Seeing history when crossing event horizon

[Hill]

TL;DR

An observer crossing event horizon sees everything that has ever crossed the event horizon at that spatial point as they ('everything') were crossing the event horizon.

Looking at Kruskal diagram, it appears that light from all previous events of something crossing event horizon at that same point, reaches the falling observer when it crosses the event horizon. Is my interpretation correct?

 

[Orodruin]

Yes. That is when those crossing events come into the past lightcone of the crossing observer. Other crossing events will come into the past lightcone at a later time.

This is no stranger than an observer in Minkowski space having a multitude of events come into their past lightcone at some point.

 

[Hill]

This is no stranger than an observer in Minkowski space having a multitude of events come into their past lightcone at some point.

With the difference being that in Minkowski space the observer does not see its past events at the same spatial point

 

[Ibix]

I would say there's a null path along the horizon, yes, but if I'm not mistaken light from older crossings will be redshifted into oblivion. So I doubt you'd actually see anything much.

 

[Orodruin]

With the difference being that in Minkowski space the observer does not see its past events at the same spatial point

Neither does the infalling observer. The horizon can hardly be considered ”a spatial point” as it is a null surface.

 

[Orodruin]

I would say there's a null path along the horizon, yes, but if I'm not mistaken light from older crossings will be redshifted into oblivion. So I doubt you'd actually see anything much.

Depends on the crossing.

 

[Hill]

light from older crossings will be redshifted into oblivion

I don't see why it would be redshifted if both events are on the event horizon.

 

[Hill]

The horizon can hardly be considered ”a spatial point” as it is a null surface.

Both events are at the same $r$ and $\Omega$. This is what I mean by the same spatial point.

 

[Ibix]

I don't see why it would be redshifted if both events are on the event horizon.

Remember that spacetime is no longer stationary at the horizon, so "they're at the same altitude" doesn't work.

My argument for identical crossings is below. Orodruin is correct that you can pick very different horizon crossings (e.g. me approaching the black hole from rest at infinity, you from near lightspeed at infinity) and get a different result, but I think the vast majority of stuff you can see drops in to a black hole at relatively low energy-at-infinity values.

For identical infall trajectories, the proper time from horizon to singularity is the same. The one that drops in from region I first strikes the singularity to the left of the second. Draw the past lightcone of the second object's impact and you will see it intersects the first object before impact. Thus the second object sees less time on the first one's clock than on its own between horizon crossing and impact. Thus light must be redshifted.

 

[Hill]

Draw the past lightcone of the second object's impact and you will see it intersects the first object before impact.

I don't see this. Both the horizon and the lightcone are at 45 degrees and thus the lightcone of the second object intersects the first at the impact rather than before the impact.

 

[Orodruin]

Both events are at the same $r$ and $\Omega$. This is what I mean by the same spatial point.

Having the same coordinates is just the artefact of a particular coordinate system.

In a stationary spacetime, such as the exterior Schwarzschild solution, it makes sense to talk about ”same spatial point” as those related by the flows of the timelike Killing field. The timelike Killing field outside the horizon ($\partial_t$) is however not timelike on the horizon. It therefore makes little sense to refer to this as the ”same spatial point”.

 

[Hill]

Having the same coordinates is just the artefact of a particular coordinate system.

In a stationary spacetime, such as the exterior Schwarzschild solution, it makes sense to talk about ”same spatial point” as those related by the flows of the timelike Killing field. The timelike Killing field outside the horizon ($\partial_t$) is however not timelike on the horizon. It therefore makes little sense to refer to this as the ”same spatial point”.

Got it. Thank you.

 

[Ibix]

I don't see this. Both the horizon and the lightcone are at 45 degrees and thus the lightcone of the second object intersects the first at the impact rather than before the impact.

Impact on the singularity, I meant. You don't impact on the event horizon - there's nothing there to hit.

 

[Hill]

Impact on the singularity, I meant. You don't impact on the event horizon - there's nothing there to hit.

I see. I thought you use "impact" figuratively. (Albeit I don't think there is anything to hit at the singularity either.)

 

[PeterDonis]

For identical infall trajectories, the proper time from horizon to singularity is the same. The one that drops in from region I first strikes the singularity to the left of the second. Draw the past lightcone of the second object's impact and you will see it intersects the first object before impact. Thus the second object sees less time on the first one's clock than on its own between horizon crossing and impact. Thus light must be redshifted.

This argument is not valid.

First, the second object doesn't see the first object's impact. The last it sees of the first object is the intersection of the past light cone of the second object's impact and the first object's worldline, which will be before the first object's impact (how much before depends on how widely separated the two impacts are on the singularity). So the second object cannot draw the conclusion you draw in your next to last sentence above.

Second, the OP was asking about light emitted and received on the horizon. Everything you say in the quote above is irrelevant to that case. You have to look at light on the horizon and see whether it is redshifted at reception as compared to emission. An obvious heuristic argument strongly suggests that it is not: (1) the horizon is composed of null geodesics, which parallel transport the wave vectors of light rays along them; (2) for identical infalls, the "spacetime angle" between the infaller's timelike worldline and the horizon null geodesics will be the same; (3) therefore, for a family of identical infall worldlines, each crossing the horizon at different events, the measured frequency of horizon null geodesics will be constant.

 

[Ibix]

First, the second object doesn't see the first object's impact.

That's exactly the point. The lapse of proper time between event horizon and singularity is the same for both objects - call it $\Delta\tau$. So the second one sees the first one's clock advance by less than $\Delta\tau$ while its own advances by $\Delta\tau$. So the average is a redshift at least, even if I'm wrong about the horizon crossing event. I'll think about your argument for that.

