B Possibility of Black Hole research

PeterDonis said:
It is perfectly possible for objects inside the horizon to exchange light signals.
How so?
 

Ibix

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You can see it's possible from a Kruskal diagram. It obeys the same rules as a Minkowski diagram - light travels on 45° lines and timelike paths are above 45°. There's a fairly narrow window for communication between horizon and singularity, but it can happen. And one way communication is always possible, at least in principle.

It's also obvious from the locally Minkowski nature of spacetime. If we are close enough together then we see spacetime between us as flat, so obviously we can communicate. That can't cut off dead outside "local" distances unless something catastrophic (like the singularity) happens to one partner.
 

pervect

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A practical example. Suppose you are on the stern of a somewhat long spaceship, and you're looking at an object in the bow of the spaceship. While you are looking, the front of the space-ship crosses the event horizon of a very large black hole. You snap snapshots just before, at the same time, and just after the front of the spaceship crosses the event horizon of a very large black hole. The black hole is so large that the curvature at the event horizon is negligible.

What do the photos show?

In all three cases, the photos show the object as normal, as if the black hole were not present. The assumptions we made that the black hole is large enough that curvature / tidal effects can be ignored is important for this result to be correct.

For completeness, I should add an explanation of what I mean by "at the same time". Due to the fact that space-time curvature can be neglected, there is essentially a local inertial frame of reference for the space-ship. It is in this local inertial frame that we judge the idea of "at the same time", via the usual convetions we apply to an inertial frame in Special relativity. You might be familiar with the Schwarzschild time coordinate "t". We are NOT using that time coordinate to judge "at the same time", we are using the notion of time associated with the local inertial frame of the ship. Due to the relativity of simultaneity, we need to specify a particular inertial frame to define the particular notion of simultaneity we are using.

To be specific, consider the photograph taken after the bow of the ship falls through the horizon. The r-coordinate of the light is decreasing with time, but slowly. The r-coordiate of the back of the ship is increasing more quickly. At some point, the light ray from the front of the ship hits the back of the ship. This will only happen after the stern of the ship has crossed the event horizon. But it is an example of "objects inside the horizon to exchange light signals."

Technically, we've only analyzed one signal of the two-way exchange. But the analysis is similar for the case of an exchange of signals in the other direction.,
 

timmdeeg

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The r-coordiate of the back of the ship is increasing more quickly.
You seem to mean 'is decreasing more quickly', right?

At some point, the light ray from the front of the ship hits the back of the ship. This will only happen after the stern of the ship has crossed the event horizon.
And is it correct that this happens in the moment the stern is crossing the event horizon as this light ray is "frozen" at ##r=2m## because it was emitted in the moment the bow was crossing the event horizon?
 
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pervect

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You seem to mean 'is decreasing more quickly', right?
Oops, yes.

And is it correct that this happens in the moment the stern is crossing the event horizon as this light ray is "frozen" at ##r=2m## because it was emitted in the moment the bow was crossing the event horizon?
I had three cases in the original - just before, just at, and just after the crossing. For the middle case, yes, the r-coordinate of the light signal emitted from the bow as it crosses the stern hangs there, and it is actually photographed when the camera reaches the horizon.
 
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timmdeeg

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I had three cases in the original - just before, just at, and just after the crossing.
To follow this up I think after the crossing the stern will receive light from the bow for a certain time only depending on the length of the ship. Light emitted later at the bow will no more hit the stern but the singularity.
 

PAllen

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To follow this up I think after the crossing the stern will receive light from the bow for a certain time only depending on the length of the ship. Light emitted later at the bow will no more hit the stern but the singularity.
Yes, but the stern will continue to “see” the bow until it reaches the singularity. That is, at any point at all in the stern’s world line, there will be a bow emission event just becoming visible.
 

timmdeeg

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Technically, we've only analyzed one signal of the two-way exchange. But the analysis is similar for the case of an exchange of signals in the other direction.,
Could you please elaborate a bit on the "other direction". By means of an Eddington-Finkelstein diagram I understand that the stern receives light from the bow but I fail to see the other way round. From the diagram it seems that the bow is in the past of the stern.
 

timmdeeg

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Yes, but the stern will continue to “see” the bow until it reaches the singularity.
Yes, until the stern "reaches the singularity". Just to avoid any misunderstanding.
 

PAllen

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Could you please elaborate a bit on the "other direction". By means of an Eddington-Finkelstein diagram I understand that the stern receives light from the bow but I fail to see the other way round. From the diagram it seems that the bow is in the past of the stern.
Just think light cones. There are stern world line events in the bow past light cone until the bow reaches the singularity. Thus there is a window of time in which two way signals can be exchanged between the bow and stern, with all relevant emission events occurring inside the horizon. And, of course, the bow therefore sees the stern until it hits the singularity.

