Crossing the event horizon of a black hole

[coelho]

Well... from a philosophycal point of view, to deny the existence of the inside of the EH for a physicist is the same that to deny the existence of a Heaven for a religion person.

The religious says that there is a Heaven some people goes after dying. We who are alive never see it. We see the dead corpse rotting, nothing else. And nobody ever returned from this so-called Heaven to tell us it actually exists. They say only when we die we will know it.

The physicists says that there is spacetime, and a singularity inside the EH. But we, outside observers, can't observe it, nor it can influence the outside of the EH, so we have not any means to know what is like inside there. If we throw something into a black hole, we only see it going nearer and nearer the EH, but never crossing it. For one were able to see how is inside the EH, one must fall into it.

The likeness is striking. And, while i don't say i disagree with all the physical "knowledge" about GR, i must admit that this knowledge is no more philosophically "solid" than the belief in the existence of Heaven is for the religious people.

Science must make falsifiable predictions. But if we can never know what happens inside the EH, how can we test any predictions concerning there? The argument that the spacetime must follow the same laws of the physics anywhere will be only an (probably methaphysical) assumption if we were not able to test it or its effects. And as we can't know for sure if it is so, we can't assume it is.

 

[DaveC426913]

The religious says that there is a Heaven

The physicists says that there is spacetime

The likeness is striking.

Well, except for the fact that we can create mathematically-consistent models from what we do know. The same can't be said in both cases.

 

[JesseM]

The thing is that the singularity doesn't occur at a point in space-time. GR doesn't demand that the inside of a black hole have the topology of a ball -- instead, a Schwartzchild black hole has the topology of the 3-dimensional analog to an (open) annulus. The r=0 points are inside the hole, rather than being parts of space-time.

(If we interpret Schwartzchild coordinates as if spacelike slices gave spherical coordinates on 3-space, the hole consists of a single point)


Yes, if we so desired, we can remove this interesting topological feature by filling it with a point and relaxing the condition that the metric be everywhere defined. But I don't believe that to be a good idea, since it gives up your ability to prove things by reasoning topologically. (And, of course, it is a bad idea if you're someone who absolutely insists that the metric be everywhere defined and differentiable)

So when you say singularities don't have to exist in GR, do you just mean that we can remove the exact point of the singularity from spacetime, while leaving the tensors at every other point in spacetime unchanged? If we lived in a universe exactly described by GR, presumably there'd be no empirical difference between these two models, since all measurements would be made at observers who are at some finite separation from the point where the singularity would be, and all the observations associated with it--the way objects get crushed as they approach it and the curvature approaches infinity, and the way worldlines in its neighborhood end after a finite proper time--would still hold true.

 

[Hurkyl]

So when you say singularities don't have to exist in GR, do you just mean that we can remove the exact point of the singularity from spacetime, while leaving the tensors at every other point in spacetime unchanged?

Right. And as far as I know, that's how things are usually treated. (But that could just be a function of my reading interests) I personally cannot think of any real advantage to adding that point to space-time. (Actually, there's no obvious reason why it should just be a single point that's added in...)

 

[aranoff]

Okay, you claim that you won the argument, and I have sour grapes. Let's clarify exactly what you say.

We are discussing GR only, not QM.

The solution of GR from the point of view of the falling observer is valid everywhere except at the mathematical point of the center of the BH. At this point, there is a singularity, which means that the solution is not valid here.

Although it is not possible to observe the falling of the observer through the event horizon, as it takes forever, this solution is still meaningful, for the observer himself is capable of observing the event. The analogy with the meaninglessness of Heaven, a concept that no observer except a dead person, can observe, is not relevant. The fact that the observer loses all thoughts once he crosses the horizon, because the only things meaningful about a BH are mass, angular momentum, and charge, does not detract from the validity and meaningfulness of the solution. The fact that nothing changed as the object fell from one point inside to another point inside still permits us to describe coordinates of the falling observer inside the BH.

Is this your position?

I still did not get how you can justify this.

 

[JesseM]

Okay, you claim that you won the argument, and I have sour grapes. Let's clarify exactly what you say.

Who are you addressing here? Mentz114, who made the "sour grapes" comment? Me? Hurkyl? Someone else?

We are discussing GR only, not QM.

If we are purely discussing what is predicted theoretically by GR, then all discussions of whether GR is empirically valid are off-limits, no?

The solution of GR from the point of view of the falling observer is valid everywhere except at the mathematical point of the center of the BH. At this point, there is a singularity, which means that the solution is not valid here.

What do you mean by the word "valid" mean here, if it has nothing to do with empirical validity?

Although it is not possible to observe the falling of the observer through the event horizon, as it takes forever, this solution is still meaningful, for the observer himself is capable of observing the event. The analogy with the meaninglessness of Heaven, a concept that no observer except a dead person, can observe, is not relevant. The fact that the observer loses all thoughts once he crosses the horizon, because the only things meaningful about a BH are mass, angular momentum, and charge, does not detract from the validity and meaningfulness of the solution.

