Chain of lights falling into a black hole

[timmdeeg]

TL;DR

How would an observer who is fixed at the upper end of a light chain see the lights above the horizon, at crossing the horizon and below the horizon?

My simple reasoning is that he can see all lights as long as he is above the horizon , during crossing the horizon an overlap of all lights and below the horizon all lights again until the lowest light vanishes because it hits the singularity before it reaches him.

How would an Eddington-Finkelstein diagram show this situation? Would the worldlines of the lights be upwards parallel in the diagram?
It is disregarded that the chain could break and that only the central part of the chain is in free fall.

 

[Ibix]

Exactly what he sees depends on details of how the chain behaves under stress, but he sees nothing unusual at the horizon. Yes, he will see the light each element of the chain emitted as it crossed the horizon when he crosses, but remember that the horizon is a null surface, not a place. You always see what's on your past lightcone, and this is no different. I think he will see light from links below him refshifted. I can't see an immediate visualisation for what he sees from links above, but I would guess a blueshift.

 

[jbriggs444]

I think he will see light from links below him refshifted. I can't see an immediate visualisation for what he sees from links above, but I would guess a blueshift.

I would guess redshift in both cases. Locally we have tidal gravity stretching the chain. There is tension in both directions. Inbound light pulses from either direction are fighting their way "upward" toward the viewer.

The OP specified that the observer is at the top of the chain, but I agree that putting the observer at the midpoint is more illuminating.

 

[timmdeeg]

Yes, he will see the light each element of the chain emitted as it crossed the horizon when he crosses, but remember that the horizon is a null surface, not a place. You always see what's on your past lightcone, and this is no different.

If I understand it correctly at this very moment the past light cone of the observer is parallel to the horizon and hence he will see the lights simultaneously and nothing else, right?

 

[Ibix]

If I understand it correctly at this very moment the past light cone of the observer is parallel to the horizon and hence he will see the lights simultaneously and nothing else, right?

Well yeah, but he'll always see the lights simultaneously, assuming they're always-on and not blinking or something.

 

[timmdeeg]

Well yeah, but he'll always see the lights simultaneously, assuming they're always-on and not blinking or something.

The geodesics of the lights beyond the observer are increasingly tilted downwards after the light directly beyond him has passed the horizon, because they crossed the horizon at different times. So I assumed that at an instant of time the observers sees the lights from different directions. Wrong?

 

[Ibix]

Wrong?

If the chain is hanging straight down, the light is always going to be coming straight up.

I think you're thinking of the inner product of the light's and the observer's tangent 4-vectors, but in terms of "where you look" it's the three vectors in the observer's local inertial frame that's of interest.

 

[timmdeeg]

If the chain is hanging straight down, the light is always going to be coming straight up.

I think you're thinking of the inner product of the light's and the observer's tangent 4-vectors, but in terms of "where you look" it's the three vectors in the observer's local inertial frame that's of interest.

Ah, got it. Thanks for clarifying this matter.

 

[timmdeeg]

Sorry, still another question. Would for the observer the intensity of the accumulated light decrease if he comes close to the horizon and the lower end of the light chain starts to cross the horizon?

 

[jbriggs444]

Sorry, still another question. Would for the observer the intensity of the accumulated light decrease if he comes close to the horizon and the lower end of the light chain starts to cross the horizon?

Is this observer free falling? Or is he hovering?

If he is hovering, there is certainly a decrease in intensity of the light as he slowly approaches the horizon and, consequently, increases his proper acceleration while putting more and more tension on the chain. The lower end of the chain will red shift into imperceptibility. Pound-Rebka on steroids.

 

[timmdeeg]

Is this observer free falling?

Yes, observer plus chain are in free fall.

 

[jbriggs444]

Yes, observer plus chain are in free fall.

Then there is nothing locally significant about the event horizon.

 

[timmdeeg]

Then there is nothing locally significant about the event horizon.

Yes, as long as all lights of the chain are above the horizon. But I think its different once crossing has started. Please see #9.

