Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford / 2018 / S16 (Plasma physics)

Assume a dilute nondegenerate hydrogen gas in thermal and chemical equilibrium, with Maxwell-Boltzmann translational partition functions, ground-state neutral atoms, single ionization and charge neutrality $n_p=n_e$. For $\mathrm{H}\rightleftharpoons p+e$, chemical equilibrium gives ${\mu }_H={\mu }_p+{\mu }_e$. The ideal-gas relation ${\mu }_i=E_{i0}+k_{B}T\log (n_i{\lambda }_i^3/g_i)$, with $E_{H0}=-R$, givesNeglecting the electron mass in $m_H/m_p$ and using the ground-state electronic statistical weights, $g_e=2$ and $g_H=2$ with cancelling nuclear-spin factors, this becomesExcited neutral states have an energy gap of at least $3R/4$ above the ground state. At the temperatures relevant to the plotted maximum their Boltzmann factors are very small despite the finite degeneracy increase. Highly excited levels also require a physical plasma cutoff; the untruncated infinite isolated-atom partition sum is not a valid finite-density model. Ground-state dominance is an approximation in this regime, not a theorem at all temperatures.

Take the heating process to keep the total hydrogen-nucleus density fixed, so $n_H+n_e=n_0$. This fixed-density assumption is needed to determine $n_0$ from the curve; freely expanding gas would require its density history. The excited-to-ground ratio is $n_2/n_H\simeq 4e^{-3R/(4k_{B}T)}$. ThereforeAt low temperature excitation is suppressed. At higher temperature excitation grows, but eventually ionization removes neutral atoms and the excited-neutral abundance decreases. To locate the maximum quantitatively, write $u=R/(k_{B}T)$. From $S/n_0=x^2/(1-x)$,Setting $d\log f/d\log T=0$ givesReading the original graph gives $T_{\max }\simeq (7.9\text{–}8.0)\times 10^3$ K and $f_{\max }\simeq 2.3\times 10^{-7}$. With $R=13.6$ eV, this implies $u\simeq 20$ and $\boxed{x_{\max }\simeq 0.82}$. Substitution intogives approximately $10^{18}{\mathrm{m}}^{-3}$. The result is particularly temperature-sensitive because of the Saha exponential; using $7900$ K gives about $9\times 10^{17}{\mathrm{m}}^{-3}$, while the coarse $8000$ K reading gives about $1.2\times 10^{18}{\mathrm{m}}^{-3}$. The plotted height is consistent with this range and should not be used to claim more precision than the source graph provides.

The plasma parameter is the mean number of electrons in a Debye sphere. With electron Debye length ${\lambda }_D=({\epsilon }_0{k}_{B}T/n_e{e}^2{)}^{1/2}$,Large $N_D$ describes weakly coupled collective screening; a single electron does not dominate its screening cloud. For a small uniform electron displacement $\xi$ relative to fixed ions, the restoring field gives $m_e\ddot{\xi }=-n_e{e}^2\xi /{\epsilon }_0$, henceThis cold-electron oscillation estimate neglects thermal dispersion and ion motion.

Using the stipulated collision-time approximation,With $C\sim 3n_e{k}_B/2$ and $v^2\sim k_{B}T/m_e$, the kinetic estimate $K=C{v}^2\tau /3$ givesThe elementary single-speed estimate corresponds to $B=1$; velocity averaging and the stated collision model are represented by the supplied $B$. A detailed Coulomb transport calculation also contains a Coulomb logarithm and cannot be inferred precisely from this simplified expression.

The electron thermal energy density is $3n_e{k}_{B}T/2$. Transporting all of it at the thermal speed gives the free-streaming scaleFor heat flowing in the positive $x$ direction, $-dT/dx>0$. Inserting $K$ into $K(-dT/dx)>A{\dot{q}}_{\max }$ givesThis flux-limiting criterion expresses the failure of a local diffusive description when an electron samples a substantial change in temperature over a transport path.

For the stipulated carbon plasma and $B\simeq 10$, setting the gradient equal to $10^{11}\mathrm{K}{\mathrm{m}}^{-1}$ gives a critical flux fraction of order $A_{\mathrm{crit}}\simeq 0.05$. Fourier transport is invalid according to this test when $\boxed{A<A_{\mathrm{crit}}}$; a still larger actual gradient raises the critical value. Thus the source's request for a minimum $A$ has the inequality reversed: the calculation gives an upper threshold, not a positive minimum. An arbitrarily smaller positive $A$ makes the criterion easier to satisfy. The exact numerical threshold is evaluated in the independent check.

Use positive angular frequencies and abbreviate $p={\omega }_{pe}$, $b={\omega }_{ce}$. The left-handed index is $n_L^2=1-p^2/[\omega (\omega +b)]$. It is negative below its positive zero,For the right-handed mode, $n_R^2=1-p^2/[\omega (\omega -b)]$. Below the cyclotron resonance $0<\omega <b$ its denominator is negative and the index square is positive. Just above the resonance it is negative untilThis is the lossless electron-only cold-plasma model. Ions, collisions and resonance absorption matter outside its applicable range.

For a whistler with $\omega \ll b$ and $p^2/(\omega b)\gg 1$,The travel time through a uniform slab is $t-t_s=Lp{\omega }^{-1/2}/(2c\sqrt{b})$. In terms of ordinary frequency,An arbitrary emission time changes the intercept, not the slope. The original figure labels a dispersion constant $D=81\pm 0.5\mathrm{H}{\mathrm{z}}^{1/2}\mathrm{s}$: it is the reciprocal of the plotted slope, not the slope itself. ThereforeFor $B=80$ nT and $L=6.0\times 10^7$ m this gives $b\simeq 1.41\times 10^4{\mathrm{s}}^{-1}$, $p\simeq 2.41\times 10^5{\mathrm{s}}^{-1}$, andThe resulting forbidden right-handed frequency width isThe lower edge is about $2.24$ kHz and the upper edge about $39.5$ kHz. In the real nonuniform plasma, the travel-time coefficient measures a path integral proportional to $\int {\omega }_{pe}/\sqrt{{\omega }_{ce}}ds$. The inferred density is consequently an effective uniform-slab estimate, not a unique local density at the spacecraft. Uncertainty in the path and magnetic field likely exceeds the formal error in $D$.