# Hawking Radiation and The Equivalence Principle

Important warning: This is just an artist’s depiction of a black hole, and what’s coming out of it certainly isn’t Hawking Radiation

Black Holes are awesome. On this, I think we can all agree. But even though they are insatiable eating machines, black holes won’t be around forever. Ultimately — and that ultimate day will be a long day off — black holes will evaporate via a process known as “Hawking Radiation.” You are probably already familiar with the idea of Hawking Radiation. If not, let me give you the simple version of it.

While we normally think of empty space as being, well, empty, in truth there’s a lot more going on there than you might suppose. The vacuum of the universe is a constantly bubbling cauldron of particles and antiparticles, coming into existence and quickly annihilating with one another. This is a consequence of the famous Uncertainty Principle. Sometimes, particles are created sufficiently close to the event horizon of a black hole that one particle falls in, and the other escapes, ultimately observable as radiation.

This leaves lots of unanswered questions, like: How does the Black Hole evaporate when it has particles falling into it? I tried to address this in an old “Ask a Physicist” but there are some who still aren’t satisfied.

Today, I wanted to give you a different way of thinking about — and a handwaving way of deriving — Hawking Radiation, one based on symmetry arguments, and in particular, on Einstein’s Equivalence Principle that I talked about earlier in the week. I’ll even use some terrible MS Paint diagrams to illustrate! At the end, for those interested, I’ll even give you a few equations to show how everything falls into place.

But first, a reminder. In one of its many, many forms, the Equivalence Principle states that:

Any local physical experiment not involving gravity will have te same result if performed in a freely falling inertial frame as if it were performed in the flat spacetime of special relativity.

Consider the International Space Station, currently in orbit about 400 km above the earth’s surface. We think of the ISS as being “in space,” but in a sense it’s very much in earth’s grasp. Gravity 400km up is only about 11% weaker than on the surface of the earth. And yet, if you’ve ever seen footage, the astronauts float around as if they’re weightless.

It’s no trick. In the case of the ISS, the free fall actually takes the form of a nearly circular orbit, but that’s still falling as far as gravity is concerned. The only reason that the ISS needs to be in space is to avoid colliding into mountains and so that air resistance won’t cause it to crash down to earth. The weightlessness is caused by the simple fact that it’s in free fall. A physicist on board should be able to do all sorts of experiments with the speed of light or muons or moving observers and find all of the effects anticipated by Special Relativity. That is the whole point of the Equivalence Principle.

Or very nearly. There’s a little bit of fine print that I feel compelled to mention. The gravitational pull of the earth gets weaker the further you get from the earth. That means that the earth side of the space station feels a slightly higher gravity than the space side, and that consequently, there’s a very, very tiny tidal effect that acts to “stretch” the station. The total tidal force on the entire 450 ton space station is only about a pound. You won’t get points off for ignoring it, and we will.

The Equivalence Principle sounds really trivial, but it’s not. In a few short steps, we’re going to go from Equivalence Principle to Hawking Radiation. And I suspect you haven’t thought of it in this way before.

If you take a charged particle and accelerate it, it will give off radiation. A radio transmitter works by jittering a bunch of electrons at the source and emitting radiation with a particular frequency. On the other hand, stationary particles — or even particles moving at a constant speed and direction — don’t emit any radiation at all.

2. Inside the ISS, freely-falling charges will appear to be standing still.

Imagine yourself as an astronaut inside the ISS. You’re in orbit around the earth, and apparently weightless. In order to do an experiment you take an electron and just let it go. If you’re careful, the electron will just appear to float in the middle of the station. After all, it’s orbiting the earth with the exact same trajectory as the rest of the station, so as far as you’re concerned, it’ll just sit still.

See what I said about the ghetto MS Paint Images?

Keep just one thing in mind: from your perspective, the doesn’t appear to accelerate. Ergo, no radiation.

3. To an outsider, both the ISS and the electron are accelerated.

Suppose I want to go to extreme measures to keep an eye on you. I build a giant 400km pedestal from the surface of the earth all the way up to the orbit of the ISS. Remember, I do not feel weightless. As I already pointed out, I only lose about 11% of my weight at this altitude.

From my perspective, you, the space station, and the electron speed by at nearly 17000mph. But even more importantly, I see all of you accelerated toward the center of the earth. Yes, even though the electrons are moving in a constant speed, because they are constantly changing directions — moving in a circle, in this case — they are absolutely being accelerated, and accelerated electrons will radiate. This is known as “synchrotron radiation,” and is a very useful experimental light source.

Of course, the direction of the acceleration doesn’t matter. All that matters is that I see a charge being accelerated and giving off light.

4. The universe is filled with random charge fluctuations.

One of the most surprising predictions of Quantum Field Theory is that the vacuum isn’t completely empty. Particles-antiparticle pairs are created out of nothing all the time. This shouldn’t cause you too much anxiety. I brought it up earlier, and you’ve had time to acclimate. We don’t encounter these particles because they typically get annihilated very, very quickly. To put things in perspective, electrons-positron pairs only last for about \$10^{-21}s\$, enough time for them to travel, at most, a bit more than the nuclear radius of an atom.

