# Why do smaller black holes emit more Hawking radiation?

1. Dec 14, 2012

### tensor33

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. Dec 14, 2012

### Staff: Mentor

I can't really answer your question but wikipedia has article that does attempt to:

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. Dec 14, 2012

### 94JZA80

^ 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 particle-aintiparticle 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. Dec 14, 2012

### phinds

Totally aside from the weirdness of black holes, the statement above does not make sense based on mass alone.

Do you suppose that a sun-sized 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. Dec 14, 2012

### Drakkith

Staff Emeritus
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. Dec 14, 2012

### phinds

Right. That's the "weirdness of black holes" that I mentioned. The statement I was addressing was one just about mass, which is why I don't think it makes sense.

7. Dec 14, 2012

### 94JZA80

no, of course not. i went back and read what i typed, and now see how silly it sounds. thanks for bringing it to my attention. my intention was to talk about luminosity/temperature per unit surface area, and i accidentally started talking about overall energy output. i meant to say that smaller BH's must emit more Hawking radiation per unit surface area than larger BH's. i was also trying to convey the latter part of what you said (about BH temps vs the CMBR temp) and that a BH would have to be quite small in diameter in order to radiate more energy than it absorbs and stand out against the CMBR.

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.

Last edited: Dec 14, 2012
8. Dec 15, 2012

### jartsa

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. Dec 15, 2012

### Chalnoth

A quick glance here shows that the temperature scales inversely with the Schwarzschild radius:

That is,
$$T_H \propto {1 \over R_s}$$

Total luminosity scales as the fourth power of temperature times the second power of the radius:

$$L \propto R^2T^4$$

Therefore, the luminosity of a black hole scales with the inverse square of the radius:

$$L_H \propto {1 \over R_s^2}$$

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/anti-particle pairs popping into existence and rapidly annihilating with one another. Throughout most of the universe, these zero-point 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:
$$T_H = {\kappa \over 2\pi}$$

(here $\kappa$ is the surface gravity)

10. Dec 15, 2012

### tensor33

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

11. Dec 24, 2012

### Naty1

A closely related description:

[I did not record the source in my notes.]