Peeter Joot's (OLD) Blog.

Math, physics, perl, and programming obscurity.

PHY452H1S Basic Statistical Mechanics. Lecture 19: Bosons. Taught by Prof. Arun Paramekanti

Posted by peeterjoot on March 28, 2013

[Click here for a PDF of this post with nicer formatting (especially if my latex to wordpress script has left FORMULA DOES NOT PARSE errors.)]

Disclaimer

Peeter’s lecture notes from class. May not be entirely coherent.

Fermions summary

We’ve considered a momentum sphere as in fig. 1.1, and performed various appromations of the occupation sums fig. 1.2.

Fig 1.1: Summation over momentum sphere

Fig 1.2: Fermion occupation

\begin{aligned}\epsilon \sim T^2\end{aligned} \hspace{\stretch{1}}(1.0.1.1)

\begin{aligned}C \sim T\end{aligned} \hspace{\stretch{1}}(1.0.1.1)

\begin{aligned}P \sim \text{constant}\end{aligned} \hspace{\stretch{1}}(1.0.1.1)

The physics of Fermi gases has an extremely wide range of applicability. Illustrating some of this range, here are some examples of Fermi temperatures (from E_{\mathrm{F}} = k_{\mathrm{B}} T_{\mathrm{F}})

  1. Electrons in copper: T_{\mathrm{F}} \sim 10^4 \mbox{K}
  2. Neutrons in neutron star: T_{\mathrm{F}} \sim 10^7 - 10^8 \mbox{K}
  3. Ultracold atomic gases: T_{\mathrm{F}} \sim (10 - 100) \mbox{n K}

Bosons

We’d like to work with a fixed number of particles, but the calculations are hard, so we move to the grand canonical ensemble

\begin{aligned}n_{\mathrm{B}}(\mathbf{k}) = \frac{1}{{ e^{\beta(\epsilon_\mathbf{k} - \mu)} - 1 }}\end{aligned} \hspace{\stretch{1}}(1.2)

Again, we’ll consider free particles with energy as in fig. 1.3, or

\begin{aligned}\epsilon_\mathbf{k} = \frac{\hbar^2 k^2}{2 m}.\end{aligned} \hspace{\stretch{1}}(1.3)

Fig 1.3: Free particle energy momentum distribution

 

Again introducing fugacity z = e^{\beta \mu}, we have

\begin{aligned}n_{\mathrm{B}}(\mathbf{k}) = \frac{1}{{ z^{-1} e^{\beta \epsilon_\mathbf{k}} - 1 }}\end{aligned} \hspace{\stretch{1}}(1.4)

We’ll consider systems for which

\begin{aligned}N = \sum_\mathbf{k} n_{\mathrm{B}}(\mathbf{k}) = \text{fixed}\end{aligned} \hspace{\stretch{1}}(1.5)

Observe that at large energies we have

\begin{aligned}n_{\mathrm{B}}(\text{large} \, \mathbf{k}) \sim z e^{-\beta \epsilon_\mathbf{k}}\end{aligned} \hspace{\stretch{1}}(1.6)

For small energies

\begin{aligned}n_{\mathrm{B}}(\mathbf{k} \rightarrow 0) \sim \frac{1}{{z^{-1} - 1}} = \frac{z}{1 - z}\end{aligned} \hspace{\stretch{1}}(1.7)

Observe that we require z < 1 (or \mu < 0) so that the number distribution is strictly positive for all energies. This tells us that the fugacity is a function of temperature, but there will be a point at which it must saturate. This is illustrated in fig. 1.4.

Fig 1.4: Density times cubed thermal de Broglie wavelength

 

Let’s calculate this density (assumed fixed for all temperatures)

\begin{aligned}\rho &= \frac{N}{V} \\ &= \int \frac{d^3 \mathbf{k}}{(2 \pi)^3} \frac{1}{{z^{-1} e^{\beta \epsilon_\mathbf{k}} -1 }} \\ &= \frac{2}{(2 \pi)^2} \int_0^\infty k^2 dk \frac{1}{{z^{-1} e^{\beta \hbar^2 k^2/2m} -1 }} \\ &= \frac{2}{(2 \pi)^2} \left( \frac {2 m} {\beta \hbar^2} \right)^{3/2}\int_0^\infty \left( \frac {\beta \hbar^2} {2 m} \right)^{3/2}k^2 dk \frac{1}{{z^{-1} e^{\beta \hbar^2 k^2/2m} -1 }}\end{aligned} \hspace{\stretch{1}}(1.8)

