Can you be in stable orbit inside a blcak hole?

[TCS]

If you are in free fall inside a black hole, shouldn't it feel like normal space even though people on Earth saw you enter a black hole?

 

[Dmitry67]

Yes, locally you don't feel anything strange when you cross the event horizon.
However, no stable orbits are possible inside - it fact, singularity (of non-rotating BH) for you will be not in the center but IN THE FUTURE.

 

[dx]

If you are in free fall inside a black hole, shouldn't it feel like normal space even though people on Earth saw you enter a black hole?

They not only saw you enter the black hole, they also saw you get thermalized by the horizon! Yet the infalling observer notices nothing special at the horizon.

 

[DrGreg]

They not only saw you enter the black hole, they also saw you get thermalized by the horizon! Yet the infalling observer notices nothing special at the horizon.

Actually, an observer outside the event horizon will never see anyone pass through the horizon. They will see them approach the horizon, but slow down and come almost to a stop (and the light red-shifted to almost zero frequency).

There's only one way to see someone fall through the horizon, and that's to follow them through yourself (in which case you won't notice anything unusual as either of you pass through, as long as you're close enough to each other).

 

[dx]

Actually, an observer outside the event horizon will never see anyone pass through the horizon. They will see them approach the horizon, but slow down and come almost to a stop

Yes that's what I meant. To an observer outside, the infalling observer appears to asymptotically approach the horizon. This is in classical general relativity. When we take quantum mechanics into account, the picture changes a little bit: when the infalling observer is very close to the horizon, around the order of the plank length, the outside observer would see the infalling observer thermalized and re-emitted as Hawking radiation.

 

[Finbar]

Yes that's what I meant. To an observer outside, the infalling observer appears to asymptotically approach the horizon. This is in classical general relativity. When we take quantum mechanics into account, the picture changes a little bit: when the infalling observer is very close to the horizon, around the order of the plank length, the outside observer would see the infalling observer thermalized and re-emitted as Hawking radiation.

If you believe Susskind and black hole complementarity. It seems to suggest that there are two different realities one for the in falling observer and one for the outside observer.

I never really got why the in falling observer has to fall through the horizon. If the horizon is dynamical surely it can just reseed as the black hole evaporates. Too much logic seems to be derived from the stationary black hole causal structure. Before we consider that an observer falls into a black hole shouldn't we consider the entire history of the space-time outside the black hole? I can except that the observer is absorbed into the stretched horizon in a finite time but why doesn't he just then reseed with the horizon rather than moving through it?

 

[Dmitry67]

I never really got why the in falling observer has to fall through the horizon. If the horizon is dynamical surely it can just reseed as the black hole evaporates.

Because there are different types of horizon: apparent and absolute.
Observer can fall thru the absolute horizon (at shw. radius) but he has his personal, apparent horizon, which receded in from of him when he approach the black hole.

Regarding the evaporation, the infalling observer falls down very quickly, much faster than absolute horizon evaporates, so of course if crosses it before it evaporates. How it is visible from the outside is a different story.

 

[Finbar]

Because there are different types of horizon: apparent and absolute.
Observer can fall thru the absolute horizon (at shw. radius) but he has his personal, apparent horizon, which receded in from of him when he approach the black hole.

Regarding the evaporation, the infalling observer falls down very quickly, much faster than absolute horizon evaporates, so of course if crosses it before it evaporates. How it is visible from the outside is a different story.

The apparent horizon for an observer outside a black hole which can be defined locally r=2GM. In order to know the position of the event horizon(also defined for the observer outside the hole) one needs to know the entire history of the space-time. An in falling observer may cross the apparent horizon within a finite amount of time measured by an outside observer.

But I don't understand how the in falling observer crosses the event horizon quicker than the black hole evaporates? This makes no sense to me. It takes an infinite time(measured by the outside observer) for the in falling observer to cross the event horizon but only a finite time for the black hole to evaporate, to say a mass M ~ a few Planck masses. The in falling observer may disagree with what time events happen but he cannot disagree that he falls through the event horizon after the hole has evaporated to a few Planck masses.

I really don't see how the in falling observer can cross the event horizon while the black hole continues to evaporate.

 

[JesseM]

But I don't understand how the in falling observer crosses the event horizon quicker than the black hole evaporates? This makes no sense to me. It takes an infinite time(measured by the outside observer) for the in falling observer to cross the event horizon but only a finite time for the black hole to evaporate

It only takes an infinite time for the outside observer to see the infalling one reach the horizon in the case of an eternal black hole, in the case of an evaporating black hole the light from the crossing would reach the outside observer at the exact same moment he sees the black hole evaporate completely--see the "What about Hawking radiation?" question here:

But I just said that an outside observer would never observe an object actually entering the horizon! If I jump in, will you see the black hole evaporate out from under me, leaving me intact but marooned in the very distant future from gravitational time dilation?

