- #1
Orion1
- 973
- 3
In order for the black hole to evaporate it must have a temperature greater
than that of the present-day black-body radiation of the Universe.
Cosmic microwave background radiation temperature:
[tex]T_u = 2.725 \; \text{K}[/tex]
Hawking radiation temperature:
[tex]T_H = \frac{\hbar c^3}{8 \pi G M k_B}[/tex]
Hawking radiation temperature is greater than or equal to cosmic microwave background radiation temperature:
[tex]T_H \geq T_u[/tex]
[tex]\frac{\hbar c^3}{8 \pi G M k_B} \geq T_u[/tex]
Hawking total black hole mass:
[tex]M_H \leq \frac{\hbar c^3}{8 \pi G k_B T_u} \leq 1.226 \cdot 10^{23} \; \text{kg}[/tex]
[tex]\boxed{M_H \leq 1.226 \cdot 10^{23} \; \text{kg}}[/tex]
Earth total mass:
[tex]M_{\oplus} = 5.9722 \cdot 10^{24} \; \text{kg}[/tex]
[tex]\frac{M_H}{M_{\oplus}} = 0.007539 = 0.754 \; \text{%}[/tex]
According to reference 4 - p. 2, eq. 1:
[tex]\frac{M_H}{M_{\oplus}} = 0.8 \; \text{%}[/tex]
Are these equations correct?
Reference:
Cosmic microwave background radiation - temperature - Wikipedia
Hawking radiation - black hole evaporation - Wikipedia
Earth mass - Wikipedia
The Last Eight Minutes Of A Primordial Black Hole - Joseph Kapusta