Statistical Mechanics and Thermodynamics

Entropy change in thermodynamic processes

Entropy is a state function, so the entropy change of a system between two equilibrium states depends only on the initial and final states and not on the path taken. For a body with constant heat capacity C taken reversibly from T_1 to T_2, \Delta S = \int C\,dT/T = C\ln(T_2/T_1). Cooling n moles of an ideal gas at constant pressure gives \Delta S = nC_p\ln(T_2/T_1), while a reversible isothermal phase change at temperature T with latent heat L gives a magnitude |\Delta S| = L/T, with the sign set by whether heat leaves or enters the system. A free expansion of an ideal gas is irreversible, yet its temperature stays fixed and entropy depends only on T and V, so the entropy change equals that of a reversible isothermal expansion between the same volumes, \Delta S = nR\ln(V_2/V_1), independent of the temperature at which it occurs. When C_p is given as a function of temperature, the change is found by integrating C_p/T; for example, C_p = AT + BT^3 gives \Delta S = \int (A + BT^2)\,dT at constant pressure. For a reservoir at temperature T_{res} that supplies heat Q, its entropy change is -Q/T_{res}, and the entropy of the universe is the sum of the body's change and the reservoirs' changes. Because the body's entropy change is path-independent, a sequence of reservoirs with temperatures close to the body's temperature at each step generates less total entropy, whereas a single step through a reservoir far from the body's temperature generates the most. When the hot and cold sources are finite bodies with equal heat capacity C, a reversible engine runs until both reach the common temperature T_f = \sqrt{T_1T_2}, where the total entropy change is zero, and the maximum work is W = C(T_1 + T_2 - 2\sqrt{T_1T_2}). For instance, bodies at 400 K and 100 K give T_f = 200 K and W = 100C J.

Partition functions of quantum systems

The canonical partition function is Z = \sum_s e^{-\beta E_s} with \beta = 1/k_BT, and the Helmholtz free energy follows from F = -k_BT\ln Z, while the internal energy is U = -\partial \ln Z/\partial\beta. For a one-dimensional harmonic oscillator the levels are E_n = (n + \tfrac{1}{2})\hbar\omega, so Z is a geometric series with ratio e^{-\beta\hbar\omega}, giving Z = e^{-\beta\hbar\omega/2}/(1 - e^{-\beta\hbar\omega}) and F = \hbar\omega/2 + k_BT\ln(1 - e^{-\beta\hbar\omega}). The term \hbar\omega/2 is the zero-point energy and must be kept in F. When \beta\hbar\omega = 2, the same sum reduces to Z = e/(e^2-1), so the value of the ratio is found by multiplying numerator and denominator by e^2. For identical particles, the counting depends on statistics: distinguishable particles give Z = z^2 for single-particle partition function z, classical indistinguishable particles give z^2/2!, and quantum particles restrict occupation, so fermions and bosons give different sums of Boltzmann factors over pairs of single-particle states. A two-level system with ground energy \epsilon_1 and gap \Delta has U(T) that starts from the nonzero plateau N\epsilon_1 at T \to 0, rises monotonically with T, and saturates at N(\epsilon_1 + \Delta/2) as T \to \infty.

Two-level paramagnet populations

For a two-level system, the number of microstates is a binomial coefficient: with N spins of which n point up, there are \binom{N}{n} equally probable arrangements, and the net moment is (n_\uparrow - n_\downarrow)\mu_B. The probability of a given net moment is therefore the number of arrangements with that moment divided by the total number of arrangements. Entropy is S = k_B\ln\Omega, where \Omega counts the accessible microstates. For a lattice of N spin-\tfrac{1}{2} ions, the paramagnetic state allows all 2^N configurations, whereas a fully aligned ferromagnetic state and an alternating antiferromagnetic state each allow only two configurations in the idealized counting, so S_P = Nk_B\ln 2 exceeds the ordered entropies for N > 1. For levels at \pm\varepsilon with E = \varepsilon(N_+ - N_-), Stirling's approximation gives S = k_B(N\ln N - N_+\ln N_+ - N_-\ln N_-), and differentiation at fixed N yields \partial S/\partial E = \dfrac{k_B}{2\varepsilon}\ln(N_-/N_+). Since 1/T = \partial S/\partial E, with N_\pm = \tfrac{1}{2}(N \pm E/\varepsilon) up to the level assignment, the temperature is positive when the lower level is more populated and becomes negative when the upper level is more populated.

