Ricci Tensor Proportional to Divergence of Christoffel Symbol?

[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?

 

[Bill_K]

Working backwards, the Riemann tensor is

R^μ_νστ = Γ^μ_ντ,σ - Γ^μ_νσ,τ + (quadratic terms)

Dropping the quadratic terms, the Ricci tensor is

R_ντ = Γ^μ_ντ,μ - Γ^μ_μν,τ

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

Γ^μ_μν = (1/2) g^μλg_μλ,ν = (1/2) g_,ν

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

 

[center o bass]

Working backwards, the Riemann tensor is

R^μ_νστ = Γ^μ_ντ,σ - Γ^μ_νσ,τ + (quadratic terms)

Dropping the quadratic terms, the Ricci tensor is

R_ντ = Γ^μ_ντ,μ - Γ^μ_μν,τ

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

Γ^μ_μν = (1/2) g^μλg_μλ,ν = (1/2) g_,ν

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

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?

 

[Bill_K]

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?

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

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

 

[sardar]

The expression for the Ricci tensor given by Kaluza is commonly known as the "weak field approximation" and is often used in theoretical physics as a simplification of the Einstein field equations. It is true in the weak field limit, where the gravitational field is weak and can be described by a small perturbation of flat space-time. In this limit, the Christoffel symbols can be written in terms of the metric perturbation, and the Ricci tensor can be expressed as a proportionality to the divergence of the Christoffel symbols. This approximation is commonly used in calculations related to general relativity and is a useful tool for understanding the behavior of the gravitational field in the weak field limit. However, it should be noted that this expression is only valid in certain gauges and may not hold in all situations. Further analysis and calculations would be needed to fully understand the implications and limitations of this expression.