Trip To An Event Horizon

Spin-Analyser

At the very least they would create an optic boom from there own light as they break the light barrier. In the case of a small super massive black hole near the end of its life with lots of light from all the matter that had ever crossed the event horizon anywhere near that side of the black hole piling up in front of the event horizon, a falling object wouldn't see the light waves building up but would see a blinding flash when reaching the event horizon and then everything would be pitch black because no light from further in can reach them. That contradicts the idea that nothing special happens locally for a falling object when crossing the event horizon.

PeterDonis

They can't "break the light barrier"; nothing can go faster than light. It's true that the correct formulation of that rule is more complicated in curved spacetime, but none of the complications make an "optic boom" possible.

(It's also true that in a material medium, something like an "optic boom" is possible, when objects travel faster than the speed of light in the medium, which can be slower than the speed of light in vacuum. This is called Cerenkov radiation. But here we're talking about vacuum and the speed of light in vacuum, so none of that applies.)

Yes, this is correct, but it is not a "local" phenomenon. See below.

But light from further out still can, so it wouldn't be pitch black. It's true that the light coming in would be distorted, but again, that's not a "local" phenomenon, because the light has to come from distant locations to be distorted. Light from locations close to the infalling observer is not distorted, and the infalling observer can't tell from it that he is at or inside the horizon.

No, it doesn't. I already explained why: all the light in the "blinding flash" was built up at the horizon over a long period of time. That is not a "local" phenomenon; it requires the horizon and the hole to exist for a long period of time. "Local" means local in space *and* time. The same goes for the other items above, such as light coming in from distant locations being distorted--they're not local because they require a large extent of space, or a long period of time, or both. Your repetition of incorrect statements after they've already been corrected is part of what got your last thread locked.

Spin-Analyser

If I get an answer that doesn't make sense to me then I'm going to have to go over it again, sorry. So you agree that the fallers view of the interior is dark and all the light coming to them after passing the event horizon is coming from outside, and the second object would see the first object and all other previously falling objects disappear after reaching the event horizon?

If the light builds up in front of a falling object until it reaches the horizon then how is the flash not a local phenomenon? If not and light always moves away normally from the falling object locally, even when and after crossing the event horizon then the light from previously falling objects is crossing the event horizon from the falling objects perspective before the falling object reaches the event horizon. You keep switching between the two. Either there is a local optic boom or information is escaping from inside the horizon. It can't be both, and it definitely can't be neither.

DrGreg

Actually it is neither.

Imagine you are travelling down a straight road, and you are following a convoy of trucks in front of you. All you can see is the back of the last truck. Light from the other trucks never reaches you because it hits another truck. If the last truck were semi-transparent, you'd see light from the last truck and the one in front of it both hitting your eyes at the same time, although the light from the front truck would be delayed more than the light from the back truck. With standard opaque trucks, you just see the back truck.

It's no different if the convoy of trucks were falling into a black hole. At the exact moment you reach the event horizon, light from the last truck, emitted earlier when it reached the event horizon, will hit your eye. You won't see light from the other trucks, because that light would already have been absorbed by the front of another truck.

So you won't get a huge flash of light as you cross the event horizon. You will just see the most recent thing to have crossed that part of the event horizon. Light from anything else that crossed the horizon earlier would have been absorbed by something else crossing the horizon. And the thing you see as you cross the horizon (e.g the back of the last truck) is the same thing you would have already been seeing before you crossed the horizon and will continue to see after crossing the horizon.

Light from objects already inside the horizon never reaches the horizon; light at the horizon is only from the most recent object to have crossed it, emitted when it crossed it.

PeterDonis

I have no problem with this per se, but many of the "answers" that don't make sense to you are not answers I have given you; they are things you have added in yourself. For example:

No, I do not agree, and if you are deducing any of this from what I've said, then you are deducing incorrectly.

Here is what you can correctly deduce from what I have said: any light reaching an infaller after he has crossed the horizon and is inside it, must have been *emitted* from a larger radius than the infaller is at when he *receives* it. So, for example, if the second observer receives, inside the horizon, a light signal emitted by the first observer, then the first observer must have *emitted* that signal when he was at a larger radius than the second observer is at when he *receives* it. This is perfectly possible, so the second observer will *not* see the first observer disappear once he has passed inside the horizon.

This means that even outgoing light, inside the horizon, moves inward--but it moves inward more slowly than the infallers do (the first and second observers), so an infaller can "catch up" to a light signal that was emitted by another infaller who crossed the horizon before him and remains below him as he continues to fall.

