# Distance of event horizon from surface of black hole.

As a black hole grows. Does the event horizon move outward in larger steps than the surface of the black hole? If surface is even a proper concept. Would that hold true for any increase in size? Or is there a point where the horizon begins to get closer to the surface?

Another thought. Is the event horizon continuous? If you were to go into it. Would it always be that their is a point ahead of you that would be the event horizon from your point of view?

The event horizon is the surface of the black hole. And it’s a boundary, so it must by definition be continuous.

A classical (Schwarzschild) black hole will have an event horizon whose radius is proportional to the hole’s mass. Bigger hole = bigger radius.

Relativistic objects without a surface are called “naked singularities”; it is not clear how physical they are.

Eta- is it clear that a mathematically calculated boundary must be smooth? Seems to me you would need to prove that, taking into account the hole’s rotation, etc

An event horizon need not be smooth in the short term, but if it isn’t, it’ll become smooth very quickly. That’s part of the no-hair theorem.

Yep and it must be a closed surface too.

Idle question - smooth very quickly from the point of view of a distant observer, or smooth very quickly from an observer just outside the event horizon (or both)?

About this, if you are unfortunate to fall into a black hole, then, yes, no matter where you go you shall keep falling inward, and never back radially outward. You can’t see the singularity ahead by penetrating the hole; it’s not like the mass goes away as soon as you cross the event horizon.

Wouldn’t you never “cross” the event horizon, in that you never notice that anything is different. The event horizon will always be between you and the singularity. You might seem to always be getting asymptotically closer, but you would never actually “hit” it (unless the firewall speculation is correct), at least until you actually reached the singularity, but at that point, well no one knows what happens.

I think I saw someone mention that they did the math on a SMBH, and it was something like 30 seconds of proper time till you reached the singularity, so you won’t be wondering long (not that you find out, you just stop wondering), smaller ones will be much shorter times.

I think of it as easier to think of a black hole’s event horizon as the surface at which space itself is receding from the outside at greater than the speed of light. Just like the cosmological event horizon, where anything past that is receding faster than light, anything that falls into a black hole will be as well. Making things travel faster than light seems tricky, but that is because relativity gets mad if two things approach each other at greater than C. Relativity seems to have little issue with two things moving away from each other at greater than C, as long as neither of those things is approaching anything else at that speed.

There’s no particular indication when you cross an event horizon, and no particular way to detect one. All you can actually detect is whether you’re inside of it or outside, and even that (in principle) only if you’re willing to wait an infinite amount of time: You shine a light away from where you are, and if that light eventually reaches an infinite distance away, then you were outside. If it doesn’t, then you were inside.

Now, of course, you might be able to detect a black hole in other ways, and determine its mass and other parameters, and from that calculate where the event horizon ought to be. But it still won’t look any different from anywhere else.

So we might all be inside a black hole right now?

So many options here in terms of dirty jokes… my god.

Yup, though it would take a cosmic conspiracy for that to be the case without us noticing any clues. But hey, there aren’t any laws against cosmic conspiracies. Heck, it’s possible that we could hit the singularity before I finished this post, without us ever noticing anythi

So the event horizon is a surface in the same way the equator of Earth is a line?

Yes in the sense that both are mathematically-defined loci. The equator is defined in terms of the coordinates of latitude and longitude on the surface of the Earth; the event horizon of a black hole is defined in terms of the metric of space-time around the black hole.

The name “event horizon” is supposed to be suggestive of the definition, that events beyond the horizon cannot affect you. That’s why your buddy can never see you fall past the event horizon (despite the fact, as has been stated, that you can cross an event horizon just fine. Although you would not be fine after your atoms are ripped into spaghetti and/or bombarded by relativistic jets, whether that occurs inside or outside the horizon.)

I am curious…

If time slows as you approach the event horizon then, presumably, it stops when you reach the event horizon so you never actually cross it.

Put another way, as you approach the event horizon the outside universe seems to progress more rapidly. When you are very, very close to the event horizon you should see the outside universe race to its end in the blink of an eye.

Wouldn’t the black hole even evaporate from your perspective before you crossed the EH?

This is not true if you are just in free-fall, as opposed to trying to hover at a fixed radius from the hole by going near the speed of light.

The usual exercise is to consider a classical static Schwarzschild-type black hole and calculate how long (as in measured by your wristwatch) it takes you to fall in, and the answer is that it really does not take all that long, so there would not be time for the hole to evaporate. This does not contradict the fact that, by definition, nobody outside could see you cross the horizon.

If my understanding is correct, that’s essentially the firewall interpretation. That, as you approach the black hole, all of the hawking radiation that will be emitted in the future is emitted right now, in your face. The firewall interpretation has gotten quite a bit of traction, but is certainly not settled.

Either way, the black hole would “seem” to be evaporating in front of you, because it would be getting “smaller” as you get “closer”. If you are traveling towards a spherical surface, and that surface is always retreating from you, then that means the sphere is getting smaller.

Why not?

As I fall towards the BH you, a distant observer, see my clock tick ever more slowly than yours. The flip side of that is I see your clock tick more quickly than mine as I fall in. Presumably this time differential increases as I get closer to the event horizon.

If our universe were going to have a Big Crunch then it would be true that we were inside a black hole the entire time, and the entire universe would have been inside the event horizon.

However, it seems very unlikely that we live in a universe where a Big Crunch is going to happen, and therefore that we don’t live inside a black hole.

You mean, why doesn’t it take an infinite amount of time to fall into the black hole? Because I can calculate how long it takes, and the answer is a finite amount of time.This is an exercise worth working out in detail.

I do not know how to display proper formulas here, but in the simple case where you just start to free-fall radially inward from far away, your motion is given by dr/dt = -sqrt(2GM/r). Simple integration gives t = 2/3 (R[sup]3/2[/sup] - r[sup]3/2[/sup])/sqrt(2GM), where M is the mass of the black hole, G is the gravitational constant, r = 2GM/c[sup]2[/sup] is the (Schwarzschild) radius of the event horizon, and R is the distance at which you start your watch, say 1000 km or whatever.

No, you also see my clock ticking more slowly! Compare to the special-relativistic case where two observers are moving away from each other; they both observe a redshift.

ETA the redshift you observe from far away, as you cross the horizon, is not infinite, but the photons rising up from near the horizon to escape far away turn up infinitely redshifted. So the situation is not symmetric.