# 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.

• 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 Nov 26 '18 at 16:07
• 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.
– user4552
Nov 26 '18 at 16:14
• 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. Nov 26 '18 at 17:12
• 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? Nov 26 '18 at 19:45
• @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. Nov 27 '18 at 0:30

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 $$u^0=1$$.

• Actually, shouldn't the +/- sign for Pg_{\mu\nu} come naturally, instead of being inserted artificially? Nov 27 '18 at 3:07
• 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. Nov 27 '18 at 3:11
• 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}). Nov 27 '18 at 3:45
• There is no change in sign, when metric changes sign. Nov 27 '18 at 3:46
• 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 Nov 27 '18 at 5:20

$$\def\bg{\mathbf g} \def\bT{\mathbf T} \let\G=\Gamma$$ 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.

• Thanks G. Smith, Elio, others, for the answers and clarifications. Nov 26 '18 at 23:31