B Physics near the event horizon

Staticboson

Summary
Is there a consensus on whether the laws of relativity can be applied up to the event horizon of a black hole, or does the theory start falling apart some distance above it.
Understanding that I might be pushing the limits into heuristic territory, I'm wondering how much agreement exists on whether the theory holds up in the proximity of an event horizon.

This came up during a recent discussion about matter falling into the black hole and the Schwarzschild solution predicting that due to time dilation time slows down for the falling matter as it approaches the black hole in relation to a far away observer's frame of reference. The formula ideally predicts that time slows down to a stop at the event horizon (again, in relation to a far away observer's frame of reference).

The meat of the discussion was about collateral implications: During the matter collapse leading to the formation of a black hole, once the mass density is such that the radius of the body becomes equal to the Schwarzschild radius, to an outside observer any further development of the black hole stops.

This would also also means that outside observers cannot witness any mass crossing the event horizon, because to the observer, the event will not take place until time approaches infinity.
So for all intents and purposes, from our perspective, we cannot state that any matter is actually crossing any event horizon of any black hole anywhere in the universe, only approaching it.
The actual crossing of the event horizon, of any matter falling into any black hole in the universe, as long as the Schwarzschild solution holds up, would occur at the end of times.

And that is the reason for the question, is there a consensus on whether the present theory holds up very near to the event horizon.

Thank you for your time... and patience.

Related Special and General Relativity News on Phys.org

.Scott

Homework Helper
The presumption is that Special and General Relativity (GR) apply in all situations.
Much of what we know about Black Holes is based on models that are based on GR.

It's pretty much the definition of an event horizon that you cannot track something through it without approaching it yourself.

To an outsider, the "actual" crossing would not happen at the "end of time", it doesn't happen at all.

Staticboson

Understood, my "end of time" reference was facetious, to mean it doesn't happen.

There some pop talk about matter falling into black holes, I was looking for reaffirmation that it doesn't happen at all.

Taking it another step, during the matter collapse leading to the formation of a black hole, once the mass density is such that the radius of the body becomes equal to the Schwarzschild radius, to an outside observer any further development of the black hole stops. The singularity defined as an infinitely small point of infinite density does not exist, the diameter of the black hole mass remains equal to the Schwarzschild radius until the... "end of times".

Thank you for responding

Dale

Mentor
Understanding that I might be pushing the limits into heuristic territory, I'm wondering how much agreement exists on whether the theory holds up in the proximity of an event horizon.
General relativity is expected to hold up at and beyond the event horizon. It is only expected to fail very near the singularity itself. Exactly how near is currently unknown until a solid theory of quantum gravity is developed, but we would expect it to be somewhere around the Planck scale.

All the things that you mention are predictions by the theory made under the assumption that the theory holds up at and well beyond the horizon.

Staticboson

General relativity is expected to hold up at and beyond the event horizon. It is only expected to fail very near the singularity itself. Exactly how near is currently unknown until a solid theory of quantum gravity is developed, but we would expect it to be somewhere around the Planck scale.

All the things that you mention are predictions by the theory made under the assumption that the theory holds up at and well beyond the horizon.
Thank you!

Mike

.Scott

Homework Helper
There some pop talk about matter falling into black holes, I was looking for reaffirmation that it doesn't happen at all.
A Black Hole is generally taken to be everything on the far side of the event horizon.
It is not inaccurate to say that something falls into a Black Hole. The event horizon is not a material object.
With careful navigation, someone crossing the event horizon could determine at what moment the crossing would occur. But when that moment happened, they would see no abrupt changes.

For a very large black hole, the visitor might last some time before becoming "spaghettiized". TON 618 has a Schwarzschild radius of 1300 AU. That works out to 1 light-week. Although I don't know how to convert a black hole radius into its interior transit time.

pervect

Staff Emeritus
Summary: Is there a consensus on whether the laws of relativity can be applied up to the event horizon of a black hole, or does the theory start falling apart some distance above it.

