Peeter Joot's (OLD) Blog.

Math, physics, perl, and programming obscurity.

Posts Tagged ‘ideal gas’

Final version of my phy452.pdf notes posted

Posted by peeterjoot on September 5, 2013

I’d intended to rework the exam problems over the summer and make that the last update to my stat mech notes. However, I ended up studying world events and some other non-mainstream ideas intensively over the summer, and never got around to that final update.

Since I’m starting a new course (condensed matter) soon, I’ll end up having to focus on that, and have now posted a final version of my notes as is.

Since the last update the following additions were made

September 05, 2013 Large volume fermi gas density

May 30, 2013 Bernoulli polynomials and numbers and Euler-MacLauren summation

May 09, 2013 Bose gas specific heat above condensation temperature

May 09, 2013 A dumb expansion of the Fermi-Dirac grand partition function

April 30, 2013 Ultra relativistic spin zero condensation temperature

April 30, 2013 Summary of statistical mechanics relations and helpful formulas

April 24, 2013 Low temperature Fermi gas chemical potential

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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)

leads to

\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)

spin addition

\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)

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An updated compilation of notes, for ‘PHY452H1S Basic Statistical Mechanics’, Taught by Prof. Arun Paramekanti

Posted by peeterjoot on March 27, 2013

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

March 27, 2013 Fermi gas

March 26, 2013 Fermi gas thermodynamics

March 26, 2013 Fermi gas thermodynamics

March 23, 2013 Relativisitic generalization of statistical mechanics

March 21, 2013 Kittel Zipper problem

March 18, 2013 Pathria chapter 4 diatomic molecule problem

March 17, 2013 Gibbs sum for a two level system

March 16, 2013 open system variance of N

March 16, 2013 probability forms of entropy

March 14, 2013 Grand Canonical/Fermion-Bosons

March 13, 2013 Quantum anharmonic oscillator

March 12, 2013 Grand canonical ensemble

March 11, 2013 Heat capacity of perturbed harmonic oscillator

March 10, 2013 Langevin small approximation

March 10, 2013 Addition of two one half spins

March 10, 2013 Midterm II reflection

March 07, 2013 Thermodynamic identities

March 06, 2013 Temperature

March 05, 2013 Interacting spin

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

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PHY452H1S Basic Statistical Mechanics. Lecture 16: Fermi gas. Taught by Prof. Arun Paramekanti

Posted by peeterjoot on March 27, 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.

Fermi gas

Review

Continuing a discussion of [1] section 8.1 content.

We found

\begin{aligned}n_{\mathbf{k}} = \frac{1}{{e^{\beta(\epsilon_k - \mu)} + 1}}\end{aligned} \hspace{\stretch{1}}(1.2.1)

With no spin

\begin{aligned}\int n_\mathbf{k} \times \frac{d^3 k}{(2\pi)^3} = \rho\end{aligned} \hspace{\stretch{1}}(1.2.2)

Fig 1.1: Occupancy at low temperature limit

 

Fig 1.2: Volume integral over momentum up to Fermi energy limit

 

\begin{aligned}\epsilon_{\mathrm{F}} = \frac{\hbar^2 k_{\mathrm{F}}^2}{2m}\end{aligned} \hspace{\stretch{1}}(1.2.3)

gives

\begin{aligned}k_{\mathrm{F}} = (6 \pi^2 \rho)^{1/3}\end{aligned} \hspace{\stretch{1}}(1.2.4)

\begin{aligned}\sum_\mathbf{k} n_\mathbf{k} = N\end{aligned} \hspace{\stretch{1}}(1.2.5)

\begin{aligned}\mathbf{k} = \frac{2\pi}{L}(n_x, n_y, n_z)\end{aligned} \hspace{\stretch{1}}(1.2.6)

This is for periodic boundary conditions \footnote{I filled in details in the last lecture using a particle in a box, whereas this periodic condition was intended. We see that both achieve the same result}, where

\begin{aligned}\Psi(x + L) = \Psi(x)\end{aligned} \hspace{\stretch{1}}(1.2.7)

Moving on

\begin{aligned}\sum_{k_x} n(\mathbf{k}) = \sum_{p_x} \Delta p_x n(\mathbf{k})\end{aligned} \hspace{\stretch{1}}(1.2.8)

with

\begin{aligned}\Delta k_x = \frac{2 \pi}{L} \Delta p_x\end{aligned} \hspace{\stretch{1}}(1.2.9)

this gives

\begin{aligned}\sum_{k_x} n(\mathbf{k}) = \sum_{n_x} \frac{L}{2\pi} \Delta k_x \rightarrow \frac{L}{2\pi} \int d k_x\end{aligned} \hspace{\stretch{1}}(1.2.10)

Over all dimensions

\begin{aligned}\sum_{\mathbf{k}} n_\mathbf{k} = \left( \frac{L}{2\pi} \right)^3 \left( \int d^3 \mathbf{k} \right)n(\mathbf{k})=N\end{aligned} \hspace{\stretch{1}}(1.2.11)

so that

\begin{aligned}\rho = \int \frac{d^3 \mathbf{k}}{(2 \pi)^3}\end{aligned} \hspace{\stretch{1}}(1.2.12)

Again

\begin{aligned}k_{\mathrm{F}} = (6 \pi^2 \rho)^{1/3}\end{aligned} \hspace{\stretch{1}}(1.2.13)

Example: Spin considerations

{example:basicStatMechLecture16:1}{

\begin{aligned}\sum_{\mathbf{k}, m_s} = N\end{aligned} \hspace{\stretch{1}}(1.2.14)

\begin{aligned}\sum_{\mathbf{k}, m_s} \frac{1}{{e^{\beta(\epsilon_k - \mu)} + 1}} = (2 S + 1)\left( \int \frac{d^3 \mathbf{k}}{(2 \pi)^3} n(\mathbf{k}) \right)L^3\end{aligned} \hspace{\stretch{1}}(1.2.15)

This gives us

\begin{aligned}k_{\mathrm{F}} = \left( \frac{ 6 \pi^2 \rho }{2 S + 1} \right)^{1/3}\end{aligned} \hspace{\stretch{1}}(1.2.16)

and again

\begin{aligned}\epsilon_{\mathrm{F}} = \frac{\hbar^2 k_{\mathrm{F}}^2}{2m}\end{aligned} \hspace{\stretch{1}}(1.2.17)

}

High Temperatures

Now we want to look at the at higher temperature range, where the occupancy may look like fig. 1.3

Fig 1.3: Occupancy at higher temperatures

 

\begin{aligned}\mu(T = 0) = \epsilon_{\mathrm{F}}\end{aligned} \hspace{\stretch{1}}(1.2.18)

\begin{aligned}\mu(T \rightarrow \infty) \rightarrow - \infty\end{aligned} \hspace{\stretch{1}}(1.2.19)

so that for large T we have

\begin{aligned}\frac{1}{{e^{\beta(\epsilon_k - \mu)} + 1}} \rightarrow e^{-\beta(\epsilon_k - \mu)}\end{aligned} \hspace{\stretch{1}}(1.2.20)

\begin{aligned}\rho &= \int \frac{d^3 \mathbf{k}}{(2 \pi)^3} e^{\beta \mu} e^{-\beta \epsilon_k} \\ &= e^{\beta \mu} \int \frac{d^3 \mathbf{k}}{(2 \pi)^3} e^{-\beta \epsilon_k} \\ &= e^{\beta \mu} \int dk \frac{4 \pi k^2}{(2 \pi)^3} e^{-\beta \hbar^2 k^2/2m}.\end{aligned} \hspace{\stretch{1}}(1.2.21)

