8
$\begingroup$

Mazur-Mottola gravastars would in many ways appear identical to black holes.

Does the new radio-photograph of M87* taken by the Event Horizon Telescope eliminate M87* as a gravastar candidate, and discredit the gravastar model?

Or would a gravastar still be consistent with the observed data?

Improve this question
$\endgroup$
2
  • $\begingroup$ The Mazur-Mottola model. I'll update the question. $\endgroup$
    – cowlinator
    May 2, 2019 at 19:20
  • $\begingroup$ It is not a photo by any definition. $\endgroup$
    – my2cts
    May 5, 2019 at 14:06

1 Answer 1

Reset to default
16
$\begingroup$

The observations of M87* probe the spacetime geometry in the region around the photon sphere i.e. around a few Schwarzschild radii. A gravastar has the same spacetime geometry in this region, so the EHT observations cannot distinguish between a gravastar and a Kerr or Schwarzschild geometry.

If we want to dive a little deeper into this things start getting complicated. From the perspective of an external observer the event horizon takes an infinite time to form so our perspective M87* is not a fully grown black hole. It is on its way to forming one, but that process will take an infinite time to complete.

So the spacetime geometry is actually something like a rotating Oppenheimer-Snyder metric (though I don't think any analytic solution is known for this geometry) and it is neither a gravastar nor a Kerr geometry. The only way you could resolve the question would be to jump into M87*, but aside from terminating your academic career this also has the problem that you would be unable to tell the rest of us what you found.

Improve this answer
$\endgroup$
10
  • 2
    $\begingroup$ @safesphere No it is not. In the OS metric the horizon takes an infinite time to form for the external observer. $\endgroup$
    – John Rennie
    May 2, 2019 at 6:29
  • 2
    $\begingroup$ @safesphere The EHT photo shows lensing of the light from the accretion disk around M87*, and that is all it shows. It does not show the horizon. The photo is dark in the middle because the lensing cannot deflect the light in such a manner as to make it appear to come from the centre. I guess it shows that M87* itself emits much less light than the accretion disk, but that is not the same as showing there is an event horizon there. However we should note that for any size black hole the appearance quickly becomes indistinguishable from a true horizon even though technically it's not there. $\endgroup$
    – John Rennie
    May 2, 2019 at 6:49
  • 1
    $\begingroup$ @safesphere this is getting complicated and should probably be continued elsewhere, but consider this. We normally consider the observer on the surface of the collapsing ball and ask how long it takes that observer to pass through the horizon in the reference frame of the external observer. But we can consider an observer inside the collapsing ball at some distance $r$ and ask the same question. I think, but would not swear to it on oath, that regardless of the value of $r$ the observer inside the collapsing ball takes an infinite time to cross the horizon regardless of the value of $r$. $\endgroup$
    – John Rennie
    May 2, 2019 at 7:55
  • 2
    $\begingroup$ @my2cts yes, though unless we badly misunderstand the equation of state any mass concentrated inside this radius could not resist further collapse so will inevitably form a black hole (eventually, or in a few milliseconds depending on which observer you believe :-). $\endgroup$
    – John Rennie
    May 2, 2019 at 10:01
  • 8
    $\begingroup$ Silver lining, you're probably only really terminating your academic career in your own reference frame. For external observers, I assume it would appear more like an indefinite hiatus. $\endgroup$
    – JMac
    May 2, 2019 at 14:57

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.