Is there an inside to a Black Hole?

[DrDu]

A particle falling into a black hole appears to us to fall the slower the nearer it gets to the event horizon. When it's de Broglie wavelength becomes comparable to the radius of the event horizon, we can't say for sure whether it has not already fallen into the black hole?

 

[phinds]

If one of two synchronized clocks far away from a mass M is brought close to M, hovers there for some time it will show less elapsed time depending on M (sun versus earth) compared to the far away clock after being brought back.

Which is exactly what I pointed out but it is not correct to say that time passes at different rates, as I also pointed out.

 

[russ_watters]

Energy/matter tends to spiral around a black hole rather than fall directly towards it.

[separate post]

I was wondering whether it ends up spiralling around the black hole rather than plowing straight into it.

This can and does happen yes. Look up "accretion disc".
[emphasis added]

Just to be clear, whether something spirals-in or falls straight in depends on the initial trajectory. There's a lot of spinning in the universe because of matter distributions not being uniform, and any indirect trajectory is going to involve a rotational/orbital component. So it's common but not required.

As long as the disk is closer to the hole than you are, that is the case, yes. However, we are talking about material that falls into the hole. That material will disappear from your view (because light coming from it, well before it actually crosses the horizon, will become so redshifted that you can't detect it any more), but its gravitational influence will still be detectable by you as part of the overall mass of the hole.

I would assume this sort of thing has been modeled/simulated? E.G., as a star spirals into a black hole it could disappear from view, yet we would still detect gravitational waves as it continues to spiral towards the event horizon? And once behind the event horizon are there still gravitational wave emanations as it continues to spiral toward the singularity or at that point does the field settle down and become static again?

 

[timmdeeg]

Yes, but that's not because time passes at different rates for them, it's because they took different paths through space-time. Look up "differential aging".

I know "differential aging" from the SR/twin paradox. Does it include gravitational time dilation too concerning the scenario of post #30?

 

[PeroK]

I know "differential aging" from the SR/twin paradox. Does it include gravitational time dilation too concerning the scenario of post #30?

Yes, differential ageing is just the difference in spacetime length of different timelike paths between two points in spacetime.

 

[phinds]

I know "differential aging" from the SR/twin paradox. Does it include gravitational time dilation too concerning the scenario of post #30?

@PeroK has answered your question but just for clarity let me add that the IS a difference between time dilation due to speed and that due to gravity.

If you and I are moving at very high speed relative to each other, you see my clock as slow and I see your clock and slow. But if I'm deep in a gravity well and you are farther out then you see my clock run slow but I see your clock run fast. In both case, of course, we each see our own clock running at normal speed.

The GPS system has to take both of those into account, else we'd end up running into the sides of buildings and off into corn fields.

 

[stevil]

@PeroK has answered your question but just for clarity let me add that the IS a difference between time dilation due to speed and that due to gravity.

If you and I are moving at very high speed relative to each other, you see my clock as slow and I see your clock and slow. But if I'm deep in a gravity well and you are farther out then you see my clock run slow but I see your clock run fast. In both case, of course, we each see our own clock running at normal speed.

The GPS system has to take both of those into account, else we'd end up running into the side of building and off into corn fields.

From what I have read the twin paradox has nothing to do with acceleration but instead a change in inertial frames.
Being positioned on the surface of large gravity source (experiencing gravity) is equivalent to accelerating.

 

[PeterDonis]

I would assume this sort of thing has been modeled/simulated? E.G., as a star spirals into a black hole it could disappear from view, yet we would still detect gravitational waves as it continues to spiral towards the event horizon?

The gravitational waves would be redshifted (and time delayed) just like light coming from the star as it got closer to the hole's horizon. So by the time the light from the star is redshifted (and time delayed) into unobservability, so would gravitational waves coming from the star.

(This is assuming that the star is much less massive than the hole so the perturbation to the hole's horizon is small. For cases like black hole mergers, where the two holes are of roughly similar mass, the time delay doesn't work the same, since LIGO can detect GWs from black hole mergers.)

 

[PeterDonis]

From what I have read the twin paradox has nothing to do with acceleration but instead a change in inertial frames.

I strongly suggest reading the Usenet Physics FAQ article on the twin paradox:

https://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_paradox.html

It addresses a number of common misconceptions, including the one you state here.

Being positioned on the surface of large gravity source (experiencing gravity) is equivalent to accelerating.

In the sense that you have nonzero proper acceleration (i.e., you feel weight) in both cases, yes. But the equivalence is only local, and the "time dilation" that we are discussing in this thread is not a local phenomenon.

 

[PeroK]

Being positioned on the surface of large gravity source (experiencing gravity) is equivalent to accelerating.

Not to an outside observer!

 

[Nugatory]

Anyone interested in following the mathematical details underpinning the explanations here may want to check out this thread: https://www.physicsforums.com/threads/mathematical-treatment-of-approaching-event-horizon.1007653/

 

[PAllen]

@PeroK has answered your question but just for clarity let me add that the IS a difference between time dilation due to speed and that due to gravity.

