Past exam of the thermal and statistical physics course of the University of Cambridge

The 2022 paper is omitted because the released copy includes worked solutions. 2022 paper with solutions, PDF pages 448–455

The fixed common temperature, rigid external container and fixed particle numbers on either side make the total Helmholtz free energy the appropriate thermodynamic potential. Let the barrier have cross-sectional area $A$, and let increasing $x$ expand the left compartment. Then $d{V}_1=Adx$ and $d{V}_2=-Adx$. At fixed temperature and particle number, $d{F}_j=-p_{j}d{V}_j$, soThus mechanical equilibrium, $p_1=p_2$, makes $F$ stationary. The physical force on the barrier towards increasing $x$ is $f=A(p_1-p_2)=-F^{\prime }(x)$. If the barrier is displaced slightly from $x_{\ast }$, strict mechanical stability requires a restoring force:ConsequentlyThe Taylor expansion $F(x_{\ast }+\delta x)=F(x_{\ast })+\frac{1}{2}{F}^{\prime \prime }(x_{\ast })\delta x^2+O(\delta x^3)$ proves that thermodynamic equilibrium is a strict local minimum of the Helmholtz free energy, rather than merely a stationary point.

For individually stable gases, the positive isothermal compressibilities ${\kappa }_{T,j}=-(1/V_j)(\partial V_j/\partial p_j{)}_{T,N_j}$ make the sign especially transparent:If these inequalities hold throughout the accessible interval, $F$ is a strictly convex function and an interior stationary point is the unique global minimum of the Helmholtz free energy. Stability at one point alone proves only the local statement. A vanishing restoring derivative describes marginal stability and requires examining higher-order terms.

An additive-looking energy expression does not by itself establish two noninteracting subsystems. Their allowed microstates must also form a Cartesian product of independently available states. Here the second label's allowed values depend on the first. The four allowed pairs and their energies areThe canonical partition function, with $\beta =(k_{B}T{)}^{-1}$, is thereforeThe conditional sum over $m_J$ is $2\cosh (\beta ϵJ)$ and depends on $J$; it cannot be pulled out as the partition function of an independent second subsystem. Summing independently over $J\in \{1,2\}$ and all four values $m\in \{-2,-1,1,2\}$ would invent four forbidden pairs and give eight states instead of four.

One can make the state labels independent by writing $s=m_J/J=\pm 1$, but then $E(J,s)=ϵJ(1+s)$ contains a coupling between them. For genuinely additive independent energies, the difference between the $s=+1$ and $s=-1$ states would be independent of $J$; here it is $2ϵJ$. Hence the proposed $J$ and $m_J$ subsystems cannot be treated as noninteracting independent subsystems. The literal two-value reading of the PDF's endpoint notation has been used. If that notation were instead intended as shorthand for full angular-momentum multiplets, including every integer between the endpoints, the eight allowed states would giveThe conclusion about the failure of independent-subsystem factorization is unchanged. The two state-counting conventions should not be mixed.

A nonzero superconducting energy gap suppresses the population of thermally excited quasiparticles. When $k_{B}T\ll \Delta$, their Boltzmann factor produces an approximately activated electronic heat capacity,where the prefactor $A(T)$ varies algebraically rather than exponentially. For the rough estimate requested, neglect its change between the two quoted temperatures. Taking a ratio eliminates its unknown magnitude:ThereforeThe corresponding excitation energy is $\Delta \simeq 2.19\times 10^{-22}\mathrm{J}\simeq 1.36\mathrm{meV}$. The units $\mathrm{mJ}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$ of the two heat-capacity readings cancel in their ratio.

This is a heat-capacity activation estimate, not an exact determination of a gap from two data points. More generally, if $C_e=A{T}^p{e}^{-\Delta /(k_{B}T)}$, the same measurements giveFor example, the low-temperature BCS theory form for a fully gapped isotropic superconductor has $p=-3/2$ and would give about $19.5\mathrm{K}$. The stated approximate data and constant-gap assumption support the simpler $16\mathrm{K}$ estimate; they do not specify a detailed density of states or heat-capacity prefactor. Here $\Delta$ means the minimum quasiparticle excitation energy, rather than automatically twice that value.

