Einstein GR and metric signature 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.
 A: 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 $u^0=1$.
A: $\def\bg{\mathbf g} \def\bT{\mathbf T} \let\G=\Gamma$ 
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. 


*

*Connection coefficients are defined through $g_{\mu\nu}$'s derivatives
and an index lifting via $g^{\lambda\mu}$. So the $\G$'s are
invariant.

*Riemann tensor is built with $\G$'s alone ($\bg$ doesn't enter). So
Riemann is invariant too.

*Ricci tensor is a trace of Riemann, again with no $\bg$.

*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.
