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

In summary: But if the hole is smaller, the tidal forces from the surrounding matter could squish the unlucky astronaut flat.In summary, the outside 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. This could potentially mitigate against the firewall paradox.
  • #1
apocope
2
0
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 won't 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.
 
Physics news on Phys.org
  • #2
apocope said:
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 won't to cross in thought experiments rather unremarkably (perhaps to be incinerated immediately, or not, but at least getting there is unremarkable).

Well, the poor chap does get Spaghettified first...
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.
http://en.wikipedia.org/wiki/Black_hole#Gravitational_collapse
While most of the energy released during gravitational collapse is emitted very quickly, an outside observer does not actually see the end of this process. Even though the collapse takes a finite amount of time from the reference frame of infalling matter, a distant observer sees the infalling material slow and halt just above the event horizon, due to gravitational time dilation. Light from the collapsing material takes longer and longer to reach the observer, with the light emitted just before the event horizon forms delayed an infinite amount of time. Thus the external observer never sees the formation of the event horizon; instead, the collapsing material seems to become dimmer and increasingly red-shifted, eventually fading away.[75]
This help?
 
  • #3
Thanks Enigman,


Enigman said:
Well, the poor chap does get Spaghettified first...

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

Enigman said:

Enigman said:
This help?

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.
 
  • #4
apocope said:
It seems runaway time dilation should occur the moment the seed singularity forms.

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

apocope said:
I assume the singularity starts in the dense core and begins absorbing successive outer layers with the event horizon expanding outward.

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).
 
  • #5
apocope said:
I was under the impression that spaghettification would happen after our unlucky astronaut crossed the event horizon...

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.

apocope said:
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?

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.

apocope said:
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.

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.

apocope said:
As he got closer, the shell would probably get brighter and he would be exposed to more and more of the radiation...

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.

apocope said:
To an outside an observer, the collision of two stellar black holes would never be complete

If the observer says outside the horizon, yes.

apocope said:
and the event horizon would remain covered in a shell.

Not really. See above.

apocope said:
I suppose an exception could be black holes arising from particle collisions...

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.
 

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

1. Why do black holes appear to have no surface while neutron stars do?

Black holes and neutron stars are both incredibly dense objects, but their differences lie in their size and mass. Black holes are formed when a massive star collapses under its own gravity, resulting in a singularity with infinite density and zero size. This means that there is no surface or boundary that can be observed from the outside. In contrast, neutron stars are the remnants of smaller stars that have undergone a supernova explosion. They have a solid, dense surface made up of tightly packed neutrons, which can be observed from the outside.

2. Why do black holes have such strong gravitational pull compared to neutron stars?

The strength of an object's gravitational pull is determined by its mass and size. Black holes have a much larger mass than neutron stars, but their size is practically zero. This results in a gravitational pull that is so strong that not even light can escape, giving black holes their characteristic appearance.

3. Can we see the surface of a black hole from the outside?

No, we cannot see the surface of a black hole from the outside because it does not have a surface. The event horizon, which is the point of no return for anything that enters the black hole, marks the boundary between the observable and unobservable regions. Anything that crosses the event horizon is pulled into the singularity, making it impossible to observe the surface of a black hole from the outside.

4. Why do black holes have such extreme distortions in spacetime?

Black holes have extremely strong gravitational pull due to their mass and size, which causes a distortion in the fabric of spacetime. This distortion becomes more extreme as one gets closer to the singularity, resulting in time dilation and gravitational lensing effects that can be observed from the outside. In contrast, the gravitational pull of a neutron star is not strong enough to cause such extreme distortions.

5. Can black holes and neutron stars coexist in the same system?

Yes, black holes and neutron stars can coexist in the same system. In fact, they are often found in binary systems where one star has collapsed into a black hole and the other has become a neutron star. These systems are known as black hole-neutron star binaries and are potential sources of gravitational waves that can be detected by advanced instruments such as the Laser Interferometer Gravitational-Wave Observatory (LIGO).

Similar threads

  • Special and General Relativity
Replies
11
Views
736
  • Special and General Relativity
2
Replies
39
Views
2K
  • Special and General Relativity
2
Replies
57
Views
1K
  • Special and General Relativity
Replies
6
Views
1K
  • Special and General Relativity
Replies
3
Views
836
  • Special and General Relativity
2
Replies
67
Views
3K
  • Special and General Relativity
Replies
29
Views
398
  • Special and General Relativity
Replies
34
Views
815
  • Special and General Relativity
Replies
3
Views
1K
  • Special and General Relativity
Replies
20
Views
873
Back
Top