# Black hole horizon confusion

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1. Apr 13, 2015

### wabbit

I am confused about black hole horizons and such common statements as "light cannot escape from inside the horizon".

The way I currently understand it is as follows :

1. Horizons are always relative to an observer, and what is called "the black hole horizon" is just a shorthand for "the black hole horizon relative to a hovering observer", i.e. one with fixed spatial Schwarzchild coordinates. For a large back hole, this is essentially his Rindler horizon.

2. For other observers, the black hole horizon is different and may not exist.

3. For a free falling observer headed directly towards the singularity (constant angular Schwarzchild coordinates), I am guessing there is a naked singularity - and that light can in fact escape from the Schwarzchild interior region (as defined by the hovering observer) from his viewpoint, as the usual descrption of freefall across the horizon suggests.

4. For an oberver in free fall on a circular orbit, I wonder if there is a horizon, perhaps the same as for the hovering observer ?

I am looking for clarification as to which of the above statements are correct, and if not in which way they are wrong.

2. Apr 13, 2015

### ShayanJ

The horizon of a BH is different from the Rindler Horizon. The Rindler Horizon is not a property of spacetime but an effect of being in an accelerated frame. A BH horizon is a property of spacetime and its existence is not depended on the frame of reference. Its just that in Schwarzschild coordinates, the horizon appears as a singularity but in other coordinates, its not a singularity.

3. Apr 13, 2015

### Staff: Mentor

The horizon is a property of the spacetime geometry, so it is not observer-dependent. Some events can be connected by lightlike and/or timelike geodesics and others cannot, and all observers in all frames will agree about which are which. So the answer to #1 and #2 is "no".

For #3, you have to remember that in the interior region (region II in Kruskal coordinates, which are way more convenient than Schwarzchild coordinates for understanding the geometry here) the Schwarzchild $r$ coordinate is continually decreasing along any timelike geodesic. Thus, the singularity is in the future of all observers and there are no hovering observers in this region - everyone is an infaller. A light signal from the singularity cannot reach any observer (although light signals from some infallers can reach other infallers in region II).

4. Apr 13, 2015

### Mentz114

The worldline of this observer is, for some constant $r$

$\frac{r-2\,M}{\sqrt{r}\,\sqrt{r-3\,M}}\partial_t + \frac{r\,\sqrt{M}}{\sqrt{r-3\,M}}\partial_\phi$

from which one may make certain inferences about the horizon.

From

Gravity: the inside story
Gen Relativ Gravit (2008) 40:2031–2036

5. Apr 13, 2015

### Staff: Mentor

I would prefer to say that it is a property of spacetime, just one that's not very interesting to anyone but the accelerated observer. There is a region of spacetime that has the the property that a lightlike geodesic from that region will not intersect the worldline of the accelerating observer, and the Rindler horizon is the edge of that region.

6. Apr 13, 2015

### wabbit

This is what I am doubting. I may well be wrong but it seems to me that horizon is very real for some observers, and inexistent for others.

7. Apr 13, 2015

### wabbit

But then this means that as I cross the horizon feet first in free fall, I cannot see my feet for at least a brief time. Weird given that it is said that nothing special happens.

8. Apr 13, 2015

### Staff: Mentor

You can save yourself much grief by identifying the invariant properties of the spacetime first... and then deciding what if anything they mean to various observers.

9. Apr 13, 2015

### ShayanJ

I think we're using different definitions of "property of spacetime". I meant its only the accelerated observer that sees such an effect and if there is no accelerated observer, there is no Rindler horizon. But I somehow see what you mean.

10. Apr 13, 2015

### CalcNerd

You can think of the event horizon as a (Black) balloon surface. Imagining this balloon as the actual surface of the event horizon of a non rotating black hole. On the outside, we can all look down and we will see something (not actually in or from the hole though). We will see matter as it falls in and if the tidal forces are great, the material will be ripped apart. If it doesn't fall straight in, it will begin a spiral in fall into the hole with some of it emitting energy (estimates are as high as 40%) via radiation (making this a very efficient way to generate energy, more so than nuclear fusion, in fact and why Quasars are so bright).

However, as the material approaches the surface (of this event horizon, it is NOT a membrane like our visual black balloon), it becomes closer to a higher gravitational field ie it slows down it TWO distinct ways from us. First it is in a much higher gravitational field (which does actually inflict a time dilatation on the objects) and second, the light coming is actually being red shifted and dimmed by this same gravitational field as well.

If we jumped in and if we could avoid spaghettification, (we would probably choose a huge hole), we would probably see at the horizon, a large radiation flash right at that imaginary membrane that hasn't actually fallen into the hole, but can't escape either. As we pass through, we are now at a gravitational well that would affect our own time dilatation as well. And we would be falling towards a naked singularity. It would likely be black, because everything would be moving at it, nothing coming out (nothing, meaning nothing, not even light). If we could fire rockets, strong enough to stabilize us, which isn't feasible and most texts indicate any energy expended would actually propel us faster down into the singularity at this point, NO MATTER which way you aim your thrusters, technically, there is only one direction, DOWN; we would see light coming at us, being sucked in too. But since we can't slow down, we probably would see darkness all the way down to our spaghettification point and hit the singularity sometime thereafter. We would be in an extreme relativistic environment, extreme speeds and gravity in a hole, headed for a singularity.