Mathematica (or integration by parts) tells us that

\begin{aligned}\frac{1}{{(2 \pi)^3}} \int 4 \pi^2 k^2 dk e^{-a k^2} = \frac{1}{{(4 \pi a )^{3/2}}},\end{aligned} \hspace{\stretch{1}}(1.2.22)

so we have

\begin{aligned}\rho &= e^{\beta \mu} \left( \frac{2m}{ 4 \pi \beta \hbar^2} \right)^{3/2} \\ &= e^{\beta \mu} \left( \frac{2 m k_{\mathrm{B}} T 4 \pi^2 }{ 4 \pi h^2} \right)^{3/2} \\ &= e^{\beta \mu} \left( \frac{2 m k_{\mathrm{B}} T \pi }{ h^2} \right)^{3/2}\end{aligned} \hspace{\stretch{1}}(1.2.23)

Introducing \lambda for the thermal de Broglie wavelength, \lambda^3 \sim T^{-3/2}

\begin{aligned}\lambda \equiv \frac{h}{\sqrt{2 \pi m k_{\mathrm{B}} T}},\end{aligned} \hspace{\stretch{1}}(1.2.24)

we have

\begin{aligned}\rho = e^{\beta \mu} \frac{1}{{\lambda^3}}.\end{aligned} \hspace{\stretch{1}}(1.2.25)

Does it make any sense to have density as a function of temperature? An inappropriately extended to low temperatures plot of the density is found in fig. 1.4 for a few arbitrarily chosen numerical values of the chemical potential \mu, where we see that it drops to zero with temperature. I suppose that makes sense if we are not holding volume constant.

Fig 1.4: Density as a function of temperature

 

We can write

\begin{aligned}\boxed{e^{\beta \mu} = \left( \rho \lambda^3 \right)}\end{aligned} \hspace{\stretch{1}}(1.2.26)

\begin{aligned}\frac{\mu}{k_{\mathrm{B}} T} = \ln \left( \rho \lambda^3 \right)\sim -\frac{3}{2} \ln T\end{aligned} \hspace{\stretch{1}}(1.2.27)

or (taking \rho (and/or volume?) as a constant) we have for large temperatures

\begin{aligned}\mu \propto -T \ln T\end{aligned} \hspace{\stretch{1}}(1.2.28)

The chemical potential is plotted in fig. 1.5, whereas this - k_{\mathrm{B}} T \ln k_{\mathrm{B}} T function is plotted in fig. 1.6. The contributions to \mu from the k_{\mathrm{B}} T \ln (\rho h^3 (2 \pi m)^{-3/2}) term are dropped for the high temperature approximation.

Fig 1.5: Chemical potential over degenerate to classical range

Fig 1.6: High temp approximation of chemical potential, extended back to T = 0

Pressure

\begin{aligned}P = - \frac{\partial {E}}{\partial {V}}\end{aligned} \hspace{\stretch{1}}(1.2.29)

For a classical ideal gas as in fig. 1.7 we have

Fig 1.7: Ideal gas pressure vs volume

 

\begin{aligned}P = \rho k_{\mathrm{B}} T\end{aligned} \hspace{\stretch{1}}(1.2.30)

For a Fermi gas at T = 0 we have

\begin{aligned}E &= \sum_\mathbf{k} \epsilon_k n_k \\ &= \sum_\mathbf{k} \epsilon_k \Theta(\mu_0 - \epsilon_k) \\ &= \frac{V}{(2\pi)^3} \int_{\epsilon_k < \mu_0} \frac{\hbar^2 \mathbf{k}^2}{2 m} d^3 \mathbf{k} \\ &= \frac{V}{(2\pi)^3} \int_0^{k_{\mathrm{F}}} \frac{\hbar^2 \mathbf{k}^2}{2 m} d^3 \mathbf{k} \\ &= \frac{V}{(2\pi)^3} \frac{\hbar^2}{2 m} \int_0^{k_{\mathrm{F}}} k^2 4 \pi k^2 d k\propto k_{\mathrm{F}}^5\end{aligned} \hspace{\stretch{1}}(1.2.31)

Specifically,

\begin{aligned}E(T = 0) = V \times \frac{3}{5} \underbrace{\epsilon_{\mathrm{F}}}_{\sim k_{\mathrm{F}}^2}\underbrace{\rho}_{\sim k_{\mathrm{F}}^3}\end{aligned} \hspace{\stretch{1}}(1.2.32)

or

\begin{aligned}\frac{E}{N} = \frac{3}{5} \epsilon_{\mathrm{F}}\end{aligned} \hspace{\stretch{1}}(1.2.33)

\begin{aligned}E = \frac{3}{5} N \frac{\hbar^2}{2 m} \left( 6 \pi^2 \frac{N}{V} \right)^{2/3} = a V^{-2/3},\end{aligned} \hspace{\stretch{1}}(1.2.34)

so that

\begin{aligned}\frac{\partial {E}}{\partial {V}} = -\frac{2}{3} a V^{-5/3}.\end{aligned} \hspace{\stretch{1}}(1.2.35)

\begin{aligned}P &= -\frac{\partial {E}}{\partial {V}}  \\ &= \frac{2}{3} \left( a V^{-2/3} \right)V^{-1} \\ &= \frac{2}{3} \frac{E}{V} \\ &= \frac{2}{3} \left( \frac{3}{5} \epsilon_{\mathrm{F}} \rho \right) \\ &= \frac{2}{5} \epsilon_{\mathrm{F}} \rho.\end{aligned} \hspace{\stretch{1}}(1.2.36)

We see that the pressure ends up deviating from the classical result at low temperatures, as sketched in fig. 1.8. This low temperature limit for the pressure 2 \epsilon_{\mathrm{F}} \rho/5 is called the degeneracy pressure.

Fig 1.8: Fermi degeneracy pressure

 

References

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

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PHY452H1S Basic Statistical Mechanics. Problem Set 4: Ideal gas

Posted by peeterjoot on March 3, 2013

[Click here for a PDF of this post with nicer formatting]

Disclaimer

This is an ungraded set of answers to the problems posed.

Question: Sackur-Tetrode entropy of an Ideal Gas

The entropy of an ideal gas is given by

\begin{aligned}S = N k_{\mathrm{B}}\left( \ln \left( \frac{V}{N} \left( \frac{4 \pi m E}{3 N h^2} \right) ^{3/2} \right) + \frac{5}{2} \right)\end{aligned} \hspace{\stretch{1}}(1.1.1)

Find the temperature of this gas via (\partial S/ \partial E)_{V,N} = 1/T. Find the energy per particle at which the entropy becomes negative. Is there any meaning to this temperature?

Answer

Taking derivatives we find

\begin{aligned}\frac{1}{{T}} &= \frac{\partial {}}{\partial {E}}\left( \not{{ N k_{\mathrm{B}} \ln \frac{V}{N} }} + N k_{\mathrm{B}} \frac{3}{2} \ln \left( \frac{4 \pi m E}{3 N h^2} \right) + \not{{N k_{\mathrm{B}} \frac{5}{2} }} \right) \\ &= \frac{3}{2} N k_{\mathrm{B}} \frac{1}{{E}}\end{aligned} \hspace{\stretch{1}}(1.1.2)

or

\begin{aligned}\boxed{T = \frac{2}{3} \frac{E}{N k_{\mathrm{B}} }}\end{aligned} \hspace{\stretch{1}}(1.1.3)

The energies for which the entropy is negative are given by

\begin{aligned}\left( \frac{4 \pi m E}{3 N h^2} \right)^{3/2}\le \frac{N}{V} e^{-5/2},\end{aligned} \hspace{\stretch{1}}(1.1.4)

or

\begin{aligned}E &\le \frac{3 N h^2}{4 \pi m} \left( \frac{N}{V e^{5/2}} \right)^{2/3} \\ &= \frac{3 h^2 N^{5/3}}{4 \pi m V^{2/3} e^{5/2}}.\end{aligned} \hspace{\stretch{1}}(1.1.5)