It really becomes obvious in the Kruskal chart.
 

pervect

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Could you please elaborate a bit on the "other direction". By means of an Eddington-Finkelstein diagram I understand that the stern receives light from the bow but I fail to see the other way round. From the diagram it seems that the bow is in the past of the stern.
As I recall, the ingoing Eddington-Finkelstein diagram is different than the outgoing one. I'd recommend Kruskal-Szerkeres coordinates and/or a Penrose diagram, where light always travels at a 45 degree angle.

To quote Wiki
Wiki said:
Kruskal–Szekeres coordinates have a number of useful features which make them helpful for building intuitions about the Schwarzschild spacetime. Chief among these is the fact that all radial light-like geodesics (the world lines of light rays moving in a radial direction) look like straight lines at a 45-degree angle when drawn in a Kruskal–Szekeres diagram (this can be derived from the metric equation given above. ... All timelike world lines of slower-than-light objects will at every point have a slope closer to the vertical time axis (the T coordinate) than 45 degrees. So, a light cone drawn in a Kruskal–Szekeres diagram will look just the same as a light cone in a Minkowski diagram in special relativity.
It shouldn't be too hard to see this that for a short enough space-ship, light from the stern can thus reach the bow, as well as the reverse, because locally the light is always travelling "faster" than any material object (the slope of the infalling object is more vertical).

This is only true in the appropriate limit of a "short enough" ship or "big enough" Schwarzschild radius.

Unfortunately, I didn't find any good diagrams of an infalling object in KS coordinates to illustrate this graphically.

The intuitive argument here is that if light from the bow couldn't reach the stern, or vica versa, an observer would obviously know that they crossed the event horizon when they lost sight of the other end of the ship. This just doesn't happen - in the limit of a short ship (which is the same as the limit of a large blac hole), one can't tell from inside the ship that one has crossed the horizon.

As one approaches the singularity , the space-time curvature keeps going, and it gets harder and harder to meet the requirements of a "short enough" ship.
 

Ibix

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Unfortunately, I didn't find any good diagrams of an infalling object in KS coordinates to illustrate this graphically.
Your wish is my command:
1564093280537.png

This is the top right quadrant of a Kruskal diagram like the one (by @DrGreg, I believe) at Wikipedia. The fine grey line is the event horizon and the dark grey curve is the singularity (the shaded area doesn't correspond to anywhere in spacetime). The red lines show the final moments of three objects free-falling from rest at infinity. The two trailing objects emit blue light pulses as they cross the horizon - you can see that they catch up to the object in front, which has time to reply and receive a second reply (well, object one doesn't quite get to receive the second reply).

I actually set this up with a solar mass black hole (##R_S=3km##) and at about ##1.5R_S## the objects had Schwarzschild ##r## coordinates 100m apart.
 
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the ingoing Eddington-Finkelstein diagram is different than the outgoing one.
That's correct. However, the ingoing one is enough to analyze the experiment in question; it just isn't as easy to see the result as it is in a Kruskal diagram since only ingoing light rays go on 45 degree lines in the ingoing Eddington-Finkelstein diagram.
 

pervect

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I did want to say a bit more about the "short space ship" condition. Material objects must follow a timelike worldline, and because of this no such object can ever go as fast as a light beam. So if a light beam and a material object have a race, if they both start at the same time and place, the lightbeam will always "be ahead of" the material object in a race.

However, if the accelerating object has a head start, even though it's slower than the lightbeam, it's possible (though a bit surprising) that the ligihtbeam will never be able to catch up. This is a feature of hyperbolic motion, for instance.

In the "short ship", limit, though, the space-time diagram will be linear, because terms of the second order will be negligible compared to terms of the first order. And for a linear diagram, the faster object will catch the slower one.
 

Ibix

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Very nice! How did you calculalte and make the diagram?
I took the geodesic equations in Schwarzschild coordinates and substituted ##u## and ##v## for ##r## and ##t##. This works even at the horizon because the geodesic equation remains valid at and below the horizon - it's just its expression in Schwarzschild coordinates that goes wrong. I fed the result to one of scipy's numerical integrators and plotted results in pyplot. Confession: the light pulses were added by hand. My code can handle null geodesics, but I couldn't be bothered to do all the initialisation and intercept calculations in order to generate 45° straight lines.

