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It's weird that they had the idea of firewall just in 2012. I had this exact idea maybe ten years prior while being a school student. A fairly obvious one if yo

It's weird that they had the idea of firewall just in 2012. I had this exact idea maybe ten years prior while being a school student. A fairly obvious one if you think of that - some photons will orbit the black hole.

privong 10y ago
> A fairly obvious one if you think of that - some photons will orbit the black hole. That some photons would orbit black holes is fairly well-known. There's an "innermost stable circular orbit"[0] ("photon sphere" for rotating black holes [1]), inside which stable (circular) orbits do not exist. These are at larger radii than the event horizon and so would be unrelated to the firewall proposal. [0] https://en.wikipedia.org/wiki/Innermost_stable_circular_orbit https://en.wikipedia.org/wiki/Innermost_stable_circular_orbi... [1] https://en.wikipedia.org/wiki/Photon_sphere https://en.wikipedia.org/wiki/Photon_sphere
zeroer 10y ago
Where photons orbit a black hole is called a Photon Sphere. https://en.m.wikipedia.org/wiki/Photon_sphere https://en.m.wikipedia.org/wiki/Photon_sphere It's at 1.5 times the radius of the event horizon for a non-charged, non-rotating black hole. But it's more of a theoretical than practical construct. You wouldn't expect to actually find any photons there because the orbit is unstable and any deviance from a perfect orbit would accumulate exponentially.
danbruc 10y ago
Photons orbit the black hole in the photon sphere which is located outside of the event horizon, at 1.5 times the Schwarzschild radius in case of a Schwarzschild black hole. As far as I know such orbits are also usually not stable. For a photon to orbit the black hole in the photon sphere it has to move tangentially, to reach the photon sphere the photon needs to move with at least some radial component and a photon can of course not just fire its rockets when reaching the photon sphere to enter the orbit. The idea of the firewall is very different from the idea of photons orbiting a black hole which was studied a long time ago.
- david-given 10y ago
  Wouldn't you get orbiting photons simply by having them emitted from falling bodies, i.e. hot gas? Any photons emitted in the right direction by incoming material in the right place would enter an orbit; and given that hot material emits photons in all directions continuously, and there's a steady supply of infalling material, there's going to be a continuous source of orbiting photons.
  - danbruc 10y ago
    IANAP so don't take my word for it, but I would assume the photons would have to be emitted EXACTLY tangentially at EXACTLY the photon sphere. Just a tiny bit earlier or later or with a tiny bit of radial momentum and the photon would eventually escape or fall into the black hole. And a real black hole is not a perfect Schwarzschild black hole, it will almost certainly have at least some charge or angular momentum, surrounding matter will change the metric somewhat. But I can not tell whether that makes it more or less likely for a photon to stay in orbit but I tend towards less likely because this seems to add more possibilities to randomly push the photon out of its orbit one way or another.
    - Retra 10y ago
      All orbits should 'eventually' decay as more matter comes in and perturbs the orbit. Stability is a matter of how quickly it decays, not if it decays.
      - MaulingMonkey 10y ago
        My understanding is that photons are uniquely unstable. By virtue of traveling at a constant (the speed of light), they lack the self-regulating orbital mechanics of matter (where going deeper into the gravity well speeds you up, and going farther away slows you down.) Dealing with spherical cows, before accounting for drag, you can describe the possible space for valid orbits as a volume, and you could describe the bounds of the velocity vector via another volume, for all positions and velocities at those positions that describe - at least on paper - mathematically perfectly stable orbits. Dealing with a photonic cow, before accounting for drag, you can describe the possible space for mathematically stable orbits as not a volume but an area - the surface of a sphere. The velocity vector has a very specific magnitude (c), and a very limited range of possible directions (exactly perpendicular to the normal of the sphere's surface). The shape you would use to describe the bounds of this velocity vector have 0 volume, 0 area. It only has a length. Both cows can be further perturbed by drag, tidal effects, non-point gravity sources, etc. to further worsen the problem. Individual collisions with particles of space dust will push the spherical cow from one mathematically stable orbit to another, until eventually they add up enough to most likely deorbit it. Individual collisions with particles of space dust will instead take the photonic cow from one mathematically unstable orbit to another, with a decent chance of it ending up on an escape trajectory instead.
TheOtherHobbes 10y ago
It's potentially more complicated than that. Someone on Reddit asked a brilliant question a while back - what happens to quantum fields at the event horizon? In QFT, fields are everywhere. But to support a field, you need a mechanism that allows causal propagation - which is exactly what isn't allowed across an event horizon. So at the very least you have a discontinuity where three and possibly all four fundamental forces stop working, and which is separate to any hypothetical relativistic singularity. Whatever is left is going to be some kind of unimaginably weird sub-quantum soup. I don't know if that's the same firewall that was invented in 2012. But the takeaway is that relativity isn't complete enough to model black holes. You absolutely need to include quantum effects - and when you do, things get very strange indeed.
- danbruc 10y ago
  In QFT, fields are everywhere. The fields in quantum field theory are mathematical tools, they are not physical entities. But to support a field, you need a mechanism that allows causal propagation - which is exactly what isn't allowed across an event horizon. But that is only a one-way thing - the future light cones of events inside the event horizon are contained inside the event horizon but the future light cones of events outside the event horizon certainly overlap the inside of the black hole.
  - vanderZwan 10y ago
    > The fields in quantum field theory are mathematical tools, they are not physical entities. Still, they only maintain physical relevance as long as they are continuous, no? Otherwise you literally have a break in reality.
    - danbruc 10y ago
      Do you mean that they can not have holes?
    - guitarbill 10y ago
      > a break in reality. Such as a singularity (e.g. gravitational)? I think in Physics (just as when analysing functions), the interesting things happen when you approach such limits.
losvedir 10y ago
Perhaps, then, you should consider that you're misunderstanding the situation? A firewall is not "orbiting photons". See https://en.wikipedia.org/wiki/Firewall_(physics) https://en.wikipedia.org/wiki/Firewall_(physics).

