I Can the Size of the Event Horizon change faster than c?

1. Aug 18, 2016

Algren

Basically, if the size of the event horizon(EH) of a blackhole is R (radius) can the rate of change of this R or dR/dt be faster than the speed of light(c)?

When it comes to expansion of an EH of a blackhole: Does the event horizon immediately expand? For example, if one black hole consumes another, how long does it take for the new horizon to form? or for example, if a blackhole swallows a massive object very fast, can its event horizon expand faster than c?

WHen it comes to contraction of the EH: We know that only hawking radiation as of yet contracts the blackhole EH. So if the blackhole is small enough, will the decrease in the blackhole EH radius be maxed at c? How does it work?

For those of you who may say that it depends on the charge and rotation of the blackhole, assume a simple static one. I don't know the math.

2. Aug 18, 2016

tionis

I think the event horizon only travels at c when viewed locally. From afar, the event horizon appears to stay still.

3. Aug 18, 2016

Staff: Mentor

First, the "radius" of the horizon is not an actual physical radius; it's $\sqrt{A / 4 \pi}$, where $A$ is the area of the horizon. The hole does not have a "center" in the usual sense.

Second, the rate of change of the "radius" (or equivalently of the horizon area) depends on the coordinates you choose. It is not an invariant quantity and doesn't have any physical meaning. It is certainly possible to choose coordinates in which this rate is "faster than light", but this rate does not correspond to any physical measurement that anyone can actually make. It's just a calculated number.

The same is true for your other questions; there is no invariant number that expresses the "rate" at which the horizon expands as matter falls in or contracts as Hawking radiation is emitted. The horizon is not a "place" with a "position" that can move. It's an outgoing lightlike surface, a global feature of the geometry of spacetime.

4. Aug 18, 2016

Staff: Mentor

More precisely, the horizon will appear to be moving outward at $c$ to someone falling through it. This is because the horizon is an outgoing lightlike surface.

In appropriate coordinates, if the hole has nothing falling into it (and if we ignore emission of Hawking radiation), yes. But that's a restricted statement, as you can see; it's not completely general and isn't always true.

5. Aug 18, 2016

tionis

yes, the apparent horizon can jump faster than c, but I thought it was unphysical, is it not?

6. Aug 18, 2016

Staff: Mentor

More precisely, the apparent horizon can be spacelike in certain cases, whereas the event horizon is always null. But to translate "spacelike" into "jump faster than c" requires a choice of coordinates; it isn't an invariant statement.

No, it isn't. An apparent horizon is a 3-surface made up of trapped 2-surfaces. A "trapped 2-surface" is a 2-surface (a sphere) at which, locally, radially outgoing light does not move outward but stays trapped at the surface. This can be measured by local measurements; it's not an artifact of coordinates.

7. Aug 18, 2016

tionis

Thanks Peter. My understanding is that when the BH is not evolving, the horizon and the apparent horizon coincide, but even when there is a collision, like the OP is asking, the event horizon is always outside the apparent horizon. How can the apparent horizon be treated as something physical? This is probably more than I can chew, but hey lol.

8. Aug 18, 2016

Staff: Mentor

Yes.

It is when things are falling into the hole, yes.

Because trapped surfaces are physical; you can locally measure how the light rays behave. An apparent horizon is just a set of trapped surfaces.

9. Aug 20, 2016

MattRob

One of the things that helped me understand a horizon best is an example of a spherically symmetric "shell" of light rays all converging on a single point.

I couldn't find exactly what I was looking for, so I made this based off of what I remember and reason;

It's a spacetime diagram with increasing height representing later moments in time, and increasing distance to the right representing increased radial distance "r" from the center of a spherical coordinate system.

The blue line slanted towards the left at 45* represents a spherical shell of infalling light rays, with r = 2GM/c2 being its associated Schwarzchild radius. The red dashed line is the event horizon of the black hole that will form by the convergence of the "blue" light rays.