This would also also means that outside observers cannot witness any mass crossing the event horizon, because to the observer, the event will not take place until time approaches infinity.
So for all intents and purposes, from our perspective, we cannot state that any matter is actually crossing any event horizon of any black hole anywhere in the universe, only approaching it.
So far, so good.
The actual crossing of the event horizon, of any matter falling into any black hole in the universe, as long as the Schwarzschild solution holds up, would occur at the end of times.
This is a rather bad way of looking at things, because, for an infalling observer, the crossing does not occur at "the end of time" at all. It occurs in a finite amount of time acccording to a clock carried by an infalling observer. . This is a well-accepted result for the Schwarzschild geometry.

It may be difficult to appreciate that the time is infinite for one observer, and finite for another. But that's what the classical theory predicts.

This isn't to say that there aren't some non-classical peer-reviewed theories that explore the idea that there could be something strange going on at the event horizon - the "firewall" theories, for instance, propose that this could happen.

But the motivations for these non-classical theories are different than the "end of time" argument you propose. Understanding the relativity of simultaneity is key to understanding how the time can be infinite for the outside observer, but finite for the infalling observer.

Staticboson

So far, so good.

This is a rather bad way of looking at things, because, for an infalling observer, the crossing does not occur at "the end of time" at all. It occurs in a finite amount of time acccording to a clock carried by an infalling observer. . This is a well-accepted result for the Schwarzschild geometry.

It may be difficult to appreciate that the time is infinite for one observer, and finite for another. But that's what the classical theory predicts.

This isn't to say that there aren't some non-classical peer-reviewed theories that explore the idea that there could be something strange going on at the event horizon - the "firewall" theories, for instance, propose that this could happen.

But the motivations for these non-classical theories are different than the "end of time" argument you propose. Understanding the relativity of simultaneity is key to understanding how the time can be infinite for the outside observer, but finite for the infalling observer.

Completely understanding that from the perspective of the falling observer time passes normally and there will be no apparent event horizon "boundary" crossing, and he will find himself reaching the singularity in what I would assume would be a fairly quick process.

However note every aspect of my argument is from the perspective of the far away observer, and for that matter, the perspective of the universe outside of the event horizon.

Thank you for the input!

Staticboson

A Black Hole is generally taken to be everything on the far side of the event horizon.
It is not inaccurate to say that something falls into a Black Hole. The event horizon is not a material object.
With careful navigation, someone crossing the event horizon could determine at what moment the crossing would occur. But when that moment happened, they would see no abrupt changes.

For a very large black hole, the visitor might last some time before becoming "spaghettiized". TON 618 has a Schwarzschild radius of 1300 AU. That works out to 1 light-week. Although I don't know how to convert a black hole radius into its interior transit time.
Understood, note that you're taking the perspective of the falling object, my argument is from the perspective of the outside observer only.

Thanks!

Ibix

Although I don't know how to convert a black ho
The maximum survival time is 15$\mu$s per solar mass (divide your Schwarzschild radius by 3km to get the mass in solar masses) after you cross the event horizon.

Caveats abound: paths with shorter survival times are possible, and unless you are a point particle tidal forces will kill you early. And I'm assuming that GR holds all the way to the singularity, which probably isn't true.

Ibix

Understood, note that you're taking the perspective of the falling object, my argument is from the perspective of the outside observer only.

Thanks!
The problem is that if you use Schwarzschild coordinates (which I suspect is what you mean by the perspective of an outside observer), those coordinates do not cover the event horizon. Like asking where a Mars rover is but insisting the answer be in terms of UK grid references, you can't get a meaningful answer because you are choosing to use a coordinate system that can't describe what you want to study.

Nothing stops a distant observer adopting something like Gullstrand-Painleve coordinates that do cross the horizon. Then they have no problem describing things crossing the horizon in finite time.

Staticboson

The problem is that if you use Schwarzschild coordinates (which I suspect is what you mean by the perspective of an outside observer), those coordinates do not cover the event horizon. Like asking where a Mars rover is but insisting the answer be in terms of UK grid references, you can't get a meaningful answer because you are choosing to use a coordinate system that can't describe what you want to study.

Nothing stops a distant observer adopting something like Gullstrand-Painleve coordinates that do cross the horizon. Then they have no problem describing things crossing the horizon in finite time.

