Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford / 2015 / S16

Assume ideal, nondegenerate particles in local thermodynamic equilibrium, ground-state atomic hydrogen, no molecules, and a common temperature. Let ${\lambda }_j=h/\sqrt{2\pi m_j{k}_{B}T}$ be the thermal de Broglie wavelength. Choose zero energy for a separated stationary proton and electron, so a neutral atom has ground-state energy $-R$. The classical chemical potentials areChemical equilibrium for $H\rightleftharpoons p+e$ requires ${\mu }_H={\mu }_p+{\mu }_e$. Rearranging givesIncluding electron and proton spin degeneracy, $g_e=2$, $g_p=2$ and $g_H=4$ when hyperfine splittings are neglected. Hence the statistical factor is one. Also $m_H\simeq m_p$, so ${\lambda }_H^3/{\lambda }_p^3\simeq 1$. Charge neutrality gives $n_p=n_e$, and thereforeThis is the specified ground-state Saha equation. Keeping excited atomic partition functions or a different spin-counting convention can change a prefactor, so its convention must be stated rather than silently adding a factor of two. The left side has dimensions of a number density, not a dimensionless electron-to-neutral ratio.

Let $N=n_H+n_p$ be the conserved number density of hydrogen nuclei, whether ionized or bound, and $x=n_p/N=n_e/N$ the ionization fraction. The ideal gas pressure counts neutral atoms, free protons and free electrons:At low temperature $x\simeq 0$ and $P\simeq N{k}_{B}T$. Heating through the ionization region increases the particle count as well as their thermal energy, giving a steeper rise. At high temperature $x\simeq 1$, so $P\simeq 2N{k}_{B}T$ and the asymptotic slope is twice the low-temperature slope. There is a smooth ionization crossover, not a pressure discontinuity.

The held graph is legible. Representative readings are $P(5000\mathrm{K})\simeq 6.8\mathrm{Pa}$ and $P(15000\mathrm{K})\simeq 41.4\mathrm{Pa}$, with readout uncertainties of roughly $0.3$–$0.5\mathrm{Pa}$ and about $100\mathrm{K}$. They give $N\simeq P/(k_{B}T)$ on the neutral branch and $N\simeq P/(2k_{B}T)$ on the ionized branch, respectively. Both are consistent withThe error is a conservative graph-reading estimate, not a fitted experimental confidence interval.

For that density, the Saha equation gives $x\simeq 0.50$ at $9000\mathrm{K}$, $x\simeq 0.81$ at $10000\mathrm{K}$, and $x\simeq 0.9992$ at $15000\mathrm{K}$. Differentiating the ionization relation givesThe calculated maximum slope occurs near $9.25\times 10^3\mathrm{K}$, consistent with the source's visibly steepest region at roughly $(9.3\pm 0.4)\times 10^3\mathrm{K}$. This checks the actual slope rather than identifying it without justification with exactly half ionization.

Taking the neutral ground-state gas as the zero of internal binding energy, the matter energy density is translational kinetic energy plus ionization energy:At $15000\mathrm{K}$, with $R=13.6\mathrm{eV}$, these terms are about $62$ and $218\mathrm{J}{\mathrm{m}}^{-3}$, givingRest energies are excluded. A separate equilibrium radiation field, if the gas were in a blackbody cavity, would add its own radiation energy density; it is not part of this gas-only estimate.

Increasing the conserved nucleus density to $10N$ multiplies the low-temperature neutral-gas slope and high-temperature ionized-gas slope by ten. However, at a fixed intermediate temperature, $x^2/(1-x)=S(T)/(10N)$ is smaller, so a smaller fraction is ionized. The curved transition therefore moves to higher temperature; the new graph is not simply ten times the old graph throughout the crossover.

At a fixed chosen ionization fraction, $S(T)$ must increase tenfold. Locally this requireswhich is of order $10^3\mathrm{K}$ in this region. Directly evaluating the same Saha equation at $10^{21}{\mathrm{m}}^{-3}$ moves the maximum pressure slope from about $9250$ to $10560\mathrm{K}$. The two asymptotic slopes still have ratio two, because each hydrogen nucleus still contributes one particle when neutral and two when fully ionized.

The electron Debye length is the scale over which a weak electrostatic disturbance is screened. Take the ions as an immobile neutralizing background and the electrons as an isothermal Boltzmann distribution. For a small potential $\varphi$,Poisson's equation then becomes ${\nabla }^2\varphi =\varphi /{\lambda }_D^2$ away from the perturbing source, withA test charge has a screened potential proportional to $e^{-r/{\lambda }_D}/r$. For fully static screening with mobile ions, inverse squared Debye lengths from all species add; the displayed expression is the electron scale used below.

