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Let us take the einstein Equation $R_{\mu\nu} -\frac{1}{2}g_{\mu\nu}R = T_{\mu\nu}$. I'm just ignoring all the constants. For a perfect fluid, $$T_{\mu\nu} = (\rho + P)u_{\mu}u_{\nu} - Pg_{\mu\nu}.$$

If one swaps between the two metric sign $(-,+,+,+)$ and $(+,-,-,-)$, $g_{\mu\nu}$ changes sign. i.e. $g_{\mu\nu} \rightarrow -g_{\mu\nu}$; $R_{\mu\nu}$ changes sign; $R, \rho, P$ do not change sign. This means that the LHS, changes sign form $$R_{\mu\nu} -\frac{1}{2}g_{\mu\nu}R \rightarrow -(R_{\mu\nu} -\frac{1}{2}g_{\mu\nu}R) .$$ However, on the RHS, only $Pg_{\mu\nu}$ changes sign. The other term $(\rho + P)u_{\mu}u_{\nu}$ does not. Thus, this is the odd man out, and due to this term, the RHS, gets a different value.

Does this not show an inconsistency of the GR equation w.r.t change in the signature of the metric from $(+,-,-,-)$ to $(-,+,+,+)$?

Is something incorrect in what I have done here? For the laws of physics to be consistent, the equation should not depend upon what signature one takes.

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  • $\begingroup$ Equations depend on the convention chosen, laws don't. You're making confusion between a law that is fixed and equations that depends on what you like more $\endgroup$ Nov 26, 2018 at 16:07
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    $\begingroup$ Lots of equations have these factors in them that depend on the signature. IIRC there's a nice table of this kind of thing in the back of Misner, Thorne, and Wheeler. $\endgroup$
    – user4552
    Nov 26, 2018 at 16:14
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    $\begingroup$ You also have to ask how $T_{\mu \nu}$ is defined. In this context, it is defined by the variation of some "matter" part in the action by the metric. So the defintion of $T_{\mu \nu}$ would negate when negating the metric. $T_{\mu \nu}$ is not some random thing with a totally independent definition. $\endgroup$ Nov 26, 2018 at 17:12
  • $\begingroup$ The real question is: whether the east coast convention (+,-,-,-) and west coast convention (−,+,+,+) are just a matter of convention or there is something more about it? $\endgroup$
    – MadMax
    Nov 26, 2018 at 19:45
  • $\begingroup$ @MadMax You have the coasts backwards. The sign conventions are just conventions and nothing more. They reflect the fact that the metric is neither positive-definite nor negative-definite. Both conventions have pros and cons. $\endgroup$
    – G. Smith
    Nov 27, 2018 at 0:30

2 Answers 2

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For a (+,-,-,-) metric, the energy-momentum-stress tensor of a perfect fluid is

$$T^{\mu\nu}=(\rho+P)u^\mu u^\nu-P g^{\mu\nu}$$

but for a (-,+,+,+) metric it is

$$T^{\mu\nu}=(\rho+P)u^\mu u^\nu+P g^{\mu\nu}.$$

One way to remember which is which is that you want $T^{00}=\rho$ in the rest frame where $u^0=1$.

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  • $\begingroup$ Actually, shouldn't the +/- sign for Pg_{\mu\nu} come naturally, instead of being inserted artificially? $\endgroup$
    – Angela
    Nov 27, 2018 at 3:07
  • $\begingroup$ I don't know what "come naturally" means. It isn't "inserted artificially". Some equations simply take different forms with different sign conventions. They express identical physics. $\endgroup$
    – G. Smith
    Nov 27, 2018 at 3:11
  • $\begingroup$ I meant while deriving itself. I've not seen the derivation of the above stress energy tensor, but as a guess, if I take the variation of P.d = Pg_{\mu\nu}g^{\mu\nu} w.r.t g^{\mu\nu}, I get Pg_{\mu\nu}. (Here d = constant = g_{\mu\nu}g^{\mu\nu}). $\endgroup$
    – Angela
    Nov 27, 2018 at 3:45
  • $\begingroup$ There is no change in sign, when metric changes sign. $\endgroup$
    – Angela
    Nov 27, 2018 at 3:46
  • $\begingroup$ You can't derive it that way! You try to find a tensor which in the rest frame looks like a diagonal matrix with elements $(\rho,p,p,p)$. See Wikipedia on "Perfect fluid": en.wikipedia.org/wiki/Perfect_fluid $\endgroup$
    – G. Smith
    Nov 27, 2018 at 5:20
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$\def\bg{\mathbf g} \def\bT{\mathbf T} \let\G=\Gamma$ G. Smith:

One way to remember which is which is that you want $T^{00}=\rho$ in the rest frame where $u^0=1$.

I agree. And would add: In that frame $\bT$ is diagonal and all its components are non-negative. On the other hand, the components of $\bT$ don't depend on sign of $\bg$.

Second. I don't agree with the changes OP assigns to various objects. To me Ricci tensor is invariant, its trace changes sign.

Proof.

  1. Connection coefficients are defined through $g_{\mu\nu}$'s derivatives and an index lifting via $g^{\lambda\mu}$. So the $\G$'s are invariant.
  2. Riemann tensor is built with $\G$'s alone ($\bg$ doesn't enter). So Riemann is invariant too.
  3. Ricci tensor is a trace of Riemann, again with no $\bg$.
  4. The trace of $R_{\mu\nu}$ is $R=g^{\mu\nu} R_{\mu\nu}$, thus it changes sign.

Conclusion: both members of Einstein equations are invariant wrt $\bg$'s sign.

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  • $\begingroup$ Thanks G. Smith, Elio, others, for the answers and clarifications. $\endgroup$
    – Angela
    Nov 26, 2018 at 23:31

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