Put $\beta =1/(k_{B}T)$ and define the thermal de Broglie wavelength of species $\alpha$ byIgnoring the internal degrees of freedom as stipulated, each molecule's translational canonical partition function is $V/{\lambda }_{\alpha }^3$. The two species do not interact, so their partition functions multiply; the treatment of indistinguishable particles requires a separate factorial for each species, not a factorial of the total particle number. ThusUsing $F=-k_{B}T\ln Z$ and Stirling's approximation gives the extensive Helmholtz free energyThese are ideal gas expressions, valid for $n_{\alpha }{\lambda }_{\alpha }^3\ll 1$, with $n_{\alpha }=N_{\alpha }/V$. The energy zero has been chosen so that separated CO and oxygen molecules have zero internal reference energy.

The given binding energy $ϵ$ belongs to the formation of two carbon-dioxide molecules. Relative to the reference used in (a), the internal energy offset of one carbon-dioxide molecule is therefore $-ϵ/2$, not $-ϵ$. Its mass is $m_c=m_{\mathrm{CO}}+m_{{\mathrm{O}}_2}/2$ and its thermal de Broglie wavelength is ${\lambda }_c=h/\sqrt{2\pi m_c{k}_{B}T}$, where the subscript $c$ denotes $\mathrm{C}{\mathrm{O}}_2$.

The single-molecule canonical partition function is $(V/{\lambda }_c^3)e^{\beta ϵ/2}$. For $N_c$ indistinguishable noninteracting molecules,The corresponding Helmholtz free energy, using Stirling's approximation, isThe negative sign in the last term expresses the stabilizing binding energy and ensures that low temperature favors association.

At fixed temperature, volume and the numbers of all other species, the chemical potential is ${\mu }_{\alpha }=(\partial F/\partial N_{\alpha }{)}_{T,V,N_{{\alpha }^{\prime }\ne \alpha }}$. For the combined noninteracting mixture, add the free energies from (a) and (b). Since the thermal de Broglie wavelengths are independent of particle number at fixed temperature, differentiation givesEach logarithm has a dimensionless argument, and each chemical potential has units of energy per molecule. The binding offset must be kept consistently when comparing species.

Introduce an extent of reaction $\xi$ increasing in the dissociation direction. The allowed particle-number changes conserve both carbon and oxygen atoms:At fixed temperature and volume, the resulting Helmholtz free energy change isAt an interior chemical equilibrium, both forward and backward changes are allowed, so the coefficient must vanish:Equality of the three individual chemical potentials would be wrong: a reaction exchanges species in its stoichiometric proportions rather than molecule for molecule. Stability is also explicit here, sincewhen all three species are present. Thus the equilibrium composition minimizes the free energy along the conserved-composition constraint.

Substituting the chemical potentials into the equilibrium relation givesExponentiation and the positive square root give the law of mass action in number-density form:All three thermal de Broglie wavelengths have the same $T^{-1/2}$ dependence. Define the temperature-independent mass factorIf $f_{\alpha }=n_{\alpha }/n$, this exact equilibrium relation can be writtenIt already has the requested dimensional structure: $n^{3/2}{\lambda }_c^{3/2}$ has units ${\mathrm{m}}^{-3}$. Its coefficient, however, depends on the conserved elemental composition through the species fractions.

For the intended case of a gas obtained by dissociating initially pure carbon dioxide, conservation of the elemental ratio gives $n_{\mathrm{CO}}=2n_{{\mathrm{O}}_2}$ throughout. Henceand the exact relation within this ideal-mixture model isAt high temperature with $k_{B}T\gg ϵ$ and in the dilute ideal gas regime, $n{\lambda }_c^3\ll 1$, this equation is self-consistently dissociation dominated:Thus $n-n_c\simeq n$, and to leading orderUsing nominal molecular masses $28,32,44$ in common mass units gives $A\simeq 0.963$, confirming that it is of order unity. The exponential has been retained as the leading binding correction, although it tends to one as $T\to \infty$.

The printed question does not explicitly fix the elemental composition. Without the initially-pure-$\mathrm{C}{\mathrm{O}}_2$ assumption, the universal result is the boxed mass-action relation, and $A=B{f}_{\mathrm{CO}}\sqrt{f_{{\mathrm{O}}_2}}$ approaches a composition-dependent constant when both dissociation products have nonzero high-temperature fractions. It need not be of order unity for an arbitrarily unbalanced mixture. Also, $k_{B}T\gg ϵ$ alone does not justify replacing $n-n_c$ by $n$ at arbitrarily high density; the dilute classical regime is part of the model.

