Why can't a gas of photons reach a Bose-Einstein condensate?

Question

I have read in many places that as the gas of photons has a chemical potential $\mu=0$ it can't reach a Bose-Einstein condensate (BEC), but I don't understand why.

I am reading Greiner's "Thermodynamics and Statistical Mechanics" and in chapter 13 "The Ideal Bose Gas", he first derives the Bose-Einstein conditions for an ideal Bose gas and then, as a particular example, he derives some relations for the gas of photons. I think I understand the case for the ideal gas, but when I try to apply that same approach to the gas of photons, I can't seem to reach the conclusion that the BEC is not possible.

So, a quick summary of how I understand that an ideal Bose gas condensates:

Ideal bose gas

The mean occupation number is given by: $$ N = \sum_k \frac 1 {e^{(E_k-\mu)/kT}-1} \tag{1} $$ And, to compute this summation, we can approximate it as an integral which results in: $$ N_E = \frac V {\lambda^3} g_{3/2}(z) \tag{1.1}$$ where the $z=e^{\mu/kT}$ is called the fugacity and: $$ g_n(z) = \frac 1 {\Gamma(n)} \int_0^\infty \frac {x^{n-1}dx} {z^{-1}e^x-1} \tag{1.2} $$ The reason why I am labeling this as $N_E$ is because the approximation leaves out the case for the ground state (when $E_k=0$), so the mean occupation number is in reality: $$ N = N_e + N_0 \tag{1.3}$$ where $$ N_0 = \frac z {1-z} \tag{2} $$ which is just $(1)$ evaluated for $E_k=0$. We can see from here that $N_0$ diverges when $z=1$ (which is when $\mu=0$), for which $N_E$ has a finite value (as all $g_n(1)$ are finite for $n>1$). This means that, in this scenario, almost all particles are in the ground state which results in the Bose-Einstein condensate.

Now, for the gas of photons (using Greiner's equations):

Gas of photons

Equation $(1)$ still holds if we set $\mu=0$, which must mean that equation $(2)$ also holds, so (considering de density of states $g(E)$ as @SuperCiocia noticed were missing and adding more details): $$ N_0 = \lim_{\mu \to 0} \frac {e^{-\mu/kT}} {1 - e^{-\mu/kT}} \to \infty \tag{3} $$ And approximating $N$ as an integral, we get: $$ N_E = \int \langle n_E \rangle g(E)dE = \frac {8\pi V} {h^3 c^3} \int_0^\infty \frac {E^{2}dE} {e^{E/kT}-1} = \frac {8\pi V k^3 T^3} {h^3 c^3} g_3(1) \tag{4} $$ which is finite. Then this must mean that the gas of photons is always in the Bose-Einstein condensate phase, which clearly doesn't make any sense.

Can someone tell me, what am I doing wrong?

I have checked some questions about this (like what-is-condensed-light, can-a-system-entirely-of-photons-be-a-bose-einsten-condensate) and the explanations given are more of physical intuition about photons popping out of existence (as $\mu=0$) which implies there is never a saturation of states (either ground state or any excited state), but I can't seem to relate these equations to that argument. The closest I have found is why-zero-chemical-potential-does-not-allow-the-bose-einstein-condensation-of-phonons which performs a similar analysis, except that instead of getting $N_0$ from $(1)$ (as I did here and as Greiner does), he calculates the limit of $N_E$ when $E \to 0$ which yields a correct result, but then I don't get why Greiner doesn't do it that way.

Answer 1

It can, it's just hard as you have to engineer it.
See, for instance, here.

A Bose-Einstein Condensate is more than just a state described by a macroscopic wavefunction. It's the result of a phase transition. Indeed, it's the phase transition that defines the BEC phase, as the breaking of the $U(1)$ symmetry is what guarantees the choice of a specific phase and all that. In this context, Bose-Einstein condensation is different from "Bose-Einstein proliferation", i.e. just a lot of bosons in the ground state.

So you have to show that your system can go from a thermal state to a BEC as you decrease the temperature (or increase particle number).

Crash course in Bose-Einstein condensation

A BEC is essentially a saturation effect; the occupancy of bosons $$f(E) = \frac{1}{\mathrm{e}^\frac{E-\mu}{k_{\mathrm{B}}T}-1}$$ has to be positive, which means that $E-\mu \geqslant 0 \quad \forall E$. So you if fix $\mu$ and choose your ground energy to be $E_0 = 0$ (hence $\mu \leqslant 0$), any energy level is capped at $\mathrm{max}[f(E)] = (\mathrm{e}^\frac{E}{k_{\mathrm{B}}T}-1)^{-1}$.

Because the occupancy is capped, energy levels may run out of places to accommodate particles. Decreasing the temperature $T$ helps bringing this cap sooner. So as you decrease the temperature $T$, the $n^{\mathrm{th}}$ energy level doesn't have any free spots any longer. Etc. At some point (critical temperature $T_{\mathrm{c}}$), all excited states ($E>0$) are full.
If particle number is conserved, then particles cannot disappear. They have to go somewhere. Indeed, they go to the only state with infinite acceptance, i.e. the ground state with $E_0 = 0$ and hence $f(E_0) \rightarrow \infty$. It is this saturation that triggers the macroscopic occupation of the ground state.

Ok so what about the chemical potential?

Essentially, the chemical potential $\mu$ is the change in Helmholtz free energy $F = U-TS$ when a particle is added to the system. Adding a particle at a particular temperature increases the internal energy $U$, but this extra particle results in many more possible arrangements of the particles in the system, which in turn increases the entropy $S$. In the thermal phase, the entropy change is larger than the energy term, hence the chemical potential is negative $\mu < 0$. This agrees with what found above from the mathematical requirement of $f(E) > 0$.

When you hit condensation, then new particles can only be allocated in the ground state. Which has energy zero, so $U=0$. The certainty of the state where it ends up in, also, means that entropy does not increase$^\dagger$. Hence $\mu = 0$ only for $T \leqslant T_{\mathrm{c}}$.

Photons

For (free) photons, $\mu = 0$ always. It is not a function of temperature. It does not entail any interesting dynamics.

The Planck's distribution, indeed, tends to zero for $T \rightarrow 0$. Which is the same thing, really: photons just "vanish". Objects do not radiate as much when cold.

So how do you make a gas photons undergo Bose-Einstein condensation?
You force them to adpot $\mu \neq 0$. For example by placing them in a cavity, where different modes interact via a dye -- as done in the reference in the first line.

Addenda:

1) Is a laser a BEC?
No. While both a laser and a BEC are coherent states, the latter is an equilibrium state of matter while the former is a "steady-state" -- meaning pumping and stimulated emission are balanced but both need to be non-zero. And pumping is external.

2) In your equations you forgot an essential part; the density of states $g(E)$. Its functional dependence with dimensionality $d$ is what determines which trap geometries can have a BEC.

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$^\dagger$: Indeed, the BEC is a coherent state with zero entropy. Experimentally reaching a BEC therefore not only requires loss of energy via cooling, but also, more importantly, the removal of entropy. This dictates which cooling mechanisms are useful and which are not (e.g. adiabatic relaxations).