The electron plasma frequency is the natural collective oscillation frequency when electrons are displaced from a nearly fixed ion background. Displace a uniform electron fluid by $\xi$ along a slab normal. The charge separation produces the restoring electric force $-n_e{e}^2\xi /{ϵ}_0$ on an electron. Hence $m_e\ddot{\xi }=-n_e{e}^2\xi /{ϵ}_0$ andIt sets a response time $1/{\omega }_p$ and the cold unmagnetized electromagnetic cutoff frequency. These collective effects arise from long-range electric fields, even when individual binary collisions are infrequent.

Define the plasma parameter here as the number of electrons in an electron Debye sphere,Some texts instead use $n_e{\lambda }_D^3$, or its inverse as a coupling parameter. A weakly coupled collective plasma has $N_D\gg 1$; fluctuations from individual nearby particles are small compared with the collective description.

A Coulomb collision means the cumulative deflection or relaxation produced by many small-angle encounters, not necessarily a hard contact between two particles. The Coulomb logarithm $\ln {\Lambda }_C\simeq \ln (b_{\max }/b_{\min })$ summarizes the impact-parameter range, with $b_{\max }$ of order ${\lambda }_D$. A process is collisionless when its timescale is much shorter than the appropriate collisional relaxation time, or its length scale is much smaller than a corresponding mean free path. It still contains collective interactions.

For fully ionized ions of charge $Ze$, a thermal electron–ion momentum-relaxation rate has the scalingup to numerical factors depending on the Debye-sphere and collision-time conventions. More explicitly, the Maxwellian rate used below gives ${\nu }_{ei}/{\omega }_p\simeq 0.089Z\ln {\Lambda }_C/N_D$, or $t_{\mathrm{col}}{\omega }_p\simeq 11.3N_D/(Z\ln {\Lambda }_C)$. Specifying this factor prevents mixing a small coupling parameter with a large Debye-sphere population.^[ref]

Use collisional inverse bremsstrahlung in a weakly coupled, unmagnetized plasma. The specified density is $n_e=10^{26}{\mathrm{m}}^{-3}$, and $k_{B}T=1000\mathrm{eV}$. The electron Debye length is $2.35\times 10^{-8}\mathrm{m}$, the plasma frequency is $5.64\times 10^{14}{\mathrm{s}}^{-1}$, and the laser angular frequency is $\omega =2\pi c/\lambda =1.79\times 10^{15}{\mathrm{s}}^{-1}$. Thus $X={\omega }_p^2/{\omega }^2\simeq 0.099<1$: the plasma is underdense and a propagating wave exists.

Estimate $b_{\min }$ as the larger of the classical Coulomb-deflection length $Z{e}^2/(4\pi {ϵ}_0{k}_{B}T)$ and the electron quantum length $\hslash /\sqrt{m_e{k}_{B}T}$. Both are about $9\times 10^{-12}\mathrm{m}$ here, giving $\ln {\Lambda }_C\simeq 8$. With the usual thermal collision convention,The $Z$ factor follows from $n_i{Z}^2=n_{e}Z$.^[ref]

The weakly collisional Drude model gives ${ϵ}_r=1-{\omega }_p^2/[\omega (\omega +i{\nu }_{ei})]$ for fields proportional to $e^{-i\omega t}$. Expanding at ${\nu }_{ei}\ll \omega$ yields a real refractive index $n_r=\sqrt{1-X}\simeq 0.949$ and intensity attenuation coefficientConsequently the intensity e-folding absorption length isi.e. of order a centimetre. Taking a Coulomb logarithm between about 5 and 10 gives roughly $5$–$10\mathrm{mm}$. The electric-field amplitude decay length is twice this intensity length. This estimate assumes a uniform slab thick enough to define an absorption length and neglects gradients, nonlinear quiver heating and other absorption mechanisms.

For a sinusoidal laser field of peak amplitude $E_0$, an electron's peak nonrelativistic quiver speed is $v_q=e{E}_0/(m_e\omega )$. Choose the one-component electron thermal speed $v_T=\sqrt{k_{B}T/m_e}=1.33\times 10^7\mathrm{m}{\mathrm{s}}^{-1}$. The vacuum-equivalent laser intensity is $I={ϵ}_{0}c{E}_0^2/2$, so the onset $v_q\sim v_T$ occurs atUsing $\sqrt{2k_{B}T/m_e}$ as the thermal-speed convention doubles this estimate; its order of magnitude is the physical content. Once the quiver is comparable to the thermal speed, a collision rate computed for an undriven thermal distribution is no longer adequate.

