Let's say I have a metric in radial coordinates such that at $r \to \infty$ we find flat spacetime:

$$ds^2 \sim -c^2 dt^2 + dr^2 + r^2 d \Omega^2$$

where $ds^2$ is the line element and $t$ is time cordinate and $r$ is the radial coordinate? What constraint does this impose on the stress-energy tensor?

For example

The Schwarzschild metric is given by:

$$ ds^2 = - (1- \frac{r_s}{r})c^2 dt^2 + (1-\frac{r_s}{r})^{-1}dr^2 + r^2 d\Omega^2 $$

Now, for large $r$ we see:

$$ ds^2 \sim -c^2 dt^2 + dr^2 + r^2 d \Omega^2$$

  • $\begingroup$ When you take a limit like $r \to \infty$ the metric should not depend on $dr$ anymore. $\endgroup$
    – JamalS
    Commented Aug 9, 2020 at 11:56
  • $\begingroup$ @JamalS I mean't it was asymptotic to that that. To be more precise: $ \lim_{r \to \infty} \frac{-c^2 dt^2 + dr^2 +r^2 d \Omega^2}{ds^2} = 1$ $\endgroup$ Commented Aug 9, 2020 at 12:01
  • 3
    $\begingroup$ One obvious constraint is that since the Einstein tensor goes to zero, so does the energy-momentum tensor. $\endgroup$
    – Javier
    Commented Aug 9, 2020 at 15:15
  • $\begingroup$ @Javier I agree however I'm unable to show this is the only constraint it implies $\endgroup$ Commented Aug 9, 2020 at 16:37

2 Answers 2


The only constraint is that the stress-energy tensor vanishes at infinity. To see this, try approaching the problem from the other side: Given any arbitrary stress-energy tensor which vanishes at infinity, you can show that there exists a corresponding metric which satisfies your condition.


When deriving the Schwartzschild metric, we make the assumption that the stress energy tensor is zero. Additionally, the field must have a spherical symmetry and be independent of the time.

From the mathematical point of view, it is not necessary to postulate that there is a region where that tensor is different from zero.

But to match the Newtonian gravity of our known experience, it is assumed that there is a mass in the center of symmetry.


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