Past exam of the short options paper shared by Parts A and B of the physics course of the University of Oxford / 2019 / S16

Assume a dilute, nondegenerate, nonrelativistic ideal gas in local thermodynamic equilibrium, with ground-state atomic hydrogen, protons and electrons, and neglect excited atomic levels and interactions. Let the conserved number of hydrogen nuclei per volume be $n_0$, so $n_e=n_p=\zeta n_0$ and $n_H=(1-\zeta )n_0$. Equality of chemical potentials for $H\rightleftharpoons p+e$ gives the Saha ionization equationThe factor two is electron spin; $g_p$ and $g_H$ must be defined consistently with nuclear spin. The small difference between the atomic and proton translational partition functions has been neglected. ThereforeFor the ground-state convention $g_p=1$, $g_H=2$, the statistical prefactor $2g_p/g_H$ is one. At fixed volume $n_0$ is constant, whereas an isobaric heating process would not have a constant $A$.

One mole of original atoms contains $(1-\zeta )$ moles of neutral atoms, $\zeta$ moles of protons and $\zeta$ moles of electrons. All are monatomic translational species, so their combined thermal energy is $\frac{3}{2}RT(1+\zeta )$. Taking the neutral ground state as zero internal energy, ionization stores $\zeta N_A{E}_I$. ThusRadiation, excited states and nonideal interactions have been excluded consistently with the elementary Saha ionization equation.

At fixed volume, logarithmic differentiation of the Saha ionization equation givesDifferentiation the internal energy then yieldsAt low temperature $\zeta \to 0$ and $C_V\to 3R/2$. At high nonrelativistic temperature the gas is fully ionized, $\zeta \to 1$, and $C_V\to 3R$. The large intermediate excess reflects energy spent increasing ionization, as well as heating and creating additional translational particles. “Very high” here remains below regimes where relativistic, radiative or pair-creation contributions invalidate the model.

Put $x=E_I/(k_{B}T)$. With $\zeta =1/2$, the heat capacity formula becomesTaking $E_I=13.6\mathrm{eV}$ gives $\boxed{T=1.02\times 10^4\mathrm{K}}$. At half ionization, $n_e{n}_p/n_H=n_e$, so the Saha ionization equation directly givesThis uses $2g_p/g_H=1$ as specified in part (a); a different internal-state convention must be carried consistently into the numerical electron density. The original nuclei density is $n_0=2n_e\simeq 10^{21}{\mathrm{m}}^{-3}$.

Population can move between the levels by electron-impact excitation and de-excitation, spontaneous emission, radiative absorption and stimulated emission. Other collision partners and radiative cascades can matter in a fuller atom. In the elementary coronal approximation, the plasma is dilute, optically thin, steady, and weakly illuminated at the transition frequency. Electron impacts predominantly excite the lower level, while spontaneous emission predominantly empties the upper level; collisional de-excitation and photon trapping are negligible. The population balance isA key condition is $n_e{C}_{21}\ll A_{21}$, together with insignificant external pumping and competing channels. A Maxwellian electron Maxwell-Boltzmann distribution lets the tabulated rate coefficient be expressed using one temperature.

At $7000\mathrm{K}$, $k_{B}T=0.603\mathrm{eV}$ andThe $2p$ ground term has fine-structure levels $J=1/2$ and $J=3/2$, with energy-level degeneracy $g_1=2$ and $g_2=4$. Detailed balance gives $C_{21}=C_{12}(g_1/g_2)e^{\Delta E/(k_{B}T)}=7.17\times 10^{-14}{\mathrm{m}}^3{\mathrm{s}}^{-1}$. The critical density is thereforeWith collisions and spontaneous decay both retained, $N_2/N_1=n_e{C}_{12}/(A_{21}+n_e{C}_{21})$. A Boltzmann distribution, $N_2/N_1\to 2e^{-\Delta E/(k_{B}T)}\simeq 1.97$, requires $n_e\gg n_{\mathrm{crit}}$, not merely equality. For example $n_e\gtrsim 3\times 10^8{\mathrm{m}}^{-3}$ brings the ratio within about ten per cent of that limit.