Bose and Fermi statistics of identical particles

Identical particles are classified by spin: integer spin gives bosons, which obey symmetric wavefunctions and may share a state, and half-integer spin gives fermions, which obey antisymmetric wavefunctions and obey the Pauli exclusion principle. A composite atom is a boson or fermion according to the parity of its number of fermionic constituents, so ^4He with six such constituents is a boson, while ^3He with five is a fermion. For two particles in g non-degenerate levels, the symmetric (bosonic) states number \binom{g+1}{2}, of which \binom{g}{2} have the particles in different levels, while the antisymmetric (fermionic) states number \binom{g}{2}, all of which have different levels. Comparing these counts gives the probabilities of finding the two particles in distinct levels. A photon has zero rest mass but finite energy h\nu and momentum \hbar k, its spin is 1 so its spin angular momentum is not zero, and it has two independent polarization states corresponding to helicities \pm 1. For two identical particles in a harmonic trap with fixed total energy, the number of states depends on spin: spin-\tfrac{1}{2} fermions have two spin states per orbital, so a pair in two distinct orbitals gives four states, whereas spinless bosons in the same orbitals give a single symmetric state, and S = k_B\ln\Omega follows from these counts.

Fermi gas chemical potential

The Fermi-Dirac occupation f(E) = 1/(e^{(E-\mu)/k_BT}+1) equals exactly 1/2 at E = \mu for every temperature, so the chemical potential is the energy at which the occupation probability is one half. The Sommerfeld expansion gives \mu(T) = E_F - \dfrac{\pi^2}{6}(k_BT)^2\dfrac{g'(E_F)}{g(E_F)} for a density of states g(E). Therefore, when the density of states increases with energy, g'(E_F) > 0, and \mu decreases as the temperature rises; the sign of the temperature dependence is set by the slope of g(E). For a Bose gas in a trap at T = 0, all particles occupy the lowest single-particle state, so the chemical potential approaches that ground-state energy from below. For an isotropic three-dimensional harmonic trap the ground-state energy is E_0 = \tfrac{3}{2}\hbar\omega, and this is the value of \mu at T = 0.

Helmholtz free energy thermodynamic relations

The Helmholtz free energy is defined as F = U - TS, with dF = -S\,dT - P\,dV, so the entropy at constant volume is S = -(\partial F/\partial T)_V. Substituting into U = F + TS gives U = F - T(\partial F/\partial T)_V, which can be written compactly as U = -T^2\,\partial(F/T)/\partial T at constant volume. The chemical potential is \mu = (\partial F/\partial N)_{T,V}. For an ideal gas the canonical partition function is Q = (V/\lambda^3)^N/N!, and using \ln N! \simeq N\ln N - N gives F = -Nk_BT\left[\ln\dfrac{V}{N\lambda^3} + 1\right] and \mu = k_BT\ln\dfrac{N\lambda^3}{V}. Hence e^{\mu/k_BT}/\lambda^3 = N/V, so this ratio equals the number density.

Joule-Thomson effect

The Joule-Thomson process is a steady throttling of a gas through a valve or porous plug. It is adiabatic and does no work on the outside, so the enthalpy is conserved, h_1 = h_2, and the process is isenthalpic. Throttling is irreversible and entropy increases, so the process is not isentropic. The temperature change is governed by the coefficient \mu_{JT} = (\partial T/\partial P)_H, which is positive below the inversion temperature, where the gas cools, and negative above it, where the gas heats; hydrogen at room temperature, for example, warms on throttling. For an ideal gas, h depends only on T, so \mu_{JT} = 0 and throttling produces no temperature change.

Clausius-Clapeyron triple point

The Clausius-Clapeyron relation dP/dT = L/(T\,\Delta V) describes the slope of every first-order coexistence curve, including melting, sublimation and vaporization. The melting line is usually much steeper than the vaporization line because the volume change on melting is small, which makes |dP/dT| large. The critical point is the endpoint of the liquid-vapour coexistence curve only, and the melting and sublimation curves do not terminate at a critical point. The triple point is where the solid and liquid vapour-pressure curves intersect. If \ln P_s = a - b_s/T and \ln P_l = a' - b_l/T, equating them gives T_{tp} = (b_l - b_s)/(a' - a). With a = 23.03, b_s = 3754, a' = 19.49 and b_l = 3063, this yields T_{tp}\approx 195.2 K.