A phase transition is a qualitative change of thermodynamic phase as control parameters such as temperature, pressure or chemical potential are varied. A genuine equilibrium singularity is a nonanalyticity of the appropriate thermodynamic potential in the thermodynamic limit; finite systems generally show rounded crossovers instead.

An order parameter is a macroscopic quantity that distinguishes the phases. It often vanishes in a symmetric disordered phase and becomes nonzero after spontaneous symmetry breaking, although density-driven transitions need not have that simple symmetry interpretation. It can be a scalar, vector or tensor. At a first-order phase transition an appropriate first derivative of the free energy is discontinuous, and the order parameter commonly jumps; at a continuous phase transition the order parameter can approach zero continuously while a susceptibility or a higher derivative becomes singular.

Three examples are:

These examples also illustrate why an order parameter must reflect the physical distinction between phases rather than merely be any quantity that changes with temperature.

For $N$ indistinguishable classical particles the canonical partition function is an integral over phase space,The factor $1/N!$ removes overcounting under particle permutations, and $h^{3N}$ makes the classical phase-space measure dimensionless. The pair potential depends only on positions, so all momentum integrations separate. The three-dimensional Gaussian integral for each particle givesThe grand canonical partition function is $\Xi ={\sum }_{N=0}^{\infty }{e}^{\beta \mu N}{Z}_N$, where chemical potential $\mu$ controls the mean number of particles. ThereforeFor $N=0$, the empty products and configurational integral equal one. The factor $1/2$ in the interaction prevents double counting of unordered particle pairs.

Write $M=\mathcal{N}=V/b$ for the number of cells; this is fixed even though the particle number $N$ fluctuates. Approximate each position integral by the cell sum $\int d^3{r}_i\simeq b{\sum }_{{\alpha }_i=1}^M$, evaluating the slowly varying intercell pair potential at cell centres. With the assumed single-occupancy restriction, the fixed-$N$ contribution becomesAn occupation configuration $\{n_{\alpha }\}$ with $N={\sum }_{\alpha }{n}_{\alpha }$ selects an unordered set of $N$ occupied cells. There are exactly $N!$ ways to assign the temporary particle labels to that set. They all have the same interaction energy, cancelling the prefactor $1/N!$.

Summing over all particle numbers is therefore equivalent to summing each cell occupation independently over $0,1$:The activity per cell, $e^{\beta \mu }b/{\lambda }^3$, is dimensionless. This is a lattice gas, with each cell either empty or occupied.

The exclusion of multiple occupation is an additional physical modeling assumption, explicitly imposed by the question. Making a cell small does not by itself forbid two point particles from entering it; a repulsive core or a dilute-occupancy approximation would be needed to justify that restriction starting from a general continuum pair potential. In the remaining parts it is part of the defined cell model.

Let $q$ be the coordination number of a lattice, the number of nearest neighbours of each cell away from boundaries. Define an effective chemical potential incorporating the cell's translational entropy:Use $⟨\alpha {\alpha }^{\prime }⟩$ for an unordered nearest-neighbour bond, counted once. The occupation Hamiltonian including the chemical-potential term isUnder $n_{\alpha }=(1+{\sigma }_{\alpha })/2$, one bond contributesIn a regular lattice with periodic boundary conditions, the number of bonds is $Mq/2$, and each spin occurs in $q$ linear bond terms. ThereforeThe pair sum equals one half of the ordered sum used in the paper. Consequently the effective field isThus $\Xi ={\sum }_{\{\sigma \}}{e}^{-\beta {\mathcal{H}}_{\mathrm{eff}}}$ is the Ising model partition sum, including the displayed spin-independent constant if its absolute normalization is required. Dropping that constant changes $\Xi$ by a multiplicative factor but leaves spin probabilities and the phase-transition argument unchanged. For simple cubic cells in three dimensions, $q=6$ and the first term in $h$ is $6J$. Open boundaries replace $q$ by the local coordination near the surface; the stated uniform field is the bulk or periodic-lattice result.

For attractive nearest-neighbour interactions, $J>0$, the transformed model is a ferromagnetic Ising model. In three dimensions and at $T<T_c$, its infinite-volume spontaneous magnetization has two branches. As the effective field crosses zero, the equilibrium mean spin changes discontinuously from a negative value to a positive value. Since $(\partial h/\partial \mu {)}_T=1/2$, varying the physical chemical potential crosses this same first-order phase transition.