In terms of the temperature T this negative entropy condition is given by

\begin{aligned}\not{{\frac{3 N}{2}}} k_{\mathrm{B}} T \le \not{{\frac{3 N}{2}}} \left( \frac{ N}{V} \right)^{2/3} \frac{h^2}{e^{5/2}},\end{aligned} \hspace{\stretch{1}}(1.1.6)

or

\begin{aligned}\boxed{\frac{\sqrt{2 \pi m k_{\mathrm{B}} T}}{h} \le \left( \frac{N}{V} \right)^{1/3} \frac{1}{{e^{5/4}}}.}\end{aligned} \hspace{\stretch{1}}(1.1.7)

There will be a particle density V/N for which this distance h/\sqrt{2 \pi m k_{\mathrm{B}} T} will start approaching the distance between atoms. This distance constrains the validity of the ideal gas law entropy equation. Putting this quantity back into the entropy eq. 1.1.1 we have

\begin{aligned}\frac{S}{N k_{\mathrm{B}}} = \ln \frac{V}{N} \left( \frac{\sqrt{2 \pi m k_{\mathrm{B}} T}}{h} \right)^3 + \frac{5}{2}\end{aligned} \hspace{\stretch{1}}(1.1.8)

We see that a positive entropy requirement puts a bound on this distance (as a function of temperature) since we must also have

\begin{aligned}\frac{h}{\sqrt{2 \pi m k_{\mathrm{B}} T}} \ll \left( \frac{V}{N} \right)^{1/3},\end{aligned} \hspace{\stretch{1}}(1.1.9)

for the gas to be in the classical domain. I’d actually expect a gas to liquefy before this transition point, making such a low temperature nonphysical. To get a feel for whether this is likely the case, we should expect that the logarithm argument to be

\begin{aligned}\frac{V}{N} \left( \frac{\sqrt{2 \pi m k_{\mathrm{B}} T}}{h} \right)^3\end{aligned} \hspace{\stretch{1}}(1.1.10)

at the point where gasses liquefy (at which point we assume the ideal gas law is no longer accurate) to be well above unity. Checking this for 1 liter of a gas with 10^23 atoms for hydrogen, helium, and neon respectively we find the values for eq. 1.1.10 are

\begin{aligned}173.682, 130.462, 23993.\end{aligned} \hspace{\stretch{1}}(1.1.11)

At least for these first few cases we see that the ideal gas law has lost its meaning well before the temperatures below which the entropy would become negative.

Question: Ideal gas thermodynamics

An ideal gas starts at (V_0, P_0) in the pressure-volume diagram (x-axis = V, y-axis = P), then moves at constant pressure to a larger volume (V_1, P_0), then moves to a larger pressure at constant volume to (V_1, P_1), and finally returns to (V_0, P_0), thus undergoing a cyclic process (forming a triangle in P-V plane). For each step, find the work done on the gas, the change in energy content, and heat added to the gas. Find the total work/energy/heat change over the entire cycle.

Answer

Our process is illustrated in fig. 1.1.

Fig 1.1: Cyclic pressure volume process

Step 1
This problem is somewhat underspecified. From the ideal gas law, regardless of how the gas got from the initial to the final states, we have

\begin{aligned}P_0 V_0 = N_0 k_{\mathrm{B}} T_0\end{aligned} \hspace{\stretch{1}}(1.0.12a)

\begin{aligned}P_0 V_1 = N_1 k_{\mathrm{B}} T_1\end{aligned} \hspace{\stretch{1}}(1.0.12b)

So a volume increase with fixed P implies that there is a corresponding increase in N T. We could have for example, an increase in the number of particles, as in the evaporation process illustrated of fig. 1.2, where a piston held down by (fixed) atmospheric pressure is pushed up as the additional gas boils off.

Fig 1.2: Evaporation process under (fixed) atmospheric pressure

Alternately, we could have a system such as that of fig. 1.3, with a fixed amount of gas is in contact with a heat source that supplies the energy required to induce the required increase in temperature.

Fig 1.3: Gas of fixed mass absorbing heat

Regardless of the source of the energy that accounts for the increase in volume the work done on the gas (a negation of the positive work the gas is performing on the system, perhaps a piston as in the picture) is

\begin{aligned}d W_1 = - \int_{V_0}^{V_1} p dV = -P_0 (V_1 - V_0).\end{aligned} \hspace{\stretch{1}}(1.0.13)

Let’s now assume that we have the second sort of configuration above, where the total amount of gas is held fixed. From the ideal gas relations of eq. 1.0.12.12, and with \Delta V = V_1 - V_0, \Delta T = T_1 - T_0, and N_1 = N_0 = N, we have

\begin{aligned}P_0 \Delta V = N k_{\mathrm{B}} \Delta T.\end{aligned} \hspace{\stretch{1}}(1.0.14)

The change in energy of the ideal gas, assuming three degrees of freedom, is

\begin{aligned}d U = \frac{3}{2} N k_{\mathrm{B}} \Delta T = \frac{3}{2} P_0 \Delta V.\end{aligned} \hspace{\stretch{1}}(1.0.15)

The energy balance then requires that the total heat absorbed by the gas must include that portion that has done work on the system, plus the excess kinetic energy of the gas. That is

\begin{aligned}d Q_1 &= \frac{3}{2} P_0 \Delta V + P_0 \Delta V \\ &= \frac{5}{2} P_0 \Delta V.\end{aligned} \hspace{\stretch{1}}(1.0.16)

Step 2

For this leg of the cycle we have no work done on the gas

\begin{aligned}d W_2 = -\int_{V_1}^{V_1} P dV = 0.\end{aligned} \hspace{\stretch{1}}(1.0.17)

We do, however have a change in energy. The energy of the gas is

\begin{aligned}U = \frac{3}{2} N k_b T = \frac{3}{2} P V.\end{aligned} \hspace{\stretch{1}}(1.0.18)

With \Delta P = P_1 - P_0, the change of energy of the gas, the total heat absorbed by the gas, is

\begin{aligned}dU_2 = d Q_2 = \frac{3}{2} V_1 \Delta P.\end{aligned} \hspace{\stretch{1}}(1.0.19)

Step 3

For the final leg of the cycle, the work done on the gas is

\begin{aligned}d W_3 &= -\int_{V_1}^{V_0} P dV \\ &= \int_{V_0}^{V_1} P dV \\ &= \Delta V \frac{P_0 + P_1}{2}.\end{aligned} \hspace{\stretch{1}}(1.0.20)

This is positive this time
Unlike the first part of the cycle, the work done on the gas is positive this time (work is being done on the gas to both compress it). The change in energy of the gas, however, is negative, with the difference between final and initial energy being

\begin{aligned}dU_3 &= \frac{3}{2} (P_0 V_0 - P_1 V_1)  \\ &= -\frac{3}{2} (P_1 V_1 - P_0 V_0) <0.\end{aligned} \hspace{\stretch{1}}(1.0.21)

The simultaneous compression and the pressure reduction require energy to be removed from the gas. We must have a negative change in heat d Q < 0, with heat emitted in this phase of the cycle. This can be verified explicitly

\begin{aligned}d Q_3 &= dU - d W \\ &= -\frac{3}{2} (P_1 V_1 - P_0 V_0) - \frac{1}{{2}} \Delta V (P_1 + P_0)< 0.\end{aligned} \hspace{\stretch{1}}(1.0.22)

Changes over the complete cycle.