Here you once again jump to conclusions about a subject you have obviously not studied beyond a superficial level. The only observable attributes of a BH as viewed from the outside the event horizon are mass, angular momentum, and charge (according to GR), but this does not mean there can't be all sorts of interesting stuff going on inside the horizon, like observers making measurements and having thoughts as they fall towards the singularity.

 

[Hurkyl]

The solution of GR from the point of view of the falling observer is valid everywhere except at the mathematical point of the center of the BH. At this point, there is a singularity, which means that the solution is not valid here.

Why do you continue to insist that there a point at the 'mathematical center' of a BH? GR does not demand that the interior of a black hole have the topology of a ball -- in fact, it demands exactly the opposite if you use a formulation that rejects singularities. Colloquially speaking, there must be at least one point 'missing'. (Or something else much more exotic, like a wormhole)

You seem to have no problem with a space-time manifold for which the entire black hole is 'missing'. Why do you have a problem with a space-time manifold for which a single point is 'missing'?


As an aside, if you are allowing 'generalized' fields (e.g. distributional fields) that have singularities and other bad behavior... then you really ought to be using the corresponding 'generalized' notion of a partial differential equation (e.g. distributional derivatives). In that case, a field can still satisfy the EFE, even if it has singularities.

The fact that the observer loses all thoughts once he crosses the horizon, because the only things meaningful about a BH are mass, angular momentum, and charge, does not detract from the validity and meaningfulness of the solution. The fact that nothing changed as the object fell from one point inside to another point inside still permits us to describe coordinates of the falling observer inside the BH.

You have misunderstood the no hair theorem -- while such information is inaccessible to observers outside the event horizon, it is by no means inaccessible to observers inside the event horizon. In fact, in order to even make the quoted statement, you had to assume that there were physically distinct states inside the black hole!


There is another thing; physical theories routinely allow us to infer the existence of things we cannot directly observe. This is, in fact, necessary. There is merit to waxing philosophically about which things we should infer exist and which things we shouldn't -- but IMHO you're being far too absolute about things.


Incidentally, I feel compelled to state that the notion of a point is not physically meaningful in GR (really, it isn't meaningful even in Newtonian mechanics!). I assume you already know this and are referring to something that is meaningful (e.g. the point where the infalling observer's wristwatch read 3:00 PM), but I want to make absolutely sure that there wasn't a conceptual problem being obscured by this abuse of language.

 

[coelho]

Well, except for the fact that we can create mathematically-consistent models from what we do know. The same can't be said in both cases.

Yes... but mathematical consistence does not mean anything by itself. The mere fact one theory is mathematically consistent does not imply its a valid theory to describe a physical phenomena, if it were not testable in the "real world".
Testing hypotheses derived from the theorethical framework against the experiences is a necessary step of the scientific method, and if this one can't be done in some class of phenomena, then this phenomena can't be completely "scientifically" treated, and some degree of "faith" is required to admit the validity of any theory used to describe this phenomena, as we don't have the proof of the experimental facts to ensure its validity.

 

[aranoff]

You have misunderstood the no hair theorem -- while such information is inaccessible to observers outside the event horizon, it is by no means inaccessible to observers inside the event horizon. In fact, in order to even make the quoted statement, you had to assume that there were physically distinct states inside the black hole!

What do you mean by an observer inside the BH? The only property the observer has is mass. Furthermore, as the observer falls, the angular momentum of the BH does not change. When we speak about observers inside a BH is like talking about observers who observe Heaven after death. As the dead observers move to different parts of Heaven, nothing changes for the external observers (the living).

I am sick and tired of mixing religion and science. With science, look at the theory. The math must be consistent. The results must be capable of verification or falisfaction.

Indeed, I am suspicious that one reason for the popularity of the Big Bang is the idea that maybe God created the universe.

Your insistence on talking about the inside of the BH may be due to some type of religious thinking.

The BH is not an excluded point. It is simply an end to the universe. The spherical universe has no end. Travel out at a constant speed in a straight line for a time t. Wait longer, and you are still in the universe. Same if we approach a BH. Just that the geometry is not Euclidean, and this confuses you.

 

[aranoff]

There is another thing; physical theories routinely allow us to infer the existence of things we cannot directly observe. This is, in fact, necessary. There is merit to waxing philosophically about which things we should infer exist and which things we shouldn't -- but IMHO you're being far too absolute about things.

We can infer things that we cannot observe, but must be capable of observing in principle.

E.g., superfluid helium. We cannot observe the positions of the atoms, except to say that they are in the container. Therefore, the positions do not exist, because we cannot observe it in principle.

 

[aranoff]

Who are you addressing here? Mentz114, who made the "sour grapes" comment? Me? Hurkyl? Someone else?

If we are purely discussing what is predicted theoretically by GR, then all discussions of whether GR is empirically valid are off-limits, no?

Someone made the comment about sour grapes. Let's drop it!

I do not follow your comment about GR. We discuss whatever we can about GR. My point is that we cannot discuss what is inside the BH.

 

[aranoff]

Right. And as far as I know, that's how things are usually treated. (But that could just be a function of my reading interests) I personally cannot think of any real advantage to adding that point to space-time. (Actually, there's no obvious reason why it should just be a single point that's added in...)

My understanding is that there is no transformation that can remove the singularity from the BH.