You won't, and the reason is that the discussion above only applies to a black hole that is not shrinking to nil from evaporation. Remember that the apparent slowing of my fall is due to the paths of outgoing light rays near the event horizon. If the black hole does evaporate, the delay in escaping light caused by proximity to the event horizon can only last as long as the event horizon does! Consider your external view of me as I fall in.

If the black hole is eternal, events happening to me (by my watch) closer and closer to the time I fall through happen divergingly later according to you (supposing that your vision is somehow not limited by the discreteness of photons, or the redshift).

If the black hole is mortal, you'll instead see those events happen closer and closer to the time the black hole evaporates. Extrapolating, you would calculate my time of passage through the event horizon as the exact moment the hole disappears! (Of course, even if you could see me, the image would be drowned out by all the radiation from the evaporating hole.) I won't experience that cataclysm myself, though; I'll be through the horizon, leaving only my light behind. As far as I'm concerned, my grisly fate is unaffected by the evaporation.

All of this assumes you can see me at all, of course. In practice the time of the last photon would have long been past. Besides, there's the brilliant background of Hawking radiation to see through as the hole shrinks to nothing.

Also note that even in the case of an eternal black hole, it's a little ambiguous to say that the outside observer "measures" an infinite time for the infalling observer to cross the horizon--it's true that it takes an infinite time for the outside observer to receive a signal from this event, but the outside observer can still use a coordinate system (like Kruskal-Szekeres coordinates) where it only takes a finite coordinate time to reach the horizon.

 

[Dmitry67]

No, infalling observer falls to BH in the matter of minutes or seconds, so there is no question that BH can evaporate during his fall.

However, when BH slowly evaporates, it shrinks, and some of the light 'trapped' very close to the horizon, but outside of the BH, manages to escape. For an observer in infinity, it creates an illusion that falling object falls inside evaporating BH. But it is nothing more than an illusion: we observe a mixture of light from falling object, emitted long time ago, and hawking radiation, emitted recently, and both reach us at the same time.

 

[Finbar]

Ok I think your right Dmitry. But let's get it straight if I have a black hole space time that is simply losing mass and I neglect back reaction then a test particle couldn't get to the event horizon before the black hole evaporates. This is true you can do the calculation to show it. However if the black hole is gaining mass from the observer crossing the horizon then the radius increases such that the falling observer does cross the horizon.

But even so he will not reach the singularity until the black hole evaporates. He still remains in the proximity of the event horizon until the end.

 

[Trenton]

The short answer must be yes you can be in a stable orbit inside the event horizon. The horizon is only the radius at which the escape velocity is light. The velocity required for orbit is lower than this and is thus legal. The raduis at which one travel at the speed of light in a circular orbit is quite a bit less than that of the horizon. It would also be possible to describe an eliptical orbit that was part inside and part outside the horizon.

This gets easier to comprehend when one considers massive black holes. The gravity at the horizon is much lower with these as the radius is proportional to the mass while the gravity is subject to the inverse square law. A black hole of one solar mass would have gravity of 1.5 trillion times that of Earth at the horizon but a hole of a billion solar masses would only have 1500 times Earth gravity at the horizon.

Another obvious point to make is that the debate as to wether or not the universe has enough matter to eventually arrest it's expansion (if it really is expanding and not contracting) - is really the same as debating if the universe is one giant black hole.

 

[DrGreg]

The short answer must be yes you can be in a stable orbit inside the event horizon. The horizon is only the radius at which the escape velocity is light. The velocity required for orbit is lower than this and is thus legal. The raduis at which one travel at the speed of light in a circular orbit is quite a bit less than that of the horizon. It would also be possible to describe an eliptical orbit that was part inside and part outside the horizon.

That logic would be correct under Newtonian gravity, but under general relativity it isn't. In fact, for a static non-rotating black hole, there are no stable orbits below a radius of 1½ times the event horizon. See photon sphere.

 

[Dmitry67]

But even so he will not reach the singularity until the black hole evaporates. He still remains in the proximity of the event horizon until the end.

Ok, let's define what do you mean first. Do you understand that in GR there are 3 *dfferent* things which can be described by what you had said:

1. In the frame of freely falling observer
2. For an observer in infinity, based on his/her calculations
3. For an observer in infinity, based on the image of light he/she receives

For 2, you should also specify type of coordinates
Do you undertsand that these 3 approaches give DIFFERENT result?
It is important to understand that there is no clear meaning of "remains in the proximity" until you specify method - 1, 2 or 3.

