3 ms·

That doesn't make sense. Everywhere is the edge of the observable universe to someone.

XorNot 4y ago
Yes, but because space is expanding - i.e. there's a value of meters per meter per second for the expansion of space, then every point in space has a sphere around it of locations which are sufficiently far away that they are growing more distant at the speed of light. Any virtual particle pair which appears along that horizon can potentially be caught on the wrong side of it - i.e. one virtual particle is in a part of space which is now far enough way to be expanding faster then light, whereas the other particle is at a location close enough that it is not. At the moment that happens, there's a particle or photon which is now inside the light cone of a distant object, paired with a particle that it will never meet again because it's outside of it. As a result, you get Hawking radiation: because one particle can go off and interact with your universe, but it's partner will never be able to causally effect anything inside that horizon again. So the virtual particle has to become real.
- dmbche 4y ago
  Thanks for writing this, I hadn't realised that. How cool!
- fallingknife 4y ago
  Ok, but we are at such a horizon right now. Why can't we observe the hawking radiation from it?
  - XorNot 4y ago
    It's a very large horizon, so the Hawking radiation is very weak (it's always a weak effect). We can't see it because the cosmic microwave drowns it out. Broadly Hawking radiation intensity goes with curvature - i.e. a more curved surface has a better chance of separating a particle pair then a flatter one. This is because a sharper curve means more vectors which carry you away from the event horizon. This is also why small black holes evaporate faster then big ones - as the black hole shrinks, the Hawking radiation intensity increases because it's curving more and more (hence why microscopic black holes don't devour everything).
antonvs 4y ago
Correct, that's a key fact about the underlying effects in question, of which Hawking radiation is just a consequence: different observers see different particles in the vacuum. In 1976, Bill Unruh published "Notes on black-hole evaporation"[1], in which he showed that "an accelerated detector even in flat spacetime will detect particles in the vacuum" - now known as the Unruh effect This means that an observer in an accelerated reference frame will observe particles in the vacuum where an inertial observer will observe none. The presence of certain particles - the ones we call Hawking radiation in the context of a black hole - is a relative phenomenon. This is known as the Unruh effect. The equation for the Hawking temperature is essentially the same as the equation for the Unruh temperature, where the acceleration value is the acceleration due to gravity of the black hole. Then in 1977, Gibbons and Hawking published "Cosmological event horizons, thermodynamics, and particle creation"[2], which showed that "the close connection between event horizons and thermodynamics which has been found in the case of black holes can be extended to cosmological models with a repulsive cosmological constant" and that "An observer with a particle detector will indeed observe a background of thermal radiation coming apparently from the cosmological event horizon." This is known as the Gibbons-Hawking effect. There's a fairly complex relationship between the two effects which I won't try to describe, but if you're interested then [3] discusses it. The abstract itself gives some sense of the connection. [1] https://journals.aps.org/prd/abstract/10.1103/PhysRevD.14.870 https://journals.aps.org/prd/abstract/10.1103/PhysRevD.14.87... [2] https://journals.aps.org/prd/abstract/10.1103/PhysRevD.15.2738 https://journals.aps.org/prd/abstract/10.1103/PhysRevD.15.27... [3] https://arxiv.org/abs/2211.14747 https://arxiv.org/abs/2211.14747

Yes, but because space is expanding - i.e. there's a value of meters per meter per second for the expansion of space, then every point in space has a sphere around it of locations which are sufficiently far away that they are growing more distant at the speed of light. Any virtual particle pair which appears along that horizon can potentially be caught on the wrong side of it - i.e. one virtual particle is in a part of space which is now far enough way to be expanding faster then light, whereas the other particle is at a location close enough that it is not. At the moment that happens, there's a particle or photon which is now inside the light cone of a distant object, paired with a particle that it will never meet again because it's outside of it. As a result, you get Hawking radiation: because one particle can go off and interact with your universe, but it's partner will never be able to causally effect anything inside that horizon again. So the virtual particle has to become real.

dmbche 4y ago
Thanks for writing this, I hadn't realised that. How cool!
fallingknife 4y ago
Ok, but we are at such a horizon right now. Why can't we observe the hawking radiation from it?
- XorNot 4y ago
  It's a very large horizon, so the Hawking radiation is very weak (it's always a weak effect). We can't see it because the cosmic microwave drowns it out. Broadly Hawking radiation intensity goes with curvature - i.e. a more curved surface has a better chance of separating a particle pair then a flatter one. This is because a sharper curve means more vectors which carry you away from the event horizon. This is also why small black holes evaporate faster then big ones - as the black hole shrinks, the Hawking radiation intensity increases because it's curving more and more (hence why microscopic black holes don't devour everything).

antonvs 4y ago

Correct, that's a key fact about the underlying effects in question, of which Hawking radiation is just a consequence: different observers see different particles in the vacuum. In 1976, Bill Unruh published "Notes on black-hole evaporation"[1], in which he showed that "an accelerated detector even in flat spacetime will detect particles in the vacuum" - now known as the Unruh effect This means that an observer in an accelerated reference frame will observe particles in the vacuum where an inertial observer will observe none. The presence of certain particles - the ones we call Hawking radiation in the context of a black hole - is a relative phenomenon. This is known as the Unruh effect. The equation for the Hawking temperature is essentially the same as the equation for the Unruh temperature, where the acceleration value is the acceleration due to gravity of the black hole. Then in 1977, Gibbons and Hawking published "Cosmological event horizons, thermodynamics, and particle creation"[2], which showed that "the close connection between event horizons and thermodynamics which has been found in the case of black holes can be extended to cosmological models with a repulsive cosmological constant" and that "An observer with a particle detector will indeed observe a background of thermal radiation coming apparently from the cosmological event horizon." This is known as the Gibbons-Hawking effect. There's a fairly complex relationship between the two effects which I won't try to describe, but if you're interested then [3] discusses it. The abstract itself gives some sense of the connection. [1] https://journals.aps.org/prd/abstract/10.1103/PhysRevD.14.870 https://journals.aps.org/prd/abstract/10.1103/PhysRevD.14.87... [2] https://journals.aps.org/prd/abstract/10.1103/PhysRevD.15.2738 https://journals.aps.org/prd/abstract/10.1103/PhysRevD.15.27... [3] https://arxiv.org/abs/2211.14747 https://arxiv.org/abs/2211.14747