Bose gas

A Bose gas is a model for a system of many non-interacting bosons, the simplest example of which is a bosonic ideal gas. A notable property is that three-dimensional Bose gases exhibit a phenomenon known as Bose-Einstein condensation at near-zero temperatures. The equation of state of a Bose gas is

\frac{P}{k_{B}T}=\frac{1}{\lambda ^{3}}g_{5/2}(z)

where $g_{5/2}(z)$ is a Bose function. Meanwhile, the particle density is

n=\frac{g_{3/2}(z)}{\lambda ^{3}}+ \frac{1}{V}\frac{z}{1-z}

The additional term is found by extracting the $p=0$ term from $N=\sum_{p}n_{p}$ before committing to integration. This is done because Bose-Einstein condensation causes a significant buildup of particle density at the ground state and this term (being a single point) would be neglected by the integral, causing a wrong result. For a derivation of these equations, see Ideal gas > In the quantum grand canonical ensemble.

Unlike fermions, there exist massless bosons on top massive ones (although electronic neutrinos are good approximations of massless fermions). A Bose gas of massless bosons has somewhat different properties to a gas of massive bosons, since ensembles of massless particles always have zero chemical potential. For an example of a Photon gas, see Black body. For an example a massless boson quasiparticle, see Harmonic oscillator > Phonons.

Critical condensation temperature#

Condensation can be illustrated through an gas of $N$ bosons. Assume this gas is enclosed in a cubic volume $V=L^{3}$ , where

N=\sum_{\mathbf{p}}\langle n_{\mathbf{p}} \rangle =\sum_{\mathbf{p}} \frac{1}{e^{\beta(\varepsilon_{\mathbf{p}}-\mu)}-1}

and $\varepsilon_{\mathbf{p}}=\mathbf{p}^{2}/2m$ . In the Thomas-Fermi approximation, it becomes

N=\int_{0}^{\infty}g(\varepsilon) \frac{1}{e^{\beta(\varepsilon-\mu)}-1}d\varepsilon=\int_{0}^{\infty} \frac{C_{\alpha}\varepsilon^{\alpha-1}}{e^{\beta(\varepsilon-\mu)}-1}

We can find an approximate solution for this integral¹, for instance with the Sommerfeld expansion. The more interesting question is: is there a nonzero critical temperature $T_{C}$ where the chemical potential vanishes, that is $\mu=\min\varepsilon_{p}=0$ ²? To check this, we try to solve the integral with $\mu=0$ and see where that leads us, if it's even possible.

The original solution is due to Einstein. Since at the condensation point the number of particles in the ground state blows up, we can split $N$ into $N_\text{exc}=\langle n_{p>0} \rangle$ and $N_\text{ground}=\langle n_{p=0} \rangle$ . The sum of these two makes $N$ . For most temperatures, $N_\text{ground}\simeq0$ and $\text{N}_\text{exc}\simeq N$ , which is why this split makes little sense in most conditions. The integral to solve is

N_\text{exc}(T=T_{C},\mu=0)=\int_{0}^{\infty} \frac{C_{\alpha}\varepsilon^{\alpha-1}}{e^{\beta_{C} \varepsilon_{p}}-1}d\varepsilon=N

where $\beta_{C}=1/k_{B}T_{C}$ . If there is a solution to this, then $T_{C}$ exists and is not zero.

To start, we make the substitution $\beta_{C}\varepsilon=x$ . This turns the integral into

N=k_{B}T_{C}\int_{0}^{\infty} \frac{C_{\alpha}(k_{B}T_{C})^{\alpha-1}x^{\alpha-1}}{e^{x}-1}=(k_{B}T_{C})^{\alpha}\int_{0}^{\infty} \frac{x^{\alpha-1}}{e^{x}-1}\ dx