Summarizing the results from each of the phases, we have

\begin{aligned}d W_1 = -P_0 \Delta V\end{aligned} \hspace{\stretch{1}}(1.0.23a)

\begin{aligned}d Q_1 = \frac{5}{2} P_0 \Delta V \end{aligned} \hspace{\stretch{1}}(1.0.23b)

\begin{aligned}d U_1 = \frac{3}{2} P_0 \Delta V \end{aligned} \hspace{\stretch{1}}(1.0.23c)

\begin{aligned}d W_2 = 0 \end{aligned} \hspace{\stretch{1}}(1.0.24a)

\begin{aligned}d Q_2 = \frac{3}{2} V_1 \Delta P \end{aligned} \hspace{\stretch{1}}(1.0.24b)

\begin{aligned}d U_2 = \frac{3}{2} V_1 \Delta P \end{aligned} \hspace{\stretch{1}}(1.0.24c)

\begin{aligned}d W_3 = \Delta V \frac{P_0 + P_1}{2} \end{aligned} \hspace{\stretch{1}}(1.0.25a)

\begin{aligned}d Q_3 = -\frac{1}{{2}} ( 3(P_1 V_1 - P_0 V_0) + \Delta V (P_1 + P_0)) \end{aligned} \hspace{\stretch{1}}(1.0.25b)

\begin{aligned}d U_3 = -\frac{3}{2} ( P_1 V_1 - P_0 V_0 )\end{aligned} \hspace{\stretch{1}}(1.0.25c)

Summing the changes in the work we have

\begin{aligned}\sum_{i = 1}^3 d W_i = \frac{1}{{2}} \Delta V \Delta P > 0.\end{aligned} \hspace{\stretch{1}}(1.0.26)

This is the area of the triangle, as expected. Since it is positive, there is net work done on the gas.

We expect the energy changes to sum to zero, and this can be verified explicitly finding

\begin{aligned}\sum_{i = 1}^3 d U_i &= \frac{3}{2} P_0 \Delta V -\frac{3}{2} ( P_1 V_1 - P_0 V_0 ) \\ &= 0.\end{aligned} \hspace{\stretch{1}}(1.0.27)

With net work done on the gas and no change in energy, there should be no net heat absorption by the gas, with a total change in heat that should equal, in amplitude, the total work done on the gas. This is confirmed by summation

\begin{aligned}\sum_{i = 1}^3 d Q_i &= \frac{5}{2} P_0 \Delta V +\frac{3}{2} V_1 \Delta P -\frac{1}{{2}} ( 3(P_1 V_1 - P_0 V_0) + \Delta V (P_1 + P_0)) \\ &= -\frac{1}{{2}} \Delta P \Delta V.\end{aligned} \hspace{\stretch{1}}(1.0.28)

Question: Adiabatic process for an Ideal Gas

Show that when an ideal monoatomic gas expands adiabatically, the temperature and pressure are related by

\begin{aligned}\frac{dT}{dP}=\frac{2}{5}\frac{T}{P}\end{aligned} \hspace{\stretch{1}}(1.0.29)

Answer

From (3.34b) of [1], we find that the Adiabatic condition can be expressed algebraically as

\begin{aligned}0 = d Q = T dS = dU + P dV.\end{aligned} \hspace{\stretch{1}}(1.0.30)

With

\begin{aligned}U = \frac{3}{2} N k_{\mathrm{B}} T = \frac{3}{2} P V,\end{aligned} \hspace{\stretch{1}}(1.0.31)

this is

\begin{aligned}0 &= \frac{3}{2} V dP + \frac{3}{2} P dV + P dV \\ &= \frac{3}{2} V dP + \frac{5}{2} P dV.\end{aligned} \hspace{\stretch{1}}(1.0.32)

Dividing through by P V, this becomes a perfect differential, and we can integrate

\begin{aligned}0 &= 3 \int \frac{dP }{P}+ 5 \int \frac{dV}{V} \\ &= 3 \ln P + 5 \ln V + \ln C \\ &= 3 \ln PV + 2 \ln V + \ln C \\ &= \ln (N k_{\mathrm{B}} T)^3 + \ln \left( \frac{N k_{\mathrm{B}} T}{P} \right)^2 + \ln C.\end{aligned} \hspace{\stretch{1}}(1.0.33)

Exponentiating yields

\begin{aligned}T^5 = C' P^2.\end{aligned} \hspace{\stretch{1}}(1.0.34)

The desired relation follows by taking derivatives

\begin{aligned}2 C' P &= 5 T^4 \frac{dT}{dP} \\ &= 5 C' \frac{P^2}{T} \frac{dT}{dP},\end{aligned} \hspace{\stretch{1}}(1.0.35)

or

\begin{aligned}\frac{dT}{dP} =\frac{2}{5} \frac{T}{P},\end{aligned} \hspace{\stretch{1}}(1.0.36)

as desired.

References

[1] C. Kittel and H. Kroemer. Thermal physics. WH Freeman, 1980.

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An updated compilation of notes, for ‘PHY452H1S Basic Statistical Mechanics’, Taught by Prof. Arun Paramekanti

Posted by peeterjoot on March 3, 2013

In A compilation of notes, so far, for ‘PHY452H1S Basic Statistical Mechanics’ I posted a link this compilation of statistical mechanics course notes.

That compilation now all of the following too (no further updates will be made to any of these) :

February 28, 2013 Rotation of diatomic molecules

February 28, 2013 Helmholtz free energy

February 26, 2013 Statistical and thermodynamic connection

February 24, 2013 Ideal gas

February 16, 2013 One dimensional well problem from Pathria chapter II

February 15, 2013 1D pendulum problem in phase space

February 14, 2013 Continuing review of thermodynamics

February 13, 2013 Lightning review of thermodynamics

February 11, 2013 Cartesian to spherical change of variables in 3d phase space

February 10, 2013 n SHO particle phase space volume

February 10, 2013 Change of variables in 2d phase space

February 10, 2013 Some problems from Kittel chapter 3

February 07, 2013 Midterm review, thermodynamics

February 06, 2013 Limit of unfair coin distribution, the hard way

February 05, 2013 Ideal gas and SHO phase space volume calculations

February 03, 2013 One dimensional random walk

February 02, 2013 1D SHO phase space

February 02, 2013 Application of the central limit theorem to a product of random vars

January 31, 2013 Liouville’s theorem questions on density and current

January 30, 2013 State counting

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Rotation of diatomic molecules

Posted by peeterjoot on February 28, 2013

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Question: Rotation of diatomic molecules ([2] problem 3.6)

In our first look at the ideal gas we considered only the translational energy of the particles. But molecules can rotate, with kinetic energy. The rotation motion is quantized; and the energy levels of a diatomic molecule are of the form

\begin{aligned}\epsilon(j) = j(j + 1) \epsilon_0\end{aligned} \hspace{\stretch{1}}(1.0.1)

where j is any positive integer including zero: j = 0, 1, 2, \cdots. The multiplicity of each rotation level is g(j) = 2 j + 1.

a

Find the partition function Z_R(\tau) for the rotational states of one molecule. Remember that Z is a sum over all states, not over all levels — this makes a difference.

b

Evaluate Z_R(\tau) approximately for \tau \gg \epsilon_0, by converting the sum to an integral.

c

Do the same for \tau \ll \epsilon_0, by truncating the sum after the second term.

d

Give expressions for the energy U and the heat capacity C, as functions of \tau, in both limits. Observe that the rotational contribution to the heat capacity of a diatomic molecule approaches 1 (or, in conventional units, k_{\mathrm{B}}) when \tau \gg \epsilon_0.

e

Sketch the behavior of U(\tau) and C(\tau), showing the limiting behaviors for \tau \rightarrow \infty and \tau \rightarrow 0.

