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Choose the drift-kinetic equation in $(\mathbf{r},\mu ,v_{\parallel })$ with $v_{\parallel }$ the laboratory parallel velocity along the exact $\mathbf{b}$. To first order, before discarding terms small in the stated steep-gradient ordering, it isThe last term is the linear acceleration associated with the moving curved field. Terms quadratic in the perturbations have already been removed. A formulation using parallel velocity relative to the common flow gives the same physical result after transforming back to these variables.

The two mirror terms and the curvature acceleration are subleading here. Their relative sizes are controlled by $h\ll \delta p/p_0$ and $L_p/L_B,L_p/R\ll 1$. The field-direction perturbation multiplying the steep equilibrium gradient must remain:Using $u_x=-i\omega {\xi }_x$ and $\delta b_x=i{k}_{\parallel }{\xi }_x$ therefore gives the leading equationThe displaced branch satisfying quasineutrality has $\delta E_{\parallel }=0$ to this order: its species number densities are $\delta n_{\alpha }=-{\xi }_x{n}_{0\alpha }^{\prime }$, so ${\sum }_{\alpha }{q}_{\alpha }\delta n_{\alpha }=-{\xi }_x{\partial }_x{\sum }_{\alpha }{q}_{\alpha }{n}_{0\alpha }=0$. More explicitly, putting $\mathcal{D}=-i\omega +i{k}_{\parallel }{v}_{\parallel }$ givesAway from an independent longitudinal eigenfrequency, the summed susceptibility is nonzero, giving $\delta E_{\parallel }=0$. An independent longitudinal eigenmode is not forced by the interchange displacement. The required solution is thusThe apparent streaming denominator cancels identically, including at $\omega =k_{\parallel }{v}_{\parallel }$; this response has no leading Landau damping singularity.

To see the temperature and density contributions explicitly, differentiate at fixed $\mu ,v_{\parallel }$ and neglect $B_0^{\prime }$ at this order:The gyrotropic velocity-space measure is $d^{3}v=2\pi B_{0}d\mu d{v}_{\parallel }$. The species perpendicular plasma pressure and parallel plasma pressure are the momentsExplicit perturbations of the measure and of $\mu B$ produce terms proportional to $h$, which are subleading in these pressure moments; derivatives of the equilibrium measure produce the discarded $B_0^{\prime }$ terms. The remaining Maxwell-Boltzmann distribution integrals giveInsertion into b proves

The incident X-ray drives the electrons of an atom. Away from an absorption edge, and when the photon energy greatly exceeds the relevant electronic binding energies but is small compared with $m_e{c}^2$, each electron has approximately the same Thomson scattering amplitude. The much heavier nucleus makes a negligible contribution because its acceleration is much smaller.

For elastic scattering, the amplitudes from different bound electrons add coherently. In units of the single-electron amplitude, the atomic scattering factor is the Fourier transform of the bound-electron number density:Consequently, when the electron cloud is not well resolved by the scattering vector, its internal phase differences are small andThis is an approximation to the coherent scattering amplitude, not an assertion that independently scattered intensities add as $N_e$. At larger momentum transfer, destructive interference within the cloud reduces $f$; near absorption edges, the atomic scattering factor acquires dispersive corrections. Inelastic scattering is a separate contribution. In an ionic crystal, use the ion's bound-electron count rather than automatically using the neutral atom's atomic number.

For one electron per primitive unit cell, the areal number density is $n_s=1/(ab)$. The Fermi circle has radius $k_F=\sqrt{2\pi /(ab)}$. The nearer first-zone boundary is $k_x=\pm \pi /a$, because $a>b$. The circle protrudes precisely whenAt equality the Fermi surface only touches the boundary. Crossing an extended-zone boundary alone does not make a gap: the free electron gas remains metallic when the ionic periodic potential is neglected.

