Past exam of the thermal and statistical physics course of the University of Cambridge / 2014 / Paper 1 (THERMAL AND STATISTICAL PHYSICS, PHYSICS (1))

Original paper: Public original PDF, physical pages 287–290; cover on page 287 and questions on pages 288–290.

For one oscillator, set $x=\hslash \Omega /(k_{B}T)$. Its quantum-oscillator energy and heat capacity areAt low temperature, $U\to \hslash \Omega /2$ and $C_V\sim k_B{x}^2{e}^{-x}\to 0$: excitations freeze out but zero-point energy remains. At high temperature, $U\sim k_{B}T$ and $C_V\to k_B$, recovering equipartition. The zero-point term does not contribute to heat capacity.

Using $C_V=T(\partial S/\partial T{)}_V$ and the stated Maxwell relation, interchange derivatives within a smooth single phase:Therefore $C_V$ depends only on temperature for the given temperature-linear equation of state. It need not be constant in temperature.

For $N$ identical classical noninteracting particles, use $Z_N=Z_1^N/N!$. At fixed $N$,so the canonical partition-function identities giveThus, for one mole,The supplied $T^{3/2}$ dependence includes only three translational degrees of freedom; additional thermally active internal modes would change the heat capacity.

For one Cartesian velocity component of a free Brownian particle, the inertial Langevin equation isThe random force and viscous drag obey the fluctuation-dissipation relation. With a thermal equilibrium velocity distribution, the relaxation time is $\tau =m/\zeta$ andIntegrating velocity twice givesFor $t\gg \tau$, this is Einstein's diffusion law, $⟨\Delta x^2⟩=2Dt$, with $D=k_{B}T/\zeta$. In three dimensions $⟨|\Delta \mathbf{r}{|}^2⟩=6Dt$. At $t\ll \tau$ the motion is ballistic, $⟨\Delta x^2⟩\simeq (k_{B}T/m)t^2$. A sphere in the Stokes-drag regime has $\zeta =6\pi \eta a$, giving $D=k_{B}T/(6\pi \eta a)$.

Landau theory expands the free energy in an order parameter. For a scalar with sign-reversal symmetry and zero conjugate field, a useful sextic even Landau potential isEquilibrium selects the global minimum; other local minima can be metastable.

For $b>0$, the minimum changes continuously from $m=0$ to $m\simeq \pm \sqrt{-r/b}$ as $r$ passes below zero. This is a second-order transition at $T_0$: the order parameter is continuous, there is no latent heat, and thermodynamic derivatives such as heat capacity can be discontinuous.

For $b<0$, the sixth-order term stabilizes the free energy. Equality of the ordered and disordered minima, together with $f^{\prime }(m)=0$, givesThe order parameter jumps at $T_t=T_0+r_t/a$. The entropy generally jumps too, giving latent heat; metastable branches and barriers permit hysteresis, though equilibrium coexistence itself does not require hysteresis. The special case $b=0$ is tricritical. Mean-field exponents can fail where fluctuations dominate.

At $T\ll T_F$, Pauli exclusion restricts thermally changed occupations to an energy interval of order $k_{B}T$ about the Fermi energy. The affected fraction is of order $T/T_F$; their typical excitation energy is of order $k_{B}T$, so the thermal energy correction is proportional to $T^2$ and the heat capacity is linear in $T$.

More precisely, with $g(E_F)$ the total density of one-particle states per unit energy, including spin,For a three-dimensional ideal gas with quadratic dispersion, $g(E_F)=3N/(2E_F)$, henceThe fixed-particle-number Sommerfeld expansion gives $U=3N{E}_F/5+({\pi }^2/4)N(k_{B}T{)}^2/E_F+\cdots$. The temperature dependence of the chemical potential must be included when obtaining this expression; holding it artificially fixed changes the intermediate energy correction.

For a single quantum state, the equilibrium occupations areBoth reduce to $e^{(\mu -ϵ)/(k_{B}T)}$ when $ϵ-\mu \gg k_{B}T$.

For one three-dimensional quadratic branch, the density of states per unit volume isWith ${\lambda }_T=\sqrt{2\pi {\hslash }^2/(m{k}_{B}T)}$, changing integration variable to $x=ϵ/(k_{B}T)$ givesA degeneracy factor multiplies both densities if additional equivalent branches are present. The first expression is the noncondensed density: a macroscopically occupied ground state must be counted separately.