 

[Finbar]

Ok, let's define what do you mean first. Do you understand that in GR there are 3 *dfferent* things which can be described by what you had said:

1. In the frame of freely falling observer
2. For an observer in infinity, based on his/her calculations
3. For an observer in infinity, based on the image of light he/she receives

For 2, you should also specify type of coordinates
Do you undertsand that these 3 approaches give DIFFERENT result?
It is important to understand that there is no clear meaning of "remains in the proximity" until you specify method - 1, 2 or 3.

Yes ok let's say that the proper time of the observer at infinity is t and she has "correct" equations to describe the physics up to some point where the mass of the black hole, defined at some point in time M(t) by a surface integral just outside the horizon, is of order the Planck mass. An important point is that given that at some point in time her theory breaks down she cannot actually define an event horizon since this evolves a knowledge of the entire spacetime. So I'm working with number 2. Then let's say we have some matter that crosses the apparent horizon at some time t. Now let's parametrize the geodesic of the in falling matter by its proper time v say. so there is a curve x(v), in our observer at infinities coordinates, say x(v) = (t(v) r(v)). We can then define the function M(v) = M(t(v)). What I'm saying is that M(v)=M_pl before r(v)=0. Although the proper time along the geodesic v may be very small when M(v)=M_pl the proper time for the outside observer t can reach the time needed for the black hole to loses almost all of its mass. The point is that dt(v)/dv diverges when r(v)=2GM.

What I'm getting at is that we cannot really know what happens to information, whether it is lost or not until we know the final stages of the evaporation process. I think that this is a little appreciated point in black hole literature. Maybe I'm wrong but I haven't seen anywhere that a geodesic that classically ends at a singularity will do so in a finite time t of the observer or at least before the mass reaches the Planck scale.

 

[Dmitry67]

Then let's say we have some matter that crosses the apparent horizon at some time t.

I am losing you here
1. what apparent horizon
A. apparent horizon of an infalling observer? he never crosses it as it always recedes in front of his as he progresses inside the BH
B. When falling observer crosses Schwarzschild radius (based on his calculations, as he does not feel anything strange when it happens)
C. or something different from perspective of observer at infinity?

2. some time - in what metrics? How 'some time t1' for falling observer is mapped to 'some time t2' for an observer at infinity?

 

[Finbar]

I am losing you here
1. what apparent horizon
A. apparent horizon of an infalling observer? he never crosses it as it always recedes in front of his as he progresses inside the BH
B. When falling observer crosses Schwarzschild radius (based on his calculations, as he does not feel anything strange when it happens)
C. or something different from perspective of observer at infinity?

2. some time - in what metrics? How 'some time t1' for falling observer is mapped to 'some time t2' for an observer at infinity?

t is the proper time of the observer at infinity so the in falling matter crosses the apparent horizon defined in coordinates (r,t, etc.) at some time t=t1 say.

I have two diffeomorphic manifolds so I can map any points (r, t etc.) to (v, etc.)


In any case physics is diffeomorphism invariant so if the matter does not reach the singularity before the black hole reaches the black hole evaporates in one set of coordinates it must be true in all coordinates.

 

[Dmitry67]

if t is the proper time of the observer at infinity - ok
now you should chose the coordinates which are not singular at Schwarzschild radius.
I like Eddington–Finkelstein coordinates, for example:

[PLAIN]http://www.valdostamuseum.org/hamsmith/DFblackIn.gif

 

[Finbar]

if t is the proper time of the observer at infinity - ok
now you should chose the coordinates which are not singular at Schwarzschild radius.
I like Eddington–Finkelstein coordinates, for example:

[PLAIN]http://www.valdostamuseum.org/hamsmith/DFblackIn.gif[/QUOTE]

Yeah that's what I use in practice. Ok so you know the Vaidya metric for an evaporating black hole? I'm basically talking about the causal structure of that metric in Eddington–Finkelstein coordinates. So just the schwarzschild metric in Eddington–Finkelstein coordinates with M->M(t) where t is the "advanced time". In this case the apparent horizon is just R=2GM(t). M(t) is found via the stephan Boltzmann law for the hawking radiation.

So then what I say is I can't get a photon that starts at r>2GM(t) to get to r=0 before M ->0.

 

[Dmitry67]

Yes, we agree on that, so we are on the same page
but I now don't the original question/paradox you had.

As BH evaporates, the cylinder becomes smaller in size.
I can draw the worldline of freely falling observer. In fact, it is on the diagram.
I can draw worldlines of light emitted from him at different times.
I can draw the trajectories of hawking radiation.
So where is a problem?

