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Archive for April, 2013

Ultra relativisitic spin zero condensation temperature

Posted by peeterjoot on April 30, 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.)]

Here’s a bash at one of the exam questions, where I get the time to think things through properly. I think I did something like this on the exam itself, but may have also made some arithmetic errors.

Question: Ultra relativisitic spin zero condensation temperature (2013 final exam pr 2)

Consider a Bose gas with particles having no spin and obeying an ultra relativisitic dispersion $E_\mathbf{k} = c \left\lvert {\mathbf{k}} \right\rvert$. Unlike photons or phonons, these particles are {\bf conserved}, and hence we must determine the chemical potential $\mu$ which fixes their density. Working in three dimensions, show whether or not these particles will exhibit Bose condensation, and find $T_c$ if it is nonzero.

For the number of particles in the gas, as with photons, we still have

\begin{aligned}\left\langle{{N}}\right\rangle = \sum_\mathbf{k} \frac{1}{{z^{-1} e^{\beta \epsilon_\mathbf{k}} - 1}}= \frac{1}{{z^{-1} - 1}}+ \sum_{\mathbf{k} \ne 0} \frac{1}{{z^{-1} e^{\beta \epsilon_\mathbf{k}} - 1}}.\end{aligned} \hspace{\stretch{1}}(1.1)

As in the discussion of low velocity particles in [1] section 7.1, the ground state term has been split out, before making any continuum approximation of the sum over the energetic states.

Writing

\begin{aligned}\left\langle{{N}}\right\rangle = N_0 + N_e,\end{aligned} \hspace{\stretch{1}}(1.0.2)

where the number of particles in the ground state is chemical potential and temperature dependent

\begin{aligned}N_0 = \frac{z}{1 - z}.\end{aligned} \hspace{\stretch{1}}(1.0.3)

We proceed with the continuum approximation for the number of particles in the energetic states

\begin{aligned}N_e &= \sum_{\mathbf{k} \ne 0} \frac{1}{{z^{-1} e^{\beta \epsilon_\mathbf{k}} - 1}} \\ &\sim V \int \frac{d^3 \mathbf{k}}{(2 \pi)^3}\frac{1}{{z^{-1} e^{\beta \epsilon_\mathbf{k}} - 1}} \\ &= \frac{4 \pi V}{(2 \pi)^3} \int_0^\infty k^2 dk\frac{1}{{z^{-1} e^{\beta c k} - 1}} \\ &= \frac{V}{2 \pi^2} \left( { \frac{1}{{\beta c}} } \right)^3\int_0^\infty x^2 dx\frac{1}{{z^{-1} e^{x} - 1}} \\ &= \frac{V}{2 \pi^2} \left( { \frac{1}{{\beta c}} } \right)^3\Gamma(3) g_3(z).\end{aligned} \hspace{\stretch{1}}(1.0.3)

So we have

\begin{aligned}N_e=\frac{V}{\pi^2} \left( { \frac{k_{\mathrm{B}} T}{c} } \right)^3g_3(z)\le \frac{V}{\pi^2} \left( { \frac{k_{\mathrm{B}} T}{c} } \right)^3\zeta(3).\end{aligned} \hspace{\stretch{1}}(1.0.5)

Note that $\zeta(3) \approx 1.20206$, a fixed number. The key feature of Bose condensation remains. There is a finite limit to the number of particles that can be in the energetic state at a given temperature and volume. Any remaining particles are forced into the ground state.

In general the number of particles in the ground state is

\begin{aligned}N_0 = N - \frac{V}{\pi^2} \left( { \frac{k_{\mathrm{B}} T}{c} } \right)^3g_3(z),\end{aligned} \hspace{\stretch{1}}(1.0.6)

and we will necessarily have particles in this state if

\begin{aligned}N - \frac{V}{\pi^2} \left( { \frac{k_{\mathrm{B}} T}{c} } \right)^3\zeta(3) > 0.\end{aligned} \hspace{\stretch{1}}(1.0.7)

That temperature threshold $T \le T_c$ is the Bose condensation temperature

\begin{aligned}\boxed{k_{\mathrm{B}} T_c = c \left( { \frac{n \pi^2}{\zeta(3)} } \right)^{1/3}.}\end{aligned} \hspace{\stretch{1}}(1.0.8)

With $n = N/V$, $n_0 = N_0/V$, we have for the ground state average number density

\begin{aligned}n_0 = n\left( 1 - \frac{g_3(z)}{\zeta(3)} \left( { \frac{T}{T_c} } \right)^3 \right)\end{aligned} \hspace{\stretch{1}}(1.0.9)

This is plotted in fig. 1.1.

Fig 1.1: Ratio of ground state number density to total number density

From the figure it appears that the notion of any sort of absolute condensation temperature is an approximation. We can start having particles go into the ground state at higher temperatures than $T_c$, but once the chemical potential starts approaching zero, that temperature for which we start having particles in the ground state approaches $T_c$. The key takeout idea appears to be, once the temperature does drop below $T_c$, we necessarily start having a non-zero ground state population, and as the temperature drops more and more, the ratio of the number of particles in the ground state relative to the total approaches unity (all particles are forced into the ground state).

References

[1] RK Pathria. Statistical mechanics. Butterworth Heinemann, Oxford, UK, 1996.

Summary of statistical mechanics relations and helpful formulas (cheat sheet fodder)

Posted by peeterjoot on April 29, 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.)]

Central limit theorem

If $\left\langle{{x}}\right\rangle = \mu$ and $\sigma^2 = \left\langle{{x^2}}\right\rangle - \left\langle{{x}}\right\rangle^2$, and $X = \sum x$, then in the limit

\begin{aligned}\lim_{N \rightarrow \infty} P(X)= \frac{1}{{\sigma \sqrt{2 \pi N}}} \exp\left( - \frac{ (x - N \mu)^2}{2 N \sigma^2} \right)\end{aligned} \hspace{\stretch{1}}(1.0.1a)

\begin{aligned}\left\langle{{X}}\right\rangle = N \mu\end{aligned} \hspace{\stretch{1}}(1.0.1b)

\begin{aligned}\left\langle{{X^2}}\right\rangle - \left\langle{{X}}\right\rangle^2 = N \sigma^2\end{aligned} \hspace{\stretch{1}}(1.0.1c)

Binomial distribution

\begin{aligned}P_N(X) = \left\{\begin{array}{l l}\left(\frac{1}{{2}}\right)^N \frac{N!}{\left(\frac{N-X}{2}\right)!\left(\frac{N+X}{2}\right)!}& \quad \mbox{if X and N have same parity} \\ 0 & \quad \mbox{otherwise} \end{array},\right.\end{aligned} \hspace{\stretch{1}}(1.0.2)

where $X$ was something like number of Heads minus number of Tails.

Generating function

Given the Fourier transform of a probability distribution $\tilde{P}(k)$ we have

\begin{aligned}{\left.{{ \frac{\partial^n}{\partial k^n} \tilde{P}(k) }}\right\vert}_{{k = 0}}= (-i)^n \left\langle{{x^n}}\right\rangle\end{aligned} \hspace{\stretch{1}}(1.0.2)

Handy mathematics

\begin{aligned}\ln( 1 + x ) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4}\end{aligned} \hspace{\stretch{1}}(1.0.2)

\begin{aligned}N! \approx \sqrt{ 2 \pi N} N^N e^{-N}\end{aligned} \hspace{\stretch{1}}(1.0.5)

\begin{aligned}\ln N! \approx \frac{1}{{2}} \ln 2 \pi -N + \left( N + \frac{1}{{2}} \right)\ln N \approx N \ln N - N\end{aligned} \hspace{\stretch{1}}(1.0.6)

\begin{aligned}\text{erf}(z) = \frac{2}{\sqrt{\pi}} \int_0^z e^{-t^2} dt\end{aligned} \hspace{\stretch{1}}(1.0.7)

\begin{aligned}\Gamma(\alpha) = \int_0^\infty dy e^{-y} y^{\alpha - 1}\end{aligned} \hspace{\stretch{1}}(1.0.8)

\begin{aligned}\Gamma(\alpha + 1) = \alpha \Gamma(\alpha)\end{aligned} \hspace{\stretch{1}}(1.0.9)

\begin{aligned}\Gamma\left( 1/2 \right) = \sqrt{\pi}\end{aligned} \hspace{\stretch{1}}(1.0.10)

\begin{aligned}\zeta(s) = \sum_{k=1}^{\infty} k^{-s}\end{aligned} \hspace{\stretch{1}}(1.0.10)

\begin{aligned}\begin{aligned}\zeta(3/2) &\approx 2.61238 \\ \zeta(2) &\approx 1.64493 \\ \zeta(5/2) &\approx 1.34149 \\ \zeta(3) &\approx 1.20206\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.12)

\begin{aligned}\Gamma(z) \Gamma(1-z) = \frac{\pi}{\sin(\pi z)}\end{aligned} \hspace{\stretch{1}}(1.0.12)

\begin{aligned}P(x, t) = \int_{-\infty}^\infty \frac{dk}{2 \pi} \tilde{P}(k, t) \exp\left( i k x \right)\end{aligned} \hspace{\stretch{1}}(1.0.14a)

\begin{aligned}\tilde{P}(k, t) = \int_{-\infty}^\infty dx P(x, t) \exp\left( -i k x \right)\end{aligned} \hspace{\stretch{1}}(1.0.14b)

Heavyside theta

\begin{aligned}\Theta(x) = \left\{\begin{array}{l l}1 & \quad x \ge 0 \\ 0 & \quad x < 0\end{array}\right.\end{aligned} \hspace{\stretch{1}}(1.0.15a)

\begin{aligned}\frac{d\Theta}{dx} = \delta(x)\end{aligned} \hspace{\stretch{1}}(1.0.15b)

\begin{aligned}\sum_{m = -l}^l a^m=\frac{a^{l + 1/2} - a^{-(l+1/2)}}{a^{1/2} - a^{-1/2}}\end{aligned} \hspace{\stretch{1}}(1.0.16.16)

\begin{aligned}\sum_{m = -l}^l e^{b m}=\frac{\sinh(b(l + 1/2))}{\sinh(b/2)}\end{aligned} \hspace{\stretch{1}}(1.0.16b)

\begin{aligned}\int_{-\infty}^\infty q^{2 N} e^{-a q^2} dq=\frac{(2 N - 1)!!}{(2a)^N} \sqrt{\frac{\pi}{a}}\end{aligned} \hspace{\stretch{1}}(1.0.17.17)

\begin{aligned}\int_{-\infty}^\infty e^{-a q^2} dq=\sqrt{\frac{\pi}{a}}\end{aligned} \hspace{\stretch{1}}(1.0.17.17)

\begin{aligned}\binom{-\left\lvert {m} \right\rvert}{k} = (-1)^k \frac{\left\lvert {m} \right\rvert}{\left\lvert {m} \right\rvert + k} \binom{\left\lvert {m} \right\rvert+k}{\left\lvert {m} \right\rvert}\end{aligned} \hspace{\stretch{1}}(1.0.18)

\begin{aligned}\int_0^\infty d\epsilon \frac{\epsilon^3}{e^{\beta \epsilon} - 1} =\frac{\pi ^4}{15 \beta ^4},\end{aligned} \hspace{\stretch{1}}(1.0.18)

volume in mD

\begin{aligned}V_m= \frac{ \pi^{m/2} R^{m} }{ \Gamma\left( m/2 + 1 \right)}\end{aligned} \hspace{\stretch{1}}(1.0.20)

area of ellipse

\begin{aligned}A = \pi a b\end{aligned} \hspace{\stretch{1}}(1.0.21)

Radius of gyration of a 3D polymer

With radius $a$, we have

\begin{aligned}r_N \approx a \sqrt{N}\end{aligned} \hspace{\stretch{1}}(1.0.21)

Velocity random walk

Find

\begin{aligned}\mathcal{P}_{N_{\mathrm{c}}}(\mathbf{v}) \propto e^{-\frac{(\mathbf{v} - \mathbf{v}_0)^2}{2 N_{\mathrm{c}}}}\end{aligned} \hspace{\stretch{1}}(1.0.23)

Random walk

1D Random walk

\begin{aligned}\mathcal{P}( x, t ) = \frac{1}{{2}} \mathcal{P}(x + \delta x, t - \delta t)+\frac{1}{{2}} \mathcal{P}(x - \delta x, t - \delta t)\end{aligned} \hspace{\stretch{1}}(1.0.23)

\begin{aligned}\frac{\partial {\mathcal{P}}}{\partial {t}}(x, t) =\frac{1}{{2}} \frac{(\delta x)^2}{\delta t}\frac{\partial^2 {{\mathcal{P}}}}{\partial {{x}}^2}(x, t) = D \frac{\partial^2 {{\mathcal{P}}}}{\partial {{x}}^2}(x, t) = -\frac{\partial {J}}{\partial {x}},\end{aligned} \hspace{\stretch{1}}(1.0.25)

The diffusion constant relation to the probability current is referred to as Fick’s law

\begin{aligned}D = -\frac{\partial {J}}{\partial {x}}\end{aligned} \hspace{\stretch{1}}(1.0.25)

with which we can cast the probability diffusion identity into a continuity equation form

\begin{aligned}\frac{\partial {\mathcal{P}}}{\partial {t}} + \frac{\partial {J}}{\partial {x}} = 0 \end{aligned} \hspace{\stretch{1}}(1.0.25)

In 3D (with the Maxwell distribution frictional term), this takes the form

\begin{aligned}\mathbf{j} = -D \boldsymbol{\nabla}_\mathbf{v} c(\mathbf{v}, t) - \eta \mathbf{v} c(\mathbf{v}, t)\end{aligned} \hspace{\stretch{1}}(1.0.28a)

\begin{aligned}\frac{\partial {}}{\partial {t}} c(\mathbf{v}, t) + \boldsymbol{\nabla}_\mathbf{v} \cdot \mathbf{j}(\mathbf{v}, t) = 0\end{aligned} \hspace{\stretch{1}}(1.0.28b)

Maxwell distribution

Add a frictional term to the velocity space diffusion current

\begin{aligned}j_v = -D \frac{\partial {c}}{\partial {v}}(v, t) - \eta v c(v).\end{aligned} \hspace{\stretch{1}}(1.0.29)

For steady state the continity equation $0 = \frac{dc}{dt} = -\frac{\partial {j_v}}{\partial {v}}$ leads to

\begin{aligned}c(v) \propto \exp\left(- \frac{\eta v^2}{2 D}\right).\end{aligned} \hspace{\stretch{1}}(1.0.30)

We also find

\begin{aligned}\left\langle{{v^2}}\right\rangle = \frac{D}{\eta},\end{aligned} \hspace{\stretch{1}}(1.0.30)

and identify

\begin{aligned}\frac{1}{{2}} m \left\langle{{\mathbf{v}^2}}\right\rangle = \frac{1}{{2}} m \left( \frac{D}{\eta} \right) = \frac{1}{{2}} k_{\mathrm{B}} T\end{aligned} \hspace{\stretch{1}}(1.0.32)

Hamilton’s equations

\begin{aligned}\frac{\partial {H}}{\partial {p}} = \dot{x}\end{aligned} \hspace{\stretch{1}}(1.0.33a)

\begin{aligned}\frac{\partial {H}}{\partial {x}} = -\dot{p}\end{aligned} \hspace{\stretch{1}}(1.0.33b)

SHO

\begin{aligned}H = \frac{p^2}{2m} + \frac{1}{{2}} k x^2\end{aligned} \hspace{\stretch{1}}(1.0.34a)

\begin{aligned}\omega^2 = \frac{k}{m}\end{aligned} \hspace{\stretch{1}}(1.0.34b)

Quantum energy eigenvalues

\begin{aligned}E_n = \left( n + \frac{1}{{2}} \right) \hbar \omega\end{aligned} \hspace{\stretch{1}}(1.0.35)

Liouville’s theorem

\begin{aligned}\frac{d{{\rho}}}{dt} = \frac{\partial {\rho}}{\partial {t}} + \dot{x} \frac{\partial {\rho}}{\partial {x}} + \dot{p} \frac{\partial {\rho}}{\partial {p}}= \cdots = \frac{\partial {\rho}}{\partial {t}} + \frac{\partial {\left( \dot{x} \rho \right)}}{\partial {x}} + \frac{\partial {\left( \dot{x} \rho \right)}}{\partial {p}} = \frac{\partial {\rho}}{\partial {t}} + \boldsymbol{\nabla}_{x,p} \cdot (\rho \dot{x}, \rho \dot{p})= \frac{\partial {\rho}}{\partial {t}} + \boldsymbol{\nabla} \cdot \mathbf{J}= 0,\end{aligned} \hspace{\stretch{1}}(1.0.35)

Regardless of whether we have a steady state system, if we sit on a region of phase space volume, the probability density in that neighbourhood will be constant.

