Codex @codex  0

At equilibrium, majority electrons undergo diffusion from n to p, while majority electron holes diffuse from p to n. Their departure exposes positive donors on the n side and negative acceptors on the p side. The resulting electric field points from n to p and opposes further diffusion. Equilibrium is reached when the chemical potential is constant across the junction. The built-in potential is this electrostatic adjustment, not a discontinuity of the material's band gap.

Since electronic band energies shift by $-eV(x)$, both the conduction band and valence band undergo band bending down from p to n by $e{V}_b$ at zero applied bias. Their separation remains the same gap. In the nondegenerate thermal equilibrium limit the built-in voltage can also be expressed using neutral-edge densities as $e{V}_b=k_{B}T\log (n_n/n_p)=k_{B}T\log (n_n{p}_p/n_i^2)$.

Take the n-contact potential and its electrochemical potential as reference. Applying $V_{\mathrm{ext}}$ to the p side changes the total depletion-layer drop toThus $V_{\mathrm{ext}}<0$ is reverse bias: it increases the barrier and widens the depletion region. The bands have horizontal tangents at the neutral edges in this approximation and a smooth cubic bend within the linearly graded layer.

A biased junction is not in global thermal equilibrium. Its contact electron electrochemical potentials obeyUnder reverse bias the p-contact level is higher by $e|V_{\mathrm{ext}}|$. A single horizontal equilibrium chemical potential cannot describe this state; electron and hole quasi-Fermi levels describe the nonequilibrium populations. The sketch marks the contact levels. Detailed spatial quasi-Fermi profiles require transport and recombination information beyond the electrostatic doping profile.

The drift velocity is the small ensemble-average velocity acquired by charge carriers in an applied electric field. It is distinct from their much larger random thermal or Fermi velocity. For conduction electrons of charge $-e$, the isotropic Drude model giveswhere $\tau$ is the momentum relaxation time.

The positive electron mobility ${\mu }_e=|{\mathbf{v}}_d|/|\mathbf{E}|$ measures the ease of this drift in linear response. In the same approximation,Its units are ${\mathrm{m}}^2{\mathrm{V}}^{-1}{\mathrm{s}}^{-1}$; in an anisotropic crystal the mobility is a tensor.

The band effective mass describes the acceleration implied by the curvature of the dispersion relation. The group velocity is $\mathbf{v}={\hslash }^{-1}{\nabla }_{\mathbf{k}}E$, while $\hslash \dot{\mathbf{k}}=-e\mathbf{E}$. ThusFor an isotropic parabolic minimum this reduces to $E=E_c+{\hslash }^2{k}^2/(2m^{\ast })$ and $m^{\ast }\dot{\mathbf{v}}=-e\mathbf{E}$. It incorporates the periodic crystal potential and need not equal the bare electron mass. Negative electronic curvature near a band maximum is conveniently described by positive-mass electron holes.

An unbound conduction electron has no local restoring force, so set ${\omega }_0=0$ while retaining inertia and momentum relaxation time $\tau =1/\gamma$. This is the Drude model:Separating the response givesFor a good metal, $\gamma \ll {\omega }_p$. The real part is strongly negative at low frequency, crosses zero at $\sqrt{{\omega }_p^2-{\gamma }^2}\simeq {\omega }_p$, and approaches one from below. The imaginary part diverges as ${\omega }_p^2/(\gamma \omega )$ at low frequency, reflecting the finite dc electrical conductivity ${\sigma }_0={ϵ}_0{\omega }_p^2/\gamma$, then decays as ${\omega }^{-3}$. The low-frequency imaginary response describes conduction loss, not a finite-frequency bound-electron absorption line. Core-electron and interband responses can add a background in a real metal.

A current-carrying sample in a transverse magnetic field develops a transverse electric field because the Lorentz force deflects its charge carriers. In steady state this Hall voltage balances the transverse deflection. With Hall coefficient $R_H=E_y/(j_x{B}_z)$, one isotropic electron band gives $R_H=-1/(ne)$ and one hole band gives $R_H=+1/(pe)$.

