$\require{cancel}$ $\require{physics}$ $\require{mathtools}$

Statmech Notes

Table of Contents

  1. Classical Thermo
  2. Microcanonical Ensemble
  3. Canonical Ensemble
  4. Grand Canonical Ensemble
  5. Quantum Information Perspective
  6. Classical Gases
  7. Quantum Gases
  8. Debye Model of Solids
  9. Electronic Gas in a Magnetic Field
  10. Phase Transitions and Van der Waals gas
  11. Ising Model
  12. Polymers
  13. Brownian motion

Classical Thermo


  1. equilibrium is transitive (gives us idea of temperature)
  2. amount of work required to change isolated system's state is independent of how work is performed. for nonisolated systems, change of energy includes a heat term $\Delta E = Q + W$
  3. entropy increases
    • (Kelvin:) no process exists whose sole effect is to extract heat from a resevoir and turn it into work
    • (Clausius:) no process exists whose sole effect is to transfer heat from cold to hot
  4. not really as important as the others, but it's
    \begin{equation} \lim_{T\to 0}S(T) = 0 \end{equation}
    • the ground state entropy shouldnt grow extensively
    • heat capacities tend to zero


isothermal expansion at hot temperature $\to$ adiabatic expansion (cools) $\to$ isothermal contraction at cold temperature $\to$ adiabatic contraction (heats up)

\begin{equation} \eta = \frac{W}{Q_H} = \frac{Q_H-Q_C}{Q_H} = 1 - \frac{Q_C}{Q_H} \stackrel{\text{Carnot}}{=} 1 - \frac{T_C}{T_H} \end{equation}


\begin{equation} S = \int \frac{\var{Q}}{T} \end{equation}

is a function of state – doing the integral in a (REVERSIBLE) loop gives you zero, otherwise we have clausius inequality

\begin{equation} \oint \frac{\var{Q}}{T} \leq 0 \end{equation}

(tells us that entropy of an isolated system never decreases)

Thermodynamic potentials

at fixed energy, the entropy doesnt decrease. other extremization principles follow:

free energy is a measure of the amount of energy free to do work at a finite temperature – at constant temp and volume, free energy can never increase

at fixed temperature and pressure, minimize $G$.

Note: by extensivity, $G(p,\,T,\,N) = \mu(p,\,T)N$; cf. $\Phi = -p(T,\,\mu)V$

at fixed energy and pressure, consider enthalpy

Maxwell relations

rewrite derivatives that you dont know in terms of things you do!

when looking for something of the form

\begin{equation} \pdv{A}{B}\eval{}_C \end{equation}

the idea is to find $A$ as a first derivative of some function of state that has $\dd{B}$ and $\dd{C}$ as differentials; this lets us swap $A$ for the $B$ derivative. more explicitly,

find a thermodynamic potential of the form $\dd{X} = A\dd{\alpha} + \beta\dd{B} + \gamma\dd{C}$. Then

\begin{equation} \pdv{X}{B}{\alpha} = \pdv{A}{B}\eval{}_{C,\,\alpha} = \pdv{\beta}{\alpha}\eval{}_{C,\,B} \end{equation}

as an example, consider

\begin{equation} \pdv{\mu}{p}\eval{}_T \end{equation}

our function of state is

\begin{equation} \dd{G} = -S\dd{T} + V\dd{p} + \mu\dd{N} \implies \pdv{G}{p}{N} = \pdv{\mu}{p}\eval{}_{N,\,T} = \pdv{V}{N}\eval{}_{p,\,T} \end{equation}

Heat capacities this does nice things for them; recalling

\begin{equation} C_\bullet = T\pdv{S}{T}\eval{}_\bullet \end{equation}

we find

\begin{gather*} \pdv{C_V}{V}\eval{}_T = T\pdv[2]{p}{T}\eval{}_V\qq{and}\pdv{C_p}{p}\eval{}_T = -T\pdv[2]{V}{T}\eval{}_p \implies C_p-C_V = T\pdv{V}{T}\eval{}_p\pdv{p}{T}\eval{}_{V} \end{gather*}

