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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 $g’=\left(\det\frac{\partial x^{\mu’}}{\partial x^{\mu}}\right)^{-2}g$, 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?

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  • $\begingroup$ What do you mean? The determinant is squared, so no sign. $\endgroup$
    – Qmechanic
    Sep 18 '20 at 15:28
  • $\begingroup$ Taking the square root then gives the absolute value of the determinant, this leaves a sgn(dx’/dx) so the new object still does not transform like a tensor unless the new coordinates are orientation preserving $\endgroup$
    – Leon231000
    Sep 18 '20 at 15:51
  • $\begingroup$ What 'new object' are you referring to? $\endgroup$
    – Qmechanic
    Sep 18 '20 at 15:54
  • $\begingroup$ Carroll states that when we multiply a tensor density by this metric-dependent factor we get an “actual” tensor. This is supposed to transform according to “normal” tensor transformation rules but it does only with a sgn() factor unless the coordinate change is orientation preserving $\endgroup$
    – Leon231000
    Sep 18 '20 at 15:56
  • $\begingroup$ What definition of a tensor density of weight $w$ do you use? What definition does Carroll use? Which page? $\endgroup$
    – Qmechanic
    Sep 18 '20 at 15:59
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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 $det\Big(\frac{\partial x^{\mu'}}{\partial x^{\mu}}\Big)$ just substitute it in place of $\sqrt{|g|}$.

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  • $\begingroup$ Yes, an orientation preserving coordinate change would be the only way this makes sense in my mind. Is this all there’s to it? $\endgroup$
    – Leon231000
    Sep 18 '20 at 15:54

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