"Due to time dilation from general relativity, an object from our universe, for an outside observer, would never fall into a black hole. It would simply move slower and slower (and reflect light more and more dimly) towards the event horizon."

Ronny, this is a common misconception, found even in textbooks. An in-falling object will be observed to accelerate toward and into a black hole, as it would toward any massive object. It's clock-speed will appear to slow toward zero as it crosses the horizon and disappears. Also, the wavelength of any radiation it emits will stretch to infinity at the horizon, which means that it will actually seem to disappear at some distance from the horizon, when frequencies fall below the visible spectrum.

Various factors can cause a difference between what you see when you look at things, and what is really going on. In the case of a black hole, as pointed out above, even though you might never be able to see an object fall into the hole because of time dilation, the falling object reaches the singularity at the center and becomes part of it in very short order. And the gravitational field of the hole does not reflect what we see, but where the material actually is, which is in the center, not at the event horizon.

A similar situation occurs when dealing with objects that are very far away. Objects whose light has taken most of the age of the Universe to reach us are said to be the best part of 14 billion light years away; but that only means it took nearly that long for their light to reach us. However, at the time their light left them they were actually very close to us, and simply going away from us so fast (at nearly the speed of light) because of the expansion of the Universe, that it took till now for their light to move from the region where the speed of light was very close to the rate of expansion to regions close enough for the light to make significant headway in our direction. So if we could see such objects clearly, they would look as if they were very close. But just because they would appear to be close doesn't mean they ARE close. In fact, because of the expansion of the Universe, they are now tens of billions of light years away, and the cumulative expansion of the space between us and them is now several times the speed of light, so any of their light coming in our direction is actually getting further away from us than the object was when the light was emitted, and that light will never reach us. (Hence the term "Observable Universe", referring to what we can see, rather than all of the Universe.)

"you might never be able to see an object fall into the hole because of time dilation"

Courtney, as I've pointed out in my recent question about black holes, that is a fundamental misconception. As with relative uniform motion, where one will observe the clock of a body moving at high speed to be ticking slowly, when an observer sees a body fall into a black hole, its clock moves ever more slowly as it accelerates toward the event horizon.

Bodies are observed to accelerate toward massive objects; they accelerate more toward more massive objects. As they accelerate, their clocks are observed to tick more slowly. Acceleration and time dilation are inversely related.

Dear Ronny,

you are right that to an outside observer, once the black hole has formed, no mass would ever be seen to cross the horizon, exactly as you describe.

The key point is "once the black hole has formed", however. Before that, mass accretion would be seen to take place until the critical mass has accumulated inside the respective Schwarzschild radius.

From a purely practical point of view, however, to an outside observer, the difference between a black hole and an extremely large mass aggregation is difficult to tell apart. Therefore the difference is mostly a theoretical one, albeit a fundamental one.

Also, we must bear in mind that this scenario is all "classical general relativity", i.e. no quantum effects whatsoever are taken into consideration. As soon as semiclassical effects come into play, the whole scenario changes fundamentally. And it is needless to say, a real theory of quantum gravity is completely unknown as of yet.

Best regards

Oliver

Dear Ronny,

James Arnold is partly correct. It depends on the point of view of the observer. Consider two observers, one traveling with the infalling object and the second some distance away from but at rest with respect to the black hole. Thus the infalling object is in motion with respect to the distant observer who will see the abject accelerating as it falls toward the black hole.As the object's velocity approaches relativistic values relative to the distant observer, the distant observer will observe not only that the clocks traveling with the falling object begin to slow as a result of Einstein's time dilation but also that the meter sticks traveling with the moving object shrink such that the instantaneous velocity and, in particular, the time rate of change of the velocity doesn't change. Hence the infalling object's acceleration as seen by the distant observer continues until the object disappears from view beyond the event horizon. All the while the observer on the infalling object doesn't notice anything untoward until tidal forces rent him asunder!

Dear Dwight,

let's ignore for a while that your and James's answer is more in connection to another discussion on black holes we are both involved in (https://www.researchgate.net/post/In_Black_hole_theory_have_mathematicians_finally_lost_all_connection_with_physics) and doesn't really refer to Ronny's question: you are contradicting what you already (correctly) once replied to that very other question.

A distant observer will not *see* an acceleration of an infalling object, and I am wondering how you would define "acceleration" and "velocity" in the first place, as both are locally measured quantities. All he sees is a spherical formation of a black hole, both visually decelerating and optically fading in this process.

The rest of my answer will be posted to the above-mentioned discussion where it belongs.

Best regards

Oliver

Oliver Tennert, thank you for your reply. I agree that my scenario probably only relates to objects who have fallen into a black hole after it has become one. I also agree that it does not take quantum effects into account, I however was very uncertain on to which an extent they would change a subject of e.g.very macroscopic rotation much, but I am sure many know much more than I do on this.

