Time apparently slows down to a stop at the event horizon (or appears to) of a black hole, from the perspective of an outside observer, right? Which means that any matter which has “fallen” into the black hole would end up “hovering” right on the edge of the event horizon, including any matter which made up the original object before it collapsed. Thus the singularity hasn’t actually formed yet, and may not for many billions of years. Or am I missing something here?
The nature of time and space inside a black hole is such that it’s really difficult to define terms like “not yet”, and the nature of singularities is such that it’s really difficult to define them “existing”. You’re not exactly wrong, but I don’t think any detailed description is possible in words.
Well there would already be matter in the region bounded by the event horizon by the time it formed, but that’s not really the answer to your question.
This where relative nature of time in relativity comes in to play. For Schwarzchild observer hovering far away from the black hole the region bounded by the event horizon always lies infinitely far in their future, i.e. no matter how long they wait nothing that happens in this region can affect them.
In Schwarzchild cooridnates nothing can reach the event horizon in a finite amount of coordinate time. The event horizon is a coordinate singularity in Schwarzchild coordinates.
However you shouldn’t set too much score by coordinate time as it’s entirely dependent on the cooridnates you use. Non-local coordinate systems in general relativity often bring their own sets of problems.
Infact even in special relatvity spatially-extended non-inertial coordinate systems can lead to simalir problems. For example in Rindler coordinates which is a set of cooridnate systems on Minkowski space (i.e. the flat spacetime of special relativity) there occurs a coordinate singularity called the Rindler horizon which is very simlair to the event horizon in Schwarzchild coordinates.
An observer falling in to a black hole the region bounded by the event horizon and the singualrity are at a finite point in the future as they will reach the singularity in a finite proper time.
Even in special relativity there’s no single correct notion of universal simultaneity, so there isn’t going to be a single correct answer to this question. However, it is possible to consistently define a “time” coordinate (i.e., one for which the surfaces of constant time are spacelike) in such a way that as long as you stay outside of the event horizon, the singularity is always on a later timeslice than you are. It’s also possible to define a time coordinate so that the singularity sometimes is on an earlier timeslice than you are (though it obviously can never be in your past).
Yep, this is it. Coordinate systems in relativity are usually defined by a class of observers. In Schwarzchild coordinates the observers are essentially maintain a constant local distance from each other, this is impossible in the region bounded by the Schwarzchild horizon as all observers must be moving towards the singularity. The local proper acceleration of a Schwarzchild observer diverges at the Schwarzchild horizon.
This is the same in Rindler cooridnates as the Rindler observers are trying to maintain a constant distance from each other in all instantaneous co-moving inertial frames. This is impossible beyond the Rindler horizon as the proper acceleration of Rindler observers diverges at the Rindler horizon.
However if you allow at least some (or enoguh) of your observers to fall in to the black hole then it’s possible to map the inside of the event horizon and thus assign events in the region a coordinate time.
But John in the OP did specify that he’s talking about the observer who is out where we are, those not falling into the hole, I would assume at a far distance from the hole so that we’re not significantly affected by its warping.
So given that, is it right that we would observe something which is on its way to falling in, getting closer and closer to the event horizon, but never reaching it? And I guess the light from those objects, which we observe, would be red-shifted? Or would it not, since time has sped up for those objects from our POV, and that would offset the red shift?
IOW, if the person falling in had a clock, then to the distant observer it would appear that his clock was running faster and faster as it got near the horizon, but on the other hand the light from it would be farther and farther red-shifted, so maybe that would offset the time dilation?
I’m throwing these out not because I’m convinced they’re right but because I’d like a knowledgeable person to correct where I may have gone wrong.
I’ve heard this too, but I’ve never understood it. If this is true, then why would the hole be black? Wouldn’t we perceive instead an agglutination of everything that’s ever fallen into the hole?
The problem is though for us to measure time between two events that our worldine is not coincident to (i.e. two events which we are not ‘physically’ at) we need to construct some sort of coordinate system. When the two events are close enough to us this is not difficult, but when the distance becomes far enough that certain effects (depending on our motion and the geometry of spacetime) come in to play
A clock falling in to a black hole would appear to run slower and slower as it moved towards the event horizon asympotically coming to a halt as it reaches the event horizon, for this reason light emitted by the object would appear to be red-shifted beyond the non-relativistic doppler shift. Their motion would slow in such a way that they appear to never reach the event horizon.
