Varying determinant of a metric

[pleasehelpmeno]

Hi does anyone know how to calculate:

$\delta (det|g_{\mu\nu}|) or simply \delta g$

 

[elfmotat]

You can calculate it using Jacobi's forumula:
http://en.wikipedia.org/wiki/Determinant#Derivative
http://en.wikipedia.org/wiki/Jacobi's_formula

You should get $\delta g=gg_{\mu \nu}\delta(g^{\mu \nu})$

 

[Bill_K]

Or equivalently,

δg = g g^μν δg_μν

 

[Bill_K]

You should get δg=gg_μνδ(g^μν)

Actually I believe this answer has a sign error, and does not correspond to what I wrote, since δg^μν = - g^μα g^νβ δg_αβ.

A quick way to check the sign is to consider a special case, g_μν = λ η_μν under the variation δλ. For this case,

g^μν = λ^-1 η^μν and g = -λ⁴, so δg_μν = η_μν δλ and δg = -4 λ³ δλ.

The minus sign agrees with the expression I gave, namely g g^μν δg_μν = (-λ⁴)(λ^-1η^μν)(η_μν δλ).

However the other expression g g_μν δ(g^μν) = (-λ⁴)(λ η_μν)(-λ^-2) (η^μν δλ) comes out positive, which is incorrect.

 

[elfmotat]

Actually I believe this answer has a sign error, and does not correspond to what I wrote, since δg^μν = - g^μα g^νβ δg_αβ.

A quick way to check the sign is to consider a special case, g_μν = λ η_μν under the variation δλ. For this case,

g^μν = λ^-1 η^μν and g = -λ⁴, so δg_μν = η_μν δλ and δg = -4 λ³ δλ.

The minus sign agrees with the expression I gave, namely g g^μν δg_μν = (-λ⁴)(λ^-1η^μν)(η_μν δλ).

However the other expression g g_μν δ(g^μν) = (-λ⁴)(λ η_μν)(-λ^-2) (η^μν δλ) comes out positive, which is incorrect.

You're right. Thanks for the correction.

 

[pleasehelpmeno]

yeah thanks i knew the answer but had absolutely no idea how to get it, I end up with $g^{\nu\rho}g_{p\mu}\delta g^{\mu p}=-g^{\nu\rho}g^{\mu p} \delta g_{p\mu}$ and am not sure how to proceed from here, there is clearly relabelling but I am still a bit stuck.