Answer

a. Partition function Z_R(\tau)

To understand the reference to multiplicity recall (section 4.13 [1]) that the rotational Hamiltonian was of the form

\begin{aligned}H = \frac{\mathbf{L}^2}{2 M r^2},\end{aligned} \hspace{\stretch{1}}(1.0.2)

where the \mathbf{L}^2 eigenvectors satisfied

\begin{subequations}

\begin{aligned}\mathbf{L}^2 {\left\lvert {l m} \right\rangle} = l (l + 1) \hbar^2 {\left\lvert {l m} \right\rangle}\end{aligned} \hspace{\stretch{1}}(1.0.3a)

\begin{aligned}L_z {\left\lvert {l m} \right\rangle} = m \hbar {\left\lvert {l m} \right\rangle}\end{aligned} \hspace{\stretch{1}}(1.0.3b)

\end{subequations}

and -l \le m \le l, where l \ge 0 is a positive integer. We see that \epsilon_0 is of the form

\begin{aligned}\epsilon_0 = \frac{\hbar^2}{2 M R_l(r)},\end{aligned} \hspace{\stretch{1}}(1.0.4)

and our partition function is

\begin{aligned}Z_R(\tau) = \sum_{l = 0}^\infty \sum_{m = -l}^l e^{-l (l + 1)\epsilon_0/\tau}= \sum_{l = 0}^\infty (2 l + 1) e^{-l (l + 1)\epsilon_0/\tau}.\end{aligned} \hspace{\stretch{1}}(1.0.5)

We have no dependence on m in the sum, and just have to sum terms like fig 1, and are able to sum over m trivially, which is where the multiplicity comes from.

Fig 1: Summation over m

 

To get a feel for how many terms are significant in these sums, we refer to the plot of fig 2. We plot the partition function itself in, truncation at l = 30 terms in fig 3.

Fig 2: Plotting the partition function summand

 

Fig 3: Z_R(tau) truncated after 30 terms in log plot

 

b. Evaluate partition function for large temperatures

If \tau \gg \epsilon_0, so that \epsilon_0/\tau \ll 1, all our exponentials are close to unity. Employing an integral approximation of the partition function, we can somewhat miraculously integrate this directly

\begin{aligned}Z_R(\tau) &\approx \int_0^\infty dl (2 l + 1) e^{-l(l+1)\epsilon_0/\tau} \\ &= \int_0^\infty dl \frac{d}{dl} \left( -\frac{\tau}{\epsilon_0} e^{-l(l+1)\epsilon_0/\tau} \right) \\ &= \frac{\tau}{\epsilon_0}\end{aligned} \hspace{\stretch{1}}(1.0.6)

c. Evaluate partition function for small temperatures

When \tau \ll \epsilon_0, so that \epsilon_0/\tau \gg 1, all our exponentials are increasingly close to zero as l increases. Dropping all the second and higher order terms we have

\begin{aligned}Z_R(\tau) \approx 1 + 3 e^{-2 \epsilon_0/\tau}\end{aligned} \hspace{\stretch{1}}(1.0.7)

d. Energy and heat capacity

In the large \epsilon_0/\tau domain (small temperatures) we have

\begin{aligned}U &= \tau^2 \frac{\partial {}}{\partial {\tau}} \ln Z \\ &= \tau^2 \frac{\partial {}}{\partial {\tau}} \ln \left( 1 + 3 e^{-2 \epsilon_0/\tau} \right) \\ &= \tau^2 \frac{3 (-2\epsilon_0)(-1/\tau^2)}{1 + 3 e^{-2 \epsilon_0/\tau}} \\ &= \frac{6 \epsilon_0}{1 + 3 e^{-2 \epsilon_0/\tau}} \\ &\approx 6 \epsilon_0.\end{aligned} \hspace{\stretch{1}}(1.0.8)

The specific heat in this domain is

\begin{aligned}C_{\mathrm{V}} = \frac{\partial {U}}{\partial {\tau}}=\left( \frac{6 \epsilon_0/\tau}{1 + 3 e^{-2 \epsilon_0/\tau}} \right)^2\approx \left( \frac{6 \epsilon_0}{\tau} \right)^2\end{aligned} \hspace{\stretch{1}}(1.0.9)

For the small \epsilon_0/\tau (large temperatures) case we have

\begin{aligned}U = \tau^2 \frac{\partial {}}{\partial {\tau}} \ln Z= \tau^2 \frac{\partial {}}{\partial {\tau}} \ln \frac{\tau}{\epsilon_0}= \tau^2 \frac{1}{{\tau}}= \tau\end{aligned} \hspace{\stretch{1}}(1.0.10)

The heat capacity in this large temperature region is

\begin{aligned}C_{\mathrm{V}} = \frac{\partial {U}}{\partial {\tau}} = 1,\end{aligned} \hspace{\stretch{1}}(1.0.11)

which is unity as described in the problem.

e. Sketch

The energy and heat capacities are roughly sketched in fig 4.

Fig 4: Energy and heat capacity

It’s somewhat odd seeming that we have a zero point energy at zero temperature. Plotting the energy (truncating the sums to 30 terms) in fig 5, we don’t see such a zero point energy.

Fig 5: Exact plot of the energy for a range of temperatures (30 terms of the sums retained)

That plotted energy is as follows, computed without first dropping any terms of the partition function

\begin{aligned}U &= \tau^2 \frac{\partial}{\partial \tau} \ln\left( \sum_{l = 0}^\infty (2 l + 1) e^{-l (l + 1)\epsilon_0/\tau} \right) \\ &= \epsilon_0\frac{\left( \sum_{l = 1}^\infty l (l + 1)(2 l + 1) e^{-l (l + 1)\epsilon_0/\tau} \right)}{\left( \sum_{l = 0}^\infty (2 l + 1) e^{-l (l + 1)\epsilon_0/\tau} \right)} \\ &= \epsilon_0\frac{\left( \sum_{l = 1}^\infty l (l + 1)(2 l + 1) e^{-l (l + 1)\epsilon_0/\tau} \right)}{Z}\end{aligned} \hspace{\stretch{1}}(1.0.12)

To avoid the zero point energy, we have to use this and not the truncated partition function to do the integral approximation. Doing that calculation (which isn’t as convenient, so I cheated and used Mathematica). We obtain

\begin{aligned}U \approx \frac{\int_1^\infty l (l + 1)(2 l + 1) e^{-l (l + 1)\epsilon_0/\tau}}{\int_0^\infty (2 l + 1) e^{-l (l + 1)\epsilon_0/\tau}}=\epsilon_0 e^{2 \epsilon_0/\tau} \left( 2 + \frac{\tau}{\epsilon_0} \right).\end{aligned} \hspace{\stretch{1}}(1.0.13)

This approximation, which has taken the sums to infinity, is plotted in fig 6.

Fig 6: Low temperature approximation of the energy

From eq. 1.0.12, we can take one more derivative to calculate the exact specific heat

\begin{aligned}C_{\mathrm{V}} &= \epsilon_0\frac{\partial {}}{\partial {\tau}}\left(\frac{\left( \sum_{l = 1}^\infty l (l + 1)(2 l + 1) e^{-l (l + 1)\epsilon_0/\tau} \right)}{\left( \sum_{l = 0}^\infty (2 l + 1) e^{-l (l + 1)\epsilon_0/\tau} \right)}\right) \\ &= \left( \frac{\epsilon_0}{\tau} \right)^2\left(\frac{\left( \sum_{l = 1}^\infty l^2 (l + 1)^2 (2 l + 1) e^{-l (l + 1)\epsilon_0/\tau} \right)}{\left( \sum_{l = 0}^\infty (2 l + 1) e^{-l (l + 1)\epsilon_0/\tau} \right)}+\frac{\left( \sum_{l = 1}^\infty l (l + 1)(2 l + 1) e^{-l (l + 1)\epsilon_0/\tau} \right)^2}{\left( \sum_{l = 0}^\infty (2 l + 1) e^{-l (l + 1)\epsilon_0/\tau} \right)^2}\right) \\ &= \left( \frac{\epsilon_0}{\tau} \right)^2\left(\frac{\left( \sum_{l = 1}^\infty l^2 (l + 1)^2 (2 l + 1) e^{-l (l + 1)\epsilon_0/\tau} \right)}{Z}+ \frac{U^2}{\epsilon_0^2}\right) \\ &= \frac{U^2}{\epsilon_0^2}+\left( \frac{\epsilon_0}{\tau} \right)^2\frac{\left( \sum_{l = 1}^\infty l^2 (l + 1)^2 (2 l + 1) e^{-l (l + 1)\epsilon_0/\tau} \right)}{Z}.\end{aligned} \hspace{\stretch{1}}(1.0.14)

This is plotted to 30 terms in fig 7.