 

[DaveC426913]

When we speak about observers inside a BH is like talking about observers who observe Heaven after death. As the dead observers move to different parts of Heaven, nothing changes for the external observers (the living).

I am sick and tired of mixing religion and science.

Your insistence on talking about the inside of the BH may be due to some type of religious thinking.

The only one here talking about religion is you.

You have not made your case that these two examples are equivalent, so your attempt to accuse your opponents of talking religion is premature.

You can however, get this thread shut down by repeatedly making ad hominem and well poisoning arguments.

 

[aranoff]

The only one here talking about religion is you.

You have not made your case that these two examples are equivalent, so your attempt to accuse your opponents of talking religion is premature.

You can however, get this thread shut down by repeatedly making ad hominem and well poisoning arguments.

Ok. The only reason I mentioned religion is that I simply do not see any other logic in your arguments.

 

[DaveC426913]

Ok. The only reason I mentioned religion is that I simply do not see any other logic in your arguments.

This statement has been made more than once:

"...while such information is inaccessible to observers outside the event horizon, it is by no means inaccessible to observers inside the event horizon. "

If false, your logic follows. If true the logic that others are purporting follows.

Until satisfactorily shown to be false, your refutation of others' arguments holds little water.

 

[JesseM]

What do you mean by an observer inside the BH? The only property the observer has is mass.

Nonsense. The only properties of the observer that contribute to what the BH looks like from outside the event horizon are mass, charge, and angular momentum, but according to GR the observer can still have all sorts of complex properties from the perspective of anyone inside the event horizon.

I am sick and tired of mixing religion and science. With science, look at the theory. The math must be consistent. The results must be capable of verification or falisfaction.

I suppose it's no surprise that you avoided answering my question about whether you were just talking about the theoretical predictions of GR or the larger question of whether these theoretical predictions are actually true in the real world, because here you are going back to talking about empirical verification. On a purely theoretical level, it's clear that the theory of GR does make predictions about what is going on inside the black hole. On an empirical level, GR's predictions about the inside of a BH may be wrong, but if so most physicists think that it's because GR is just an approximation for a true theory of quantum gravity, and yet you got all huffy earlier when I tried to bring quantum gravity into the discussion (also note that quantum gravity might actually allow information from inside the horizon to escape, removing your objection about falsifiability). So which is it--do you want to talk about empirical questions about whether GR's predictions about the inside of the event horizon are actually correct (in which case you have absolutely no excuse for ignoring quantum gravity), or do you just want to talk about what the mathematical theory of GR predicts without worrying about whether these predictions are true?

Indeed, I am suspicious that one reason for the popularity of the Big Bang is the idea that maybe God created the universe.

The popularity of the Big Bang has to do with two main factors:

1. The theory of GR has plenty of experimental verification on a local scale (the orbit of Mercury, gravitational lensing of light passing near the sun, gravitational time dilation at different heights on Earth, etc.), and when GR is applied to the universe as a whole, it's almost impossible to get a universe consistent with GR that isn't expanding or contracting.

2. The Big Bang theory makes plenty of predictions about things like galaxy redshifts and the spectrum of the cosmological microwave background radiation and the relative amounts of different elements in the universe, and the predictions have been verified empirically (see here for a quick summary). Of course, these predictions just depend on the idea that space has been expanding from an extremely hot and dense state in the past, they don't prove that the density actually approaches infinity at some finite time in the past as predicted by GR, and indeed this notion of the Big Bang as the "beginning of time" would probably be altered by a theory of quantum gravity.

Your insistence on talking about the inside of the BH may be due to some type of religious thinking.

You said earlier that you specifically wanted to talk about the theoretical predictions of GR, and then when people explain GR's theoretical predictions, you accuse them of religious thinking. Your argument is pretty incoherent.

The BH is not an excluded point. It is simply an end to the universe. The spherical universe has no end. Travel out at a constant speed in a straight line for a time t. Wait longer, and you are still in the universe. Same if we approach a BH.

Except that an observer will reach the event horizon in finite proper time according to GR, whereas you could travel for infinite proper time and never reach any "end of the universe" in ordinary space.

I do not follow your comment about GR. We discuss whatever we can about GR. My point is that we cannot discuss what is inside the BH.

What don't you follow? Are you unable to understand the concept of discussing the predictions of a given mathematical theory like GR independently of questions about the empirical truth of these predictions? And if you want to talk about what's empirically true, what's your excuse for ignoring the issue of quantum gravity? And if you want to talk about empirical questions, your response to this previous comment of mine is completely at odds with that:

I'm sorry, but I am at a loss of how to communicate with you guys. You are supposed to be physicists, either students or professors. I said very clearly that I was discussing the meaning of GR, and you mention quantum gravity!

As I understand it, you are not discussing purely theoretical questions about what GR predicts (it predicts singularities, and there is nothing a priori impossible about the idea that these could be real physical entities), but rather the empirical question of whether GR's predictions are correct (you reject its predictions everywhere inside the event horizon for reasons that aren't really clear, while most physicists have physical reasons for thinking its predictions are only likely to go significantly wrong at the Planck scale). Am I misunderstanding something here?