The particle-hole-symmetric coexistence line isThe mean Ising spin is a rescaled occupation density. Writing $n=⟨N⟩/V$ for the physical number density,The negative-spin phase is a dilute gas and the positive-spin phase is a dense, liquid-like phase. If the spontaneous spin magnitude at coexistence is $m_0(T)>0$, their number densities areThus the physical order parameter is a density or occupation imbalance, not an actual magnetic moment of these particles. In a closed system with an intermediate mean density, the same transition appears as phase coexistence between the two densities. At $T=T_c$ the density contrast vanishes continuously; above $T_c$, sweeping the chemical potential gives no coexistence jump. A true discontinuity requires the thermodynamic limit; a finite sum over finitely many cells remains analytic.

Let each step have length $a$, and let $K$ be the number of right steps among $N$ independent steps. Then $K$ has a binomial distribution and the physical displacement is $x=a[K-(N-K)]=a(2K-N)$. Therefore $K=(N+x/a)/2$, and the position probability mass isThis is zero unless $x/a$ is an integer, $|x/a|\le N$, and $N+x/a$ is even. The parity condition matters because the walk alternates between the two sublattices. If distance is measured in lattice spacings, replace $x/a$ by $x$.

Normalization follows from the binomial theorem. Its first two moments provide useful checks:These are exact discrete-time statements; a smooth continuum probability density requires coarse-graining over the lattice and its parity oscillation.

The one-step master equation for the biased random walk isExpand in a Taylor series and retain the first time derivative and the first two space derivatives. Since $w_{+}+w_{-}=1$ and $w_{-}-w_{+}=2ϵ$,Thus the usual continuum truncation gives the requested Fokker-Planck equation,The diffusion coefficient has units $\mathrm{lengt}{\mathrm{h}}^2/time$, while $C$ has units of velocity. The physical drift velocity is $v=-C=a(w_{+}-w_{-})/\tau$, because the standard advection term is $-v{\partial }_{x}P$. A rightward bias therefore means $ϵ<0$, $C<0$, and $v>0$.

The approximation must be stated carefully. The truncation is a controlled diffusion limit of a weakly biased random walk when $a\to 0$, $\tau \propto a^2$, and $ϵ\propto a$, keeping $D_0$ and $C$ finite. It is also the spatial diffusion approximation for a continuous-time jump process with exponentially distributed waiting times of mean $\tau$.

If the original fixed-step-time walk has a finite bias, its exact variance from (a) instead gives the diffusion coefficient of the Gaussian envelope about the drifting mean:One obtains the same correction by retaining $\frac{1}{2}{\tau }^2{\partial }_t^{2}P$ in the time expansion and using the leading advection equation to replace ${\partial }_t^{2}P$ by $v^2{\partial }_x^{2}P$. The result is $D_{\mathrm{fixed}}=D_0-v^2\tau /2$. In particular, a deterministic one-way walk has zero variance, as this expression requires. Thus $D_0$ is the usual intended first-time-derivative approximation, not an exact variance-growth coefficient at arbitrary bias. The later continuum estimates use that convention, with the fixed-time alternative identified where relevant.

For this radial random walk on a regular tree, the drawing is an infinite regular tree of degree $z$, not a planar lattice with closed loops. The radial coordinate is the number of edges in the unique path to the root. A non-root vertex has one neighbour inward and $z-1$ neighbours outward, so equal probabilities for its $z$ edges giveLet $r$ be radial distance measured in edge units. For $r>1$, probability arriving in shell $r$ comes from shell $r-1$ by an outward step or from shell $r+1$ by an inward step:This has precisely the same bulk master equation as the one-dimensional biased random walk, withThe outward bias is geometric: beyond the first shell, the number of vertices is $M_r=z(z-1{)}^{r-1}$, so there are more outward than inward choices even though no edge direction is dynamically preferred. $P(r,N)$ is the probability summed over an entire shell, not the probability at one specified vertex; no additional shell-size factor belongs in the boxed recurrence.

The equality concerns the evolution equation away from the root, not equality to the unrestricted full-line binomial distribution at all positions. The root imposes a different boundary rule.

At the root all $z$ neighbours are in shell one, so the probability of leaving the root for that shell is one. Conversely, a walker anywhere in shell one chooses its unique rootward edge with probability $1/z$. The two boundary recurrences are thereforeTogether with $P(0,0)=1$ and $P(r>0,0)=0$, these equations determine the distribution. In particular $P(1,1)=1$ and $P(0,2)=1/z$, which check the special root coefficient. For the original indexing, replace $N+1$ by $N$ and $N$ by $N-1$ on the right-hand sides. Summing these equations and the bulk recurrence conserves ${\sum }_{r\ge 0}P(r,N)=1$.