Fig 7: Specific heat to 30 terms

References

[1] BR Desai. Quantum mechanics with basic field theory. Cambridge University Press, 2009.

[2] C. Kittel and H. Kroemer. Thermal physics. WH Freeman, 1980.

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Some problems from Kittel chapter 3

Posted by peeterjoot on February 10, 2013

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Question: Classical gas partition function

[1] expresses the classical gas partition function (3.77) as

\begin{aligned}Z_1 \propto \int \exp\left( - \frac{p_x^2 + p_y^2 + p_z^2 }{2 M \tau}\right) dp_x dp_y dp_z\end{aligned} \hspace{\stretch{1}}(1.0.1)

Show that this leads to the expected 3 \tau/2 result for the thermal average energy.

Answer

Let’s use the adjustment technique from the text for the N partition case and write

\begin{aligned}Z_N = \frac{1}{{N!}} Z_1^N,\end{aligned} \hspace{\stretch{1}}(1.0.2)

with Z_1 as above. This gives us

\begin{aligned}U &= \tau^2 \frac{\partial {}}{\partial {\tau}} \ln Z_N \\ &= \tau^2 \frac{\partial {}}{\partial {\tau}} \left(N \ln Z_1 - \ln N!\right) \\ &= N \tau^2 \frac{\partial {\ln Z_1 }}{\partial {\tau}} \\ &= N \tau^2 \frac{\partial {}}{\partial {\tau}} \sum_{k = 1}^{3} \ln\int \exp\left( - \frac{p_k^2 }{2 M \tau}\right) dp_k \\ &= N \tau^2\sum_{k = 1}^{3}\frac{\frac{\partial {}}{\partial {\tau}} \int \exp\left( - \frac{p_k^2 }{2 M \tau} \right) dp_k}{\int \exp\left( - \frac{p_k^2 }{2 M \tau} \right) dp_k} \\ &= N \tau^2\sum_{k = 1}^{3}\frac{\frac{\partial {}}{\partial {\tau}} \sqrt{ 2 \pi M \tau }}{\sqrt{ 2 \pi M \tau}} \\ &= 3 N \tau^2\frac{\frac{1}{{2}} \tau^{-1/2}}{\sqrt{ \tau}} \\ &= \frac{3}{2} N \tau \\ &= \frac{3}{2} N k_{\mathrm{B}} T\end{aligned} \hspace{\stretch{1}}(1.0.3)

Question: Two state system

[1] problem 3.1.

Find an expression for the free energy as a function of \tau of a system with two states, one at energy 0 and one at energy \epsilon. From the free energy, find expressions for the energy and entropy of the system.

Answer

Our partition function is

\begin{aligned}Z = 1 + e^{-\epsilon /\tau}\end{aligned} \hspace{\stretch{1}}(1.0.4)

The free energy is just

\begin{aligned}F = -\tau \ln Z = -\tau \ln (1 + e^{-\epsilon/\tau})\end{aligned} \hspace{\stretch{1}}(1.0.5)

The entropy follows immediately

\begin{aligned}\sigma \\ &= -\frac{\partial {F}}{\partial {\tau}} \\ &= \frac{\partial {}}{\partial {\tau}}\left( \tau \ln \left( 1 + e^{-\epsilon/\tau} \right) \right) \\ &= \ln \left( 1 + e^{-\epsilon/\tau} \right)-\tau \epsilon \frac{-1}{\tau^2} \frac{1}{{1 + e^{-\epsilon/\tau}}} \\ &= \ln \left( 1 + e^{-\epsilon/\tau} \right)+\frac{\epsilon}{\tau} \frac{e^{-\epsilon/\tau}}{1 + e^{-\epsilon/\tau}}\end{aligned} \hspace{\stretch{1}}(1.0.6)

The energy is

\begin{aligned}U \\ &= F + \tau \sigma \\ &= -\tau \ln (1 + e^{-\epsilon/\tau}) + \tau \sigma \\ &= \tau\left( \not{{\ln \left( 1 + e^{-\epsilon/\tau} \right)}} + \frac{\epsilon}{\tau} \frac{e^{-\epsilon/\tau}}{1 + e^{-\epsilon/\tau}} -\not{{\ln (1 + e^{-\epsilon/\tau}) }} \right)\end{aligned} \hspace{\stretch{1}}(1.0.7)

This is

\begin{aligned}U=\frac{\epsilon e^{-\epsilon/\tau}}{1 + e^{-\epsilon/\tau}}=\frac{\epsilon}{1 + e^{\epsilon/\tau}}.\end{aligned} \hspace{\stretch{1}}(1.0.8)

These are all plotted in (Fig 1).

Fig1: Plots for two state system

Fig1: Plots for two state system

 

\imageFigure{kittelCh3Problem1PlotsFig1}{Plots for two state system}{fig:kittelCh3Problem1Plots:kittelCh3Problem1PlotsFig1}{0.2}

Question: Magnetic susceptibility

[1] problem 3.2.

Use the partition function to find an exact expression for the magnetization M and the susceptibility \chi = dM/dB as a function of temperature and magnetic field for the model system of magnetic moments in a magnetic field. The result for the magnetization, found by other means, was M = n m \tanh( m B/\tau), where n is the particle concentration. Find the free energy and express the result as a function only of \tau and the parameter x = M/nm. Show that the susceptibility is \chi = n m^2/\tau in the limit m B \ll \tau.

Answer

Our partition function for a unit volume containing n spins is

\begin{aligned}Z=\frac{\left( e^{-m B/\tau} +e^{m B/\tau} \right)^n}{n!}=2 \frac{\left( \cosh\left( m B/\tau \right) \right)^n}{n!},\end{aligned} \hspace{\stretch{1}}(1.0.9)

so that the Free energy is

\begin{aligned}F = -\tau\left( \ln 2 - \ln n! + n \ln \cosh\left( m B/\tau \right) \right).\end{aligned} \hspace{\stretch{1}}(1.0.15)

The energy, magnetization and magnetic field were interrelated by

\begin{aligned}- M B &= U \\ &= \tau^2 \frac{\partial {}}{\partial {\tau}}\left( -\frac{F}{\tau} \right) \\ &= \tau^2 n\frac{\partial {}}{\partial {\tau}}\ln \cosh\left( m B/\tau \right) \\ &= \tau^2 n \frac{ -m B/\tau^2\sinh\left( m B/\tau \right)}{\cosh\left( m B/\tau \right)} \\ &= - m B n \tanh \left( m B/\tau \right).\end{aligned} \hspace{\stretch{1}}(1.0.11)

This gives us

\begin{aligned}M = m n \tanh \left( m B/\tau \right),\end{aligned} \hspace{\stretch{1}}(1.0.12)

so that

\begin{aligned}\chi = \frac{dM}{dB}= \frac{m^2 n}{\tau \cosh^2 \left( m B/\tau \right)}.\end{aligned} \hspace{\stretch{1}}(1.0.13)

For m B/\tau \ll 1, the cosh term goes to unity, so we have

\begin{aligned}\chi \approx= \frac{m^2 n}{\tau},\end{aligned} \hspace{\stretch{1}}(1.0.14)

as desired.