My discussions were what does GR predict.

That last response clearly indicated you were picking the first option, "discussing purely theoretical questions about what GR predicts". But I'm beginning to feel at this point that you don't really have a coherent position of any kind and are just basing your responses on various knee-jerk reactions like "I don't like talking about the inside of black holes" and "I don't like talking about quantum gravity".

 

[JesseM]

My understanding is that there is no transformation that can remove the singularity from the BH.

What Hurkyl is talking about is simply defining the singularity to not be part of spacetime, so instead you have a spacetime with a "hole" where the singularity should be. But the curvature of spacetime would still approach infinity in the neighborhood of this hole, and this model would be experimentally indistinguishable from a model which considers the singularity to be part of spacetime, even for experimenters inside the event horizon. So it seems like a cosmetic change with no real physical significance.

 

[PaulDent]

There is no physical singularity at the event horizon of a black hole. Schwarzschild coordinates do go to infinity there, but you can pick different coordinate systems (like some of the ones mentioned on this page) where there is no coordinate singularity at the event horizon, and you can show it only takes the infalling observer a finite proper time to pass the event horizon. The singularity at the center of the black hole is a real physical one though, since infinities appear there no matter what coordinate system you choose.

Those claiming there is no singularity at the event horizon should be asked the following question:

If your head was inside the event horizon and your feet were outside, could you wiggle your toes?

If not, I suspect you would find that rather disconcerting, and some catastrophe would probably have befallen you. I do not believe that anything from our universe can survive crossing the event horizon. I think Stephen Hawking was of that belief too.

Yes, there are coodinate transformations of the interior metric that make it look rather normal, but the coordinates do not join up smoothly with those we are coming from on the outside. In particular, the radial dimension acquires the role of time inside, while time becomes a spatial coordinate. That switch is pretty catastrophic to matter or radiation entering from the outside.

 

[PaulDent]

Where is this alledged discontinuity? Is it at a point of space-time? Or is it merely at some point in a faux-coordinate chart that doesn't correspond to a point of space-time?

And no, continuity really isn't a hard requirement of the theory -- it (and sufficient differentiability) only a requirement for the use of the simplest mathematical framework.


This is patently absurd.

Yes, it is absurd, but not stupid. It may be absurd AND correct.

I have tried to imagine a convoy of spacecraft heading directly towards a BH.
Far from the hole, they were equispaced and traveling at the same speed, one behind another. If you were in one of the middle ones, what would you see?

Firstly, the singularity at the event horizon means that it's not like the thin surface of a balloon, but in fact it is very, maybe infinitely thick/deep. So I think you see the spaceships ahead of you accelerate and get apparently further and further from you, increasingly redshifted, and ultimately seeming to disappear infinitely far from you, AND they STILL HAVEN'T REACHED THE EVENT HORIZON. In otherwords, we seem to be able to prick the black hole's surface way down deep it and we still haven't got to the event horizon.

Another curiosity of the event horizon its that it is the envelope of orbits of light rays. In otherwords, every great circle is a null geodesic. In fact, the whole darn surface is a null-geodesic surface, there being zero tangential distance between any two points. This jibes with what the spaceships see, as they all seem to be disappearing towards a point somewhere in the depths - and that point is still on the surface, not the interior.

So, from this perspective, the "spherical balloon skin" notion of the event horizon becomes
something quite different : Its area becomes zero, while its thickness becomes infinite - i.e.
it becomes a radial straight line to infinity, which is the "absurd" conclusion. - but maybe still correct.

 

[JesseM]

Those claiming there is no singularity at the event horizon should be asked the following question:

If your head was inside the event horizon and your feet were outside, could you wiggle your toes?

Sure, the signal sent from your brain would reach your toes, and assuming your head is falling into the event horizon after the toes (rather than being kept at a constant distance above the horizon by a rocket or something, which would mean your body would get ripped apart) then sensory signals from the toes could get back to the brain too. The only events on the worldline of the brain that do not have events on the worldline of the toes in their future light cone are events from when the brain is so close to hitting the singularity that there is no time for light signals from the brain to reach the toes before they, too, hit the singularity.

I do not believe that anything from our universe can survive crossing the event horizon. I think Stephen Hawking was of that belief too.

Hawking has never said anything like that. In fact, in http://64.233.169.104/search?q=cache:_yDFPsUuWbgJ:irealitylib.hit.bg/Stephen%2520Hawking/A%2520Brief%2520History%2520of%2520Time/e.html+%22would+not,+in+fact,+feel+anything+special+as+he+reached+the+critical+radius%22&hl=en&ct=clnk&cd=1&gl=us of A Brief History of Time he explicitly says that if the black hole were large enough so that tidal forces were not too great at the horizon, the astronaut would notice nothing unusual upon crossing it:

Gravity gets weaker the farther you are from the star, so the gravitational force on our intrepid astronaut’s feet would always be greater than the force on his head. This difference in the forces would stretch our astronaut out like spaghetti or tear him apart before the star had contracted to the critical radius at which the event horizon formed! However, we believe that there are much larger objects in the universe, like the central regions of galaxies, that can also undergo gravitational collapse to produce black holes; an astronaut on one of these would not be torn apart before the black hole formed. He would not, in fact, feel anything special as he reached the critical radius, and could pass the point of no return without noticing it. However, within just a few hours, as the region continued to collapse, the difference in the gravitational forces on his head and his feet would become so strong that again it would tear him apart.

