I Seeing history when crossing event horizon

[Hill]

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.

I got it.
Thank you very much to all the helpers in this thread!

 

[PeterDonis]

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. And that corresponds to what I said about curves of constant $r$.

But that is not the same as saying this:

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

 

[Hill]

Yes. And that corresponds to what I said about curves of constant $r$.

But that is not the same as saying this:

Yes, it is not.
As I just said above,

I got it.
Thank you very much to all the helpers in this thread!

 

[Orodruin]

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, but those are surfaces of constant r, not the r coordinate lines (which result from keeping everything but r constant).

In Painleve coordinates, all coordinates are spacelike inside the horizon.

 

[Hill]

Yes, but those are surfaces of constant r, not the r coordinate lines (which result from keeping everything but r constant).

In Painleve coordinates, all coordinates are spacelike inside the horizon.

Understood.
I'll say again what I've posted above:

I got it.
Thank you very much to all the helpers in this thread!

 

[pervect]

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.

I'd agree that the image is faint, but I don't think it's due to redshift. By my analysis, light emitted right at the instant of the event horizon should just stay there, without gaining or losing any energy. As I see it, the problem is that the number of photons with just the right timing to stay on the horizon gets lower and lower. Unfortunately, I don't have a rigorous calculation, so my intuition could be incorrect.

If we consider a very long spaceship falling into an ultra-massive black hole, consider looking at a light on the bow of the ship from the stern. Nothing special happens to the light in the bow as we consider longer and longer ships (no redshift for an ultra-massive blackhole), but it gets fainter and fainter due to the distance from the bow to the stern. For a black hole with finite mass, there will be some redshift due to tidal forces, but that redshif will be present through the entire falling process, it won't suddenly start as the ship crosses the horizon.

 

[PeterDonis]

the problem is that the number of photons with just the right timing to stay on the horizon gets lower and lower.

Why do you think so? In a classical model (or, if you like, in the geometric optics approximation, which should be sufficient here), the light is just null worldlines that coincide with the generators of the horizon. There is no issue of "timing" to stay there; they just stay there.

 

[PeterDonis]

Nothing special happens to the light in the bow as we consider longer and longer ships (no redshift for an ultra-massive blackhole), but it gets fainter and fainter due to the distance from the bow to the stern.

If we consider the distance from the bow to the stern to be constant (which is not quite exact--see below), then this statement is incorrect; there is no redshift from bow to stern. The simplest way to see this is just to adopt Fermi normal coordinates centered on the ship. But if you insist on a global analysis, heuristically, the "gravitational redshift" from bow to stern is exactly cancelled by the Doppler blueshift due to the increase in the falling speed of the stern during the travel time of the light.

In actuality, the bow-stern distance is not constant due to tidal effects, which, since the ship cannot be infinitely rigid, will eventually stretch the ship radially. But for a sufficiently massive hole and a reasonably sized ship this correction is negligible until the ship is well below the horizon.

 

[pervect]

I had two different thought pictures in mind. But, I think that my first thought picture was confused. I was thinking that pulses emitted near the event horizon would be "smeared out" in time, but from a coordinate independent point of view, I don't think that makes any sense. The coordinate dependent viewpoint starts to get confused by the Schwarzchild coordinate singularities, so I think I'll stick with coordinate independent thinking.

The second thought picture is to re-frame the problem of "previous travelers" by imagining that all the travellers are on a very long spaceship falling into the black hole together.

Then, in the limit of an ultra-massive black hole, which requires that the length of the ship is less than the Schwarzschild radius of the black hole, the previous image of the bow of the ship falling in first is nothing mysterious, it's just like watching the bow of the ship from the stern. So - no redshift, for the ultra-massive case. But for any given black hole, eventually the analogy starts to get strained. For Sag A, that would be about a minute between travellers, if my numbers are not too far off.

For the case where the analogy makes sense, though, if we watch a beacon falling into a 4 million solar mass black hole with a 10 second head start, it should be the same as watching the same beacon in Minkowskii space from about 3 million kilometers (3 billion meters). There'll be no redshift, but it'll be fainter in intensity because of the inverse square law.

 

[PeterDonis]

it'll be fainter in intensity because of the inverse square law.

Ah, I see, you were thinking of a spherical wave front instead of a light ray (or "laser pulse", or whatever you want to call it--basically a tight beam that can be represented as a null worldline). Treating it that way means you can no longer use the geometric optics approximation, so the whole problem becomes more complicated. I think the complications don't matter as long as we are restricting consideration to a single local inertial frame in which the wave front can travel from bow to stern.

 

[Orodruin]

Why do you think so? In a classical model (or, if you like, in the geometric optics approximation, which should be sufficient here), the light is just null worldlines that coincide with the generators of the horizon. There is no issue of "timing" to stay there; they just stay there.

This is no stranger than the optical size of an object decreasing as you go further away. Except described in a rather roundabout manner.

 

[PeterDonis]

This is no stranger than the optical size of an object decreasing as you go further away. Except described in a rather roundabout manner.

Yes, after post #39 I understand what he meant--no redshift, just optical dimming due to distance apart.