Monday, 13 July 2015

homework and exercises - Contracting Indices


Does anyone know how to get from (1) to (2) in the system


gμν,ρ+gσνΓμσρ+gμσΓνρσ12(Γσρσ+Γσσρ)gμν=0,g[μν],ν+12(ΓρρνΓρνρ)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: 12gρνΓaaρ12gρνΓaρa+gaνΓρaρ+gρaΓνρa+gρν,ρ=0

and when contracting with respect to nu and ρ I get: 12gμρΓaaρ12gμρΓaρa+gaρΓμaρ+gμaΓρρa+gμρ,ρ=0
when subtracting these two equations I get: 12gμρΓaaρ12gρνΓaaρ+12gμρΓaρa12gρνΓaρa+gaνΓρaρgaρΓμaρgμaΓρρa+gρaΓνρagμρ,ρ+gρν,ρ=0
I cant see how this is equal to equation (2)



Answer



The attempt of obtaining


g[μν],ν+12(ΓρρνΓρνρ)g(μν)=0,

was almost right! The only thing missing was a little care with relabeling indices. We will proceed in three main steps.


1.) So when contracting Eq. (1) with respect to μ and ρ, we get the identity:


12gρνΓaaρ12gρνΓaρa+gaνΓρaρ+gρaΓνρa+gρν,ρ=0.


Now by relabeling the dummy indices in the 3rd term as aρ, 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 νμ, 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:



12gμρΓaaρ12gμρΓaρa+gaρΓμaρ+gμaΓρρa+gμρ,ρ=0.


Let us also rename the dummy indices in the 4th term as aρ. 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 aρ "dummy index relabeling", yielding gρaΓμρa for the 3rd term. After these manipulations our identity reads as


12gμρΓaaρ12gμρΓaρa+gρaΓμρa+gμρ,ρ=0.(B)


3.) Now taking (B)-(A), we obtain:


gμρ,ρgρμ,ρ+12(gμρ+gμρ)(ΓaaρΓaρa)=0,

which after the aρ and ρν relabeling is exactly the same as the desired Eq. (2).


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