Yes, there are coodinate transformations of the interior metric that make it look rather normal, but the coordinates do not join up smoothly with those we are coming from on the outside. In particular, the radial dimension acquires the role of time inside, while time becomes a spatial coordinate. That switch is pretty catastrophic to matter or radiation entering from the outside.

No, this is just an artifact of the way Schwarzschild coordinates work, you're free to use a coordinate system where there is no "switch" and the time coordinate continues to be physically timelike inside the event horizon, such as Kruskal-Szekeres coordinates.

 

[Hurkyl]

Those claiming there is no singularity at the event horizon should be asked the following question:

If your head was inside the event horizon and your feet were outside, could you wiggle your toes?

Yes -- but the event horizon will have passed by your feet before the electrical impulse from your brain would have reached them.

Unless you meant that your feet were undergoing sufficiently strong acceleration to remain outside of the event horizon -- in which case they would have already been ripped off of your body.

Yes, there are coodinate transformations of the interior metric that make it look rather normal, but the coordinates do not join up smoothly with those we are coming from on the outside.

Thus the term "coordinate singularity": you've merely given one explanation of why Schwarzschild 'coordinates' are unsuitable for talking about the event horizon.

 

[Hurkyl]

Yes, it is absurd, but not stupid. It may be absurd AND correct.

Since your previous quotation is misleading, allow me to recall that this is all in reference to aranoff's claim:

Therefore, the event horizon is geometrically the same as a straight line.​

The event horizon, for example, is not one-dimensional, so it's clearly not correct.

(Incidentally, if the statement was correct, then it would not be absurd)

 

[PaulDent]

Sure, the signal sent from your brain would reach your toes, and assuming your head is falling into the event horizon after the toes (rather than being kept at a constant distance above the horizon by a rocket or something, which would mean your body would get ripped apart) then sensory signals from the toes could get back to the brain too. The only events on the worldline of the brain that do not have events on the worldline of the toes in their future light cone are events from when the brain is so close to hitting the singularity that there is no time for light signals from the brain to reach the toes before they, too, hit the singularity.

I think you must be talking about an observer in free fall into the BH.

However, consider a BH of mass 1.237e43 Kg, which is still only about 1e-15 the mass of our universe. The gravitational acceleration at the event horizon is c^4/GM, which you will find is our regular 9.81 m/s^2 with the above mass. That means hovering above it with rocket power is no big trick. It would be exactly the same as hovering above the surface of a lake or ocean on earth.
Now I can decide to descend slowly headfirst and dunk my head into the event horizon, and the rockets on my feet hold me there without catastrophic stretching. Now can I wiggle my toes, given that no signal can cross the event horizon in the direction inside to outside?

 

[PaulDent]

Since your previous quotation is misleading, allow me to recall that this is all in reference to aranoff's claim:

Therefore, the event horizon is geometrically the same as a straight line.​

The event horizon, for example, is not one-dimensional, so it's clearly not correct.

(Incidentally, if the statement was correct, then it would not be absurd)

What I was saying is that, from a certain perspective, the event horizon can appear one-dimensional; if you imagine its area appearing to be zero (instead of a sphere) while its thickness or depth appears to be infinite instead of zero, then it morphs to a straight line in the radial direction. Aranoff is not alone in coming to the conclusion that it might appear to look like this from some perspective. I imagine it might TEND to appear that way as you approached the event horizon closer and closer.

But these things are hard to imagine, and my reason for corresponding here is to find help in getting my head round it, so thank you for your take on it.

 

[JesseM]

However, consider a BH of mass 1.237e43 Kg, which is still only about 1e-15 the mass of our universe. The gravitational acceleration at the event horizon is c^4/GM, which you will find is our regular 9.81 m/s^2 with the above mass.

Where did you get that formula for gravitational acceleration? Assuming it's correct, I would guess that it's talking about something like the coordinate acceleration for a freefalling observer in some specific coordinate system like Schwarzschild coordinates. The proper acceleration needed to maintain a constant height above the event horizon (i.e. the G-force felt by someone who is maintaining that height) always approaches infinity as you approach the event horizon--see pervect's post #4 on this thread.

Now I can decide to descend slowly headfirst and dunk my head into the event horizon, and the rockets on my feet hold me there without catastrophic stretching. Now can I wiggle my toes, given that no signal can cross the event horizon in the direction inside to outside?

You most definitely will experience catastrophic stretching if you try to keep your feet at a constant height above the horizon while your head is below it.

 

[PaulDent]

No, this is just an artifact of the way Schwarzschild coordinates work, you're free to use a coordinate system where there is no "switch" and the time coordinate continues to be physically timelike inside the event horizon, such as Kruskal-Szekeres coordinates.