The black vertical line along points A and B is the worldline of some unfortunate free-falling observer. Now, I hope I'm not misleading here - I'm actually not entirely sure what happens after A along this line, but I presume that because this observer remains outside the light-cone of the infalling light rays, they are unaffected by the light rays, and thus aren't even aware that they're in an event horizon - not seeing the incoming light rays or feeling any gravitational influence at all. Any time before B, this region of space inside an event horizon is impossible to differentiate from the region of space outside of one, as it rests outside the light-cone of the infalling light. Only when impacted by the incoming light at B does the observer get any indication whatsoever of being inside an E.H.

In fact, he "falls through" the E.H. at A, since A is inside a trapped surface - nothing, not even a ray or light emitted outwards, can now escape from the singularity that hasn't even formed yet. In fact, there's not even a "black hole" yet (the light has not yet come within its Schwarzchild radius at time A), but there's an E.H.

If this observer lets out a light ray at time A, then it will arrive at the Schwarzchild radius at the exact moment the infalling light forms a black hole, trapping the light ray at r = 2GM/c2.

I could be wrong about the observer's worldline at T = c, though, where c is the point in this coordinate time where the infalling light passes r = 2GM/c2 - perhaps the curved spacetime makes my "outside the light's light-cone" reasoning fall apart here, but I don't think so? I hope/think Peter will correct me if I'm wrong, here.

In any case, the important takeaway is that this horizon forms before the black hole does - before the light has converged within r = 2GM/c2. The observer is trapped at A, despite the light rays not having even converged within the Schwarzchild radius yet. This really helped me understand that an event horizon isn't a sort of thing you can touch - it's just as real and physical as anything you can touch, but what it is is a set of coordinates that defines a region of spacetime that light cannot escape from. It's physical, but it's made entirely of spacetime geometry and a set of points in space, not any sort of solid or fluid structure or anything.

So, a spacelike interval is one with a greater spatial interval than temporal, so if something is "moving" along a spacelike interval, wouldn't that imply faster than c motion, and if something is spacelike in one frame, isn't it spacelike in all frames? Or is the invariance of the type of interval just a property of Minkowski spacetime?

10. Aug 20, 2016

Staff Emeritus
Let me come at this from another angle. The horizon is not a physical thing. It's a line of demarcation, like the equator or an orbit. There is nothing wrong with a non-physical thing moving faster than light: the intersection of a pair of scissors, for example.

11. Aug 20, 2016

Staff: Mentor

You have it correct. Since the only stress-energy present is the infalling shell of light rays, spacetime inside the shell is flat (by the shell theorem). So the free-falling observer will sit happily at rest, not falling inward at all, as long as he is inside the shell.

Yes, exactly.

No, it isn't. There is no trapped surface anywhere in this spacetime until the point at which the shell crosses the horizon.

How do we know that A is not inside a trapped surface? Remember the definition of a trapped surface: radially outgoing light stays at the same radius. But the red dotted line through A is radially outgoing light--and it's not staying at the same radius, it's moving outward. (In fact it's moving outward in a flat region of spacetime, as noted above.) So there is no trapped surface at A--and in fact there can't be one anywhere in the flat region, i.e., inside the infalling shell.

Yes, this is true. But it doesn't mean what you are thinking it does. See below.

He is in the sense of being below the horizon, but he is not in the sense of not being able to move outward. Nothing stops him from firing rockets and moving outward from A. He won't be able to catch up with the outgoing light ray through A, but he can still move outward. He can move outward and still be inside the horizon because the horizon itself is moving outward at A--the outgoing light ray through A shown by the red dotted line is the horizon, and it's moving outward.

The key thing you have left out of your analysis is: what happens when the free-falling observer passes the shell? You have shown what happens to the outgoing light ray that passes through A: it becomes trapped at $r = 2M$ and stays there forever. But note that that means the light ray is outside the shell. A timelike observer can similarly pass through the shell, although he won't be able to keep up with an outgoing light ray. So the free-falling observer will be able to pass through the shell; but after that point he will be forced to fall inward towards the singularity. (An outgoing light ray would also be forced to fall inward at that point, since it is well inside the horizon; but it would do so more slowly than the free-falling observer.)

