Looking at Wikipedia articles about solutions for the Einstein Field Equation (fluid solution, dust solution) I hoped to find a table relating simple forms of the stress-energy tensor to the metric tensor.

For example, given the simplest non-zero stress energy tensor I can think of, with $T^{00}(x^0,x^1,x^2,x^3)=\rho,$ $\forall (x^0,x^1,x^2,x^3)\in\mathbf{R}^4$ and $T^{\mu\nu}=0$ for $(\mu,\nu)\neq (0,0)$, how does the metric tensor $g_{\mu\nu}(x^0,x^1,x^2,x^3)$ look like (in some basis)?

  • $\begingroup$ Do you want the energy density $\rho$ to be a constant, independent of $t$ and any spatial coordinates? $\endgroup$
    – G. Smith
    Oct 27, 2019 at 17:40
  • $\begingroup$ @G.Smith wouldn't that just be a pressure-less dust solution? $\endgroup$
    – Kyle Kanos
    Oct 27, 2019 at 19:13
  • $\begingroup$ @KyleKanos It seems so to me. Dusts have energy density but no pressure. But they can have time-varying and, I think, space-varying energy density, which is why I asked the OP about conditions on $\rho$. $\endgroup$
    – G. Smith
    Oct 27, 2019 at 19:34
  • $\begingroup$ @G.Smith: yes, that's what I tried to state with the "everywhere" now replaced with a $\forall$. $\endgroup$
    – Harald
    Oct 29, 2019 at 12:38
  • $\begingroup$ Related. $\endgroup$
    – J.G.
    Jan 21, 2022 at 18:58

2 Answers 2


You are not going to find such a table. There are a several problems with this concept.

(1) You're talking about expressing the stress-energy in terms of its components in a certain coordinate system. But the same spacetime can be described using infinitely many different coordinate systems, none of which is preferred.

(2) This also isn't going to work because we don't normally have a unique solution for a given stress-energy tensor. For example, the simplest stress-energy is the vacuum, and this happens to be one that we can write in a coordinate-independent way, $T=0$. But there are many vacuum solutions, including a flat spacetime, a flat spacetime with a nontrivial topologies, the Schwarzschild spacetime (maximally extended or not), spacetimes containing $n$ black holes, and all kinds of gravitational wave spacetimes.

(3) There are very few closed-form solutions to the field equations, so we wouldn't normally be able to write down the metric.

To get more of a feel for this, you might want to look at Stefani et al., Exact Solutions of Einstein's Field Equations.

  • 1
    $\begingroup$ I would like to argue with the answer in point (2). A solution of Einstein's field equations is unique when it is subjected to the corresponding boundary conditions and if the solution is global and maximal. It means that the metric 'spans' the whole spacetime. $\endgroup$
    – JanG
    Jan 21, 2022 at 18:51

In static spherically symmetric case one has to solve Einstein's field equations (EFE) as shown in https://physics.stackexchange.com/a/679431/281096 . The requirement of zero pressure spacetime results in metric with $g_{00}=g_{rr}^{-1}=1-\kappa c^2~1/r\int_0^r\rho r^2~dr$, where $\kappa=8\pi G/c^4$. For example, in case of constant density $\rho=M/(4/3~\pi R^3)$ the interior metric would yield

\begin{equation} \label{metric} {\rm d}s^2=\Bigg(1-\frac{r_{S}}{R} \frac{r^2}{R^2}\Bigg)^{ }~c^2{\rm d}t^2-\Bigg(1-\frac{r_{S}}{R} \frac{r^2}{R^2}\Bigg)^{-1}{\rm d}r^2-r^2{\rm d}\Omega^2~,\tag{1} \end{equation} with curvature radius $r$ and the infinitesimal surface element ${\rm d}\Omega$ of a $2$-sphere and $r_{S}=2GM/c^2$.

However, the above solution has been generated by imposing of additional constrain ($p(r)=0$) on the corresponding differential equation, whereas the proper way of solving EFE is to apply only two boundary conditions, in our case $p(R)=0$ and $\rho(R)=M/(4/3~\pi R^3)$. Using them one obtains the correct solution known as Schwarzschild interior solution \begin{equation} \label{schwarzschild interior metric} {\rm d}s^2=\Bigg(\frac{3}{2}\sqrt{1-\frac{r_{S}}{R}}-\frac{1}{2}\sqrt{1-\frac{r_{S}}{R} \frac{r^2}{R^2}}\Bigg)^2~c^2{\rm d}t^2-\Bigg(1-\frac{r_{S}}{R} \frac{r^2}{R^2}\Bigg)^{-1}{\rm d}r^2-r^2{\rm d}\Omega^2~.\tag{2} \end{equation} Notably, both solutions have the same $g_{rr}$ but different $g_{00}$ metric component. In my opinion metric \eqref{metric} is not a true solution of EFE.


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