Can One Observe Anything Falling Into a Black Hole?

[Replusz]

TL;DR

Statements 1-3, are they correct?
Questions a and b, please answer : )

1. So for an asymptotically far-away observer, something falling towards a black hole will never reach it
2. However, the thing falling in will reach the event horizon is finite affine parameter
3. The Universe has a finite age for an asymptotically far away observer

a) Does this mean that one can never observe anything falling into a black hole?
b) My lecturer noted that 'one cannot observe a black hole forming', did I hear this correctly? But then why are there black holes?

 

[Ibix]

1 is coordinate dependent. Certainly they will never see an object fall into a black hole, and certainly the usual Schwarzschild coordinates reflect this, but the distant observer may adopt other coordinates in which the object falls in in finite time.

3 is true for everybody, so far as I am aware, not just the distant observer.

Indeed you cannot observe anything falling into a black hole, although if you wait long enough you can (in principle) see it approach arbitrarily close.

A forming black hole is a somewhat different spacetime from a pre-existing one, but it is still the case that you cannot observe anything crossing an event horizon - by definition, since light cannot escape it. So no, you cannot observe a black hole form. However you can observe the mass of the maybe-black-hole from the orbits of test objects, and you can probe with light rays to show that the object is compact within its Schwarzschild radius plus an arbitrarily small distance - small enough that the matter cannot possibly resist collapse.

I suppose there's a philosophical sense in which black holes "don't yet exist" because "now" their mass hasn't crossed the horizon according to the obvious simultaneity criterion for a distant observer. I'd say it's just one of those cases where everyday language isn't really capable of handling the concepts, and not worry about it otherwise.

 

[vanhees71]

2. is also correct. Using the particle's proper time there's nothing special at all when it crosses the Schwarzschild horizon. There's also nothing special at all, i.e., the singularity at the Schwarzschild radius is a coordinate singularity of the usual Schwarzschild coordinates. The only true singularity is the center of the black hole itself (i.e., $r=0$ of the usual Schwarzschild coordinates).

 

[PeterDonis]

@Replusz you should read the series of Insights articles on the Schwarzschild geometry, starting with this one:

https://www.physicsforums.com/insights/schwarzschild-geometry-part-1/

It addresses the issues you raise.