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Problems with a complex stress-energy tensor

  1. Jun 25, 2011 #1
    Hi , im working with the following stress-energy tensor:

    T[itex]\mu\nu[/itex]=[itex]\partial[/itex][itex]\mu[/itex][itex]\phi[/itex][itex]\partial[/itex][itex]\nu[/itex][itex]\phi[/itex]* - g[itex]\mu\nu[/itex]([itex]\nabla[/itex][itex]\mu[/itex][itex]\phi[/itex][itex]\nabla[/itex][itex]\mu[/itex][itex]\phi[/itex]* - m2[itex]\phi\phi[/itex]*)

    Where [itex]\phi[/itex] is a complex scalar of the form:
    [itex]\phi[/itex](r,t) = [itex]\psi[/itex](r)eiwt

    that obeys the Klein-gordon equation and [itex]\phi[/itex]* is the complex conjugate, and g the metric tensor.

    My problem is that i think this tensor is not symmetric as Ttr /= Trt by a minus sign on the term of the partial derivates.

    Thanks a lot for reading.
  2. jcsd
  3. Jun 26, 2011 #2
    I think my problem wasn't really clear, this tensor is supposed to be a stress energy tensor and therefore it should be symmetric but I'm finding that it isn't Ttr /= Trt. I'm not sure if I'm differentiating wrong the complex scalar field or I think originally instead of partial derivatives they were covariant derivatives but since it is a scalar quantity I thought it wouldn't matter.

    Hope this gives a better idea of the problem at hand.

    Thanks a lot.
  4. Jun 26, 2011 #3


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    So why don't you just symmetrize it.

    Or better yet, forget the "canonical" stress-energy tensor and use the correct and much simpler definition, Tμν = 2δL/δgμν whose calculation typically involves no taking of derivatives and is guaranteed to come out symmetrical. All you have to remember is that the Lagrangian density contains a factor √(-g) which must also be varied, and δ√(-g)/δgμν = 1/2 √(-g) gμν.
  5. Jun 28, 2011 #4
    Thanks Bill_K, with that and the Klein-Gordon Lagrangian I found a Tensor which is indeed symmetric.

    Thanks a lot.
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