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Microcanonical Ensemble

Fixed energy $E$ gives us a notion of $S$, $T$

\begin{gather*} P(n) = \frac{1}{\Omega(E_n)}\\ S(E) = k_B\log\Omega(E)\\ \frac{1}{T} = \pdv{S}{E} \qquad \pdv{S}{T} = \frac{C}{T}\qquad p = T \pdv{S}{V}\\ C = \pdv{E}{T} \qquad C_V = \pdv{E}{T}\eval{}_V = T\pdv{S}{T}\eval{}_V \qquad C_p = T\pdv{S}{T}\eval{}_p\\ \dd{E} = T\dd{S} - p\dd{V} \end{gather*}

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Canonical Ensemble

Fixed $T$ gives us an $\ev{E}$ (“softly” fixed energy by tuning $\beta$)

Boltzmann distrib:

\begin{equation} P(n) = \frac{e^{-\beta E_n}}{Z} \qq{for} Z = \sum_\text{states}e^{-\beta E_n} \end{equation}

$Z$ multiplicative for independent systems

\begin{gather*} \ev{E} = -\partial_\beta \log Z \qquad \Delta E^2 = \partial_\beta^2 \log Z = k_BT^2 C_V \sim \sqrt{N}\\ S = -k_B\sum_n P(n) \log P(n) = k_B \partial_T(T\log Z) \end{gather*}

where the last equality holds for the Boltz dist

reduces to microcanon def if $E = E_\star$ (most likely energy) $= \ev{E}$

Free energy

\begin{gather} F = \ev{E} - TS = -\frac{\log Z}{\beta} \\ \dd{F} = -S\dd{T} - p\dd{V} (+ \mu \dd{N})\\ \implies S = -\pdv{F}{T}\eval{}_V \qquad p = -\pdv{F}{V}\eval{}_T \end{gather}

with particle number,

\begin{equation} \mu = -T\pdv{S}{N}\eval{}_{E,V}=\pdv{F}{N}\eval{}_{T,V} \end{equation}

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Grand Canonical Ensemble

no longer fix particle number

\begin{equation} \mathcal{Z} = \sum e^{-\beta(E_n - \mu N_n)} \qquad P(n) = \frac{e^{-\beta E + \beta\mu N}}{\mathcal{Z}} \end{equation}

Entropy has the same as in CE, $k_B\partial_T(T\log \mathcal{Z})$. $E$ picks up an extra term:

\begin{gather*} \ev{E} - \mu\ev{N} = -\partial_\beta \log \mathcal{Z}\\ \ev{N} = \frac{1}{\beta}\partial_\mu \log \mathcal{Z} \qquad \Delta N^2 = {\large(}\frac{1}{\beta}\partial_\beta{\large)}^2 \log \mathcal{Z} \end{gather*}

grand potential

\begin{gather} \Phi = F - \mu N = E - TS - \mu N = -\frac{1}{\beta}\log \mathcal{Z} = -p(T,\,\mu) V\\ \dd{\Phi} = -S\dd{T} - p\dd{V} - N\dd{\mu} \end{gather}

we have a pairing of intensive-extensive: $TS$ $pV$ $\mu N$, gives $E$ extensive

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Quantum Information Perspective

have a density matrix instead of probability distribution:

\begin{align} \hat{\rho}_C &= \frac{1}{Z}\exp(-\beta\hat{H}) & Z &= \tr(e^{-\beta\hat{H}})\\ \hat{\rho}_{GC} &= \frac{1}{\mathcal{Z}}\exp(-\beta\hat{H} + \beta\mu\hat{N}) & \mathcal{Z} &= \tr(e^{-\beta\hat{H}+\beta\mu\hat{N}}) \end{align}

