# Varying determinant of a metric

1. Feb 6, 2013

Hi does anyone know how to calculate:

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

2. Feb 6, 2013

3. Feb 6, 2013

### Bill_K

Or equivalently,

δg = g gμν δgμν

4. Feb 7, 2013

### Bill_K

Actually I believe this answer has a sign error, and does not correspond to what I wrote, since δgμν = - gμα gνβ δgαβ.

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

gμν = λ-1 ημν and g = -λ4, so δgμν = ημν δλ and δg = -4 λ3 δλ.

The minus sign agrees with the expression I gave, namely g gμν δgμν = (-λ4)(λ-1ημν)(ημν δλ).

However the other expression g gμν δ(gμν) = (-λ4)(λ ημν)(-λ-2) (ημν δλ) comes out positive, which is incorrect.

5. Feb 7, 2013

### elfmotat

You're right. Thanks for the correction.

6. Feb 8, 2013

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 im still a bit stuck.