Ergodic

A system for which all accessible phase space is swept out by the trajectories. This and Liouville’s threorm allows us to assume that we can treat any given small phase space volume as if it is equally probable to the same time evolved phase space region, and switch to ensemble averaging instead of time averaging.

Thermodynamics

\begin{aligned}dE = T dS - P dV + \mu dN\end{aligned} \hspace{\stretch{1}}(1.0.37.37)

\begin{aligned}\frac{1}{{T}} = \left({\partial {S}}/{\partial {E}}\right)_{{N,V}}\end{aligned} \hspace{\stretch{1}}(1.0.37.37)

\begin{aligned}\frac{P}{T} = \left({\partial {S}}/{\partial {V}}\right)_{{N,E}}\end{aligned} \hspace{\stretch{1}}(1.0.37.37)

\begin{aligned}-\frac{\mu}{T} = \left({\partial {S}}/{\partial {N}}\right)_{{V,E}}\end{aligned} \hspace{\stretch{1}}(1.0.37.37)

\begin{aligned}P = - \left({\partial {E}}/{\partial {V}}\right)_{{N,S}}= - \left({\partial {F}}/{\partial {V}}\right)_{{N,T}}\end{aligned} \hspace{\stretch{1}}(1.0.37e)

\begin{aligned}\mu = \left({\partial {E}}/{\partial {N}}\right)_{{V,S}} = \left({\partial {F}}/{\partial {N}}\right)_{{V,T}}\end{aligned} \hspace{\stretch{1}}(1.0.37e)

\begin{aligned}T = \left({\partial {E}}/{\partial {S}}\right)_{{N,V}}\end{aligned} \hspace{\stretch{1}}(1.0.37e)

\begin{aligned}F = E - TS\end{aligned} \hspace{\stretch{1}}(1.0.37e)

\begin{aligned}G = F + P V = E - T S + P V = \mu N\end{aligned} \hspace{\stretch{1}}(1.0.37i)

\begin{aligned}H = E + P V = G + T S\end{aligned} \hspace{\stretch{1}}(1.0.37j)

\begin{aligned}C_{\mathrm{V}} = T \left({\partial {S}}/{\partial {T}}\right)_{{N,V}} = \left({\partial {E}}/{\partial {T}}\right)_{{N,V}} = - T \left( \frac{\partial^2 {{F}}}{\partial {{T}}^2} \right)_{N,V}\end{aligned} \hspace{\stretch{1}}(1.0.37k)

\begin{aligned}C_{\mathrm{P}} = T \left({\partial {S}}/{\partial {T}}\right)_{{N,P}} = \left({\partial {H}}/{\partial {T}}\right)_{{N,P}}\end{aligned} \hspace{\stretch{1}}(1.0.37l)

\begin{aligned}\underbrace{dE}_{\text{Change in energy}}=\underbrace{d W}_{\text{work done on the system}}+\underbrace{d Q}_{\text{Heat supplied to the system}}\end{aligned} \hspace{\stretch{1}}(1.0.38)

Example (work on gas): $d W = -P dV$. Adiabatic: $d Q = 0$. Cyclic: $dE = 0$.

Microstates

\begin{aligned}\beta = \frac{1}{k_{\mathrm{B}} T}\end{aligned} \hspace{\stretch{1}}(1.0.38)

\begin{aligned}S = k_{\mathrm{B}} \ln \Omega \end{aligned} \hspace{\stretch{1}}(1.0.40)

\begin{aligned}\Omega(N, V, E) = \frac{1}{h^{3N} N!} \int_V d\mathbf{x}_1 \cdots d\mathbf{x}_N \int d\mathbf{p}_1 \cdots d\mathbf{p}_N \delta \left(E - \frac{\mathbf{p}_1^2}{2 m} \cdots - \frac{\mathbf{p}_N^2}{2 m}\right)=\frac{V^N}{h^{3N} N!}\int d\mathbf{p}_1 \cdots d\mathbf{p}_N \delta \left(E - \frac{\mathbf{p}_1^2}{2m} \cdots - \frac{\mathbf{p}_N^2}{2m}\right)\end{aligned} \hspace{\stretch{1}}(1.0.40)

\begin{aligned}\Omega = \frac{d\gamma}{dE}\end{aligned} \hspace{\stretch{1}}(1.0.42)

\begin{aligned}\gamma=\frac{V^N}{h^{3N} N!}\int d\mathbf{p}_1 \cdots d\mathbf{p}_N \Theta \left(E - \frac{\mathbf{p}_1^2}{2m} \cdots - \frac{\mathbf{p}_N^2}{2m}\right)\end{aligned} \hspace{\stretch{1}}(1.0.43)

quantum

\begin{aligned}\gamma = \sum_i \Theta(E - \epsilon_i)\end{aligned} \hspace{\stretch{1}}(1.0.44)

Ideal gas

\begin{aligned}\Omega = \frac{V^N}{N!} \frac{1}{{h^{3N}}} \frac{( 2 \pi m E)^{3 N/2 }}{E} \frac{1}{\Gamma( 3N/2 ) }\end{aligned} \hspace{\stretch{1}}(1.0.45)

\begin{aligned}S_{\mathrm{ideal}} = k_{\mathrm{B}} \left(N \ln \frac{V}{N} + \frac{3 N}{2} \ln \left( \frac{4 \pi m E }{3 N h^2} \right) + \frac{5 N}{2} \right)\end{aligned} \hspace{\stretch{1}}(1.0.46)

Quantum free particle in a box

\begin{aligned}\Psi_{n_1, n_2, n_3}(x, y, z) = \left( \frac{2}{L} \right)^{3/2} \sin\left( \frac{ n_1 \pi x}{L} \right)\sin\left( \frac{ n_2 \pi x}{L} \right)\sin\left( \frac{ n_3 \pi x}{L} \right)\end{aligned} \hspace{\stretch{1}}(1.0.47a)

\begin{aligned}\epsilon_{n_1, n_2, n_3} = \frac{h^2}{8 m L^2} \left( n_1^2 + n_2^2 + n_3^2 \right)\end{aligned} \hspace{\stretch{1}}(1.0.47b)

\begin{aligned}\epsilon_k = \frac{\hbar^2 k^2}{2m},\end{aligned} \hspace{\stretch{1}}(1.0.47b)

Spin

magnetization

\begin{aligned}\mu = \frac{\partial {F}}{\partial {B}}\end{aligned} \hspace{\stretch{1}}(1.0.48)

moment per particle

\begin{aligned}m = \mu/N\end{aligned} \hspace{\stretch{1}}(1.0.49)

spin matrices

\begin{aligned}\sigma_x = \begin{bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix}\end{aligned} \hspace{\stretch{1}}(1.0.50a)

\begin{aligned}\sigma_y = \begin{bmatrix} 0 & -i \\ i & 0 \\ \end{bmatrix}\end{aligned} \hspace{\stretch{1}}(1.0.50b)

\begin{aligned}\sigma_z = \begin{bmatrix} 1 & 0 \\ 0 & -1 \\ \end{bmatrix}\end{aligned} \hspace{\stretch{1}}(1.0.50c)

$l \ge 0, -l \le m \le l$

\begin{aligned}\mathbf{L}^2 {\left\lvert {lm} \right\rangle} = l(l+1)\hbar^2 {\left\lvert {lm} \right\rangle}\end{aligned} \hspace{\stretch{1}}(1.0.51a)

\begin{aligned}L_z {\left\lvert {l m} \right\rangle} = \hbar m {\left\lvert {l m} \right\rangle}\end{aligned} \hspace{\stretch{1}}(1.0.51b)

\begin{aligned}S(S + 1) \hbar^2\end{aligned} \hspace{\stretch{1}}(1.0.51b)

Canonical ensemble

classical

\begin{aligned}\Omega(N, E) = \frac{ V }{ h^3 N} \int d\mathbf{p}_1 e^{\frac{S}{k_{\mathrm{B}}}(N, E)}e^{-\frac{1}{{k_{\mathrm{B}}}} \left( \frac{\partial {S}}{\partial {N}} \right)_{E, V} }e^{-\frac{\mathbf{p}_1^2}{2m k_{\mathrm{B}}}\left( \frac{\partial {S}}{\partial {E}} \right)_{N, V}}\end{aligned} \hspace{\stretch{1}}(1.0.53)

quantum

\begin{aligned}\Omega(E) \approx\sum_{m \in \text{subsystem}} e^{\frac{1}{{k_{\mathrm{B}}}} S(E)}e^{-\beta \mathcal{E}_m}\end{aligned} \hspace{\stretch{1}}(1.0.54.54)

\begin{aligned}Z = \sum_m e^{-\beta \mathcal{E}_m} = \text{Tr} \left( e^{-\beta \hat{H}_{\text{subsystem}}} \right)\end{aligned} \hspace{\stretch{1}}(1.0.54b)

\begin{aligned}\left\langle{{E}}\right\rangle = \frac{\int He^{- \beta H }}{\int e^{- \beta H }}\end{aligned} \hspace{\stretch{1}}(1.0.55a)

\begin{aligned}\left\langle{{E^2}}\right\rangle = \frac{\int H^2e^{- \beta H }}{\int e^{- \beta H }}\end{aligned} \hspace{\stretch{1}}(1.0.55b)

\begin{aligned}Z \equiv \frac{1}{{h^{3N} N!}}\int e^{- \beta H }\end{aligned} \hspace{\stretch{1}}(1.0.55c)

\begin{aligned}\left\langle{{E}}\right\rangle = -\frac{1}{{Z}} \frac{\partial {Z}}{\partial {\beta}} = - \frac{\partial {\ln Z}}{\partial {\beta}} =\frac{\partial {(\beta F)}}{\partial {\beta}}\end{aligned} \hspace{\stretch{1}}(1.0.55d)

\begin{aligned}\sigma_{\mathrm{E}}^2= \left\langle{{E^2}}\right\rangle - \left\langle{{E}}\right\rangle^2 =\frac{\partial^2 {{\ln Z}}}{\partial {{\beta}}^2} = k_{\mathrm{B}} T^2 \frac{\partial {\left\langle{{E}}\right\rangle}}{\partial {T}}= k_{\mathrm{B}} T^2 C_{\mathrm{V}} \propto N\end{aligned} \hspace{\stretch{1}}(1.0.55e)

\begin{aligned}Z = e^{-\beta (\left\langle{{E}}\right\rangle - T S) } = e^{-\beta F}\end{aligned} \hspace{\stretch{1}}(1.0.55f)

\begin{aligned}F = \left\langle{{E}}\right\rangle - T S = -k_{\mathrm{B}} T \ln Z\end{aligned} \hspace{\stretch{1}}(1.0.55g)

Grand Canonical ensemble

\begin{aligned}S = - k_{\mathrm{B}} \sum_{r,s} P_{r,s} \ln P_{r,s}\end{aligned} \hspace{\stretch{1}}(1.0.56)

\begin{aligned}P_{r, s} = \frac{e^{-\alpha N_r - \beta E_s}}{Z_{\mathrm{G}}}\end{aligned} \hspace{\stretch{1}}(1.0.57a)

\begin{aligned}Z_{\mathrm{G}} = \sum_{r,s} e^{-\alpha N_r - \beta E_s} = \sum_{r,s} z^{N_r} e^{-\beta E_s} = \sum_{N_r} z^{N_r} Z_{N_r}\end{aligned} \hspace{\stretch{1}}(1.0.57b)

\begin{aligned}z = e^{-\alpha} = e^{\mu \beta}\end{aligned} \hspace{\stretch{1}}(1.0.57c)

\begin{aligned}q = \ln Z_{\mathrm{G}} = P V \beta\end{aligned} \hspace{\stretch{1}}(1.0.57d)

\begin{aligned}\left\langle{{H}}\right\rangle = -\left({\partial {q}}/{\partial {\beta}}\right)_{{z,V}} = k_{\mathrm{B}} T^2 \left({\partial {q}}/{\partial {\mu}}\right)_{{z,V}} = \sum_\epsilon \frac{\epsilon}{z^{-1} e^{\beta \epsilon} \pm 1}\end{aligned} \hspace{\stretch{1}}(1.0.57e)

\begin{aligned}\left\langle{{N}}\right\rangle = z \left({\partial {q}}/{\partial {z}}\right)_{{V,T}} = \sum_\epsilon \frac{1}{{z^{-1} e^{\beta\epsilon} \pm 1}}\end{aligned} \hspace{\stretch{1}}(1.0.57f)

\begin{aligned}F = - k_{\mathrm{B}} T \ln \frac{ Z_{\mathrm{G}} }{z^N}\end{aligned} \hspace{\stretch{1}}(1.0.57g)

\begin{aligned}\left\langle{{n_\epsilon}}\right\rangle = -\frac{1}{{\beta}} \left({\partial {q}}/{\partial {\epsilon}}\right)_{{z, T, \text{other} \epsilon}} = \frac{1}{{z^{-1} e^{\beta \epsilon} \pm 1}}\end{aligned} \hspace{\stretch{1}}(1.0.57h)

