Does anyone know how to get from (1) to (2) in the system
gμν,ρ+gσνΓμσρ+gμσΓνρσ−12(Γσρσ+Γσσρ)gμν=0,g[μν],ν+12(Γρρν−Γρνρ)g(μν)=0,
by contracting equation (1) once with respect to (μ,ρ), then with respect to (ν,ρ)?
Where Γ is not symmetric with respect to the lower Indices.
My attempt so far to solve this problem is: Well, when contracting with respect to μ and ρ I get: −12gρνΓaaρ−12gρνΓaρa+gaνΓρaρ+gρaΓνρa+gρν,ρ=0
Answer
The attempt of obtaining
g[μν],ν+12(Γρρν−Γρνρ)g(μν)=0,
1.) So when contracting Eq. (1) with respect to μ and ρ, we get the identity:
−12gρνΓaaρ−12gρνΓaρa+gaνΓρaρ+gρaΓνρa+gρν,ρ=0.
Now by relabeling the dummy indices in the 3rd term as a↔ρ, we get that the 3rd term can be written as gρνΓaρa. Moreover, we can see that now the 2nd and 3rd term can be simplified: adding them together gives +12gρνΓaρa. By a final index label change ν→μ, we get that:
−12gρμΓaaρ+12gρμΓaρa+gρaΓμρa+gρμ,ρ=0.(A)
2.) When contracting Eq. (1) with respect to ν and ρ, we get the identity:
−12gμρΓaaρ−12gμρΓaρa+gaρΓμaρ+gμaΓρρa+gμρ,ρ=0.
Let us also rename the dummy indices in the 4th term as a↔ρ. We can see now that the 4th term is simply gμρΓaaρ, and the 1st and 4th term hence together give 12gμρΓaaρ. Moreover, let us perform also the a↔ρ "dummy index relabeling", yielding gρaΓμρa for the 3rd term. After these manipulations our identity reads as
12gμρΓaaρ−12gμρΓaρa+gρaΓμρa+gμρ,ρ=0.(B)
3.) Now taking (B)-(A), we obtain:
gμρ,ρ−gρμ,ρ+12(gμρ+gμρ)(Γaaρ−Γaρa)=0,
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