Farther than a few edge lengths from the root, replace the radial coordinate $r$ by a physical coordinate $x=ar$ and smooth the master equation. It has the same constant-coefficient Fokker-Planck equation as in (b). Under the conventional first-time-derivative truncation,The radial drift $v=-C=a(z-2)/(z\tau )$ is outward. For $z=2$ the tree is the infinite line, its radial distance is an unbiased reflected walk, and $C=0$.

The root cannot absorb probability or allow negative radial distance. Write the equation as a continuity equation, using the Fokker-Planck probability currentThe appropriate continuum boundary condition is reflecting, or zero flux:It is not generally ${\partial }_{x}P(0,t)=0$: that simpler condition is correct only when $C=0$. With the finite-bias fixed-step Gaussian convention from (b), use $D=D_{\mathrm{fixed}}=2a^2(z-1)/(z^2\tau )$ instead; the drift sign and zero-current boundary condition are unchanged. A continuum boundary condition represents the microscopic root region after coarse-graining, rather than reproducing each discrete root probability at individual step times.

For the infinite system with outward drift, total probability remains one, but probability at a fixed distance from the root eventually decreases as the walker moves away. The question concerns this local decay, not a loss of total probability.

The independent dimensions of the continuum coefficients are $[D]=L^2/t$ and $[C]=L/t$. The drift-diffusion decay time and its associated length are, up to numerical factors:Equivalently, equating the drift distance $|C|t$ to the diffusive width of order $\sqrt{Dt}$ gives this crossover scale. Rescaling $x=(D/|C|)y$ and $t=(D/C^2)s$ removes the dimensional coefficients from the equation.

With the conventional coefficients used in the requested continuum derivation,where the last expression suppresses the numerical factor as requested. The elementary drifted Gaussian distribution on the full line hasAt fixed $x$, its exponential decay factor is $e^{-C^{2}t/(4D)}$, so an exponential e-folding time would conventionally be called $4D/C^2$, consistent with the same dimensional scale. The reflecting half-line changes the spatial form and algebraic prefactor. The substitution $P=e^{-Cx/(2D)}\psi$ gives ${\partial }_t\psi =D{\partial }_x^2\psi -(C^2/4D)\psi$, displaying the same exponential scale for outward drift.

For $z=2$, $C=0$ and no finite drift-induced exponential decay time exists: the probability near the root decays only algebraically. For the finite-bias, deterministic-step Gaussian envelope discussed in (b), the corresponding dimensional expression is instead $D_{\mathrm{fixed}}/C^2=2\tau (z-1)/(z-2{)}^2$. These continuum estimates are not exact return-probability decay rates for the lattice walk at arbitrary $z$; in particular, a diffusion approximation need not describe far-tail return events accurately at strong bias.

Use the Gibbs-Duhem equation for a homogeneous, extensive, single-component thermodynamic phase:At fixed temperature, this becomes $dp=nd\mu$, with number density $n=N/V$. Separately, at fixed particle number,The chain rule along the same isothermal process in the equation of state now givesConsequentlyThis is isothermal compressibility from chemical-potential response. Since chemical potential has units of energy per particle, the right-hand side has units of volume per energy, or $\mathrm{P}{\mathrm{a}}^{-1}$, as required. The derivation assumes a smooth bulk equation of state; changing the fractions of coexisting thermodynamic phases at fixed pressure needs separate treatment.

Measure single-particle energy from the lowest state and write ${\epsilon }_k=c{k}^{\alpha }$, with $c>0$. In a homogeneous system, the number of wavevector states in a thin spherical shell is proportional to $k^{d-1}dk$. Changing variables from $k$ to $\epsilon$ gives the density of states per unit spatial volume,where $S_{d-1}$ is the area of the unit sphere in $d$ dimensions; an additional internal state degeneracy would multiply this expression by a constant.

The Bose-Einstein distribution has $\mu \le 0$. At saturation, $\mu \to 0^{-}$, the maximum excited-state number density isOnly the lower endpoint can cause divergence. There the integrand is proportional toThe improper integral converges precisely when its power exceeds $-1$. ThusWhen $d=\alpha$, the divergence is logarithmic; when $d<\alpha$, it is a power divergence. In either case the excited states can accommodate any fixed finite density at every positive temperature, so there is no bulk Bose-Einstein condensation at nonzero temperature.