Let $x=g{\mu }_{B}B/(k_{B}T)$. The single-ion partition function and mean projected magnetic moment areFor $J>0$, the geometric sum also givesIn the high-temperature, weak-polarization regime $|x|J\ll 1$, expand the Boltzmann factors. The numerator's zero-order term vanishes because $\sum m_J=0$, and the first-order term uses $\sum m_J^2=J(J+1)(2J+1)/3$. The denominator is $2J+1+O(x^2)$. ThereforeThis fixed-field per-ion Curie coefficient describes mean moment per ion at fixed field, so it includes $B$. It is not the field-independent volume-susceptibility Curie constant. For number density $n$ and $B\simeq {\mu }_{0}H$, the latter is $C_{\chi }=n{\mu }_0(g{\mu }_B{)}^{2}J(J+1)/(3k_B)$, giving ${\chi }_H=C_{\chi }/T$. At $J=0$, the specified moment is zero and both expressions vanish.

In the spin-degenerate free electron model, a monovalent metal has conduction-electron number density $n$. The density of wave-vector states is $1/(2\pi {)}^3$ per unit real-space volume, with two spin states at each wave vector. Filling a sphere at zero temperature givesFor the Debye model, there are three acoustic polarizations and $3n$ modes per unit volume in total. Replacing their dispersions by $\omega =ck$ with the effective Debye sound speed, mode counting givesThe factor two in the electron count is spin, whereas the factor three in the phonon count is polarization; the two counts are not interchangeable. If branch speeds differ, the Debye effective average obeys $c^{-3}=(c_l^{-3}+2c_t^{-3})/3$ in the isotropic approximation.

Dividing gives $T_F/{\theta }_D=\lambda (3{\pi }^{2}n/2{)}^{1/3}$, where $\lambda =\hslash /(2m_{e}c)$. A monatomic FCC conventional cell contains four atoms, so $n=4/a^3$. ThereforeHere and below $c$ is the speed of sound, not the speed of light. The Debye cutoff counts acoustic modes in a spherical approximation to the Brillouin zone; it is not an electron-state cutoff.

In the simple transport model, ${\mu }_e=e{\tau }_e/m_e^{\ast }$ and ${\mu }_h=e{\tau }_h/m_h^{\ast }$. Conduction bands in many common semiconductors have lighter transport masses than the dominant valence hole bands. For comparable momentum-relaxation times, this gives larger electron mobility. It is a common trend, not a theorem: band anisotropy, light and heavy hole populations and different scattering times can alter the comparison. The hole's mass is not the mass of an atom, and crystal translational symmetry does not make its mobility infinite.

For the intrinsic germanium sample, electrons and holes both contribute positive conventional conductivity:ThereforeThis is the density of each species; the combined carrier number density is $2n_i$. Assigning separate densities from the two mobilities would violate intrinsic charge neutrality.

The SI volume magnetic susceptibility is the weak-field response ${\chi }_M=\partial M/\partial H$, or $M/H$ for a linear isotropic sample with no spontaneous moment. It is dimensionless; a molar susceptibility would be a different quantity. For weak magnetization, $B\simeq {\mu }_{0}H$. Neglect local-field and demagnetizing corrections in this independent-spin model.

A spin-$1/2$ ion has moment projections $\pm g{\mu }_B/2$. The requested coefficient assumes a spin-only $g\simeq 2$, giving projections $\pm {\mu }_B$ and Zeeman energies $\mp {\mu }_{B}B$. With $x={\mu }_{B}B/(k_{B}T)$, the one-ion partition function and magnetization areFor $|x|\ll 1$, $\tanh x\simeq x$, soHere $N$ is a number density, as specified, rather than total ion number. The small-field condition ismeaning the Zeeman splitting $2{\mu }_{B}B$ is small compared with the thermal scale up to an irrelevant order-one convention. It is not a temperature-independent absolute upper bound on $B$. For a different $g$ value the coefficient contains $(g{\mu }_B/2{)}^2$ instead of ${\mu }_B^2$.

Take one conduction electron per atom and a three-dimensional free electron gas, with spin degeneracy two and electron mass $m_e$. State counting inside the occupied Fermi sphere givesThe highest occupied energy at zero temperature is $E_F={\hslash }^2{k}_F^2/(2m_e)$, soThis relates the atomic number density to the Fermi temperature only because monovalency and the free-electron dispersion have been assumed. A real energy band can require a different density of states and band effective mass.

