Past exam of the thermal and statistical physics course of the University of Cambridge / 2014 / Paper 1 / 2

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.