Entropy, heat death and black holes

  • Thread starter narrator
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  • #1
narrator
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Talking about the big bounce and black holes in another thread made me wonder about the life span of black holes, especially in the heat death scenario.

If the U went through heat death, does that mean that black holes would also exhaust their energy? Do black holes die, and if so, what happens? I seem to remember a theory about black holes emitting radiation and eventually evaporating, depending on their size.
 

Answers and Replies

  • #2
Chalnoth
Science Advisor
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Yes, black holes evaporate very slowly and eventually evaporate entirely. But it takes a long, long time. As in around 10^100 years for a large supermassive black hole.
 
  • #3
edgepflow
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One can 'experiment' and possibly amuse oneself with a blackhole's lifetime using the following formula:

[itex]t_{life}[/itex] = [itex]10^{66}[/itex] years ([itex]M_{bh}[/itex] / [itex]M_{sun}[/itex])[itex]^{3}[/itex]

The mass of the sun is about 2 X 10^30 kg

Reference: Black Holes A Traveler's Guide, Pickover, P.112
 
  • #4
alex2515
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I seem to remember a theory about black holes emitting radiation and eventually evaporating, depending on their size.



Your talking about 'Hawking Radiation" What happens is that particle/anti particle pairs that come into existense at or near the BH event horizon become seperated. One particle gets sucked in due to the BH, and the other escapes, therefore the BH slowly losses mass over time. It has temperature, It radiates.
 
  • #5
edgepflow
688
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Your talking about 'Hawking Radiation" What happens is that particle/anti particle pairs that come into existense at or near the BH event horizon become seperated. One particle gets sucked in due to the BH, and the other escapes, therefore the BH slowly losses mass over time. It has temperature, It radiates.
And this temperature is usually very cold but will get very hot just before the black hole totally evaporates. From the earlier reference I listed, the temperature at the black hole event horizon is:

[itex]T_{horizon}[/itex] = [itex]\frac{6 X 10^{-8} Kelvin}{M_{bh} / M_{sun}}[/itex]
 

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