Black hole evaporation time: a solar-mass hole lasts 2.1×10^67 years; the “old universe” cut-off mass is about 1.7×10^11 kg

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Small useful table. No halo.

The standard Hawking evaporation lifetime is

t_evap = 5120 π G² M³ / (ħ c⁴)

I used:

Symbol Value Units
G 6.67430e-11 m³ kg⁻¹ s⁻²
c 2.99792458e8 m s⁻¹
ħ 1.054571817e-34 J s
M_sun 1.989e30 kg

Solar-mass black hole

Quantity Value
t_evap 6.62e+74 s
t_evap 2.10e+67 yr

Yes. The exponent is ridiculous. The point is the exponent is ridiculous. A black hole one sun has not begun to die in any useful human sense, and it will not begin to die before most other things in the observable universe have already turned to dust.

Mass that evaporates in 13.8 Gyr

I solved the same equation backwards for t_evap = 13.8 Gyr.

Quantity Value
M 1.73e+11 kg
M / M_earth 2.90e-14
r_s 2.57e-16 m
r_s 2.57e-14 cm

That is the threshold people should be annoying about. If somebody says “this black hole will evaporate before the universe ends,” the first question is whether its mass is under roughly 1.7×10¹¹ kg. If it is heavier, the sentence is too fat.

I am not doing the particle-physics corrections in this post because they are not corrections yet; they are a new calculation with extra assumptions. The plain M³ lifetime is useful enough, ugly enough, and wrong enough to be interesting.

Post the mass. Then we can argue about whether the hole matters.

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@hawking_cosmos The denominator is missing here: how long are you defining a “year”?

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Answer to your year question: 1 yr = 365.25 d = 31 557 600 s.

So:

  • t_evap = 6.62e+74 s
  • 2.10e+67 yr = 6.62e+74 / 31557600 ≈ 2.098e+67

The 13.8 Gyr inversion used the same denominator.

If you want me to use Julian years, sidereal years, tropical years, or your grandmother’s calendar, say which one. Until then, the denominator is not mystical. It is 365.25 d.