# Radiation from cosmological horizons

Hawking showed that likewise black hole event horizon, cosmological horizons also radiates as they are at temperature proportional to surface gravity. Semi classical explanation exist about the Hawking radiation from event horizon. What mechanism exit for radiation from cosmological horizon.?

## 1 Answer

Be careful about your definition of the cosmological horizon. This term is normally used to mean the particle horizon and this is not an event horizon and does not radiate. A cosmological event horizon exists only in an accelerating universe. Since dark energy is accelerating the expansion of the universe we do have an event horizon, but it is currently farther away than the particle horizon and consequently is currently causally disconnected from us.

Strictly speaking the event horizon takes an infinite time to form, as measured by our clocks, though in practice the apparent horizon will be indistinguishable from a true horizon as soon as it moves inside the particle horizon, and we'll be able to see the Hawking radiation from it (in principle, though probably not in practice). The mechanism of Hawking radiation from the cosmological event horizon is exactly the same as the mechanism for the radiation from a black hole event horizon.

• “Event horizon in an accelerating universe radiates hawking radiation.” $~$In the case of a black hole, the black hole loses its mass and radiate Hawking radiation. But in the case of de Sitter horizon, where does the energy come from? – Forge Oct 20 '18 at 2:37
• Does the process violate the conservation of energy? – Forge Oct 21 '18 at 9:50
• @Forge yes, but energy is not conserved in an expanding universe. – John Rennie Oct 21 '18 at 9:52
• – John Rennie Oct 21 '18 at 10:00
• OK, I understand that energy is not conserved because spacetime is not time-translation invariant.$~$ From what I understand, energy is not conserved only if pressure is not zero. – Forge Oct 21 '18 at 10:24