
#1
Dec1412, 12:17 PM

P: 52

Why is it that the smaller a black hole is, the more Hawking radiation it emits? It seems counterintuitive to me. I would think that a larger black hole with a larger surface area would trap more antiparticles from virtual particle antiparticle pairs, hence emitting more Hawking radiation.




#2
Dec1412, 01:03 PM

P: 2,477

I can't really answer your question but wikipedia has article that does attempt to:
http://en.wikipedia.org/wiki/Hawking_radiation near the very end of the article it says that as the hole shrinks the temp increases exponentially until poof there a release of gamma radiation. (just above the Large Extra Dimensions section) So I guess the answer comes out of the mathematics that models the black hole and explains its thermodynamics. 



#3
Dec1412, 02:13 PM

P: 113

^ that is how i understand it as well. the smaller the black hole, the higher its temperature is. a stellar mass BH has a higher temperature than a supermassive BH...it therefore radiates more energy than the supermassive BH. as i understand it, we have not yet directly identified Hawking radiation b/c the temperature of stellar mass BH's is an infinitesimal fraction of a degree Kelvin. the temperature of a supermassive BH is therefore even lower. the ~2.7°K cosmic microwave background temperature/radiation therefore overwhelms and covers up any Hawking radiation in the foreground.
it makes for quite the paradox. on the one hand, there is no doubt that more quantum fluctuations (and therefore more particleaintiparticle pairs) occur near the event horizon of a supermassive BH than a stellar mass BH due to the substantial difference in the surface areas of their event horizons. but given that we know that smaller BH's have higher temperatures than larger BH's, they therefore must radiate more energy. i don't know how to resolve this one... 



#4
Dec1412, 02:26 PM

PF Gold
P: 5,690

Why do smaller black holes emit more Hawking radiation?Do you suppose that a sunsized object at n degrees and would radiate less energy than a basketball sized object at n + .01 degrees ? Really ? I'm not talking about the fact that in space if n < 2.7 then the net radiation of each is negative (they take in more than they give out) 



#5
Dec1412, 03:08 PM

PF Gold
P: 11,042

I believe it comes down the smaller black holes have a much steeper gravitational gradient. The force goes from "not so bad" to "trapping all light" much quicker as you approach a smaller black hole. I think this has the effect of causing more radiation output, but I don't know how.




#6
Dec1412, 03:16 PM

PF Gold
P: 5,690





#7
Dec1412, 10:20 PM

P: 113

so w/ regard to the original question, smaller BH's aren't necessarily more luminous (i.e. emit more Hawking radiation) than larger BH's  they're just more luminous per unit surface area than larger BH's, and therefore have higher "surface" (event horizon) temperatures than larger BH's. 



#8
Dec1512, 12:41 PM

P: 440

Let's say there is a photon factory, which produces one photon every second.
This photon factory is in space. A spacecraft delivers a massive load of fuel to the factory. This causes the photon production rate to drop, because of gravitational time dilation, and also the photons lose energy when climbing up from the gravity well. Let's say the production rate halved, and energy of each photon halves during the climbing, this causes the power of the photon stream to become 1/4 of its previous value. Dropping stuff into a black hole, so that its mass doubles, halves the temperature, makes the area of the event horizon 4 times larger, and causes the power of Hawking radiation to become 1/4 of the previous value. 



#9
Dec1512, 12:57 PM

Sci Advisor
P: 4,721

http://en.wikipedia.org/wiki/Hawking...ission_process That is, [tex]T_H \propto {1 \over R_s}[/tex] Total luminosity scales as the fourth power of temperature times the second power of the radius: [tex]L \propto R^2T^4[/tex] Therefore, the luminosity of a black hole scales with the inverse square of the radius: [tex]L_H \propto {1 \over R_s^2}[/tex] So yes, the smaller a black hole is, the brighter it is. And yes, this is precisely because the smaller a black hole is, the stronger its surface gravity. To understand this, remember the field theory picture of Hawking radiation: The field theory picture is that all throughout the vacuum, there are particle/antiparticle pairs popping into existence and rapidly annihilating with one another. Throughout most of the universe, these zeropoint field fluctuations don't have any effect. However, very near a black hole, it is possible for one of the two particles to be sucked into the black hole, while the other escapes. The one that is sucked into the black hole ends up subtracting from the black hole's mass, while the one that escapes becomes Hawking radiation. How efficient this process is depends entirely upon the gravity gradient at the horizon: if the gradient is low, then most of the time both particles in the pair will experience the same forces, and either both fall in or both escape, with no net effect either way. But if the surface gravity is high, then the closer of the two particles will be pulled towards the black hole that much more strongly than the other, leading to that much higher of a probability that only one will escape. In the end, the temperature of the black hole is simply given by its surface gravity: [tex]T_H = {\kappa \over 2\pi}[/tex] (here [itex]\kappa[/itex] is the surface gravity) 



#10
Dec1512, 10:55 PM

P: 52

Thanks to all those who replied. Now it makes sense.




#11
Dec2412, 04:07 PM

P: 5,634

A closely related description:



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