# General Relativity: Prove symmetry of Einstein tensor

1. Apr 19, 2012

### ck99

1. The problem statement, all variables and given/known data

Show that Gij = Gji using the Riemann tensor identity (below)

2. Relevant equations

Gij = Rij - 1/2(gijR)

Rabcd + Rbcad + Rcabd = 0

R = gmrRmr

Rmr = Rmnrn

3. 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

Rij = Rinjn = gnaRinja

gijR = gia gjb gab R = gia gjb Rab = gia gjb Ranbn = gia gjb gnd Ranbd

So Gij = ( gnaRinja - 1/2 gia gjb gnd Ranbd )

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

2. Apr 19, 2012

### George Jones

Staff Emeritus
Can you show that the Ricci tensor is symmetric?

3. Apr 19, 2012

### 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

Rabcd = -Rbacd and that Rabcd = -Rabdc

But I'm not sure if

Rabcd = -Rbacd and Rabcd = -Rabdc

Looking at the Ricci tensor, in order for Rmr = Rrm I would need

Rmnrn = Rrnmn

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!

4. Apr 19, 2012

### George Jones

Staff Emeritus
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.

5. Apr 19, 2012

### ck99

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

gbd(Rabcd + Rbcad + Rcabd) = 0

Rabcb + Rbcab + Rcabb = 0

Rabcb - Rcbab + Rcabb = 0

Rac - Rca + Rcabb = 0

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

6. Apr 19, 2012

### George Jones

Staff Emeritus
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.

7. Apr 24, 2012

### 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 :(

gbd(Rabcd + Rbcad + Rcabd) = 0

Rabcb + Rbcab + gbdRcabd = 0

Rabcb + Rbcab - gdbRcadb = 0

Rabcb - Rcbab - Rcadd = 0

Rac - Rca - Rcadd = 0

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

8. Apr 24, 2012

### George Jones

Staff Emeritus
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 $g^{db} R_{cadb} = 0$.

9. May 7, 2012

### 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.