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Maxwell Equations in Tensor notation

  1. Feb 19, 2009 #1
    http://en.wikipedia.org/wiki/Formul...s_in_special_relativity#Maxwell.27s_equations

    Why does
    [tex]0 = \epsilon^{\alpha \beta \gamma \delta} \frac{\partial F_{\alpha \beta}}{\partial x^\gamma}
    [/tex]
    reduce to
    [tex]
    0 = {\partial F_{\alpha\beta}\over\partial x^\gamma} + {\partial F_{\beta\gamma}\over\partial x^\alpha} + {\partial F_{\gamma\alpha}\over\partial x^\beta}
    [/tex]
    ?
     
  2. jcsd
  3. Feb 19, 2009 #2
    [tex]
    0
    = \epsilon^{\alpha \beta \gamma 0} \frac{\partial F_{\alpha \beta}}{\partial x^\gamma}
    = {\partial F_{\alpha\beta}\over\partial x^\gamma} + {\partial F_{\beta\gamma}\over\partial x^\alpha} + {\partial F_{\gamma\alpha}\over\partial x^\beta} - {\partial F_{\beta\alpha}\over\partial x^\gamma} - {\partial F_{\gamma\beta}\over\partial x^\alpha} - {\partial F_{\alpha\gamma}\over\partial x^\beta}
    = 2 \left( {\partial F_{\alpha\beta}\over\partial x^\gamma} + {\partial F_{\beta\gamma}\over\partial x^\alpha} + {\partial F_{\gamma\alpha}\over\partial x^\beta} \right)
    [/tex]
    because [tex]F_{\alpha\beta}[/itex] is antisymmetric.
     
  4. Feb 21, 2009 #3
    Why was delta set to 0?
     
  5. Feb 21, 2009 #4

    HallsofIvy

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    What "delta" are you talking about? There was no "delta" in your original question nor in Adriank's response.
     
  6. Feb 21, 2009 #5
    Actually, delta is the fourth index on the anti-symmetric epsilon tensor in both Karliski's post and in the wikipedia equation he linked to. It is basically an error, we should have delta = 0 as in adriank's post, but if delta is not zero then we just get four copies of the same equation.
     
  7. Feb 21, 2009 #6

    Fredrik

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    The "equation" in #1 is actually four equations, one for each value of [itex]\delta[/itex]. The one with [itex]\delta=0[/itex] is the scalar equation

    [tex]\nabla\cdot\vec B=0[/tex]

    The one with [itex]\delta=i\neq 0[/itex] is the ith component of the vector equation

    [tex]\nabla\times\vec E+\frac{\partial\vec B}{\partial t}=0[/tex]

    (Edit: ...except for a factor of two).
     
    Last edited: Feb 21, 2009
  8. Feb 21, 2009 #7
    The equation
    [tex]0 = \epsilon^{\alpha \beta \gamma \delta} \frac{\partial F_{\alpha \beta}}{\partial x^\gamma}[/tex]
    is written with a [tex]\delta[/tex] only on one side, which means we can "plug in" any specific value for it. So I put [tex]\delta = 0[/tex] and explicitly wrote out the sum [tex]\epsilon_{\alpha\beta\gamma0}F_{\alpha\beta,\gamma}[/tex].
     
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