By an outside observer I mean far away from the black hole and nowhere near the event horizon. Looking into Gullstrand-Painleve coordinates to see if I'm missing something.

Thanks

.Scott

Homework Helper
The maximum survival time is 15$\mu$s per solar mass (divide your Schwarzschild radius by 3km to get the mass in solar masses) after you cross the event horizon.

Caveats abound: paths with shorter survival times are possible, and unless you are a point particle tidal forces will kill you early. And I'm assuming that GR holds all the way to the singularity, which probably isn't true.
So for TON 618 at 66 billion solar masses, you would have 990,000 seconds - or 275 hours - or about 11.5 days.
Of course, TON 618 is probably spinning - so it could be anything.

PeterDonis

Mentor
The formula ideally predicts that time slows down to a stop at the event horizon (again, in relation to a far away observer's frame of reference).
No, it doesn't. When properly interpreted, it says that radially outgoing light rays from the infalling object take longer and longer to reach the faraway observer as the infalling object gets closer and closer to the horizon, and a radially outgoing light ray emitted by the infalling object exactly at the horizon never reaches the faraway observer at all.

for all intents and purposes, from our perspective, we cannot state that any matter is actually crossing any event horizon of any black hole anywhere in the universe, only approaching it.
You can say that a faraway observer will never see any matter crossing the horizon. But unless you are willing to take the position that anything the faraway observer will never see cannot exist, you cannot make the stronger statement you are making here.

The actual crossing of the event horizon, of any matter falling into any black hole in the universe, as long as the Schwarzschild solution holds up, would occur at the end of times.
This is not correct. You might want to read my series of Insights articles on the Schwarzschild geometry, which discuss and clarify a number of misconceptions such as the ones you are stating:

The link is to the first article in a 4-part series.

To an outsider, the "actual" crossing would not happen at the "end of time", it doesn't happen at all.
This is not correct. See my response to @Staticboson above. Things either happen or they don't; there is no such thing as an event happening according to some observers but not others. Events are part of the spacetime geometry, and the spacetime geometry is what it is, even though in some geometries, such as this one, not all observers can receive light signals from all events.

during the matter collapse leading to the formation of a black hole, once the mass density is such that the radius of the body becomes equal to the Schwarzschild radius, to an outside observer any further development of the black hole stops.
None of this is correct.

The singularity defined as an infinitely small point of infinite density does not exist
That is the wrong definition of the singularity in Schwarzschild spacetime. In Schwarzschild spacetime, the singularity itself (the locus $r = 0$) is not even part of spacetime at all. But considered as a limit or boundary of the spacetime, it is a spacelike line, i.e., it's a moment of time, which is to the future of all other times inside the horizon, but is not visible or accessible from outside the horizon.

It is not inaccurate to say that something falls into a Black Hole.
I have no idea what you are trying to say here. It is perfectly possible for objects to fall through the horizon and reach the black hole region of spacetime inside.

I don't know how to convert a black hole radius into its interior transit time.
If you're just looking for a rule of thumb, divide the radius by $c$. Or in geometric units where $G = c = 1$, they're the same.

It may be difficult to appreciate that the time is infinite for one observer, and finite for another. But that's what the classical theory predicts.
Not quite. What classical GR says is that, in the faraway observer's coordinates (i.e., the Schwarzschild coordinate patch on the region exterior to the horizon), there is no well-defined "time" assigned at all to events on or inside the horizon; those coordinates are only valid outside the horizon. The common statement that "time is infinite" at the horizon in these coordinates is a sloppy and inaccurate way of saying it, and in this thread I think it's important not to do that since we are trying to clear up a number of common misconceptions.

every aspect of my argument is from the perspective of the far away observer, and for that matter, the perspective of the universe outside of the event horizon
But you are not making claims that are limited to that perspective. You are claiming that matter crossing the horizon doesn't happen. That's wrong. It does. The faraway observer just can't see it happen.

