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Rewriting Maxwell's Equations in Tensor Form

  1. Sep 10, 2012 #1
    1. The problem statement, all variables and given/known data

    From Sean Carroll's notes on general relativity (chapter 1, pg. 20):

    Show that [itex]F_{[\alpha\beta,\gamma]} = 0[/itex] is equivalent to half of the Maxwell equations.

    2. Relevant equations

    [itex]F_{\mu\nu}[/itex] is the electromagnetic tensor

    [itex]\Phi_{,\nu} \equiv \partial_{\nu}\Phi[/itex]

    [itex]F_{i0}= E_{i}[/itex]

    [itex]F_{ij}= \epsilon^{ijk}B_{k}[/itex]

    3. The attempt at a solution

    I'm specifically looking to turn Maxwell's (homogeneous) equations into tensor form, not just show that they fall out of [itex]F_{[\alpha\beta,\gamma]} = 0[/itex]. I sort of have a solution, but I feel like I'm missing a step.

    [itex]
    \begin{eqnarray*}
    \nabla×\textbf{E} + \partial_{t}\textbf{B} = 0 \\
    \nabla\cdot \textbf{B} = 0
    \\
    \\
    \epsilon^{ijk}\partial_{j}E_{k} + \partial_{0}B^{i} = 0 \\
    \partial_{i}B^{i} = 0
    \\
    \\
    \epsilon^{ijk}\partial_{j}F_{k0} + \frac{1}{2}\epsilon^{ijk}\partial_{0}F_{jk} = 0 \\
    \frac{1}{2}\epsilon^{ijk}\partial_{i}F_{jk} = 0
    \end{eqnarray*}
    [/itex]

    which can be rewritten as:

    [itex]\epsilon^{\mu\nu\rho\sigma}\partial_{\rho}F_{\mu \nu} = 0[/itex]

    which, up to a normalization constant, is just:

    [itex]F_{[\alpha\beta,\gamma]} = 0[/itex]

    My question is about going from step 3 to step 4. I sort of pulled it out of my hat and checked that it was correct (term by term). I'm looking for some kind of justification for this step, or a nudge in the right direction if I'm approaching this all wrong.

    Also, I'm still getting the hang of tensor notation, and I feel like equation 4 doesn't make sense. Only 3 indices are contracted, leaving the right side a vector, not a scalar. On the other hand, trying to use the Levi-Civita tensor with 3 indices here seems wrong too, as the indices run from 0 to 4 leaving you with stuff like [itex]\epsilon^{013}[/itex].
     
  2. jcsd
  3. May 26, 2015 #2
    Bump. : (
     
  4. May 29, 2015 #3

    Fredrik

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    The first equation in your step 3 is actually three equations, one for each value of i. And the second equation is just one. That's why the final result is a set of four equations.

    I think the final step will be easier to understand if you make sure that you understand every step of the following rewrites: (Here A is anything with three indices from 0 to 3).

    \begin{align}
    &\varepsilon^{ijk} A_{ijk} =\varepsilon^{ijk0} A_{ijk} =\varepsilon^{\mu\nu\rho 0} A_{\mu\nu\rho}\\
    &\varepsilon^{ijk} A_{0jk} =\varepsilon^{0jki} A_{0jk} =\varepsilon^{0\nu\rho i} A_{0\nu\rho} =\varepsilon^{\mu\nu\rho i} A_{\mu\nu\rho}
    \end{align}
     
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