I am just getting round to taking another look at the Kruskal extension. The last time I looked, I had many problems with it, at least the way it was presented, as it had quantities under square root signs that became negative. And that wasn't the only problem. It basically transforms only the r and t terms, and leaves the terms in theta and phi as is to get "Mixed Edington-Finkelstein coordinates". However that does not allow construction of a 4d metric that I can recognize as supporting our physics. To be able to unequivocally say that a metric represents a universe like ours, it should be possible to get it in the form

a^2(t)(dx^2 + dy^2 + dz^2) - (cdt)^2

i.e. Minkowski with some tolerable rate of expansion or contraction

 

[PaulDent]

You most definitely will experience catastrophic stretching if you try to keep your feet at a constant height above the horizon while your head is below it.

Then it seems you agree that there IS a singularity at the event horizon, at least to an observer who is not in free fall. And the catastrophic stretching is nothing to do with the rocket thrust, as in my example, that would only have to be g=9.81m/s^2, as here on earth.

 

[PaulDent]

Where did you get that formula for gravitational acceleration? Assuming it's correct, I would guess that it's talking about something like the coordinate acceleration for a freefalling observer in some specific coordinate system like Schwarzschild coordinates. The proper acceleration needed to maintain a constant height above the event horizon (i.e. the G-force felt by someone who is maintaining that height) always approaches infinity as you approach the event horizon--see pervect's post #4 on this thread.

gravitational pull = GM/r^2 at static radius r outside the event horizon

substitute schwartzschild radius r = GM/c^2

and you get c^4/GM as the gravitational pull at the event horizon.

For modest BHs (small M) this is enormous. But for ginormous M, it becomes tolerable.

 

[JesseM]

gravitational pull = GM/r^2 at static radius r outside the event horizon

But that's a Newtonian formula--you can't expect it to work in cases where GR diverges significantly from Newtonian gravity, and the vicinity of a black hole is definitely one of these cases! As I said, see pervect's post #4 here for the actual formula for gravitational acceleration near the event horizon, calculated from GR, which does go to infinity as you approach the horizon.

 

[JesseM]

You most definitely will experience catastrophic stretching if you try to keep your feet at a constant height above the horizon while your head is below it.

Then it seems you agree that there IS a singularity at the event horizon, at least to an observer who is not in free fall. And the catastrophic stretching is nothing to do with the rocket thrust, as in my example, that would only have to be g=9.81m/s^2, as here on earth.

Again, you're mistaken in using Newtonian formulas here, the proper acceleration definitely does have to do with rocket thrust, and the thrust needed to maintain a given height goes to infinity as you approach zero height above the horizon. I don't think most physicists would consider this a physical singularity though, as it is only a theoretical requirement that reaches infinity at the horizon rather than any quantity actually measured by anyone (there are no actual objects maintaining zero height above the horizon)--as an analogy, the theoretical requirement "energy required to increase one's velocity by 0.1c" also goes to infinity at 0.9c, but no one would say that means there's a physical singularity at 0.9c.

 

[stevebd1]

I think PaulDent might be confusing gravity with gravity gradient which, as far as I know, remains unaffected by Schwarzschild coordinates (or any kind of metric), the gravity gradient being a measure of how much gravity increases over 1 metre and in the case of stellar black holes, results in 'spaghettification' at the event horizon. The equation for gravity gradient (or tidal forces) being-

\frac{2Gm}{r^3}

where the results are in m/s^2/m. Multiply the answer by 2 to get an approx figure for the change in gravity from head to foot. This is why if you plan to cross the event horizon, a large black hole is favoured over a small one (from looking at various sources, the most a human can endure is a gradient of 15 Earth g's from head to toe).

 

[JesseM]

\frac{2Gm}{r^3}

This is also a Newtonian equation--as I said, you can't assume Newtonian equations are still correct in GR, GR is a very different theory of gravity which only looks like Newtonian gravity in certain limits.

 

[George Jones]

This is also a Newtonian equation

Actually,

\frac{2Gm}{r^3}L,

where L is the small radial spatial separation between two test masses, is both the Newtonian and general relativistic expression for tidal acceleration outside a spherically symmetric mass.

 

[George Jones]

I am just getting round to taking another look at the Kruskal extension. The last time I looked, I had many problems with it, at least the way it was presented, as it had quantities under square root signs that became negative. And that wasn't the only problem. It basically transforms only the r and t terms, and leaves the terms in theta and phi as is to get "Mixed Edington-Finkelstein coordinates".

I think have misinterpreted Kruskal-Szekeres coordinates.

However that does not allow construction of a 4d metric that I can recognize as supporting our physics. To be able to unequivocally say that a metric represents a universe like ours, it should be possible to get it in the form

a^2(t)(dx^2 + dy^2 + dz^2) - (cdt)^2

i.e. Minkowski with some tolerable rate of expansion or contraction

It's not supposed to look like a flat space Friedmann-Robertson-Walker spacetime.

 

[stevebd1]

Hi JesseM.

While I understand your argument that Newtonian equations cannot be assumed to be correct beyond the event horizon, I have seen stated a number of times 'While Newton's equation for gravity, g=Gm/r^2, is affected by coordinate acceleration, diverging at the event horizon, the Newtonian equation for tidal forces remains unchanged up to and possibly beyond the event horizon'. Another equation I've seen for calculating the gravity gradient at the event horizon is-

\frac{c^6\ dr}{4 (Gm)^2}

which looks like it's derived from the Newtonian equation and the Schwarzschild radius.

