29 ms·

I recently heard somewhere that the cosmic background radiation (CMR, 2.7 kelvin or so) is so much hotter than the "temperature" generated by Hawking radiation

I recently heard somewhere that the cosmic background radiation (CMR, 2.7 kelvin or so) is so much hotter than the "temperature" generated by Hawking radiation of black holes (apparently on the order of a billionth of a kelvin) such that they effectively do not radiate, and are not expected to do so for a loooong time (until expansion of the universe drops the background temp to an incredibly cold temperature). What I have not been able to determine is when they might be expected to occur? How far in the future will black holes start evaporating? (I believe the answer depends on the size of the black hole as well)

cdumler 5y ago
According to this, on the order of million trillion trillion trillion trillion trillion years [1]. [1] https://www.youtube.com/watch?v=uD4izuDMUQA https://www.youtube.com/watch?v=uD4izuDMUQA
mr_toad 5y ago
Some of the larger black holes could take 10^100 years to evaporate even after the universe grows cold. I don’t know how long that would take, but the whole process will take an unimaginably long time.
T-A 5y ago
You can easily find out yourself. First use a black hole temperature calculator like https://www.omnicalculator.com/physics/black-hole-temperature https://www.omnicalculator.com/physics/black-hole-temperatur... or https://www.vttoth.com/CMS/physics-notes/311-hawking-radiation-calculator https://www.vttoth.com/CMS/physics-notes/311-hawking-radiati... to find the Hawking temperature T_BH of your black hole. Then use the fact that CMB temperature T_CMB is inversely proportional to the cosmological scale factor a(t), where t is time: https://physics.stackexchange.com/questions/76241/cmbr-temperature-over-time https://physics.stackexchange.com/questions/76241/cmbr-tempe... Set a(t_now) = 1 for convenience; then T_CMB(t) = T_CMB(t_now) / a(t) and you want to find the time t when T_BH = T_CMB(t). That depends on how the scale factor will change over time: https://en.wikipedia.org/wiki/Scale_factor_(cosmology) https://en.wikipedia.org/wiki/Scale_factor_(cosmology) We don't know that, but if we assume for simplicity that dark energy will keep dominating, a(t) will grow exponentially, i.e. a(t) = a(t) / a(t_now) ~ exp(H_0 * (t - t_now)) where H_0 is Hubble's constant. So T_CMB(t) = T_CMB(t_now) / a(t) ~ T_CMB(t_now) / exp(H_0 * (t - t_now)) Numbers: T_CMB(t_now) ~ 2.7 K H_0 ~ 71 km/s/Mpc ~ 2.3e-18 s^-1 t_now ~ 14 Gyr ~ 3.2e16 s so, with t expressed in seconds, T_CMB(t) ~ 2.7 / exp(2.3e-18 * (t - 3.2e16)) K Setting this equal to T_BH and solving for t, I get t = 3.2e16 + ln(2.7 / T_BH) / 2.3e-18 s So let's say you have a solar mass black hole. Then T_BH ~ 6.16871e-8 K and so t ~ 7.7e18 s ~ 2.4e11 years i.e. 240 billion years (17 times the current age of the universe). That's actually not a humongous number, thanks to the exponential expansion of a(t). It would take a lot longer with the expansion rate of a matter-dominated universe (exercise for the reader!)
pcj-github 5y ago
Thanks, all great responses!
FeepingCreature 5y ago
Does that mean you can build a Dyson sphere around a black hole, then use the black hole as a thermal sink?
- T-A 5y ago
  http://www.weidai.com/black-holes.txt http://www.weidai.com/black-holes.txt
  - FeepingCreature 5y ago
    God damn, nice.

cdumler 5y ago

According to this, on the order of million trillion trillion trillion trillion trillion years [1]. [1] https://www.youtube.com/watch?v=uD4izuDMUQA https://www.youtube.com/watch?v=uD4izuDMUQA

mr_toad 5y ago

Some of the larger black holes could take 10^100 years to evaporate even after the universe grows cold. I don’t know how long that would take, but the whole process will take an unimaginably long time.

T-A 5y ago

You can easily find out yourself. First use a black hole temperature calculator like https://www.omnicalculator.com/physics/black-hole-temperature https://www.omnicalculator.com/physics/black-hole-temperatur... or https://www.vttoth.com/CMS/physics-notes/311-hawking-radiation-calculator https://www.vttoth.com/CMS/physics-notes/311-hawking-radiati... to find the Hawking temperature T_BH of your black hole. Then use the fact that CMB temperature T_CMB is inversely proportional to the cosmological scale factor a(t), where t is time: https://physics.stackexchange.com/questions/76241/cmbr-temperature-over-time https://physics.stackexchange.com/questions/76241/cmbr-tempe... Set a(t_now) = 1 for convenience; then T_CMB(t) = T_CMB(t_now) / a(t) and you want to find the time t when T_BH = T_CMB(t). That depends on how the scale factor will change over time: https://en.wikipedia.org/wiki/Scale_factor_(cosmology) https://en.wikipedia.org/wiki/Scale_factor_(cosmology) We don't know that, but if we assume for simplicity that dark energy will keep dominating, a(t) will grow exponentially, i.e. a(t) = a(t) / a(t_now) ~ exp(H_0 * (t - t_now)) where H_0 is Hubble's constant. So T_CMB(t) = T_CMB(t_now) / a(t) ~ T_CMB(t_now) / exp(H_0 * (t - t_now)) Numbers: T_CMB(t_now) ~ 2.7 K H_0 ~ 71 km/s/Mpc ~ 2.3e-18 s^-1 t_now ~ 14 Gyr ~ 3.2e16 s so, with t expressed in seconds, T_CMB(t) ~ 2.7 / exp(2.3e-18 * (t - 3.2e16)) K Setting this equal to T_BH and solving for t, I get t = 3.2e16 + ln(2.7 / T_BH) / 2.3e-18 s So let's say you have a solar mass black hole. Then T_BH ~ 6.16871e-8 K and so t ~ 7.7e18 s ~ 2.4e11 years i.e. 240 billion years (17 times the current age of the universe). That's actually not a humongous number, thanks to the exponential expansion of a(t). It would take a lot longer with the expansion rate of a matter-dominated universe (exercise for the reader!)

pcj-github 5y ago

Thanks, all great responses!

FeepingCreature 5y ago

Does that mean you can build a Dyson sphere around a black hole, then use the black hole as a thermal sink?

T-A 5y ago
http://www.weidai.com/black-holes.txt http://www.weidai.com/black-holes.txt
- FeepingCreature 5y ago
  God damn, nice.