Grand canon nice in second quant where we have ladder operators for $\hat{N}$

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Classical Gases

Monatomic gas

\begin{equation} Z_1 = \frac{1}{(2\pi\hbar)^3}\int\dd[3]{q}\dd[3]{p}e^{-\beta H} \end{equation}

in the monatomic case,

\begin{equation} Z_1 = V\qty(\sqrt{\frac{2\pi \hbar^2}{mk_B T}})^{-3} = V/\lambda^3 \end{equation}

and we get the $N$-particle gas by $Z_N = Z_1^N = V^N\lambda^{-3N}$

Ideal gas law EoS from $p=-\partial_VF$

equipartition: for each kinetic DoF we have $E\mathbin{+\kern-0.5ex=} \frac{1}{2}k_BT$, (3D = 3$N$ DoF)

note: need to account for indistinguishability in the ideal gas partition function:

\begin{equation} Z_N = \frac{1}{N!}Z_1^N \implies S = Nk_B\qty[\log(\frac{V}{N\lambda^3}) + \frac{5}{3}] \end{equation}

(sackur-tetrode equation)

adding in a chemical potential, (remember to sum over all $N$ – gives an exp)

\begin{equation} \mathcal{Z} = \exp(e^{\beta \mu}V/\lambda^3) \implies \text{(rearranging $N$) }\mu = k_BT\log(\lambda^3 N/V) \qquad \Delta N^2 = N \end{equation}

maxwell-boltz distrib (from viewing $Z_1$ as sum over states of probability):

\begin{equation} P(v) = 4\pi \qty(\frac{m}{2\pi k_BT})^{3/2}v^2 e^{-mv^2/2k_BT} \end{equation}

gives velocity distribution of a classical gas

Diatomic gas

\begin{equation} Z_1 = Z_\text{trans}Z_\text{rot}Z_\text{vib} \end{equation}

get these new $Z$s from a phase space integral for the various parts of the hammy

\begin{gather} Z_\text{rot} = \frac{2Ik_BT}{\hbar^2}\implies E_\text{rot} = \frac{2}{2}k_BT\\ Z_\text{vib} = \frac{k_BT}{\hbar\omega}\implies E_\text{vib} = \frac{2}{2}k_BT\\ \end{gather}

oscillation “freezes out” first, then rotation – limitations of classical equipartition theory (also think about how a deep potential well gives same mechanics as rigid connection, but different degrees of freedom counting. we need the full quantum explanation)

Interacting Gas

virial expansion

\begin{equation} \beta p = \frac{N}{V} + B_2(T) \frac{N^2}{V^2} + B_3(T)\frac{N^3}{V^3} + \dots \end{equation}

define the mayer f function

\begin{equation} f(r) = e^{-\beta U(r)} - 1 \end{equation}

allows us to rewrite partition

\begin{align} Z_N = \frac{V^N}{N!\lambda^{3N}}\qty(1 + \frac{N}{2V}\int\dd[3]{r}f(r) + \dots)^N\\ F = F_\text{ideal} - Nk_BT\log(1 + \frac{N}{2V}\int f(r)) \end{align}

and we find the pressure is

\begin{equation} p = -\partial_VF = \frac{\rho}{\beta} - \frac{\rho^2}{2\beta}\int f(r) \end{equation}

at which point we must pick a $U$ and perform the $f$ integral. typical choice:

\begin{equation} U(r) = \begin{cases} \infty & r< r_0\\ -U_0 \qty(\frac{r_0}{r})^6 & r\geq r_0 \end{cases} \end{equation}

which gives

\begin{gather} \frac{pV}{Nk_BT} = 1 - \frac{N}{V}\qty(\frac{a}{k_BT}-b) \iff k_BT = \qty(p + \frac{N^2}{V^2}a)\qty(\frac{V}{N}-b)^{-1} \\\implies p = \frac{Nk_BT}{V-bN} - a \frac{N^2}{V^2} \end{gather}

at low density and high temperatures for parameters

\begin{equation} a = \frac{2\pi r_0^3 U_0}{3} \text{ (attractive $p$ reduction)} \qquad b = \frac{2\pi r_0^3}{3} \text{ (excluded volume)} \end{equation}

higher corrections by cluster expansion

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Quantum Gases

DENSITY OF STATES: “if instead of integrating over states, i want to integrate over energies, what do i need as a prefactor?”