For the isotropic Debye model, count three acoustic polarizations up to one spherical wavevector cutoff $q_D$. The number of vibrational modes per volume isWith $\omega =cq$ and $\hslash {\omega }_D=k_B{\theta }_D$,The effective $c$ is the Debye sound speed: matching the low-frequency mode density of one longitudinal and two transverse branches requires $3/c^3=1/c_L^3+2/c_T^3$. It is not generally the arithmetic average of their speeds. The cutoff preserves the mode count; it need not coincide with the actual nonspherical Brillouin zone boundary.

The specified two-direction constraint permits magnetic moments ${\mu }_z=\pm \mu$, withwhere ${\mu }_B$ is the Bohr magneton. There are four magnetic gadolinium ions per conventional face-centered cubic lattice cell; the closed-shell nitrogen ions add no local spin moment. Hence the magnetic-ion number density isAt high temperature, neglect interactions and use $B_z\simeq {\mu }_0{H}_z$. The one-ion partition function and mean magnetic dipole moment areIn the weak-magnetic field limit, $\tanh x\simeq x$, so $M_z=n⟨{\mu }_z⟩={\mu }_{0}n{\mu }^2{H}_z/(k_{B}T)$. Thus the dimensionless SI magnetic susceptibility is $\chi =C/T$, withThe coefficient has units of kelvin. The unrestricted-spin expression proportional to $S(S+1)/3$ is inappropriate here: the source explicitly retains only the two orientations with $S_z=\pm S$. In this two-state Curie law, the squared axial moment is $(g_{J}S{\mu }_B{)}^2$.

Assume uncompensated donor impurities, negligible intrinsic electron holes, and a nondegenerate conduction band. Then charge neutrality equates the conduction electron number density to the ionized-donor number density:The low-temperature condition relative to the band gap supports neglect of intrinsic carriers; one also needs $n_i\ll n$ and the Boltzmann approximation for the conduction electrons. Combining part (b)(i) with the effective density of states result givesEliminate the chemical potential by writing $e^{\beta (\mu -{ϵ}_D)}=(n/N_c)e^{\beta ({ϵ}_c-{ϵ}_D)}$. With the donor binding energy $E_i={ϵ}_c-{ϵ}_D>0$,Writing $A(T)=N_c{e}^{-\beta E_i}$, the physical root of the resulting quadratic equation isThis lies between zero and $N_D$. At semiconductor carrier freeze-out, $A\ll N_D$, it becomes $n\simeq \sqrt{N_D{N}_c}{e}^{-E_i/(2k_{B}T)}$; at essentially complete ionization, $A\gg N_D$, it tends to $N_D$. Including donor charge-state degeneracy would replace $A$ by $A{g}_{\mathrm{empty}}/g_{\mathrm{full}}$.

Normalize the present scale factor to $a_0=1$. Conservation of the number of electrons associated with the baryons makes their physical number density scale as $a^{-3}$. At this early epoch they are free rather than bound in neutral atoms, soHere the specified present density is interpreted as the conserved total electron abundance, including electrons that might be bound today. A present free-electron density would require an ionization-fraction correction.

Density alone does not determine whether a gas is relativistic; its random kinetic energy must be compared with the rest energy. In the usual thermal Hot Big Bang model, use the measured cosmic microwave background temperature $T_0\simeq 2.7\mathrm{K}$ and the temperature scaling of thermal photon number and energy densities, $T_{\gamma }\propto a^{-1}$. Tight coupling then givesSince $m_e{c}^2\simeq 5.1\times 10^5\mathrm{eV}$,Their thermal speed is of order $\sqrt{3k_B{T}_e/m_e}\simeq 0.04c$. The gas is fully ionized because its temperature is far above the recombination temperature, but ionization does not imply relativistic electrons. Thermal electron-positron pairs are exponentially suppressed at this temperature.^[ref]

The given age and scale-factor ratio should be treated as independent supplied estimates in the following rate comparison. They do not describe the same epoch in standard radiation-era expansion: $a/a_0=10^{-6}$ normally corresponds to an age of order a year, not $10^4$ years. The density and temperature estimates above depend on the scale-factor ratio, not on assigning it the stated age.

