The Hawking wattage of a black hole

  • Thread starter marcus
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What's the radiant power of a BH with surface area 10[sup]50[/sup]?

  • 0.36 nanowatts

    Votes: 1 100.0%
  • 0.36 microwatts

    Votes: 0 0.0%
  • 0.36 milliwatts

    Votes: 0 0.0%
  • 0.36 watts

    Votes: 0 0.0%

  • Total voters
    1
  • #1
marcus
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Assume the surface area of the event horizon of an ordinary black hole is 1050,
ordinary meaning uncharged and nonrotating.
The hole glows with Hawking radiation----what is the radiant power?


I gave the event horizon area in natural units but it's easy to convert to square meters if you wish it that way.
The area 1050 is 2.6 x 10-20 square meters, or 2.6 square angstroms.

In natural units (c=G=hbar=k=1) the radiant power of a BH
with area A is simply

1/960A

So having the area 1050 means that the luminosity is essentially 10-53. To which of the four wattage figures given in the poll is this equivalent?
 

Answers and Replies

  • #2
kyle_soule
240
1
For some reason I'm getting 1.0110686E-103 for the radiant power...is this right?
 
  • #3
marcus
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Originally posted by kyle_soule
For some reason I'm getting 1.0110686E-103 for the radiant power...is this right?

Could you have squared the area by mistake?

the area is E50 in Planck units
the formula for the overall power is 1/960A

so you should take reciprocal of 960E50
and it comes to about 1.04E-53 in units of Planck power

Planck power (c5/G) is an awesome amount of power.
It could supply the mass-energy of a whole galaxy in a few days.
You can easily calculate what it is in watts, if you look up the metric system values for G and c. It is around 3.6E52 watts---you can easily check this.

WAIT! Kyle you have calculated the radiant power PER UNIT AREA approximately correctly! The total power from the whole ball (namely 1.04E-53) divided by the area of the ball (namely E50) is indeed 1.04E-103.

Bravo Kyle! And thank you for responding.
 

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