Particle Creation and Annihilation in the Vacuum. This is most definitely not to scale.

More to the point, since particles and antiparticles always have the exact opposite charge from one another, vacuum fluctuations won’t cause an excess of either positive or negative charges. There are reasons to be very wary about the vacuum energy density in the universe. For instance, a naive calculation of the total energy density yields something like $10^{120}$ times larger than the observed energy density of the universe. This, to my mind, is the worst problem in physics.

On the other hand, we can’t simply assume that there is no vacuum energy density. In 1948, Hendrik Casimir noticed that if you take two neutral metal plates and place them close together, they will attract one another. This is known as the Casimir Effect, and it basically only works if you imagine that there are a swarm of virtual charged particles between the plates — exactly with the properties predicted by our vacuum energy density.

5. If you accelerate through a vacuum, you’ll see radiation.

Now imagine yourself in a rocket ship flying through the vacuum. From your perspective, each of those virtual charged particles appears to be accelerating, and if the equivalence principle is right (it is), then those particles will appear to be accelerating according to you. And, as you now know, accelerating particles radiate. This was discovered by a number of people in the 1970’s, including William Unruh for whom the effect is named.

Very much not to scale.

I should point out that under normal circumstances, this is a tiny effect. Accelerating at “g,” you’d only see Unruh radiation of about 2 billionths of a degree Kelvin.

6. Being suspended outside a black hole is just like acceleration

Now we finally get to the main event. I’m not going to review all of the properties of black holes at this time, but for the moment, you need only consider one: there is an event horizon inside of which nothing can escape, not even light.

Outside the black hole, the gravity is very, very strong, and normally, the tidal forces near a black hole are sufficient to spaghettify you to death. However, supposing the black hole is big enough — at minimum about 10,000 times the mass of a sun — a freely-falling observer won’t even notice that he’s crossed the event horizon.

If you’re an observer dangling outside, you will absolutely notice the gravitational force; you can tell that you’re being accelerated. By the Equivalence Principle, there shouldn’t be any local observational differences between dangling in a gravitational field (or standing on the surface of the earth, for instance), or being accelerated in a rocket ship at the same rate.

You feel as though you’re being accelerated, and as far as Relativity is concerned, you are. Just as you would see Unruh radiation in an accelerating rocket ship, by the Equivalence Principle, you should see the same thing if you’re “really” in a gravitational field. In this case, we call it Hawking Radiation.

The amazing thing is that you don’t need to know much about gravity in order to get the correct result. Students of GR (including me, when I was one) are often troubled by how weirdly everything conspires in Hawking Radiation to get everything correct. Unruh radiation, on the other hand, only requires a bit of electromagnetic knowledge. That plus the Equivalence Principle and blammo! You have Hawking Radiation.

7. Some technical notes

I didn’t mark this overall entry as “technical” because while it’s detailed, there haven’t been any equations yet. Still, I imagine a few of you might want to see how this works out in practice.

First, Unruh radiation. If you’re accelerating at a rate, $g$ through the vacuum then you’ll appear to see a temperature:

$T_{Unruh}=frac{hbar g}{2pi c k_B}$

where $k_B$ is the Boltzman constant, and $hbar$ is the reduced Planck’s constant, as usual. Note that the gravitational constant, $G$ doesn’t appear anywhere in here.

Now, consider the gravity near the event horizon of a black hole. This gets a little confusing (and is far beyond the current discussion), since you need to worry about whether we’re talking about the gravitational force as seen using local coordinates near the black hole, or “global coordinates” seen from far away. We’re going to use the far away coordinates, since those are going to tell us about what the black hole gives off to someone far away.

Near the black hole, the acceleration is:

$g=frac{GM}{r_s^2}$

where $r_s$ is the Schwarzschild radius:

$r_s=frac{2GM}{c^2}$

If you plug all of this in, you find that the apparent temperature of a black hole is:

$T_{Black Hole}=frac{hbar}{2pi c k_B}frac{GM}{(2GM/c^2)^2}=frac{hbar c^3}{8pi k_B}frac{1}{GM}$

This is exactly the result that you get from doing a detailed GR analysis! Pretty cool, no?

As a side note, notice that the more massive the black hole is, the cooler it will be. If our sun were to turn into a black hole, it would radiate at a temperature of about 60 nano-Kelvin, and more massive black holes would be even cooler. This is insanely cold, about fifty million times cooler than the background temperature of the universe. Because heat flows from hot to cold, the radiation of the universe actually feeds a typical black hole. Only incredibly puny ones — less massive than the moon — are actually shrinking these days. Stellar mass black holes won’t actually start evaporating until the universe gets fifty million times cooler (and thus fifty million times bigger) than it is now. That won’t be for a few hundred billion years or so.

In other words, there’s no chance we could actually use Hawking Radiation to see black holes today. They’re just too cool. We do, however, see hot material falling on to black holes in the form of quasars, but that’s not the same thing as seeing the black hole, itself, and at any rate is a topic for another day.

-Dave

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