Bose-Einstein condensation occurs when the excited states can no longer hold all particles and the ground-state occupation becomes macroscopic. At fixed total density,For field-induced triplon condensation, write $\Delta =\nu n-\eta (H-H_g)$, so $ϵ(k)={\hslash }^2{k}^2/(2m)+\Delta$. At fixed reference $n$, the three requested dispersions have gaps $\nu n+\eta H_g$, $\nu n$ and zero respectively. Their curvature is unchanged. Because the actual thermodynamic chemical potential is zero, the Bose distribution requires $\Delta \ge 0$. At $H_c=H_g+\nu n/\eta$, the minimum reaches zero and the noncondensed population saturates, permitting macroscopic occupation.

For the triplon condensation field scaling, the critical density at onset is $n_T=\zeta (3/2)/{\lambda }_T^3$. ThereforeThe low-temperature phase boundary rises from $(T,H)=(0,H_g)$; the condensed phase lies on the higher-field side.

For $H>H_c$, simply continuing the gap below zero would give unphysical Bose occupations. For the condensate density in a shifted-parabola triplon model, added condensate density keeps the minimum at zero: $\nu (n_0+n_T)=\eta (H-H_g)$. At fixed temperature this giveswith $n_0=0$ below onset. At zero temperature it becomes $n_0=\eta (H-H_g)/\nu$ within that model.

This prefactor is conditional on continuing the stated single-density mean-field shift into the condensed phase. A microscopic interacting Bose condensate has collective Bogoliubov excitations $E_k=\sqrt{{ϵ}_k^0({ϵ}_k^0+2g{n}_0)}$, with ${ϵ}_k^0={\hslash }^2{k}^2/(2m)$, which are linear at small $k$. The normal-state shift alone does not fix the condensed interaction closure or identify $g$ with $\nu$. The gap closes and the condensate density grows linearly with excess field in the simple mean-field description; the detailed collective dispersion needs that additional interaction model. A dilute-boson treatment of field-induced magnetic ordering makes this normal/condensed distinction explicitly.^[ref]

Use $v$ for molar volume to distinguish it from the fluctuating parcel volume below. The equation is $p=RT/(v-b)-a/v^2$. The parameter $b>0$ models excluded volume, while $a>0$ models the reduction in pressure from attractive interactions.

Below the critical temperature, a raw isotherm has a loop. The positive-slope segment is mechanically unstable and surrounding metastable portions do not give the stable equilibrium mixture. Replace the loop by a constant coexistence pressure $p_s$ satisfying the Maxwell equal-area condition,The endpoints are liquid and vapour molar volumes; intermediate mean volumes correspond to phase mixtures with changing proportions. Above the critical temperature, liquid-like and gas-like states connect continuously without a coexistence plateau.

In the pressure-temperature diagram, the liquid-vapour coexistence line ends at the critical point. Below it, the low-pressure side is vapour and the high-pressure side liquid. “Gas” conventionally denotes the region above the critical temperature, where compression alone cannot cross a liquid-vapour transition. The model has no solid phase.

The critical isotherm has an inflection: $p_v=p_{vv}=0$. Thuswhich giveFor fluctuations about a stable equilibrium value $x_0$, the system plus reservoir has probability proportional to $e^{-A(x)/(k_{B}T)}$. Expanding availability,gives a Gaussian distribution and, to this order, $⟨x⟩=x_0$ withThis requires positive curvature and sufficiently small fluctuations; it is not exact when higher-order terms matter.

The whole cylinder's volume is fixed. For the volume fluctuations of a small fluid parcel, interpret the small quantity as a material parcel containing $s$ moles, whose volume can fluctuate against the much larger remainder. Its mean volume at the critical density is $\bar{V}=s{v}_c=3sb$. At fixed temperature its availability is $A(V)=F_s(V,T)+p_{\mathrm{bath}}V$, soOn the critical isochore, approached from $T>T_c$,ConsequentlyEquivalently, $⟨\delta V^2⟩=k_{B}T{\kappa }_T\bar{V}$, with ${\kappa }_T=4b/[3R(T-T_c)]$. The mean-field variance diverges as $(T-T_c{)}^{-1}$, so its critical exponent is $1$; the root-mean-square fluctuation has exponent $1/2$. A fixed spatial observation box instead has number fluctuations, not fluctuating geometric volume. Finite size and non-Gaussian terms cut off the literal divergence; the approximation fails exactly at the critical point.