\begin{aligned}\text{var}(N) = \frac{1}{{\beta}} \left({\partial {\left\langle{{N}}\right\rangle}}/{\partial {\mu}}\right)_{{V, T}} = - \frac{1}{{\beta}} \left({\partial {\left\langle{{n_\epsilon}}\right\rangle}}/{\partial {\epsilon}}\right)_{{z,T}} = z^{-1} e^{\beta \epsilon}\end{aligned} \hspace{\stretch{1}}(1.0.57h)

\begin{aligned}\mathcal{P} \propto e^{\frac{\mu}{k_{\mathrm{B}} T} N_S}e^{-\frac{E_S}{k_{\mathrm{B}} T} }\end{aligned} \hspace{\stretch{1}}(1.0.59.59)

\begin{aligned}Z_{\mathrm{G}}= \sum_{N=0}^\infty e^{\beta \mu N}\sum_{n_k, \sum n_m = N} e^{-\beta \sum_m n_m \epsilon_m}=\prod_{k} \left( \sum_{n_k} e^{-\beta(\epsilon_k - \mu) n_k} \right)\end{aligned} \hspace{\stretch{1}}(1.0.59b)

\begin{aligned}Z_{\mathrm{G}}^{\mathrm{QM}} = {\text{Tr}}_{\{\text{energy}, N\}} \left( e^{ -\beta (\hat{H} - \mu \hat{N} ) } \right)\end{aligned} \hspace{\stretch{1}}(1.0.59b)

\begin{aligned}P V = \frac{2}{3} U\end{aligned} \hspace{\stretch{1}}(1.0.60a)

\begin{aligned}f_\nu^\pm(z) = \frac{1}{{\Gamma(\nu)}} \int_0^\infty dx \frac{x^{\nu - 1}}{z^{-1} e^x \pm 1}\end{aligned} \hspace{\stretch{1}}(1.0.60a)

\begin{aligned}f_\nu^\pm(z \approx 0) =z\mp\frac{z^{2}}{2^\nu}+\frac{z^{3}}{3^\nu}\mp\frac{z^{4}}{4^\nu}+ \cdots \end{aligned} \hspace{\stretch{1}}(1.0.60a)

\begin{aligned}z \frac{d f_\nu^{\pm}(z) }{dz} = f_{\nu-1}^{\pm}(z)\end{aligned} \hspace{\stretch{1}}(1.0.61)

\begin{aligned}\frac{d f_{3/2}^{\pm}(z) }{dT} = -\frac{3}{2T} f_{3/2}^{\pm}(z)f_{\nu-1}^{\pm}(z)\end{aligned} \hspace{\stretch{1}}(1.0.62)

Fermions

\begin{aligned}\sum_{n_k = 0}^1 e^{-\beta(\epsilon_k - \mu) n_k}=1 + e^{-\beta(\epsilon_k - \mu)}\end{aligned} \hspace{\stretch{1}}(1.0.62)

\begin{aligned}N = (2 S + 1) V \int_0^{k_{\mathrm{F}}} \frac{4 \pi k^2 dk}{(2 \pi)^3}\end{aligned} \hspace{\stretch{1}}(1.0.64)

\begin{aligned}k_{\mathrm{F}} = \left( \frac{ 6 \pi^2 \rho }{2 S + 1} \right)^{1/3}\end{aligned} \hspace{\stretch{1}}(1.0.65.65)

\begin{aligned}\epsilon_{\mathrm{F}} = \frac{\hbar^2}{2m} \left( \frac{6 \pi \rho}{2 S + 1} \right)^{2/3}\end{aligned} \hspace{\stretch{1}}(1.0.65.65)

\begin{aligned}\mu = \epsilon_{\mathrm{F}} - \frac{\pi^2}{12} \frac{(k_{\mathrm{B}} T)^2}{\epsilon_{\mathrm{F}}} + \cdots \end{aligned} \hspace{\stretch{1}}(1.0.65.65)

\begin{aligned}\lambda \equiv \frac{h}{\sqrt{2 \pi m k_{\mathrm{B}} T}}\end{aligned} \hspace{\stretch{1}}(1.0.65.65)

\begin{aligned}\frac{N}{V}=\frac{g}{\lambda^3} f_{3/2}(z)=\frac{g}{\lambda^3} \left( e^{\beta \mu} - \frac{e^{2 \beta \mu}}{2^{3/2}} + \cdots \right) \end{aligned} \hspace{\stretch{1}}(1.0.68)

(so $n = \frac{g}{\lambda^3} e^{\beta \mu}$ for large temperatures)

\begin{aligned}P \beta = \frac{g}{\lambda^3} f_{5/2}(z)\end{aligned} \hspace{\stretch{1}}(1.0.69a)

\begin{aligned}U= \frac{3}{2} N k_{\mathrm{B}} T \frac{f_{5/2}(z)}{f_{3/2}(z) }.\end{aligned} \hspace{\stretch{1}}(1.0.69a)

\begin{aligned}f_\nu^+(e^y) \approx\frac{y^\nu}{\Gamma(\nu + 1)}\left( 1 + 2 \nu \sum_{j = 1, 3, 5, \cdots } (\nu-1) \cdots (\nu - j) \left( 1 - 2^{-j} \right) \frac{\zeta(j+1)}{ y^{j + 1} } \right)\end{aligned} \hspace{\stretch{1}}(1.0.69a)

\begin{aligned}\frac{C}{N} = \frac{\pi^2}{2} k_{\mathrm{B}} \frac{ k_{\mathrm{B}} T}{\epsilon_{\mathrm{F}}}\end{aligned} \hspace{\stretch{1}}(1.0.71.71)

\begin{aligned}A = N k_{\mathrm{B}} T \left( \ln z - \frac{f_{5/2}(z)}{f_{3/2}(z)} \right)\end{aligned} \hspace{\stretch{1}}(1.0.71.71)

Bosons

\begin{aligned}Z_{\mathrm{G}} = \prod_\epsilon \frac{1}{{ 1 - z e^{-\beta \epsilon} }}\end{aligned} \hspace{\stretch{1}}(1.0.72)

\begin{aligned}P \beta = \frac{1}{{\lambda^3}} g_{5/2}(z)\end{aligned} \hspace{\stretch{1}}(1.0.73)

\begin{aligned}U = \frac{3}{2} k_{\mathrm{B}} T \frac{V}{\lambda^3} g_{5/2}(z)\end{aligned} \hspace{\stretch{1}}(1.0.74)

\begin{aligned}N_e = N - N_0 = N \left( \frac{T}{T_c} \right)^{3/2}\end{aligned} \hspace{\stretch{1}}(1.0.75)

For $T < T_c$, $z = 1$.

\begin{aligned}g_\nu(1) = \zeta(\nu).\end{aligned} \hspace{\stretch{1}}(1.0.76)

\begin{aligned}\sum_{n_k = 0}^\infty e^{-\beta(\epsilon_k - \mu) n_k} =\frac{1}{{1 - e^{-\beta(\epsilon_k - \mu)}}}\end{aligned} \hspace{\stretch{1}}(1.0.76)

\begin{aligned}f_\nu^-( e^{-\alpha} ) = \frac{ \Gamma(1 - \nu)}{ \alpha^{1 - \nu} } + \cdots \end{aligned} \hspace{\stretch{1}}(1.0.76)

\begin{aligned}\rho \lambda^3 = g_{3/2}(z) \le \zeta(3/2) \approx 2.612\end{aligned} \hspace{\stretch{1}}(1.0.79.79)

\begin{aligned}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}}(1.0.79.79)

BEC

\begin{aligned}\rho= \rho_{\mathbf{k} = 0}+ \frac{1}{{\lambda^3}} g_{3/2}(z)\end{aligned} \hspace{\stretch{1}}(1.0.80.80)

\begin{aligned}\rho_0 = \rho \left(1 - \left( \frac{T}{T_{\mathrm{c}}} \right)^{3/2}\right)\end{aligned} \hspace{\stretch{1}}(1.0.80b)

\begin{aligned}\frac{E}{V} \propto \left( k_{\mathrm{B}} T \right)^{5/2}\end{aligned} \hspace{\stretch{1}}(1.0.81.81)

\begin{aligned}\frac{C}{V} \propto \left( k_{\mathrm{B}} T \right)^{3/2}\end{aligned} \hspace{\stretch{1}}(1.0.81.81)

\begin{aligned}\frac{S}{N k_{\mathrm{B}}} = \frac{5}{2} \frac{g_{5/2}}{g_{3/2}} - \ln z \Theta(T - T_c)\end{aligned} \hspace{\stretch{1}}(1.0.81.81)

Density of states

Low velocities

\begin{aligned}N_1(\epsilon)=V \frac{m \hbar}{\hbar^2 \sqrt{ 2 m \epsilon}}\end{aligned} \hspace{\stretch{1}}(1.0.82a)

\begin{aligned}N_2(\epsilon)=V \frac{m}{\hbar^2}\end{aligned} \hspace{\stretch{1}}(1.0.82b)

\begin{aligned}N_3(\epsilon)=V \left( \frac{2 m}{\hbar^2} \right)^{3/2} \frac{1}{{4 \pi^2}} \sqrt{\epsilon}\end{aligned} \hspace{\stretch{1}}(1.0.82c)

relativistic

\begin{aligned}\mathcal{D}_1(\epsilon)=\frac{2 L}{ c h } \frac{ \sqrt{ \epsilon^2 - \left( m c^2 \right)^2} }{\epsilon}\end{aligned} \hspace{\stretch{1}}(1.0.83.83)

\begin{aligned}\mathcal{D}_2(\epsilon)=\frac{2 \pi A}{ (c h)^2 } \frac{ \epsilon^2 - \left( m c^2 \right)^2 }{ \epsilon }\end{aligned} \hspace{\stretch{1}}(1.0.83.83)

\begin{aligned}\mathcal{D}_3(\epsilon)=\frac{4 \pi V}{ (c h)^3 } \frac{\left( \epsilon^2 - \left( m c^2 \right)^2 \right)^{3/2}}{\epsilon}\end{aligned} \hspace{\stretch{1}}(1.0.83.83)

Low temperature Fermi gas chemical potential

Posted by peeterjoot on April 24, 2013

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Question: Low temperature Fermi gas chemical potential

[1] section 8.1 equation (33) provides an implicit function for $\mu \equiv k_{\mathrm{B}} T \ln z$

\begin{aligned}n = \frac{4 \pi g}{3} \left( \frac{2m}{h^2} \right)^{3/2}\mu^{3/2}\left( 1 + \frac{\pi^2}{8} \frac{ (k_{\mathrm{B}} T)^2 }{ \mu^2 } + \cdots \right),\end{aligned} \hspace{\stretch{1}}(1.0.1)

or

\begin{aligned}E_{\mathrm{F}}^{3/2} = \mu^{3/2} \left( 1 + \frac{\pi^2}{8} \frac{ (k_{\mathrm{B}} T)^2 }{ \mu^2 } + \cdots \right).\end{aligned} \hspace{\stretch{1}}(1.0.2)

In class, we assumed that $\mu$ was quadratic in $k_{\mathrm{B}} T$ as a mechanism to invert this non-linear equation. Without making this quadratic assumption find the lowest order, non-constant approximation for $\mu(T)$.

To determine an approximate inversion, let’s start by multiplying eq. 1.0.2 by $\mu^{1/2}/E_{\mathrm{F}}^2$ to non-dimensionalize things

\begin{aligned}\left( \frac{\mu}{E_{\mathrm{F}}} \right)^{1/2} = \left( \frac{\mu}{E_{\mathrm{F}}} \right)^2 + \frac{\pi^2}{8} \left( \frac{k_{\mathrm{B}} T}{E_{\mathrm{F}}} \right)^2,\end{aligned} \hspace{\stretch{1}}(1.0.3)

or

\begin{aligned}\left( \frac{\mu}{E_{\mathrm{F}}} \right)^{1/2} =\frac{1}{{ 1 - \left( \frac{\mu}{E_{\mathrm{F}}} \right)^{3/2} }}\frac{\pi^2}{8} \left( \frac{k_{\mathrm{B}} T}{E_{\mathrm{F}}} \right)^2.\end{aligned} \hspace{\stretch{1}}(1.0.4)

If we are looking for an approximation in the neighborhood of $\mu = E_{\mathrm{F}}$, then the LHS factor is approximately one, whereas the fractional difference term is large (with a corresponding requirement for $k_{\mathrm{B}} T/E_{\mathrm{F}}$ to be small. We must then have

\begin{aligned}\left( \frac{\mu}{E_{\mathrm{F}}} \right)^{3/2} \approx 1 - \frac{\pi^2}{8} \left( \frac{k_{\mathrm{B}} T}{E_{\mathrm{F}}} \right)^2,\end{aligned} \hspace{\stretch{1}}(1.0.4)

or

\begin{aligned}\mu\approx E_{\mathrm{F}}\left(1 - \frac{\pi^2}{8} \left( \frac{k_{\mathrm{B}} T}{E_{\mathrm{F}}} \right)^2\right)^{2/3}\approx E_{\mathrm{F}}\left(1 - \frac{2}{3} \frac{\pi^2}{8} \left( \frac{k_{\mathrm{B}} T}{E_{\mathrm{F}}} \right)^2\right).\end{aligned} \hspace{\stretch{1}}(1.0.4)

This gives us the desired result

\begin{aligned}\boxed{\mu \approx E_{\mathrm{F}}\left(1 - \frac{\pi^2}{12} \left( \frac{k_{\mathrm{B}} T}{E_{\mathrm{F}}} \right)^2\right).}\end{aligned} \hspace{\stretch{1}}(1.0.7)

References

[1] RK Pathria. Statistical mechanics. Butterworth Heinemann, Oxford, UK, 1996.

A final pre-exam update of my notes compilation for â€˜PHY452H1S Basic Statistical Mechanicsâ€™, Taught by Prof. Arun Paramekanti

Posted by peeterjoot on April 22, 2013

Here’s my third update of my notes compilation for this course, including all of the following:

April 21, 2013 Fermi function expansion for thermodynamic quantities

April 20, 2013 Relativistic Fermi Gas

April 10, 2013 Non integral binomial coefficient

April 10, 2013 energy distribution around mean energy

April 09, 2013 Velocity volume element to momentum volume element

April 04, 2013 Phonon modes

April 03, 2013 BEC and phonons

April 03, 2013 Max entropy, fugacity, and Fermi gas

April 02, 2013 Bosons

April 02, 2013 Relativisitic density of states

March 28, 2013 Bosons

plus everything detailed in the description of my previous update and before.

Fermi-Dirac function expansion for thermodynamic quantities

Posted by peeterjoot on April 21, 2013

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In section 8.1 of [1] are some Fermi-Dirac \index{Fermi-Dirac function} expansions for $P$, $N$, and $U$. Let’s work through these in detail.

