Is Symmetry on μ and α Valid for the Derivative of the Metric Tensor?

[kent davidge]

I was thinking about the metric tensor. Given a metric g_μν we know that it is symmetric on its two indices. If we have g_μν,α (the derivative of the metric with respect to x^α), is it also valid to consider symmetry on μ and α? i.e. is the identity g_μν,α = g_αν,μ valid?

 

[andrewkirk]

It is not valid. A counterexample is the FLRW metric, in which $g_{11,0}$ is nonzero because the metric changes over time with the expansion of the cosmos, while $g_{01,1}$ is zero because $g_{01}$ is uniformly zero.

 

[kent davidge]

It is not valid. A counterexample is the FLRW metric, in which $g_{11,0}$ is nonzero because the metric changes over time with the expansion of the cosmos, while $g_{01,1}$ is zero because $g_{01}$ is uniformly zero.

How would it look if we symmetrize / antisymmetrize it on its first and third indices (μ and α)?

 

[andrewkirk]

What do you mean by symmetrise/antisymmetrise? In my understanding those are operations one performs on a tensor, and $g_{ab,c}$ is not a tensor, because partial differentiation (what the 'comma' does) is not a valid tensor operation.

 

[kent davidge]

What do you mean by symmetrise/antisymmetrise? In my understanding those are operations one performs on a tensor, and $g_{ab,c}$ is not a tensor, because partial differentiation (what the 'comma' does) is not a valid tensor operation.

I'm sorry. Actually I mean how could we obtain g_μα,β + g_μβ,α - g_αβ,μ from g_αβ,μ. (I'm trying to derive the Christoffel Symbol to put it in the geodesic equation.)

 

[Orodruin]

I'm sorry. Actually I mean how could we obtain g_μα,β + g_μβ,α - g_αβ,μ from g_αβ,μ. (I'm trying to derive the Christoffel Symbol to put it in the geodesic equation.)

You are being unclear. These are not the same things.