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Maxwell Tensor Symmetries Problem

  1. Jul 4, 2012 #1
    Hi community:

    I'm Federico and I'm new user here!

    I'm trying to show that the Electromegnetic Field Tensor

    [itex]F_{ab}[/itex] = 2A(r) [itex](e_{0})_{[a}(e_{1})_{b]}[/itex] + 2B(r) [itex](e_{2})_{[a}(e_{3})_{b]}[/itex]

    where [itex](e_{0},e_{1},e_{2},e_{3})[/itex] is the tetrad basis associated with the metric

    [itex]ds^2= -f(r)dt^2+h(r)dr^2+r^2dθ^2+r^2sin^2(θ)d\varphi^2[/itex]

    has the same symmetries that this metric (static and spherical symmetry).

    I`ve tried using the Lie Derivative in the direction of the Killing fields of this metric, but the algebra becomes a little complicated.

    Any ideas on this issue?

    Thanks a lot!
     
  2. jcsd
  3. Jul 4, 2012 #2
    Hello Federico, I'm familiar with tetrads, but could you elaborate in the meaning of the square brackets you used in your expression for the EM tensor?
     
  4. Jul 4, 2012 #3
    Yes, no problem: the square brackets means antisymmetric

    [itex](e_{0})_{[a}(e_{1})_{b]}[/itex]=[itex]\frac{1}{2}[(e_{0})_{a}(e_{1})_{b} - (e_{0})_{b}(e_{1})_{a}][/itex]

    thanks!
     
  5. Jul 4, 2012 #4
    All right, that makes sense. U can't give an in-depth analysis right this moment, but do you have an idea of the tensor that converts between the tetrad and coordinate bases? I suspect if you choose this to be have a specific form, the properties you want will hold.
     
  6. Jul 4, 2012 #5
    yes, looking the metric, it let me know that a good choice is:

    [itex](e_{0})_{a}=\sqrt{f}(dt)_{a}[/itex]
    [itex](e_{1})_{a}=\sqrt{h}(dr)_{a}[/itex]
    [itex](e_{2})_{a}=r(d\theta)_{a}[/itex]
    [itex](e_{3})_{a}=rsin(\theta)(d\varphi)_{a}[/itex]

    I mean, the tensor is diagonal.
     
  7. Jul 4, 2012 #6
    Yeah, I mean, I know that's not the only gauge choice that gives the metric, but it's easy and it works. Nothing is a function of time, so I think you're okay there. How about applying a rotation matrix in theta or phi and verifying by hand that spherical symmetry is still manifest?
     
  8. Jul 4, 2012 #7
    Ok, I'll try with that and let you know later. Thanks a lot for the idea!
     
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