Our starting point is the relations

\begin{aligned}P V \beta = \ln Z_{\mathrm{G}} = \sum \ln \left( 1 + z e^{-\beta \epsilon} \right)\end{aligned} \hspace{\stretch{1}}(1.0.1a)

\begin{aligned}N = \sum \frac{1}{{ z^{-1} e^{\beta \epsilon} + 1 }}.\end{aligned} \hspace{\stretch{1}}(1.0.1b)

Recap. Density of states

We’ll employ the 3D non-relativisitic density of states

\begin{aligned}\mathcal{D}(\epsilon) &= \sum_\mathbf{k} \delta(\epsilon - \epsilon_\mathbf{k}) \\ &\sim V \int \frac{d^3 \mathbf{k}}{(2 \pi)^3}\delta(\epsilon - \epsilon_\mathbf{k}) \\ &= \frac{4 \pi V}{(2 \pi)^3}\int dk k^2 \delta\left( \epsilon - \frac{\hbar^2 k^2}{2 m} \right) \\ &= \frac{4 \pi V}{(2 \pi)^3}\int dk k^2 \frac{ \delta\left( k - \sqrt{2 m \epsilon}/\hbar \right)}{ \frac{\hbar^2}{m} \frac{\sqrt{2 m \epsilon}}{\hbar}} \\ &= \frac{2 V}{(2 \pi)^2 }\frac{m}{\hbar^2}\sqrt{\frac{2 m \epsilon}{\hbar^2}},\end{aligned} \hspace{\stretch{1}}(1.0.1b)

or

\begin{aligned}\boxed{\mathcal{D}(\epsilon)=\frac{V}{(2 \pi)^2 }\left( \frac{2 m}{\hbar^2} \right)^{3/2}\epsilon^{1/2}.}\end{aligned} \hspace{\stretch{1}}(1.0.1b)

Density

Now let’s make our integral approximation of the sum for $N$. That is

\begin{aligned}N &= g \int d\epsilon \mathcal{D}(\epsilon) \frac{1}{{ z^{-1} e^{\beta \epsilon} + 1 }} \\ &= g \frac{V}{(2 \pi)^2 }\left( \frac{2 m}{\hbar^2} \right)^{3/2}\int_0^\infty d\epsilon \frac{\epsilon^{1/2}}{ z^{-1} e^{\beta \epsilon} + 1 } \\ &= g \frac{V}{(2 \pi)^2 \beta^{3/2}}\left( \frac{2 m}{\hbar^2} \right)^{3/2}\int_0^\infty du \frac{u^{1/2}}{ z^{-1} e^{u} + 1 } \\ &= g \frac{V}{(2 \pi)^2 \beta^{3/2}}\left( \frac{2 m}{\hbar^2} \right)^{3/2}\Gamma(3/2) f_{3/2}(z) \\ &= g \frac{V}{(2 \pi)^2 \beta^{3/2}}\frac{\left( 2 m k_{\mathrm{B}} T \right)^{3/2}}{\hbar^3}\frac{1}{{2}} \sqrt{\pi}f_{3/2}(z)\\ &= g V \not{{2}} \pi\frac{\left( 2 m k_{\mathrm{B}} T \right)^{3/2}}{h^3}\frac{1}{{\not{{2}}}} \sqrt{\pi}f_{3/2}(z),\end{aligned} \hspace{\stretch{1}}(1.0.1b)

or

\begin{aligned}\frac{N}{V} = g \frac{\left( 2 \pi m k_{\mathrm{B}} T \right)^{3/2}}{h^3}f_{3/2}(z).\end{aligned} \hspace{\stretch{1}}(1.0.5)

With

\begin{aligned}\lambda = \frac{h}{\sqrt{ 2 \pi m k_{\mathrm{B}} T }},\end{aligned} \hspace{\stretch{1}}(1.0.6)

this gives us the desired density result from the text

\begin{aligned}\boxed{\frac{N}{V}=\frac{g}{\lambda^3} f_{3/2}(z).}\end{aligned} \hspace{\stretch{1}}(1.0.7)

Pressure

For the pressure, we can do the same, but have to integrate by parts

\begin{aligned}P V \beta &= g \sum \ln \left( 1 + z e^{-\beta \epsilon} \right) \\ &\sim g \frac{V}{(2 \pi)^2 }\left( \frac{2 m}{\hbar^2} \right)^{3/2}\int_0^\infty d\epsilon \epsilon^{1/2} \ln \left( 1 + z e^{-\beta \epsilon} \right) \\ &= - g \frac{V}{(2 \pi)^2 }\left( \frac{2 m}{\hbar^2} \right)^{3/2}\int_0^\infty d\epsilon \frac{2}{3} \epsilon^{3/2} \frac{-\beta z e^{-\beta \epsilon} }{ 1 + z e^{-\beta \epsilon} } \\ &= g\frac{V}{(2 \pi)^2 }\left( \frac{2 m}{\hbar^2} \right)^{3/2}\frac{2}{3} \frac{1}{{\beta^{3/2}}}\int_0^\infty dx\frac{x^{3/2}}{z^{-1} e^{x} + 1 } \\ &= g\frac{2}{3} 2 \pi V\frac{\left( 2 m k_{\mathrm{B}} T \right)^{3/2}}{h^3 }\Gamma(5/2)f_{5/2}(z) \\ &= g\frac{2}{3} 2 \pi V\frac{\left( 2 m k_{\mathrm{B}} T \right)^{3/2}}{h^3 }\frac{3}{2} \frac{1}{2} \sqrt{\pi}f_{5/2}(z) \\ &= g V\frac{\left( 2 \pi m k_{\mathrm{B}} T \right)^{3/2}}{h^3 }f_{5/2}(z),\end{aligned} \hspace{\stretch{1}}(1.0.7)

or

\begin{aligned}\boxed{P \beta = \frac{g}{\lambda^3} f_{5/2}(z).}\end{aligned} \hspace{\stretch{1}}(1.0.9)

Energy

The average energy is the last thermodynamic quantity to come very easily. We have

\begin{aligned}U &= - \frac{\partial {}}{\partial {\beta}} \ln Z_{\mathrm{G}} \\ &= - \frac{\partial {T}}{\partial {\beta}} \frac{\partial {}}{\partial {T}} \ln Z_{\mathrm{G}} \\ &= - \frac{\partial {(1/k_{\mathrm{B}} T)}}{\partial {\beta}} \frac{\partial {}}{\partial {T}} P V \beta \\ &= \frac{1}{{k_{\mathrm{B}} \beta^2}}\frac{\partial {}}{\partial {T}} \frac{g V}{\lambda^3} f_{5/2}(z) \\ &= g V k_{\mathrm{B}} T^2f_{5/2}(z)\frac{\partial {}}{\partial {T}} \frac{\left( 2 \pi m k_{\mathrm{B}} T \right)^{3/2}}{h^3} \\ &= \frac{3}{2} \frac{g V k_{\mathrm{B}} T}{\lambda^3}f_{5/2}(z).\end{aligned} \hspace{\stretch{1}}(1.0.9)

From eq. 1.0.7, we have

\begin{aligned}\frac{g V}{\lambda^3} = \frac{N}{f_{3/2}(z) },\end{aligned} \hspace{\stretch{1}}(1.0.11)

so the energy takes the form

\begin{aligned}\boxed{U = \frac{3}{2} N k_{\mathrm{B}} T \frac{f_{5/2}(z)}{f_{3/2}(z) }.}\end{aligned} \hspace{\stretch{1}}(1.0.11)

We can compare this to the ratio of pressure to density

\begin{aligned}\frac{P \beta}{n} = \frac{f_{5/2}(z)}{f_{3/2}(z) },\end{aligned} \hspace{\stretch{1}}(1.0.11)

to find

\begin{aligned}U= \frac{3}{2} N k_{\mathrm{B}} T \frac{P V \beta}{N}= \frac{3}{2} P V,\end{aligned} \hspace{\stretch{1}}(1.0.11)

or

\begin{aligned}\boxed{P V = \frac{2}{3} U.}\end{aligned} \hspace{\stretch{1}}(1.0.11)

References

[1] RK Pathria. Statistical mechanics. Butterworth Heinemann, Oxford, UK, 1996.

Relativisitic Fermi gas

Posted by peeterjoot on April 20, 2013

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Question: Relativisitic Fermi gas ([1], pr 9.3)

Consider a relativisitic gas of $N$ particles of spin $1/2$ obeying Fermi statistics, enclosed in volume $V$, at absolute zero. The energy-momentum relation is

\begin{aligned}\epsilon = \sqrt{(p c)^2 + \epsilon_0^2},\end{aligned} \hspace{\stretch{1}}(1.1)

where $\epsilon_0 = m c^2$, and $m$ is the rest mass.

Find the Fermi energy at density $n$.

With the pressure $P$ defined as the average force per unit area exerted on a perfectly-reflecting wall of the container.

Set up expressions for this in the form of an integral.

Define the internal energy $U$ as the average $\epsilon - \epsilon_0$.
Set up expressions for this in the form of an integral.

Show that $P V = 2 U/3$ at low densities, and $P V = U/3$ at high densities. State the criteria for low and high densities.

There may exist a gas of neutrinos (and/or antineutrinos) in the cosmos. (Neutrinos are massless Fermions of spin $1/2$.) Calculate the Fermi energy (in eV) of such a gas, assuming a density of one particle per $\text{cm}^3$.

Attempt exact evaluation of the various integrals.

We’ve found [3] that the density of states associated with a 3D relativisitic system is

\begin{aligned}\mathcal{D}(\epsilon) = \frac{4 \pi V}{(c h)^3} \epsilon \sqrt{\epsilon^2 -\epsilon_0^2},\end{aligned} \hspace{\stretch{1}}(1.0.2)

For a given density $n$, we can find the Fermi energy in the same way as we did for the non-relativisitic energies, with the exception that we have to integrate from a lowest energy of $\epsilon_0$ instead of $0$ (the energy at $\mathbf{p} = 0$). That is

\begin{aligned}n &= \frac{N}{V} \\ &= \left( 2 \frac{1}{{2}} + 1 \right)\frac{4 \pi}{(c h)^3} \int_{\epsilon_0}^{\epsilon_{\mathrm{F}}}d\epsilon \epsilon \sqrt{ \epsilon^2 -\epsilon_0^2} \\ &= \frac{8 \pi}{(c h)^3}\frac{1}{{3}} {\left.{{\left( x^2 - \epsilon_0^2 \right)^{3/2}}}\right\vert}_{{\epsilon_0}}^{{\epsilon_{\mathrm{F}}}} \\ &= \frac{8 \pi}{3 (c h)^3}\left( \epsilon_{\mathrm{F}}^2 - \epsilon_0^2 \right)^{3/2}.\end{aligned} \hspace{\stretch{1}}(1.0.2)

Solving for $\epsilon_{\mathrm{F}}/\epsilon_0$ we have

\begin{aligned}\frac{\epsilon_{\mathrm{F}}}{\epsilon_0} =\sqrt{\left( \frac{3 (c h)^3 n}{8 \pi \epsilon_0^3} \right)^{2/3}+ 1}.\end{aligned} \hspace{\stretch{1}}(1.0.2)

We’ll see the constant factor above a number of times below and designate it

\begin{aligned}n_0 = \frac{8 \pi}{3} \left( \frac{\epsilon_0}{c h} \right)^3,\end{aligned} \hspace{\stretch{1}}(1.0.2)

so that the Fermi energy is

\begin{aligned}\frac{\epsilon_{\mathrm{F}}}{\epsilon_0} =\sqrt{\left( \frac{n}{n_0} \right)^{2/3}+ 1}.\end{aligned} \hspace{\stretch{1}}(1.0.2)

For the pressure calculation, let’s suppose that we have a configuration with a plane in the $x,y$ orientation as in fig. 1.1.

Fig 1.1: Pressure against x,y oriented plane

It’s argued in [4] section 6.4 that the pressure for such a configuration is

\begin{aligned}P = n \int p_z u_z f(\mathbf{u}) d^3 \mathbf{u},\end{aligned} \hspace{\stretch{1}}(1.7)

where $n$ is the number density and $f(\mathbf{u})$ is a normalized distribution function for the velocities. The velocity and momentum components are related by the Hamiltonian equations. From the Hamiltonian eq. 1.1 we find \footnote{ Observe that by squaring and summing one can show that this is equivalent to the standard relativisitic momentum $p_x = \frac{m v_x}{\sqrt{ 1 - \mathbf{u}^2/c^2}}$.} (for the x-component which is representative)

\begin{aligned}u_x \\ &= \frac{\partial {\epsilon}}{\partial {p_x}} \\ &= \frac{\partial {}}{\partial {p_x}}\sqrt{(p c)^2 +\epsilon_0^2} \\ &= \frac{ p_x c^2 }{\sqrt{(p c)^2 +\epsilon_0^2}}.\end{aligned} \hspace{\stretch{1}}(1.8)

For $\alpha \in \{1, 2, 3\}$ we can summarize these velocity-momentum relationships as

\begin{aligned}\frac{u_\alpha}{c} = \frac{ c p_\alpha }{ \epsilon }.\end{aligned} \hspace{\stretch{1}}(1.9)

Should we attempt to calculate the pressure with this parameterization of the velocity space we end up with convergence problems, and can’t express the results in terms of $f^+_\nu(z)$. Let’s try instead with a distribution over momentum space

\begin{aligned}P=n \int \frac{(c p_z)^2}{\epsilon} f(c \mathbf{p}) d^3 (c \mathbf{p}).\end{aligned} \hspace{\stretch{1}}(1.10)

Here the momenta have been scaled to have units of energy since we want to express this integral in terms of energy in the end. Our normalized distribution function is

\begin{aligned}f(c \mathbf{p})\propto \frac{\frac{1}{{ z^{-1} e^{\beta \epsilon} + 1 }}}{\int \frac{1}{{ z^{-1} e^{\beta \epsilon} + 1 }} d^3 (c \mathbf{p})},\end{aligned} \hspace{\stretch{1}}(1.11)

but before evaluating anything, we first want to change our integration variable from momentum to energy. In spherical coordinates our volume element takes the form

\begin{aligned}d^3 (c \mathbf{p}) &= 2 \pi (c p)^2 d (c p) \sin\theta d\theta \\ &= 2 \pi (c p)^2 \frac{d (c p)}{d \epsilon} d \epsilon \sin\theta d\theta.\end{aligned} \hspace{\stretch{1}}(1.12)

Implicit derivatives of

\begin{aligned}c^2 p^2 = \epsilon^2 - \epsilon_0^2,\end{aligned} \hspace{\stretch{1}}(1.13)

gives us

\begin{aligned}\frac{d (c p)}{d\epsilon}= \frac{\epsilon}{c p}=\frac{\epsilon}{\sqrt{\epsilon^2 -\epsilon_0^2}}.\end{aligned} \hspace{\stretch{1}}(1.0.14)

Our momentum volume element becomes

\begin{aligned}d^3 (c \mathbf{p}) \\ &= 2 \pi (c p)^2 \frac{\epsilon}{\sqrt{\epsilon^2 - \epsilon_0^2 }}d \epsilon \sin\theta d\theta \\ &= 2 \pi \left( \epsilon^2 - \epsilon_0^2 \right)\frac{\epsilon}{\sqrt{\epsilon^2 - \epsilon_0^2 }}d \epsilon \sin\theta d\theta \\ &= 2 \pi \epsilon \sqrt{ \epsilon^2 - \epsilon_0^2} d \epsilon \sin\theta d\theta.\end{aligned} \hspace{\stretch{1}}(1.0.14)

