Peeter Joot's (OLD) Blog.

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Posts Tagged ‘Liouville’s theorem’

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 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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Liouville’s theorem questions on density and current

Posted by peeterjoot on February 5, 2013

[Click here for a PDF of this post with nicer formatting and figures if the post had any]

Liouville’s theorem questions on density and current

In the midterm we were asked to state and prove Liouville’s theorem. I couldn’t remember the proof, having only a recollection that it had something to do with the continuity equation

\begin{aligned}0 = \frac{\partial {\rho}}{\partial {t}} + \frac{\partial {j}}{\partial {x}},\end{aligned} \hspace{\stretch{1}}(1.1.1)

but unfortunately couldn’t remember what the j was. Looking up the proof, it’s actually really simple, just the application of chain rule for a function \rho that’s presumed to be a function of time, position and momentum variables. It didn’t appear to me that this proof has anything to do with any sort of notion of density, so I posed the following questions.

Context

The core of the proof can be distilled to one dimension, removing all the indexes that obfuscate what’s being one. For that case, application of the chain rule to a function \rho(t, x, p), we have

\begin{aligned}\frac{d{{\rho}}}{dt} &= \frac{\partial {\rho}}{\partial {t}} + \frac{\partial {x}}{\partial {t}} \frac{\partial {\rho}}{\partial {x}} + \frac{\partial {p}}{\partial {t}} \frac{\partial {\rho}}{\partial {p}} \\ &= \frac{\partial {\rho}}{\partial {t}} + \dot{x} \frac{\partial {\rho}}{\partial {x}} + \dot{p} \frac{\partial {\rho}}{\partial {p}} \\ &= \frac{\partial {\rho}}{\partial {t}} + \frac{\partial {\left( \dot{x} \rho \right)}}{\partial {x}} + \frac{\partial {\left( \dot{x} \rho \right)}}{\partial {p}} - \rho \left(  \frac{\partial {\dot{x}}}{\partial {x}} + \frac{\partial {\dot{p}}}{\partial {p}}  \right) \\ &= \frac{\partial {\rho}}{\partial {t}} + \frac{\partial {\left( \dot{x} \rho \right)}}{\partial {x}} + \frac{\partial {\left( \dot{x} \rho \right)}}{\partial {p}} - \rho \underbrace{\left(\frac{\partial {}}{\partial {x}} \left(  \frac{\partial {H}}{\partial {p}}  \right)+\frac{\partial {}}{\partial {p}} \left(  -\frac{\partial {H}}{\partial {x}}  \right)\right)}_{= 0}\end{aligned} \hspace{\stretch{1}}(1.1.2)

Wrong interpretation

From this I’d thought that the theorem was about steady states. If we do have a steady state, where d\rho/dt = 0 we have

\begin{aligned}0 = \frac{\partial {\rho}}{\partial {t}} + \frac{\partial {\left( \dot{x} \rho \right)}}{\partial {x}} + \frac{\partial {\left( \dot{p} \rho \right)}}{\partial {p}}.\end{aligned} \hspace{\stretch{1}}(1.1.3)

That would answer the question of what the current is, it’s this tuple

\begin{aligned}\mathbf{j} = \rho (\dot{x}, \dot{p}),\end{aligned} \hspace{\stretch{1}}(1.1.4)

so if we introduce a “phase space” gradient

\begin{aligned}\boldsymbol{\nabla} = \left(  \frac{\partial {}}{\partial {x}}, \frac{\partial {}}{\partial {p}}  \right)\end{aligned} \hspace{\stretch{1}}(1.1.5)

we’ve got something that looks like a continuity equation

\begin{aligned}0 = \frac{\partial {\rho}}{\partial {t}} + \boldsymbol{\nabla} \cdot \mathbf{j}.\end{aligned} \hspace{\stretch{1}}(1.1.6)

Given this misinterpretation of the theorem, I had the following two questions

  • This function \rho appears to be pretty much arbitrary. I don’t see how this connects to any notion of density?
  • If we pick a specific Hamiltonian, say the 1D SHO, what physical interpretation do we have for this “current” \mathbf{j}?

The clarification

Asking about this, the response was “Actually, equation 1.1.3 has to be assumed for the proof. This equation holds if \rho is the phase space density and since the pair in 1.1.4 is the current density in phase space. The theorem then states that d\rho/dt = 0 whether or not one is in the steady state. This means even as the system is evolving in time, if we sit on a particular phase space point and follow it around as it evolves, the density in our neighborhood will be a constant.”

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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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PHY452H1S Basic Statistical Mechanics. Lecture 5: Motion in phase space. Liouville and Poincar’e theorems. Taught by Prof. Arun Paramekanti

Posted by peeterjoot on January 22, 2013

[Click here for a PDF of this post with nicer formatting and figures if the post had any (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.

Motion in phase space

Classical system: \mathbf{x}_i, \mathbf{p}_i with dimensionality

\begin{aligned}\underbrace{2}_{x, p}\underbrace{d}_{\text{space dimension}}\underbrace{N}_{\text{Number of particles}}\end{aligned} \hspace{\stretch{1}}(1.2.1)

Hamiltonian H is the “energy function”

\begin{aligned}H = \underbrace{\sum_{i = 1}^N \frac{\mathbf{p}_i^2}{2m} }_{\text{Kinetic energy}}+ \underbrace{\sum_{i = 1}^N V(\mathbf{x}_i) }_{\text{Potential energy}}+ \underbrace{\sum_{i < j}^N \Phi(\mathbf{x}_i - \mathbf{x}_j)}_{\text{Internal energy}}\end{aligned} \hspace{\stretch{1}}(1.2.2)

\begin{aligned}\mathbf{\dot{p}}_i = \mathbf{F} = \text{force}\end{aligned} \hspace{\stretch{1}}(1.0.3a)

\begin{aligned}\mathbf{\dot{x}}_i = \frac{\mathbf{p}_i}{m}\end{aligned} \hspace{\stretch{1}}(1.0.3b)

Expressed in terms of the Hamiltonian this is

\begin{aligned}\dot{p}_{i_\alpha} = - \frac{\partial {H}}{\partial {x_{i_\alpha}}}\end{aligned} \hspace{\stretch{1}}(1.0.4a)

\begin{aligned}\dot{x}_{i_\alpha} = \frac{\partial {H}}{\partial {p_{i_\alpha}}}\end{aligned} \hspace{\stretch{1}}(1.0.4b)

In phase space we can have any number of possible trajectories as illustrated in (Fig1).

