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Can a system entirely of photons be a Bose-Einsten condensate?

+ 3 like - 0 dislike
32 views

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?

This post imported from StackExchange Physics at 2014-04-01 12:41 (UCT), posted by SE-user metzgeer
asked Jan 5, 2012 in Theoretical Physics by metzgeer (15 points) [ no revision ]
Thanks metzgeer :-)

This post imported from StackExchange Physics at 2014-04-01 12:41 (UCT), posted by SE-user David Z
Ah, I've finally discovered a page on wiki en.wikipedia.org/wiki/Photon_gas which explains everything to me. Thanks all! Wish I had found it earlier.

This post imported from StackExchange Physics at 2014-04-01 12:41 (UCT), posted by SE-user metzgeer

2 Answers

+ 7 like - 0 dislike

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)

This post imported from StackExchange Physics at 2014-04-01 12:41 (UCT), posted by SE-user Chris Gerig
answered Jan 5, 2012 by Chris Gerig (540 points) [ no revision ]
Your system has matter in the form of walls. I'm thinking about a star bound by gravity made up entirely of photons, although I didn't write that above. In your system the temperature is determined by the interaction between the walls of chamber and the photons. I'm trying to understand the temperature purely in terms of a photon gas.

This post imported from StackExchange Physics at 2014-04-01 12:41 (UCT), posted by SE-user metzgeer
@metzgeer: There are no stable gravitating photon systems. These were searched for and studied by Wheeler, under the name "Geons", and as a result of Wheeler's research, it is now believed that any Geons are unstable.

This post imported from StackExchange Physics at 2014-04-01 12:41 (UCT), posted by SE-user Ron Maimon
+ 3 like - 0 dislike

It is possible to think of a Bose Einstein condensate is simply matter in a situation where it is described by a classical field. Any classical field is a BEC of its particle, so electromagnetic radiation is the BEC for photons.

Your question is whether there are electromagnetic fields which are thermally stable. This is not true, because there is no photon number conservation law in general, so the thermal equilibrium state is described by Plackian statistics. Chris Gerig described situations where you can have photon number conservation anyway, and the experimental realization of BEC in such systems is a more tranditional phase transition notion of BEC of photons.

But ignoring the number conservation issue, the electromagnetic field is a BEC of photons, although it does not normally make a statistical equilibrium state. Historically, Bose was thinking about photon statistics, and Einstein just generalized the situation of photons making a classical wave to find the matter condensate. So the photon statisitcs were the direct inspiration for the condensation (although the lack of particle number conservation means that you do not have a chemical potential, the aggregation of photons into a coherent state in electromagnetic radiation is physically the same as in any other BEC)

From your comments, it seems you were interested in a BEC of photons (an electromagnetic wave) making a gravitationally stable configuration. This possibility was extensively studied by Wheeler, and any such configuration is called a Geon. Geons are all believed to be unstable, much like a black hole with light orbiting unstably in circles at the smallest stable orbit radius. I am not aware of a proof of this, but I think it is widely accepted (and I also think it is true).

This post imported from StackExchange Physics at 2014-04-01 12:41 (UCT), posted by SE-user Ron Maimon
answered Jan 5, 2012 by Ron Maimon (7,295 points) [ no revision ]
Wait, isn't a group of coherent photons in thermal equilibrium just another way of saying laser. So if we imagine a pulse of coherent light not interacting with matter moving along a vector then we can ignore any gravitational component and still have a BEC. Wait this is Ron I'm talking to, oh why can't I get Bunny of Love out of my head, it burns! it Burns! This is right isn't it? A laser pulse is a BEC?

This post imported from StackExchange Physics at 2014-04-01 12:41 (UCT), posted by SE-user metzgeer
@Metzgeer: A laser is not, and cannot, be in thermal equilibrium. It is a "BEC of photons" because a BEC of matter is just like a matter laser, but the laser can't be in thermal equilibrium without making the laser go away. There is no thermal equilibrium BEC of photons, because their number is not conserved. Why does the bunny of love make people flip out? I like it very much. It is a post-apocalyptic Mad-Max style song, I don't hear that now that the cold war is over.

This post imported from StackExchange Physics at 2014-04-01 12:41 (UCT), posted by SE-user Ron Maimon

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