### Signs in Einstein’s Equation

13Aug10

Wow, months passed since I’ve moved Equatorial Frequencies to wordpress. Well, here’s a long overdue post that I meant to do right after EQuaLS4 but I left thinking about it until very recently. During that event, my student Hassan presented our work with Pradhan and showed the following form of Einstein’s equation (actually it’s the mixed tensor variant)

$G_{ab}=-\kappa T_{ab}$

A question was raised regarding the negative sign (unfortunately I wasn’t there) since many books use

$G_{ab}=\kappa T_{ab}$

There was a slight confusion then, but really the whole thing is very much dependent on how we define things. You might be surprised by this since  one negative sign can make a whole lot of difference in its physics. Well there are different sign conventions followed by physicists but ultimately (and I’ve checked) they all boiled down to the same physics. Almost immediately after the event, Hassan showed me the notes by M. Haehnelt at DAMTP, Cambridge, where a table of different sign conventions made by different authors is shown (and he himself adopted a different sign convention for some objects). I decided to follow up this study of signs and show that all results point to the same physics. I included all the references mentioned by Haehnelt (d’Inverno, Rindler, Morris-Thorne-Wheeler, Weinberg) including Haehnelt himself, all the standard text books (Schutz, Stephani, Wald, Hawking & Ellis) including the new ones by Jim Hartle and Sean Carroll, the textbook by Hervik (who was here) and peculiar ones that I use like Carmeli and deSabbata & Gasperini.

I will use d’Inverno as the standard reference here since I have always recommended this book to students for learning general relativity due to its friendly approach. As such, I will use an overbar for all quantities defined in d’Inverno and later relate them with corresponding quantities from the other books (without the overbar).

First and foremost is the signature of the metric adopted by the books. Many books adopted the $(-,+,+,+)$ signature (Morris-Thorne-Wheeler, Weinberg, Hervik, Hartle, Schutz, Carroll,, Stephani, Hawking-Ellis). This signature can be thought of as a natural extension from the Euclidean signature i.e. for example in the flat case

$ds^2 = dx^2 + dy^2 + dz^2 \longrightarrow ds^2= -dt^2+ dx^2 + dy^2 + dz^2$

The other convention is to pick the signature $(+,-,-,-)$ and this is adopted by d’Inverno, Rindler, Haehnelt and Carmeli. The signature is perhaps considered natural when one talks about time-like events for which

$ds^2 = dt^2 -dx^2 -dy^2-dz^2>0$

Now there are authors like de Sabbata & Gasperini who puts the time coordinate as the fourth coordinate and not the zeroth coordinate for which the signature will look like $(-,-,-,+)$. But I will consider this is equivalent to $(+,-,-,-)$ since it is simply a permutation. Note however some care is needed when shuffling indices of tensors to make tensors of different permutation of signature look equivalent. Thus, now if we denote $g$ for the metric of the first signature and $\bar{g}$ for the metric of the second signature (of d’ Inverno), then

$g_{ab} = -\bar{g}_{ab}\quad .$

The following quantities are defined in the same way in (almost) all books. Namely, the Christoffel symbols

$\Gamma^a_{bc}= g^{ad}(\partial_b g_{dc} + \partial_c g_{bd} - \partial_d g_{bc})\equiv \bar{\Gamma}^a_{bc}\quad ,$

the Ricci tensor $R_{ab} = R^c_{acb}\equiv \bar{R}_{ab}$ and the Ricci scalar $R = g^{ab} R_{ab}\equiv \bar{R}$.

Do note that Rindler uses a different definition for the Ricci tensor, namely $R_{ab}=R^c_{abc}$. From symmetries of the Riemann tensor, one finds that

$R_{ab}=-\bar{R}_{ab}\quad .$

I purposely mention the Ricci tensor first before the Riemann tensor since there are two definitions available in the literature. The common one, used by d’Inverno (and almost many others), is

$\bar{R}^a_{bcd}=\partial_c\Gamma^a_{bd} - \partial_d\Gamma^a_{bc} +\Gamma^a_{ec}\Gamma^e_{bd} - \Gamma^a_{ed}\Gamma^e_{bc}\quad .$

Weinberg and Haehnelt however use

$R^a_{bcd}=\partial_d\Gamma^a_{bc} - \partial_c\Gamma^a_{bd} +\Gamma^a_{ed}\Gamma^e_{bc} - \Gamma^a_{ec}\Gamma^e_{bd}\quad ,$

which implies $R^a_{bcd} = -\bar{R}^a_{bcd}$. It also implies $R_{ab} = -\bar{R}_{ab}$ and $R=-\bar{R}$ for the case of Weinberg and Haehnelt.

Next, the Einstein tensor, whose usual definition is $\bar{G}_{ab}=\bar{R}_{ab}-\frac{1}{2}\bar{g}_{ab}\bar{R}$. Haehnelt differs from the rest in his definition by putting $G_{ab}=\frac{1}{2}g_{ab} R- R_{ab}$ but because his ‘negative’ definition of the Riemann tensor, one has $G_{ab}=\bar{G}_{ab}$ (from double negative). Weinberg’s case is also the same; there is double negative arising from the different signature of the metric, which I will explain below.