This integral can be solved using the geometric series:

\begin{align} \int_{0}^{\infty} \frac{x^{\alpha-1}}{e^{x}-1}\ dx&=\int_{0}^{\infty} \frac{x^{\alpha-1}}{e^{x}(1-e^{-x})}\ dx \\ \text{(geometric series)}&=\int_{0}^{\infty}x^{\alpha-1}e^{-x}\sum_{j=0}^{\infty}(e^{-x})^{j} \\ &=\sum_{j=0}^{\infty}\int x^{\alpha-1}e^{-x}e^{-jx}\ dx \\ (j+1\to j')&=\sum_{j'=1}^{\infty} \int_{0}^{\infty} e^{-j'x}x^{\alpha-1}\ dx \\ (j'x=y)&=\sum_{j'=1}^{\infty} \int_{0}^{\infty} \frac{1}{j'}e^{-y}\left( \frac{y}{j} \right)^{\alpha-1}= \\ &=\underbrace{ \sum_{j'=1}^{\infty} \frac{1}{(j')^{\alpha}} }_{ \zeta(\alpha) }\underbrace{ \int_{0}^{\infty}e^{-y}y^{\alpha-1}dy }_{ \Gamma(\alpha) } \\ &=\zeta(\alpha)\Gamma(\alpha) \end{align}

using both the Riemann Zeta function and the Gamma function. Back to $N$ , we have

N=(k_{B}T_{C})^{\alpha}\Gamma(\alpha)\zeta(\alpha)\simeq N_\text{exc}

Curious selection of functions aside, we can invert the formula to find the critical temperature

\boxed{T_{C}=\frac{1}{k_{B}}\left[ \frac{N}{C_{\alpha}\Gamma(\alpha)\zeta(\alpha)} \right]^{1/\alpha}}

In the 3D case $d=3\to \alpha=3/2$ we get

T_{C}=\frac{1}{k_{B}} \frac{N^{2/3}}{\left[ \frac{Vm^{3/2}}{\sqrt{ 2 }\pi ^{2}\hbar ^{3}}\Gamma\left( \frac{3}{2} \right)\zeta\left( \frac{3}{2} \right) \right]^{2/3}}=\frac{1}{k_{B}}\frac{2\pi}{\left[ \zeta\left( \frac{3}{2} \right) \right]^{2/3}} \frac{\hbar^{2}n^{2/3}}{m}

using the density $n=N/V$ . This is a finite number, which proves that a free boson gas can condense. With realistic numbers for an atomic gas, we get a value of around $\sim 100$ nanokelvins³.

In both the 1D and 2D case, $\alpha\leq 1$ and since $\zeta(\alpha)\to \infty$ for $\alpha\leq 1$ , this gives $T_{C}\to0$ for these dimensions. Evidently, bosons cannot condense in 1 or 2 dimensions.

Condensed to not condensed ratio#

It is interesting to look at the ratio of bosons that are in the condensed phase compared to ones that aren't, i.e. the number $N_{0}/N$ . The number of uncondensed particles at the critical temperature $N_\text{exc}$ can be found as we did above. Then, the number of condensed ones is just $N_{0}=N-N_\text{exc}$ . However, an easier way to find the ratio is to calculate the number of uncondensed particles at or below the critical temperature. This leads to the same result, except not dependent on the critical temperature:

N_\text{exc}(T\leq T_{C})=C_{\alpha}\Gamma(\alpha)\zeta(\alpha)(k_{B}T)^{\alpha}

so the ratio of excited particles below the critical temperature is

\frac{N_\text{exc}}{N}=\frac{C_{\alpha}\Gamma(\alpha)\zeta(\alpha)(k_{B}T)^{\alpha}}{C_{\alpha}\Gamma(\alpha)\zeta(\alpha)(k_{B}T_{C})^{\alpha}}=\left( \frac{T}{T_{C}} \right)^{\alpha}

The ground state particles are then

\boxed{\frac{N_{0}}{N}=1-\left( \frac{T}{T_{C}} \right)^{\alpha}}

which is valid only when $T_{C}\leq T$ . In three dimensions it goes like this:

From Pathria & Beale's Statistical Mechanics, 3rd Ed., page 185

Also see Chapter 1 of Feynman's Statistical Mechanics. ↩
Remember that $\mu <\varepsilon_{p}$ for all $p$ , so if $\varepsilon_{p}\to 0$ then $\mu\to 0$ . ↩
Fun fact: when Einstein first solved this problem, he thought this was mostly a moot answer. Not because it was wrong, but because the temperatures you get are so tiny, he was convinced we would never reach them experimentally. He wasn't completely wrong: most methods of cooling, even modern ones, can't go below $\sim 10$ millikelvins, several order of magnitudes too high. But with the discovery of methods like laser cooling about 70 years later, it became possible to reach temperatures this low for small systems. What do you know, experiments showed that condensation actually happens. Also, side note: atomic helium gases have a critical temperature of about 3 kelvins and in fact go superfluid around that point. ↩

Critical condensation temperature#

Condensed to not condensed ratio#

Footnotes#