If you and I are moving at very high speed relative to each other, you see my clock as slow and I see your clock and slow. But if I'm deep in a gravity well and you are farther out then you see my clock run slow but I see your clock run fast. In both case, of course, we each see our own clock running at normal speed.

The GPS system has to take both of those into account, else we'd end up running into the sides of buildings and off into corn fields.

Just to add that these different cases also arise in special relativity alone, without gravity.

Two inertial observers in relative motion each see and model the other's clock the same way. That is, by relativistic Doppler, both visually see the other as slow, or both see fast. After accounting for signal delay, each concludes the other clock is slow (there is no 'fast' case once signal delay is accounted for).

Then consider an inertial observer at the center of rotating disc (space station, merry go round, whatever) and an observer on the periphery. In this case you have exactly the same situation as described for gravitational time dilation (the central observer both sees and models the peripheral clock as slow while the peripheral observer both sees and models the central clock as fast), with the amount being proportional to the work done moving a body from the periphery to the center. Similarly for fore and aft of an accelerating rocket.

 

[Quester]

The outside observer can only see the infaller actually fall in if they wait around until the BH evaporates. Since this would take and amount of time that would make the current age of the universe effectively zero (a rounding error at about the 50th or 60th decimal place) it's not likely, no.

Do objects actually "fall into" a black hole, or are they "pushed into" a black hole by gravity?

 

[meekerdb]

So from the observer's perspective if they can never see anything ever fall beyond the BH's EH, not even given an almost infinite amount of time, then how are we expected to accept that anything can ever go inside?
Why do we speculate about what is on the inside if it seems impossible for anything to go inside? (Impossible because there isn't enough time for this ever to happen).

Is it possible that there is no inside and that the black hole is just a consequence of all the stuff on the outside?

Almost everything we think we know about a black hole is from solutions of Einstein's equations. We have every reason to believe this well tested theory except in the neighborhood of where it predicts singularities. So the existence of an event horizon and the appearance to a distant observer all come from the theory. And that same theory says there is nothing unusual at the event horizon. It's just a place in in space defined by which way light cones are tilted. An infalling observer wouldn't even notice crossing the event horizon. If you deny that, then you're contradicting the whole theory and you have no reason to even believe there's such a thing as black hole. GR is a consistent mathematical theory. You can't just throw out part of it.

 

[Dale]

Do objects actually "fall into" a black hole, or are they "pushed into" a black hole by gravity?

They fall into, meaning there is 0 proper acceleration. Being pushed would involve proper acceleration.

 

[phinds]

Do objects actually "fall into" a black hole, or are they "pushed into" a black hole by gravity?

Unlike some celebrities, gravity is not pushy.

 

[Tomas Vencl]

It is possible to adopt a convention ……..-for example, to assign "times" to events on the worldline of an object falling into a black hole, according to an observer far away. But any such assignment is, as I have just said, a convention; it has no physical meaning. It's just a way of organizing the data.

I know the Einstein convention of simultaneity, using light signals, which we can use also in curved spacetimes. Can you please show some other example how to meaningfully connect propper times of falling object to times of distant observer (at rest)? Thank you.

 

[PeterDonis]

Can you please show some other example how to meaningfully connect propper times of falling object to times of distant observer (at rest)?

For Schwarzschild spacetime, Painleve coordinates, Eddington-Finkelstein coordinates, and Kruskal-Szekeres coordinates all do this. The first of the three is the one I would start with if you're not familiar with any of them.

 

[stevil]

An infalling observer wouldn't even notice crossing the event horizon.

I accept this.

Even from the infalling observer the event horizon would have no light come at them from the inside. But the EH isn't a physical barrier, the person falling in doesn't realize that time slows down, their own clock appears to them to tick normally.
If they were looking outward though, they would see the universe speed up and hot into heat death before they are able to cross the event horizon. (if they could see clearly outwards, i understand people have said there are red shift issues with being able to see outward).

I'd actually be interested to know, how fast they would see the black hole evaporate. Would it evaporate before they are able to cross the EH? i.e. from the perspective of someone close to the Event horizon (experiencing that time dilated perspective) how quickly will they view the evaporation of the BH?

Anyway, sorry if my questions are stupid.

 

[CelHolo]

Summary:: If time slows down as you approach the Event Horizon does it get so slow that time stops?

Is it possible that there is no inside?

Yes, in some models of quantum gravity Black holes have no interior. This is related to something called black hole complementarity, but I don't think an explanation would be possible for a beginner.

Consider it an open question.

 

[PAllen]

I accept this.

Even from the infalling observer the event horizon would have no light come at them from the inside. But the EH isn't a physical barrier, the person falling in doesn't realize that time slows down, their own clock appears to them to tick normally.
If they were looking outward though, they would see the universe speed up and hot into heat death before they are able to cross the event horizon. (if they could see clearly outwards, i understand people have said there are red shift issues with being able to see outward).
.