\begin{align} \sum_n \sim \int \dd[3]{n} = \int \frac{\dd[3]{x}\dd[3]{k}}{(2\pi)^3} &= \frac{4\pi V}{(2\pi)^3}\int_0^\infty \dd{k}\,k^2\\ &= \frac{V}{2\pi^2}\int\dd{E}\sqrt{\frac{2mE}{\hbar^2}}\frac{m}{\hbar^2} = \int \dd{E}g(E) \end{align}

for the usual dispersion relation

\begin{equation} E = \frac{\hbar^2 k^2}{2m} \implies g(E) = \frac{V}{4\pi^2}\qty(\frac{2m}{\hbar^2})^{3/2}\sqrt{E} \end{equation}

or relativistic

\begin{align} E = \sqrt{k^2 + m^2} \implies g(E) &= \frac{VE}{2\pi^2 \hbar^3 c^3}\sqrt{E^2 - m^2c^4}\\ &\stackrel{E\gg m}{\approx} \frac{V}{\pi^2\hbar^3 c^3}\qty(E^2 - \frac{m^2c^4}{2} + \dots) \end{align}

photon Gas

photons: idea is to have $Z_\omega$ for each frequency, sum over occupation:

\begin{equation} Z_\omega = \sum_{n=0}^\infty e^{-\beta n\hbar\omega} = \frac{1}{1-e^{-\beta\hbar\omega}} \end{equation}


\begin{equation} \log Z = \int_0^\infty \dd{\omega}g(\omega)\log Z_\omega = -\frac{V}{\pi^2c^3}\int_0^\infty\dd{\omega}\omega^2\log(1-e^{-\beta\hbar\omega}) \end{equation}

whence we find the Planck distribution of energy,

\begin{equation} E = -\partial_\beta \log Z = \frac{V\hbar}{\pi^2 c^3}\int_0^\infty \dd{\omega}\frac{\omega^3}{e^{\beta\hbar\omega}-1} = \frac{\pi^2V(k_BT)^4}{15\hbar^3c^3} \end{equation}

and wien’s law, $\omega_\text{max} \sim 1/\beta\hbar$. we also get stefan-boltz,

\begin{equation} \text{energy flux} = \frac{Ec}{4V} = \qty(\frac{\pi^2 k_B^4}{60\hbar^3 c^2}) T^4 \end{equation}

free energy gives us pressure, entropy, heat capacity

Bose Gas

\begin{equation} \mathcal{Z} = \prod_r \frac{1}{1-e^{-\beta (E_r - \mu)}} \implies \ev{n_r} = \frac{1}{e^{\beta(E_r-\mu)}-1} \end{equation}

only makes sense when $\mu < 0$, or fugacity $z = e^{\beta\mu} \in (0,\,1)$

doing the usual,

\begin{align} N = \int \dd{E} \frac{g(E)}{z^{-1}e^{\beta E}-1} \qquad E = \int \dd{E} \frac{Eg(E)}{z^{-1}e^{\beta E}-1} \\ pV = -F = -\frac{1}{\beta}\int\dd{E}g(E)\log(1-ze^{-\beta E}) = \frac{2}{3}E = \frac{Vk_BT}{\lambda^3}g_{5/2}(z) \end{align}

where we integrate the log using an IBP: $\dd{E}g(E) \sim \dd{(E^{3/2})} \sim \dd{(Eg(E))}$

high-temp (small $z$) expansion of density:

\begin{align} \frac{N}{V} &= \frac{z}{\lambda^3}\qty(1 + \frac{z}{2\sqrt{2}} + \dots )\\ &\xRightarrow{invert} z = \frac{\lambda^3 N}{V}\qty(1 - \frac{1}{2\sqrt{2}}\frac{\lambda^3 N}{V} + \dots) \end{align}

gives equation of state

\begin{equation} pV = Nk_BT \qty(1 - \frac{\lambda^3 N}{4\sqrt{2}V} + \dots) \end{equation}

bosons reduce pressure!


our $\int\dd{E}\sqrt{E}$ kills $E =0$ states when we try to sum over momenta; manually add in

\begin{equation} N = \frac{V}{\lambda^3}g_{3/2}(z) \to N = \frac{V}{\lambda^3}g_{3/2}(z) + \underbrace{\frac{z}{1-z}}_{\ev{n_0}} \end{equation}

($g$ is a polylog – numerical integration factor. $g_n(1) = \zeta(n)$). Fix parameters st

\begin{equation} \rho > \lambda^{-3}\zeta(3/2) \geq \rho_{\text{excited}} \end{equation}

which lets $\rho_\text{gs}$ make up for the difference; leads to the above expression for $N$ so long as $\rho\geq \rho_c = \lambda^{-3}\zeta(3/2)$ (“critical density”). below this density, $\mu < 0$ strictly and we have the usual bose gas form. at and above, however, $\mu = 0$ and we get ground state occupancy

GS occupancy has

\begin{equation} \frac{n_0}{N} = 1-\qty(\frac{T}{T_c})^{3/2} \end{equation}

for $T_c$ the temp when $z=1$. let’s see $C_V$:

\begin{equation} C_V = \frac{15V k_B}{4\lambda^3}g_{5/2}(z) - b\qty(\frac{T-T_c}{T_c}) \end{equation}

after a lot of approximations. $C_V$ continuous but its derivative is not – first order pt

Fermi Gas

\begin{equation} \mathcal{Z} = \prod_r(1 + ze^{-\beta E_r}) \implies n_r = \frac{1}{z^{-1}e^{\beta E} + 1} \end{equation}

no restrictions on $\mu$ anymore. $g(E)$ carries spin degeneracy $g_s = 2s+1$

\begin{equation} g(E) = \frac{g_sV}{4\pi^2}\qty(\frac{2m}{\hbar^2})^{3/2}\sqrt{E} \end{equation}

and we have the usual

\begin{align} N = \int \dd{E} \frac{g(E)}{z^{-1}e^{\beta E}+1} \qquad E = \int \dd{E} \frac{Eg(E)}{z^{-1}e^{\beta E}+1} \\ pV = \frac{1}{\beta}\int\dd{E}g(E)\log(1+ze^{-\beta E}) = \frac{2}{3}E %= \frac{Vk_BT}{\lambda^3}g_{5/2}(z) \end{align}

with the small $z$ EoS

\begin{equation} pV = Nk_BT \qty(1 + \frac{\lambda^3 N}{4\sqrt{2}g_sV} + \dots) \end{equation}

fermions increase the pressure (by the same factor!)

in the $T\to 0$ limit, we have states filled until the fermi energy $E_F=\mu(T=0)$ — though $\mu$ isnt really a function of $T$, the condition on keeping $N$ fixed allows us to write one in terms of the other (write $N$ as integral up to the surface)

\begin{equation} E_F = \frac{\hbar^2}{2m}\qty(\frac{6\pi^2}{g_s}\frac{N}{V})^{2/3} \end{equation}

and we can compute

\begin{equation} pV= \frac{2}{3}E = \frac{2}{3}\int_0^{E_F}\dd{E}Eg(E) = \frac{2}{3}\qty(\frac{3}{5}NE_F) \end{equation}

which is a nonzero “degeneracy” pressure at $T=0$

in $T\ll T_F$, we can take the integrals to infinity instead of cutting them off. Only states within $k_BT$ of the fermi surface are affected by the temperature, so we can evaluate derivatives of the distribution at $E_F$; this is the only place it changes.

\begin{equation} C_V = \pdv{E}{T}\eval{}_{N,V}\sim Tg(E_F) = Nk_B \frac{\pi^2}{2}\frac{T}{T_F} \end{equation}

(idea: we have $g(E_F)k_BT$ particles contributing to the physics, each of which has $E\sim k_BT$ – this gives linear heat capacity)

we often combine this linear electronic contribution with the cubic phononic contribution (see here) to get the full heat capacity of metals.

to do this low temp expansion rigorously, we sommerfeld expand some polylogs

\begin{equation} \frac{N}{V} = \frac{g_s}{\lambda^3}f_{3/2}(z) \qq{and} \frac{E}{V} = \frac{3}{2}\frac{g_s}{\lambda^3}f_{5/2}(z) \end{equation}

the expansion tells us the low-temp expansion in $1/\log(z) = 1/\beta\mu$

\begin{equation} f_n(z) = \frac{(\log z)^n}{\Gamma(n+1)}\qty(1 + \frac{\pi^2}{6}\frac{n(n-1)}{(\log z)^2} + \dots) \end{equation}

whence we can find

\begin{equation} \frac{E}{N} = \frac{3E_F}{5}\qty(1 + \frac{5\pi^2}{12}\qty(\frac{k_BT}{E_F})^2 + \dots) \end{equation}

and get the heat capacity above.

