Background:
In Bose-Einstein stats the quantum concentration $N_q$ (particles per volume) is proportional to the total mass M of the system:
$$ N_q = (M k T/2 \pi \hbar^2)^{3/2} $$
where k Boltzmann constant, T temperature
Questions:
A) For a B-E system "entirely of Photons" – what is the total mass of the system? (answered, see below)
B) Does an ensemble of photons have a temperature? (answered, see below)
C) Is this a Bose-Einstein condensate?
I've found a paper here (like the paper put forward by Chris Gerig below) which finds a BEC, but it is within a chamber filled with dye, and the interaction of the photons with the dye molecules makes it a dual system, as to one purely of photons. I think in this case there is a coupling between the dye molecules and the photons that is responsible for the chemical potential in the partition equation
$$N_q = \frac{g_i}{e^{\left.\left(\epsilon _i- \mu \right)\right/\text{kT} – 1}}$$
where $g_i$ is the degeneracy of state i, $\mu$ is the chemical potential, $\epsilon_i$ is the energy of the ith state.
I suspect an Ansatz along the lines of $\mu$ = 0, and $\epsilon_i$ = $\hbar \nu_i$, where $\nu_i$ is the frequency of the i photon.
another edit:
After going for a walk, I've realized the Ansatz is almost identical to Planck's Radiation Law but the degeneracy = 1 and chemical potential = 0.
So, in answer to my own questions:
A) is a nonsensical question, as photons have no mass, noting from wiki on Quantum Concentration: "Quantum effects become appreciable when the particle concentration is greater than or equal to the quantum concentration", but this shouldn't apply to non-coupling bosons.
B) yes the ensemble has a temperature, but I was too stupid to remember photons are subject to Planck's Law.
C) Is this a Bose-Einstein condensate? No, as photons have no coupling or chemical potential required for a BEC.
So, for an exotic star composed entirely photons, all the photons should sit in their lowest energy levels and the star will do nothing more than disperse.
Is this right?
Best Answer
Well, photons are massless.
The key is the confinement of photons and molecules in an optical cavity long enough for them to reach thermal equilibrium.
A BEC is a state of matter that spontaneously emerges when a system of bosons becomes cold enough that a significant fraction of them condenses into a single quantum state to minimize the system's free energy. These particles act collectively as a coherent wave.
Blackbody photons (those in thermal equilibrium with the cavity walls) do not go through the phase transition. Unlike atoms, as photons are cooled in a cavity they simply diminish in number by disappearing into its walls.
By confining laser light within a thin cavity filled with dye at room temperature and bounded by two concave mirrors, it is possible to create the conditions required for light to thermally equilibrate as a gas of conserved particles. The photons exchange energy with the dye molecules through multiple scattering. The canonical condition for BEC is that the de Broglie wavelength of the bosons is comparable to the distance between them. Lowering their temperature is the usual approach. But for cavity photons, whose effective masses are so small that quantum effects emerge even at room temperature, density is the more convenient knob to turn.
So yes, a BEC of photons has been obtained:
BEC of Photons in an Optical Microcavity (Jan Klaers, et. al., doi:10.1038/nature09567)