#### marcus

Science Advisor

Gold Member

Dearly Missed

- 24,713

- 783

all black holes are non-rotating uncharged.

I think everybody here probably has a qualitative notion of how the Hawking radiation arises from the event horizon of a black hole---there has been tons of pop-sci journalism about this. What may be lacking is a corresponding quantitative feel---an ability to estimate magnitudes.

If a black hole surface area is on the order of a square angstrom, for instance, what is the temperature compared with things you know about----a hot sidewalk, the surface of the sun, the core of the sun, the big bang, etc?

Again, if the surface area is a square angstrom or so, what is the wattage roughly speaking? One should have a rough quantitative grasp to fill out the intuitive picture.

History: in 1974 Bekenstein said the entropy S is given by

S = A/4, where A is the area

And according to Hawking (1975) the temperature T is given by

1/sqrt(4piA)

We should do an example to get a feel for that----say area is 10

^{50}planck, which is about a square angstrom.

then 4piA is 12.6E50 and sqrt(4piA) is 3.5E25 and the temperature (one over that) is 0.28E-25

This is about one quarter of the temperature at the core of the sun----suncore temp is 10

^{-25}, roughly 1200 eevee.

Being a quarter of that one can think of it as soft xray or hard UV or whatever. So it isnt big bang temperature which would be just

*one*, or even suncore, but about a quarter of that.

What is the luminosity or wattage of this thing? Well the Boltzmann law says take the fourth power of the temp and multiply by pi

^{2}/60. That gives the radiant power per unit area and then you have to multiply by A to get the overall power.

The Hawking temperature, remember, is 1/sqrt(4piA). And so

the fourth power is 1/(4piA)

^{2}

And multiplying by the Stefan-Boltzmann const pi

^{2}/60

gives us (1/960)(1/A

^{2}) which is power per unit area

And then multiplying by A to get the total power gives

(1/960)(1/A) which if you want you can write 1/(960A)

*****THAT'S ALL, the tutorial is over*****

But there is one more little thing having to do with SI units.

Suppose you love metric units and want to put back in all the hbar and cees and stuff and do a purely metric calculation of the same thing. Here is how to get the metric formula back:

Planck power is c

^{5}/G

Planck area is Ghbar/c

^{3}

That is all you need to know! You just take the 1/(960A) formula and instead of A you put in (A'/planckarea) where A' is the area in metric. Then the formula becomes

1/(960 A') x (planckarea in metric) x (planckpower in metric),

in other words (1/960A') x hbar c

^{2}.

This hbar ceesquared is just what is needed to

make the units agree, if you choose to work in SI metric.