With the substitution

\begin{aligned}x^2 = \beta \frac{\hbar^2 k^2}{2m},\end{aligned} \hspace{\stretch{1}}(1.9)

we find

\begin{aligned}\rho \lambda^3 &= \frac{2}{(2 \pi)^2} \left( \frac {2 \not{{m}}} {\not{{\beta \hbar^2}}} \right)^{3/2}\left( \frac{ 2 \pi \not{{\hbar^2 \beta}}}{\not{{m}}} \right)^{3/2}\int_0^\infty x^2 dx \frac{1}{{z^{-1} e^{x^2} -1 }} \\ &= \frac{4}{\sqrt{\pi}} \int_0^\infty dx \frac{x^2}{z^{-1} e^{x^2} - 1 } \\ &\equiv g_{3/2}(z).\end{aligned} \hspace{\stretch{1}}(1.10)

This implicitly defines a relationship for the fugacity as a function of temperature z = z(T).

It can be shown that

\begin{aligned}g_{3/2}(z) = z + \frac{z^2}{2^{3/2}}+ \frac{z^3}{3^{3/2}}+ \cdots\end{aligned} \hspace{\stretch{1}}(1.11)

As z \rightarrow 1 we end up with a zeta function, for which we can look up the value

\begin{aligned}g_{3/2}(z \rightarrow 1) = \sum_{n = 1}^\infty \frac{1}{{n^{3/2}}} = \zeta(3/2) \approx 2.612\end{aligned} \hspace{\stretch{1}}(1.12)

where the Riemann zeta function is defined as

\begin{aligned}\zeta(s) = \sum_{ n = 1 } \frac{1}{{n^s}}.\end{aligned} \hspace{\stretch{1}}(1.13)

\begin{aligned}g_{3/2}(z) = \rho \lambda^3\end{aligned} \hspace{\stretch{1}}(1.14)

At high temperatures we have

\begin{aligned}\rho \lambda^3 \rightarrow 0\end{aligned} \hspace{\stretch{1}}(1.15)

(as T does down, \rho \lambda^3 goes up)

Looking at g_{3/2}(z = 1) = \rho \lambda^3(T_{\mathrm{c}}) leads to

\begin{aligned}\boxed{k_{\mathrm{B}} T_{\mathrm{c}} = \left( \frac{\rho}{\zeta(3/2)} \right)^{2/3} \frac{ 2 \pi \hbar^2}{m}.}\end{aligned} \hspace{\stretch{1}}(1.16)

How do I satisfy number conservation?

We have a problem here since as T \rightarrow 0 the 1/\lambda^3 \sim T^{3/2} term in \rho above drops to zero, yet g_{3/2}(z) cannot keep increasing without bounds to compensate and keep the density fixed. The way to deal with this was worked out by

  1. Bose (1924) for photons (examining statistics for symmetric wave functions).
  2. Einstein (1925) for conserved particles.

To deal with this issue, we (somewhat arbitrarily, because we need to) introduce a non-zero density for \mathbf{k} = 0. This is an adjustment of the approximation so that we have

\begin{aligned}\sum_{\mathbf{k}} \rightarrow \int \frac{d^3 \mathbf{k}}{(2 \pi)^3} \qquad \mbox{Except around k = 0},\end{aligned} \hspace{\stretch{1}}(1.17)

as in fig. 1.5, so that

Fig 1.5: Momentum sphere with origin omitted

 

\begin{aligned}\sum_\mathbf{k} = \left( \mbox{Contribution at k = 0} \right)+ V \int \frac{d^3 \mathbf{k}}{(2 \pi)^3}.\end{aligned} \hspace{\stretch{1}}(1.18)

Given this, we have

\begin{aligned}N= N_{\mathbf{k} = 0}+ V \int \frac{d^3 \mathbf{k}}{(2 \pi)^3} n_{\mathrm{B}}(\mathbf{k})\end{aligned} \hspace{\stretch{1}}(1.19)

We can illustrate this as in fig. 1.6.

Fig 1.6: Boson occupation vs momentum

 

\begin{aligned}\rho= \rho_{\mathbf{k} = 0}+ \frac{1}{{\lambda^3}} g_{3/2}(z)= \rho_{\mathbf{k} = 0}+ \frac{ \lambda(T_{\mathrm{c}}) }{ \lambda(T)}\frac{1}{{ \lambda^3(T_{\mathrm{c}})}}g_{3/2}(z)\end{aligned} \hspace{\stretch{1}}(1.20)

At T > T_{\mathrm{c}} we have \rho_{\mathbf{k} = 0}, whereas at T < T_{\mathrm{c}} we must introduce a non-zero density if we want to be able to keep a constant density constraint.

Advertisements

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

 
%d bloggers like this: