4 ms·

It is argued that Bob sees light from Alice's crossing of the horizon at the same instant Bob himself crosses. Isn't this true of all matter that enters? When B

by cvoss 2y ago

It is argued that Bob sees light from Alice's crossing of the horizon at the same instant Bob himself crosses. Isn't this true of all matter that enters? When Bob enters, he sees everything that ever fell into the black hole "before" him, at all once? Is it blinding? Does it fry and scramble Bob? Or is it so redshifted that Bob survives?

kobalsky 2y ago
this is my uninformed guess. why would bob see anything? I understood that the event horizon is a threshold, not a shell that you cross and suddenly can see inside. to see something photons have to bounce on something and reach our eyes, we stop seeing stuff inside the horizon because those photons don't bounce back and they are pulled into the singularity. my logic said that if light can't escape the horizon, then it can't escape alice to reach bob, even if he's inside the horizon, photons can't suddenly go backwards from alice until bob, they are being pulled further inside.
- Sharlin 2y ago
  No, we're talking about images of things, photons emitted by everything that has fallen in before, before they crossed the horizon.
- cyberax 2y ago
  The objects can emit photons by themselves. The problem is that (classically) when you cross the event horizon, the photons that you emit at just that moment will _stay_ _in_ _place_ forever.
  - raattgift 2y ago
    > (classically) ... photons Uhhh... one of those words should go. Let's keep it fully classical and drop "photon": we're interested in gravitational effects rather than quantum ones (uncertainty, fluctuations, tunnelling, details about scattering and more). Really what we want is something to illuminate (pardon the pun) interesting null geodesics, so a thin collimated beam -- a pencil of light -- will do. The relevant surface here is the apparent horizon, which can be measured by infalling apparatuses, and not the event horizon, the location of which is determined by the configuration of the entire spacetime. (See Visser PRD 2014 <https://journals.aps.org/prd/abstract/10.1103/PhysRevD.90.127502 https://journals.aps.org/prd/abstract/10.1103/PhysRevD.90.12...> or the corresponding arxiv version <https://arxiv.org/abs/1407.7295 https://arxiv.org/abs/1407.7295>). > stay in place forever Note the region inside the shell and the downward-pointing wedge in Fig 1. of Visser 2014 is flat Minkowski spacetime. Everything in that region will work like Special Relativity, as one would expect from the shell theorem. In particular, a pencil of light directed outwards through the apparent horizon in Fig. 1. will ride the AH down to the singularity, but a receiver intercepting the pencil of light just inside the shell would notice nothing unusual: spacetime is flat there. Eventually the collapsing shell collides with observers floating weightlessly inside it, and they have a bad time. But they can direct a pencil of light inwards just before the shell hits them.
    - cyberax 2y ago
      I mean, we can say "electromagnetic wave" instead of a "photon". It doesn't really change much in this case. I don't quite understand their diagram and their point. Are they looking at a region inside the collapsing shell of material? Then there's no contradiction here, the observer won't see anything until the reach the singularity (where the space-time stops being locally flat). Even if we consider an observer in the center of the shell, they'll only encounter the singularity after they get hit by the infalling matter.
mtdewcmu 2y ago
I'm going to guess. From Bob's perspective, Alice's ship would still be able to block light. So he wouldn't be able to see what was ahead of him through the back of Alice's ship; Alice's ship would occlude his view.
insapio 2y ago
https://en.wikipedia.org/wiki/Firewall_(physics) https://en.wikipedia.org/wiki/Firewall_(physics)
cyberax 2y ago
> Isn't this true of all matter that enters? Not quite. He will see the light emitted by _all_ of the matter that has fallen in before him, but only in an infinitely small area.
- mtdewcmu 2y ago
  A single photon can't be seen multiple times, right? So, if photon A goes into Alica's retina, then Bob can't see photon A. If a big, opaque object passes through the event horizon right in front of you, it would absorb or scatter the photons in its path, and you would not see them.
  - cyberax 2y ago
    Yep, I explained it a bit more here: https://news.ycombinator.com/item?id=42299891 https://news.ycombinator.com/item?id=42299891
    - mtdewcmu 2y ago
      So you don't see _everything_ that went in before you, mostly just the _last_ thing.
      - cyberax 2y ago
        You'll see kind of a "cone" where the light emitted from all the objects just ahead of you can be seen, but as you look further away, you can see only more and more recent objects.
daxfohl 2y ago
The article is missing on a couple points. The analysis assumes all Alice's light is emitted radially exactly outward from the center of the black hole. In reality, light is emitted in all directions, and anything emitted at even slightly different angles would get sucked into the black hole. But, Bob might still see it, because he can catch up with it. So when the article says Bob sees Alice cross the horizon when Bob crosses the horizon, it really means that Bob won't see any photons they emitted from inside the BH until Bob crosses into the BH. But similarly, one second prior Bob will be encountering photons Alice emitted roughly 0.99 seconds prior to crossing the horizon, and so on, because those are all getting redshifted too. It's similar to if you and the car in front of you are accelerating at the same rate, they have a small head start, and they've got someone throwing fastballs at you at 100mph. When you get to the point where you are going 100mph relative to the ground, you'll hit the ball that was thrown from exactly that spot relative to the ground. So, sure, crossing the 100mph barrier implies something interesting mathematically, but it's not something that the observer would particularly notice. The math for GR isn't exactly the same (baseballs won't redshift), and in particular there isn't even a piece of dirt to compare the photon's motion to, as the horizon is just a mathematically-defined "place", but to a first-order approximation, it's the same thing going on with photons and spacetime distortion. It's a continuous function. There's a bit more to it than can be explained in a comment, but the main thing to know is that (as far as we know) nothing special happens at the horizon if you're falling in.

by cvoss 2y ago

kobalsky 2y ago

this is my uninformed guess. why would bob see anything? I understood that the event horizon is a threshold, not a shell that you cross and suddenly can see inside. to see something photons have to bounce on something and reach our eyes, we stop seeing stuff inside the horizon because those photons don't bounce back and they are pulled into the singularity. my logic said that if light can't escape the horizon, then it can't escape alice to reach bob, even if he's inside the horizon, photons can't suddenly go backwards from alice until bob, they are being pulled further inside.