Another example of you adding in things I have not said:

Light does *not* "build up in front of a falling object". I have not only not said this, I have explicitly *denied* it. I have already explained, twice, how the flash is a "local phenomenon". Please re-read my explanation (plus what I've added below about the precise meaning of "local"), and then if you still have questions, by all means ask them; but ask them about what I actually wrote, not about things I have not written and have specifically said already that I have not written. If you simply can't let go of this picture you have of light "building up in front of an falling object", then you're going to have to explain, in detail, how you are coming up with that picture. Just re-asserting it won't do.

It's very important to understand precisely what "locally" means. I haven't given a precise statement yet in this thread, so I'll do so now. Strictly and precisely speaking, "locally" means "in the local inertial frame of a specific observer, centered on a specific event on that observer's worldline".

I don't know how familiar you are with the terms I just used, so let me expand on the precise definition I just gave. An "event" is a point in spacetime--a particular point in space at a particular instant of time. Physically, we identify events by what happens at them; for example, "the second observer crosses the horizon" is an event. If we want to be more precise, we can locate events in spacetime by the intersection of two curves: for example, the event I just gave could also be described as "the intersection of the second observer's worldline with the horizon". Both the worldline and the horizon are curves in spacetime, and their intersection is a point, so that is sufficient to pin down a specific event.

A local inertial frame is a way of making a small patch of an arbitrary curved spacetime look like a small patch of flat Minkowski spacetime--the spacetime of Special Relativity. It works like this: first we pick a particular event on a particular observer's worldline, such as the one described above--the second observer crosses the horizon. Then we restrict ourselves to a small enough range in space and time around that event that the effects of spacetime curvature--tidal gravity--can be ignored. Then we set up standard inertial coordinates, the ones we use in flat Minkowski spacetime, with the event we picked as the origin, and with the observer's worldline as the time or "t" axis. We can in principle orient the spatial axes any way we like; here the easiest thing is to orient our "x" axis radially; i.e., positive "x" points radially outward, and the line x = 0 is the second observer's worldline. However, bear in mind that the surfaces of simultaneity--the surfaces of constant time--in this local frame are those of the infalling observer, so they are really small pieces of surfaces of constant Painleve time; they are *not* surfaces of constant Schwarzschild time (which would be the surfaces of constant time for observers "hovering" above the horizon).

Now, within this local inertial frame, physics works the same as it does in SR; we can basically ignore gravity (except that of course being "at rest" in this local frame means free-falling into the hole). So suppose that the first observer was just a little in front of us when he crossed the horizon. Then, since the second observer is at x = 0, the first observer will be at some slightly negative value of x; say x = -1, where we scale the x coordinate appropriately so 1 unit is a small enough distance. Then, what does the horizon look like in this local frame? Well, it is an outgoing lightlike surface, and it has to pass through the origin (since that's the event where the second observer crosses it), so it will be the line x = t (sloping up and to the right at 45 degrees--we are using units where the speed of light is 1).

So in this local frame, the first observer will cross the horizon at x = -1, t = -1. If he emits a light signal radially outward (towards us) at that event, it will have the same worldline as the horizon, so it will reach the second observer at the origin--x = 0, t = 0. The light moves away from the first observer at speed 1, in this local frame, just as in flat spacetime; and it moves towards the second observer at the same speed, just as in flat spacetime. Everything works just the same as if both were in flat spacetime--within this small local frame.

However, remember what I said above about the limits in space and time of this local frame; it has to be small enough that tidal gravity can be ignored. Suppose there were some other observer that crossed the horizon a long while before the first observer--long enough that tidal gravity can no longer be ignored. Then we could not fit that observer's worldline into the local frame we constructed above, and we could not use that frame to predict what would happen to light signals from it. We would have to use our global knowledge of the spacetime, such as the fact that the horizon is an outgoing lightlike surface, globally, to tell us that light emitted by that previous observer, at the instant he crossed the horizon, would be received by the first observer at the instant when *he* crossed the horizon.

I agree with DrGreg's post on this. In particular, he brings up a good point that I had not considered: if we have multiple objects crossing the horizon, each one absorbs the light from the previous one.

Spin-Analyser

Hold on, let me just get this straight. You're both saying that light doesn't build up at the event horizon because it catches a lift from the matter that falls in behind it and move quicker than it would do otherwise preventing a flash of light when an object crosses an event horizon?


