Could the chemical potential of a Bose gas be zero ?
If it was the case, we will have an infinite number of particles in the ground state ! No ?
But I've heard that for $T < T_c$, $\mu = 0$, so I don't understand...
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$\begingroup$ The chemical potential for a Bose gas can be zero, sure. There are many examples: photons in blackbody radiation have $\mu=0$, and so do phonons in the Debye model. Any Bose gas where $N$ is not conserved has zero chemical potential. $\endgroup$– Chay PatersonCommented Jun 25, 2013 at 21:10
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$\begingroup$ So, $n_0 = 1/(e^{-\beta \mu} - 1)$ should be infinite, no ? $\endgroup$– ArnaudCommented Jun 25, 2013 at 21:27
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$\begingroup$ @ChayPaterson : If you are in grand-canonical ensemble, you can only speak of conservation of the average number of particles. In grand-canonical ensemble, the number of particles (for a microscopical configuration )is not fixed. $\endgroup$– TrimokCommented Jun 26, 2013 at 8:15
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1$\begingroup$ @Trimok Sure, $N$ there is expected number of particles at a given temperature. $\endgroup$– Chay PatersonCommented Jun 26, 2013 at 14:10
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2 Answers
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Throughout, let's assume that the ground state energy of the system under consideration is zero.
Chay Paterson has addressed your question in the case of a gas of bosons in which the number of particles is not conserved, but from the wording of your question, it seems that you're concerned about the case in which the total number of particles is fixed.
For a system of bosons with a fixed number $N$ of particles, the answer to your question is
No. The chemical potential is nonzero for all $T>0$.
You point out, however, that it is often said that below the critical temperature $T_c$, the chemical potential is zero, so what's going on? The resolution is essentially that
the chemical potential very well-approximated by zero for almost all temperatures below the critical temperature, but it is never exactly zero.
As you cool the system down from above the critical temperature, the number $N_e$ of particles in excited states get's lower and lower. On the other hand, the number of particles in the excited states at any given temperature is bounded above, and this bound decreases as a function of temperature like $T^{3/2}$ (for non-relativistic systems in three dimensions). At a particular sufficiently low temperature (the critical temperature), this upper bound is lower than the total number of particles in the system and particles are forced into the ground state. But at this point, the chemical potential does not drop exactly to zero. It does, however, get very small very quickly as the temperature decreases and the number of particles in the ground state increases.
In fact, If we look at the number of particles in the ground state as a function of temperature $$ N_0 = \frac{1}{e^{-\mu(T)/kT}-1} $$ which gives $$ \mu(T) = -kT\ln\left(1+\frac{1}{N_0}\right) $$ then we see that the chemical potential is always strictly less than zero because the argument of the log is always strictly greater than $1$. But as you dial the temperature down below the critical temperature, and as the number of particles in the ground state increases towards $N$, the total number of particles in the system, which is presumably very large, and the chemical potential decreases since the argument of the log approaches $1$.
answered Jun 25, 2013 at 21:53
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$\begingroup$ Thank you very much, you've clarified my ideas (1 hour before my exam!). $\endgroup$– ArnaudCommented Jun 26, 2013 at 6:08
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2$\begingroup$ @joshphysics : If you are in grand-canonical ensemble, you can only speak of conservation of the average number of particles. In grand-canonical ensemble, the number of particles (for a microscopical configuration )is not fixed $\endgroup$– TrimokCommented Jun 26, 2013 at 8:18
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$\begingroup$ @Trimok I completely agree. The system we're considering in my response is not a system in the grand canonical ensemble. In fact, the chemical potential is determined as a function of temperature precisely by the constraint that the total number of particles $N$ is fixed. This is the standard situation in which Bose-Einstein condensation occurs. $\endgroup$ Commented Jun 26, 2013 at 15:11
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$\begingroup$ @Arnaud Sure thing. So how was the exam? $\endgroup$ Commented Jun 26, 2013 at 15:41
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2$\begingroup$ I never make predictions about my exams! $\endgroup$– ArnaudCommented Jun 26, 2013 at 16:44
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For a bose gas where particle number is not conserved, e.g. blackbody photons, indeed $\mu=0$. How does that work?
Well, as you approach zero frequency, the number of blackbody photons gets higher and higher -- yes, you bet, it approaches infinity. BUT nevertheless the total collective energy of those low-frequency photons gets lower and lower. (As you approach zero frequency, the energy per photon decreases faster than the photon population increases.)
So these ultra-low-frequency photons store negligible energy, even taking into account their immense (theoretically infinite) number. You can think the same way about whether these photons have noticeable effects on matter. The answer is No, despite their immense number, because as frequency decreases the interactions get weaker and weaker faster than the photon population increases.
So the number of photons is infinite, but nobody cares because they're totally undetectable in any way. Remember, we're talking about, say, photons in space with a period of 1 year and wavelength of 1 light-year.
For more details look up "infrared divergence". :-D
answered Jun 26, 2013 at 2:42