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Why don't black holes look like neutron stars to outside observer?

  1. Jan 28, 2014 #1
    Yesterday, I read about Hawking's new proposal regarding the firewall paradox.

    A more general thought about standard black holes occurred to me. Black holes including stellar black holes are of course always presented as if the event horizon is an invisible barrier, which the unfortunate astronaut is wont to cross in thought experiments rather unremarkably (perhaps to be incinerated immediately, or not, but at least getting there is unremarkable). In the thought experiments of course the outside of observer never sees the astronaut actually get to the event horizon because of gravitational time dilation. It is unclear to me why the same would not hold for the original collapsing mass itself. It seems runaway time dilation should occur the moment the seed singularity forms.

    I assume the singularity starts in the dense core and begins absorbing successive outer layers with the event horizon expanding outward. Why would an outside observer ever be able to see the end result and not a degenerate star just starting to implode?

    I'm not a specialist, though I do have a BS in physics, so perhaps I am missing something.
     
  2. jcsd
  3. Jan 28, 2014 #2
    Well, the poor chap does get Spaghettified first...
    http://en.wikipedia.org/wiki/Black_hole#Gravitational_collapse
    This help?
     
  4. Jan 28, 2014 #3
    Thanks Enigman,


    I was under the impression that spaghettification would happen after our unlucky astronaut crossed the event horizon...

    Yes. So it does appear that, to an outside observer, at least a stellar black hole would appear as a solid mass, with a layer of the degenerate star just above the event horizon. Is that correct? If so, he could never reach the event horizon because he would be in a reference frame outside that of the layer appearing to remain. As he got closer, the shell would probably get brighter and he would be exposed to more and more of the radiation... Is this a more realistic version of the unlucky astronaut?

    Of course, this would apply to the case of stellar collapse, but it seems the same would be true of almost any scenario leading to a singularity. To an outside an observer, the collision of two stellar black holes would never be complete, and the event horizon would remain covered in a shell. I suppose an exception could be black holes arising from particle collisions...

    It strikes me that this brick wall around at least stellar black holes could mitigate against the firewall paradox, but of course I suppose that would have already been noticed, but it certainly helps me understand what a black hole might look like from the outside.
     
  5. Jan 28, 2014 #4

    PeterDonis

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    There is no "seed singularity"; the singularity is not the same as the event horizon. See below.

    No; the singularity doesn't form until the entire collapsing star has reached zero radius and infinite density. When the event horizon forms at the center of the star, the density there is still finite. The horizon reaches the surface of the star just as that surface is falling inward past the star's Schwarzschild radius (2GM/c^2).
     
  6. Jan 28, 2014 #5

    PeterDonis

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    It depends on the size of the hole. For a stellar-mass black hole, tidal gravity at the horizon is strong enough that any normal material would be torn apart by the time it reached there. For a supermassive black hole (like the one at the center of our galaxy), tidal gravity at the horizon is small enough that an astronaut wouldn't get spaghettified until he was well inside the horizon.

    No. To an outside observer that remains "hovering" at the same altitude, high above the star, the star will seem to collapse more and more slowly as it gets closer and closer to its Schwarzschild radius; it will never appear static. Its material will still appear as ordinary star material (but collapsing, of course).

    To an observer who falls inward towards the horizon after it forms, things are different. See below.

    No; it's perfectly possible for an outside observer to fall through the horizon after the star has done so. If he does, then when he crosses the horizon, he will see light from the star when it crossed the horizon. As he gets further and further inside the horizon, he will see light from further (meaning, further inside the horizon) portions of the star's collapse, up to a point; at some point, he himself will hit the singularity, before light from the final portion of the star's collapse to the singularity reaches him.

    Not classically, no. Classically, there is nothing special about the horizon. The "firewall" controversy is about how quantum effects change the classical picture; but whatever those changes are, they don't amount to a classical "shell" around the horizon, since there isn't one, as above.

    If the observer says outside the horizon, yes.

    Not really. See above.

    According to our best current understanding, there can't be any; the smallest mass a black hole can have is the Planck mass, which is much larger than any elementary particle mass.
     
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