No. Coordinates aren't physical things. The EH is a perfectly physical thing--it's marked out by outgoing light rays.

What the EH is not is local. It's impossible to tell just from local measurements where the EH is. For example, the observer that sees the outgoing light ray pass him at A cannot tell just from that local measurement that that outgoing light ray is an event horizon, so that he is now trapped inside one and can never escape to infinity.

The EH is usually viewed as a feature of the spacetime geometry, yes; but it's not a "set of points in space", because it's a null surface--a surface made up of outgoing null geodesics. A null surface cannot be viewed as a "set of points in space"; only a timelike surface can.

The apparent horizon is not a "thing" that "moves". It's just a 3-surface made up of a series of trapped 2-surfaces. Parts of that 3-surface are spacelike. (Other parts can be null or timelike.)

Yes.

12. Aug 20, 2016

MattRob

I think I was confused - mis-remembering a "trapped surface" as something where outgoing light rays will eventually re-converge, not as a point where they won't travel outward at all. Thanks for correcting that.

As for coordinates being physical, I think we're using different ideas of what constitutes "physical"? One of the reasons I call it that is because - shouldn't they be contra-variant? ie, something that will have a clear relationship with an event in spacetime. If a physical object emits a flash of light, this defines a particular physical event, and all observers will agree, regardless of choice of coordinates, whether this event is inside or outside the horizon, and thus there's an empirical distinction in-between events inside and outside the horizon, so I call it physical.

But the sense of "physical" you mean here is something more related to "carries information" / can or cannot go faster than light without tachyonic paradoxes?

So, if it's spacelike in one coordinate system it's spacelike in all, so no matter your choice of coordinates, if it's spacelike, it's moving faster than light, yes? So even though it's true in any coordinate system, without a coordinate system it's not true because you haven't defined a concept of distances/times, thus it's not invariant? Or am I missing something else? That's the only reasoning I could come up with for how something could be true in every possible coordinate system but not be invariant. Or is it just invariant?

13. Aug 21, 2016

Staff: Mentor

This is all correct. And it does not mean that coordinates are "physical". You even say that all observers agree "regardless of choice of coordinates". So whatever is "physical" in what you're describing here, it isn't coordinates, by your own statement.

No. The apparent horizon doesn't "move" at all, as I said in my previous post. More generally, a spacelike curve does not describe the "motion" of anything.

The statement of mine that you quoted was misstated, however. I should have said what I just said above. Sorry for the confusion.

14. Aug 21, 2016

MattRob

Going back to the start of this, I guess I should've specified better. The trouble is that calling it a "surface" is true, but that sort of creates the wrong picture in people's minds - they tend to envision something like the surface of water - but the event horizon isn't tangible in that sense that you could touch it - it's just a region in space, perhaps is the best way to describe it without accidentally creating the wrong mental image? A region in space made of nothing but spacetime itself, so there's nothing really "there".

The issue is that "the event horizon 'moving' faster than c" is surprising to someone, which seems to indicate that they're envisioning the event horizon as a tangible something, as opposed to spacetime geometry that can exist in a vacuum, so I guess the real issue is trying to convey that the E.H. is a real something, but is a result of spacetime geometry itself that can exist in a vacuum.

15. Aug 21, 2016

Staff: Mentor

The event horizon never "moves faster than c". You are confusing it with the apparent horizon. The event horizon is always a null surface.

No, the event horizon is a region in spacetime--a 3-D surface in 4-D spacetime. It is not a "region in space"; that is not an appropriate term for a null surface.

The apparent horizon, in portions where it happens to be timelike, could be viewed as a "region in space". But not in portions where it is null or spacelike. (A spacelike surface is more appropriately described as a "moment of time" if you have to find an ordinary language description.)