Diatomic gas

rotation: (recall $2j+1$ degeneracy, sum over all $j$)

\begin{equation} E_\text{rot} = \frac{\hbar^2}{2I}j(j+1) \implies Z_\text{rot} \approx \begin{cases} \frac{2I}{\beta\hbar^2} & T \gg \hbar^2/2Ik_B \\ 1 & T \ll \hbar^2/2Ik_B \end{cases} \end{equation}


\begin{equation} E_\text{vib} = \hbar\omega(n+1/2) \implies Z_\text{vib} = \frac{1}{2\sinh(\beta\hbar\omega/2)} \approx \begin{cases} 1/\beta\hbar\omega & \text{high $T$} \\ \exp(-\beta\hbar\omega/2) & \text{low $T$} \end{cases} \end{equation}

where the low $T$ gives zero-point energy of QHO and doesnt contribute to $C_V$

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Debye Model of Solids

basically just follows from a linear dispersion (and polarization degeneracy)

\begin{equation} E = \hbar\omega = \hbar k c_s \implies g(\omega) = \frac{3V}{2\pi^2 c_s^3}\omega^2 \end{equation}

integrals taken up to a cutoff frequency $\omega_D$. To determine the cutoff, consider

\begin{equation} 3N \text{atomic dof} \implies \text{3N phonon dof} = \#\text{one-phonon states} = \int_0^{\omega_D}\dd{\omega}g(\omega) \end{equation}

which gives

\begin{equation} \omega_D = \qty(\frac{6\pi^2N}{V})^{1/3}c_s \end{equation}

and we can find energy and heat capacity the usual ways.

\begin{equation} C_V = \begin{cases} Nk_B\frac{12\pi^4}{5}\qty(\frac{T}{T_D})^3 & T\ll T_D \\ 3Nk_B & T\gg T_D \end{cases} \end{equation}

in low temp limit, integrate to infinity; in high temp limit expand integrand

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Electronic Gas in a Magnetic Field

pauli paramagnetism

effect from spin coupling to $B$:

\begin{equation} E \to E + \underbrace{\frac{|e|\hbar}{2m}}_{\mu_B}Bs \end{equation}

we can compute high temp ($z\sim 0$) magnetization

\begin{equation} M = -\pdv{E}{B} = -\mu_B(N_\uparrow - N_\downarrow) \approx \frac{2\mu_B Vz}{\lambda^3}\sinh(\beta\mu_BB) \approx \mu_BN\tanh(\beta\mu_BB) \end{equation}

and susceptibility

\begin{equation} \chi = \pdv{M}{B}\eval{}_{B=0} = \frac{N\mu_B^2}{k_BT} \end{equation}

at low temps, use expansion of $f_n(z)$ to find

\begin{gather} M \approx \frac{\mu_B^2 V}{2\pi^2}\qty(\frac{2m}{\hbar^2})^{3/2}\sqrt{E_F}B \approx \mu_B^2g(E_F)B\\ \chi \approx \mu_B^2g(E_F) > 0 \end{gather}

idea: only the $g(E_F)$ electrons on the surface are free to flip

Landau diamagnetism

effect from lorentz force (taking $B$ in the $+z$ direction)

\begin{equation} H = \frac{1}{2m}\qty(p + eA)^2 \end{equation}

solving the eigenvalue problem says energy states come in landau levels

\begin{equation} E = \hbar\omega_c\qty(n + \frac{1}{2}) + \frac{\hbar^2 k_z^2}{2m} \qq{for} n \in \mathbb{Z} \end{equation}

which have degeneracy

\begin{equation} \frac{L^2B}{2\pi\hbar/e} = \frac{\phi}{\phi_0} = \frac{\text{total flux}}{\text{flux quantum}} \end{equation}

