Past exam of the thermal and statistical physics course of the University of Cambridge / 2017 / 3

Component chemical potentials. Define the per-spin quantum concentration $n_Q={\lambda }_T^{-3}=(m{k}_{B}T/(2\pi {\hslash }^2){)}^{3/2}$. For a classical component with one-particle potential energy $U_{\sigma }(\mathbf{r})$, the local Helmholtz free energy density is $k_{B}T{n}_{\sigma }[\ln (n_{\sigma }/n_Q)-1]+n_{\sigma }{U}_{\sigma }$. Its derivative with respect to number density gives the full local chemical potential, including the magnetic energy. With the up state defined as the lower level,The Bohr magneton ${\mu }_B$ is a magnetic moment; ${\mu }_{B}B$ is an energy. The factor two for the two spin states is not already included in $n_Q$.

Total equilibrium chemical potential. If spin conversion is allowed, a particle can be added in either spin state at the same equilibrium free-energy cost. Thusnot their sum or zero. The total particle number, or a reservoir, fixes this common value; the supplied information does not specify a numerical chemical potential. A useful local expression is $\mu =k_{B}T\ln [n/(2n_Q\cosh x)]$, with $x={\mu }_{B}B/(k_{B}T)$, whose density dependence is found below.

Why it is uniform and common. Moving a small number $dN$ of one species from ${\mathbf{r}}_1$ to ${\mathbf{r}}_2$ changes the free energy by $[{\mu }_{\sigma }({\mathbf{r}}_2)-{\mu }_{\sigma }({\mathbf{r}}_1)]dN$. If this difference were nonzero, a transfer could lower the free energy. Equilibrium under diffusion therefore requires each full chemical potential to be spatially constant. A spin flip transferring $dN$ from down to up similarly changes free energy by $({\mu }_{\uparrow }-{\mu }_{\downarrow })dN$, proving equality when spin relaxation is permitted. This is chemical equilibrium of a classical spin gas. Mere thermal contact at fixed separate spin populations would not enforce the latter equality; full spin equilibrium is the necessary assumption. A spatially varying magnetic field is balanced by a spatially varying density, not by a varying full chemical potential.

Magnetization. Invert the component equations at the common $\mu$:The two levels have moments $+{\mu }_B$ and $-{\mu }_B$ along the field. The magnetization, magnetic moment per unit volume, is thereforeIts SI units are $\mathrm{A}{\mathrm{m}}^{-1}$. This expression neglects interactions and magnetic feedback of the gas on the specified external field.

High-temperature response. For $|x|\ll 1$, $\sinh x\simeq x$ and $\cosh x\simeq 1$, yieldingAt fixed number density, this is the Curie law: the susceptibility with respect to $B$ is $n{\mu }_B^2/(k_{B}T)$. The conventional dimensionless susceptibility with respect to $H$ is approximately ${\chi }_v={\mu }_{0}n{\mu }_B^2/(k_{B}T)$ when $B\simeq {\mu }_{0}H$. At fixed $\mu$, however, $n_Q{e}^{\beta \mu }$ changes with temperature, so $M$ itself need not scale simply as $1/T$. The moment per particle has the Curie behaviour; the density constraint must be stated.

Total density. Adding the two populations givesFor a finite container and fixed $N$, normalization gives $e^{\beta \mu }=N/[2n_Q\int \cosh x(\mathbf{r})d^{3}r]$. In a weak field the leading density is uniform, with changes only of order $x^2$. Particles preferentially occupy regions where the magnitude of the field lowers one of the spin energies, even though the common chemical potential remains uniform.

Gibbs free energy and its minimum. For fixed atom number $N$, there are $N+n$ sites andconfigurations with $n$ indistinguishable vacancies. Their configurational entropy is $k_B\ln W$. Relative to the vacancy-free Gibbs free energy $G_0$, creating $n$ vacancies costs energy $nϵ$ and pressure-volume work $npv$. Put $A=ϵ+pv>0$. ThenAt leading thermodynamic order, Stirling's approximation givesThe minimum obtained from the stationary point below isIf the perfect crystal's internal energy and other entropy are chosen as zero, its pressure term is $G_0=pNv$. More generally the information given fixes the vacancy contribution rather than the absolute bulk free energy. The pressure-volume term must be retained because the number of sites, and hence the volume, changes.

Equilibrium vacancy population. Differentiating the extensive free energy at fixed $N,T,p$ givesThe stationary point is consequently the unique interior minimum. Write $q=e^{-A/(k_{B}T)}$; its condition is $n/(N+n)=q$, giving the requested thermodynamic result for equilibrium vacancies with variable lattice size,Substitution back into $G$ cancels the terms containing $n_{\ast }$ and yields the boxed minimum. In the dilute-defect limit $A\gg k_{B}T$, $n_{\ast }\simeq N{e}^{-A/(k_{B}T)}$. The Bose-like denominator is a consequence of the changing number of lattice sites and combinatorial entropy; it does not assert that vacancies here are an ideal gas of bosonic particles.

There is a finite-size correction to a variable-site vacancy model if “average” is interpreted exactly. The microscopic count written above gives the relative isothermal-isobaric sumsoIts relative equilibrium free energy is $-k_{B}T\ln \mathcal{Y}=(N+1)k_{B}T\ln (1-q)$. Thus the source's $N$ numerator is correct to leading extensive order, but is not the exact finite-$N$ mean for that stated configuration count. A free-energy minimum is likewise not generally identical to the mean of a finite distribution. These distinctions vanish relatively as $N\to \infty$.

Thermal volume. Each atom or vacancy occupies volume $v$, so at the same thermodynamic accuracyFor the exact counting model, replace $n_{\ast }$ by $(N+1)q/(1-q)$ in $v(N+⟨n⟩)$. The vacancy-free limit is $Nv$ as $T\to 0$. This model isolates expansion from vacancies, taking $v$ and $ϵ$ constant; it omits ordinary phonon thermal expansion. Its growth at high temperature should not be extrapolated to arbitrarily large vacancy fractions in a stable crystal, and $A>0$ is essential for a normalizable variable-site ensemble.