I am wondering how to apply the usual linear algebra to the rather unfamiliar case of 'matrices' with indices in special relativity or even general relativity. In particular, consider$$f=\sqrt{-\det\gamma_{\alpha\beta}}$$ where $\gamma_{\alpha\beta}$ represents a metric so that it has inverse $\gamma^{\beta\gamma}$. My understanding is that the $\alpha$ and $\beta$ here are not real indices, they are just there to let you know that this matrix we calculate determinant from, is the metric with both indices down. I would like to calculate $\frac{\partial f}{\gamma^{ab}}$ so the first step is$$\frac{\partial f}{\partial \gamma^{ab}}=-\frac{1}{2f}\frac{\partial (\det \gamma_{\alpha\beta})}{\partial \gamma^{ab}}$$ Proceed with Jacobi's formula for determinant:\begin{align}\frac{\partial\text{det}\gamma_{\alpha\beta}}{\partial \gamma^{ab}}&=\text{Tr}(\det\gamma_{\alpha\beta}\gamma^{-1}_{\alpha\beta}\frac{\partial\gamma_{\alpha\beta}}{\partial\gamma^{ab}})\end{align}

Here is the reason I fail to proceed. To calculate the quantity $\frac{\gamma_{\alpha\beta}}{\gamma^{ab}}$, one needs to treat $\alpha$ and $\beta$ as real indices and in particular they cannot be arbitrary $\alpha $ and $\beta$ if we want the derivative to be non-zero, as we will use $\gamma_{ab}\gamma^{bc}=\delta_a~^c$ to calculate this quantity. It would be extremely helpful if someone could write out the steps after this as an example for me to follow.

  • $\begingroup$ Are you specifically trying to calculate the partial wrt the metric or are you wanting to vary the metric determinant? $\endgroup$
    – aygx
    Nov 16, 2022 at 5:32
  • $\begingroup$ I am hoping to see how indices can be used properly in this case (i.e. instead of usual physics way of varying metric in determinant, which could certainly arrive at the same correct answer, use the more proper maths identity Jacobi gives us). $\endgroup$
    – Rescy_
    Nov 16, 2022 at 5:36
  • $\begingroup$ Did you mean to use gamma again as an index to the metric inverse, and is there supposed to be a partial symbol in the denominator under $\partial f$? $\endgroup$
    – Triatticus
    Nov 16, 2022 at 12:05

1 Answer 1


Let $\gamma\equiv -\det \gamma_{\alpha\beta}$ where $\alpha\beta$ is used to indicate the covariant metric tensor. We are interested in $$\frac{\partial\sqrt{-\gamma}}{\partial\gamma^{ab}}=\frac{1}{2\sqrt{-\gamma}}\frac{\partial\gamma}{\partial \gamma^{ab}}$$ and we are left to obtain a form of $\partial \det \gamma_{\alpha\beta}/\partial \gamma^{ab}$. Referring to Jacobi's formula, \begin{align} \frac{\partial \det \gamma_{\alpha\beta}}{\partial\gamma^{ab}}&=\gamma~\text{Tr}((\gamma_{\alpha\beta})^{-1}\cdot \frac{\partial\gamma_{\alpha\beta}}{\partial\gamma^{ab}})\\&=\gamma (\gamma^{\alpha\beta})(-1)\gamma_{\beta a}\gamma_{b\gamma}\\&=-\gamma\gamma_{ab} \end{align} and in particular when we let $a=\alpha, b=\beta$ we get $-\gamma \gamma_{\alpha\beta}$ so $$\frac{\partial \sqrt{-\gamma}}{\partial \gamma^{\alpha\beta}}=-\frac{1}{2}\sqrt{-\gamma}\gamma_{\alpha\beta}$$


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