Einstein Probabilistic Units/Black Holes

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Black Hole

A black hole can be defined by the Schwarzschild radius

$r_s=\frac{2G{M}_{bh}}{c^2}$

Expressing this using Einstein Probabilistic Units

$r_{bh}=2B{M}_{bh}$

Solving for B give us

$B_{bh}=\frac{r_{bh}}{2M_{bh}}$

The following table shows some properties of Black Hole using Einstein's Probabilistic Units (EPU); Template:Center top

Property	Equation	EPU
Mass	$$M_s=2\frac{1}{r_s}\frac{G}{c^2}$$	$$M_s=\frac{2c}{A_s{B}_s}$$	$$M_s=\frac{c}{B_0}\frac{A_s}{A_0^2}$$
Schwarzschild radius	$$r_s=2M_s\frac{G}{c^2}$$	$$r_s=2M_s{B}_0$$	$$B_0=\frac{G}{c^2}=\frac{1}{2}{B}_s$$
Area	$$A_s=16\pi M_s^2\frac{G^2}{c^4}$$	$$A_s=16\pi M_s^2{B}_0^2$$	$$B_0^2=\frac{1}{16\pi }{B}_s^2$$
Density	$${\rho }_s=\frac{3}{32\pi }\frac{c^6}{M_s^2{G}^3}$$	$${\rho }_s=\frac{3}{32\pi }\frac{1}{M_s^2}\frac{1}{B_0^3}$$	$${\rho }_s=\frac{3}{8\pi }\frac{1}{B_0}\frac{1}{r_s^2}$$
Entropy	$$S_{BH}=\frac{1}{4}{k}_B\frac{A_s}{l_P^2}$$	$$S_{BH}=\frac{1}{4}{k}_B\frac{B_s}{B_0}$$	$$B_0=\frac{1}{4}{k}_B\frac{B_s}{S_{BH}}$$
Temperature	$$T=\frac{1}{8\pi }\frac{1}{M_s}\frac{\hslash c}{k_B}\frac{c^2}{G}$$	$${\displaystyle }$$	$${\displaystyle }$$

Entropy

The Bekenstein–Hawking formula for black-hole entropy is proportional to the area of its event horizon A.

$S_{BH}=\frac{k_B{A}_s}{4l_P^2}=\frac{k_B{c}^3{A}_s}{4G\hslash }\frac{J}{K}$

Using Einstein's Probabilistic units to express Bekenstein–Hawking formula;

$S_{BH}=k_B\frac{c^3}{\hslash }\frac{A_s}{c^2}\frac{c^2}{G}=k_B\frac{c^3}{\hslash }\frac{1}{\underset{s}{\overset{2}{\upsilon }}}\frac{1}{B_0}=k_B\frac{B_s}{B_0}=Constant\frac{1}{\underset{s}{\overset{2}{\upsilon }}}=-k_B<lnW>$

$S_{BH}=k_{B}ln(W)=k_B\frac{B_s}{B_0}=k_B\frac{A_0^2}{A_S^2}$

$W=e^{S/k_B}=e^{B_s/B_0}$

Hawking radiation temperature

$T_H=\frac{\hslash c^3}{8\pi GM{k}_B}$

$T_H=\frac{\hslash c^3}{8\pi {\mu }_M{k}_B}=\frac{\hslash c}{8\pi k_B}\times \frac{m^2}{s^2}\frac{s^2}{m^3}=\frac{1}{8\pi k_B}\frac{Q^2}{m}$

$T_H=\frac{1}{8\pi }\frac{\hslash c}{M}\frac{1}{B_0}\frac{1}{k_B}=\frac{1}{8\pi }\frac{\mu }{B_0}\frac{1}{k_B}$

$B_0=\frac{1}{8\pi }\frac{{\mu }_M}{T_H}\frac{1}{k_B}$

$B=\frac{hc}{\left(T_H{k}_B\right)M_{\odot }}$

$T_{\mathrm{H}}=\frac{\hslash c^3}{8\pi GM{k}_{\mathrm{B}}}$