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Homework Help: A problem with Wald's General Relativity

  1. Jun 24, 2011 #1
    1. The problem statement, all variables and given/known data

    This problem is Problem 5 in Chapter 4. It is that [itex]T_{ab}[/itex] is a symmetric, conserved field ([itex]T_{ab}=T_{ba}, \partial ^aT_{ab}=0[/itex]) in Minkowski spacetime. Show that there is a tensor field [itex]U_{acbd}[/itex] with the symmetries [itex]U_{acbd}=U_{[ac]bd}=U_{ac[bd]}=U_{bdac}[/itex] such that [itex]T_{ab}=\partial^c\partial^dT_{acbd}[/itex].

    Wald gave a hint: For any vector field [itex]v^a[/itex] in Minkowski spacetime satisfying [itex]\partial_av^a=0[/itex] there is a tensor field [itex]s^{ab}=-s^{ba}[/itex] such that [itex]v^a=\partial_bs^{ab}[/itex]. Use this fact to show that [itex]T_{ab}=\partial^cW_{cab}[/itex] with [itex]W_{cab}=W_{[ca]b}[/itex]. The use the fact that [itex]\partial^cW_{c[ab]}=0[/itex] to derive the desired result.



    3. The attempt at a solution

    Based on his hint, I got a solution [itex]T_{ab}=\partial^c\partial^dU_{acbd}[/itex]. Like [itex]s^{ab}=-s^{ba}[/itex], I required that [itex]U_{acbd}=-U_{adbc}[/itex], but this condition would lead to the result [itex]T_{ab}=0[/itex]!!!!

    So what is wrong with my solution? I need your help, Thank you!
     
  2. jcsd
  3. Jun 24, 2011 #2
    Could you show your calculation in a bit more detail, please? Otherwise it is difficult seeing where you went wrong, as I haven't done the calculation.
     
  4. Jun 24, 2011 #3
    Thanks for your reply! I will show my calculation.

    To derive [itex]T_{ab}=\partial^cW_{cab}[/itex], just consider the components [itex]{T^a}_{\mu}={T^a}_b(\partial_\mu)^b[/itex] which is a vector satisfying [itex]\partial_a{T^a}_\mu=0[/itex]. So according to the hint, there exits a tensor [itex]{W^{ca}}_\mu[/itex] such that [itex]{T^a}_\mu=\partial_c{W^{ca}}_\mu[/itex]. Therefore, [itex]T_{ab}=\partial^cW_{cab}[/itex]. Of course, [itex]{W^{ca}}_\mu={W^{[ca]}}_\mu[/itex].

    Then, to get the final result, we just use this fact [itex]\partial^cW_{c[ab]}=0[/itex] or [itex]\partial_c{W^c}_{\mu\nu}=0[/itex]. So there is a tensor [itex]{{{U^c}_{\mu \nu}}}^d=-{{U^d}_{\mu \nu}}^c[/itex] with [itex]{W^c}_{\mu \nu}=\partial_d{{U^c}_{\mu \nu}}^d[/itex]. Then, [itex]{W^c}_{ab}=\partial_d{{U^c}_{ab}}^d[/itex].

    In the end, [itex]T_{ab}=\partial^cW_{cab}=\partial^c\partial^dU_{cabd}=-\partial^c\partial^dU_{acbd}[/itex].

    That's all. The only thing about my final result is that there is an overall minus sign, which is trivial.
     
  5. Jun 25, 2011 #4
    The original theorem was that for any v^a with d_a v^a = 0, there is an antisymmetric tensor s^{ab} such that v^a = d_b s^{ab} = - d_b s^{ba}. So you should have T^a_μ = ∂_c W^{ac}_μ, which is minus the result you gave. Of course you can always redefine W to be -W, but let's stick with mine for now.

    Evidently. So T^{ab} = ∂_c W^{acb} with W^{acb} = W^{[ac]b}.

    I don't see it that clearly. Let's see, from the symmetry of the energy-momentum tensor we get
    ∂_c W^{cab} = - ∂_c W^{acb} = - ∂_c W^{bca} = ∂_c W^{cba}
    so, yes, check.

    Yes. So, T^{ab} = ∂_c W^{acb} = ∂_c ∂_d U^{acbd}, and we found your minus sign. For the symmetries, we know that W^{acb} = - W^{cab} = - W^{cba}. Let's check the symmetries of U. First, it is antisymmetric in the first and last index from the definition, so U^{acbd} = - U^{dcba}. From the W symmetries it gets U^{acbd} = - U^{cabd} = - U^{cbad}, and that should be all. So we still need that U^{acdb} = - U^{acbd} and U^{acbd} = U^{bdac} according to Wald, let's see if we can achieve this.
    U^{bdac} = U^{badc} = - U^{cadb} = U^{acdb}
    No! Doesn't work. I think you should set W^c_{μν} = ∂_d U^{cd}_{μν} and not ∂_d U^c_{μν}^d , because you don't get the required symmetries otherwise. With that choice, T^{ab} also shouldn't vanish anymore :)
     
  6. Jun 25, 2011 #5
    Thanks for your reply! But you see, we agree that [itex]{U^{cd}}_{\mu \nu}[/itex](your notation) or [itex]{{U^c}_{\mu \nu}}^d[/itex] (my notation) should be antisymmetric in indices c and d, then [itex]T_{ab}=\partial^c \partial^d U_{cdab}[/itex] or [itex]T_{ab}=\partial^c \partial^dU_{cabd}[/itex] is zero since [itex]\partial^c\partial^d=\partial^d\partial^c[/itex]. Is that right?
     
  7. Jun 27, 2011 #6
    Yes, with your notation T^{ab} comes out zero. With the notation I proposed also, I see it now. But note that in the original problem the derivatives should act on the first and third index of U, so that T doesn't vanish. So you should take W^c_{μν} = ∂_d U^c_μ^d_ν, sorry for the confusion before.
     
  8. Jun 29, 2011 #7
    Sorry. I was busy those days. Hmmm, I do not think we can put the indices in arbitrary position. And even if you take W^c_{μν} = ∂_d U^c_μ^d_ν, we still have to assume that U_{cadb}=-U_{dacb}, that is, antisymmetric in the first and the third indices. So still, we get a vanishing T_{ab}.
     
  9. Jun 30, 2011 #8
    But yes, you can, since μ and ν are not Lorentz indices, and the first theorem (that v^a = ∂_b s^{ab}) only makes a statement about Lorentz indices.

    U is antisymmetric in the first and second index, and separately in the third and fourth, but not in the first and third. So only in that case T^{ab} is not zero.
     
  10. Jul 1, 2011 #9
    Well, I really cannot agree with you, but can you introduce some material on tensor, especially on indices? I will appreciate it!
     
  11. Dec 12, 2013 #10
    Then, to get the final result, we just use this fact [itex]\partial^cW_{c[ab]}=0[/itex] or [itex]\partial_c{W^c}_{\mu\nu}=0[/itex].

    Pretty sure what you said above was wrong, [itex] \partial ^cW_{c[ab]}=0 [/itex] does not imply that [itex]\partial d ^cW_{cab}=0[/itex]. If it did it would mean then the stress energy tensor is zero from the step before.
     
    Last edited: Dec 12, 2013
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