General Relativity: Prove symmetry of Einstein tensor

[ck99]

Homework Statement

Show that G_ij = G_ji using the Riemann tensor identity (below)

Homework Equations

G_ij = R_ij - 1/2(g_ijR)

R_abcd + R_bcad + R_cabd = 0

R = g^mrR_mr

R_mr = R_mnrⁿ

The Attempt at a Solution

I have tried to put the Ricci tensor and Ricci scalar (from the Gij equation) into full Riemann tensor form using the metric. For Gij I get

R_ij = R_injⁿ = g^naR_inja

g_ijR = g_ia g_jb g^ab R = g_ia g_jb R_ab = g_ia g_jb R_anbⁿ = g_ia g_jb g^nd R_anbd

So G_ij = ( g^naR_inja - 1/2 g_ia g_jb g^nd R_anbd )

I have done the same for G_ji but can't see how to link the two equations using the required Riemann identity. Is this the right approach?

 

[George Jones]

Can you show that the Ricci tensor is symmetric?

 

[ck99]

I know that the Riemann tensor is skew-symmetric in the first and second pair of indexes, when it is in fully covariant form. I'm not sure if that holds when it is in mixed form though?

I know that

R_abcd = -R_bacd and that R_abcd = -R_abdc

But I'm not sure if

R_abc^d = -R_bac^d and R_abc^d = -R_abd^c

Looking at the Ricci tensor, in order for R_mr = R_rm I would need

R_mnrⁿ = R_rnmⁿ

And I'm not sure how to swap indices between the first pair and second pair on Riemann. Do I need to go all the way back to the equation for Riemann in terms of Christoffels, and Christoffels from the metric? There must be a quicker way, this is a tiny bit of an exam question that should take 5 minutes!

 

[George Jones]

I know that

R_abcd = -R_bacd and that R_abcd = -R_abdc

But I'm not sure if

R_abc^d = -R_bac^d

Yes.

and R_abc^d = -R_abd^c

No, indices need to have the same upstairs/downstairs locations on the both sides of an equation, i.e., if you raise the d on the left, then then d on the right should be raised. More explicitly,

$$R_{abcd} g^{de} = -R_{abdc}g^{de}$$

Looking at the Ricci tensor, in order for R_mr = R_rm I would need

R_mnrⁿ = R_rnmⁿ

And I'm not sure how to swap indices between the first pair and second pair on Riemann. Do I need to go all the way back to the equation for Riemann in terms of Christoffels, and Christoffels from the metric? There must be a quicker way, this is a tiny bit of an exam question that should take 5 minutes!

Raise the d in each term of

$$R_{abcd} + R_{bcad} + R_{cabd} = 0$$

and contract with one of the other indices.

 

[ck99]

Yes.


No, indices need to have the same upstairs/downstairs locations on the both sides of an equation, i.e., if you raise the d on the left, then then d on the right should be raised. More explicitly,

$$R_{abcd} g^{de} = -R_{abdc}g^{de}$$



Raise the d in each term of

$$R_{abcd} + R_{bcad} + R_{cabd} = 0$$

and contract with one of the other indices.

Thanks George, that helps a bit but I am still missing something. I tried

g^bd(R_abcd + R_bcad + R_cabd) = 0

R_abc^b + R_bca^b + R_cab^b = 0

R_abc^b - R_cba^b + R_cab^b = 0

R_ac - R_ca + R_cab^b = 0

So I would be OK if that last term were to vanish somehow! Can someone suggest where I go from here?

 

[George Jones]

So I would be OK if that last term were to vanish somehow! Can someone suggest where I go from here?

Put the g back in the last term, and look at it more carefully. g is symmetric in b and d, while the last term is antisymmetric in b and d.

 

[ck99]

Sorry for the delay, I had to work on some other stuff. I have tried your suggestion but I'm still not getting this :(

g^bd(R_abcd + R_bcad + R_cabd) = 0

R_abc^b + R_bca^b + g^bdR_cabd = 0

R_abc^b + R_bca^b - g^dbR_cadb = 0

R_abc^b - R_cba^b - R_cad^d = 0

R_ac - R_ca - R_cad^d = 0

I'm not sure how that helps me. Have I done it right?

 

[George Jones]

You need to show $R_{cad}{}^d = 0$, i.e, you need to show $g^{db} R_{cadb} = 0$.

g is symmetric in b and d, while $R_{cadb}$ is antisymmetric in b and d. b and d are both dummy indices, and so can be relabeled. Use these facts to show $g^{db} R_{cadb} = 0$.

 

[ck99]

I forgot to thank you for this George, I figured it out in the end (although it didn't come up in my exam!). Your help is much appreciated.