Detection of red-shifted photons near an event horizon

[Grinkle]

TL;DR Summary

Why can one detect red shifted photons even after one knows the emitter has crossed the EH?

I realize that something I thought I understood about a beacon approaching a black hole I am unsure of.

My hypothetical -

I am at a safe distance from as simple a black hole as it makes sense to discuss. I launch a beacon at it, and I calculate that in one hour of my own proper time the beacon will cross the event horizon. I understand that I will never observe this crossing, for me its an unobservable event that I have to believe happens because I believe the theory my calculations are based on is correct.

The beacon is emitting a signal at some frequency, say 1kHz but it can be any number if picking a specific number helps the discussion.

As the beacon gets closer to the EH, I will see the signal frequency decrease because the signal is Doppler shifted as it makes its way out of the gravity well.

Assuming my calculations are correct, after one hour, the beacon has emitted the last photon any observer outside the EH will ever be able to detect.

So, is this correct, or at least not flatly wrong -

The reason I continue to detect photons from the beacon even after I am sure the beacon has crossed the EH is because light travels along spacetime geodesics and as one approaches the EH the geodesics between my viewpoint and the beacon are getting longer and longer, and right at the EH they become closed (and never reach me) and just infinitesimally prior to this they are essentially infinitely long. So the path length the beacon signal has to travel to reach me has been increasing towards infinity as the beacon has approached the EH.

 

[Ibix]

The simplest explanation is that the horizon crossing leaves your future light cone but never enters your past lightcone. So there is never an absolute sense in which it is in the past, so you should never be surprised to receive a signal from it. There is never a time when the last light signal would reach you.

In terms of geodesics, the point is that an event horizon is an outgoing null surface. Radially outgoing light signals emitted on the horizon hover there, they don't travel in loops.

 

[PeterDonis]

The beacon is emitting a signal at some frequency

If signals are emitted at a finite frequency, then there will be a last signal emitted before the beacon falls through the horizon since signals are emitted at discrete points on the beacon's worldline, and there will be some such point that is the last one above the horizon. This is why you receive a last signal in your scenario: because there is a last signal emitted.

The effect of redshift as the signal climbs back up to you from the beacon is just to make the signal harder for you to detect; in practice signals from the beacon will probably become undetectable by you well before the last signal the beacon emits above the horizon.

The reason I continue to detect photons from the beacon even after I am sure the beacon has crossed the EH is because light travels along spacetime geodesics and as one approaches the EH the geodesics between my viewpoint and the beacon are getting longer and longer, and right at the EH they become closed (and never reach me) and just infinitesimally prior to this they are essentially infinitely long.

Lightlike geodesics have spacetime length zero, so that standard concept of "longer" does not work here.

You could say that the elapsed time by your clock gets longer and longer between releasing the beacon and receiving successive light signals emitted by it as it gets closer and closer to the horizon.

 

[Grinkle]

Lightlike geodesics have spacetime length zero, so that standard concept of "longer" does not work here.

This may be a place where if I don't know the math, I am as far as I can come. I'll try to articulate my conceptual problems, acknowledging that many folks say "GR is hard" and "at some point you just need to understand the math".

Light travels at the same speed regardless of frequency, so the red shifted light is moving at c. If it doesn't in some sense need to cover more distance to reach me, how can it be taking longer and longer to do so?

If I can calculate the time on my clock when the last signal was emitted, and it was emitted at beacon-release + one hour, how is it taking the very red-shifted light so long to reach me, is my struggle. I thought I had it resolved with my incorrect geodesic picture.

You could say that the elapsed time by your clock gets longer and longer between releasing the beacon and receiving successive light signals emitted by it as it gets closer and closer to the horizon.

Isn't this to some degree circular reasoning? What I mean is one sees time slow down as objects approach the event horizon, and one interprets that as time dilation. Still, one calculates that despite the time dilation being observed, a last photon was indeed already emitted one hour after beacon release.

If my interpretation of my observations is that elapsed time by my clock is getting longer and longer between the beacon pings, then how can I also interpret my calculations as correct that the beacon is no longer emitting signals?

 

[Grinkle]

The simplest explanation is that the horizon crossing leaves your future light cone but never enters your past lightcone. So there is never an absolute sense in which it is in the past, so you should never be surprised to receive a signal from it. There is never a time when the last light signal would reach you.

This might be the best way to look at it. My attempts to relate it all back to things I am familiar with may be a dead end.

 

[jbriggs444]

I am at a safe distance from as simple a black hole as it makes sense to discuss. I launch a beacon at it, and I calculate that in one hour of my own proper time the beacon will cross the event horizon. I understand that I will never observe this crossing, for me its an unobservable event that I have to believe happens because I believe the theory my calculations are based on is correct.

The beacon is emitting a signal at some frequency, say 1kHz but it can be any number if picking a specific number helps the discussion.

As the beacon gets closer to the EH, I will see the signal frequency decrease because the signal is Doppler shifted as it makes its way out of the gravity well.

Assuming my calculations are correct, after one hour, the beacon has emitted the last photon any observer outside the EH will ever be able to detect.

So, is this correct, or at least not flatly wrong -

It is flatly wrong. It assumes a framework of absolute simultaneity -- that everything that happens does so at a particular knowable time.

Or, at least, it makes the erroneous assumption that specifying an observer is enough to take the ambiguity out of the notion of the time at which remote events occur. Not so. It takes more than just an observer. You also need a "foliation" -- an agreement on what events are simultaneous with what other events. Like full blown coordinate systems, foliations are arbitrary.

It is hard. Very difficult to remove the idea of a universe where things evolve in lock step as a single global absolute time passes. It takes work to expunge that pernicious idea from one's thinking.

 

[PeterDonis]

Light travels at the same speed regardless of frequency, so the red shifted light is moving at c.

This works in flat spacetime, but not in curved spacetime; there is no well-defined global concept of "speed" in curved spacetime. What you can say is that light rays travel on null/lightlike worldlines. You can figure out which worldlines those are from knowledge of the spacetime geometry.

If I can calculate the time on my clock when the last signal was emitted

Any such calculation requires assuming a simultaneity convention and is dependent on that convention. There is no invariant that corresponds to what you are describing here.

Isn't this to some degree circular reasoning?

No. It's the result of a calculation that is often assigned as a homework problem in GR textbooks.

one sees time slow down as objects approach the event horizon

No, that's not what one sees. What one sees is the light signals arriving, and taking longer and longer to arrive and being more and more redshifted. "Time slowing down" or "time dilation" is a coordinate-dependent interpretation of what you are seeing. And because it is coordinate-dependent, it's not something you should rely on to actually understand the physics.

If my interpretation of my observations is that elapsed time by my clock is getting longer and longer between the beacon pings

That's not an "interpretation". It is your direct observation. You are thinking of it backwards.

You directly observe, first, that you released the beacon in free fall at some time on your clock, and second, the arrival times of the beacon's pings emitted back to you. From that knowledge, and your knowledge of the spacetime geometry and the finite frequency of the pings (note that this frequency is in terms of the beacon's own clock, i.e., if the frequency is 1 kHz, the beacon emits a ping every millisecond by its own clock), you can calculate that there will be some last ping emitted by the beacon before it falls through the horizon, and you can calculate when it is emitted according to the beacon's clock and when you will receive it according to your own clock.

All of the things I just described are invariants. And notice that I never talked about "time dilation" at all. I didn't have to. It's not an invariant and it's not needed for any actual analysis of the scenario.

 

[Grinkle]

Thanks - very helpful and gives me some hooks for further study.