privong 10y ago

> A fairly obvious one if you think of that - some photons will orbit the black hole. That some photons would orbit black holes is fairly well-known. There's an "innermost stable circular orbit"[0] ("photon sphere" for rotating black holes [1]), inside which stable (circular) orbits do not exist. These are at larger radii than the event horizon and so would be unrelated to the firewall proposal. [0] https://en.wikipedia.org/wiki/Innermost_stable_circular_orbit https://en.wikipedia.org/wiki/Innermost_stable_circular_orbi... [1] https://en.wikipedia.org/wiki/Photon_sphere https://en.wikipedia.org/wiki/Photon_sphere

zeroer 10y ago

Where photons orbit a black hole is called a Photon Sphere. https://en.m.wikipedia.org/wiki/Photon_sphere https://en.m.wikipedia.org/wiki/Photon_sphere It's at 1.5 times the radius of the event horizon for a non-charged, non-rotating black hole. But it's more of a theoretical than practical construct. You wouldn't expect to actually find any photons there because the orbit is unstable and any deviance from a perfect orbit would accumulate exponentially.

danbruc 10y ago

Photons orbit the black hole in the photon sphere which is located outside of the event horizon, at 1.5 times the Schwarzschild radius in case of a Schwarzschild black hole. As far as I know such orbits are also usually not stable. For a photon to orbit the black hole in the photon sphere it has to move tangentially, to reach the photon sphere the photon needs to move with at least some radial component and a photon can of course not just fire its rockets when reaching the photon sphere to enter the orbit. The idea of the firewall is very different from the idea of photons orbiting a black hole which was studied a long time ago.