With x = M/nm, or m = M/nx, the free energy is

\begin{aligned}F =-\tau\left( \ln 2/n! + n \ln \cosh\left( \frac{M B}{n x \tau} \right) \right)\end{aligned} \hspace{\stretch{1}}(1.0.15)

That last expression isn’t particularly illuminating. What was the point of that substitution?

Question: Free energy of a harmonic oscillator

[1] problem 3.3.

A one dimensional harmonic oscillator has an infinite series of equally spaced energy states, with \epsilon_s = s \hbar \omega, where s is a positive integer or zero, and \omega is the classical frequency of the oscillator. We have chosen the zero of energy at the state s = 0. Show that for a harmonic oscillator the free energy is

\begin{aligned}F = \tau \ln\left( 1 - e^{-\hbar \omega/\tau} \right).\end{aligned} \hspace{\stretch{1}}(1.0.16)

Note that at high temperatures such that \tau \gg \hbar \omega we may expand the argument of the logarithm to obtain F \approx \tau \ln (\hbar \omega/\tau). From 1.0.16 show that the entropy is

\begin{aligned}\sigma = \frac{\hbar\omega/\tau}{e^{\hbar \omega/\tau} - 1} -\ln\left( 1 - e^{-\hbar \omega/\tau} \right)\end{aligned} \hspace{\stretch{1}}(1.0.17)

Answer

I found it curious that this problem dropped the factor of \hbar\omega/2 from the energy. Including it we have

\begin{aligned}\epsilon_s = \left( s + \frac{1}{{2}} \right) \hbar \omega,\end{aligned} \hspace{\stretch{1}}(1.0.18)

So that the partition function is

\begin{aligned}Z= \sum_{s = 0}^\infty e^{-\left( s + \frac{1}{{2}} \right) \hbar \omega/\tau }=e^{-\hbar \omega/2\tau}\sum_{s = 0}^\infty e^{-s \hbar \omega/\tau}.\end{aligned} \hspace{\stretch{1}}(1.0.19)

The free energy is

\begin{aligned}F &= -\tau \ln Z \\ &= -\tau\left( -\frac{\hbar \omega}{2\tau} + \ln \left( \sum_{s = 0}^\infty e^{-s \hbar \omega/\tau} \right) \right) \\ &= \frac{\hbar \omega}{2} +\ln\left( \sum_{s = 0}^\infty e^{-s \hbar \omega/\tau} \right)\end{aligned} \hspace{\stretch{1}}(1.0.20)

We see that the contribution of the \hbar \omega/2 in the energy of each state just adds a constant factor to the free energy. This will drop out when we compute the entropy. Dropping that factor now that we know why it doesn’t contribute, we can complete the summation, so have, by inspection

\begin{aligned}F = -\tau \ln Z=\tau \ln\left( 1 - e^{-\hbar \omega/\tau} \right).\end{aligned} \hspace{\stretch{1}}(1.0.21)

Taking derivatives for the entropy we have

\begin{aligned}\sigma &= -\frac{\partial {F}}{\partial {\tau}} \\ &= -\ln\left( 1 - e^{-\hbar \omega/\tau} \right)+\tau\frac{\hbar \omega}{\tau^2} \frac{e^{-\hbar \omega/\tau}}{1 - e^{-\hbar \omega/\tau}} \\ &= -\ln\left( 1 - e^{-\hbar \omega/\tau} \right)+\frac{\frac{\hbar \omega}{\tau}}{e^{\hbar \omega/\tau} - 1}\end{aligned} \hspace{\stretch{1}}(1.0.22)

Question: Energy fluctuation

[1] problem 3.4.

Consider a system of fixed volume in thermal contact with a reservoir. Show that the mean square fluctuation in the energy of the system is

\begin{aligned}\left\langle{{ (\epsilon - \left\langle{\epsilon}\right\rangle)^2 }}\right\rangle = \tau^2\left( \frac{\partial {U}}{\partial {\tau}} \right)_V\end{aligned} \hspace{\stretch{1}}(1.0.23)

Here U is the conventional symbol for \left\langle{{\epsilon}}\right\rangle. Hint: Use the partition function Z to relate {\partial {U}}/{\partial {t}} to the mean square fluctuation. Also, multiply out the term (\cdots)^2.

Answer

With a probability of finding the system in state s of

\begin{aligned}P_s = \frac{e^{-\epsilon_s/\tau}}{Z}\end{aligned} \hspace{\stretch{1}}(1.0.24)

the average energy is

\begin{aligned}U &= \left\langle{{\epsilon}}\right\rangle \\ &= \sum_s P_s \epsilon_s \\ &= \sum_s \epsilon_s \frac{e^{-\epsilon_s/\tau}}{Z} \\ &= \frac{1}{{Z}} \sum_s \epsilon_s e^{-\epsilon_s/\tau}\end{aligned} \hspace{\stretch{1}}(1.0.25)

So we have

\begin{aligned}\tau^2 \frac{\partial {U}}{\partial {\tau}} \\ &= -\frac{\tau^2}{Z^2} \frac{dZ}{d\tau}\sum_s \epsilon_s e^{-\epsilon_s/\tau}+ \frac{\tau^2}{Z}\sum_s \frac{\epsilon_s^2}{\tau^2} e^{-\epsilon_s/\tau} \\ &= -\frac{\tau^2}{Z^2} \frac{dZ}{d\tau}\sum_s \epsilon_s e^{-\epsilon_s/\tau}+ \frac{1}{{Z}}\sum_s \epsilon_s^2 e^{-\epsilon_s/\tau}.\end{aligned} \hspace{\stretch{1}}(1.0.26)

But

\begin{aligned}\frac{dZ}{d\tau}=\frac{d}{d\tau} \sum_s e^{-\epsilon_s/\tau}=\sum_s \frac{\epsilon_s}{\tau^2} e^{-\epsilon_s/\tau},\end{aligned} \hspace{\stretch{1}}(1.0.27)

giving

\begin{aligned}\tau^2 \frac{\partial {U}}{\partial {\tau}} &= \frac{1}{Z^2}\sum_s \epsilon_s e^{-\epsilon_s/\tau} \sum_s \epsilon_s e^{-\epsilon_s/\tau}+ \frac{1}{{Z}}\sum_s \epsilon_s^2 e^{-\epsilon_s/\tau} \\ &= -\left\langle{{ \epsilon }}\right\rangle^2 + \left\langle{{\epsilon^2}}\right\rangle,\end{aligned} \hspace{\stretch{1}}(1.0.28)

which shows 1.0.23 as desired.

References

[1] C. Kittel and H. Kroemer. Thermal physics. WH Freeman, 1980.

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PHY452H1S Basic Statistical Mechanics. Lecture 6: Volumes in phase space. Taught by Prof. Arun Paramekanti

Posted by peeterjoot on January 29, 2013

[Click here for a PDF of this post with nicer formatting]

Disclaimer

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

Liouville’s theorem

We’ve looked at the continuity equation of phase space density

\begin{aligned}0 = \frac{\partial {\rho}}{\partial {t}} + \sum_{i_\alpha} \left(\frac{\partial {}}{\partial {p_{i_\alpha}}} \left( \rho \dot{p}_{i_\alpha} \right) + \frac{\partial {\left( \rho \dot{x}_{i_\alpha} \right) }}{\partial {x_{i_\alpha}}}\right)\end{aligned} \hspace{\stretch{1}}(1.2.1)

which with

\begin{aligned}\frac{\partial {\dot{p}_{i_\alpha}}}{\partial {p_{i_\alpha}}} + \frac{\partial {\dot{x}_{i_\alpha}}}{\partial {x_{i_\alpha}}} = 0\end{aligned} \hspace{\stretch{1}}(1.2.2)

led us to Liouville’s theorem

\begin{aligned}\\ boxed{\frac{d{{\rho}}}{dt}(x, p, t) = 0}.\end{aligned} \hspace{\stretch{1}}(1.2.3)

We define Ergodic, meaning that with time, as you wait for t \rightarrow \infty, all available phase space will be covered. Not all systems are necessarily ergodic, but the hope is that all sufficiently complicated systems will be so.