By an outside observer I mean far away from the black hole and nowhere near the event horizon.
Yes, and that means if he is going to make any statements at all about what happens near or at the horizon, he has only two ways of doing it:

(1) Draw inferences from his direct observations, like seeing light signals emitted by objects falling towards the black hole. But those inferences will be limited by the light signals he can receive, and he cannot draw the inferences that events he can't see light signals from don't happen.

(2) Adopt some system of coordinates that allows him to calculate what happens at locations distant from him. What calculations he can make, and what he can infer from them, will depend on what coordinates he adopts and what region of spacetime they are valid in. If he adopts coordinates that are only valid outside the horizon, he can only use them to calculate what happens outside the horizon, and he cannot infer that things he cannot calculate using those coordinates don't happen.

Dale

Mentor
However note every aspect of my argument is from the perspective of the far away observer, and for that matter, the perspective of the universe outside of the event horizon
Relativity covers more than just the perspective of far away observers. So if you want to talk about where the laws of relativity are valid you cannot limit your argument that way.

Staticboson

@PeterDonis,
Thank you for the input and I will definitely spend the time to read your Insight articles on the subject.

I just want to clarify that it was not my intent to suggest that the object will not fall past the event horizon because the outside observer doesn't see it. What I meant to say/ask is that in spite of the fact that the object falls in, the event is not accessible to the observer because of the time dilation experienced by the falling object in relation to the observer.

@Dale,
My intent was to limit the perspective, not the argument, my bad for the poor choice of words. Basically just asking the question, if I'm looking at a black hole can I see stuff falling in and disappearing past the event horizon, or does it appear to freeze at the boundary.

Thank you again for your patience.

pervect

Staff Emeritus
However note every aspect of my argument is from the perspective of the far away observer, and for that matter, the perspective of the universe outside of the event horizon.

Thank you for the input!
Probably the underlying issue is that you think the two perspectives are unrelated. But they're not unrelated. A crude way of saying this is that we can't get qualitatively different answers to physical problems just by changing our perspective. We can write physics in a manner that's independent of perspective, via the techniques of tensor algebra.

For example, we can work physics problems on the plane in cartesian coordinates (x,y), or polar coordinates (r, theta). And because we know how to convert the answers from one set of coordinates to the otherm we can work the problem in either set of coordinates.

The more formal way of saying this is that physics is covariant.

There is one modest requirement for our general statement that "we can use any coordinates we like and then know that we have the general answer for any viewpoint because we know how to convert from one set of coordiantes to another set". That requirement is that the coordinate systems be well behaved.

It's possible to come up with a set of coordinates that is not well behaved. For instance, polar coordinates are not 'well behaved" near the origin, where r=0. So we typically wind up with mathematical difficulties in analyzing the behavior of something in polar coordinates at the origin, though it's fine everywhere else. Such problems are not due to the physics being bad at the origin, the origin is a human choice that lacks physical significance. The problem is the behavior of polar coordinates at the origin. This well understood, we say that the polar coordinates are singular at the origin.

When we look at black hole, we can say that Schwarzschild coordinates are singular near the event horizon.

Your argument basically boils down to "I want to adopt a point of view that uses Schwarzschild coordinates", because those are the sort of coordinates that naturally describe the viewpoint of the observer at infinity that you favor.

You are noting, correctly, that there are some difficulties at the event horizon when you do this. Where you're going wrong is assuming that the difficulty lies with the physics. There's no problem with the physics, the problem is with the coordinate choice. This can be demonstrated by the existence of coordinate systems in which the problem is well behaved and in which the mathematical solution is finite everywhere.

But you're apparently not willing to adopt any other coordinate system that is well behaved, because you are mentally wedded to one particular point of view that happens to be associated with an ill-behaved coordinate system.

Basically, you want to blame physics for the issue, but the issue isn't physical at all. It's all related to the singular behavior of the coordinate system that you want to use.

In particular, when you use any of the well-behaved sets of coordinates (such as Kruskal-Szerrkes coordinates), <<link>> you'll find that the time coordinates associated with particles crossing the event horizon are finite, not infinite.

In well-behaved coordinates, the issue of "infinite future" does not appear. There are bunch of other interesting thigns that do happen, but it would be too unfocussed to go into them.

