations, so 2 degrees of freedom remain.
  6. Jun 14, 2010 #5

    I was thinking that the since the current must satisfy the continuity equation, we cannot arbitrarily set all four components of j. Once we set [tex]\vec{J}(\vec{x},t)[/tex] (and the initial value of [tex]\rho[/tex]), then the charge density [tex]\rho(\vec{x},t)[/tex] is determined for all x and t.

    But if we only have two degrees of freedom does that mean we are only free to arbitrarily set two components of [tex]j^\mu[/tex]?
    Last edited: Jun 14, 2010
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