Assume a cold, unmagnetized, weakly collisional plasma, a small fractional bandwidth and $\omega \gg {\omega }_p$. The dispersion relation is ${\omega }^2=c^2{k}^2+{\omega }_p^2$, with ${\omega }_p^2=n_e{e}^2/({ϵ}_0{m}_e)$. The group velocity and propagation time areThe lower frequencies arrive later. Taking the magnitude of the frequency derivative over $\Delta \omega$ givesFor a nonuniform line of sight, replace $n_{e}L$ by the dispersion measure $\int n_{e}dl$. The light travels at the group velocity, not at the phase velocity.

Use the preceding electron density as the line-of-sight mean, $L/c=2000\mathrm{yr}$, $\omega =2\pi (1.4\times 10^9){\mathrm{s}}^{-1}$ and $\Delta \omega =2\pi (4.0\times 10^7){\mathrm{s}}^{-1}$. ThenThis estimate corresponds to dispersion measure about $69.5\mathrm{pc}\mathrm{c}{\mathrm{m}}^{-3}$. A cloud density inferred from absorption need not equal the actual path average, so a real pulsar may differ even at the same distance.

Linearize the electron Boltzmann distribution, $n_e(\varphi )\simeq n_{e0}[1+e\varphi /(k_{B}T)]$, about a uniform fixed-ion background. Poisson's equation gives ${\nabla }^2\varphi =\varphi /{\lambda }_D^2$, where the electron Debye length isDefine the plasma parameter as the number of electrons in a Debye sphere:Some authors omit $4\pi /3$ in this definition. Mobile ions alter the total screening length by contributing their own susceptibility; the electron-only convention is used consistently here.

Use quasineutrality, $n_e=Z{n}_i$. Multiplying the supplied collision time by the electron plasma frequency and using the preceding definition givesThus ${\tau }_{ei}\gg 2\pi /{\omega }_p$ requires $N_D\gg (2\pi /9.6)Z\ln \Lambda$; for ordinary $Z$ and $\ln \Lambda \sim 1$–$30$, this is a large plasma parameter. Many particles then collectively screen a charge, fluctuations in a Debye sphere are relatively small, and mean-field electrostatics descriptions become meaningful. A large plasma parameter is also associated with weak Coulomb coupling, well-separated screening and close-collision scales, and collective plasma oscillations surviving many oscillations before collisionally relaxing. Roughly, relative counting fluctuations scale as $N_D^{-1/2}$ and the ratio of interaction to thermal energy decreases as $N_D$ increases. Small $N_D$ signals that strong correlations can invalidate the simple weakly coupled plasma description.

Introduce the electron momentum-relaxation rate $\nu$ through $m_e(\dot{\mathbf{v}}+\nu \mathbf{v})=-e\mathbf{E}$. For a harmonic field $\mathbf{E}=\mathrm{Re}({\mathbf{E}}_0{e}^{-i\omega t})$, the real part of the electrical conductivity is $\mathrm{Re}\sigma =n_e{e}^2\nu /[m_e({\omega }^2+{\nu }^2)]$. At weak damping, the mean heating is $Q=\frac{1}{2}\mathrm{Re}\sigma |E_0{|}^2$ and the beam intensity is $I=\frac{1}{2}{ϵ}_{0}cn|E_0{|}^2$, where $n=\sqrt{1-{\omega }_p^2/{\omega }^2}$. ThereforeWith the relaxation convention $\nu =1/(2{\tau }_{ei})$, this becomes the printed resultConvention qualification: a $90^{\circ }$ deflection time does not by itself specify a momentum-relaxation rate without a collision model. If instead ${\tau }_{ei}$ is defined by $\nu =1/{\tau }_{ei}$, the intensity absorption coefficient is twice the printed one. Amplitude and intensity damping also differ by two. The result above states the extra convention needed rather than silently identifying these quantities. It assumes $\omega >{\omega }_p$, $\nu \ll \omega$ and sufficiently weak damping away from the cutoff; the apparent divergence near cutoff is outside that approximation.

Quasineutrality gives $n_e=4\times 10^{26}{\mathrm{m}}^{-3}$. Using the supplied $90^{\circ }$ collision-time expression and $\ln \Lambda =10$ givesThe printed absorption coefficient then yields $\kappa =1.62\times 10^2{\mathrm{m}}^{-1}$. Since the unabsorbed fraction is $e^{-\kappa L}$,This uses exactly the coefficient specified by the paper and neglects reflection and changing plasma conditions. With $\nu =1/{\tau }_{ei}$ instead, the corresponding thickness is $2.14\mathrm{mm}$, illustrating the convention dependence discussed in part (c).