Use $dz=d^{3}x{d}^{3}v$ and setFor unit particle mass, $\rho$ is both the number density and the mass density in the number convention of c. A fixed reference scale inside the logarithm changes $\mathcal{H}$ by a constant multiple of the conserved total mass, leaving its derivative unchanged.

Both prefactors in the source's part (d) need correction. Multiplication of the mean-field alignment equation by $|\mathbf{v}{|}^2/2$ and integration by parts givesExchange $z$ and $z_{\star }$, use the symmetry of $K$, and average the two expressions:The printed right-hand side omits $1/2$. This corrected velocity-alignment energy dissipation agrees with the pair-counting calculation in a.

For velocity-alignment entropy production, let $A(\mathbf{x},t)=\int K(\mathbf{x},{\mathbf{x}}_{\star })\rho ({\mathbf{x}}_{\star },t)d^3{x}_{\star }$. The velocity-space divergence is ${\nabla }_{\mathbf{v}}\cdot \mathbf{L}[f]=-3A$. Assuming $f>0$ and sufficient decay, integration by parts yieldsThe printed formula omits the factor $3$ from the three velocity components. In $d$ spatial dimensions, that factor would be $d$. Zeros of $f$ can be treated by a regularized entropy calculation when the required integrability holds.

A concrete smooth counterexample to both printed coefficients is $K\equiv 1$ and $f(z,0)=g(\mathbf{x})(2\pi {)}^{-3/2}{e}^{-|\mathbf{v}{|}^2/2}$, where $g$ is a positive normalized Gaussian distribution. The total mass is one, the total momentum is zero, and $\mathbf{L}[f]=-\mathbf{v}$ initially. The kinetic energy derivative is $-\int |\mathbf{v}{|}^{2}fdz=-3$, whereas the printed formula gives $-6$. The entropy functional derivative is $3$, whereas the printed formula gives $1$.

The signs describe cooling of relative velocities and contraction in phase space: $\mathcal{H}$ increases, so the fine-grained Gibbs entropy $-\mathcal{H}$ decreases. These dissipative particle dynamics are not an isolated Hamiltonian system; there is no contradiction with Liouville's theorem or the Boltzmann equation's usual entropy increase. The microscopic mechanism removing kinetic energy is outside this alignment model.

The step in the source must be at zero. We use the Heaviside step function $H(x)=1$ for $x>0$, zero for $x<0$, with its value at zero immaterial here. The printed threshold $1$ is inconsistent with both $H^{\prime }=\delta (x)$ and the stated number density. For example, measuring velocities in fixed units and taking $v_F=2$, the literal printed cutoff fills only $v<1$ and gives one eighth of the stated number density. Its distributional derivative is $\delta (x-1)$, not $\delta (x)$. The zero-threshold convention restores the stated Fermi sphere and its normalization of a probability distribution.

Let $k=|\mathbf{k}|>0$, choose $u=\mathbf{v}\cdot \mathbf{k}/k$, and abbreviate $V_{\alpha }=v_{\mathrm{F}\alpha }$. The three-dimensional phase-space distribution function is $f_{0\alpha }=C_{\alpha }{1}_{v<V_{\alpha }}$, whereIntegrate over the transverse velocity disk to obtain the parallel marginal distributionThere are no edge Dirac delta functions in $F_{\alpha }^{\prime }$ because $F_{\alpha }$ itself vanishes continuously at the endpoints. In particular, the three-dimensional constant occupation does not imply a constant parallel marginal distribution.

Use time dependence $e^{pt+i\mathbf{k}\cdot \mathbf{x}}$, with $p=\gamma -i\omega$ and $\gamma <0$ for damping. The given dielectric function becomesThe Landau contour supplies causal continuation. For a real phase velocity $a=\omega /k>V_{\alpha }$, there is no resonant pole in the populated interval, andHence this species contributes

For comparable species number densities and fixed charge ratios, $m_e\ll m_i$ implies $V_i/V_e=O(m_e/m_i)$ and ${\omega }_{\mathrm{p}i}^2/{\omega }_{\mathrm{p}e}^2=O(m_e/m_i)$. Since $\omega >k{V}_e$, the ions have $s_i\gg 1$ and ${\chi }_i=-{\omega }_{\mathrm{p}i}^2/{\omega }^2+O(k^2{V}_i^2{\omega }_{\mathrm{p}i}^2/{\omega }^4)$. Their response is negligible to leading order in the mass ratio. The real dispersion relation is thereforeThis degenerate-plasma dielectric response treats a completely degenerate, nonrelativistic Fermi gas with classical particle trajectories. It does not include quantum recoil or a collision term. A finite ion response shifts the root slightly; the boxed equation is the stated leading electron approximation.

