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I enjoy the derivation of $T_{\mu\nu} = \rho u_\mu u_\nu$ by Dirac, reproduced in this Phys.SE question. But when I try to rewrite it with metric convention $(-+++)$ (used by MTW), I find a rather puzzling result.

The action for the relativistic dust should now be: $$ S_M = -\int \rho c \sqrt{-v_\mu v^\mu} \sqrt{ |\det g| } d^4 x = -\int c \sqrt{-p_\mu p^\mu} d^4 x $$ since $v_\mu v^\mu < 0,\, p_\mu p^\mu < 0$, while the rest of the derivation remains almost the same, which is: $$ \delta S_M = - \delta \int c \sqrt{-{p}_\mu {p}^\mu} d^4 x = $$ $$ = - \int c {\delta(-g^{\mu\nu} {p}_\mu {p}_\nu) \over 2\sqrt{-{p}_\alpha {p}^\alpha}} d^4 x = $$ $$ = \int c { {p}_\mu {p}_\nu \over 2\sqrt{-{p}_\alpha {p}^\alpha}} \delta(g^{\mu\nu}) d^4 x = $$ $$ = \int c { \rho v_\mu \rho v_\nu \sqrt{ |\det g| }^2 \over 2 \rho c \sqrt{ |\det g| } } \delta(g^{\mu\nu}) d^4 x = $$ $$ = \int {1\over2} \rho v_\mu v_\nu \delta(g^{\mu\nu}) \sqrt{ |\det g| } d^4 x $$ From which we calculate the stress energy tensor using the standard GR formula for it: $$ T_{\mu\nu} = - {2\over\sqrt{ |\det g| }}{\delta S_M\over\delta g^{\mu\nu}} = $$ $$ = - {2\over\sqrt{ |\det g| }} \left( {1\over2} \rho v_\mu v_\nu \sqrt{ |\det g| } \right)= $$ $$ = - \rho v_\mu v_\nu $$ As you can see, it gives an extra minus sign. Clearly I've done something stupid, but I just couldn't spot it. Please help me find that mistake (or mistakes) in the above derivation.

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It's a matter of convention. You'll find that this question and the first answer are relevant. Essentially your formula is consistent with the requirement that $T^{00}$ is positive, which is one possible way of fixing the sign of the stress-energy tensor. For this convention to be maintained when flipping the metric signature, one must introduce a minus sign in the definition of the stress-energy tensor, which the person in the question you're quoting didn't do. This doesn't mean their derivation is wrong: they are just following a different convention.

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