For our distribution function, we can now write

\begin{aligned}f(c \mathbf{p}) d^3 (c \mathbf{p})= C \frac{\epsilon \sqrt{ \epsilon^2 - \epsilon_0^2} d \epsilon }{ z^{-1} e^{\beta \epsilon} + 1 }\frac{ 2 \pi \sin\theta d\theta }{ 4 \pi \epsilon_0^3 },\end{aligned} \hspace{\stretch{1}}(1.0.14)

where $C$ is determined by the requirement $\int f(c \mathbf{p}) d^3 (c \mathbf{p}) = 1$

\begin{aligned}C^{-1} = \int_{0}^\infty \frac{(y + 1)\sqrt{ (y + 1)^2 - 1} dy }{ z^{-1} e^{\beta \epsilon_0 (y + 1)} + 1 }.\end{aligned} \hspace{\stretch{1}}(1.0.14)

The z component of our momentum can be written in spherical coordinates as

\begin{aligned}(c p_z)^2= (c p)^2 \cos^2\theta= \left( \epsilon^2 - \epsilon_0^2 \right)\cos^2\theta,\end{aligned} \hspace{\stretch{1}}(1.0.18)

Noting that

\begin{aligned}\int_0^\pi \cos^2\theta \sin\theta d\theta =-\int_0^\pi \cos^2\theta d(\cos\theta)= \frac{2}{3},\end{aligned} \hspace{\stretch{1}}(1.0.19)

all the bits come together as

\begin{aligned}P &= \frac{C n}{3 \epsilon_0^3 } \int_{\epsilon_0}^\infty\left( \epsilon^2 - \epsilon_0^2 \right)^{3/2} \frac{1}{{ z^{-1} e^{\beta \epsilon} + 1 }} d \epsilon \\ &= \frac{n \epsilon_0}{3} \int_{1}^\infty\left( x^2 - 1 \right)^{3/2} \frac{1}{{ z^{-1} e^{\beta \epsilon_0 x} + 1 }} dx.\end{aligned} \hspace{\stretch{1}}(1.0.19)

Letting $y = x - 1$, this is

\begin{aligned}P= \frac{C n \epsilon_0}{3} \int_{0}^\infty \frac{ \left( (y + 1)^2 - 1 \right)^{3/2} } { z^{-1} e^{\beta \epsilon_0 (y + 1)} + 1 } dy.\end{aligned} \hspace{\stretch{1}}(1.0.19)

We could conceivable expand the numerators of each of these integrals in power series, which could then be evaluated as a sum of $f^+_\nu(z e^{-\beta \epsilon_0})$ terms.

Note that above the Fermi energy $n$ also has an integral representation

\begin{aligned}n &= \left(2\left( \frac{1}{{2}} \right) + 1\right)\int_{\epsilon_0}^\infty d\epsilon \mathcal{D}(\epsilon) \frac{1}{{ z^{-1} e^{\beta \epsilon} + 1}} \\ &= \frac{8 \pi}{(c h)^3} \int_{\epsilon_0}^\infty d\epsilon\frac{\epsilon \sqrt{\epsilon^2 - \epsilon_0^2} }{ z^{-1} e^{\beta \epsilon} + 1} \\ &= \frac{8 \pi \epsilon_0^3}{(c h)^3} \int_{0}^\infty dy\frac{(y + 1)\sqrt{(y + 1)^2 - 1} }{ z^{-1} e^{\beta \epsilon_0 (y + 1)} + 1},\end{aligned} \hspace{\stretch{1}}(1.0.19)

or

\begin{aligned}\boxed{n = \frac{3 n_0}{C}.}\end{aligned} \hspace{\stretch{1}}(1.0.23)

Observe that we can use this result to remove the dependence of pressure on this constant $C$

\begin{aligned}\boxed{\frac{P}{n_0 \epsilon_0}= \int_{0}^\infty dy \frac{ \left( (y + 1)^2 - 1 \right)^{3/2} } { z^{-1} e^{\beta \epsilon_0 (y + 1)} + 1 }.}\end{aligned} \hspace{\stretch{1}}(1.0.24)

Now for the average energy difference from the rest energy $\epsilon_0$

\begin{aligned}U &= \left\langle{{\epsilon - \epsilon_0}}\right\rangle \\ &= \int_{\epsilon_0}^\infty d\epsilon \mathcal{D}(\epsilon) f(\epsilon) (\epsilon - \epsilon_0) \\ &= \frac{8 \pi V}{(c h)^3}\int_{\epsilon_0}^\infty d\epsilon \frac{ \epsilon(\epsilon - \epsilon_0) \sqrt{ \epsilon^2 - \epsilon_0 } }{ z^{-1} e^{\beta \epsilon} + 1} \\ &= \frac{8 \pi V \epsilon_0^4}{(c h)^3}\int_{0}^\infty dy\frac{ y ( y - 1 ) \sqrt{ (y + 1)^2 - 1 }}{ z^{-1} e^{\beta \epsilon} + 1}.\end{aligned} \hspace{\stretch{1}}(1.0.24)

So the average energy density difference from the rest energy, relative to the rest energy, is

\begin{aligned}\boxed{\frac{\left\langle{{\epsilon - \epsilon_0}}\right\rangle}{V \epsilon_0} =3 n_0 \int_{0}^\infty dy \frac { y (y + 1)\sqrt{(y + 1)^2 - 1} } { z^{-1} e^{\beta \epsilon_0 (y + 1)} + 1 }.}\end{aligned} \hspace{\stretch{1}}(1.0.26)

From eq. 1.0.24 and eq. 1.0.26 we have

\begin{aligned}\frac{1}{{n_0}} &= 3 \frac{V \epsilon_0} {\left\langle{{\epsilon - \epsilon_0}}\right\rangle} \int_{0}^\infty \frac { y (y + 1)\sqrt{(y + 1)^2 - 1} dy } { z^{-1} e^{\beta \epsilon_0 (y + 1)} + 1 } \\ &= \frac{\epsilon_0}{P} \int_{0}^\infty \frac{ \left( (y + 1)^2 - 1 \right)^{3/2} } { z^{-1} e^{\beta \epsilon_0 (y + 1)} + 1 } dy,\end{aligned} \hspace{\stretch{1}}(1.0.26)

or

\begin{aligned}P V =\frac{U}{3}\frac{ \int_{0}^\infty \frac{ \left( (y + 1)^2 - 1 \right)^{3/2} } { z^{-1} e^{\beta \epsilon_0 (y + 1)} + 1 } dy}{ \int_{0}^\infty \frac { y (y + 1)\sqrt{(y + 1)^2 - 1} dy } { z^{-1} e^{\beta \epsilon_0 (y + 1)} + 1 }}.\end{aligned} \hspace{\stretch{1}}(1.0.26)

This ratio of integrals is supposed to resolve to 1 and 2 in the low and high density limits. To consider this let’s perform one final non-dimensionalization, writing

\begin{aligned}\begin{aligned} \\ x &= \beta \epsilon_0 y \\ \theta &= \frac{1}{{\beta \epsilon_0}} = \frac{k_{\mathrm{B}} T}{\epsilon_0} \\ \bar{\mu} &= \mu - \epsilon_0 \\ \bar{z} &= e^{\beta \bar{\mu}}.\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.29)

The density, pressure, and energy take the form

\begin{aligned}\frac{n}{n_0}= 3 \theta\int_{0}^\infty dx\frac{(\theta x + 1)\sqrt{(\theta x + 1)^2 - 1} }{ \bar{z}^{-1} e^{x} + 1}\end{aligned} \hspace{\stretch{1}}(1.0.30a)

\begin{aligned}\frac{P}{n_0 \epsilon_0}= \theta \int_{0}^\infty dx \frac{ \left( (\theta x + 1)^2 - 1 \right)^{3/2} } { \bar{z}^{-1} e^{x} + 1 }\end{aligned} \hspace{\stretch{1}}(1.0.30b)

\begin{aligned}\frac{\left\langle{{\epsilon - \epsilon_0}}\right\rangle}{V \epsilon_0 n_0} =3 \theta^2 \int_{0}^\infty dx \frac { x (\theta x + 1)\sqrt{(\theta x + 1)^2 - 1} } { \bar{z}^{-1} e^{x} + 1 }.\end{aligned} \hspace{\stretch{1}}(1.0.30c)

We can rewrite the square roots in the number density and energy density expressions by expanding out the completion of the square

\begin{aligned}(1 + \theta x) \sqrt{ (1 + \theta x)^2 - 1}=(1 + \theta x) \sqrt{ 1 + \theta x + 1 }\sqrt{ 1 + \theta x - 1 }= \sqrt{2 \theta} x^{1/2} (1 + \theta x) \sqrt{ 1 + \frac{\theta x}{2}},\end{aligned} \hspace{\stretch{1}}(1.0.30c)

Expanding the distribution about $\bar{z} e^{-x} = 0$, we have

\begin{aligned}\frac{1}{ \bar{z}^{-1} e^{x} + 1}=\frac{\bar{z} e^{-x}}{ 1 + \bar{z} e^{-x}}=z e^{-x} \sum_{s = 0}^\infty (-1)^s \left( \bar{z} e^{-x} \right)^s,\end{aligned} \hspace{\stretch{1}}(1.0.32)

allowing us to write, in the low density limit with respect to $\bar{z}$

\begin{aligned}\frac{n}{n_0}= 3 \sqrt{2}\theta^{3/2} \sum_{s=0}^\infty(-1)^s\bar{z}^{s + 1}\int_{0}^\infty dx x^{1/2}(1 + \theta x) \sqrt{ 1 + \frac{\theta x}{2}} e^{-x(1 + s)} \end{aligned} \hspace{\stretch{1}}(1.0.33a)

\begin{aligned}\frac{P}{n_0 \epsilon_0}= \theta\sum_{s=0}^\infty(-1)^s\bar{z}^{s + 1} \int_{0}^\infty dx\left( (\theta x + 1)^2 - 1 \right)^{3/2} e^{-x(1 + s)} \end{aligned} \hspace{\stretch{1}}(1.0.33b)

\begin{aligned}\frac{\left\langle{{\epsilon - \epsilon_0}}\right\rangle}{V \epsilon_0 n_0} =3 \sqrt{2} \theta^{5/2} \sum_{s=0}^\infty(-1)^s\bar{z}^{s + 1} \int_{0}^\infty dx x^{3/2} (1 + \theta x) \sqrt{ 1 + \frac{\theta x}{2}} e^{-x(1 + s)} .\end{aligned} \hspace{\stretch{1}}(1.0.33c)

Low density result

An exact integration of the various integrals above is possible in terms of special functions. However, that attempt (included below) introduced an erroneous extra factor of $\theta$. Given that this end result was obtained by tossing all but the lowest order terms in $\theta$ and $\bar{z}$, let’s try that right from the get go.

For the pressure we have an integrand containing a factor

\begin{aligned}\left( (\theta x + 1)^2 -1 \right)^{3/2}&= \left( \theta x + 1 - 1 \right)^{3/2}\left( \theta x + 1 + 1 \right)^{3/2} \\ &= \theta^{3/2} x^{3/2} 2^{3/2} \left( 1 + \frac{\theta x}{2} \right)^{3/2} \\ &= 2 \sqrt{2} \theta^{3/2} x^{3/2} \left( 1 + \frac{\theta x}{2} \right)^{3/2}\approx2 \sqrt{2} \theta^{3/2} x^{3/2} \end{aligned} \hspace{\stretch{1}}(1.0.33c)

Our pressure, to lowest order in $\theta$ and $\bar{z}$ is then

\begin{aligned}\frac{P}{\epsilon_0 n_0} = 2 \sqrt{2} \theta^{5/2} \bar{z} \int_0^\infty x^{3/2} e^{-x} dx= 2 \sqrt{2} \theta^{5/2} \bar{z} \Gamma(5/2).\end{aligned} \hspace{\stretch{1}}(1.0.33c)

Our energy density to lowest order in $\theta$ and $\bar{z}$ from eq. 1.0.33c is

\begin{aligned}\frac{U}{V \epsilon_0 n_0} &= 3 \sqrt{2} \theta^{5/2} \bar{z} \int_{0}^\infty dx x^{3/2} e^{-x} \\ &= 3 \sqrt{2} \theta^{5/2} \bar{z} \Gamma(5/2).\end{aligned} \hspace{\stretch{1}}(1.0.33c)

Comparing these, we have

\begin{aligned}\frac{1}{{\epsilon_0 n_0\sqrt{2} \theta^{5/2} \bar{z} \Gamma(5/2)}} &= 3 \frac{V}{U} \\ &= \frac{2}{P},\end{aligned} \hspace{\stretch{1}}(1.0.37)

or in this low density limit

\begin{aligned}\boxed{P V = \frac{2}{3} U.}\end{aligned} \hspace{\stretch{1}}(1.0.38)

High density limit

For the high density limit write $\bar{z} = e^y$, so that the distribution takes the form

\begin{aligned}f(\bar{z}) &= \frac{1}{ \bar{z}^{-1} e^{x} + 1} \\ &= \frac{1}{ e^{x - y} + 1}.\end{aligned} \hspace{\stretch{1}}(1.0.39)

This can be approximated by a step function, so that

\begin{aligned}\frac{P}{n_0 \epsilon_0}\approx \int_{0}^y \theta dx\left( (\theta x + 1)^2 - 1 \right)^{3/2} \end{aligned} \hspace{\stretch{1}}(1.0.40a)

\begin{aligned}\frac{U}{V \epsilon_0 n_0} \approx3 \int_{0}^\infty \theta dx \theta x (\theta x + 1)\sqrt{(\theta x + 1)^2 - 1} \end{aligned} \hspace{\stretch{1}}(1.0.40b)

With a change of variables $u = \theta x + 1$, we have

\begin{aligned}\begin{aligned}\frac{P}{n_0 \epsilon_0} &\approx \int_{1}^{\theta y + 1x} du\left( u^2 - 1 \right)^{3/2} \\ &=\frac{1}{8} \left((2 \theta y (\theta y+2)-3) \sqrt{\theta y (\theta y+2)} (\theta y+1)+3 \ln \left(\theta y+\sqrt{\theta y (\theta y+2)}+1\right)\right) \\ &\approx\frac{1}{4} \left( \theta \ln \bar{z} \right)^4\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.41a)

\begin{aligned}\begin{aligned}\frac{U}{V \epsilon_0 n_0} &\approx3 \int_{1}^{\theta y + 1x} (u^2 - u)\sqrt{u^2 - 1} \\ &=\frac{3}{24} \left(\sqrt{\theta y (\theta y+2)} (\theta y (2 \theta y (3 \theta y+5)-1)+3)-3 \left(\ln \left(\theta y+\sqrt{\theta y (\theta y+2)}+1\right)\right)\right) \\ &\approx\frac{3}{4} \left( \theta \ln \bar{z} \right)^4\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.41b)

Comparing both we have

\begin{aligned}\frac{4}{\epsilon_0 n_0 \left( \theta \ln \bar{z} \right) } = \frac{1}{{P}} = \frac{3 V}{U},\end{aligned} \hspace{\stretch{1}}(1.0.42)

or

\begin{aligned}\boxed{P V = \frac{1}{{3}} U.}\end{aligned} \hspace{\stretch{1}}(1.0.43)

\begin{aligned}{\left.{{\epsilon_{\mathrm{F}}}}\right\vert}_{{n = 1/(0.01)^3}} = 6.12402 \times 10^{-35} \text{J} \times 6.24150934 \times 10^{18} \frac{\text{eV}}{\text{J}} = 3.82231 \times 10^{-16} \text{eV}\end{aligned} \hspace{\stretch{1}}(1.0.43)

Wow. That’s pretty low!