I am not thinking we are going to see anything like "Interstellar". I was disappointed with that movie.

Last edited: Apr 13, 2015
11. Apr 13, 2015

### wabbit

But then what is this invariant definition of the horizon ? After all, horizons are observer dependent, and the Rindler horizon occurs in Minkowski spacetime ?

12. Apr 13, 2015

### Staff: Mentor

That's not what happens. There is a never a moment when light that's reflected off your feet isn't reaching your eyes. It's just that the last light to reach your eyes before they fall through the horizon came from your feet just before they passed through the horizon, and the first light to reach your eyes after they pass through the horizon came from your feet just after they passed through the horizon. (We have to be very careful with the word "after" in this context - it's best to identify the the points on the worldlines of your feet and of your eyes at which the light is emitted and absorbed, speak about the relationships of these points).

13. Apr 13, 2015

### wabbit

Hmmm... I was thinking of this solution but could not muster the courage to calculate and check it. But this seems hard to reconcile with the near-flat spacetime across the horizon of a very large black hole, and the "nothing special happens". After all if the horizon gravity is 1g and I just jumped off my hovering spacecraft, right above the horizon, I must be passing through it at well below the speed of light, no ?

But that "after" and the changing nature of the radial coordinate is perhaps throwing my intuition off here. Still, don't both feet and head have decreasing $r(\tau)$ where $\tau$ is my proper time - both for feet and head since I am in nearly flat spacetime. But even if not, my feet still are essentially stationary wrt my head, in my frame.

Maybe what I am missing is that the horizon must be moving towards me at light speed, whatever my velocity relative to say that spacecraft.

There is something similar in the Rindler case. The horizon suddenly disappears when he jumps off his spacecraft, although he then sees that spacecraft accelerating slowly away say at 1g or less. Something drastic happens to the horizon with no change of velocity, just a change in acceleration.

The Rindler analogy (and nearly indistiguishable spacetime metric from that of a small region crossing a very large black hole horizon) is what leads me to imagine that the back hole horizon simply disappears (or at least recedes away very fast) for me when I jump off my spacecraft...

Last edited: Apr 13, 2015
14. Apr 13, 2015

### wabbit

I wish I knew how to do it :) though the discussion of the observer jumping straight into the black hole is already quite tough so I'm happy to leave the orbiting one alone for now...
This "always" would seem to support my assumption that like any other, the blackhole horizon is observer dependent.

15. Apr 13, 2015

### Mentz114

The orbiter must have $r>3M$ so they never can get to the horizon. But what is important is what happens to light sent by them to the center.

Padmanabhan argues the proper acceleration causes a local entropy change that must be balanced by a horizon. Laws of thermodynamics.

I am not 100% convinced.

16. Apr 13, 2015

### wabbit

@CalcNerd, so you are saying that despite the nearly flat metric, crossing the event horizon of a very large black hole is in fact very eventful and not like cruising in Minkowski spacetime ?

17. Apr 13, 2015

### wabbit

Seems reasonable to me. In any case the entropy is also relative to the observer so it may well change between hovering and free fall. I believe this does happen in the Rindler case, no ?

18. Apr 13, 2015

### bcrowell

Staff Emeritus
No, that doesn't happen. One way of stating the equivalence principle is that spacetime is always locally Minkowski. Therefore a local experiment like the one you're describing would not give an unusual result. As a free-falling observer, you see nothing special happening. You can see your feet at all times. An observer using Schwarzschild coordinates would explain this by saying that although the light from your feet didn't escape the event horizon, your eyes crossed the horizon in time to see that light.

19. Apr 13, 2015

### bcrowell

Staff Emeritus
There is a local definition of a naked singularity and a global definition. The local definition is basically that it's timelike, http://physicsforums.com/showthread.php?t=635641 , and this is observer-independent. The global definition is defined in terms of intrinsic properties of the whole spacetime, and is therefore also observer-independent.

The Schwarzschild singularity is spacelike, and this is an observer-independent property. It's also surrounded by an event horizon, and this is also an observer-independent property. Therefore by both definitions it's not a naked singularity.

20. Apr 13, 2015

### wabbit

Right... This makes sense. From his viewpoint, it does. But from my viewpoint, I see my feet just fine, light travels from them to my eyes unimpeded. Which I still find hard to interpret otherwise than, from my viewpoint, I amnot crossing a horizon, just cruising in nearly flat space. And if I am still accelerating sightly (below what's required for hovering), then I "see" yet another horizon, which neither freefalling nor hovering observers will agree with me "exists" - all three are correct, within their own frame/perspective.