<math>\begin{align}   & \left\{ \begin{align}   & m{\alpha }'\tau \hbar \varepsilon \mu \alpha \mathbf{T}ic\varsigma  \\   & \text{      }\And  \\   & \text{  }\mathsf{\mathcal{P}}\pi y\sigma \mathbf{I}\subset \mathbf{S} \\  \end{align} \right\} \\   & \text{    }Wikia \\  \end{align}</math>

Hawking Radiation

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Quantum Field Theory
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Relativistic Quantum Mechanics Klein-Gordon Equation
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From a framework to a model Yang-Mills Theory
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Hawking Radiation refers to radiation that is emitted from a black hole.

Black Hole Temperature EquationEdit this section

We start with a standard expression that relates to Black Hole Entropy:

E=\frac{\hbar g}{2\pi c_0}

Here, g refers to the Gravitational Field at the surface, given by the following expression: \


Substituting the above in:

E=\frac{\hbar c_0}{8\pi GM}

We know that E=k_BT, and therefore,

Black Hole Temperature Equation

E=\frac{\hbar c_0}{8\pi GM}

Calculating the Black Hole EntropyEdit this section

From the equation for Black Hole Temperature, we may calculate the Entropy of a Black Hole, as follows.


\mathrm{d}S=\frac{8\pi Gk_BM }{\hbar c_0^3} \mathrm{d}Q

\mathrm{d}S=\frac{8\pi Gk_BM }{\hbar c_0} \mathrm{d}M

\mathrm{d}S=\frac{2\pi k_B r_S c_0^3}{\hbar G} \mathrm{d}r_s

Integrating both sides,

\int \mathrm{d}S=\frac{2\pi k_B  c_0^3}{\hbar G} \int r_S \mathrm{d}r_s

S  =           \frac{\pi r_s^2 c_0^3     }{\hbar G   } S  =           \frac{Ak_Bc_0^3}{4   \hbar G   }

Here, A is the surface area of the black hole, and is obviously equal to 4\pi r_s^2  , and therefore,

Entropy of a Black Hole

\frac{S}{k_B}=\frac14   \frac{A}{A_P}

Where A_P is the Planck Area.

Power radiated by a black holeEdit this section

The Stefan-Boltzmann Law states that:

P=A\epsilon \sigma T^4

A Black Hole is a perfect black body, so it's radiativity \epsilon   =    1. Therefore,

P=A\epsilon \sigma \frac{\hbar^4 c_0^4  }{5096 \pi^4 G^4 M^4 k_B^4}

  & P=\frac{16\pi {{G}^{2}}{{M}^{2}}}{c_{0}^{4}}\sigma \frac{{{\hbar }^{4}}c_{0}^{12}}{5096{{\pi }^{4}}{{G}^{4}}{{M}^{4}}k_{B}^{4}} \\ 
 & P=\frac{16\pi {{G}^{2}}{{M}^{2}}}{c_{0}^{4}}\frac{{{\pi }^{2}}k_{B}^{4}}{60{{\hbar }^{3}}c_{0}^{2}}\frac{{{\hbar }^{4}}c_{0}^{12}}{5096{{\pi }^{4}}{{G}^{4}}{{M}^{4}}k_{B}^{4}} \\ 
 & P=\frac{\hbar c_{0}^{6}}{15360\pi {{G}^{2}}{{M}^{2}}} \\ 

  & P=\frac{16\pi {{G}^{2}}{{M}^{2}}}{c_{0}^{4}}\sigma \frac{{{\hbar }^{4}}c_{0}^{12}}{5096{{\pi }^{4}}{{G}^{4}}{{M}^{4}}k_{B}^{4}} \\ 
 & P=\frac{16\pi {{G}^{2}}{{M}^{2}}}{c_{0}^{4}}\frac{{{\pi }^{2}}k_{B}^{4}}{60{{\hbar }^{3}}c_{0}^{2}}\frac{{{\hbar }^{4}}c_{0}^{12}}{5096{{\pi }^{4}}{{G}^{4}}{{M}^{4}}k_{B}^{4}} \\   

Power radiated from a black hole

 P=\frac{\hbar c_{0}^{6}}{15360\pi {{G}^{2}}{{M}^{2}}}

Countdown to Black Hole EvaporationEdit this section

The Power is the rate of Hawking Radiation, i.e.


Let us define a constant term \kappa =\frac{\hbar {{{c}_{0}}^{6}}}{15360\pi {{G}^{2}}} so that P=\frac{\kappa }{{{M}^{2}}}. Then,


& \frac{\kappa }{{{M}^{2}}}=-\frac{\text{d}E}{\text{d}t}=-c_{0}^{2}\frac{\text{d}M}{\text{d}t} \\

& {{M}^{2}}\text{d}M=-\frac{\kappa }{c_{0}^{2}}\text{d}t \\

& \int_{{{M}_{i}}}^{0}{{{M}^{2}}\text{d}M}=-\frac{\kappa }{c_{0}^{2}}\int_{0}^{{{t}_{f}}}{\text{d}t} \\

& -\frac{M_{i}^{3}}{3}=-\frac{\kappa {{t}_{f}}}{c_{0}^{2}} \\

& {{t}_{f}}=\frac{c_{0}^{2}}{3\kappa }M_{i}^{3} \\


Time for Black Hole Evaporation

{{t}_{f}}=\frac{5120\pi {{G}^{2}}M_{i}^{3}}{\hbar c_{0}^{4}}

This is the time taken for the black hole to completely evaporate. However, taking into consideration the cosmic background microwave radiation, the temperature radiated must be greater than 2.7 K, and thus the mass of the black hole must be lesser than 1.226\times {{10}^{23}} , or \frac{3}{400} of the mass of the Earth, or else, it will not evaporate at all.

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