Pressure integral

Of these the pressure integral is yields directly to Mathematica

\begin{aligned}\begin{aligned} \int_{0}^\infty & dx\left( (\theta x + 1)^2 - 1 \right)^{3/2} e^{-x(1 + s)} \\ &=\frac{3 \theta e^{(s+1)/\theta}}{(s + 1)^2} K_2\left( \frac{s+1}{\theta } \right) \\ &=\frac{3 \sqrt{\frac{\pi }{2}} \theta ^{3/2}}{(s+1)^{5/2}}+\frac{45 \sqrt{\frac{\pi }{2}} \theta ^{5/2}}{8 (s+1)^{7/2}}+\frac{315 \sqrt{\frac{\pi }{2}} \theta ^{7/2}}{128 (s+1)^{9/2}}-\frac{945 \sqrt{\frac{\pi }{2}} \theta ^{9/2}}{1024 (s+1)^{11/2}}+\frac{31185 \sqrt{\frac{\pi }{2}} \theta ^{11/2}}{32768 (s+1)^{13/2}} + \cdots\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.45)

where $K_2(z)$ is a modified Bessel function [5] of the second kind as plotted in fig. 1.2.

Fig 1.2: Modified Bessel function of the second kind

Plugging this into the series for the pressure, we have

\begin{aligned}\frac{P}{n_0 \epsilon_0}= 3 \left( \frac{k_{\mathrm{B}} T}{\epsilon_0} \right)^2\sum_{s=0}^\infty(-1)^s\frac{\left( \bar{z} e^{\epsilon_0/k_{\mathrm{B}} T} \right)^{s + 1}}{(s + 1)^2}K_2\left( (s+1) \epsilon_0/k_{\mathrm{B}} T \right).\end{aligned} \hspace{\stretch{1}}(1.0.46)

Plotting the summands $3 (-1)^s \frac{\theta^2}{(s + 1)^2} \left( \bar{z} e^{ 1/\theta} \right)^{s + 1} K_2\left((s+1)/\theta\right)$ for $\bar{z} = 1$ in fig. 1.4 shows that this mix of exponential Bessel and quadratic terms decreases with $s$.

Plotting this sum in fig. 1.3 numerically to 10 terms, shows that we have a function that appears roughly polynomial in $\bar{z}$ and $\theta$.

Fig 1.3: Pressure to ten terms in z and theta

Fig 1.4: Pressure summands

For small $\bar{z}$ it can be seen graphically that there is very little contribution from anything but the $s = 0$ term of this sum. An expansion in series for a few terms in $\bar{z}$ and $\theta$ gives us

\begin{aligned}\begin{aligned}\frac{P}{\epsilon_0 n_0}&=\sqrt{\pi} \theta^{5/2} \left(\frac{3 \bar{z}}{\sqrt{2}}-\frac{3 \bar{z}^2}{8}+\frac{\bar{z}^3}{3 \sqrt{6}}-\frac{3 \bar{z}^4}{32 \sqrt{2}}+\frac{3 \bar{z}^5}{25 \sqrt{10}}\right) \\ &+\sqrt{\pi} \theta^{7/2} \left(\frac{45 \bar{z}}{8 \sqrt{2}}\right) -\frac{45 \bar{z}^2}{128}+\frac{5 \bar{z}^3}{24 \sqrt{6}}-\frac{45 \bar{z}^4}{1024 \sqrt{2}}+\frac{9 \bar{z}^5}{200 \sqrt{10}}\\ &+\sqrt{\pi} \theta^{9/2} \left(\frac{315 \bar{z}}{128 \sqrt{2}}-\frac{315 \bar{z}^2}{4096}+\frac{35 \bar{z}^3}{1152 \sqrt{6}}-\frac{315 \bar{z}^4}{65536 \sqrt{2}}+\frac{63 \bar{z}^5}{16000 \sqrt{10}}\right).\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.47)

This allows a $k_{\mathrm{B}} T \ll m c^2$ and $\bar{z} \ll 1$ approximation of the pressure

\begin{aligned}\frac{P}{\epsilon_0 n_0} = \frac{3}{2} \sqrt{2 \pi} \bar{z} \theta^{5/2}.\end{aligned} \hspace{\stretch{1}}(1.0.48)

Number density integral

For the number density, it appears that we can evaluate the integral using integration from parts applied to eq. 1.0.30.30

\begin{aligned}\frac{n}{n_0}= \theta\int_{0}^\infty dx\frac{3 (\theta x + 1)\sqrt{(\theta x + 1)^2 - 1} }{ \bar{z}^{-1} e^{x} + 1}=\theta\int_{0}^\infty dx\left( \frac{d}{dx} \left( (\theta x + 1)^2 - 1 \right) ^{3/2} \right)\frac{1}{ \bar{z}^{-1} e^{x} + 1}={\left.{{\theta\left( (\theta x + 1)^2 - 1 \right)^{3/2}\frac{1}{ \bar{z}^{-1} e^{x} + 1}}}\right\vert}_{{0}}^{{\infty}}-\theta\int_{0}^\infty dx\left( (\theta x + 1)^2 - 1 \right)^{3/2}\frac{ -\bar{z}^{-1} e^{x} }{ \left( \bar{z}^{-1} e^{x} + 1 \right)^2}=\theta\int_{0}^\infty dx\left( (\theta x + 1)^2 - 1 \right)^{3/2}\frac{ \bar{z} e^{-x} }{ \left( 1 + \bar{z} e^{-x} \right)^2}.\end{aligned} \hspace{\stretch{1}}(1.0.48)

Expanding in series, gives us

\begin{aligned}\frac{n}{n_0}=\theta\sum_{s = 0}^\infty\binom{-2}{s}\bar{z}^{s + 1} \int_{0}^\infty dx\left( (\theta x + 1)^2 - 1 \right)^{3/2} e^{-x(s + 1)}=3 \theta^2\sum_{s = 0}^\infty\binom{-2}{s}\frac{\left( \bar{z} e^{1/\theta} \right)^{s + 1}}{(s + 1)^2}K_2\left( \frac{s+1}{\theta } \right).\end{aligned} \hspace{\stretch{1}}(1.0.48)

Here the binomial coefficient has the meaning given in the definitions of \statmechchapcite{nonIntegralBinomialSeries}, where for negative integral values of $b$ we have

\begin{aligned}\binom{b}{s}\equiv(-1)^s \frac{-b}{-b + s} \binom{-b+s}{-b}.\end{aligned} \hspace{\stretch{1}}(1.0.51)

Expanding in series to a couple of orders in $\theta$ and $\bar{z}$ we have

\begin{aligned}\frac{n}{n_0} = \frac{\sqrt{2 \pi}}{36} \theta^{1/2} \left(\left(2 \sqrt{3} \bar{z} - 9/\sqrt{2} \right) \bar{z} +18 \right) \bar{z}+\frac{5 \sqrt{ 2 \pi}}{576} \theta^{3/2} \left(\left(4 \sqrt{3} \bar{z} - 27/\sqrt{2}\right) \bar{z} +108 \right) \bar{z}+ \cdots\end{aligned} \hspace{\stretch{1}}(1.0.52)

To first order in $\theta$ and $\bar{z}$ this is

\begin{aligned}\frac{n}{n_0} = \frac{1}{{2}} \sqrt{ 2 \pi } \bar{z} \theta^{1/2},\end{aligned} \hspace{\stretch{1}}(1.0.53)

which allows a relation to pressure

\begin{aligned}P V = 3 N (k_{\mathrm{B}} T)^2 /\epsilon_0.\end{aligned} \hspace{\stretch{1}}(1.0.54)

It’s kind of odd seeming that this is quadratic in temperature. Is there an error?

Energy integral

Starting from eq. 1.0.30c and integrating by parts we have

\begin{aligned}\frac{\left\langle{{\epsilon - \epsilon_0}}\right\rangle}{V \epsilon_0 n_0} &= 3 \theta^2 \int_{0}^\infty dx \frac { x (\theta x + 1)\sqrt{(\theta x + 1)^2 - 1} } { \bar{z}^{-1} e^{x} + 1 } \\ &= -\theta^2 \int_{0}^\infty dx\left( (\theta x + 1)^2 - 1 \right)^{3/2}\frac{d}{dx} \left( \frac{x} { \bar{z}^{-1} e^{x} + 1 } \right) \\ &= -\theta^2 \int_{0}^\infty dx\left( (\theta x + 1)^2 - 1 \right)^{3/2}\left( \frac{1} { \bar{z}^{-1} e^{x} + 1 } - \frac{x \bar{z}^{-1} e^{x} } { \left( \bar{z}^{-1} e^{x} + 1 \right)^2 } \right) \\ &= \theta^2 \int_{0}^\infty dx \left( (\theta x + 1)^2 - 1 \right)^{3/2} \frac{ (x - 1)\bar{z}^{-1} e^{x} - 1} { \left( \bar{z}^{-1} e^{x} + 1 \right)^2 } \\ &= \theta^2 \int_{0}^\infty dx \left( (\theta x + 1)^2 - 1 \right)^{3/2} \frac{ (x - 1)\bar{z} e^{-x} - \bar{z}^2 e^{-2 x}} { \left( 1 + \bar{z} e^{-x} \right)^2 } \\ &= \theta^2\sum_{s=0}^\infty \binom{-2}{s} \int_{0}^\infty dx \left( (\theta x + 1)^2 - 1 \right)^{3/2} \left( (x - 1)\bar{z} e^{-x} - \bar{z}^2 e^{-2 x} \right) (\bar{z} e^{-x})^s \\ &= \theta^2\sum_{s=0}^\infty \binom{-2}{s} \bar{z}^{s + 1} \int_{0}^\infty dx \left( (\theta x + 1)^2 - 1 \right)^{3/2} \left( (x - 1) e^{-x(s + 1)} - \bar{z} e^{-x(s + 2)} \right).\end{aligned} \hspace{\stretch{1}}(1.0.54)

The integral with the factor of $x$ doesn’t have a nice closed form as before (if you consider the $K_2$ a nice closed form), but instead evaluates to a confluent hypergeometric function [6]. That integral is

\begin{aligned}\int_0^{\infty } x \left((\theta x+1)^2-1\right)^{3/2} e^{-x (1+s)} dx = \frac{15 \sqrt{\pi } \theta^3 U\left(-\frac{3}{2},-4,\frac{2 (s+1)}{\theta }\right)}{8 (s+1)^5},\end{aligned} \hspace{\stretch{1}}(1.0.54)

and looks like fig. 1.5. Series expansion shows that this hypergeometricU function has a $\theta^{3/2}$ singularity at the origin

Fig 1.5: Plot of HypergeometricU, and with theta^5 scaling

\begin{aligned}U\left(-\frac{3}{2},-4,\frac{2 (s+1)}{\theta }\right)=\frac{2 \sqrt{2} \sqrt{s+1} s+2 \sqrt{2} \sqrt{s+1}}{\theta^{3/2}}+\frac{21 \sqrt{s+1}}{2 \sqrt{2} \sqrt{\theta }}+ \cdots\end{aligned} \hspace{\stretch{1}}(1.57)

so our multiplication by $\theta^5$ brings us to zero as seen in the plot. Evaluating the complete integral yields the unholy mess

\begin{aligned}\frac{\left\langle{{\epsilon - \epsilon_0}}\right\rangle}{V \epsilon_0 n_0} &= \sum_{s=0}^\infty \theta^2 (-1)^s (s+1) \bar{z}^{s+1} \Bigl( \\ &\frac{105 \sqrt{\pi } \theta^3 U\left(-\frac{1}{2},-4,\frac{2 (s+1)}{\theta }\right)}{16 (s+1)^5} \\ &- \frac{3 \sqrt{\pi } \theta^2 U\left(-\frac{1}{2},-2,\frac{2 (s+1)}{\theta }\right)}{2 (s+1)^3} \\ &- \frac{3 \sqrt{\pi } \theta^2 \bar{z} U\left(-\frac{1}{2},-2,\frac{2 (s+2)}{\theta }\right)}{2 (s+2)^3} \\ &+\frac{(\theta -2) (-3 \theta +2 s+2) e^{\frac{s+1}{\theta }} K_2\left(\frac{s+1}{\theta }\right)}{\theta (s+1)^2} \\ &-\frac{2 (\theta -2) e^{\frac{s+1}{\theta }} K_1\left(\frac{s+1}{\theta }\right)}{\theta (s+1)} \\ &+\frac{\bar{z} (-3 \theta +2 s+4) e^{\frac{s+2}{\theta }} K_2\left(\frac{s+2}{\theta }\right)}{(s+2)^2} \\ &-\frac{2 \bar{z} e^{\frac{s+2}{\theta }} K_1\left(\frac{s+2}{\theta }\right)}{s+2} \Bigr),\end{aligned} \hspace{\stretch{1}}(1.58)

to first order in $\bar{z}$ and $\theta$ this is

\begin{aligned}\frac{\left\langle{{\epsilon - \epsilon_0}}\right\rangle}{V \epsilon_0 n_0} =\frac{9}{4} \sqrt{2 \pi} \bar{z} \theta^{7/2}.\end{aligned} \hspace{\stretch{1}}(1.59)

Comparing pressure and energy we have for low densities (where $\bar{z} \approx 0$)

\begin{aligned}\frac{1}{{\epsilon_0 n_0 \sqrt{2 \pi} \bar{z} \theta^{5/2}}} = \frac{3}{2} \frac{1}{{P}} = \frac{9}{4} \theta \frac{V}{U},\end{aligned} \hspace{\stretch{1}}(1.0.60)

or

\begin{aligned}\theta P V = \frac{2}{3} U.\end{aligned} \hspace{\stretch{1}}(1.0.61)

It appears that I’ve picked up an extra factor of $\theta$ somewhere, but at least I’ve got the $2/3$ low density expression. Given that I’ve Taylor expanded everything anyways around $\bar{z}$ and $\theta$ this could likely have been done right from the get go, instead of dragging along the messy geometric integrals. Reworking this part of this problem like that was done above.

References

[1] Kerson Huang. Introduction to statistical physics. CRC Press, 2001.

[2] Peeter Joot. Basic statistical mechanics., chapter {Non integral binomial coefficient}. \natexlab{a}. URL http://sites.google.com/site/peeterjoot2/math2013/phy452.pdf.

[3] Peeter Joot. Basic statistical mechanics., chapter {Relativisitic density of states}. \natexlab{b}. URL http://sites.google.com/site/peeterjoot2/math2013/phy452.pdf.

[4] RK Pathria. Statistical mechanics. Butterworth Heinemann, Oxford, UK, 1996.

[5] Wolfram. BesselK, \natexlab{a}. URL http://reference.wolfram.com/mathematica/ref/BesselK.html. [Online; accessed 11-April-2013].

[6] Wolfram. HyperGeometricU, \natexlab{b}. URL http://reference.wolfram.com/mathematica/ref/HypergeometricU.html. [Online; accessed 17-April-2013].

Non integral binomial coefficient

Posted by peeterjoot on April 10, 2013

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Updated.Â  Original generalization of binomial coefficient wasn’t correct for negative exponents.

In [2] appendix section F was the use of binomial coefficients in a non-integral binomial expansion. This surprised me, since I’d never seen that before. However, on reflection, this is a very sensible notation, provided the binomial coefficients are defined in terms of the gamma function. Let’s explore this little detail explicitly.

Taylor series

\begin{aligned}f(x) = (a + x)^b.\end{aligned} \hspace{\stretch{1}}(1.1)

Our derivatives are

\begin{aligned}\begin{aligned}f'(x) &= b (a + x)^{b-1} \\ f''(x) &= b(b-1) (a + x)^{b-2} \\ f^3(x) &= b(b-1)(b-(3-1)) (a + x)^{b-3} \\ \dot{v}s & \\ f^k(x) &= b(b-1)\cdots (b-(k-1)) (a + x)^{b-k}.\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.2)

Our Taylor series is then

\begin{aligned}(a + x)^b = \sum_{k = 0}^\infty \frac{1}{{k!}} b(b-1)\cdots (b-(k-1)) a^{b-k} x^k.\end{aligned} \hspace{\stretch{1}}(1.0.2)

Note that if $b$ is a positive integer, then all the elements of this series become zero at $b = k - 1$, or

\begin{aligned}(a + x)^b = \sum_{k = 0}^k \frac{1}{{k!}} b(b-1)\cdots (b-(k-1)) a^{b-k} x^k.\end{aligned} \hspace{\stretch{1}}(1.0.2)

Gamma function

Let’s now relate this to the gamma function. From [1] section 6.1.1 we have

\begin{aligned}\Gamma(z) = \int_0^\infty t^{z-1} e^{-t} dt.\end{aligned} \hspace{\stretch{1}}(1.0.2)

Iteratively integrating by parts, we find the usual relation between gamma functions of integral separation

\begin{aligned}\Gamma(z + 1) &= \int_0^\infty t^{z} e^{-t} dt \\ &= \int_0^\infty t^{z} d \left( \frac{ e^{-t}}{-1} \right) \\ &= {\left.{{t^z \frac{e^{-t}}{-1}}}\right\vert}_{{0}}^{{\infty}}- \int_0^\infty z t^{z-1} \frac{e^{-t}}{-1} dt \\ &= z \int_0^\infty t^{z-1} e^{-t} dt \\ &= z (z - 1)\int_0^\infty t^{z-2} e^{-t} dt \\ &= z (z - 1)(z - (3-1))\int_0^\infty t^{z-3} e^{-t} dt \\ &= z (z - 1) \cdots (z - (k-1))\int_0^\infty t^{z-k} e^{-t} dt \\ &= z (z - 1) \cdots (z - (k-1))\int_0^\infty t^{(z + 1 - k) - 1} e^{-t} dt,\end{aligned} \hspace{\stretch{1}}(1.0.2)

or

\begin{aligned}\Gamma(z + 1) = z (z - 1) \cdots (z - (k-1)) \Gamma( z - (k-1) ).\end{aligned} \hspace{\stretch{1}}(1.0.2)

Flipping this gives us a nice closed form expression for the products of a number of positive unit separated values

\begin{aligned}z (z - 1) \cdots (z - k)=\frac{\Gamma(z + 1)}{\Gamma( z - (k-1) )}.\end{aligned} \hspace{\stretch{1}}(1.0.2)

Binomial coefficient for positive exponents

Considering first positive exponents $b$, we can now use this in our Taylor expansion eq. 1.0.2

\begin{aligned}(a + x)^b &= \sum_{k = 0}^\infty \frac{1}{{k!}} \frac{\Gamma(b+1)}{\Gamma(b - k + 1)}a^{b-k} x^k \\ &= \sum_{k = 0}^\infty \frac{\Gamma(b + 1)}{\Gamma(k + 1)\Gamma(b - k + 1)}a^{b-k} x^k.\end{aligned} \hspace{\stretch{1}}(1.0.2)

Observe that when $b$ is a positive integer we have

\begin{aligned}\frac{\Gamma(b + 1)}{\Gamma(k + 1)\Gamma(b - k + 1)} &= \frac{b!}{k!(b-k)!} \\ &= \binom{b}{k}.\end{aligned} \hspace{\stretch{1}}(1.0.10)

So for positive values of $b$, even non-integer values, we see that is then very reasonable to define the binomial coefficient \index{binomial coefficient} explicitly in terms of the gamma function

\begin{aligned}\binom{b}{k} \equiv\frac{\Gamma(b + 1)}{\Gamma(k + 1)\Gamma(b - k + 1)}.\end{aligned} \hspace{\stretch{1}}(1.0.11)

If we do that, then the binomial expansion for non-integral values of $b$ is simply

\begin{aligned}(a + x)^b = \sum_{k = 0}^\infty \binom{b}{k} a^{b-k} x^k.\end{aligned} \hspace{\stretch{1}}(1.0.11)

Binomial coefficient for negative integer exponents

Using the relation eq. 1.0.11 blindly leads to some trouble, since $\Gamma(-\left\lvert {m} \right\rvert)$ goes to infinity for integer values of $m > 0$. We have to modify the definition of the binomial coefficient. Let’s rewrite eq. 1.0.2 for negative integer values of $b = -m$ as

\begin{aligned}(a + x)^{-m} &= \sum_{k = 0}^\infty \frac{1}{{k!}} (-m)(-m-1)\cdots (-m-(k-1)) a^{-m-k} x^k \\ &= \sum_{k = 0}^\infty \frac{1}{{k!}} (-1)^km(m+1)\cdots (m+(k-1)) a^{-m-k} x^k.\end{aligned} \hspace{\stretch{1}}(1.0.11)

Let’s also put the ratio of gamma functions relation of eq. 1.0.2, in a slightly more general form. For $u, v > 0$, where $u - v$ is an integer, we can write

\begin{aligned}u (u - 1) \cdots (v) = \frac{\Gamma(u + 1)}{\Gamma( v )}.\end{aligned} \hspace{\stretch{1}}(1.0.11)

Our Taylor series takes the form

\begin{aligned}(a + x)^{-m} &= \sum_{k = 0}^\infty \frac{1}{{k!}} (-1)^k(m+k-1) (m+k-2) \cdots (m)a^{-m-k} x^k \\ &= \sum_{k = 0}^\infty (-1)^k \frac{ \Gamma(m + k) }{\Gamma(m)\Gamma(k+1)}a^{-m-k} x^k.\end{aligned} \hspace{\stretch{1}}(1.0.16)

We can now define, for negative integers $-m$

\begin{aligned}\binom{-m}{k}\equiv(-1)^k \frac{ \Gamma(m + k) }{ \Gamma(m)\Gamma(k+1) }.\end{aligned} \hspace{\stretch{1}}(1.0.16)

With such a definition, our Taylor series takes the tidy form

\begin{aligned}(a + x)^{-m} = \sum_{k = 0}^\infty \binom{-m}{k} a^{-m-k} x^k.\end{aligned} \hspace{\stretch{1}}(1.0.16)

For negative integer values of $b = -m$, this is now consistent with eq. 1.0.11.

Observe that we can put eq. 1.0.16 into the standard binomial form with a bit of manipulation

\begin{aligned}\binom{-m}{k} &= (-1)^k \frac{ \Gamma(m + k) }{ \Gamma(m)\Gamma(k+1) } \\ &= (-1)^k \frac{ (m + k -1)! }{ (m-1)! k! } \\ &= (-1)^k \frac{ m (m + k)! }{ (m + k) m! k! },\end{aligned} \hspace{\stretch{1}}(1.0.16)

or

\begin{aligned}\binom{-m}{k}=(-1)^k \frac{m}{m + k} \binom{m+k}{m}.\end{aligned} \hspace{\stretch{1}}(1.0.19)

Negative non-integral binomial coefficients

TODO. There will be some ugliness due to the changes of sign in the products $b(b-1)\cdots (b -k + 1)$ since $b$ and $b -k + 1$ may have different sign. A product of two ratios of gamma functions will be required to express this product, which will further complicate the definition of binomial coefficient.

References

[1] M. Abramowitz and I.A. Stegun. \emph{Handbook of mathematical functions with formulas, graphs, and mathematical tables}, volume 55. Dover publications, 1964.

[2] RK Pathria. Statistical mechanics. Butterworth Heinemann, Oxford, UK, 1996.

PHY452H1S Basic Statistical Mechanics. Problem Set 7: BEC and phonons

Posted by peeterjoot on April 10, 2013

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Disclaimer

Question: Bose-Einstein condensation (BEC) in one and two dimensions

Obtain the density of states $N(\epsilon)$ in one and two dimensions for a particle with an energy-momentum relation

\begin{aligned}E_\mathbf{k} = \frac{\hbar^2 \mathbf{k}^2}{2 m}.\end{aligned} \hspace{\stretch{1}}(1.1)

Using this, show that for particles whose number is conserved the BEC transition temperature vanishes in these cases – so we can always pick a chemical potential $\mu < 0$ which preserves a constant density at any temperature.

We’d like to evaluate

\begin{aligned}N_d(\epsilon) \equiv\sum_\mathbf{k}\delta(\epsilon - \epsilon_\mathbf{k})\approx\frac{L^d}{(2 \pi)^d} \int d^d \mathbf{k} \delta\left( \epsilon - \frac{\hbar^2 k^2}{2 m} \right),\end{aligned} \hspace{\stretch{1}}(1.2)

We’ll use

\begin{aligned}\delta(g(x)) = \sum_{x_0} \frac{\delta(x - x_0)}{\left\lvert {g'(x_0)} \right\rvert},\end{aligned} \hspace{\stretch{1}}(1.3)

where the roots of $g(x)$ are $x_0$. With

\begin{aligned}g(k) = \epsilon - \frac{\hbar^2 k^2}{2 m},\end{aligned} \hspace{\stretch{1}}(1.4)

the roots $k^{*}$ of $g(k) = 0$ are

\begin{aligned}k^{*} = \pm \sqrt{\frac{2 m \epsilon }{\hbar^2}}.\end{aligned} \hspace{\stretch{1}}(1.5)

The derivative of $g(k)$ evaluated at these roots are

\begin{aligned}g'(k^{*}) &= -\frac{\hbar^2 k^{*}}{m} \\ &= \mp \frac{\hbar^2}{m}\frac{\sqrt{2 m \epsilon}}{ \hbar } \\ &= \mp \frac{\hbar \sqrt{2 m \epsilon} }{m}.\end{aligned} \hspace{\stretch{1}}(1.6)

In 2D, we can evaluate over a shell in $k$ space

\begin{aligned}N_2(\epsilon) &= \frac{A}{(2 \pi)^2} \int_0^\infty 2 \pi k dk\left( \delta \left( k - k^{*} \right) + \delta \left( k + k^{*} \right) \right)\frac{m}{\hbar \sqrt{2 m \epsilon} } \\ &= \frac{A}{2 \pi} \not{{k^{*}}}\frac{m}{\hbar^2 \not{{k^{*}}} }\end{aligned} \hspace{\stretch{1}}(1.7)

or

\begin{aligned}\boxed{N_2(\epsilon) = \frac{2 \pi A m}{h^2}.}\end{aligned} \hspace{\stretch{1}}(1.8)

In 1D we have

\begin{aligned}N_1(\epsilon) &= \frac{L}{2 \pi} \int_{-\infty}^\infty dk\left( \delta \left( k - k^{*} \right) + \delta \left( k + k^{*} \right) \right)\frac{m}{\hbar \sqrt{2 m \epsilon} } \\ &= \frac{2 L}{2 \pi} \frac{m}{\hbar \sqrt{2 m \epsilon} }.\end{aligned} \hspace{\stretch{1}}(1.9)

Observe that this time for 1D, unlike in 2D when we used a radial shell in $k$ space, we have contributions from both the delta function roots. Our end result is

\begin{aligned}\boxed{N_1(\epsilon) =\frac{2 L}{h} \sqrt{\frac{m}{2 \epsilon}}.}\end{aligned} \hspace{\stretch{1}}(1.10)

To consider the question of the BEC temperature, we’ll need to calculate the density. For the 2D case we have

\begin{aligned}\rho = \frac{N}{A} &= \frac{1}{A} A \int \frac{d^2 \mathbf{k}}{(2 \pi)^2} f(e_\mathbf{k}) \\ &= \frac{1}{A} \frac{2 \pi A m}{h^2}\int_0^\infty d\epsilon \frac{1}{{ z^{-1} e^{\beta \epsilon} -1 }} \\ &= \frac{2 \pi m}{h^2 \beta}\int_0^\infty dx \frac{1}{{ z^{-1} e^{x} -1 }} \\ &= -\frac{2 \pi m k_{\mathrm{B}} T}{h^2} \ln (1 - z) \\ &= -\frac{1}{{\lambda^2}} \ln (1 - z).\end{aligned} \hspace{\stretch{1}}(1.11)

Recall for the 3D case that we had an upper bound as $z \rightarrow 1$. We don’t have that for this 2D density, so for any value of $k_{\mathrm{B}} T > 0$, a corresponding value of $z$ can be found. That is

\begin{aligned}z &= 1 - e^{-\rho \lambda^2} \\ &= 1 - e^{-\rho h^4/(2 \pi m k_{\mathrm{B}} T)^2}.\end{aligned} \hspace{\stretch{1}}(1.1.12)

For the 1D case we have

\begin{aligned}\rho &= \frac{N}{L} \\ &= \frac{1}{L} L \int \frac{dk}{2 \pi} f(e_\mathbf{k}) \\ &= \frac{1}{L} \frac{2 L}{h} \sqrt{\frac{m}{2}}\int_0^\infty d\epsilon \frac{1}{{\sqrt{\epsilon}}}\frac{1}{{ z^{-1} e^{\beta \epsilon} -1 }} \\ &= \frac{1}{{h}} \sqrt{\frac{2 m}{\beta}} \int_0^\infty \frac{x^{1/2 - 1}}{z^{-1} e^x - 1} \\ &= \frac{1}{{h}} \sqrt{\frac{2 m}{\beta}} \Gamma(1/2) f^-_{1/2}(z),\end{aligned} \hspace{\stretch{1}}(1.1.12)

or

\begin{aligned}\rho= \frac{1}{{\lambda}} f^-_{1/2}(z).\end{aligned} \hspace{\stretch{1}}(1.1.12)

See fig. 1.1 for plots of $f^-_\nu(z)$ for $\nu \in \{1/2, 1, 3/2\}$, the respective results for the 1D, 2D and 3D densities respectively.

