Event Horizon Radius: Is it Constant?

[DarkMattrHole]

TL;DR Summary

Or, exactly where is the event horizon now?

Let's say there is a black hole billions of miles away from earth, a hefty one such that a careless traveler could end up inside the horizon before noticing he'd been swallowed by the BH. Based on Earth observations the BH event horizon radius is r. We hop in a ship and go to a safe escape distance but so close that only our powerful engines can hover over the event horizon to take measurements before returning home - from our hovering ship we measure the radius of the event horizon again - do we read the same radius r as we did from earth? Other? thanks.

 

[PeterDonis]

The area of the event horizon is an invariant. So the radius of the horizon, defined as $r = \sqrt{A / 4 \pi}$, is also an invariant.

 

[DarkMattrHole]

Thanks, PeterDonis. That's somehow reassuring.

 

[Ibix]

Why do you ask?

Note that the radius of the event horizon depends on the mass of the hole, so measuring the hole's mass by studying orbits of test particles works. However, methods such as measuring the angular size of the hole require some care in application because of the curved paths of light.

 

[DarkMattrHole]

Thanks, Ibix. I don't recall where it was, but i had read that as you approach a black hole the event horizon recedes away from you, just as the horizon at sea recedes away from a ship. I easily could have misinterpreted what was being said. I'm still wondering how a surface where time stops can still grow if nothing ever quite reaches it.

 

[Vanadium 50]

but i had read that as you approach a black hole the event horizon recedes away from you

Where did you read this?

 

[DarkMattrHole]

It might have been a discussion on another sight somewhere as i recall. Seemed impossible so i came here to find out.

 

[PeterDonis]

I'm still wondering how a surface where time stops can still grow if nothing ever quite reaches it.

Both parts of this sentence are wrong. A black hole horizon is not "a surface where time stops", and things do reach the horizon.

See these Insights articles:

https://www.physicsforums.com/insights/schwarzschild-geometry-part-1/

https://www.physicsforums.com/insights/black-holes-really-exist/

The first link is to part 1 of a four-part series with a lot of good general information about the Schwarzschild spacetime geometry, which is the geometry that describes non-rotating black holes. The second is a more specific article that addresses the "how can things ever reach the horizon" issue.

 

[Ibix]

i had read that as you approach a black hole the event horizon recedes away from you, just as the horizon at sea recedes away from a ship.

The event horizon is an invariant property of spacetime and isn't affected by where you are. However, direct visual observation would have you see it expand and climb up to surround you as you approach. For an easy-to-calculate example, an observer hovering at 1.5 times the horizon radius would see the event horizon covering all the lower half of the sky. This isn't because the horizon has moved, but because light paths curve towards the hole. At that particular radius, all light paths reaching you below horizontal can be back-tracked to the hole - so you see black hole there.

I'm still wondering how a surface where time stops can still grow if nothing ever quite reaches it.

"Time stops at the horizon" is one of those things that's simply not true, as is "nothing ever reaches it". Imagine watching someone walking towards the north pole and noting that "northwards" and "forwards" mean the same thing. Would you conclude, therefore, that at the exact pole (where every direction is southwards) any attempt to move must be backwards and would cause that person to move simultaneously in every direction? It's a similar over-reliance on a coordinate system that doesn't work well at certain places to make predictions that leads to the "time stops, nothing crosses" claims in GR.

 

[pervect]

Schwarzschild coordinates are ill-behaved at the event horizon, a condition known as a "coordinate singularity". Basically, the Schwarzschild coordinates don't assign a finite label to the time an object crosses the event horizon. However, if an object has a clock, and it starts a finite distance away from the event horizon in free fall, a finite time will elapse on the clock carried by the object when it reaches the event horizon.

People frequently confuse the ill-behavior of the Schwarzschild coordinates with an ill-behavior of time itself, for instance thinking that time somehow stops. In many cases, the fundamental confusion relates to the notion of time itself. To address this confusion turns out to be tricky - the starting place is to distinguish between "proper time", the time one measures with a clock, and the notion of simultaneity, the notion of different clocks sharing a "now". People are used to a universal notion of now, but it turns out that this is incompatible with special relativity. The notion of "now" for an inertial frame in special relativity is well defined - but every different inertial frame has a different notion of now, a different notion of simultaneity.