we proceed to compute the magnetism

\begin{equation} M = \frac{1}{\beta}\pdv{\log\mathcal{Z}}{B} =-\frac{\mu_B^2}{3}g(E_F)B \end{equation}

using the partition function

\begin{align} \log\mathcal{Z} &= \frac{L}{2\pi}\int\dd{k_z}\sum_n\frac{2L^2B}{\phi_0}\log\qty[1 + z\exp\Big(-\frac{\beta\hbar^2k_z^2}{2m} - \beta\hbar\omega_c(n+1/2)\Big)]\\ &\approx \frac{Vm}{2\pi^2\hbar^2}\qty[(\text{const in $B$}) - \frac{(\hbar\omega_c)^2}{24}\int\dd{k}\frac{\beta}{\exp[\beta(\hbar^2k^2/2m-\mu)]+1}] \end{align}

this is comparable to pauli but of an opposite sign.

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Phase Transitions and Van der Waals gas

isotherms have that weird wiggle in a $p-v$ diagram below the critical temperature: thus the transition is marked by

\begin{equation} \dv{p}{v} = \dv[2]{p}{v} = 0 \end{equation}

below the critical point, we have weird compressibility and it’s broken: we use maxwell’s “lol just draw a straight line then” perscription (which comes from setting liquid and gas in chemical equilibrium, $\mu_\ell = \mu_g$ – can also equate GFE per particle)

clausius-clapeyron equation from looking at $p-T$ graph. coexistence region from $p-v$ squeezed into a line (think about traversing an isobar in the $p-v$ diagram and what it means in $p-T$ space). equality of gibbs gives

\begin{equation} \dv{p}{T} = \frac{s_g - s_\ell}{v_g - v_\ell} = \frac{L}{T(v_g - v_\ell)} \end{equation}

where we’ve defined the specific latent heat

\begin{equation} L = T(s_g - s_\ell) \end{equation}

this applies to any first-order transition; here we have

\begin{equation} S = -\pdv{F}{T} \qq{or} V =\pdv{G}{p} \end{equation}

as our first-derivative discontinuities

note that $S \text{ discontinuous} \implies C\sim \partial_T S$ goes to infinity – the temperature doesnt change as we pour heat into the system

We can solve the CC equation with a few assumptions (ideal gas, $v_g\gg v_\ell$, $L$ constant).

\begin{equation} \dv{p}{T} = \frac{Lp}{k_BT^2} \implies p=p_0e^{-L/k_BT} \end{equation}

really slick way of getting the critical point: start by rearrangign VdW:

\begin{equation} p = \frac{Nk_BT}{V-bN} - a \frac{N^2}{V^2} \iff pv^3 - (pb+k_BT)v^2 + av - ab = 0 \end{equation}

the critical point is defined by $\partial_vp=\partial^2_vp = 0$, so at the critical temperature, we only have this cubic term:

\begin{equation} p_c(v-v_c)^3 = 0 = p_cv^3 - (p_cb+k_BT_c)v^2 + av - ab \end{equation}

and we can compare term by term in $v$ to get

\begin{equation} k_BT_c = \frac{8a}{27b}\qquad v_c = 3b \qquad p_c = \frac{a}{27b^2} \end{equation}

another handy way of rewriting vdw is in terms of reduced variables; we divide by the critical value, and the equation takes the form

\begin{equation} \bar{p} = \frac{8}{3}\frac{\bar{T}}{\bar{v} - 1/3} - \frac{3}{\bar{v}^2} \end{equation}

which is the path toward the critical exponents

\begin{equation} v_g - v_\ell \sim (T_c - T)^{1/2} \qquad p-p_c \sim (v-v_c)^3 \qquad \kappa = -\frac{1}{v}\pdv{v}{p}\eval{}_T\sim(T-T_c)^{-1} \end{equation}

