# Derivation of Hawking Radiation

1. Jun 5, 2010

### stevebd1

I'm not quite sure how reliable http://library.thinkquest.org/C007571/english/advance/core4.htm" as a source but I thought it provided a good derivation for the Hawking Radiation equation relative to the Heisenberg uncertainty principle-

First the energy of the radiation is established-

$$E=m\cdot a \cdot d$$

where m (in this case) represents the energy of the virtual photons, a is gravitational acceleration and d is distance covered by the virtual photons.

$$d=c \cdot \Delta t$$

$\Delta t$ is derived from the Heisenberg uncertainty principle-

$$\Delta E \Delta t=\frac{h}{4 \pi}\ \Rightarrow\ \Delta t=\frac{h}{4\pi \cdot \Delta E}$$

energy of one photon is $E=hf$ where h is Planck's constant and f is frequency, for two photons- $E=2 \cdot hf$ and the equation for $\Delta t$ can be rewritten-

$$\Delta t=\frac{h}{4\pi \cdot \Delta E}=\frac{h}{4\pi \cdot 2 \cdot hf}=\frac{1}{8 \pi \cdot f}$$

and d can be rewritten-

$$d=c \cdot \Delta t=\frac{c}{8\pi \cdot f}$$

$$E=mc^2\ \Rightarrow\ m=\frac{E}{c^2}$$

as previously established, $E=2 \cdot hf$ so m can be rewritten-

$$m= \frac{2\cdot hf}{c^2}$$

and the equation for energy can be rewritten-

$$E=\frac{2\cdot hf}{c^2}\cdot a \cdot \frac{c}{8 \pi \cdot f}= \frac{ha}{4\pi c}$$

The average energy of a photon of black body radiation is-

$$E_{photon}\ \approx 2.821 \cdot k_B \cdot T$$

where 2.821 relates to Wein's law for frequency of maximal spectral emittance, kB is the Boltzmann constant and T is the temperature of the black body.

'..this only calculate(s) the energy for virtual photons that are aligned radially to the black hole and that are originated at the event horizon. But all directions have to be considered and all pairs of virtual photons of which one reaches the Schwarzschild radius within it's lifetime can submit to the radiation...' based on this, the quantity of 2.821 is replaced with $\pi$ and the equation is rewritten-

$$E= \pi \cdot k_B \cdot T\ \Rightarrow\ T=\frac{E}{\pi k_B}$$

substituting E from above and replacing h with Planck's reduced constant $(\hbar=h/2\pi)$ we get-

$$T=\frac{E}{\pi k_B}=\frac{ah}{4\pi^2 k_B c}=\frac{a \hbar}{2 \pi k_B c}$$

which is the equation for Hawking radiation where a would be replaced with $\kappa$, the killing surface gravity of the BH as observed from infinity-

$$T_H=\frac{\hbar \kappa}{2 \pi k_B c}$$

Source-
http://library.thinkquest.org/C007571/english/advance/core4.htm

Last edited by a moderator: Apr 25, 2017
2. Jun 6, 2010

### Chronos

Hawking's derivation is a little more rigorous, but, this is close enough.

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