Dale

Mentor
My intent was to limit the perspective, not the argument,
You can’t limit the perspective without limiting the argument. If you want to avoid limiting the argument then the argument needs to be based on invariants.

Ibix

Basically just asking the question, if I'm looking at a black hole can I see stuff falling in and disappearing past the event horizon, or does it appear to freeze at the boundary.
If you mean literally look, you will see it slowing down, redshifting, and dimming until it's undetectable. You will never see it freeze. Infallers never freeze, they only fade away.

You can only see something crossing the horizon by following it in, because light from the thing cannot escape the horizon once it's crossed.

PAllen

Let me add one more observation, that I didn't see in a quick scan of this thread.

Imagine you are maintaining a constant 'distance' from a BH by maintaining the appropriate proper acceleration (which approaches zero the farther you choose). You drop a test body into the BH. There is a specific time after you drop it when any further signal you send it will only be received inside the event horizon. As of this time, all of the test body's history of being at or outside the event horizon is no longer in your causal future. Any time after this, it is physically plausible to say the event horizon crossing has occurred for you because the crossing event has become spacelike separated from you, that is 'possibly now'. It is purely conventional (effectively, a coordinate choice) when you say this happens. However, invariant is that you can no longer say the horizon crossing is in your (causal) future - because this is simply wrong.

Staticboson

Let me add one more observation, that I didn't see in a quick scan of this thread.

Imagine you are maintaining a constant 'distance' from a BH by maintaining the appropriate proper acceleration (which approaches zero the farther you choose). You drop a test body into the BH. There is a specific time after you drop it when any further signal you send it will only be received inside the event horizon. As of this time, all of the test body's history of being at or outside the event horizon is no longer in your causal future. Any time after this, it is physically plausible to say the event horizon crossing has occurred for you because the crossing event has become spacelike separated from you, that is 'possibly now'. It is purely conventional (effectively, a coordinate choice) when you say this happens. However, invariant is that you can no longer say the horizon crossing is in your (causal) future - because this is simply wrong.

Before going any further, please let me say that I have carefully read the responses and realize, much more than before, that the technical aspects of this discussion are well over my head in every way, and I am humbled that you all have taken the time to explain. I promise it is not in vain and my intention is to follow the links and try my best.

But bear with me once more, and I will try to work with PAllen's example above, where I remain at a constant arbitrary distance and drop a test body into the BH. I drop the body, start my stopwatch and observe the signal from the test body, my intent being to stop the watch when I no longer receive the signal.

Question 1: Is there a method to predict/calculate the length of time my stopwatch will run for, from the moment I drop the probe, until the time the signal ends.

Question 2: Although I understand the signal I'm receiving will not end abruptly but instead redshift asymptotically towards wavelength = infinity as the probe approaches the event horizon, does the end of the signal (wave frequency = 0) coincide with the probe reaching the event horizon.

Again, thank you.

.Scott

Homework Helper
To an outsider, the "actual" crossing would not happen at the "end of time", it doesn't happen at all.
This is not correct. See my response to @Staticboson above. Things either happen or they don't; there is no such thing as an event happening according to some observers but not others. Events are part of the spacetime geometry, and the spacetime geometry is what it is, even though in some geometries, such as this one, not all observers can receive light signals from all events.
I'm not sure that someone outside the black hole would qualify as an "observer".
This is not just a matter of not receiving the light signals. If you watch a clock descend towards the event horizon, you will notice two important effects:
1) It becomes more and more difficult to read the clock. The flow rate of information from the clock to the outside approaches zero asymptotically to its proximity to the event horizon.
2) The clock will slow down as it approaches the time when the actual crossing occurs.

So if you know that the clock will read exactly 12:00:00 when it crosses the horizon, how long will it take before you see it reach 12:00:00? Answer: Don't bother waiting. That was what I meant by "never" rather than the "end of time".
Of course the clock does cross the horizon at a time that can be calculated in advance. But it will happen without being observed.

For the outside observer, what happens is that the clock becomes encoded as part of the black hole's "hair" - presumably effecting its eventual decay products.

