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Homework Help: Special relativity the EM stress energy tensor

  1. Apr 16, 2010 #1
    1. The problem statement, all variables and given/known data

    Using the expression below for the stress energy tensor of the em field, show that it has zero trace.

    2. Relevant equations

    [tex]T^{\mu\nu}=F^{\mu}_{ \alpha}F^{\alpha\nu}+\frac{1}{4}\eta^{\mu\nu}F_{\beta\gamma}F^{\beta\gamma}[/tex]

    F is the faraday tensor and eta is the minkowski metric.

    3. The attempt at a solution

    I started by trying to calculate [tex]T^{00}[/tex] with the aim of then calculating the other diagonal components i.e. 11, 22 and 33. But I did not get anywhere. I couldnt get my head around the summations.
    Last edited: Apr 16, 2010
  2. jcsd
  3. Apr 16, 2010 #2
    Well, you could definitely solve it that way, but you would be creating a lot of unnecessary work. Think of it this way, the trace of your tensor is given by:

    [tex]\eta_{\mu\nu}T^{\mu \nu} = T[/tex]

    So just multiply the right side by the minkowski metric. You will also notice that:

    [tex]\eta_{\mu \nu} \eta^{\mu \nu} = 4[/tex]

    for 4 dimensions (3 space and 1 time).
  4. Apr 17, 2010 #3
    Thanks, for your reply. I think I can get the answer but there is one bit I just dont get. Ill come back to that at the end.

    I get
    [tex]\eta_{\mu\nu}T^{\mu\nu}=\eta_{\mu\nu}F^{\mu}_{ \alpha}F^{\alpha\nu}+\frac{1}{4}\eta_{\mu\nu}\eta^{\mu\nu}F_{\beta\gamma}F^{\beta\gamma}[/tex]

    [tex]=F_{ \nu\alpha}F^{\alpha\nu}+F_{\beta\gamma}F^{\beta\gamma}[/tex]

    Then using the antisymetry properties of the faraday tensor this can be written as


    This is then zero because the the indicies are being summed over and as such are dummy indicies implying we can change [tex]\beta[/tex] and [tex]\gamma[/tex] to [tex]\alpha[/tex] and [tex]\nu[/tex], thus the LHS goes to zero.

    My issue with this is I do not understand why multiplying the stress energy tensor by the metric gives the trace of the metric.

    If you had not told me that this gives the trace i would have said that
    which I assume is another matrix as opposed to the trace (i.e. a number).

    It is the same with the metric multiplication
    Again I cannot see how this gives 4.

    I think I am missing some key thing in my understanding of the topic that will make this obvious.
  5. Apr 17, 2010 #4


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    And [itex]T_{\nu}^{\nu}=T_{0}^{0}+T_{1}^{1}+T_{2}^{2}+T_{3}^{3}[/itex], which is the summation of all diagonal entries, which is the definition of the...?
  6. Apr 17, 2010 #5
    ok, but if that is the case how do you get

    surely because [tex]\eta^{\mu\nu}=diag(1,-1,-1,-1)[/tex]


  7. Apr 17, 2010 #6


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    There's a difference between [itex]\eta^{\mu\nu}[/itex] and [itex]\eta^\mu{}_\nu = \eta^{\mu a}\eta_{a\nu}[/itex]. The latter is represented by the identity matrix.
  8. Apr 17, 2010 #7


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    See also the first comment in this post, and don't forget the definition of matrix multiplication.
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