# What's the classical limit of the pressure of the ultrarelativistic Bose gas

The pressure for the ultrarelativistic Bose gas is

$$p~=~U/(3V) ~\propto~ (kT)^4/(hc)^3.$$

It looks to me like it diverges for $h \to 0$. Looking at the derivation, it diverges because $h$ is the unit volume in phase-space and letting it go to zero allows for an infinite number of states per volume. But I don't understand the physics.

My question is threefold:

1. Is it correct that the pressure diverges in the limit $h\to 0$ or is there some hidden $h$ dependence that I've missed?

2. Is this indeed the right classical limit? If not, what is?

3. What does that mean physically?

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First of all, when we say that something is the "classical limit", it doesn't mean that all interesting observables such as the pressure have to be finite (or zero?) in the limit. It just means that we only keep the leading order terms in an expansion in $\hbar$. So if the pressure diverged, there wouldn't be a direct contradiction.

Second of all, whether the expression diverges depends on what one keeps fixed. You implicitly keep $k,T,c$ fixed. In the SI macroscopic units, it's clear that not only $\hbar$ but also Boltzmann's constant $k$ is very small. It's natural to say that $k$ – a bridge between microscopic and macroscopic physics – isn't kept constant in the classical limit; $k$ tells you the energy per Kelvin per atom.

In an answer that just appeared, Ron keeps the number density $n$ (particle number per volume) constant in the limit. You may also scale $k$ with a power of $\hbar$ so that e.g. the gas constant $R=kN_A$ (the product of Boltzmann's constant and Avogadro constant) is kept constant. In the classical limit, $N_A$ therefore goes to infinity as $1/k$. Because one mole of the gas has $p=RT/V$ (all ideal gases obey that), and also $p=(kT)^4/(hc)^3$, you may cancel divide things to see $R/V=k(kT/hc)^3$. If you want to keep $R/V$ constant in the limit, you see that $k^4/h^3$ has to be kept constant as well, so $k$ scales like $h^{3/4}$ which is the hidden $h$-scaling you were probably asking for. In this scaling, $p$ according to your formula obviously remains finite in the limit.

However, there also exist other ways how to scale quantities when you define "the" classical limit. There isn't a single classical limit in a system in which many things may go to zero or infinity in various ways. This is a common theme in modern particle physics. For example, Yang-Mills theory with the coupling $g$ and $N$ colors has many classical limits. One may e.g. keep $N$ fixed and finite and send $g\to 0$ which is a kind of $\hbar\to 0$ in field theory, the weakly coupled classical limit. Alternatively, one may keep $g^2 N$ constant and small and send to $g\to 0$, while $N$ goes to infinity in the appropriate way; $g^2 N$ is what more naturally measuring how much "quantum" the theory is. When the 't Hooft coupling $g^2 N$ is kept fixed but large, we seemingly obtain an "anticlassical" limit but this so-called 't Hooft limit may actually be equivalently described as (leading topology, planar diagrams of) gravity (type IIB string theory, more precisely) on an anti de Sitter space which is actually another limit of the original gauge theory.

An inherently quantum system may have many classical limits.

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This is the pressure of a photon gas, and the divergence is because the photon number diverges in the classical limit, this is the ultraviolet catastrophe. You get no divergence if you hold the particle number fixed and take the ultra-relativistic hot limit, for example a relativistic plasma of electrons and protons.

For a fixed particle-number relativistic gas, the classical limit is sensible, and the classical pressure is p= nT, just as for any other ideal gas, where n is the number density, and T is the thermodynamic temperature. I prefer to express the ideal gas law as $\beta p = n$, but you learn it in grade school as PV=NRT.

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