I am attempting to create a Mathematica solver for the Einstein Field Equations (EFE) that will work for SI and Natural Units and any sign convention. It works for Natural Units, but when I try working with SI, I get results that don't agree with the literature. Specifically, for the metric tensor:$$g_{\mu\nu}=\begin{bmatrix}c^2 & 0 & 0 & 0\\0 & -\frac{a[t]^2}{1-k r^2} & 0 & 0\\0 & 0 & -r^2 a[t]^2 & 0\\0 & 0 & 0 & -r^2 a[t]^2 \sin[\theta]^2 \end{bmatrix}$$the solver yields this for the first solution: $$\frac{c^2k}{a[t]^2}+\frac{a'[t]^2}{a[t]^2}=\frac{c^2\Lambda}{3}+\frac{8\pi G}{3}c^4\rho$$As you can see, it has a factor of $c^4$ in the density term that doesn't appear in the rest of the literature. I've traced the problem back to Covariant Stress Energy Momentum tensor for a perfect fluid:$$T_{\mu\nu}=g_{\mu\nu}T^{\mu}_{\nu}$$The time-time component of $T_{\mu\nu}$ is $T_{00}=c^4\rho$. This, in turn, can be traced back to the Trace of the Stress Energy Momentum Tensor:$$T^{\mu}_{\nu}=\begin{bmatrix}c^2\rho & 0 & 0 & 0\\0 & -p & 0 & 0\\0 & 0 & -p & 0\\0 & 0 & 0 & -p \end{bmatrix}$$The Stress Energy Momentum Tensor of a perfect fluid from this article is:$$T^{\mu\nu}=\left(\rho+\frac{p}{c^2}\right)u^\mu u^\nu-p g^{\mu\nu}$$ with $u=\{1,0,0,0\}$:$$T^{\mu\nu}=\begin{bmatrix}\rho & 0 & 0 & 0\\0 & \frac{p-k p r^2}{a[t]^2} & 0 & 0\\0 & 0 & \frac{p}{r^2 a[t]^2} & 0\\0 & 0 & 0 & \frac{p \csc[\theta]^2}{r^2 a[t]^2} \end{bmatrix}$$So the $c^2$ term doesn't appear to come from the Stress-Energy Tensor. The only other option I see is that the term comes directly from the metric, $g_{00}=c^2$. Could someone please tell me where the mistake is? This extra $c^4$ term in the first EFE solution appears to come directly from the metric tensor. What is the correct value for $T_{00}$ in SI units?


2 Answers 2


It is very common to set $c=1$ in G.R. and this makes it hard to interpret the literature. I think it may be helpful to note that the four-velocity of a particle whose rest mass is not zero satisfies $$ u_\mu u^\mu = c^2. $$ This means that if you have $u^a = \{1,0,0,0\}$ then you need a $c^2$ in the metric, and if you have $u^a = \{c,0,0,0\}$ then you must not have a $c^2$ in the metric. If in the first choice zeroth coordinate is $t$ then in the second choice the zeroth coordinate is $ct$.

  • $\begingroup$ Thank you. I've mentioned this ad naseum: natural units are the bane of my existence. Even if you wanted to have some unit where the speed of light is 1, then you should still include a symbol to indicate $m\space s^{-1}$ so the dimensions are correct. In any case, I believe I've properly accounted for this. If not, would you please point out the error. $\endgroup$
    – Quark Soup
    Commented Jan 3, 2022 at 16:02

Okay, so we are using coordinates $(t,r,\theta,\phi)$, where $t$ and $r$ have dimensions of time and length, and $\theta $ and $\phi$ are dimensionless. We see that $g_{\mu\nu} \mathrm dx^\mu \mathrm dx^\nu$ has consistent units of length squared, as required.

The Stress Energy Momentum Tensor of a perfect fluid from this article is $$T^{\mu\nu}=\left(\rho+\frac{p}{c^2}\right)u^\mu u^\nu-p g^{\mu\nu}$$ with $u=\{1,0,0,0\}$.

This is a mistake. From the article you quote,

[...] where $u$ is the 4-velocity of the fluid and $\eta=\mathrm{diag}(1,-1,-1,-1)$ is the metric tensor of Minkowski spacetime.

This indicates that the expression given is derived in coordinates where $x^0 = ct$ rather than just $t$. As a result, the $00$-component of the metric should be $1$, not $c^2$, and $u=(c,0,0,0)$, not $(1,0,0,0)$. From there, we would have

$$T^{\mu\nu}= \begin{bmatrix}\rho c^2 &0&0&0\\0&\frac{p(1-kr^2)}{a^2}&0&0\\0&0&\frac{p}{a^2 r^2}&0\\0&0&0&\frac{p}{a^2r^2\sin^2(\theta)}\end{bmatrix}$$ $$T_\mu^{\ \ \nu}= \begin{bmatrix}\rho c^2 &0&0&0\\0&-p&0&0\\0&0&-p&0\\0&0&0&-p\end{bmatrix}$$ $$T_{\mu\nu}= \begin{bmatrix}\rho c^2 &0&0&0\\0&\frac{pa^2}{1-kr^2}&0&0\\0&0&pa^2r^2&0\\0&0&0&pa^2r^2\sin^2(\theta)\end{bmatrix}$$

On the other side of the Einstein equations, in coordinates $(ct,r,\theta,\phi)$, the Einstein tensor takes the form

$$G_{00} = 3\left(\frac{\dot a^2+kc^2}{a^2c^2}\right)-\Lambda$$ $$G_{rr} = -\left(\frac{2a \ddot a + \dot a^2 + kc^2}{c^2(1-kr^2)}\right)+\frac{\Lambda a^2}{1-kr^2}$$ $$G_{\theta\theta} = -\frac{\left(2a\ddot a + \dot a^2 + kc^2\right)}{c^2} r^2 +\Lambda a^2 r^2$$ $$G_{\phi\phi} = -\frac{\left(2a\ddot a + \dot a^2 + kc^2\right)}{c^2}r^2 \sin^2(\theta)+ \Lambda a^2 r^2\sin^2(\theta)$$

