# Ricci tensor proportional to divergence of Christoffel symbol?

1. May 15, 2014

### center o bass

I'm reading an old article published by Kaluza "On the Unity Problem of Physics" where i encounter an expression for the Ricci tensor given by

$$R_{\mu \nu} = \Gamma^\rho_{\ \mu \nu, \rho}$$

where he has used the weak field approximation $g_{\mu \nu} = \eta_{\mu \nu} + h_{\mu \nu}$ where $h$ is a small deviation tensor. I have never seen this expression before. Is it true in some gauge in the weak field?

2. May 15, 2014

### Bill_K

Working backwards, the Riemann tensor is

Rμνστ = Γμντ,σ - Γμνσ,τ + (quadratic terms)

Dropping the quadratic terms, the Ricci tensor is

Rντ = Γμντ,μ - Γμμν,τ

The first term is what you want - how do we get rid of the second?

Γμμν = (1/2) gμλgμλ,ν = (1/2) g

To first order, g = 1 + hμμ, so I'd say a sufficient gauge condition is that h is traceless.

3. May 15, 2014

### center o bass

I agree Bill K. Now I am a little bit familiar with the transverse-traceless gauge, which is a combination of the harmonic gauge $\partial_\mu h^\mu_\nu = 0$ and the trace condition $h=\eta^{\mu \nu}h_{\mu \nu}= 0$, but as I've understood it, this gauge is only allowable in a vacuum. Is there a gauge with $h=0$ without the harmonic gauge condition for which we do not need a vacuum?

4. May 15, 2014

### Bill_K

That's probably true if you want the field equation to be ◻hμν = Tμν, since then it would work only if Tμν was traceless also.

However in general, it seems to me you can always pick the coordinates to be "isovolume", that is, det(gμν) = const. (Although I haven't a clue how you would actually do that!)