A problem with Wald's General Relativity

In summary, the conversation discusses the problem of finding a tensor field U_{acbd} with certain symmetries that satisfies a given equation involving the energy-momentum tensor T_{ab}. The original theorem and the provided hint are used to derive the solution, but a mistake is found in the final result. The mistake is traced back to the choice of notation and it is suggested to redefine W^c_{μν} to solve the problem.
  • #1
qinglong.1397
108
1

Homework Statement



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.



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!
 
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  • #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.
 
  • #3
grey_earl said:
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.

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.
 
  • #4
qinglong.1397 said:
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].

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.

qinglong.1397 said:
Of course, [itex]{W^{ca}}_\mu={W^{[ca]}}_\mu[/itex].
Evidently. So T^{ab} = ∂_c W^{acb} with W^{acb} = W^{[ac]b}.

qinglong.1397 said:
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].
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.

qinglong.1397 said:
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].
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 :)
 
  • #5
grey_earl said:
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 :)

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?
 
  • #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.
 
  • #7
grey_earl said:
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.

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}.
 
  • #8
Hmmm, I do not think we can put the indices in arbitrary position.
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.

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}.
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.
 
  • #9
grey_earl said:
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.

Well, I really cannot agree with you, but can you introduce some material on tensor, especially on indices? I will appreciate it!
 
  • #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.
 
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FAQ: A problem with Wald's General Relativity

1. What is the problem with Wald's General Relativity?

The problem with Wald's General Relativity is that it fails to provide a complete and consistent explanation for the behavior of objects near the event horizon of a black hole. This is known as the black hole information paradox.

2. How does the black hole information paradox relate to Wald's General Relativity?

The black hole information paradox arises from the inconsistency between Wald's General Relativity and quantum mechanics. According to General Relativity, information that falls into a black hole is lost forever, while quantum mechanics suggests that information cannot be destroyed.

3. Can the problem with Wald's General Relativity be solved?

There are ongoing efforts to solve the black hole information paradox and reconcile General Relativity with quantum mechanics. Some proposed solutions include the holographic principle and the firewall hypothesis.

4. Has Wald's General Relativity been tested or observed?

Yes, Wald's General Relativity has been extensively tested and observed through various experiments and observations, including the detection of gravitational waves. However, the effects of General Relativity near the event horizon of a black hole have not yet been directly observed.

5. Are there any alternative theories to Wald's General Relativity?

There are several alternative theories to General Relativity, such as modified gravity and string theory. However, these theories also face challenges in reconciling with quantum mechanics and explaining the behavior of objects near black holes.

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