Unfortunately, while it's easy to write a short stentence pointing out the difficulty, meaningfully communicating the concepts involved tends to be very hard.

 

[DarkMattrHole]

Thanks for the explanations. It makes a lot more sense that as a BH diver descends towards the horizon, the horizon would wrap itself around on all sides of you to eventually swallow you, leaving a last rays of universe shining in from a dot above before it closes it's maw around you forever - roughly speaking. I imagine that last ray of light would happen just before/as you cross the horizon? Or does light catch up to you from outside after you cross? Thanks.

 

[PeterDonis]

It makes a lot more sense that as a BH diver descends towards the horizon, the horizon would wrap itself around on all sides of you to eventually swallow you, leaving a last rays of universe shining in from a dot above before it closes it's maw around you forever

Nobody has said this, and it's not true for someone free-falling into a black hole.

It is true that, for an observer hovering at a constant altitude above the horizon, light rays coming in from the rest of the universe get confined to a smaller and smaller circle above the observer as the observer's altitude above the horizon gets lower and lower.

However, we are talking about an observer free-falling into the hole. What such an observer sees--the actual light rays reaching them from the rest of the universe--is very different. Andrew Hamilton has made some good videos of this:

https://jila.colorado.edu/hamilton/black-holes/journey-schwarzschild-black-hole

I imagine that last ray of light would happen just before/as you cross the horizon? Or does light catch up to you from outside after you cross?

Light from the rest of the universe continues to catch up with you after you cross the horizon, and all the way down to the singularity. A black hole does not prevent light from getting in; it only prevents light from getting out.

 

[Ibix]

Nobody has said this, and it's not true for someone free-falling into a black hole.

I think that depends very much on the angular momentum you have. What a free-falling observer sees is going to be an aberrated version of what a co-located hovering observer sees. Thus a radially moving observer will see the black hole expand up and surround them even more than a hovering observer. Hamilton's video shows the view of someone on an almost grazing trajectory, and the aberration appears to be countering the apparent expansion - although I suspect a glance in the rear view mirror would be a very different story.

 

[PeterDonis]

What a free-falling observer sees is going to be an aberrated version of what a co-located hovering observer sees

Only outside the horizon, since that's the only region where there are hovering observers. But the free-faller doesn't stay outside the horizon. So there are null geodesics that will reach their worldline that won't reach any of the hovering observers they passed while they were outside the horizon. So the free-faller will see things that no hovering observer he passed during his fall will see--it's just that the free-faller will see them after he's fallen inside the horizon.

(Hamilton's page used to have some videos that showed what a free-faller sees after passing beneath the horizon. I don't know if they're still there.)

Also note that there is a maximum angular momentum a free-faller can have and still fall beneath the horizon.

 

[Ibix]

Only outside the horizon, since that's the only region where there are hovering observers.

Sure. But it means that in the exterior region a radially infalling observer will see the horizon surround them as DarkMattrHole said, and I don't see that reversing as you cross the horizon. Instead you can see new structure, as you say. Past directed null paths terminate on the surface of the star infalling into the singularity (not that you'd actually be able to see it due to enormous redshift and absorption by previously infallen matter) or, in the case of an idealised maximally extended Schwarzschild black hole, the other asymptotically flat region.

Hamilton's page used to have some videos that showed what a free-faller sees after passing beneath the horizon. I don't know if they're still there.

The video you linked on Hamilton's site shows horizon crossing. It just doesn't show it for radial infall, but instead for a high angular momentum path.

 

[PAllen]

I think that depends very much on the angular momentum you have. What a free-falling observer sees is going to be an aberrated version of what a co-located hovering observer sees. Thus a radially moving observer will see the black hole expand up and surround them even more than a hovering observer. Hamilton's video shows the view of someone on an almost grazing trajectory, and the aberration appears to be countering the apparent expansion - although I suspect a glance in the rear view mirror would be a very different story.

I'm thinking the reverse. Aberration, compared to some other colocated observer relative to whom you are moving, concentrates light to the front of you (toward the direction of your motion relative to that observer). Thus, the horizon would look smaller to a radial infaller, and the rest of the universe larger, compared to a hovering observer.

 

[Ibix]

I'm thinking the reverse.

Ugh, yes, you're right. Presumably I have a sign error somewhere in the calculations I did.