Hawking Radiation: Mass & Decay Rate

[srfriggen]

Posted this on the Astrophysics page but I think it is more appropriate here...

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, so be gentle :)

 

[Bob S]

For radioactive decay, there is no correlation between the nuclear mass and the decay rate. Compare carbon-14 (half life 5730 years) to gold-198 (half life 2.7 days), or tritium (12 years) to uranium 238 (4 x 10⁹ years). There is a correlation between the nuclear mass difference (between before and after decay) and the decay rate, however. Hawking radiation is very different, and is related (I believe) to the Heisenberg uncertainty principle and virtual particles at a black hole's event horizon. In this case, the heavier the particle, the shorter the "lifetime". In any event, the Hawking radiation "lifetime" is ≈10^-22 seconds (not including General Relativity effects).

Bob S

 

[Naty1]

There is no relationship between the weak nuclear force responsible for radioactive decay and Hawking radiation from black holes. So they are no opposites in any sense.

You can read about Hawking radiation here

http://en.wikipedia.org/wiki/Hawking_radiation

and also search here on physics forums...
Hawking radiation has not been experimentally verified. Black holes in our universe accrete mass and energy as they are colder than their surroundings...someday they will presumably give off detectable Hawking radiation as they begin to diminish in size (mass) and get hotter...and as the cosmos gets colder...

Hawking radiation is a "perfect" thermal radiation...very much like black body (thermal) radiation...but not identical.

 

[Frame Dragger]

Posted this on the Astrophysics page but I think it is more appropriate here...

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, so be gentle :)

You're right actually. Black holes have an inverse relationship between mass and temperature.

EDIT: I should say, you've reached the right conclusion for the wrong reasons see: Naty1 and Bob S

 

[Galap]

To explain this one must first explain how decay really works.

Decay is random, and can only be modeled predictively when there are a lot of atoms (of Radium to use your example.)

A given radium in a sample may decay very quickly, or it may wait nearly an eternity to decay, longer than the universe has existed. Whether or not the universe is deterministic is another issue, but the only way we can guess what is going to happen is probability.

Now, if there are a lot of atoms, what is the probability that one of them decays at a given instant? Higher than if there are less. If you integrate that, you get the exponential decay we all know. We discuss decays in terms of half lives. The half life means this: "on average, how long will it take for half of the atoms to decay?". This is a specific number. What the sample actually does deviates from this, but not by much (it gets better the more atoms you have).