Feb 072009

I remain troubled by this business with black holes.

In particular, the zeroth law. Many authors, such as Wald, say that the zeroth law states that a body’s temperature is constant at equilibrium. I find this formulation less than satisfactory. Thermodynamics is about equilibrium systems to begin with, so it’s not like you have a choice to measure temperatures in a non-equilibrium system; temperature is not even defined there! A proper formulation for the zeroth law is between systems: the idea that an equilibrium exists between systems 1 and 2 expressed in the form of a function f(p1, V1, p2, V2) being zero. Between systems 2 and 3, we have g(p2, V2, p3, V3) = 0, and between systems 3 and 1, we have h(p3, V3, p1, V1) = 0. The zeroth law says that if f(p1, V1, p2, V2) = 0 and g(p2, V2, p3, V3) = 0, then h(p3, V3, p1, V1) = 0. From this, the concept of empirical temperature can be obtained. I don’t see the analog of this for black holes… can we compare two black holes on the basis of J and Ω (which take the place of V and p) and say that they are in “equilibrium”? That makes no sense to me.

On the other hand, if you have a Pfaffian in the form of dA + B dC, there always exists an integrating denominator X (in this simple case, one doesn’t even need Carathéodory’s principle and assume the existence of irreversible processes) such that X dY = dA + B dC. So simply writing down dM – Ω dJ already gives rise to an equation in the form X dY = dM – Ω dJ. That κ and A serve nicely as X and Y may be no more than an interesting coincidence.

But then there is the area theorem such that dA > 0 (just like dS > 0). Is that another coincidence?

And then there is Hawking radiation. The temperature of which is proportional to the surface gravity, T = κ/2π, which is what leads to the identification S = A/4. Too many coincidences?

I don’t know. I can see why this black hole thermodynamics business is not outright stupid, but I remain troubled.

 Posted by at 9:50 pm