I've had code lying around to do this for a while - I didn't do it all on Thursday night. It's not perfect, since you can't eliminate ##r## completely and you need to solve for it numerically. That generates numerical issues for long-lasting orbits since ##u## and ##v## grow while ##u^2-v^2## gets small, and ##r## depends on the latter (writing ##(u+v)(u-v)## postpones the problem, but ultimately doesn't solve it). And sometimes the integrator goes wrong near the singularity and objects skip merrily across its surface. But, those issues aside, the results are consistent with some analytical checking and a couple of things I found in the literature.
 

timmdeeg

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Sorry for being a bit late and thanks for your response.
It shouldn't be too hard to see this that for a short enough space-ship, light from the stern can thus reach the bow, as well as the reverse, because locally the light is always travelling "faster" than any material object (the slope of the infalling object is more vertical).
Indeed, this is very convincing.

In the meantime I've found some Eddington-Finkelstein diagrams in Robert Geroch's nice book "General Relativity from A to B", which is written for laymen. One comes close to the space-ship scenario. Observer A falls into the black hole and after a while B follows him. Their wordlines are parallel and its easy to see that B receives light from A for some time beyond the horizon. In all the diagrams light goes up as time increases, so light emitted by B can't reach A. I'm not sure how to resolve this. In contrast in KS coordinates which @Ibix shows in #38 the situation is crystal clear.
 
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In all the diagrams light goes up as time increases, so light emitted by B can't reach A. I'm not sure how to resolve this.
If B falls in long enough after A, then light signals he emits inward won't reach A before A hits the singularity. To have A see any signals from B, you have to make sure the time between them falling in isn't too large.
 

timmdeeg

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If B falls in long enough after A, then light signals he emits inward won't reach A before A hits the singularity. To have A see any signals from B, you have to make sure the time between them falling in isn't too large.
Heureka, I can see it now. This is Fig. 93 in Geroch's book I've mentioned. It shows C freely falling and outgoing light, ingoing light according to the light cones. If I paint the worldline of B who jumped first together with the past light cone very close to that of C then it turns out that indeed B can receive light of C. It's a bit tricky though and as mentioned by others it's far better to be seen in a KS diagram.
 

PAllen

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Heureka, I can see it now. This is Fig. 93 in Geroch's book I've mentioned. It shows C freely falling and outgoing light, ingoing light according to the light cones. If I paint the worldline of B who jumped first together with the past light cone very close to that of C then it turns out that indeed B can receive light of C. It's a bit tricky though and as mentioned by others it's far better to be seen in a KS diagram.
There is something wrong with that diagram. It purports to show a timelike world line that does not remain inside its own future light cone. That is impossible. Your confusion is caused by a quantitatively inaccurate diagram. Consider especially the future light cone emanating from W, and the proposed world line beyond W. This cannot be right. It would make the free fall spacelike.
 
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timmdeeg

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Consider especially the future light cone emanating from W, and the proposed world line beyond W. This cannot be right. It would make the free fall spacelike.
Yes. The diagram which I mentioned in #43 doesn't show light cones. I was drawing the two parallel timelike geodesics close to each other and inside the respective light cones, with the ingoing null geodesics having an angle of 45° with the space axis.
 

martinbn

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There is something wrong with that diagram. It purports to show a timelike world line that does not remain inside its own future light cone. That is impossible. Your confusion is caused by a quantitatively inaccurate diagram. Consider especially the future light cone emanating from W, and the proposed world line beyond W. This cannot be right. It would make the free fall spacelike.
It only looks like this. But in this diagram the light cones tilt as you go along the green/red world line. The closer to the singularity you are the more horizontal (open to the right) they are.
 

PAllen

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It only looks like this. But in this diagram the light cones tilt as you go along the green/red world line. The closer to the singularity you are the more horizontal (open to the right) they are.
I know, but don’t think that affects my critique. The black dashed lines are paths of radially outgoing light, thus they form the left boundary of light cones. Along one of these dashes lines, the width of a light cone is shown constant. Following these rules of the diagram, the light cone at event W does not enclose the purported free fall path.

[edit: I should note that the issue is that this diagram is not remotely an accurate representation of Eddinton-Finkelstein coordinates. ]
 
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martinbn

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I know, but don’t think that affects my critique. The black dashed lines are paths of radially outgoing light, thus they form the left boundary of light cones. Along one of these dashes lines, the width of a light cone is shown constant. Following these rules of the diagram, the light cone at event W does not enclose the purported free fall path.

[edit: I should note that the issue is that this diagram is not remotely an accurate representation of Eddinton-Finkelstein coordinates. ]
I see, but I thought the diagram is only schematic.
 

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