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Ising Model

\begin{equation} E = -J\sum_{\ev{ij}}s_is_j - B\sum_i s_i \end{equation}

where we’re interested in

\begin{equation} m = \frac{1}{N}\sum\ev{s_i} = \frac{1}{N\beta}\pdv{\log Z}{B} \end{equation}

mean field approximation: write spins in terms of deviation from average and assume that fluctuations are small

\begin{equation} s_is_j = \cancel{(s_i-m)(s_j-m)}+m(s_j-m)+m(s_i-m)+m^2 \end{equation}

so the energy becomes

\begin{equation} E = \frac{1}{2}JNqm^2 -\underbrace{(Jqm+B)}_{B_{\text{eff}}}\sum s_i \end{equation}

and we find, since each spin acts independently,

\begin{equation} Z = e^{-\frac{1}{2}\beta JNqm^2}2^N\cosh^N\beta B_\text{eff} \implies m = \tanh(\beta B + \beta Jqm) \end{equation}

zero magnetic field

when $\beta Jq < 1$ the only solution for $m$ is $m=0$: there is no average magnetization at high temperatures. if the temperature is low enough, however, we have an unstable solution at $m=0$ and two stable solutions at $m=\pm m_0$, and in the limit of zero temp $m\to\pm 1$ (all spins aligned). as we vary $T$, we have a singularity in $\partial_T m$:

second order transition as we vary $T$

(note: high temperature expansion gets into some stat field theory and RG stuff – possibly important to know?)

nonzero magnetic field

there is no longer a phase transition for a fixed $B$ as $T$ varies: at large temps, magnetization to zero as

\begin{equation} m \sim \frac{B}{k_BT} \end{equation}

and at small temps all spins align with the $B$ field (no choice to make). drawing an $m-T$ graph shows how turning on $B$ separates and smooths out what was a singularity in the $B=0$ case. however, if we vary $B$ and swap its direction, the magnetization (a first derivative) jumps discontinuously:

first order transition as we vary $B$ from negative to positive and $T<T_C = Jq/k_B$

the critical exponents we get

\begin{equation} m_0\sim \pm(T_c-T)^{1/2}\qquad m\sim B^{1/3}\qquad\chi\sim (T-T_c)^{-1} \end{equation}

are the same as for VdW

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Simplest model: the polymer as a random walk. You get a binomial distribution, which approaches a gaussian in the large $N$ limit. in 1D,

\begin{equation} P(x) = \frac{1}{\sqrt{2\pi}\sigma}\exp(-x^2/2\sigma^2) \xrightarrow{\sigma^2 = Na^2} \frac{1}{\sqrt{2\pi N}a}\exp(-x^2/2Na^2) \end{equation}

for $d$ dimensions each $\sigma_x^2$ gets divided by $d$ since the total $\sigma^2$ is basically the sums of the individual dimensional walks – in each dimension you only have to walk $1/d$ of the way there (this is not valid reasoning but it’s a way to remember it)

Microcanonical perspective fix $X$ and calculate $F(X)$

\begin{equation} \dd{U} = 0 = T\dd{S} + F\dd{X} \end{equation}

so if we can find the entropy we can take a derivative to find $F(X)$. The number of states that have a length $X$ is just $N\cdot P(x)$, so entropy comes immediately from the gaussian above.

Canonical perspective fixing $F$ to calculate $X$

we note that $X = \sum a\cos\theta_i$, so we can just use $E = -FX$

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Brownian motion

idea: large particle of radius $b$ suspended in a fluid. we have a stokes law velocity-dependent damping force and some random force that is time-uncorrelated, ie $\ev{F(t)F(t’)} = c\var{(t-t’)}$

\begin{equation} m\partial_t^2r + 6\pi\eta b\partial_tr = \partial_t^2r + \alpha \partial_t r = F \end{equation}

can reduce the order and get solutions

\begin{equation} v(t) = A(t)e^{-\alpha t} \implies \dot{A} = e^{\alpha t}F/m \end{equation}

where we can now pick out a $v(t)$ by integrating $\dot{A}$ back again

\begin{equation} v(t) = \frac{1}{m}\int_0^t\dd{s} e^{-\alpha (t-s)}F(s) \end{equation}

relating our random variable $F$ to a new random variable $v$. For a fixed $T$, we can determine the $c$ in $\ev{FF}$: we calculate $\ev{v^2}$ both according to this description of $v$ and from taking a boltzmann (canonical ensemble) probability distribution.

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