 

[jbriggs444]

Yes, as long as all lights of the chain are above the horizon. But I think its different once crossing has started. Please see #9.

No. There is nothing locally significant there.

You will always and everywhere along the chain have radially outward light from various points along the chain co-existing and travelling together. The event horizon is not a special "location" in that sense. Locally the event horizon is a surface that sweeps past nearby observers at the speed of light.

The direction corresponding to "radially outward" gets confusing once the horizon passes. There is no null direction that leads outward. There is, however, a null direction that leads along the chain toward where the observer will be at a moment when he will be there.

 

[timmdeeg]

No. There is nothing locally significant there.

There is no direction that leads outward. There is, however, a direction that leads toward where the observer will be at a moment when he will be there.

Lets consider the moment where the light below the observer just passes the horizon. So, he can't see this light anymore. Would you agree so far?

 

[PeterDonis]

@timmdeeg, this is one of those cases where it really, really, really helps to draw a spacetime diagram. Trying to reason about this scenario without one is like trying to do complicated mathematical calculations using Roman numerals.

 

[Ibix]

So, he can't see this light anymore

No. He will be receiving light from it that was emitted above the horizon. He'll see light that was emitted from below the horizon when he gets there.

 

[timmdeeg]

No. He will be receiving light from it that was emitted above the horizon. He'll see light that was emitted from below the horizon when he gets there.

Ah, yes that's what I've been missing. In the meantime I think, I can see that from an Eddington-Finkelstein diagram too.

 

[pervect]

TL;DR Summary: How would an observer who is fixed at the upper end of a light chain see the lights above the horizon, at crossing the horizon and below the horizon?

My simple reasoning is that he can see all lights as long as he is above the horizon , during crossing the horizon an overlap of all lights and below the horizon all lights again until the lowest light vanishes because it hits the singularity before it reaches him.

How would an Eddington-Finkelstein diagram show this situation? Would the worldlines of the lights be upwards parallel in the diagram?
It is disregarded that the chain could break and that only the central part of the chain is in free fall.

You'll always see / photograph all the lights at the same time. See is ambiguious, I'm interpreting that as "photograph", rather than some sort of mental image you have by analyzing the photograph. Pretty much by definition, the photograph takes a picture at some instant, though I suppose that's idealized and you'd have to trace the optical path lenghts of the camera for path length variations under some specific definition of "simultaneity" if you really wanted to get all the fine details exactly correct. But enough of that, I'm getting off-track.

You'd probably want your camera offset of the straight radially line to make the lights have some angular separation to get a photograph that did NOT have all the lights overlapping. A radial camera would always photograph all the lights as overlapping, without any angular separation between them, making it rather annoying to look at.

A detailed analysis would be involved esp. with non-radial lights, but for a large black hole and a short (in comparison) string of lights, you won't notice any difference in how they behave due to the black hole. . I am assuing a free-falling observer falling into a black hole. The only difference will be due to the tidal forces of the black hole, and you can ignore those for the case I am calling "short strings". If you are interested in the tidal forces, and also want to put your camera off the radial line , I suppose you could do something equivalent to ray tracing, but it'd be a lot of work.

Note that you won't be able to tell when you reach the horizon by looking at the local lights, there will be nothing that special going on, so that limiting case should guide your thinking. I think you can tell by photographing the distant stars though, though I don't recall how they looked at the hyorizon crossing.

If you are interested in an accelerating observer, the approximation without tidal forces goes from flat Minkowskii space-time to the Rindler metric. Somewhere way back when I reparameterized the Schwarzschild metric radial coordinate to illustrate the connection between the Rindler and Schwarzschild metrics. Alternatively, we can say that when the acceleration goes to zero, the approximation goes to flat space-time.

See for example https://www.physicsforums.com/threa...ies-advanced-discussion.1007892/#post-6551199. There was a motivational analysis that led me to the specific form of the transformation equations I wrote, I might be able to dig up more if there was interest but I suspect I'm getting off-tract again.

 

[timmdeeg]

@perfect, thank you very much for the detailed and enlightening analysis of my question!