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Questions about tensors in GR

  1. Aug 12, 2004 #1
    I'm reading in on the subject of General Relativity and came across a few things I don't understand. First of all I'm not sure where the following rule comes from, and maybe someone can explain or derive it for me:

    [tex] \eta^{\mu \beta} h_{\nu \sigma,\beta} = h_{\nu \sigma}^{,\mu} [/tex]

    And I also found the following things about the metric tensor I don't understand:

    [tex] g^\mu_\mu = g_{\mu \nu}g^{\mu \nu} = 4[/tex]
  2. jcsd
  3. Aug 12, 2004 #2
    What are you reading? Is it a text (which I might have) or is it an online website notes?
    This looks like something from the week field approximation whereby the metric is written as

    [tex]g_{\mu\nu} = \eta_{\mu\nu} + h_{\mu\nu} [/tex]

    where |huv << 1. In this approximation one uses the Minkowski tensor [tex]\eta_{\mu\nu}[/tex] to raise and lower indices. That is what it looks like in the expression above
    That is called a contraction of a second rank tensor, in this case the metric tensor. Contracting a second rank tensor means you first raise (or lower) and index such that there is one upper index and one lower index. Then you sum over the values.

  4. Aug 12, 2004 #3
    I'm reading "A short course in General Relativity" by J. Foster and J.D. Nightingale

    I know it is a contraction where you sum over the diagonal elements of the metric tensor. But more specific, my question is why the trace of the matrix representing the metric tensor equals four? Is this true in general?
  5. Aug 12, 2004 #4
    Excellant text.
    What do you mean by Is this true in general? Are you asking if all metric tensors in GR have a trace equal to 4? If that is what you mean then yes. Recall that you can always choose coordinates such that the metric = diag(1, -1, -1, -1) = Minkowski metric. When you take the trace you get 4.

    If you're asking whether all metric tensors have a trace which is equal to 4 then no. If the metric was the Euclidean one where

    [tex]ds^2 = dx^2 + dy^2 + dz^2[/tex]

    then it would be three.

    Different tensors have different traces. If you took the stress-energy-momentum tensor for a beam of light (i.e. all photons moving in one direction) then the trace of that tensor would be zero. The value depends on the tensor.

    I recommend actually calculating each of the two examples that I gave you. It usually clears things up better when you actually sit and calculate these things out in all the gory details.

    Last edited: Aug 12, 2004
  6. Aug 12, 2004 #5


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    Yes it is general. The full contraction of a tensor into a scalar yields an invariant. For example mass which is invariant is the full contraction of the momentum four-vector. To verify that the contraction of the metric tensor is 4 simply find what it yields for a local frame [tex]g_{\mu }_{\nu }g^{\mu }^{\nu } = \eta _{\mu }_{\nu }\eta ^{\mu }^{\nu }[/tex] By hand it is easy to show that the right hand side yields 4. Since this result is invariant and one can do it considering the metric at any arbitrary point in spacetime, it must be the result according to arbitrary coordinates. It is therefor general.
    Last edited: Aug 13, 2004
  7. Aug 12, 2004 #6
    Ok, that is nice to know. I learned a contraction is setting a subscript equal to a supersript and summing, as the summation convention requires. But how do you account for your statement that mass is the contraction of the momentum 4-vector (wich has only 1 index).

    I guess [tex]\eta^\mu_\mu[/tex] is not really the trace, but you have to sum the absolute values, because else it yields not the number 4. You also get this by working out [tex]\eta_{\mu\nu}\eta^{\mu\nu}[/tex].

    Thank you both very much. And I sure liked the "short course" very much :). But I feel I have to read another book as well; that's why I started reading "a first course" as well. It introduces the concept of tensors in another way, not relying so much on its components. I'm not sure what's the 'better' way for a physicist. But I sure find it a fascinating subject...
  8. Aug 12, 2004 #7


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    Contracting is not so arbitrary. When you contract two tensors you raise or lower indeces by the metric as appopriate and then multiply and sum. This works for contracting tensors of any rank with tensors of any rank.
    To contract a rank one tensor called a four-vector with itself you raise or lower the index with the metric and do the multiplication and sum just the same way. The contraction of the momentum four-vector with itself is
    [tex]p_{\mu }p^{\mu } = g_{\mu }_{\nu }p^{\mu }p^{\nu } = m^{2}c^{2}[/tex]

    The latter is what you are actually doing.
    Element notation is superior to all others when it comes to direct application.
  9. Aug 12, 2004 #8
    No. It is the trace.

  10. Aug 13, 2004 #9
    But the trace of a matrix diag(1,-1,-1,-1) is 1-1-1-1=-2, so not 4. And when you calculate [tex]\eta_{\mu \nu}\eta^{\mu\nu}[/tex] you do get 4. So how can [tex]\eta^\mu_\mu=\eta_{\mu \nu}\eta^{\mu\nu}[/tex] and not yield the same value??
  11. Aug 13, 2004 #10


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    [/itex] is the trace or contraction of [itex] \eta_\mu{}^\nu
    [/itex]. When [itex]
    \eta_{\mu{}\nu}[/itex] is the metric tensor, [itex] \eta_\mu{}^\nu
    [/itex] is also known as the Kronecker delta [itex]\delta_{\mu}{}^\nu[/itex].
    [/itex] is not the trace or contraction of [itex]
  12. Aug 13, 2004 #11
    The notation I used was

    [tex]\eta_{\mu\nu} = diag(1, -1, -1, -1)[/tex]

    To take the trace of this matrix you first have to raise nu. You'll then obtain

    [tex]\eta_{\mu}^{\nu} = diag(1, 1, 1, 1)[/tex]

    Then you sum over the diagonal components to obtain

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

  13. Aug 13, 2004 #12


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    ANd to raise nu you use
    [tex]\eta^{\nu}_\mu} = \eta_{\mu\alpha}\eta^{\nu\alpha}[/tex] instead of the formula you had. Note that [tex]\eta^{\mu\nu}[/tex] is the inverse of [tex]\eta_{\mu\nu}[/tex].
  14. Aug 14, 2004 #13
    K, thanks everybody! It is perfectly clear to me now.
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