The two real branches start at $\omega =\pm {\omega }_{\mathrm{p}e}$, have zero initial slope, and approach the lines $\omega =\pm k{V}_e$ from outside the populated parallel-velocity region. The original plot solves the degenerate-plasma dielectric response numerically; the shaded wedge shows where particles with resonant parallel velocities exist.

The distinction from a nondegenerate gas with a Maxwellian Maxwell-Boltzmann velocity distribution is its tail. A Maxwellian has resonant particles at every finite phase velocity. Its long-wavelength Langmuir waves have exponentially weak Landau damping, but at shorter wavelengths that damping becomes strong. The ideal degenerate cutoff instead permits this undamped branch to persist continuously into degenerate-plasma zero-point sound. The coefficient $3V_e^2/5$ in c is also different from the classical thermal dispersive coefficient.

A natural additional branch is a degenerate ion-acoustic wave. For singly charged ions with equal electron and ion number densities, degenerate electron pressure supplies a restoring force while ion mass supplies inertia. The electron equation of state has $d{P}_e/d{n}_e=m_e{V}_e^2/3$, suggesting, at long wavelengths,Both species are degenerate here, so $V_i=(m_e/m_i)V_e$. The hierarchy $V_i\ll c_s\ll V_e$ explains the damping: there are no resonant ions, but there are resonant electrons inside their parallel marginal distribution. That electron marginal distribution has a small nonzero slope at $u=c_s$, even though the full three-dimensional occupation is constant inside the Fermi sphere. One therefore expects weak electron Landau damping, of relative order $\sqrt{m_e/m_i}$, rather than exact absence of damping. Ion degeneracy pressure adds a relatively small correction to $c_s^2$. This reasoning concerns an electrostatic ion-acoustic wave; transverse or magnetized branches require additional field dynamics or a background magnetic field.

Assume equal excess electron and electron hole number densities $N_t$, with no additional doping imbalance. At zero temperature, both occupied carrier spheres have the same Fermi wavevector $k_F$ because their spin counting is the same:For $k<k_F$, the conduction band is occupied and the valence band has electron holes, so $f_c-f_v=1$. Beyond $k_F$ the occupations reverse and the medium absorbs. Thus the upper edge of the positive-optical gain interval is the supremumEquivalently $E_{\mathrm{Fc}}=E_g+{\hslash }^2{k}_F^2/(2m_c^{\ast })$ and $E_{\mathrm{Fv}}=-{\hslash }^2{k}_F^2/(2m_v^{\ast })$. With the supplied effective masses,Using the vacuum wavelength $\lambda =hc/(\hslash \omega )$ reverses the endpoints:These boundaries are idealized and open: the lower-energy edge has zero joint density of states, and the upper edge is the zero-temperature occupation discontinuity, not a finite-width gain peak. The calculation gives positive material optical gain, not the stronger requirement of exceeding the losses at laser threshold.

A double heterostructure laser sandwiches a narrow-band gap active layer between wider-band gap cladding layers. In a type-I band alignment, the conduction band edge is lower and the valence band edge higher in the active layer. Electrons therefore see an energy well, and electron holes also see a well because their excitation energy increases when the valence band edge decreases. Both carrier populations remain concentrated where stimulated emission occurs.

The active layer also normally has the higher refractive index, forming an optical waveguide. The optical normal mode overlaps the injected carriers strongly, and the wider-band gap cladding has less interband absorption at the emitted photon energy. Carrier confinement reduces leakage, while optical confinement reduces the required pump current and hence heating.

Writing $\Gamma$ for the fraction of the optical normal mode overlapping the active layer, the threshold condition is $\Gamma g={\alpha }_i+{\alpha }_m$, with internal and mirror losses ${\alpha }_i,{\alpha }_m$. A larger overlap and a higher carrier number density at a given current both lower the threshold current. The figure shows composition-induced offsets; a working junction also has band bending under bias, which is omitted to make the confinement mechanism visible.