 

[JesseM]

Hi JesseM.

While I understand your argument that Newtonian equations cannot be assumed to be correct beyond the event horizon, I have seen stated a number of times 'While Newton's equation for gravity, g=Gm/r^2, is affected by coordinate acceleration, diverging at the event horizon, the Newtonian equation for tidal forces remains unchanged up to and possibly beyond the event horizon'.

My apologies, I jumped to conclusions there--since I knew the equation worked in Newtonian physics I thought you were just assuming it would work in GR (as PaulDent was doing in his gravitational acceleration calculation), but since you say you've seen it specifically stated that this equation works in GR (and George Jones backs this up) I'll take your word for it that it does.

 

[George Jones]

'the Newtonian equation for tidal forces remains unchanged up to and possibly beyond the event horizon'.

Yes and no. The expression does hold inside the event horizon, but, inside the event horizon, r has a different meaning than it has outside the event horizon.

 

[stevebd1]

I have heard r in relation to black holes being referred to as the 'reduced circumference' or the 'coordinate radius' but even then, they don't give the true distance to the singularity. One other definition I've seen in relation to Schwarzschild black holes is the free-float horizon to crunch distance which is expressed as \tau_{max}\text{[metres]}=\pi M where M is the gravitational radius (Gm/c^2) which implies the 'true' distance to the singularity is marginally longer than coordinate length of 2M which establishes the event horizon.

 

[George Jones]

I have heard r in relation to black holes being referred to as the 'reduced circumference' or the 'coordinate radius'

Only outside the event horizon. Inside the event horizon, r is a timelike coordinate.

but even then, they don't give the true distance to the singularity.

I don't think a "true distance" to the singularity is definable.

One other definition I've seen in relation to Schwarzschild black holes is the free-float horizon to crunch distance which is expressed as \tau_{max}\text{[metres]}=\pi M

I think that this is the proper time taken for an observer to fall from "rest" just above the event horizon to the singularity.

 

[stevebd1]

I had look in 'Exploring Black Holes' by Edwin Taylor & John Wheeler and they give 2 quantities in relation to the fall-in time from the event horizon. The one mentioned above which when divided by c is the maximum free-float horizon to crunch wristwatch time (basically from rest at the EH) and the other which is falling from rest at infinity. Assuming in that case that you're accelerating with space and space itself approaches c at the event horizon, your fall-in time from the event horizon would be reduced to -

\tau\text{[seconds]}=\frac{4M}{3c}=\frac{4Gm}{3c^3}\equiv\ 6.568\times10^{-6}\ \times\ \text{sol mass}

\tau\text{[metres]}=\frac{4}{3}M

where M is the gravitational radius.

I'm assuming this goes some way to demonstrating the timelike properties of r inside the event horizon.

For clarity, here's the maximum fall-in time (from rest at the EH)-

\tau_{max}\text{[seconds]}=\frac{\pi M}{c}=\frac{\pi Gm}{c^3}\ \equiv\ 1.548\times10^{-5}\ \times\ \text{sol mass}

\tau_{max}\text{[metres]}=\pi M

In both cases, the distance seems relative to how you cross the event horizon.

 

[stevebd1]

Yes and no. The expression does hold inside the event horizon, but, inside the event horizon, r has a different meaning than it has outside the event horizon.

Could you possibly explain or demonstrate the implications this has on the equation for the gravity gradient inside the event horizon.

 

[George Jones]

Could you possibly explain or demonstrate the implications this has on the equation for the gravity gradient inside the event horizon.

Once inside the event horizon, r is a timelike coodinate, and passage from the past to the futures implies going from a larger r to a smaller r. This and the expression for tidal force imply that inside the event horizon, tidal forces steadily increase for all observers and eventually become unbounded.

 

[stevebd1]

Thanks for the reply George

While I understand the implications of r becoming a timelike coordinate beyond the event horizon (once past the event horizon, the singularity is in everybody's future), could you elaborate on the last sentence in your post, particularly the part about tidal forces becoming unbound-

This and the expression for tidal force imply that inside the event horizon, tidal forces steadily increase for all observers and eventually become unbounded.

 

[George Jones]

While I understand the implications of r becoming a timelike coordinate beyond the event horizon (once past the event horizon, the singularity is in everybody's future), could you elaborate on the last sentence in your post, particularly the part about tidal forces becoming unbound-

Since r is timelike with smaller r pointing to the future, r must decrease along the worldline of any observer inside the event horizon. Consequently, along the worldline of any observer, 1/r^3 increases, tidal force increases. As r \rightarrow 0 (which it must, since r must decrease), tidal force becomes unbounded.

 

[PaulDent]

I think I missed a 2 in my original post. The Schwartzschild radius is 2GM/c^2.

I do believe the Newtonian formula for Gravitational pull applies down to the event horizon, i.e. acceleration due to gravity = GM/r^2 still. Substitute r=Schwartzschild radius and you get the acceleration at the event horizon. This can't go to infinity if the gravity gradient remains finite - they either both go to infinity or neither does.

