Black holes aren't black -after Hawking they shine!
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The Core

Calculating Hawking radiation

What temperature does a black hole have?

As already said, the Hawking radiation will show the same allocation as black body radiation (See "Background" chapter, black body radiation). The average energy of a photon of this radiation is given by the following formula:

<E>=2.821*k*T
k: Boltzmann constant; k=1.38*10^-23JK^-1
T: Temperature of the black body.
A black body

This Energy should be equal to the energy a black hole puts into a pair of virtual photons. So let's bring the formulas together.

T=h*c^3/2.821*k*16*pi*G*M


Now we can substitute all of the constants and we got:

T=1.368*10^23Kkg/M

Looks great! If we put in a mass in kilograms we'll get a temperature in Kelvin. Our formula looks promising, even though we used mostly classical physics.

Well, it's pretty close. In the real formula the divisor 2.821 is replaced by a pi:

T=1.228*10^23Kkg/M


What did we do wrong?

We only calculated 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. what's wrong?
By C007571,ThinkQuest 2000

Some example calculations

How hot would a black hole with the same mass as Earth be?

T=0.02K

This black hole would be colder than the cosmic background radiation.

cosmic backgound radiation spectrum

Cosmic Backgound radiation spectrum.

Courtesy of NASA.


How hot would the black hole in "The Hole Man" be?


T=1.228*10^12K

I don't know anything to compare such a high temperature to!

 

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"Black holes aren't black - After Hawking they shine!"
Presented by Angie, Matthias and Thorsten
Team C007571,ThinkQuest Internet Challenge 2000.
Last modified: 2000-08-02.