We hope that

\begin{aligned}\rho(x, p, t \rightarrow \infty) \implies \frac{\partial {\rho}}{\partial {t}} = 0 \qquad \mbox{in steady state}\end{aligned} \hspace{\stretch{1}}(1.2.4)

In particular for \rho = \text{constant}, we see that our continuity equation 1.2.1 results in 1.2.2.

For example in a SHO system with a cyclic phase space, as in (Fig 1).

Fig 1: Phase space volume trajectory

 

\begin{aligned}\left\langle{{A}}\right\rangle = \frac{1}{{\tau}} \int_0^\tau dt A( x_0(t), p_0(t) ),\end{aligned} \hspace{\stretch{1}}(1.2.5)

or equivalently with an ensemble average, imagining that we are averaging over a number of different systems

\begin{aligned}\left\langle{{A}}\right\rangle = \frac{1}{{\tau}} \int dx dp A( x, p ) \underbrace{\rho(x, p)}_{\text{constant}}\end{aligned} \hspace{\stretch{1}}(1.2.6)

If we say that

\begin{aligned}\rho(x, p) = \text{constant} = \frac{1}{{\Omega}},\end{aligned} \hspace{\stretch{1}}(1.2.7)

so that

\begin{aligned}\left\langle{{A}}\right\rangle = \frac{1}{{\Omega}} \int dx dp A( x, p ) \end{aligned} \hspace{\stretch{1}}(1.2.8)

then what is this constant. We fix this by the constraint

\begin{aligned}\int dx dp \rho(x, p) = 1\end{aligned} \hspace{\stretch{1}}(1.2.9)

So, \Omega is the allowed “volume” of phase space, the number of states that the system can take that is consistent with conservation of energy.

What’s the probability for a given configuration. We’ll have to enumerate all the possible configurations. For a coin toss example, we can also ask how many configurations exist where the sum of “coin tosses” are fixed.

A worked example: Ideal gas calculation of \Omega

  • N gas atoms at phase space points \mathbf{x}_i, \mathbf{p}_i
  • constrained to volume V
  • Energy fixed at E.

\begin{aligned}\Omega(N, V, E) = \int_V d\mathbf{x}_1 d\mathbf{x}_2 \cdots d\mathbf{x}_N \int d\mathbf{p}_1 d\mathbf{p}_2 \cdots d\mathbf{p}_N \delta \left(E - \frac{\mathbf{p}_1^2}{2m}- \frac{\mathbf{p}_2^2}{2m}\cdots- \frac{\mathbf{p}_N^2}{2m}\right)=\underbrace{V^N}_{\text{Real space volume, not N dimensional ``volume''}} \int d\mathbf{p}_1 d\mathbf{p}_2 \cdots d\mathbf{p}_N \delta \left(E - \frac{\mathbf{p}_1^2}{2m}- \frac{\mathbf{p}_2^2}{2m}\cdots- \frac{\mathbf{p}_N^2}{2m}\right)\end{aligned} \hspace{\stretch{1}}(1.10)

With \gamma defined implicitly by

\begin{aligned}\frac{d\gamma}{dE} = \Omega\end{aligned} \hspace{\stretch{1}}(1.3.11)

so that with Heavyside theta as in (Fig 2).

\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}}(1.0.12a)

\begin{aligned}\frac{d\Theta}{dx} = \delta(x),\end{aligned} \hspace{\stretch{1}}(1.0.12b)

Fig 2: Heavyside theta

 

we have

\begin{aligned}\gamma(N, V, E) = V^N \int d\mathbf{p}_1 d\mathbf{p}_2 \cdots d\mathbf{p}_N \Theta \left(E - \sum_i \frac{\mathbf{p}_i^2}{2m}\right)\end{aligned} \hspace{\stretch{1}}(1.0.13)

In three dimensions (p_x, p_y, p_z), the dimension of momentum part of the phase space is 3. In general the dimension of the space is 3N. Here

\begin{aligned}\int d\mathbf{p}_1 d\mathbf{p}_2 \cdots d\mathbf{p}_N \Theta \left(E - \sum_i \frac{\mathbf{p}_i^2}{2m}\right),\end{aligned} \hspace{\stretch{1}}(1.0.14)

is the volume of a “sphere” in 3N– dimensions, which we found in the problem set to be

\begin{aligned}V_{m} = \frac{ \pi^{m/2} R^{m} }{ \Gamma\left( m/2 + 1 \right)}.\end{aligned} \hspace{\stretch{1}}(1.0.15a)

\begin{aligned}\Gamma(x) = \int_0^\infty dy e^{-y} y^{x-1}\end{aligned} \hspace{\stretch{1}}(1.0.15b)

\begin{aligned}\Gamma(x + 1) = x \Gamma(x) = x!\end{aligned} \hspace{\stretch{1}}(1.0.15c)

Since we have

\begin{aligned}\mathbf{p}_1^2 + \cdots \mathbf{p}_N^2 \le 2 m E\end{aligned} \hspace{\stretch{1}}(1.0.16)

the radius is

\begin{aligned}\text{radius} = \sqrt{ 2 m E}.\end{aligned} \hspace{\stretch{1}}(1.0.17)

This gives

\begin{aligned}\gamma(N, V, E) = V^N \frac{ \pi^{3 N/2} ( 2 m E)^{3 N/2}}{\Gamma( 3N/2 + 1) }= V^N \frac{2}{3N} \frac{ \pi^{3 N/2} ( 2 m E)^{3 N/2}}{\Gamma( 3N/2 ) },\end{aligned} \hspace{\stretch{1}}(1.0.17)

and

\begin{aligned}\Omega(N, V, E) = V^N \pi^{3 N/2} ( 2 m E)^{3 N/2 - 1} \frac{2 m}{\Gamma( 3N/2 ) }\end{aligned} \hspace{\stretch{1}}(1.0.19)

This result is almost correct, and we have to correct in 2 ways. We have to fix the counting since we need an assumption that all the particles are indistinguishable.

  • Indistinguishability. We must divide by N!.
  • \Omega is not dimensionless. We need to divide by h^{3N}, where h is Plank’s constant.

In the real world we have to consider this as a quantum mechanical system. Imagine a two dimensional phase space. The allowed points are illustrated in (Fig 3).

Fig 3: Phase space volume adjustment for the uncertainty principle

 

Since \Delta x \Delta p \sim \hbar, the question of how many boxes there are, we calculate the total volume, and then divide by the volume of each box. This sort of handwaving wouldn’t be required if we did a proper quantum mechanical treatment.

The corrected result is

\begin{aligned}\boxed{\Omega_{\mathrm{correct}} = \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}}(1.0.20)

To come

We’ll look at entropy

\begin{aligned}\underbrace{S}_{\text{Entropy}} = \underbrace{k_{\mathrm{B}}}_{\text{Boltzmann's constant}} \ln \underbrace{\Omega_{\mathrm{correct}}}_{\text{phase space volume (number of configurations)}}\end{aligned} \hspace{\stretch{1}}(1.0.21)

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