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

Calculating Hawking radiation

How much energy does a pair of virtual photons get from a black hole?

Somehow the energy for the Hawking radiation has to get out of the black hole, doesn't it? The only thing that gets beyond the Schwarzschild radius of a black hole from the inside is gravitation. Therefore Hawking radiation has to depend on the gravitation.

When something is constantly accelerated, for example by gravity, it gets kinetic energy. This energy is proportional to the mass and the distance on which it is accelerated. You might ask how a virtual photon can be accelerated, if they always move at the speed of light. Okay, they are not, but their frequency will change giving them a higher energy.

Visible spectrum. Courtesy of NASA.
visible spectrum
<< towards higher energy
towards lower energy >>

E = m * a * d (1)
m: mass
a: gravitational acceleration
d: distance the mass m falls
The mass is the mass of a virtual photon in our case and the distance is the distance a virtual photon moves in its lifetime. Since virtual photons have such a short lifetime they won't get very far during that (24 nanometers for a virtual photon of orange light). So we can assume the gravitation to be constant and therefore use this formula.

1st: Photons always move at the speed of light, so the distance a virtual photon travels in its lifetime is given by:

d = c * delta t

We already calculated the lifetime delta t of a virtual Photon:

delta t = 1/ 8*pi*f

Jump back to the calculation of the lifetime of a virtual photons.
(click "back" on your browser to get back here.)

Putting that formula in here we get:

d = c / (8 * pi * f)

Let's put this into our formula (1):

E=m*a*c/8*pi*f (2)

2nd: We can calculate the gravitational acceleration using Newton's gravitation formula:

F=GMm/R^2

F: gravitational force
r: distance of the two bodies
G: gravitational constant
Issac Newton
Issac Newton
Portrait by Vanderbank/
courtesy Caltech Archives.

virtual photons and black hole
Visualized picture of the
verbal discription.
(NOT to scale!)
By Team C007571, ThinkQuest 2000.

 

One of the virtual photons has to fall into the black hole in order to produce Hawking radiation. Thus they must originate close to the Schwarzschild radius. As an approximation we can take the Schwarzschild radius as the distance between the pair of photons and the black hole. Remember the formula for the Schwarzschild radius?

Jump back to the derivation of the Schwarzchild radius.
(Click "back" on your browser to get back here.)

Let's put it in here:

a = c^4/ 4*G*M

Now we put this into formula (2):

E=c^5/32*pi*G*M*f
(3)

3rd: Now let's try to substitute m:

We know Einstein's famous formula:

E = m*c^2
The energy of a photon is (as we already used):

E=h*f

The energy of a pair of virtual photons therefore is:

E-2*h*f

We put this into the formula for the mass:

m=2*h*f/C^2

Putting it into our main formula (3) we get:

E=h*c^3/16*pi*G*M

E=h*c^3/16*pi*G*M

This should be the energy of a photon of the Hawking radiation.

A pair of virtual photons gets more energy from a small black hole than from a large one! Very surprising isn't it? But this explains why nobody has ever seen a shining black hole.

small and LARGE black holes. A small and a large black hole.
By Team C007571, ThinkQuest 2000.

 
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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.