Either,

when an object is falling towards a black hole the light from all previously falling observers is building up between them and the event horizon. Light from the falling object always leaves them locally at the full speed of light and slows as it approaches the event horizon in front of them, creating a build up of light waves at the front of the falling object in the same way that an object approaching the sound barrier experiences a build up of sound waves before creating a sonic boom. As the object approaches the event horizon the extended area where special relativity still applies locally (as in the area where light is still able to move away from the falling object at the speed of light up to a certain distance away) shrinks to nothing at the event horizon, because the falling object is passing the point where an object even one mm in front of them is falling away from them faster than light. Once the falling object is passed the event horizon any object further in is falling away from it faster than light so no light from previously falling objects can ever reach an object that has crossed the event horizon because the rate that objects in front of it accelerate away only increases once inside the event horizon, and no light from outside would be able to reach you because you'd be outaccelerating it, so it would be pitch black,

or,

you can't overtake your own light and there is always an extended area where special relativity still applies. If this is the case then a falling object can see a previously falling object crossing the event horizon before reaching it themselves.

yuiop

Neither of them are saying that there is no build up of light "because it catches a lift from the matter that falls in behind it". The infalling observer does not see an unusual flash of light at the event horizon, because he always sees light from the previously infalling observers, before, during and after passing the event horizon. Please see the first attached sketch titled "past" in Kruskal-Szekeres coordinates. The green curve is the path of the primary falling observer. The blue curves are the paths of previously infalling observers. The diagonal orange lines are light rays from the infalling observers. At event (a) outside the event horizon, the observer sees light from all 3 previously infalling observers. At event (b) the primary observer is at the event horizon and sees light from all 3 previously infalling observers that was emitted at the time they passed the event horizon. At event (c) the primary observer is inside the black hole and nearly at the singularity and he still sees light from all 3 previously infalling observers from when they were also inside the event horizon. Note that the primary observer never sees light from observers inside the event horizon, while he is outside the event horizon. Especially note that he never sees an unusual flash of light as he arrives at the event horizon. This is nothing to do with the light from previously infalling observers intersecting the light from other infalling observers. There still would be no unusual flash of light at the event horizon, if all previous infalling observers were transparent, or if they were slightly offset from the same radial path so that they are not hiding behind each other.

Not true. Again see the first attached drawing.
Also not true. In the attached diagram the observer at event (c) inside the event horizon still sees light from previously infalling observers, so it is not pitch black.

Also not true. The falling observer only sees previously infalling observers crossing the event horizon, when he himself is also crossing the event horizon. This is event (b) on the attached sketch.

The second attached sketch called "future" demonstrates that light (the diagonal orange lines) from the falling observer (green curved path) does not "pile up" at the event horizon. The inwardly directed light ray goes straight through the event horizon. You can also note that once light is emitted from the infalling observer, that the infalling observer never catches up with the light because his path never again intersects with the light paths.

Just for reference, the diagonal black line going from bottom left to top right labelled r=2m is the event horizon and the curve labelled r=0 near the top is the singularity of the black hole.

Attached Thumbnails

   

Spin-Analyser

I follow you up to (c). If a falling object is just a mm away from the event horizon then a falling object just inside the horizon is moving away faster than light relative to the more distant faller. That difference in relative velocity between objects depending on how close they are to the singularity only increases inside the horizon, so a falling object wouldn't be able to catch any previously falling light and no light from the outside could catch the falling object once inside the event horizon, so it would definitely be black.

Forget the light from previously falling objects. When approaching the event horizon the light emitted towards the black hole from the falling object moves away from them at the speed of light to start with, but slowing as it nears the event horizon. As the gap between the falling object and the event horizon decreases, the length of the valid frame of reference where light is moving away from the falling object at c decreases as well, creating an optic boom in the same way a planes own sound waves create a sonic boom.

If on the other hand there's always a valid frame of reference where light is moving away from the falling object at c then there must be a point close to the horizon where a falling object can tell whether or not previously falling objects have reached the event horizon yet.

PeterDonis

I'm not sure I understand what you're trying to convey here, but I think the answer is no.

Why do you think this? Once again, if you actually want to get the whole scenario here figured out, you're going to have to do more than just state this; you're going to have to explain, in detail, how you are arriving at this picture of what's happening.

