<p>An analogy in the settings of the question would be a deceleration down to an expansion rate, such that eventually light from anywhere in the universe can reach us, meaning the cosmic horizon can vanish. A constant expansion rate of the universe above the speed of light would be the analogy of a black hole of fixed mass.</p>

<p>In an infinite universe this would mean, that its expansion needs to slow down to an asymptotic speed for infinite distance below the speed of light. For a finite <a href="http://en.wikipedia.org/wiki/N-sphere" rel="nofollow">3-spherical universe</a> for a fixed cosmic time a slow-down to below the speed of light would be sufficient. This would correspond to a massive follow-up object of the black hole without event horizon.</p>

<p>For an analogy to a completely evaporating black hole the expansion rate needs to slow down to zero. The slow down between speed of light and zero corresponds to the very last phase of black hole evaporation, when its mass drops below the <a href="http://en.wikipedia.org/wiki/Planck_particle" rel="nofollow">Planck mass</a>, then without event horizon.</p>

<p>The analogy to Hawking radiation would be radiation or matter emerging from behind the former cosmic horizon.</p>

<p>Thinking in <a href="http://en.wikipedia.org/wiki/Stereographic_projection" rel="nofollow">stereographic projections</a> may help for a better understanding: Project infinity of a plane expanding universe to the singularity of a black hole at the north pole of an expanding sphere, as an initial point of further reasoning.</p>

CC BY-SA 3.0
2014-02-14T17:38:12.433Z
2014-02-14T17:38:12.433Z
Gerald
<p>The radiation from the cosmic horizon is moving inward, so the horizon should get a little bigger. That translates to the Hubble constant and the temperature getting a little smaller. If we ignore the fact that the radiation should cross the horizon again on the other side of the Universe, the rate of radiation would keep getting lower as the temperature drops. So the horizon would grow more and more slowly.</p>
<p>No, that's wrong! The horizon gets smaller, because its entropy decreases (by the entropy of the emitted radiation; the total entropy must be constant because the process is reversible). So the rate of radiation would increase over time for as long as the emitted radiation remains inside the Universe.</p>

CC BY-SA 4.0
2021-02-21T04:42:05.303Z
2021-02-21T07:32:43.603Z
2021-02-21T07:32:43.603Z
Bennett Standeven
<p>In the same way, as a black hole emits Hawking radiation corresponding to its temperature</p>

<p>$$
T = \frac{\hbar}{r_{BH}}
$$</p>

<p>the accelerating universe can be described as all of the stuff falling towards the cosmic horizon, such that it is analogue to a black hole where the cosmic horizon takes the role of the event horizon with the black hole "interior" being outside. </p>

<p>The corresponding temperature of the accelerating universe is given by</p>

<p>$$
T = \hbar H = \frac{\hbar}{R_{H}}
$$</p>

<p>where H is the Hubble constant. Therefore, the cosmic horizon emits something similar to Hawking radiation too.</p>

<p>The fate of a black hole is to finally evaporate. My question is: What does black hole evaporation correspond to in the accelerating universe when making use of this analogy?</p>

CC BY-SA 3.0
2014-04-21T11:14:26.217Z
Dear downvoter, what is wrong with my question, can you please be a bit more specific and explicit thant just pushing the button ?
Dilaton
<black-hole><cosmology><hawking-radiation><cosmological-horizon>
What does black hole evaporation correspond to in the accelerating universe / black hole analogy?
Dilaton
# What does black hole evaporation correspond to in the accelerating universe / black hole analogy?

(Asked by: username_0 on 2014-02-13)

In the same way, as a black hole emits Hawking radiation corresponding to its temperature

$$
T = \frac{\hbar}{r\_{BH}}
$$

the accelerating universe can be described as all of the stuff falling towards the cosmic horizon, such that it is analogue to a black hole where the cosmic horizon takes the role of the event horizon with the black hole "interior" being outside.

The corresponding temperature of the accelerating universe is given by

$$
T = \hbar H = \frac{\hbar}{R\_{H}}
$$

where H is the Hubble constant. Therefore, the cosmic horizon emits something similar to Hawking radiation too.

The fate of a black hole is to finally evaporate. My question is: What does black hole evaporation correspond to in the accelerating universe when making use of this analogy?

**Tags:** black-hole, cosmology, hawking-radiation, cosmological-horizon

> **Comments:**
> **username_0**: Dear downvoter, what is wrong with my question, can you please be a bit more specific and explicit thant just pushing the button ?  

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## 2 answers

### Answer by username_1 on 2014-02-14. Score: 4

An analogy in the settings of the question would be a deceleration down to an expansion rate, such that eventually light from anywhere in the universe can reach us, meaning the cosmic horizon can vanish. A constant expansion rate of the universe above the speed of light would be the analogy of a black hole of fixed mass.

In an infinite universe this would mean, that its expansion needs to slow down to an asymptotic speed for infinite distance below the speed of light. For a finite [3-spherical universe](http://en.wikipedia.org/wiki/N-sphere) for a fixed cosmic time a slow-down to below the speed of light would be sufficient. This would correspond to a massive follow-up object of the black hole without event horizon.

For an analogy to a completely evaporating black hole the expansion rate needs to slow down to zero. The slow down between speed of light and zero corresponds to the very last phase of black hole evaporation, when its mass drops below the [Planck mass](http://en.wikipedia.org/wiki/Planck_particle), then without event horizon.

The analogy to Hawking radiation would be radiation or matter emerging from behind the former cosmic horizon.

Thinking in [stereographic projections](http://en.wikipedia.org/wiki/Stereographic_projection) may help for a better understanding: Project infinity of a plane expanding universe to the singularity of a black hole at the north pole of an expanding sphere, as an initial point of further reasoning.


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### Answer by username_2 on 2021-02-21. Score: 1

The radiation from the cosmic horizon is moving inward, so the horizon should get a little bigger. That translates to the Hubble constant and the temperature getting a little smaller. If we ignore the fact that the radiation should cross the horizon again on the other side of the Universe, the rate of radiation would keep getting lower as the temperature drops. So the horizon would grow more and more slowly.

No, that's wrong! The horizon gets smaller, because its entropy decreases (by the entropy of the emitted radiation; the total entropy must be constant because the process is reversible). So the rate of radiation would increase over time for as long as the emitted radiation remains inside the Universe.


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