Hawking Radiation

  • Thread starter srfriggen
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  • #1
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Last night in my calculus class I learned about radioactive decay, a nice example using radium. I was taught that the smaller the mass, the slower the rate of decay (or at least that's what I absorbed, I still have to go over my notes again. Or does the rate not change, but the time it takes to decay is longer?). Does hawking radiation act in the opposite way? that the smaller the mass the faster the rate of decay? (and like above, i'm a bit confused if the rate of decay changes or remains constant, but perhaps just the mass decays faster).

As you can see I'm just looking for some clarification as I'm new to this stuff.
 

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  • #2
Nabeshin
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Yes, the power radiated through the hawking mechanism is inversely proportional to mass (squared, actually).

The wiki page has some nice derivations that aren't terribly involved, so that would be a good place to start.

http://en.wikipedia.org/wiki/Hawking_Radiation
 
  • #4
sylas
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Last night in my calculus class I learned about radioactive decay, a nice example using radium. I was taught that the smaller the mass, the slower the rate of decay (or at least that's what I absorbed, I still have to go over my notes again. Or does the rate not change, but the time it takes to decay is longer?). Does hawking radiation act in the opposite way? that the smaller the mass the faster the rate of decay? (and like above, i'm a bit confused if the rate of decay changes or remains constant, but perhaps just the mass decays faster).

As you can see I'm just looking for some clarification as I'm new to this stuff.
Yes. In [post=2513964]msg #11[/post] of thread 'Do black holes "evaporate" or go "bang"?', I have given a table of different sized black holes, with their mass, power output and temperature. The formulae to calculate these is also given.

As noted above, the power output is inversely proportional to mass squared. The formula for power output in Watts by Hawking radiation for a simple non-rotating hole of mass M kg is
[tex]
\frac{\hbar c^6}{15360 \pi G^2} M^{-2}[/tex]​
 
  • #5
George Jones
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Does hawking radiation act in the opposite way? that the smaller the mass the faster the rate of decay?
Yes, and since a black hole's temperature and radiation are related, a black hole has negative specific heat, which, speaking very loosely, means the following. Place a black hole in a fridge. When the black hole is taken out of the fridge, it is hotter than when it was put into the fridge. Place a black hole in an oven. When the black hole is taken out of the oven, it is cooler than when it was put into the oven.
Yes, the power radiated through the hawking mechanism is inversely proportional to mass (squared, actually).

The wiki page has some nice derivations that aren't terribly involved, so that would be a good place to start.

http://en.wikipedia.org/wiki/Hawking_Radiation
This looks similar to a post I made here,

https://www.physicsforums.com/showthread.php?t=205711.
 
  • #6
Ich
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I was taught that the smaller the mass, the slower the rate of decay
Most probably that meant the mass difference between the decaying nucleus and the decay products.
Or, if it really meant the mass of the nucleus: that works differently than gravitation. Gravitation tends to keep things together: the bigger the things, the more gravitation, the tighter the binding and the longer the life time.
In a nucleus, the only long-reaching force is the electrostatic repulsion of the protons. The more protons, the more repulsion, the weaker the binding, the shorter the life time.
 

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