One clarification, just to be sure it's clear: outgoing light only remains at the horizon if it is emitted *exactly* at the horizon. If it's emitted just a little bit outside the horizon, it moves outward--just not very much at first, though it still moves outward faster than a timelike object that is moving outward, like a rocket that has made a huge fuel "burn" to keep from falling below the horizon. If it's emitted just a little bit inside the horizon, it moves inward--just slower than timelike objects. So unless outgoing light is emitted exactly *at* the horizon, it doesn't "build up" in the same place.

Also, from the viewpoint of the infalling observer, the horizon is not "staying in the same place"--it is moving outward at the speed of light. (That's why light emitted outward at the horizon "stays" there.) So an infalling observer does not see light "build up" at the horizon; to him, the horizon is just a surface of light that moves outward past him.

Remember we are talking about *outgoing* light here. Outgoing light emitted from a falling object outside the horizon moves outward; it does not move towards the horizon. Light emitted in other directions does not behave the same way. In particular, *ingoing* light does *not* stay at the horizon; it moves inward, faster than anything else that's moving inward. So ingoing light certainly does not "build up" anywhere; it races ahead of everything else that's falling in.

I won't comment further on this portion of your post because I think you are working from mistaken premises. I would suggest re-thinking your picture of things in the light of what I've said above, and in particular being careful to distinguish ingoing from outgoing light.

Correct.

No, there is not an "extended area" where SR applies, in any curved spacetime. You can only apply SR within the confines of a local inertial frame, as I described in a previous post.

No, they can't. Once again, if you keep on asserting this without explaining, in detail, how you are arriving at this conclusion, we aren't going to get anywhere.

PeterDonis

No, it isn't. Why do you think it is?

One could say, perhaps, that an object freely falling just inside the horizon is moving faster than light relative to a *hovering* observer just outside the horizon; but that requires putting a particular interpretation on the behavior of the coordinates there. Such an interpretation is not required, and doesn't help you in making any physical predictions anyway. In any case, in a curved spacetime, there is no unique way of defining the relative velocity of objects except within the confines of a local inertial frame. If you do that for two objects both falling into the hole, one just inside the horizon and one just outside it, and you make sure that the hole is large enough compared to the distance between them that tidal gravity is negligible, then the two objects will have a relative velocity less than that of light, as evaluated within that local inertial frame. (In fact, you can make their relative velocity as small as you like by making the hole large enough.)

This effect is due to tidal gravity, so it can only be seen outside the confines of a local inertial frame. One of the consequences of that is that the allowed size of a local inertial frame gets smaller and smaller as you approach the singularity, because tidal gravity gets larger and larger.

If by "falling light" you mean "ingoing light emitted earlier", then yes, this is true; ingoing light always falls faster than anything else.

Wrong. I have already explained why this isn't true, and it looks like yuiop has too. Since ingoing light falls faster than anything else, it can catch an infalling object.

Correct.

Wrong. Ingoing light does not slow down; it "speeds up", just as everything else does. Ingoing light always falls faster than anything else. I think you are again confusing ingoing with outgoing light; they don't behave the same in this scenario.

I don't understand how you are arriving at this conclusion. Are you talking about ingoing light here? If so, again, it behaves differently than outgoing light.

DaleSpam

This is not correct. Suppose A is 1 mm away and B is 2 mm away from the EH of a supermassive black hole so that tidal effects are negligible, and suppose that they are free falling and at rest wrt each other. Then, because tidal effects are negligible they will still be at rest wrt each other as they cross and continue to free fall within the EH.

None of this is correct. Can you explain why you think any of this?

You seem to forget that the event horizon is a null surface, meaning that in a coordinate independent sense it moves outwards at c. Since the EH moves at c, in order for information about whether or not an object has crossed the EH to reach an observer outside the EH it would have to go faster than c in a coordinate independent sense.