 

[George Jones]

I think I missed a 2 in my original post. The Schwartzschild radius is 2GM/c^2.

I do believe the Newtonian formula for Gravitational pull applies down to the event horizon, i.e. acceleration due to gravity = GM/r^2 still. Substitute r=Schwartzschild radius and you get the acceleration at the event horizon.

For which observer?

=PaulDent]This can't go to infinity if the gravity gradient remains finite - they either both go to infinity or neither does.

Again, for which observer?

 

[PaulDent]

For which observer?


Again, for which observer?

I am a bit weak on that question! But it is a stationary observer outside the event horizon, and I can't decide if it is the acceleration that a distant observer decides is being experienced at the event horizon, or if it is the oberver at the event horizon that experiences that acceleration.

 

[stevebd1]

I do believe the Newtonian formula for Gravitational pull applies down to the event horizon, i.e. acceleration due to gravity = GM/r^2 still. Substitute r=Schwartzschild radius and you get the acceleration at the event horizon. This can't go to infinity if the gravity gradient remains finite - they either both go to infinity or neither does.

Basically, what you’re saying is, is why is the Newtonian equation for gravity (g=Gm/r^2) affected by coordinate acceleration (dr'=dr(1-Rs/r)^-1/2, curved space) while the Newtonian equation for tidal forces (dg=2Gm/r^3) remains unchanged as if dealing with flat space. The way I see it is that in close proximity to the event horizon, the curvature of space increases until the velocity induced by the curve is c at the event horizon, an acceleration experienced by the local observer, (for the observer at infinity, the local observer will appear to slow down and freeze at the event horizon due to the curvature having the opposite effect on light sent from near the event horizon, redshifting into infrared and eventually into long wave radio waves). For the local observer, the rate of increase in Newtonian gravitational acceleration remains constant but appears to increase to infinity because the space it's in is curving towards the horizon (hence why gravity is multiplied by coordinate increase to represent proper acceleration). At the event horizon, gravity will appear to diverge when in fact, it's going about business as usual, increasing steadily over r and it is the space it's in that rapidly accelerates (notice how at the event horizon it's the coordinates that diverge and not the gravity) so the equation for tidal forces (the rate that gravity increases) remains correct and unchanged. While this isn't a mathematical answer and there's probably a better way of explaining it, this is a mental picture I have of the situation.

Steve

 

[PaulDent]

Basically, what you’re saying is, is why is the Newtonian equation for gravity (g=Gm/r^2) affected by coordinate acceleration (dr'=dr(1-Rs/r)^-1/2, curved space) while the Newtonian equation for tidal forces (dg=2Gm/r^3) remains unchanged as if dealing with flat space.

What about the force on an object near the event horizon though?

Imagine a huge black hole, with low gravity gradient at the event horizon. I use a rocket to hover just outside the event horizon and use a simple pendulum to measure g at my feet and g at my head. If I did that at different heights above the event horizon, getting

(ghead(1),gfeet(1)) 10000km away, (ghead(2),gfeet(2)) 9000km away...(ghead(10),gfeet(10)) 1000km away,

then I can plot the gravity gradient at 1000km steps down towards the thorizon and then integrate that curve to get the value of g versus distance.

I begin to see a problem with coordinates though. WHO says I am 1000, 2000...10000km from the event horizon? Different observers will come up with different views of that. And is that distance dr what I multiply by gravity gradient in the integration?

In fact, does the event horizon appear to me to recede as I approach it, so that I experience an infinite number of 1000km steos to get there, and my g-integral goes to infinity?

 

[stevebd1]

If you take into account the velocity induced by the curvature (basically v_{rel}=-\sqrt{r_s/r}\ c, zero at infinity, c at the event horizon) then dr remains 1 as in some alternative metrics for static black holes. In this case, the surface gravity is derived from Killing* vectors, which equals-

\kappa=\frac{c^4}{4GM}

but the above only applies when free-fall velocity into the black hole is accounted for and used in conjunction with a global rain frame metric such as http://en.wikipedia.org/wiki/Gullstrand-Painlevé_coordinates" .

An extract from wiki-

'In relativity, the Newtonian concept of acceleration turns out not to be clear cut. For a black hole, which can only be truly treated relativistically, one cannot define a surface gravity as the acceleration experienced by a test body at the object's surface. This is because the acceleration of a test body at the event horizon of a black hole turns out to be infinite in relativity.

Therefore, when one talks about the surface gravity of a black hole, one is defining a notion that behaves analogously to the Newtonian surface gravity, but is not the same thing. In fact, the surface gravity of a general black hole is not well defined. However, one can define the surface gravity for a black hole whose event horizon is a Killing horizon.

The surface gravity κ of a static Killing horizon is the acceleration, as exerted at infinity, needed to keep an object at the horizon.'

Source- http://en.wikipedia.org/wiki/Surface_gravity"


A Hawking Radiation calculator tool which also calculates the Killing surface gravity-
http://xaonon.dyndns.org/hawking/


*Named after mathematician Wilhelm Killing