Fig1: Disallowed and allowed phase space trajectories

Liouville’s theorem

We are interested in asking the question of how the density of a region in phase space evolves as illustrated in (Fig2)

Fig2: Evolution of a phase space volume

We define a phase space density

\begin{aligned}\rho(p_{i_\alpha}, x_{i_\alpha}, t),\end{aligned} \hspace{\stretch{1}}(1.0.5)

and seek to demonstrate Liouville’s theorem, that the phase space density does not change. To do so, consider the total time derivative of the phase space density

\begin{aligned}\frac{d{{\rho}}}{dt} &= \frac{\partial {\rho}}{\partial {t}} + \sum_{i_\alpha} \frac{\partial {p_{i_\alpha}}}{\partial {t}} \frac{\partial {\rho}}{\partial {p_{i_\alpha}}} + \frac{\partial {x_{i_\alpha}}}{\partial {t}} \frac{\partial {\rho}}{\partial {x_{i_\alpha}}} \\ &= \frac{\partial {\rho}}{\partial {t}} + \sum_{i_\alpha} \frac{\partial {p_{i_\alpha}}}{\partial {t}} \frac{\partial {\rho \dot{p}_{i_\alpha}}}{\partial {p_{i_\alpha}}} + \frac{\partial {x_{i_\alpha}}}{\partial {t}} \frac{\partial {\rho \dot{x}_{i_\alpha}}}{\partial {x_{i_\alpha}}} - \rho \left(\frac{\partial {\dot{p}_{i_\alpha}}}{\partial {p_{i_\alpha}}} +\frac{\partial {\dot{x}_{i_\alpha}}}{\partial {x_{i_\alpha}}} \right) \\ &= \frac{\partial {\rho}}{\partial {t}} + \underbrace{\sum_{i_\alpha} \left(\frac{\partial {p_{i_\alpha}}}{\partial {t}} \frac{\partial {(\rho \dot{p}_{i_\alpha})}}{\partial {p_{i_\alpha}}} + \frac{\partial {x_{i_\alpha}}}{\partial {t}} \frac{\partial {(\rho \dot{x}_{i_\alpha})}}{\partial {x_{i_\alpha}}} \right)}_{\equiv \boldsymbol{\nabla} \cdot \mathbf{j}}- \rho \sum_{i_\alpha} \underbrace{\left(-\frac{\partial^2 H}{\partial p_{i_\alpha} \partial x_{i_\alpha}}+\frac{\partial^2 H}{\partial x_{i_\alpha} \partial p_{i_\alpha}}\right)}_{= 0}\end{aligned} \hspace{\stretch{1}}(1.0.6)

We’ve implicitly defined a current density \mathbf{j} above by comparing to the continuity equation

\begin{aligned}\frac{\partial {\rho}}{\partial {t}} + \boldsymbol{\nabla} \cdot \mathbf{j} = 0\end{aligned} \hspace{\stretch{1}}(1.0.7)

Here we have

\begin{aligned}\rho(\dot{x}_{i_\alpha}, \dot{p}_{i_\alpha}) \rightarrow \mbox{current in phase space}\end{aligned} \hspace{\stretch{1}}(1.0.8)

Usually we have

\begin{aligned}\mathbf{j}_{\mathrm{usual}} \sim \rho \mathbf{v}\end{aligned} \hspace{\stretch{1}}(1.0.9a)

\begin{aligned}\mathbf{j} = -D \boldsymbol{\nabla} \rho\end{aligned} \hspace{\stretch{1}}(1.0.9b)

but we don’t care about this diffusion relation, just the continuity equation equivalent

\begin{aligned}v \frac{\partial {\rho}}{\partial {t}} + \boldsymbol{\nabla} \cdot \mathbf{j} = 0\end{aligned} \hspace{\stretch{1}}(1.0.10)

The implication is that

\begin{aligned}\frac{d{{\rho}}}{dt} = 0.\end{aligned} \hspace{\stretch{1}}(1.0.11)

Flow in phase space is very similar to an “incompressible fluid”.

Time averages, and Poincar\’e recurrence theorem

We want to look at how various observables behave over time

\begin{aligned}\overline{A} = \frac{1}{{T}} \int_0^T dt \rho(x, p, t) A(p, x)\end{aligned} \hspace{\stretch{1}}(1.0.12)

We’d like to understand how such averages behave over long time intervals, looking at \rho(x, p, t \rightarrow \infty).

This long term behaviour is described by the Poincar\’e recurrence theorem. If we wait long enough a point in phase space will come arbitrarily close to its starting point, recurring or “closing the trajectory loops”.

A simple example of a recurrence is an undamped SHO, such as a pendulum. That pendulum bob when it hits the bottom of the cycle will have the same velocity (and position) each time it goes through a complete cycle. If we imagine a much more complicated system, such as N harmonic oscillators, each with different periods, we can imagine that it will take infinitely long for this cycle or recurrence to occur, and the trajectory will end up sweeping through all of phase space.

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