In theory yes, we can see anything that has fallen in to the black hole since the event horizon was formed, but practically any light coming from these objects would afetr a time become increasingly red-shifted until it is undetectable.
Yes, it’s correct that you’d never see it cross, and yes, it will get increasingly redshifted, approaching an infinite redshift at the horizon. You could not, however, continue to see the infalling object indefinitely, both because it would be redshifted off the bottom end of any instrument, and because there will be a last photon emitted by the infalling object.
To the original question, there is one absolutely true thing you can say in English (rather than just math): No matter what you do, no matter what the circumstances, you will never have a singularity in your past.
Yes in Schwarzchild spacetime (or variations like Kerr spacetime) this is true.
Though I’d add Gravitational singularities can lie in an observer’s past, the most obvious one would be the intial big bang singularity in big bang cosmology which lies in the past of all observers, at a guess I’d say there’s probably other’s too, thoguh likely of a very hypothetical nature.
An agglutination of spaghettification.
Disclaimer: I am an astrophysicist.
Anything that happens inside the event horizon of a black hole is what we call “causally disconnected” from what’s outside. In other words, that information can’t get out. So there might be a singularity in there and there might not. At this point in the development of physics, we have no way of knowing whether there even is a singularity or some degenerate, pressure-supported stable structure on the interior (although there are some good arguments as to why that’s unlikely).
The idea of not having a black hole in your past led me to this thought experiment.
Let us generate a micro black hole using unobtainium and whatever else is needed. Let it be of such a size that it survives some appreciable time (say, seconds) before it evaporates.
Let it be charged so that it is controllable in an accelerator. Have it whizz around at very high speed.
Have a volunteer in a space suit wander into the path of said micro black hole.
What happens?
Does it continue to circulate, acting like a circular saw, gravitationally abrading microscopic bits off the volunteer as he walks through it’s path?
Does it come to a halt in the body of said volunteer, slowly eating him up gravitationally from the inside?
What if instead of keeping it circulating, you get it up to speed and then aim it for a one-time shot at the volunteer by turning off the control magnets and letting it fly on a tangent down a tunnel?
A black hole that small will have a very small cross-section for eating anything, though it might still do some damage tidally.
And if the next question is “what happens to the singularity when the hole evaporates completely”, the answer is we don’t know. The final stages of black hole evaporation is a problem for quantum gravity.
By the way, ethan, you don’t have to say that you’re an astrophysicist in every post. Stick around for a while, and people will remember.
I disagree with this, the same physics that says the region in the an event horizon is casually disconnected is the same physics says there can’t be a stable pressure-supported structure. i.e. both can be boiled down to behaviour of wordlines of objects contained in the region. Taking one as a given whilst allowing for the possible rejection of the other seems arbitary.
Now it may be that some new physics comes in to play, effectively barring the singualrity and even allowing for pressure-supported stable structures (or maybe even the correct procedure when encountering a singualrity is to remove it by maximally extending the spacetime in some way, though from what I gather not all singularities are given to maximal extension), but in general relativity it is clear that no such structures can exist within the event horizon.
All this talk about event horizons…
In my dumbed-down understanding of all this, there are TWO event horizons. The more important one has a smaller radius, and it is the hole from which noting can escape. But I imagine a second horizon, with a larger radius, being the area at which relativistic effects become very noticeable: Everything is spaghettified, time slows down drastically, yet some red-shifted light can still escape.
If I am anywhere close to accurate, then we can say that if a singularity is defined by the outer shell, then they certainly do already exist, as evidenced by our seeing their red-shifted light. It is only the contents of the inner horizon which might be said to exist only in the future.
Do I have a point, or can someone straighten me out?
That can’t be right. All that spaghetti waving around would make it look like the black hole has hair.
There is only one event horizon. I think you’re probably thinking of the ergosphere of a rotating black hole.
Sphagettification occurs due to tidal forces, but the tidal forces at the event horizon can actually be very small. Somone falling into a very large black hole may not even notice when they cross the event horizon.
I think it all just boils down to the use of ‘already’ in Newtonian physics this word can have a perfectly straight forward meaning,but in relatvity it’s meaning can be very subjective.
Nearly infinitely long hair, nearly infinitely curly. Like a very tight afro.
Tris