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I've been studying GR through Wald's and Carroll's books, and I've been trying to derive one expression.

$$g(x^{\mu^\prime}) = \left|\dfrac{\partial x^{\mu^\prime}}{\partial x^{\mu}}\right|^{-2} g(x^\mu) $$

I'm not going anywhere, I'm stuck in some chain rules, but it does not seem to be the right way, could I do it?

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2 Answers 2

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Let's suppose we are transforming from $x^\mu$ coordinates to $y^\alpha$ coordinates. Then define the transformation $$ J^{\alpha}_{\ \ \ \mu} = \frac{\partial y^{\alpha}}{\partial x^\mu}, \ \ [J^{-1}]^\mu_{\ \ \ \alpha} = \frac{\partial x^\mu}{\partial y^\alpha} $$ We can write the metric in terms of the $y$ coordinates in terms of the metric in the $x$ coordinates as $$ g_{\alpha\beta}(y) = [J^{-1}]^\mu_{\ \ \ \alpha} [J^{-1}]^\nu_{\ \ \ \beta}g_{\mu\nu}(x) $$ Now we take the determinant of both sizes and use $\det(AB)=\det(A)\det(B)$ and $\det(A^{-1})=\det(A)^{-1}$ $$ |g(y)| = |J|^{-2}|g(x)| $$ which is the expression you want to derive.

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Recall the transformation rule for a doubly covariant tensor, and that $$ {\rm det}[A^TGA]= {\rm det}[A^T]{\rm det}[G]{\rm det}[A] =({\rm det}[A])^2 {\rm det}[G]. $$

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