Past exam of the thermal and statistical physics course of the University of Cambridge / 2013 / Paper 1 (THERMAL AND STATISTICAL PHYSICS, EXPERIMENTAL AND THEORETICAL PHYSICS (1), PHYSICAL SCIENCES: HALF SUBJECT PHYSICS (1))

Original paper: Public original PDF, physical pages 242–245; cover on page 242 and questions on pages 243–245.

In the classical equilibrium circuit, the capacitor stores $C{V}^2/2$. Equipartition assigns this quadratic term mean energy $k_{B}T/2$, soThe mean voltage is zero and its variance is $k_{B}T/C$. Resistance determines the equilibration time and noise spectrum, not this full-band equilibrium variance. A finite-bandwidth or appreciably loading readout can measure less than the ideal result.

Let $\Delta =\hslash \Omega$ and choose the nondegenerate energy levels as 0 and $\Delta$. For N independent identical units, $Z=(1+e^{-\beta \Delta }{)}^N$; set N=1 for a single two-level system. Differentiating its logarithm givesAt low temperature, $U\sim N\Delta e^{-x}$ and $C_V\sim N{k}_B{x}^2{e}^{-x}$: the upper level is frozen out. At high temperature, the two levels approach equal populations, giving $U\to N\Delta /2$ while $C_V\sim N{k}_B{x}^2/4\to 0$. The energy therefore rises monotonically to a plateau, while the heat capacity has a Schottky peak:Unequal degeneracies would change the populations and peak; the usual equal-degeneracy model is stated here. Shifting both energy levels changes the energy baseline but not heat capacity.

Use a three-dimensional ideal gas at the same number density n for both isotopes. An unpolarized helium-3 gas has spin degeneracy two, henceHelium-4 has a single spin state, and the ideal Bose-gas saturation condition givesEliminating n, using $m_3/m_4\simeq 3/4$ and $\zeta (3/2)=2.612$, yieldsThe fermionic spin degeneracy must be included. This is an ideal-gas estimate; strong interactions in liquid helium prevent identifying it with an exact experimental transition temperature.

Equipartition gives mean energy $k_{B}T/2$ for each independent quadratic term in a classical canonical Hamiltonian. For example, $p^2/(2m)$ and $k{x}^2/2$ each contribute this amount in a harmonic oscillator. It is a thermal average, not an instantaneous equality; constraints, nonquadratic terms or quantum level spacings large compared with $k_{B}T$ require a different calculation.

The fluctuation-dissipation theorem connects equilibrium fluctuations with the dissipative part of linear response. If an applied force couples as $-f(t)x$, use response to $e^{-i\omega t}$ and define the two-sided angular-frequency spectrum $S_{xx}(\omega )=\int e^{i\omega t}⟨\delta x(t)\delta x(0)⟩dt$. The classical relation isFor a viscously damped particle, $m\dot{v}=-\zeta v+\xi (t)$, the corresponding white-noise force satisfies $⟨\xi (t)\xi (t^{\prime })⟩=2\zeta k_{B}T\delta (t-t^{\prime })$. Noise replenishes energy lost by friction; their strengths cannot be varied independently at fixed equilibrium temperature.

Electrically, the one-sided voltage-noise spectrum per hertz of an equilibrium resistor is $4k_{B}TR$. A capacitor in parallel filters it toThe same equilibrium voltage variance follows from equipartition. Increasing R makes the spectrum taller and narrower without changing its integrated area. This thermodynamic connection between dissipative conductance and thermal electrical fluctuations is the content of the classical noise calculation.^[ref]

A black body absorbs incident electromagnetic radiation and, in equilibrium, emits a spectrum determined by temperature alone. A cavity with a small aperture approximates this ideal. Photon number is not conserved in thermal equilibrium, so its chemical potential is zero. Combining the Bose occupation with the two-polarization electromagnetic mode density gives the spectral energy densityChanging variable to $x=\hslash \omega /(k_{B}T)$ shows directly why the total energy density is proportional to $T^4$:Three powers of T arise from the number of thermally populated modes and one from their characteristic energy. Isotropic radiation incident on a surface has outward hemispherical flux $cu/4$, givingThe factor 1/4 is the hemispherical angular average; multiplying energy density by c alone would overcount. The spectrum's frequency scale grows with T. Classical equipartition over all modes would instead produce the Rayleigh-Jeans ultraviolet divergence; the Planck occupation suppresses high-frequency modes. Real materials can have wavelength-dependent emissivity, and a body's net radiative exchange also includes absorbed environmental radiation.

