Einstein Field Eqns: East/West Coast Metrics

[DuckAmuck]

TL;DR

Different results unless you are careful?

My questions is:
Depending on which metric you choose "east coast" or "west coast", do you have to also mind the sign on the cosmological constant in the Einstein field equations?
$$R_{\mu\nu} - \frac{1}{2} R g_{\mu\nu} \pm \Lambda g_{\mu\nu} = \frac{8\pi G}{c^4} T_{\mu\nu}$$
For example, if you stick with a +Lambda term with both metrics, you get different results for each metric. So once you choose a metric, you have to have the right sign on Lambda it seems.

 

[PeterDonis]

if you stick with a +Lambda term with both metrics, you get different results for each metric

No, you don't. Note that the Ricci scalar term also multiplies $g_{\mu \nu}$, and the sign of the Ricci scalar term has to be negative (since that's the only way to get an Einstein tensor whose covariant divergence is zero). The relative sign of that term and the $\Lambda$ term is the key thing, physically speaking, and that isn't affected by which metric signature convention you choose.

 

[DuckAmuck]

No, you don't. Note that the Ricci scalar term also multiplies $g_{\mu \nu}$, and the sign of the Ricci scalar term has to be negative (since that's the only way to get an Einstein tensor whose covariant divergence is zero). The relative sign of that term and the $\Lambda$ term is the key thing, physically speaking, and that isn't affected by which metric signature convention you choose.

This is what I thought at first. But if you take the 00 element of the equations, and the trace of the equations, and solve for Lambda, you get different results depending on your choice of sign and metric. Would just like some clarity on the sign conventions for this.

 

[PeterDonis]

if you take the 00 element of the equations

Why would you do that? The 00 element by itself has no physical meaning.

 

[DuckAmuck]

Maybe I'm not explaining it right. I swear I have seen this done before. Here is the process:
Start with this generic form:
$$R_{\mu\nu} - \frac{1}{2} R g_{\mu\nu} \pm \Lambda g_{\mu\nu} = \frac{8 \pi G}{c^4} T_{\mu\nu}$$
Take the trace to get:
$$-R \pm 4\Lambda = \frac{8 \pi G}{c^4} T$$
Take the time element (00 element) to get:
$$R_{00} - \frac{1}{2} Rg_{00} \pm \Lambda g_{00} = \frac{8 \pi G}{c^4} T_{00}$$
Where g_00 can be +1 or -1 depending on metric choice.
Treating these equations like a system, you can arrive at:
$$R_{00} \mp \Lambda g_{00} = \frac{8 \pi G}{c^4} (T_{00} - Tg_{00}/2)$$

 

[PeterDonis]

Where g_00 can be +1 or -1 depending on metric choice.

Which doesn't matter because $g_{00}$ appears explicitly in the equations.

Treating these equations like a system, you can arrive at:
$$R_{00} \mp \Lambda g_{00} = \frac{8 \pi G}{c^4} (T_{00} - Tg_{00}/2)$$

Yes. So what?

 

[vanhees71]

Note that in GR there are even more confusing differing sign convention than in SR. In SR you have essentially only two differing sign conventions: One in the Minkowski product which can be of signature (1,3) (west coast convention, usually used in the high-energy-particle physics community, but also there are exceptions, e.g., Weinberg) or of signature (3,1) (east-coast convention). The other is in defining the 4D Levi-Civita symbol. No matter how you start, in Minkowski coordinates you have $\epsilon_{\mu \nu \rho \sigma}=-\epsilon^{\mu \nu \rho \sigma}$ and different authors start with different definitions concerning the sign. It's very confusing. Most common seems to be $\epsilon^{0123}=+1$. In any case one has to check the convention for each paper.

In GR you can get additional sign changes from different non-zero contractions of the Riemann curvature tensor to the 2nd-rank Ricci tensor, leading to sign changes. A nice review about different conventions can be found in Misner-Thorne-Wheeler (note that Landau&Lifshitz changed their sign conventions from one edition to the other of their vol. 2 ;-)).