.Scott

Homework Helper
Question 1: Is there a method to predict/calculate the length of time my stopwatch will run for, from the moment I drop the probe, until the time the signal ends.
From Hawkings Explains:
Although you wouldn't notice anything particular as you fell into a black hole, someone watching you from a distance, would never see you cross the event horizon. Instead, you would appear to slow down, and hover just outside. You would get dimmer and dimmer, and redder and redder, until you were effectively lost from sight. As far as the outside world is concerned, you would be lost for ever.
Question 2: Although I understand the signal I'm receiving will not end abruptly but instead redshift asymptotically towards wavelength = infinity as the probe approaches the event horizon, does the end of the signal (wave frequency = 0) coincide with the probe reaching the event horizon.
It would, if that observation could ever be made. Which it wouldn't.
And I believe there is more to it than red-shifting, but I will put that in a separate post so that it can be separately criticized.

pervect

Staff Emeritus
Question 1: Is there a method to predict/calculate the length of time my stopwatch will run for, from the moment I drop the probe, until the time the signal ends.

Question 2: Although I understand the signal I'm receiving will not end abruptly but instead redshift asymptotically towards wavelength = infinity as the probe approaches the event horizon, does the end of the signal (wave frequency = 0) coincide with the probe reaching the event horizon.

Again, thank you.
I'm not sure what you are asking in point 1, since in point 2 you argued (correctly), that the signal will not end. So it's unclear what you might mean by "the time until the signal ends" when the signal doesn't end.

Let's continue, though. Suppose the you are dropping a bright cesium clock source. Then you know that every 9192631770 complete cycles corresponds to 1 second.

Ideally, every cycle of the clock emitted before it reaches the event horizon will eventually be received by the hovering observer, but the time it takes to reach the hovering observer will grow without bound as the clock gets closer and closer to the event horizon.

You can say that the observer falling with the clock will observe "N" complete cycles of signals from the time the clock is released until the clock crosses the event horizon, where N is some large number.

While in the ideal case, every one of those N cycles eventually reaches the hovering observer, there is no guarantee that the hovering observer has a reciever sensitive enough to detect every one of the signals.

If you are happy with imagining an idealized receiver that can receive every emitted clock cycle, no matter how badly red shifted, then the hovering observer can get the same number N that the infalling observer measures by counting the number of clock cycles that are ever received in an infinite observation period. More formally, we might insist that they take the limit of the number of received cycles as the time approaches infinity.

Of course, if the receiver isn't sensitive enough, some could be missed in an actual measurement.

The number N can be calculated easily enough It's the proper time on the worldline of the infalling observer multipied by the cesium clock constant.

I've worked it out in the past, though I've computed the proper time rather than the number N. But I don't have a handy link to the formula. There's definitely a formula for it, though.

.Scott

Homework Helper
As I said in my last post, I believe there is another fundamental problem with the notions of observing something as it approaches the event horizon - which recently struck me in another thread on these forums:

When something crosses the photon sphere, it has at least a slightly downward trajectory and thus it is in an "orbit" that would eventually take it to the event horizon. But it can still save itself - by thrusting something towards the event horizon to propel itself upwards. Of course, that would involve reaction mass that would be propelled downward into a more dire trajectory. So there is a certain "conservation of loss" which is affected by an objects progress from the photon sphere to the event horizon and by its trajectory.

So the problem in observing an object near the event horizon is not just that there is red-shifting. It is that there is a limit to how much power the object can send to the outside - a form of information throttling. Ultimately, the cost of sending a single bit will be enough to force the object into the horizon.

It should be possible to calculate this information decay from the trajectory of the falling object, it's mass, and the black hole characteristics. Given that, it should be possible to calculate the time when there will be at most only a single bit of information left to be transmitted to the outside world.

So, is it really fair to say that an outside observer will see an object hover just above the event horizon?

This also has an effect on another description that was popular several years ago. Since an object approaching the event horizon is seen to progress in time very slowly, the argument was that Heisenberg Uncertainty would oblige it to appear spread out over ever-widening distances. But since measurements of such an object becomes increasingly difficult, it would seem that HUP can be enforced with this information throttle rather than with uncertainty over its position.

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