An equation of state relates equilibrium pressure, temperature and density. The ideal-gas form $p=n{k}_{B}T$ neglects intermolecular forces and excluded volume. A low-density real-gas expansion iswhere the virial coefficients encode interaction effects and generally depend on temperature.

A simple phenomenological example is the molar Van der Waals equationHere v is molar volume, b represents excluded volume and the attractive interaction lowers pressure through a. Its dilute expansion gives $pv/(RT)=1+[b-a/(RT)]/v+\cdots$. It recovers the ideal-gas limit but is not a quantitatively universal equation for all real gases.

Below its critical temperature, the raw isotherm contains a region with $(\partial p/\partial v{)}_T>0$, corresponding to negative isothermal compressibility and mechanical instability. Equilibrium phase separation replaces that segment by a coexistence pressure satisfying equal chemical potentials, equivalently the Maxwell equal-area construction. The critical point follows from $p_v=p_{vv}=0$:The distinction between an unstable homogeneous branch and a stable two-phase state is essential when interpreting a real-gas equation of state.

Write $z=e^{\beta (\mu -ϵ)}$. For one nondegenerate fermionic quantum state, the allowed occupations 0 and 1 give ${\Xi }_F=1+z$. For a bosonic state, summing all nonnegative occupations gives ${\Xi }_B=(1-z{)}^{-1}$, requiring $z<1$ or $\mu <ϵ$. Differentiating the logarithm,The interacting point defect has two distinct spin states, not one fermionic state. Its four microstates have grand-canonical weights $1,z,z,q{z}^2$, with $q=e^{-\beta U}$. The first two single-electron weights are separate up/down spin possibilities; the double state has one electron of each spin. ThusFor $U=0$, $\Xi =(1+z{)}^2$ and $\bar{n}=2z/(1+z)$. The sketch is a single smooth step from 0 to 2, centred on $\mu =ϵ$, where $\bar{n}=1$. As $T\to 0$ its width tends to zero.

For positive U with $\beta U\gg 1$, the lowest grand energies are successively $0$, $ϵ-\mu$ and $2ϵ+U-2\mu$. Away from the two crossings,Finite temperature smooths each step over an energy range of order $k_{B}T$. Near the lower step double occupation is exponentially suppressed, so $\bar{n}\simeq 2z/(1+2z)$. Near the upper step put $y=e^{\beta (\mu -ϵ-U)}$ and neglect the empty state, giving $\bar{n}\simeq 1+y/(2+y)$. Spin degeneracy shifts the half-step positions to $ϵ-k_{B}T\ln 2$ and $ϵ+U+k_{B}T\ln 2$.

For a grand-canonical probability $p_s\propto e^{-\beta (E_s-\mu n_s)}$, differentiation gives ${\partial }_{\mu }{p}_s=\beta (n_s-\bar{n})p_s$. Multiplying by $n_s$ and summing provesFor this defect, direct state counting givesAt $\mu =ϵ$, $z=1$ and thereforeThe same limit gives $\bar{n}=2/3$: empty, up and down states have equal weights, while the double state is suppressed. Replacing this degeneracy by a two-outcome equal-weight model would incorrectly give variance $1/4$ at this chemical potential.

Across the full chemical-potential range the variance has two well-separated narrow peaks and tends to zero on the 0, 1 and 2 plateaux. The limiting shapes are $2z/(1+2z{)}^2$ near the lower step and $2y/(2+y{)}^2$ near the upper one. Each peak has height $1/4$, at the shifted half-step position; the variance at the unshifted crossings is instead $2/9$.

For $H=0$ and $a>0$, $F(M)=a{M}^2/2+b{M}^4/4$ is a symmetric single well. Its only minimum isThe zero is relative to the smooth background omitted from the supplied expansion.