Relativistic motion begins when the normalized field $a_0=e{E}_0/(m_{e}c\omega )$ is of order one. ThusThe ratio $I_{\mathrm{rel}}/I_{\mathrm{th}}=m_e{c}^2/(k_{B}T)\simeq 511$ checks the separation of the two regimes. These estimates use peak field and cycle-averaged vacuum intensity. The small refractive-index and interface-transmission corrections in this underdense plasma change only the few-percent prefactor, below the accuracy intended here.

Take $\mathbf{B}=B\hat{\mathbf{z}}$. The magnetic Lorentz force givesThus $v_x+i{v}_y=(v_x(0)+i{v}_y(0))e^{-i\Omega t}$, while $v_z$ is constant. Integrating the transverse velocity gives a circle of radius $r_L=v_{\perp }/|\Omega |$ about a fixed guiding centre; adding $z(t)=z_0+v_{\parallel }t$ gives a helical motion in a uniform magnetic field. The magnetic force does no work, so the speed and kinetic energy remain constant. The handedness reverses with the sign of $q$. If $v_{\parallel }=0$ the helix degenerates to a circle; if $v_{\perp }=0$ it is a straight line parallel to the field.

Seek a constant transverse guiding-centre drift ${\mathbf{v}}_d$ whose magnetic Lorentz force balances the applied force. DefineSince $\mathbf{F}\cdot \mathbf{B}=0$, the vector triple-product identity gives $q{\mathbf{v}}_d\times \mathbf{B}=-\mathbf{F}$. Writing $\mathbf{v}={\mathbf{v}}_d+\mathbf{u}$ therefore reduces the full equation to $m\dot{\mathbf{u}}=q\mathbf{u}\times \mathbf{B}$, the ordinary gyromotion equation. Its transverse orbit is circular in the drifting frame, soAveraging over a gyroperiod removes the transverse oscillatory velocity and leaves this drift, together with any constant parallel velocity. For an electric force $\mathbf{F}=q\mathbf{E}$, this becomes the charge-independent $\mathbf{E}\times \mathbf{B}/B^2$ drift; a charge-independent force instead gives opposite drift directions for opposite charge signs.

A purely toroidal magnetic field is stronger on the inner side of the torus and its field lines are curved. The resulting magnetic gradient drift and curvature drift move ions and electrons vertically in opposite directions. Charge separation can then produce an electric field whose E-cross-B drift moves the plasma across the nominal toroidal confinement region. Closing field lines into circles therefore does not by itself guarantee confinement.

A tokamak combines a strong externally produced toroidal field with a poloidal field supplied largely by toroidal plasma current. Field lines wind helically around nested magnetic flux surfaces. A particle following a field line samples different vertical and radial parts of the torus, so the systematic displacements associated with the toroidal geometry can compensate over its orbit rather than accumulating as in the pure-toroidal case. The poloidal field also provides the magnetic structure needed for force balance. This improves confinement but does not eliminate all orbit losses or transport: trapped particles, collisions, turbulence and magnetohydrodynamic instabilities remain important. Maintaining the plasma current and suitable field-line winding is part of the confinement problem.

The supplied confinement criterion alone fixes a product, not a unique current or machine size. Choose a representative ion density $n_i=10^{20}{\mathrm{m}}^{-3}$. ThenFor $n_i=10^{19}$ to $10^{21}{\mathrm{m}}^{-3}$, the same criterion gives $\tau \gtrsim 10$ to $0.1\mathrm{s}$. These are illustrative density choices, not additional source data.

For a current estimate, assume a large-aspect-ratio circular tokamak with major radius $R_0=3\mathrm{m}$, minor radius $a=1\mathrm{m}$, toroidal field $B_{\varphi }=5\mathrm{T}$ and edge tokamak safety factor $q_s\simeq 3$. The geometrical winding relation and Ampère's law giveFor the stated assumptions,so toroidal plasma currents of order megaamperes are a natural scale. As a consistency check at $T\sim 10^8\mathrm{K}$, quasineutral singly charged heavy hydrogen has total thermal pressure $p\simeq 2n_i{k}_{B}T\simeq 2.8\times 10^5\mathrm{Pa}$. With $5\mathrm{T}$, $2{\mu }_{0}p/B^2\simeq 0.028$, so the assumed magnetic pressure exceeds the thermal pressure. The particular current depends on geometry and field-line pitch; it cannot be inferred from the confinement product alone.