My initial road into this line of thought had to do with time. As time is an aspect of our universe, it does not exist before big bang, and neither, I believe (at least "our time"), beyond the event horizon of a black hole. Thus from an observational point within our universe, we would never see an object fall beyond the event horizon, and this should then apply to other observational data as well - like mass.

As a simple matter of some gravity, this would not change much. The gravity from a sphere and a point at a distance beyond would be identical. But would not a model where parts of the mass is in a hollow sphere around the singularity rotate differently than a model where all tha mass was at the singularity, and would this not be possible to bserve?

Ronny, Oliver would like to ignore the point that actually responds to your question directly. It is a simple misconception: No, a body falling toward a black hole never ever appears to slow in its approach toward and crossing of an event horizon. Its clock slows as it accelerates: Clock slows, motion increases.

Consider the ultimate absurdity of the "slowing" belief: A body that begins to fall toward a large mass (a planet, a star, a black hole) will be observed to accelerate, and to continue accelerating -- at least to some point. For it to actually decelerate and "freeze" (stop) at an event horizon, there must be an elevation in the gravitational field of a black hole where the normal acceleration stops and reverses. What could cause such an odd discontinuity? According to standard Relativity, its clock speed will undergo a smooth deceleration as it falls toward the horizon; clock speed will slow in direct but inverse proportion to its acceleration -- but for some reason, at some point, the relation reverses? I'd like to see an equation that would comprehend such a thing.

It's no different with uniform relative motion, when an astronaut flying by at a speed approaching c will seem to be acting in slow motion: You may measure her relative speed at 90% of c, but her speech will resemble the sound of Hal the Computer as he sings "Daisy" for the last time. Moving fast, behaving slow. Strange as it may seem, even experts in black hole theory confuse the two.

Dear Ronny,

I also agree that it does not take quantum effects into account, I however was very uncertain on to which an extent they would change a subject of e.g.very macroscopic rotation much, but I am sure many know much more than I do on this.

I am afraid not, but all we can consider right now are semiclassical effects: if one takes the existence of Hawking radiation and the gravitational back reaction as a premise (and I am saying *if*, no matter why and how it can be explained and calculated which at the moment we cannot ab initio), then a black hole, having a finite time of existence for an asymptotic observer, cannot possess an eternal horizon. Thus, an infalling observer with a non-radial trajectory would undergo some sort of "scattering experiment" in space-time, where the universe around him has aged by an enormous amount of time, whereis his/her proper time might just have been quite short. A slingshot into the future as it were.

In a sense then, falling into a *classical black hole* is just the limit of this slingshot process with "surrounding universe time = proper time of asymptotic observer" for an asymptotic observer going to infinity.

My initial road into this line of thought had to do with time. As time is an aspect of our universe, it does not exist before big bang, and neither, I believe (at least "our time"), beyond the event horizon of a black hole.

The infalling observer would experience hitting the *timelike* singularity of the (Schwarzschild) black hole in a finite proper time. The real theoretical problem is not the horizon, it is the singularity, because it is there when general relativity breaks down.

Thus from an observational point within our universe, we would never see an object fall beyond the event horizon, and this should then apply to other observational data as well - like mass.

Correct.

As a simple matter of some gravity, this would not change much. The gravity from a sphere and a point at a distance beyond would be identical. But would not a model where parts of the mass is in a hollow sphere around the singularity rotate differently than a model where all tha mass was at the singularity, and would this not be possible to bserve?

I am not sure if I have understood you correctly. Are you asking whether there is a measurable difference in the gravitational field of a (Schwarzschild) black hole and a simple massive sphere when measured from a distance?

Answer: no there isn't. There is only one unique spherically symmetric solution to the Einstein equations. This is the essence of Birkhoff's theorem.

Best regards

Oliver

Oliver Tennert

I am not sure if I have understood you correctly. Are you asking whether there is a measurable difference in the gravitational field of a (Schwarzschild) black hole and a simple massive sphere when measured from a distance?

No, not at all. I completely agree that the gravitational field should be the same. It is the rotation I believe should be different (which it should from a classical view, but where I do not know what relativistic or quantum mechanichal effects might apply) , and I wonder whether this rotation could be observational in some way.

Hi Ronny,

I must admit I do not have the full and exact answer readily available. But I would like to refer you to this review by Matt Visser: https://arxiv.org/pdf/0706.0622.pdf.

He mentions at least in a sidenote that with non-zero rotation, there is a difference between a black-hole solution, and a non-black hole solution of the EInstein equations, and I think he gives some references. Although I must admit I do not see more than 2 parameters available as tuning dials: angular momentum J and mass M. But as I said, I have not thought this through...

Best regards

Oliver