Absolute value of Jacobian when converting tensor density

Question

I am currently reading Sean M. Carroll’s book about General Relativity. He claims that, in order to convert a tensor density of weight $w$ into a true tensor, one should multiply by $sqrt{|{g}|}^w$, where $g=det g_{mu v}$ is the determinant of the metric. Knowing that $g’=left(detfrac{partial x^{mu’}}{partial x^{mu}}right)^{-2}g$, it seems as though the transformation of the new object depends on the sign if the jacobian. This confuses me. Is there an explanation for this?

aitfel · Accepted Answer

We did take the sign problem into account by using $|g|$ when multiplying with a tensor density.
If this sounds like cheating: there is something called orientation and if the manifold we're working is orientable (which is usually what we encounter Schwarzschild, Kerr, RN) then while defining new coordinates $x^{mu'}=x^{mu'}(x^{mu})$ we go for orientation preserving function (diffeomorphism) what it means that if we define orientation in our old coordinates then it will be "kind of" the same in the new one. Then our jacobian will be positive so we don't have to use $|hspace{2pt}|$ i.e.

whatever you get by $detBig(frac{partial x^{mu'}}{partial x^{mu}}Big)$ just substitute it in place of $sqrt{|g|}$.

Absolute value of Jacobian when converting tensor density

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