3
$\begingroup$

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.

Improve this question
$\endgroup$
5
  • $\begingroup$ 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 $\endgroup$
    – RenatoRenatoRenato
    Commented Nov 26, 2018 at 16:07
  • 2
    $\begingroup$ 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
    Commented Nov 26, 2018 at 16:14
  • 1
    $\begingroup$ 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. $\endgroup$
    – user1379857
    Commented 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
    Commented 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
    Commented Nov 27, 2018 at 0:30

2 Answers 2

Reset to default
3
$\begingroup$

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 $u^0=1$.

Improve this answer
$\endgroup$
5
  • $\begingroup$ Actually, shouldn't the +/- sign for Pg_{\mu\nu} come naturally, instead of being inserted artificially? $\endgroup$
    – Angela
    Commented 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
    Commented 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
    Commented Nov 27, 2018 at 3:45
  • $\begingroup$ There is no change in sign, when metric changes sign. $\endgroup$
    – Angela
    Commented 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
    Commented Nov 27, 2018 at 5:20
3
$\begingroup$

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

Improve this answer
$\endgroup$
1
  • $\begingroup$ Thanks G. Smith, Elio, others, for the answers and clarifications. $\endgroup$
    – Angela
    Commented Nov 26, 2018 at 23:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.