 

[PeterDonis]

the second one sees the first one's clock advance by less than $\Delta\tau$ while its own advances by $\Delta\tau$.'

I see. Your use of the phrase "between horizon crossing and impact" confused me; you only meant it to apply to the second object, but I was taking it to apply to both.

 

[Dale]

With the difference being that in Minkowski space the observer does not see its past events at the same spatial point

This isn't a difference. An infalling observer does not see its past events either. If you draw the past light cone from the event where the infaller crosses the horizon, the infaller's past events are timelike separated from that event, not null separated.

 

[Hill]

This isn't a difference. An infalling observer does not see its past events either. If you draw the past light cone from the event where the infaller crosses the horizon, the infaller's past events are timelike separated from that event, not null separated.

Yes, my point was that they are timelike separated but have the same spatial coordinates. In Minkowski space, one cannot see what happened at the same location a year ago; one can rather see what happened a year ago one lightyear away.

 

[Orodruin]

Yes, my point was that they are timelike separated but have the same spatial coordinates. In Minkowski space, one cannot see what happened at the same location a year ago; one can rather see what happened a year ago one lightyear away.

Again, $r$ is not spatial on the event horizon. In fact, it is singular on the event horizon.

 

[Hill]

Again, $r$ is not spatial on the event horizon. In fact, it is singular on the event horizon.

Yes, I've already understood that. In the post above, I just wanted to clarify to Dale what was my point back then, in the post he has replied to.

 

[PeterDonis]

$r$ is not spatial on the event horizon.

This depends on the coordinates. In Painleve coordinates, $r$ is spacelike all the way down.

In fact, it is singular on the event horizon.

How is $r$ singular on the horizon? It's equal to $2M$.

 

[PeterDonis]

my point was that they are timelike separated but have the same spatial coordinates.

But the "timelike separated" part is just as important as the "same spatial coordinates" part. Different events on the horizon are not timelike separated, they are null separated. That is an invariant, so it's true even in a coordinate chart where $r$ is spacelike all the way down (like Painleve).

 

[Hill]

It's equal to 2M.

I understand that in Kruskal diagram r is spatial outside the event horizon, temporal inside the event horizon, and neither on the event horizon. Please correct.

But the "timelike separated" part is just as important as the "same spatial coordinates" part. Different events on the horizon are not timelike separated, they are null separated. That is an invariant, so it's true even in a coordinate chart where $r$ is spacelike all the way down (like Painleve).

Thank you for this clarification.

 

[Orodruin]

This depends on the coordinates. In Painleve coordinates, $r$ is spacelike all the way down.

This is true, but I bet OP is thinking of standard Schwarzschild coordinates.

How is $r$ singular on the horizon? It's equal to $2M$.

$g_{rr}$ goes to infinity as $r \to 2M$.

Regardless, the main point was that there is no timelike Killing field on the horizon, which is the only way I would use to make any sort of definition of "same place".

I understand that in Kruskal diagram r is spatial outside the event horizon, temporal inside the event horizon, and neither on the event horizon. Please correct.

Kruskal diagrams are based on Kruskal-Szekeres coordinates. Not Schwarzschild or Painleve coordinates. You can of course draw the r coordinate surfaces in a Kruskal diagram, but that is a different thing.

 

[Hill]

This is true, but I bet OP is thinking of standard Schwarzschild coordinates.

Kruskal diagrams are based on Kruskal-Szekeres coordinates. Not Schwarzschild or Painleve coordinates. You can of course draw the r coordinate surfaces in a Kruskal diagram, but that is a different thing.

I understand that $X$ and $T$ axis on the diagram are Kruskal-Szekeres coordinates, and that Schwarzschild's $r$ and $t$ are the hyperbolas and rays drawn there.

 

[Orodruin]

I understand that $X$ and $T$ axis on the diagram are Kruskal-Szekeres coordinates, and that Schwarzschild's $r$ and $t$ are the hyperbolas and rays drawn there.

The Schwarzschild r is timelike inside the horizon. The Painleve is not. They have the same level curves.

 

[PeterDonis]

I understand that in Kruskal diagram r is spatial outside the event horizon, temporal inside the event horizon, and neither on the event horizon. Please correct.

In the Kruskal diagram, $r$ isn't a coordinate at all, so it makes no sense to talk about whether it's spacelike, timelike, or null. That only makes sense for coordinates.

You can say that curves of constant $r$ are timelike outside the horizon, null on the horizon, and spacelike inside the horizon. Those statements are invariant since $r$ can be given an invariant definition as "areal radius" of 2-spheres.

 

[Hill]

The Schwarzschild r is timelike inside the horizon. The Painleve is not. They have the same level curves.

In the Kruskal diagram, $r$ isn't a coordinate at all, so it makes no sense to talk about whether it's spacelike, timelike, or null. That only makes sense for coordinates.

You can say that curves of constant $r$ are timelike outside the horizon, null on the horizon, and spacelike inside the horizon. Those statements are invariant since $r$ can be given an invariant definition as "areal radius" of 2-spheres.

Let's take this Kruskal diagram for example:

Isn't it unambiguous that events on the blue hyperbolas are timelike separated, events on the orange hyperbolas are spacelike separated, and events on the black rays are null separated?

 

[Ibix]

Let's take this Kruskal diagram for example:
View attachment 340300
Isn't it unambiguous that events on the blue hyperbolas are timelike separated, events on the orange hyperbolas are spacelike separated, and events on the black rays are null separated?

Yes. Those are lines of constant $r$. But the $r$ coordinate (in systems that have one) changes when you cross those lines, not move along them, and you can move across them in timelike, null, or spacelike directions at any point. So whether the $r$ coordinate is spacelike or not depends on how you define "keeping $t$ constant", and different coordinate systems do that in different ways.