This is completely false. They would see the universe outside proceeding at a fairly normal rate, with moderate redshift. They would not see the end of the universe. In fact, if they were viewing a distant clock there is a specific time it would show as they approached the singularity. This time on the distant clock is the time when the event of infaller reaching the singularity is no longer in the (causal) future of the distant clock, and thus the distant observer can legitimately consider that the infaller reaching the singularity is “now” true.

 

[PeterDonis]

I'd actually be interested to know, how fast they would see the black hole evaporate. Would it evaporate before they are able to cross the EH?

No. For a black hole that evaporates, the spacetime geometry is not quite the same as in the non-evaporating case. We have had previous threads on this, though it has been some time since the last one.

Heuristically, you can get a reasonable first approximation to what happens in the evaporating case by thinking of the coordinate time the black hole finally evaporates as playing the role that $t = \infty$ plays in the non-evaporating case--i.e., the coordinate time the black hole finally evaporates is when a distant observer would see an infalling object cross the horizon (in fact the distant observer would see everything that happened on the horizon in the same outgoing light signal that shows the final evaporation of the hole). Here we are assuming Schwarzschild-type coordinates in which coordinate time is the same as proper time for the distant observer. (However, the coordinates aren't quite the same as standard Schwarzschild coordinates; those cannot be used to describe black hole evaporation. I am leaving out a number of technical points that are beyond the scope of a "B" level thread.)

An infalling observer would never see the black hole evaporate; they would fall into the hole, hit the singularity, and be destroyed, and their past light cone would never include any events anywhere close to the final evaporation of the hole.

 

[PeterDonis]

in some models of quantum gravity Black holes have no interior. This is related to something called black hole complementarity

I don't think it's quite correct to say black holes have no interior in the models you refer to. It would be more correct to say that the classical GR concept of a "black hole", or even the semi-classical concept of an "evaporating black hole" that Hawking and others originally investigated, does not even exist in these models.

 

[PeterDonis]

If they were looking outward though, they would see the universe speed up and hot into heat death before they are able to cross the event horizon.

As @PAllen has already said, this is wrong. The confusion you appear to be having is that you are thinking that "time dilation" for an infalling observer works the same as it does for a hovering observer--an observer who has nonzero proper acceleration and is keeping a constant altitude above the horizon. For the latter observer, yes, he sees the outside universe "speeded up" (blueshifted) compared to him. But the infalling observer does not; he sees the outside universe "slowed down" (redshifted) compared to him; his "time dilation" is different from that of the hovering observer.

 

[f todd baker]

I have a question: when a distant black hole eats a nearby star it has to gain mass. But if, from our perspective, we see the captured star stop at the event horizon, how do we see the mass increase? And clearly from gravity waves we see a merger of two black holes but we shouldn't see a merger from our reference frame. Obviously, I am no expert at cosmology!

 

[PeterDonis]

how do we see the mass increase?

We observe the gravitational effects of the increased mass--for example, if we are orbiting the hole, we observe our orbital parameters change. Those effects don't come from inside the hole's horizon; they come from the spacetime outside the horizon, in the past, whose geometry gets changed as the star falls into the hole.

 

[stevil]

The meaning of the term "black hole" is well established (a region of spacetime that is not in the causal past of future null infinity), so using it to refer to a model that obviously does not satisfy that meaning is, to say the least, not a very good idea. The fact that "many papers and talks" do it anyway just means scientists, like the rest of us, don't always do a good job of picking terminology.

I'm probably saying something stupid here.

But I find that definition to be assuming that SpaceTime is everywhere, at the very least it assumes that SpaceTime exists inside the EH therefore if it were found that there was no SpaceTime inside the EH then by that definition, that which we currently call a Blackhole could no longer be called a "blackhole". The EH would still exist, the immense gravity emanating from that region of space would still exist, but we would have to either modify the definition of BH or come up with a new name for it.

 

[stevil]

As @PAllen has already said, this is wrong. The confusion you appear to be having is that you are thinking that "time dilation" for an infalling observer works the same as it does for a hovering observer--an observer who has nonzero proper acceleration and is keeping a constant altitude above the horizon. For the latter observer, yes, he sees the outside universe "speeded up" (blueshifted) compared to him. But the infalling observer does not; he sees the outside universe "slowed down" (redshifted) compared to him; his "time dilation" is different from that of the hovering observer.

Yes, I've finally come to realize this is my issue. I mistakenly thought a free falling observer experiences time dilation. Staticboson has set me straight on this, as well as this post of yours. Thanks.

 

[PeroK]

Yes, I've finally come to realize this is my issue. I mistakenly thought a free falling observer experiences time dilation. Staticboson has set me straight on this, as well as this post of yours. Thanks.

Nothing ever "experiences" time dilation. It's a purely coordinate effect. This is not semantics. What an object experiences is the passing of its own proper time. The object cannot experience how others may measure that time.

 

[PeterDonis]

I find that definition to be assuming that SpaceTime is everywhere

That's not an assumption, it's a definition of spacetime. There is no such thing as a classical spacetime model where spacetime is not everywhere. The idea doesn't even make sense.