# Can you be in stable orbit inside a blcak hole?

1. May 17, 2010

### 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?

2. May 18, 2010

### Dmitry67

Yes, locally you dont 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.

3. May 18, 2010

### dx

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.

4. May 18, 2010

### DrGreg

Re: Can you be in stable orbit inside a black hole?

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).

5. May 18, 2010

### dx

Re: Can you be in stable orbit inside a black hole?

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.

6. May 18, 2010

### Finbar

Re: Can you be in stable orbit inside a black hole?

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?

7. May 18, 2010

### Dmitry67

Re: Can you be in stable orbit inside a black hole?

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.

8. May 19, 2010

### Finbar

Re: Can you be in stable orbit inside a black hole?

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.

9. May 19, 2010

### JesseM

Re: Can you be in stable orbit inside a black hole?

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:
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.

10. May 19, 2010

### 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.

11. May 19, 2010

### Finbar

Ok I think your right Dmitry. But lets 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.

12. May 19, 2010

### 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.

13. May 19, 2010

### DrGreg

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.

14. May 20, 2010

### Dmitry67

Ok, lets 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.

15. May 20, 2010

### Finbar

Yes ok lets 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 lets say we have some matter that crosses the apparent horizon at some time t. Now lets 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.

16. May 20, 2010

### Dmitry67

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?

17. May 20, 2010

### Finbar

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.

18. May 20, 2010

### 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 [Broken]

Last edited by a moderator: May 4, 2017
19. May 20, 2010

### Finbar

Last edited by a moderator: May 4, 2017
20. May 20, 2010

### Dmitry67

Yes, we agree on that, so we are on the same page

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?

21. May 20, 2010

### Dmitry67

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

#### Attached Files:

• ###### bh.jpg
File size:
10.8 KB
Views:
114
22. May 20, 2010

### Finbar

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?

23. May 21, 2010

### 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 cant 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.

24. May 21, 2010

### 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.

25. May 21, 2010

### 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 modelled by an influx of negative energy so dM/dv =-L.

Does that make sense?

Actually I should make a new thread.