A problem with Wald's General Relativity

[qinglong.1397]

Homework Statement

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

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

The Attempt at a Solution

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

So what is wrong with my solution? I need your help, Thank you!

 

[grey_earl]

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.

 

[qinglong.1397]

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

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

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

That's all. The only thing about my final result is that there is an overall minus sign, which is trivial.

 

[grey_earl]

Thanks for your reply! I will show my calculation.

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

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.

Of course, ${W^{ca}}_\mu={W^{[ca]}}_\mu$.

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

Then, to get the final result, we just use this fact $\partial^cW_{c[ab]}=0$ or $\partial_c{W^c}_{\mu\nu}=0$.

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.

So there is a tensor ${{{U^c}_{\mu \nu}}}^d=-{{U^d}_{\mu \nu}}^c$ with ${W^c}_{\mu \nu}=\partial_d{{U^c}_{\mu \nu}}^d$. Then, ${W^c}_{ab}=\partial_d{{U^c}_{ab}}^d$.

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 :)

 

[qinglong.1397]

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 ${U^{cd}}_{\mu \nu}$(your notation) or ${{U^c}_{\mu \nu}}^d$ (my notation) should be antisymmetric in indices c and d, then $T_{ab}=\partial^c \partial^d U_{cdab}$ or $T_{ab}=\partial^c \partial^dU_{cabd}$ is zero since $\partial^c\partial^d=\partial^d\partial^c$. Is that right?

 

[grey_earl]

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.

 

[qinglong.1397]

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}.

 

[grey_earl]

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.

 

[qinglong.1397]

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!

 

[tommyj]

Then, to get the final result, we just use this fact $\partial^cW_{c[ab]}=0$ or $\partial_c{W^c}_{\mu\nu}=0$.

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