0
$\begingroup$

Question

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:

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

$\endgroup$
4
  • $\begingroup$ 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

1
$\begingroup$

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.

$\endgroup$
-2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.