Despite that everyone uses the same definition of the Ricci tensor (apart from Rindler), there is an implicit sign difference involved when different signature is used. Under the transformation of metric $\bar{g}_{ab}\rightarrow g_{ab}=-\bar{g}_{ab}$ to one of different signature, I claim $\bar{R}_{ab}\rightarrow R_{ab}=-\bar{R}_{ab}$. This is definitely true for Einstein manifolds, where $R_{ab} = k g_{ab}$. For the general case, I’m afraid, I do not know how to show this apart from writing explicitly the equations out. However, note that it is this crucial sign difference that later shows us that we have the same physics in the Einstein’s field equations throughout irrespective of the sign conventions taken.

To write out the field equations, one has to pick out one stress-energy tensor $T_{ab}$. For our purposes, we will pick that of the perfect fluid. The one used by d’Inverno (and the rest that uses signature $(+,-,-,-)$ is

$T_{ab} = (\rho + p) u_a u_b - p g_{ab}\quad .$

Those using signature $(-,+,+,+)$ will use

$T_{ab}=(\rho + p)u_a u_b + p g_{ab}\quad .$

Do note the difference in sign in front of the second term. Now we can come to the field equations. Whatever sign taken in $G_{ab}=\pm\kappa T_{ab}$, the sign also persists in its contracted form $G^a_a=\kappa T^a_a$. The physics that arise can be checked easily in this contracted form. The contracted Einstein tensor gives

$G^a_a = R^a_a -\frac{1}{2}\delta^a_a R=-R\quad .$

For the perfect fluid stress-energy tensor $T_{ab}=(\rho+p)u_a u_b\pm p g_{ab}$, the contracted form is

$T^a_a= (\rho+p)u^a u_a \pm p \delta^a_a=\pm(\rho-3p)\quad .$

Let us now consider the field equation $\bar{G}_{ab} = +\kappa \bar{T}_{ab}$, its contracted form is

$-\bar{R}=\kappa (\rho-3p)\quad .$

If the physics is the same in whatever conventions are taken mentioned in this post, they should reproduce this equation. Let us now consider the case given by Weinberg (which is followed by our paper) where $G_{ab}=-\kappa T_{ab}$; since Weinberg uses the negative form of Riemann tensor (see remark earlier), his use of Ricci tensor (and hence Einstein tensor) equals that of d’Inverno. But now he uses the plus sign in the stress energy tensor, and thus

$-R = -\kappa(3p-\rho)\quad\Rightarrow\quad -\bar{R} =\kappa(\rho-3p)$

which is equivalent to the one by d’Inverno. Rindler too has $G_{ab}=-\kappa T_{ab}$. But now his Ricci tensor is $R_{ab} = -\bar{R}_{ab}$ and hence $G_{ab}=-\bar{G}_{ab}$ whose negative sign cancels the one on the right hand side of the field equation. Again it reproduces the same equation.

Let us now move to those who use $(-,+,+,+)$ signature but has field equation $G_{ab} =\kappa T_{ab}$ and $T_{ab}=(\rho+p)u_a u_b + p g_{ab}$ (most authors use this). Note that in this case, the Einstein tensor is $G_{ab}=-\bar{G}_{ab}$. Writing out the contracted field equation:

$-R=\kappa(3p-\rho)\quad\Rightarrow\quad +\bar{R}=\kappa(3p-\rho)$

which is again the same equation. This should be comforting for us to know that all conventions give the same physics!

The lesson here is important that one does not simply write equations down without taking notice the convention for each step. For otherwise, one could really be doing a different physics.

#### 6 Responses to “Signs in Einstein’s Equation”

1. Hi prof,
during my fyp I was also confused with the inconsistency of the sign convention of various introductory General Relativity Books (most of which were mentioned by you at above), particularly the differences of definition of the Riemann Tensor by D’inverno’s “Introducing Einstein’s Relativity” and the “Tensor Analysis: A first course”. They do perplexed me a lots in the earlier stage of my learning. I agree with you that the mathematicians tend to use the (- + + + ) sign convention while the physicists are more clung to the ( + – – -) sign convention. Actually I also prefer the (+ – – -) convention since it’s more intuitively plausible as we always think of time as flowing forward (going positive) instead of flowing backward.

Besides, can you explain a little more detail on this equation? Because I have a little confusion with it…
T^a_a= (\rho+p)u^a u_a \pm p \delta^a_a=\pm(\rho-3p)\quad . ——- (1)
I made a deduction from this equation that the contraction of u^a and u_a should result in a coefficient of -1 if the sign convention is ( + – – -) and +1 if using the sign convention (- + + +), is this because the inner product of u is non-zero only for time coordinate (WHY?), so that for the other space coordinate the u is zero? I didn’t deal with stress-energy tensor that much (aside from the derivation of Einstein Equations) during my fyp so I am not very understood regarding the nature of contraction of this tensor. I hope I can get some clarification. Thanks.

2. Oops, forgot to mention, I am Wong.

3. 3 hishamuddinz

Hello Wong. It is a bit confusing and that’s the reason I post this up. On the $u^a$, this should be thought as the unit tangent vector to the particle’s worldline and hence it’s absolute square is $\pm 1$. You got the sign opposite for each signature. The best way to think about this is to go to the particle’s co-moving fame (and the spatial components will be zero).

4. 5 Coleman

How does the electromagnetic stress energy tensor changes when the signature chages from (+,-,-,-) to (-,+,+,+)?

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