For $H=0$ and $a<0$, the free energy is a symmetric double well. The central stationary point is a maximum; the two stable minima areThese are the magnetisations of the two symmetry-broken phases, selected by an infinitesimal field or a boundary condition. An exactly symmetric finite-system ensemble can have $⟨M⟩=0$ by averaging both wells; that does not make the central maximum a stable homogeneous phase.

For $a>0$ and $H>0$, the field tilts the single well toward positive M. Since $F^{\prime \prime }=a+3b{M}^2>0$ everywhere, there is exactly one minimum, at the positive solution $M_H$ of the equation of state. At that solution,This follows by eliminating ${\mu }_{0}H=a{M}_H+b{M}_H^3$ from F. At weak field, $M_H\simeq {\mu }_{0}H/a$ and $F_{\min }\simeq -{\mu }_0^2{H}^2/(2a)$.

At zero applied field, time-reversal symmetry interchanges $M$ and $-M$ without changing the free energy, so the analytic expansion contains only even powers. The term $-{\mu }_{0}HM$ breaks this symmetry at fixed nonzero H. Stationarity and local stability requireFor $T>T_c$ at small field, the cubic term is higher order and $M\simeq {\mu }_{0}H/a(T)$. Thus the susceptibility as defined in the question isThe three requested free-energy sketches and their minima are given separately above.

For the plotted isotherms, divide the equation of state by nonzero M and rearrange:This is an Arrott plot: its slope determines b, and the isotherm through the origin identifies the mean-field Curie temperature.^[ref] Under the paper's stated uniform-field Landau model, use H as supplied on its axis.

The ordinate unit is $(10^4\mathrm{A}{\mathrm{m}}^{-1}{)}^2=10^8{\mathrm{A}}^2{\mathrm{m}}^{-2}$. Reading the original graph gives slopes about $0.076$–$0.080$ ordinate units per unit H/M, so the physical slope is approximately $7.8\times 10^6{\mathrm{A}}^2{\mathrm{m}}^{-2}$. ConsequentlyApproximate horizontal intercepts and the corresponding $a={\mu }_0(H/M{)}_{M^2=0}$ are:

The sign change brackets the transition between 19 and 25 K. A linear fit of $a(T)=\alpha (T-T_c)$ across the four read-offs gives approximately $\alpha =4.8\times 10^{-6}\mathrm{H}{\mathrm{m}}^{-1}{\mathrm{K}}^{-1}$ andInterpolating only the two nearest isotherms also gives about 22 K. These are graph estimates, not exact recovered measurements.

Below $T_c$ at zero field, the stable branches have $M_0^2=-a/b$. Substitution gives the free-energy density relative to the omitted smooth background:The equilibrium excess heat-capacity density is $c_H=-T{\partial }_T^2{F}_{\min }=T{\alpha }^2/(2b)$ below $T_c$ and zero above it. ThereforeThe all-isotherm linear fit gives about $1.5\times 10^3\mathrm{J}{\mathrm{m}}^{-3}{\mathrm{K}}^{-1}$. Estimating the slope only between the 19 K and 25 K curves instead gives $\alpha \simeq 6.9\times 10^{-6}\mathrm{H}{\mathrm{m}}^{-1}{\mathrm{K}}^{-1}$ and a jump near $3\times 10^3\mathrm{J}{\mathrm{m}}^{-3}{\mathrm{K}}^{-1}$. The lines are not exactly parallel and the intercepts are not perfectly linear in T, so a jump of order a few $10^3\mathrm{J}{\mathrm{m}}^{-3}{\mathrm{K}}^{-1}$ is supported; finer precision is not. This is per volume because F is a density; no sample volume is given to convert it to a total heat capacity.

If $b<0$, the quartic truncation is unbounded below. Add a positive sixth-order term $c{M}^6/6$ with $c>0$. At zero field the coexistence conditions giveThus the zero-field transition is generically first order, with a discontinuous spontaneous magnetisation and possible metastability. It occurs before the quadratic coefficient reaches zero. The case $b=0$ is the tricritical boundary of this mean-field model; no value of c is invented.