# Why is temperature inversely proportional to the horizon's area?

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1. Oct 2, 2014

### David Carroll

As I posted in another thread, I'm giving the caveat that I am no physicist and have only a rudimentary knowledge of math.

Anyway, I am currently reading a book called "Three Roads to Quantum Gravity" by Lee Smolin. I came across a section of the book that confused me. Namely, Dr. Smolin showed that an accelerating observer would register temperatures that were proportional to the rate of acceleration because the greater the rate of the acceleration, the greater the area of the horizon that separates quantum fluctuation photons in her field of measurement from the corresponding pairs on the other side of the horizon, and the greater resulting indeterminateness would yield more randomness in the photons and thus a higher temperature.

What is confusing to me, is that Lee Smolin then compared this situation with that of a black hole which itself has a horizon but then proceeded to quote an equation by Stephen Hawking which stated that the temperature of a black hole is inversely proportional to its horizon. This lost me, especially since Lee Smolin showed earlier in the book that Einstein brought us the equivalence of gravity and acceleration. And wouldn't a more massive black hole have a stronger gravitational field than a less massive one?

2. Oct 2, 2014

### Staff: Mentor

To its mass, which means to the "size" of its horizon, yes. See below.

It depends on what you mean by "stronger". ;)

Here's a heuristic way to look at it: the laws of black hole thermodynamics that Hawking came up with give the following correspondences between the usual thermodynamic quantities and their black hole counterparts:

Energy -> Mass of the hole
Entropy -> Area of the hole's horizon
Temperature -> "Surface gravity" at the hole's horizon

I put "surface gravity" in quotes because it's not the actual "acceleration due to gravity" at the horizon; that's infinite (i.e., the acceleration required to hold station at a given altitude above the horizon increases without bound as the altitude goes to zero, i.e., as the horizon is approached). The "surface gravity" is better thought of as the "redshifted" acceleration at the horizon; that is, it's the force per unit mass that would have to be exerted "at infinity" to hold an object static at the horizon. Even this requires some care in definition, because you can't actually hold an object static at the horizon. But you can, for example, consider an object held static at some altitude by a rope, with the other end of the rope being held by an observer at infinity, and compute the force per unit mass the observer at infinity would have to exert as a function of altitude, and take the limit as the altitude goes to zero, i.e., as the object at the lower end of the rope approaches the horizon. That limit is finite, and is equal to the "surface gravity" of the hole.

Now the key point is this: when you compute the force per unit mass required at infinity to hold an object static at some radius $r$, you find that it is exactly equal to the Newtonian formula for acceleration, i.e., it is just $G M / r^2$! So if we plug in the horizon radius $r = 2GM / c^2$, we get a "surface gravity" of $c^4 / 4 G^2 M$. (The weird constant factor of $c^4 / G^2$ is just because we're using ordinary units instead of "natural" ones; if you work it out, you will see that everything still works out to have units of acceleration.) As you can see, the "surface gravity" is inversely proportional to the mass $M$ of the hole. And by the correspondence above, that means the temperature is too.

In other words, in at least one sense (the sense of "surface gravity"), a more massive black hole has a weaker gravitational field than a less massive one. This is indeed very different behavior from ordinary objects: for example, the planet Jupiter, with 318 times the mass of the Earth, has stronger surface gravity than the Earth does, and the Sun, about 1100 times the mass of Jupiter, has stronger surface gravity than Jupiter does. Black holes are not ordinary objects.

Last edited: Oct 2, 2014
3. Oct 2, 2014

### Staff: Mentor

Btw, you can actually play the same trick here as I did with black holes in my previous post: you can define a "horizon" for the accelerating observer (this is called a "Rindler horizon" and is the one talked about in the quantum field stuff you mentioned) and define a "surface gravity" for that horizon such that the statement quoted above is true, with "surface gravity" playing the role of "acceleration". In fact, the reason this works for black holes is that a black hole's horizon and a Rindler horizon have key features in common (the main one is that they are both "Killing horizons", which you can Google for more information on the concept).

4. Oct 2, 2014

### David Carroll

Okay. Thank you so much for taking time to explain this to me. Also, one more question, if you have the patience. Lee Smolin, in the same book, equates the concept of entropy with "missing information" about a system of particles. In other words, the more "missing information" the more entropy. In nuclear reactor school in the Navy (so I do know a little something about physics in the crash-course way that the Navy teaches), we were taught to view entropy in sort of the following way: If you were to take a "snapshot" of any subset of a given volume of the total system of particles, the more one could not tell apart that particular subset of particles from any other subset of the same volume, the more entropy the entire system had.....something along those lines. But to me, this seems that the more entropy a system has, the more information we have about it, since all we'd have to do is take a "snapshot" of a subset of an almost arbitrarily small volume and predicate the characteristics of that subset to the entire system of particles.

5. Oct 2, 2014

### bapowell

Wouldn't one's horizon shrink the greater the rate of acceleration? Intuitively, I am able to observe less and less of the universe the more rapidly I accelerate. I don't have a reference handy, but I believe the Rindler horizon goes something like $T \propto 1/a$ in Rindler coordinates.

6. Oct 2, 2014

### Staff: Mentor

This is sort of right, but not completely. Let me give a better definition.

Suppose we have a system in a given macrostate, which means we have known definite values for all the macroscopic thermodynamic parameters--temperature, pressure, density, etc. Then the entropy of the system is the logarithm of the number of microstates (where the microstate is something like the set of individual positions and velocities for all the particles in the system, for example a gas) that are consistent with the known macrostate (where "consistent" means "will give the known values for the macroscopic parameters when the appropriate computations are done using the microstate data as a starting point"). So the higher the entropy of the system, the more microstates there are that are consistent with what we know, i.e., with the macrostate. That is the sense in which higher entropy means more "missing information", as Smolin said.

Given the above, it should be evident why this is backwards. The higher the entropy of the system, the more microstates are consistent with the known macrostate. But that means that a given small subset of the system gives less information about other subsets, because there are more possible configurations of subsets that are consistent with the known macrostate. (To see this at the extreme, consider picking a single particle as the "subset" of the system: the higher the entropy, the more individual particle configurations there are that are consistent with the known macrostate, which means a particular particle chosen at random gives less information about the state of the others.)

7. Oct 2, 2014

### Staff: Mentor

Yes, it does.

8. Oct 2, 2014

Ah, I see.