Setting $G_{\mu\nu}= \frac{8\pi G}{c^4} T_{\mu\nu}$, this yields two independent equations: $$3\left(\frac{\dot a^2+kc^2}{a^2c^2}\right)-\Lambda= \frac{8 \pi G}{c^4}(\rho c^2) \iff \frac{\dot a^2}{a^2} + \frac{kc^2}{a^2}= \frac{8\pi G \rho}{3} + \frac{\Lambda c^2}{3} $$ $$-\left(\frac{2a \ddot a + \dot a^2 +k}{c^2(1-kr^2)}\right)+\frac{\Lambda a^2}{1-kr^2}=\frac{8 \pi G}{c^4}\frac{pa^2}{1-kr^2} \iff 2\frac{\ddot a}{a} + \frac{\dot a^2}{a^2} + \frac{kc^2}{a^2} = -\frac{8\pi G}{c^2}p + \Lambda c^2$$

Alternatively, we could use $x^0=t$ as you propose in the beginning, so $g_{00}=c^2$ and $u=(1,0,0,0)$. This amounts to a coordinate change $x=(ct,r,\theta,\phi)\mapsto x'=(t,r,\theta,\phi) \implies \frac{\partial x'^0}{\partial x^0} = 1/c$. Applying the standard tensor transformation rules we would have that $T'^{00}=T^{00}/c^2 = \rho$ and $T'_{00} = T_{00} c^2 = \rho c^4$, which is (if I understand correctly) what you have found.

I've mentioned this ad naseum: natural units are the bane of my existence.

Actually, the problem here is not natural units, but rather choices of coordinate. Whether you use SI or not, you may choose $(ct,r,\theta,\phi)$ or $(t,r,\theta,\phi)$. The most common convention by far is to do the former, so in that sense your choice of the latter will yield differences with most of the established literature.

The key thing to remember is that coordinates - and by extension, the components of tensors - are arbitrary. The value (and even the dimensions) of $T_{00}$ depend on which coordinates you choose, even within the SI system of units. The full expression for the tensor $T$ is $$T = T_{\mu\nu} \mathrm dx^\mu \mathrm dx^\nu$$ and it is this which must have consistent dimensions across all coordinate systems. In SI, it has units of $\mathrm J \mathrm m^{-1} \mathrm s^{-2}$, so the units of $T_{00}$ are $\frac{\mathrm J \mathrm m^{-1} \mathrm s^{-2}}{[x^0]^2}$, where $[a]$ means "units of $a$."

  • $\begingroup$ @GluonSoup I did not use natural units anywhere in my answer. We have a choice as to whether we want the first coordinate to be $t$, which has dimensions of time, or $ct$, which has dimensions of length. Contrary to your assertion, Andrew and I are saying precisely the same thing. If you choose your first coordinate to be $t$, then for a particle at rest $u^\mu=(1,0,0,0)$ and $g_{00}=c^2$ so $g_{\mu\nu}u^\mu u^\nu = c^2$. If you choose your first coordinate to be $ct$, then for a particle at rest $u^\mu=(c,0,0,0)$ and $g_{00}=1$ so $g_{\mu\nu} u^\mu u^\nu=c^2$ once again. $\endgroup$
    – J. Murray
    Commented Jan 3, 2022 at 19:41
  • $\begingroup$ Something is wrong with your derivation. If I follow your $x^0=ct$ formulas through to the EFEs, the first solution is:$$\frac{c^2k}{a[t]^2}+\frac{a'[t]^2}{a[t]^2}=\frac{c^2\Lambda}{3}+\frac{8\pi G}{3}c^2\rho$$The $c^2\rho$ comes from your $T_{00}$ element. Same problem I'm having with $x^0=t$. $\endgroup$
    – Quark Soup
    Commented Jan 3, 2022 at 20:23
  • 1
    $\begingroup$ @GluonSoup Done. The $T$ and $G$ which I explicitly wrote use coordinates $(ct,r,\theta,\phi)$ with $g=\mathrm{diag}\big(1,-a^2/(1-kr^2),-a^2r^2,-a^2r^2\sin^2(\theta)\big)$. If you use coordinates $(t,r,\theta,\phi)$, then $T_{00},G_{00},$ and $g_{00}$ will all be scaled by $c^2$. Also, please note that I noticed you use $(1+kr^2)$ in the $g_{rr}$ term - the standard convention uses the opposite sign for $k$, so I've edited that into my answer. Just something to watch out for. $\endgroup$
    – J. Murray
    Commented Jan 4, 2022 at 16:16
  • 1
    $\begingroup$ @GluonSoup I don't understand - you just said that the metric tensor components are the input to your program. The $(11)$-component of the FLRW metric is usually written $-a^2/(1-kr^2)$, but in your post you said you plugged in $-a^2/(1\color{red}{+}kr^2)$. $\endgroup$
    – J. Murray
    Commented Jan 4, 2022 at 21:08
  • 1
    $\begingroup$ @GluonSoup My apologies, the typo is mine. Some sources use the trace-reversed versions of the Einstein equations and this is what was in my notes, so I carelessly wrote down terms from $R_{ij}$ rather than $G_{ij}$. This is now fixed, and it agrees with the output of your program. $\endgroup$
    – J. Murray
    Commented Jan 5, 2022 at 6:22

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