I’m reading Robert Wald’s book, *Quantum Field Theory in Curved Spacetime and Black Hole Thermodynamics*, and I am puzzled. According to Wald, the black hole equivalent of the First Law reads (for a Kerr black hole):

(1/8π)κ*dA* = *dM* – Ω*dJ*,

where κ is the surface gravity, *A* is the area of the event horizon, *M* is the mass, Ω is the angular velocity of the event horizon, and *J* is the black hole’s angular momentum.

The analogy with thermodynamics is obvious if one write the First Law as

*TdS* = *dU* + *pdV,*

where *T* is the temperature, *S* is the entropy, *U* is the internal energy, *p* is the pressure, and *V* is the volume. Further, as per the black hole area theorem, which Wald proves, *A* always increases, in analogy with the thermodynamical entropy.

But… if I am to take this analogy seriously, then I am reminded of the fact that in a thermodynamical system the temperature is determined as a function of pressure and volume, i.e., there is a function *f* such that *T* = *f*(*p*, *V*). Is there an analogue of this in black hole physics? Is the surface gravity κ fully determined as a function of Ω and *J*? It is not obvious to me that this is the case, and Wald doesn’t say. Yet without it, there is no zeroth law and no thermodynamics. He does mention the zeroth law in the context of a single black hole having uniform surface gravity, but that’s not good enough. It doesn’t tell me how the surface gravity can be calculated from Ω and *J* alone, nor does it tell me anything about more than one black hole being involved, whereas in thermodynamics, the zeroth law is about *multiple* thermodynamical systems being in thermal equilibrium.

Another puzzling aspect is that the area theorem has often been quoted as “proof” that a black hole cannot evaporate. Yet again, if I take the analogy with thermodynamics seriously, the Second Law applies only to closed systems that exchange neither matter nor energy with their environment; it is, in fact, quite possible to reduce *S* in an open system, otherwise your fridge would not work. So if a black hole can exchange energy and matter with its environment, perhaps it can evaporate after all.

Moreover, for the analogy to be complete, we’d also be required to have

8π∂*M*/d*A* = κ,

∂*M*/∂*J* = Ω,

just as in ordinary thermodynamics, we have *T* = ∂*U*/∂*S* and *p* = –∂*U*/∂*V*. So, do these relationships hold for black holes?

I guess I’ll go to ArXiv and read some recent papers on black hole thermodynamics.

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