Sharlin 2y ago
No, we're talking about images of things, photons emitted by everything that has fallen in before, before they crossed the horizon.
cyberax 2y ago
The objects can emit photons by themselves. The problem is that (classically) when you cross the event horizon, the photons that you emit at just that moment will _stay_ _in_ _place_ forever.
- raattgift 2y ago
  > (classically) ... photons Uhhh... one of those words should go. Let's keep it fully classical and drop "photon": we're interested in gravitational effects rather than quantum ones (uncertainty, fluctuations, tunnelling, details about scattering and more). Really what we want is something to illuminate (pardon the pun) interesting null geodesics, so a thin collimated beam -- a pencil of light -- will do. The relevant surface here is the apparent horizon, which can be measured by infalling apparatuses, and not the event horizon, the location of which is determined by the configuration of the entire spacetime. (See Visser PRD 2014 <https://journals.aps.org/prd/abstract/10.1103/PhysRevD.90.127502 https://journals.aps.org/prd/abstract/10.1103/PhysRevD.90.12...> or the corresponding arxiv version <https://arxiv.org/abs/1407.7295 https://arxiv.org/abs/1407.7295>). > stay in place forever Note the region inside the shell and the downward-pointing wedge in Fig 1. of Visser 2014 is flat Minkowski spacetime. Everything in that region will work like Special Relativity, as one would expect from the shell theorem. In particular, a pencil of light directed outwards through the apparent horizon in Fig. 1. will ride the AH down to the singularity, but a receiver intercepting the pencil of light just inside the shell would notice nothing unusual: spacetime is flat there. Eventually the collapsing shell collides with observers floating weightlessly inside it, and they have a bad time. But they can direct a pencil of light inwards just before the shell hits them.
  - cyberax 2y ago
    I mean, we can say "electromagnetic wave" instead of a "photon". It doesn't really change much in this case. I don't quite understand their diagram and their point. Are they looking at a region inside the collapsing shell of material? Then there's no contradiction here, the observer won't see anything until the reach the singularity (where the space-time stops being locally flat). Even if we consider an observer in the center of the shell, they'll only encounter the singularity after they get hit by the infalling matter.

mtdewcmu 2y ago

I'm going to guess. From Bob's perspective, Alice's ship would still be able to block light. So he wouldn't be able to see what was ahead of him through the back of Alice's ship; Alice's ship would occlude his view.

insapio 2y ago

https://en.wikipedia.org/wiki/Firewall_(physics) https://en.wikipedia.org/wiki/Firewall_(physics)

cyberax 2y ago

> Isn't this true of all matter that enters? Not quite. He will see the light emitted by _all_ of the matter that has fallen in before him, but only in an infinitely small area.

mtdewcmu 2y ago
A single photon can't be seen multiple times, right? So, if photon A goes into Alica's retina, then Bob can't see photon A. If a big, opaque object passes through the event horizon right in front of you, it would absorb or scatter the photons in its path, and you would not see them.
- cyberax 2y ago
  Yep, I explained it a bit more here: https://news.ycombinator.com/item?id=42299891 https://news.ycombinator.com/item?id=42299891
  - mtdewcmu 2y ago
    So you don't see _everything_ that went in before you, mostly just the _last_ thing.
    - cyberax 2y ago
      You'll see kind of a "cone" where the light emitted from all the objects just ahead of you can be seen, but as you look further away, you can see only more and more recent objects.

daxfohl 2y ago

The article is missing on a couple points. The analysis assumes all Alice's light is emitted radially exactly outward from the center of the black hole. In reality, light is emitted in all directions, and anything emitted at even slightly different angles would get sucked into the black hole. But, Bob might still see it, because he can catch up with it. So when the article says Bob sees Alice cross the horizon when Bob crosses the horizon, it really means that Bob won't see any photons they emitted from inside the BH until Bob crosses into the BH. But similarly, one second prior Bob will be encountering photons Alice emitted roughly 0.99 seconds prior to crossing the horizon, and so on, because those are all getting redshifted too. It's similar to if you and the car in front of you are accelerating at the same rate, they have a small head start, and they've got someone throwing fastballs at you at 100mph. When you get to the point where you are going 100mph relative to the ground, you'll hit the ball that was thrown from exactly that spot relative to the ground. So, sure, crossing the 100mph barrier implies something interesting mathematically, but it's not something that the observer would particularly notice. The math for GR isn't exactly the same (baseballs won't redshift), and in particular there isn't even a piece of dirt to compare the photon's motion to, as the horizon is just a mathematically-defined "place", but to a first-order approximation, it's the same thing going on with photons and spacetime distortion. It's a continuous function. There's a bit more to it than can be explained in a comment, but the main thing to know is that (as far as we know) nothing special happens at the horizon if you're falling in.