 

[Dmitry67]

So then what I say is I can't get a photon that starts at r>2GM(t) to get to r=0 before M ->0.

I tried to draw it (sorry for the quality)
Black - shrinking BH
red - falling observer
green - hawking radiation

 

[Finbar]

I tried to draw it (sorry for the quality)
Black - shrinking BH
red - falling observer
green - hawking radiation

No I don't understand

time is going up the page? The red line shouldn't get to r=0 before R=2GM(t)=0. Why are the red lines pointing in down the page?

 

[Dmitry67]

Check the image above (not the one I posted, but for the Eddington–Finkelstein coordinates posted before). There is a line of the fallinf object on it, and it hits singularity in finite proper time.

I think your question is: we know that hawking evaporation is very slow process - at least for an observer at inifinity. But is it also very slow for a falling observer? You seem to expect that black hole would literally evaporates in front of the falling observer.

Yes, in fact, the emission is very powerful (but very red shifted) so observer at infinity sees just a weak shadow of much more powerful process inside. However, as noticed by Hawking himself (cant give you a reference) falling observer can't observe hawking radiation! Like in Unruh effect, the very existence of the particles is frame-dependent.

I believe however that falling observer should see (much more intensive) hawking radiation from his (much smaller) apparent horizon, and this might explain why singularity does not exist.

 

[Dmitry67]

Ok, I have a proof
I am observer in infinity. My friend jumps into black hole. Then I wait 1000000y until it evaporates. I take the last glimps of light from the black hole and trace it back (mathematically) to its origin thru spacetime. Is it possible that BH evaporated 10000000y ago, even before my friend hit it? If so, what prevented the light from it for 1000000 years (in my frame) to reach me? It is not possible. SO evaporation is objectively long process - it takes long in all frames.

 

[Finbar]

Ok I may be on the verge of understanding or I've gone half mad. I think the essence of my confusion has been understand how the causal structure is related to the in flow of negative energy. The luminosity is defined as the derivative of the mass with respect to the in-going time v. One could write a metric where the luminosity is defined with respect to the Schwarzschild time. But these metrics aren't the same. v = t+r*(r) is the in going time. For the in going case M(v)=M(t,r) so the mass M becomes dependent on r in the Schwarzschild coordinates but only v in the Eddington-–Finkelstein.

What I think is if I wanted to model a star radiating I would say dM/du = -L where u=t-r* is the out going time due to a positive out flux of energy. But a black hole is modeled by an influx of negative energy so dM/dv =-L.

Does that make sense?

Actually I should make a new thread.

 

[Trenton]

Dr Greg,

Thanks for your reply. I read the 'photon sphere' entry in Wikipiedia with interest, however I am not convinced by this. I have two issues.

Firstly the issue of a rotating or non-rotating black hole. It is said that a rotating singularity 'drags' spacetime around with it. If this were the case any rotating body such as the Earth or the Sun would do the same - and even though the effects would be orders of magnitude smaller, this could be measured. Has it been?

It seems to me that spacetime dragging would require quantum gravity with carrier 'particles' and I am far from convinced of this concept. It starts by failing Occam's Razor by appearing to be completely un-nessesary. But more empirically there is a measurable effect that I think contradicts this notion. It takes 500 seconds for light to reach the Earth from the Sun and thus we see the Sun subtend an angle that is 500 seconds out of date. But satelites orbiting Earth act (so I am told) as though in a gravity field that is a superimpostion of that from the Earth, the moon and the Sun subtending the angle of its actual position without the 500 second delay.

Some interpret this as 'evidence' that gravity travels faster than light. I interpret it as evidence that gravity is an omni-present field due to mass (non-quantized) and that only changes in mass and thus gravity propogate at the speed of light. If the Sun was sudendly converted to light it would take us 500 seconds to notice and 500 seconds to exit the current orbit and fly off at a tangent.

The second issue is one of energy and the application of the Lorentz contraction. Has the said contraction been properly applied?

If we fire accelerated protons for example such that their path will go through the event horizon but miss the center. If the protons are accelerated to a million times their mass (easily achived by cosmic rays) time for the protons goes a million times slower and to the protons the black hole looks a million times smaller - We see them go through the horizon but actually they don't go anywhere near it as far as they are concened.

 

[Al68]

It is said that a rotating singularity 'drags' spacetime around with it. If this were the case any rotating body such as the Earth or the Sun would do the same - and even though the effects would be orders of magnitude smaller, this could be measured. Has it been?

Yes, a major experiment is currently underway. Gravity Probe B is currently in orbit measuring the frame dragging effect around earth.

http://en.wikipedia.org/wiki/Gravity_Probe_B