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For a bose gas we can calculate the average number of particles through $$N = \int_0^\infty \rho(\varepsilon)n(\varepsilon) d\varepsilon$$ where $\rho(\varepsilon)$ is the particle density for energy $\varepsilon$ and $n(\varepsilon)$ is the Bose-Einstein-statistic.

Now we know that $\mu\leq 0$. So if we take the number of particles $N$ as a fixed number, we can see that the chemical potential $\mu(T)$ has to decrease for growing temperatures.

Now I am wondering how I can show this property mathematically?

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  • $\begingroup$ Am I wrong in saying that your question has actually nothing to do with BEC but that it concerns the chemical potential of a bosonic gas ? $\endgroup$ – Tom-Tom Dec 4 '14 at 9:19
  • $\begingroup$ I read this in a chapter about the BEC so that is why I asked it this way but yes, you are right. $\endgroup$ – dinosaur Dec 4 '14 at 18:51
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Let us suppose the gas is confined by a harmonic potential. The bosons have, in three dimensions, energy levels $\hbar\omega(n+3/2)$ with degeneracy $n(n+1)/2$. The grand-canonical partition function of level $n$ is (without degeneracy) $$ \xi_n=\sum_{p=0}^\infty \left(\mathrm e^{-\beta \hbar\omega(n+3/2)+\beta\mu(T)}\right)^p$$ where $p$ is the number of bosons occupying state $n$. This series must converge and therefore the exponential must be smaller than $1$ or if you prefer, $-\beta\hbar\omega(n+\frac32)+\beta\mu(T)<0$. Considering state $n=0$, we must have $\mu(T)<\frac32\hbar\omega$, which is a very small energy. We can shift the energies of this amount and get $\epsilon=\hbar\omega n$.

Let us now compute, as you did, the number of particles in the gas. To do this, we compute $\Xi=\prod_n \xi_n^{n(n+1)/2}$ the grand canonical partition function and find the expression of $N$ as $$N=\frac{\partial \ln \Xi}{\partial (\beta\mu)}=-\frac\partial{\partial(\beta \mu)}\sum_n\frac{n(n+1)}2\ln(1-\exp(\beta\mu-\beta\hbar\omega n)$$ $$ N=\sum_n\frac{n(n+1)}2 \frac{1}{\mathrm e^{\beta\hbar\omega n-\beta\mu}-1}$$ One recognizes the expression of the density of states and the Bose-Einstein statistics occupation number. As the separation between energy levels is small compared to the thermal energy, one can transform the sum into an integral and obtain the formula in your question, with the explicit form $$N=\frac1{\hbar^3\omega^3}\int \frac{\epsilon^2}{\mathrm e^{\beta\epsilon-\beta\mu}-1}\mathrm d\epsilon=\frac{2}{(\beta\hbar\omega)^3}\mathrm{Li}_3(\mathrm e^{-\beta\mu}).$$ $\mathrm{Li}_3$ is a trilogarithm function. Here is a plot of $\mathrm {Li}_3(\mathrm e^{x})$(upper blue curve) and of $\mathrm{Li}_3(\mathrm e^{x})/x^3$ (lower purple curve):

enter image description here

This function determines the chemical potential $\mu$ as the solution of $$\frac{\mathrm Li_3(\mathrm e^{-\beta\mu})}{(\beta\mu)^3}=N\left(\frac{\hbar\omega}{\mu}\right)^3.$$ We must therefore prove that this implicit definition of $\mu$ is increasing function of $\beta$. The graph of $\mathrm{Li}_3(\mathrm e^{x})/x^3$ provides a graphical answer. To prove it mathematically compute the derivative $$\partial_x\frac{\mathrm {Li}_3(\mathrm e^{x})}{x^3}=\frac{1}{x^4}\left(x\mathrm {Li}_2(\mathrm e^{x})-3\mathrm{Li}_3(\mathrm e^{x})\right)$$ and show that for $x<0$ this expression is negative.

I hope this answers your question, even though I have not proved it mathematically but only graphically.

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  • $\begingroup$ I was looking for a mathematical intuition and this is more than good. Thank you very much! $\endgroup$ – dinosaur Dec 5 '14 at 13:12

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