6.7 Bose-Einstein Distribution

As the previous section explained, the energy distribution of a macroscopic system of particles can be found by merely counting system energy eigenfunctions.

The details of doing so are messy but the results are simple. For a system of identical bosons, it gives the so-called:

\begin{displaymath}
\fbox{$\displaystyle
\mbox{Bose-Einstein distribution:}\...
...1}{e^{({\vphantom' E}^{\rm p}- \mu)/{k_{\rm B}}T} - 1}
$} %
\end{displaymath} (6.9)

Here $\iota^{\rm {b}}$ is the average number of bosons in a single-particle state with single-particle energy ${\vphantom' E}^{\rm p}$. Further $T$ is the absolute temperature, and $k_{\rm B}$ is the Boltzmann constant, equal to 1.380,65 10$\POW9,{-23}$ J/K.

Finally, $\mu$ is known as the chemical potential and is a function of the temperature and particle density. The chemical potential is an important physical quantity, related to such diverse areas as particle diffusion, the work that a device can produce, and to chemical and phase equilibria. It equals the so-called Gibbs free energy on a molar basis. It is discussed in more detail in chapter 11.12.

The Bose-Einstein distribution is derived in chapter 11. In fact, for various reasons that chapter gives three different derivations of the distribution. Fortunately they all give the same answer. Keep in mind that whatever this book tells you thrice is absolutely true.

The Bose-Einstein distribution may be used to better understand Bose-Einstein condensation using a bit of simple algebra. First note that the chemical potential for bosons must always be less than the lowest single-particle energy ${\vphantom' E}^{\rm p}_{\rm {gs}}$. Just check it out using the formula above: if $\mu$ would be greater than ${\vphantom' E}^{\rm p}_{\rm {gs}}$, then the number of particles $\iota^{\rm {b}}$ in the lowest single-particle state would be negative. Negative numbers of particles do not exist. Similarly, if $\mu$ would equal ${\vphantom' E}^{\rm p}_{\rm {gs}}$ then the number of particles in the lowest single-particle state would be infinite.

The fact that $\mu$ must stay less than ${\vphantom' E}^{\rm p}_{\rm {gs}}$ means that the number of particles in anything but the lowest single-particle state has a limit. It cannot become greater than

\begin{displaymath}
\iota^{\rm {b}}_{\rm max} =\frac{1}{e^{({\vphantom' E}^{\rm p}- {\vphantom' E}^{\rm p}_{\rm gs})/{k_{\rm B}}T} - 1}
\end{displaymath}

Now assume that you keep the box size and temperature both fixed and start putting more and more particles in the box. Then eventually, all the single-particle states except the ground state hit their limit. Any further particles have nowhere else to go than into the ground state. That is when Bose-Einstein condensation starts.

The above argument also illustrates that there are two main ways to produce Bose-Einstein condensation: you can keep the box and number of particles constant and lower the temperature, or you can keep the temperature and box constant and push more particles in the box. Or a suitable combination of these two, of course.

If you keep the box and number of particles constant and lower the temperature, the mathematics is more subtle. By itself, lowering the temperature lowers the number of particles $\iota^{\rm {b}}$ in all states. However, that would lower the total number of particles, which is kept constant. To compensate, $\mu$ inches closer to ${\vphantom' E}^{\rm p}_{\rm {gs}}$. This eventually causes all states except the ground state to hit their limit, and beyond that stage the left-over particles must then go into the ground state.

You may recall that Bose-Einstein condensation is only Bose-Einstein condensation if it does not disappear with increasing system size. That too can be verified from the Bose-Einstein distribution under fairly general conditions that include noninteracting particles in a box. However, the details are messy and will be left to chapter 11.14.1.


Key Points
$\begin{picture}(15,5.5)(0,-3)
\put(2,0){\makebox(0,0){\scriptsize\bf0}}
\put(12...
...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}}
\end{picture}$
The Bose-Einstein distribution gives the number of bosons per single-particle state for a macroscopic system at a nonzero temperature.

$\begin{picture}(15,5.5)(0,-3)
\put(2,0){\makebox(0,0){\scriptsize\bf0}}
\put(12...
...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}}
\end{picture}$
It also involves the Boltzmann constant and the chemical potential.

$\begin{picture}(15,5.5)(0,-3)
\put(2,0){\makebox(0,0){\scriptsize\bf0}}
\put(12...
...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}}
\end{picture}$
It can be used to explain Bose-Einstein condensation.