Assume the two thermodynamic phases have similar atomic volumes, so use $f_V\simeq 0.15$. Model the stated average pitch $L=1\mu \mathrm{m}$ by one sphere per representative cubic volume $L^3$. If the sphere diameter is $d$,Using the more literal lever-rule reading0.145 gives about0.65micrometres instead, within graphical uncertainty. The central estimate is therefore a diameter of roughly two-thirds of a micrometre.

A random distribution does not make a nearest-neighbor mean distance exactly equal to the cube root of inverse number density. The representative cubic spacing is an explicit geometric closure, not evidence that the precipitates actually form a cubic lattice. Unequal molar volumes or a different spacing statistic would change the inferred size.

For the spherical Abel deprojection, put $A_r=r^2+b^2$ and $s=(R^2+b^2)/A_r$. Then $d{R}^2=A_{r}ds$ and $\sqrt{R^2-r^2}=A_r^{1/2}\sqrt{s-1}$. Consequently the integral appearing in the reconstruction of $n$ isDifferentiate before substituting the numerical value of $I_{5/2}$:For the projected second stellar velocity moment, $N{\sigma }^2=N_0{\sigma }_0^2{b}^6(R^2+b^2{)}^{-3}$. The identical substitution in isotropic stellar pressure deprojection givesFor completeness, setting $t=1/s$ yields $I_{\alpha }=B(1/2,\alpha -1/2)$, where $B$ is the Euler beta function. The Gamma function recurrence gives $I_{5/2}=4/3$ and $I_3=3\pi /8$. This also checks the normalization of the power-law spherical projection kernel.

The radial Jeans equation is $p^{\prime }=n{\psi }^{\prime }$. Since $p^{\prime }=-105N_0{\sigma }_0^2{b}^{6}r/(16A_r^{9/2})$, division by the tracer number density givesWriting $A=315\pi {\sigma }_0^{2}b/128$, the total gravitational acceleration is thereforeTo find all gravitating mass density, apply the Poisson equation for Newtonian gravity, with the same sign convention:Equivalently, the enclosed mass is $M(r)=A{r}^3/[G(r^2+b^2{)}^{3/2}]$, and ${\rho }_{\mathrm{tot}}=M^{\prime }/(4\pi r^2)$. The inferred total mass is $A/G$ and its relative potential is $A/\sqrt{r^2+b^2}$, a Plummer model. The tracer number density has a different radial exponent from this total mass density. One must not multiply $n$ by an arbitrary stellar mass and identify it with all matter: the inferred field can include unobserved stars, gas, and dark matter.

Use the same conserved-source interpretation, comoving number density $n_0$ and normalization $a_0=1$ as in the preceding part. DefineThe open matter-dominated Friedmann solution then has $a=C(\cosh u-1)$, $t=(C/\kappa )(\sinh u-u)$ and $dt=(C/\kappa )(\cosh u-1)du$. The present parameter satisfiesThe cosmological bolometric background from conserved sources is thereforeSince ${\cosh }^{2}u=(1+\cosh 2u)/2$, the open dust bolometric background integral isThis leaves the answer in the requested parameter form and makes no use of a radial-distance approximation. For arbitrary proper number density instead, the same calculation gives

For the conserved population, compare models with the same present $n_0,L,H_0$. As ${\Omega }_0\to 0$, $u_0\to \infty$ and $\cosh u_0\sim 2/{\Omega }_0$. The leading term in the square bracket is $\sinh (2u_0)/4\sim 2/{\Omega }_0^2$; the terms linear in $\sinh u_0$ and $u_0$ vanish after multiplication by ${\Omega }_0^2$. HenceThis is also immediate in the Milne model, for which $a=t/t_0$ and $t_0=H_0^{-1}$. If one compares at fixed cosmic age instead, the results are $L{n}_0{t}_0/2$ and $3L{n}_0{t}_0/5$, so the empty open model gives $5/6$ of the flat-model flux. Specifying what is held fixed is necessary because the two expansion histories have different ages at the same Hubble constant.