Fig 1.1: Density integrals for 1D, 2D and 3D cases

We’ve found that $f^-_{1/2}(z)$ is also unbounded as $z \rightarrow 1$, so while we cannot invert this easily as in the 2D case, we can at least say that there will be some $z$ for any value of $k_{\mathrm{B}} T > 0$ that allows the density (and thus the number of particles) to remain fixed.

Question: Estimating the BEC transition temperature

Find data for the atomic mass of liquid ${}^4$ He and its density at ambient atmospheric pressure and hence estimate its BEC temperature assuming interactions are unimportant (even though this assumption is a very bad one!).

For dilute atomic gases of the sort used in Professor
Thywissen’s lab
, one typically has a cloud of $10^6$ atoms confined to an approximate cubic region with linear dimension 1 $\mu\,m$. Find the density – it is pretty low, so interactions can be assumed to be extremely weak. Assuming these are ${}^{87}$ Rb atoms, estimate the BEC transition temperature.

With an atomic weight of 4.0026, the mass in grams for one atom of Helium is

\begin{aligned}4.0026 \,\text{amu} \times \frac{\text{g}}{6.022 \times 10^{23} \text{amu}} &= 6.64 \times 10^{-24} \text{g} \\ &= 6.64 \times 10^{-27} \text{kg}.\end{aligned} \hspace{\stretch{1}}(1.15)

With the density of liquid He-4, at 5.2K (boiling point): 125 grams per liter, the number density is

\begin{aligned}\rho &= \frac{\text{mass}}{\text{volume}} \times \frac{1}{{\text{mass of one He atom}}} \\ &= \frac{125 \text{g}}{10^{-3} m^3} \times \frac{1}{{6.64 \times 10^{-24} g}} \\ &= \frac{125 \text{g}}{10^{-3} m^3} \times \frac{1}{{6.64 \times 10^{-24} g}} \\ &= 1.88 \times 10^{28} m^{-3}\end{aligned} \hspace{\stretch{1}}(1.16)

In class the $T_{\mathrm{BEC}}$ was found to be

\begin{aligned}T_{\mathrm{BEC}} &= \frac{1}{k_{\mathrm{B}}} \left( \frac{\rho}{\zeta(3/2)} \right)^{2/3} \frac{ 2 \pi \hbar^2}{M} \\ &= \frac{1}{{1.3806488 \times 10^{-23} m^2 kg/s^2/K}} \left( \frac{\rho}{ 2.61238 } \right)^{2/3} \frac{ 2 \pi (1.05457173 \times 10^{-34} m^2 kg / s)^2}{M} \\ &= 2.66824 \times 10^{-45} \frac{\rho^{2/3}}{M} K.\end{aligned} \hspace{\stretch{1}}(1.17)

So for liquid helium we have

\begin{aligned}T_{\mathrm{BEC}} &= 2.66824 \times 10^{-45} \left( 1.88 \times 10^{28} \right)^{2/3} \frac{1}{{ 6.64 \times 10^{-27} }} K \\ &= 2.84 K.\end{aligned} \hspace{\stretch{1}}(1.18)

The number density for the gas in Thywissen’s lab is

\begin{aligned}\rho &= \frac{10^6}{(10^{-6} \text{m})^3} \\ &= 10^{24} m^{-3}.\end{aligned} \hspace{\stretch{1}}(1.1.19)

The mass of an atom of ${}^{87}$ Rb is

\begin{aligned}86.90 \,\text{amu} \times \frac{10^{-3} \text{kg}}{6.022 \times 10^{23} \text{amu}} = 1.443 \times 10^{-25} \text{kg},\end{aligned} \hspace{\stretch{1}}(1.1.19)

which gives us

\begin{aligned}T_{\mathrm{BEC}} &= 2.66824 \times 10^{-45} \left( 10^{24} \right)^{2/3} \frac{1}{{ 1.443 \times 10^{-25} }} K \\ &= 1.85 \times 10^{-4} K.\end{aligned} \hspace{\stretch{1}}(1.1.19)

Question: Phonons in two dimensions

Consider phonons (quanta of lattice vibrations) which obey a dispersion relation

\begin{aligned}E_\mathbf{k} = \hbar v \left\lvert {\mathbf{k}} \right\rvert\end{aligned} \hspace{\stretch{1}}(1.1.22)

for small momenta $\left\lvert {\mathbf{k}} \right\rvert$, where $v$ is the speed of sound. Assuming a two-dimensional crystal, phonons only propagate along the plane containing the atoms. Find the specific heat of this crystal due to phonons at low temperature. Recall that phonons are not conserved, so there is no chemical potential associated with maintaining a fixed phonon density.

The energy density of the system is

\begin{aligned}\frac{E}{V} &= \int \frac{d^2 \mathbf{k}}{(2 \pi)^2} \frac{\epsilon}{ e^{\beta \epsilon} - 1 } \\ &= \int d\epsilon \frac{N(\epsilon)}{V} \frac{\epsilon}{ e^{\beta \epsilon} - 1 }.\end{aligned} \hspace{\stretch{1}}(1.23)

For the density of states we have

\begin{aligned}\frac{N(\epsilon) }{V} &= \int \frac{d^2 \mathbf{k}}{(2 \pi)^2} \delta( \epsilon - \epsilon_\mathbf{k} ) \\ &= \frac{1}{{(2 \pi)^2}} 2 \pi \int_0^\infty k dk \delta( \epsilon - \hbar v k ) \\ &= \frac{1}{{2 \pi}} \int_0^\infty k dk \delta \left( k - \frac{\epsilon}{\hbar v} \right) \frac{1}{{\hbar v}} \\ &= \frac{1}{{2 \pi}} \frac{\epsilon}{(\hbar v)^2}.\end{aligned} \hspace{\stretch{1}}(1.24)

Plugging back into the energy density we have

\begin{aligned}\frac{E}{V} &= \frac{2 \pi}{(\hbar v)^2}\int_0^\infty d\epsilon \frac{\epsilon^2}{ e^{\beta \epsilon} - 1 } \\ &= \frac{\pi \left( k_{\mathrm{B}} T \right)^3 }{(\hbar v)^2}\zeta(3),\end{aligned} \hspace{\stretch{1}}(1.25)

where $\zeta(3) \approx 2.40411$. Taking derivatives we have

\begin{aligned}\boxed{C_V = \frac{dE}{dT} = V\frac{3 \pi k_{\mathrm{B}}^3 T^2 }{(\hbar v)^2}\zeta(3).}\end{aligned} \hspace{\stretch{1}}(1.1.26)

Unresolved question about energy distribution around mean energy

Posted by peeterjoot on April 10, 2013

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In [1] is an expansion of

\begin{aligned}P(E) \propto e^{-\beta E} g(E),\end{aligned} \hspace{\stretch{1}}(1.0.1)

around the mean energy $E^{*} = U$. The first derivative part of the expansion is simple enough

\begin{aligned}\frac{\partial }{\partial E} \left( e^{-\beta E} g(E) \right) &= \left( -\beta g(E) + g'(E) \right)e^{-\beta E} \\ &= g(E) e^{-\beta E}\left( -\beta + (\ln g(E))' \right)\end{aligned} \hspace{\stretch{1}}(1.0.1)

The peak energy $E^{*}$ will be where this derivative equals zero. That is

\begin{aligned}0 = g(E^{*}) e^{-\beta E^{*}}\left( -\beta + {\left.{{(\ln g(E))'}}\right\vert}_{{E = E^{*}}} \right),\end{aligned} \hspace{\stretch{1}}(1.0.3)

or

\begin{aligned}{\left.{{\frac{\partial }{\partial E}\left( \ln g(E) \right)}}\right\vert}_{{E = E^{*}}} = \beta\end{aligned} \hspace{\stretch{1}}(1.0.4)

With

\begin{aligned}S = k_{\mathrm{B}} \ln g\end{aligned} \hspace{\stretch{1}}(1.0.5a)

\begin{aligned}\frac{1}{{k_{\mathrm{B}}}} \left( \frac{\partial S}{\partial E} \right)_{E = U} &= \frac{1}{{k_{\mathrm{B}} T}} \\ &= \beta\end{aligned} \hspace{\stretch{1}}(1.0.5b)

We have

\begin{aligned}\left( \frac{\partial \ln g(E) }{\partial E} \right)_{E = U} = \beta\end{aligned} \hspace{\stretch{1}}(1.0.6)

so that

\begin{aligned}E^{*} = U.\end{aligned} \hspace{\stretch{1}}(1.0.7)

So far so good. Reading the text, the expansion of the logarithm of $P(E)$ around $E = E^{*} = U$ wasn’t clear. Let’s write that out in full. To two terms that is

\begin{aligned}\ln e^{-\beta E} g(E)= \underbrace{\ln e^{-\beta U} g(U)}_{- \beta U + \frac{1}{{k_{\mathrm{B}}}} S}+ {\left.{{\frac{\partial }{\partial E} \left( \ln e^{-\beta E} g(E) \right)}}\right\vert}_{{E = U}}+ \frac{1}{2}{\left.{{\frac{\partial^2 }{\partial E^2} \left( \ln e^{-\beta E} g(E) \right)}}\right\vert}_{{E = U}}(E - U)^2.\end{aligned} \hspace{\stretch{1}}(1.0.7)

The first order term has the derivative of the logarithm of $e^{-\beta E}g(E)$. Since the logarithm is monotonic and the derivative of $e^{-\beta E}g(E)$ has been shown to be zero at $E = U$, this must be zero. We can also see this explicitly by computation

\begin{aligned}{\left.{{\frac{\partial }{\partial E} \ln e^{-\beta E} g(E)}}\right\vert}_{{E = U}} &= {\left.{{\frac{-\beta e^{-\beta E} g(E) + e^{-\beta E} g'(E)}{e^{-\beta E} g(E)}}}\right\vert}_{{E = U}} \\ &= {\left.{{\frac{-\beta g + g'}{g}}}\right\vert}_{{E = U}} \\ &= -\beta+{\left.{{(\ln g)'}}\right\vert}_{{E = U}} \\ &= -\beta + \frac{1}{{k_{\mathrm{B}}}} {\left.{{\frac{\partial S}{\partial E}}}\right\vert}_{{E = U}} \\ &= -\beta + \frac{1}{{k_{\mathrm{B}} T}} \\ &= -\beta + \beta \\ &= 0.\end{aligned} \hspace{\stretch{1}}(1.0.7)

For the second derivative we have

\begin{aligned}{\left.{{\frac{\partial }{\partial E} \ln e^{-\beta E} g(E)}}\right\vert}_{{E = U}} &= {\left.{{\frac{\partial }{\partial E} \left( -\beta + (\ln g)' \right)}}\right\vert}_{{E = U}} \\ &= {\left.{{\frac{\partial }{\partial E} \frac{g'}{g}}}\right\vert}_{{E = U}} \\ &= {\left.{{\frac{g''}{g} - \frac{(g')^2}{g^2}}}\right\vert}_{{E = U}} \\ &= {\left.{{\frac{g''}{g}}}\right\vert}_{{E = U}} - ((\ln g)')^2 \\ &= {\left.{{\frac{g''}{g}}}\right\vert}_{{E = U}} - \beta^2.\end{aligned} \hspace{\stretch{1}}(1.0.7)

Somehow this is supposed to come out to $k_{\mathrm{B}} T^2 C_{\mathrm{V}}$? Backing up, we have

\begin{aligned}{\left.{{\frac{\partial }{\partial E} \ln e^{-\beta E} g(E)}}\right\vert}_{{E = U}} &= {\left.{{\frac{\partial^2 }{\partial E^2} \ln g}}\right\vert}_{{E = U}} \\ &= \frac{1}{{k_{\mathrm{B}}}}{\left.{{\frac{\partial^2 S}{\partial E^2}}}\right\vert}_{{E = U}}.\end{aligned} \hspace{\stretch{1}}(1.0.7)

I still don’t see how to get $C_{\mathrm{V}} = {\partial U}/{\partial T}$ out of this? $C_{\mathrm{V}}$ is a derivative with respect to temperature, but here we have derivatives with respect to energy (keeping $\beta = 1/k_{\mathrm{B}} T$ fixed)?

References

[1] RK Pathria. Statistical mechanics. Butterworth Heinemann, Oxford, UK, 1996.

Velocity volume element to momentum volume element

Posted by peeterjoot on April 9, 2013

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Motivation

One of the problems I attempted had integrals over velocity space with volume element $d^3\mathbf{u}$. Initially I thought that I’d need a change of variables to momentum space, and calculated the corresponding momentum space volume element. Here’s that calculation.

Guts

We are working with a Hamiltonian

\begin{aligned}\epsilon = \sqrt{ (p c)^2 + \epsilon_0^2 },\end{aligned} \hspace{\stretch{1}}(1.1)

where the rest energy is

\begin{aligned}\epsilon_0 = m c^2.\end{aligned} \hspace{\stretch{1}}(1.2)

Hamilton’s equations give us

\begin{aligned}u_\alpha = \frac{ p_\alpha/c^2 }{\epsilon},\end{aligned} \hspace{\stretch{1}}(1.3)

or

\begin{aligned}p_\alpha = \frac{ m u_\alpha }{\sqrt{1 - \mathbf{u}^2/c^2}}.\end{aligned} \hspace{\stretch{1}}(1.4)

This is enough to calculate the Jacobian for our volume element change of variables

\begin{aligned}du_x \wedge du_y \wedge du_z &= \frac{\partial(u_x, u_y, u_z)}{\partial(p_x, p_y, p_z)}dp_x \wedge dp_y \wedge dp_z \\ &= \frac{1}{{c^6 \left( { m^2 + (\mathbf{p}/c)^2 } \right)^{9/2}}}\begin{vmatrix}m^2 c^2 + p_y^2 + p_z^2 & - p_y p_x & - p_z p_x \\ -p_x p_y & m^2 c^2 + p_x^2 + p_z^2 & - p_z p_y \\ -p_x p_z & -p_y p_z & m^2 c^2 + p_x^2 + p_y^2\end{vmatrix}dp_x \wedge dp_y \wedge dp_z \\ &= m^2 \left( { m^2 + \mathbf{p}^2/c^2 } \right)^{-5/2}dp_x \wedge dp_y \wedge dp_z.\end{aligned} \hspace{\stretch{1}}(1.5)

That final simplification of the determinant was a little hairy, but yielded nicely to Mathematica.

Our final result for the velocity volume element in momentum space, in terms of the particle energy is

\begin{aligned}d^3 \mathbf{u} = \frac{c^6 \epsilon_0^2 } {\epsilon^5} d^3 \mathbf{p}.\end{aligned} \hspace{\stretch{1}}(1.6)