david-given 10y ago
Wouldn't you get orbiting photons simply by having them emitted from falling bodies, i.e. hot gas? Any photons emitted in the right direction by incoming material in the right place would enter an orbit; and given that hot material emits photons in all directions continuously, and there's a steady supply of infalling material, there's going to be a continuous source of orbiting photons.
- danbruc 10y ago
  IANAP so don't take my word for it, but I would assume the photons would have to be emitted EXACTLY tangentially at EXACTLY the photon sphere. Just a tiny bit earlier or later or with a tiny bit of radial momentum and the photon would eventually escape or fall into the black hole. And a real black hole is not a perfect Schwarzschild black hole, it will almost certainly have at least some charge or angular momentum, surrounding matter will change the metric somewhat. But I can not tell whether that makes it more or less likely for a photon to stay in orbit but I tend towards less likely because this seems to add more possibilities to randomly push the photon out of its orbit one way or another.
  - Retra 10y ago
    All orbits should 'eventually' decay as more matter comes in and perturbs the orbit. Stability is a matter of how quickly it decays, not if it decays.
    - MaulingMonkey 10y ago
      My understanding is that photons are uniquely unstable. By virtue of traveling at a constant (the speed of light), they lack the self-regulating orbital mechanics of matter (where going deeper into the gravity well speeds you up, and going farther away slows you down.) Dealing with spherical cows, before accounting for drag, you can describe the possible space for valid orbits as a volume, and you could describe the bounds of the velocity vector via another volume, for all positions and velocities at those positions that describe - at least on paper - mathematically perfectly stable orbits. Dealing with a photonic cow, before accounting for drag, you can describe the possible space for mathematically stable orbits as not a volume but an area - the surface of a sphere. The velocity vector has a very specific magnitude (c), and a very limited range of possible directions (exactly perpendicular to the normal of the sphere's surface). The shape you would use to describe the bounds of this velocity vector have 0 volume, 0 area. It only has a length. Both cows can be further perturbed by drag, tidal effects, non-point gravity sources, etc. to further worsen the problem. Individual collisions with particles of space dust will push the spherical cow from one mathematically stable orbit to another, until eventually they add up enough to most likely deorbit it. Individual collisions with particles of space dust will instead take the photonic cow from one mathematically unstable orbit to another, with a decent chance of it ending up on an escape trajectory instead.

TheOtherHobbes 10y ago

It's potentially more complicated than that. Someone on Reddit asked a brilliant question a while back - what happens to quantum fields at the event horizon? In QFT, fields are everywhere. But to support a field, you need a mechanism that allows causal propagation - which is exactly what isn't allowed across an event horizon. So at the very least you have a discontinuity where three and possibly all four fundamental forces stop working, and which is separate to any hypothetical relativistic singularity. Whatever is left is going to be some kind of unimaginably weird sub-quantum soup. I don't know if that's the same firewall that was invented in 2012. But the takeaway is that relativity isn't complete enough to model black holes. You absolutely need to include quantum effects - and when you do, things get very strange indeed.

danbruc 10y ago
In QFT, fields are everywhere. The fields in quantum field theory are mathematical tools, they are not physical entities. But to support a field, you need a mechanism that allows causal propagation - which is exactly what isn't allowed across an event horizon. But that is only a one-way thing - the future light cones of events inside the event horizon are contained inside the event horizon but the future light cones of events outside the event horizon certainly overlap the inside of the black hole.
- vanderZwan 10y ago
  > The fields in quantum field theory are mathematical tools, they are not physical entities. Still, they only maintain physical relevance as long as they are continuous, no? Otherwise you literally have a break in reality.
  - danbruc 10y ago
    Do you mean that they can not have holes?
  - guitarbill 10y ago
    > a break in reality. Such as a singularity (e.g. gravitational)? I think in Physics (just as when analysing functions), the interesting things happen when you approach such limits.

losvedir 10y ago

Perhaps, then, you should consider that you're misunderstanding the situation? A firewall is not "orbiting photons". See https://en.wikipedia.org/wiki/Firewall_(physics) https://en.wikipedia.org/wiki/Firewall_(physics).