4
$\begingroup$

BE Problem

I am currently working on modelling the density of states and optical conductivity of graphene utilizing the GW algorithm. In calculating the exchange self energy of the system, the formula I am currently using is

$$f(w,T) = \frac{1}{e^{\frac{\omega}{T}}-1}$$

where the planck constant and the boltzmann constant $k$ is set to 1. To conservative bosonic particles, such as the core of helium-4, then it is believed to form a Bose-Einstein Condensate. I am dealing with the non-conservative bosonic particles, such as photons and phonons.

What would happen if I set the temperature $T$ equals to zero for the non-conservative bosons? My advisor believes that there would be no Bose-Einstein Condensate because the boson can pop in and out of the system. Is this true? If it is, what happens to the bosonic particles at the--or at least near--zero temperature?

Update and Edit

I've consulted my adviser and my colleagues, and this is the result. In no way that this is the solution, but it is one step further nonetheless. I've made a model of the BE distribution and it follows similarly to the image in the middle:

MB, BE, and FD Distributions
(source: universe-review.ca)

What I did is vary T = 1 K, 0.1 K, and 0.01 K in Scidavis (a numerical software on linux, if you are wondering). As the T decreases, the graph gets steeper and steeper; analytically, inserting T = 0 in the formula would equal to infinity. This of course, is the puzzling question.

Since I am required to put this in my calculation, my adviser suggested that at T = 0 the distribution equals to 0, where we assumed that the photons disappear after being absorbed by the electrons.

Why does this matter so much? Because I am now calculating the self energy of the system and the final formula requires a Hilbert transform integration from -inf to +inf. If there are anyone working on this problem or something similar, this would really help.

Improve this question
$\endgroup$
8
  • 3
    $\begingroup$ If you set T=0, then the third law of thermodynamics starts crying. $\endgroup$
    – CuriousOne
    Commented May 5, 2015 at 9:09
  • $\begingroup$ Do explain @CuriousOne, and perhaps shine a light on what happens on the distribution if I set T = 0.. $\endgroup$
    – M. Avicenna Naradipa
    Commented May 5, 2015 at 9:19
  • 4
    $\begingroup$ I think there's room for an interesting analysis of what happens in the limit as $T\to 0^+$. $\endgroup$
    – David Z
    Commented May 5, 2015 at 9:22
  • 1
    $\begingroup$ For T->0 naive theories break down... $\endgroup$
    – CuriousOne
    Commented May 5, 2015 at 10:11
  • 1
    $\begingroup$ @M. Avicenna Naradipa: No, the number of photons/phonons will not be conserved: when we assume that the temperature goes down, that means that the system is connected to a thermostat or something like that, so energy (and photons/phonons) can flow there. $\endgroup$
    – akhmeteli
    Commented May 5, 2015 at 10:13

2 Answers 2

Reset to default
4
$\begingroup$

It seems to me that you just cannot tell the difference between a Bose condensate and nothing in this case. What will change if you add some photons or phonons with zero energy to the system? No characteristics of the system will change. So it seems to me we have no criterion to decide if there is a Bose condensate in this case, and what's more important, it does not matter.

Improve this answer
$\endgroup$
7
  • $\begingroup$ You will get nothing all right, but it can never be a BEC because there is no way the photons can achieve equilibrium without interacting. $\endgroup$
    – SuperCiocia
    Commented Sep 25, 2019 at 9:25
  • $\begingroup$ @SuperCiocia : I am not sure I understand this. First, "You will get nothing all right, but it can never be a BEC" - but there is no difference between "nothing" and BEC in this case, so why do I get "nothing", but not BEC? Second, "there is no way the photons can achieve equilibrium without interacting" - but I would say photons interact with the thermostat and thus achieve equilibrium. $\endgroup$
    – akhmeteli
    Commented Sep 25, 2019 at 11:55
  • $\begingroup$ In order to be a BEC you need to have a macroscopic occuation of a single quantum state, and you must be in equilibrium. At $T=0$, there are no photons left in a resonator. What is the meaning of calling "nothing" a BEC? The interactions with the cavity walls do not count the thermalisation needed for a BEC -- you need interparticle thermalisation to achieve a non-zero chemical potential. Photon-photon interactions are very low, unless enhanced by a non-linear medium (this is exactly how they do achieve BEC of photons). $\endgroup$
    – SuperCiocia
    Commented Sep 25, 2019 at 18:21
  • $\begingroup$ @SuperCiocia : "At $T=0$, there are no photons left in a resonator." Of course, you can take such a position. That does not mean one cannot take an opposite position: "At $T=0$, there are a lot of photons in the ground state in a resonator." These two positions seem mutually contradictory, but neither of them (strongly) contradicts experimental results, as photons in the ground state are (pretty much) undetectable. Let me also remind you that the issue of the energy of empty space is pretty controversial (en.wikipedia.org/wiki/Cosmological_constant_problem). $\endgroup$
    – akhmeteli
    Commented Sep 26, 2019 at 2:09
  • $\begingroup$ I don't see where you're going with this. A BEC is a phase transition driven essentially by the thermal de Broglie wavelength becoming comparable with the interparticle separation. You cannot define such a thing for thermal photons in a cavity. The cosmological constant problem concerns zero point energies, so it would also affect atomic BECs. $\endgroup$
    – SuperCiocia
    Commented Sep 26, 2019 at 7:40
1
$\begingroup$

The Bose-Einstein occupancy at energy $E$ is given by: $$ f_{\mathrm{BE}} = \frac{1}{\mathrm{e}^{\frac{E-\mu}{k_BT}}-1}.$$

The chemical potential $\mu$ is the Lagrange multiplier associated with a fixed particle number $N$. What it means is that you can vary the temperature $T$, and $\mu(T$) will also vary so as to conserve your total number of particles $N$.

Photons do not have a chemical potential, and indeed your first formula is the BE occupancy with $\mu = 0$ and $E \propto \omega$.

Thermal photons do not become quantum degenerate: as $T\rightarrow 0$, photons vanish instead of forming a condensate. Indeed, black-body radiation ceases for an object at $T=0$.

A zero chemical potential means photons do not thermalise, which the key to achieve equilibrium and hence Bose-Einstein condensation. A laser isn't a BEC -- it's a coherent state all right, but it is not in equilibrium. The chemical potential stuff is derived in the context of equilibrium thermodynamics. So lasers are a completely different story.

In order to achieve true BEC in photons, you need to introduce photon-photon interactions, leading to photon thermalisation and a non-zero chemical potential. This has been achieved in nonlinear microcavities, dye-filled cavities and other forms of materials that interact with the photons thereby mediating the interaction amongst the photons.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.