yuiop

The Kruskal-Szekeres diagram I posted in my last reply clearly shows that the falling observer does catch light from previously falling objects whilst inside the event horizon so your statement here is not correct. Did you even look at the diagrams or just fail to understand them?
I have attached an amended KS diagram "past2" that includes the light rays from an object that falls after the infalling observer and light rays from an object that remains outside the event horizon at r=3m. The infalling observer continues to receive light from both these objects (and previously infalling objects) outside and inside the event horizon. All objects that he can see while outside the event horizon are still visible when he is inside the event horizon. Your statement that "it would definitely be black" inside the event horizon is also not correct.
This "optic boom" effect you describe is clearly not present if you study the attached KS diagram. The KS chart is effectively the reference frame in which the infalling observer considers himself to be stationary. In this chart, the speed of light is always constant and not just locally so there is no slowing down of light near the vent horizon. In Schwarzschild coordinates, there is a sense in which the coordinate speed of light is slowing down but in the Schwarzschild chart, the infalling observer is also slowing down as he approaches the event horizon, so it is not clear that the falling observer is catching up with his own light. Everything near the event horizon appears to be "bunched up" near the event horizon in the Schwarzschild chart so that region is difficult to analyse in that chart. This apparent bunching up disappears in the KS chart that expands the event horizon region. This occurs partly because radial distances between events are measured to be much greater by the freefalling observer, relative to the measurements by the Schwarzschild observer in the region near the event horizon.
That would be the KS chart in the diagrams I have attached ...
... but again you have reached the wrong conclusion. There is no point outside the event horizon, where the falling observer sees light from events at or below the event horizon. Again, this should be very obvious from the KS diagrams I have attached.

Attached Thumbnails

 

PeterDonis

No, it isn't. A "stationary" object in the KS chart is at a constant Kruskal X-coordinate, i.e., moving on a vertical line in your diagram; this is *not* the worldline of an infalling observer. (Your line "radial geodesic with apogee at r = 3m is one example of an "infalling observer", but it is not a vertical line.) A vertical line in the KS chart is not a geodesic.

No, it isn't; at least not if by "speed of light in this chart" you mean the coordinate speed of light, dX/dT, where X and T are the Kruskal space and time coordinates (the horizontal and vertical axes in your diagram). The coordinate speed of light varies with position on the chart. This is because the "scale" of the chart (the relationship between coordinate differentials and actual physical proper times/proper distances) is not constant; it varies with position on the chart.

I think what you are trying to say is that the worldlines of light rays are always 45 degree lines: ingoing light rays move up and to the left at 45 degrees, and ougoing light rays move up and to the right at 45 degrees. This is true, and it is one of the things that makes this chart so useful. But the *speed* of light is *not* constant; the "rate at which light moves along the 45 degree lines" varies with position on the chart, because the scale of the chart varies as above.

A better way to say this would be that light rays always move on 45 degree lines; their "direction" on the chart does not change. This also helps with better understanding the contrast with this...

A better way of saying this is that the "direction" of light rays on the Schwarzschild chart changes with radius; they aren't always 45 degree lines.

This is true; the distortion of the Schwarzschild chart near the horizon makes it practically useless for answering questions of this sort. (You can still do the computation in Schwarzschild coordinates, in principle, as long as things remain outside the horizon, but it's very tedious compared to the computation in other charts.)

Yes.

Is this true? I think it's the other way around, but it depends on how you interpret "radial distances" and how you interpret "distances measured by the freefalling observer". My interpretations are: pick two fixed radial r-coordinate values, r1 and r2. The radial distance measured between these two radial coordinates in Schwarzschild coordinates (i.e., along a line of constant Schwarzschild time) is *larger* than the distance measured between them in Painleve coordinates (i.e., a long a line of constant Painleve time). The latter is the most natural "distance" for the infalling observer (since Painleve time is the same as proper time for the infalling observer).

If you interpret "moving away at c" as "moving along a 45 degree line", yes. But I would say that light "moves away from a falling object at c" only within the local inertial frame of the falling object, at the event where it emits the light. Beyond that you get into issues with trying to compare speeds at spatially distant locations in curved spacetime; the KS chart does not avoid those issues.

Definitely agree. Please note: I agree with the main substance of your post, since the key claims you are making about light rays and infalling observers are the same ones I've been making. But I think it's important not to confuse things by making statements about the KS chart that are either untrue or, at least, require interpretation that doesn't really help with the main topic of this thread.

yuiop

I agree with the fine points you mention, but as you mention, they do not alter the main substance of the points I was trying to make. However, I was guilty of being a bit sloppy. Thanks for the correction!

I think the statements I made in the last post would be more accurate if we considered a non inertial observer in a rocket accelerating towards the black hole, in such a way that the worldine of the rocket is a vertical line in the KS chart.

Either way, the points about the ingoing observer:

1) still seeing light from objects outside the event horizon when inside the event horizon.

2) still seeing light from previously infalling objects when inside the event horizon.

3) not seeing light from events at or inside the event horizon, while outside the event horizon.

remain valid.

PeterDonis

That would be a "stationary observer" in the KS chart, yes. The acceleration profile would be a weird one.

Yes, definitely.