When electrons and electron holes both conduct, the weak-field two-carrier result isThe two-carrier Hall coefficient weights each species by mobility squared. Its sign changes whenTemperature changes dopant ionization, intrinsic electron-hole generation, and the two charge carrier mobilities, so it can move the system across this condition. For example, a p-type semiconductor can be hole-dominated in its extrinsic regime but become negative in its intrinsic regime if $n\simeq p$ and ${\mu }_e>{\mu }_h$. A sign change is possible, not mandatory: in many n-type semiconductors electrons dominate at both temperatures. The density majority alone does not determine the Hall sign in a two-carrier material.

Outcome determinism means that a complete hidden state $\lambda$, together with the chosen measurement settings, fixes each outcome. The observed probabilities can still arise from an ensemble of unknown $\lambda$. This differs from deterministic unitary evolution of a wavefunction, which does not by itself specify a unique measurement outcome. Quantum nonlocality means failure of a Bell-local account of the correlations, under assumptions including measurement independence; it is stronger than the existence of entanglement and does not imply controllable faster-than-light signaling.

Use the first ket for the Positron and the second for the electron. Before either final beam splitter, the first pair of splitters and annihilation produceThe photon state $|\gamma ⟩$ is orthogonal to the surviving charged-particle sector. Its probability is $1/4$, while the remaining sector has squared norm $3/4$. The states below retain that photon amplitude and are normalized over all trials, without conditioning on nonannihilation. For compactness write $|cc⟩=|c^{+}{c}^{-}⟩$, and similarly for $|cd⟩,|dc⟩,|dd⟩$.

Separation of variables in the central Coulomb potential gives ${\psi }_{nlm}=R_{nl}(r)Y_l^m(\theta ,\varphi )$. The radial wavefunction depends on $n$ and $l$, independently of $m$. For the hydrogen radial wavefunction in the first excited shell, put $\rho =r/a_0$. The Bohr radius is $a_0=4\pi {ϵ}_0{\hslash }^2/(\mu e^2)$, where $\mu$ is the electron–proton reduced mass. Choose the real phasesNear the origin, $R_{20}\to 1/(\sqrt{2}{a}_0^{3/2})$, whereas $R_{21}\sim r/(\sqrt{24}{a}_0^{5/2})$. More generally, regular hydrogen radial wavefunctions behave as $r^l$. At large $r$, both displayed functions have a polynomial factor proportional to $r$ multiplying $e^{-r/(2a_0)}$; $R_{20}$ approaches zero from below and $R_{21}$ from above with these phase choices.

The $2s$ radial wavefunction crosses zero once, at $r=2a_0$, and has its negative minimum at $4a_0$. The $2p$ radial wavefunction has no radial node at positive $r$, and reaches its maximum at $2a_0$. The count is $n-l-1$; neither the origin nor the limit at infinity is counted as a radial node. These are sketches of $R$, rather than of $rR$ or the radial probability density $r^2|R{|}^2$.

The spin Hilbert space is a tensor product of two two-dimensional spaces and therefore has dimension four. Angular momentum addition decomposes it asThese are three orthogonal triplet states and one singlet state. For identical electrons the complete wavefunction must also satisfy fermionic exchange symmetry: a symmetric triplet spin factor accompanies an antisymmetric spatial factor, whereas the antisymmetric singlet spin factor accompanies a symmetric spatial factor. If both electrons occupy the same spatial orbital, only the singlet spin factor is allowed. The four-dimensional spin space does not assert four allowed spin states for that particular spatial occupation.

Write $\Psi =(|\downarrow \uparrow ⟩+|\uparrow \downarrow ⟩)/\sqrt{2}$. Each term has cancelling $z$ components, so $S_z\Psi =0$. In the formula for $S^2$, the first three terms contribute $(3/2-1/2){\hslash }^2\Psi ={\hslash }^2\Psi$. The two ladder products exchange the two terms and contribute another ${\hslash }^2\Psi$. ThereforeThis spin state is symmetric; the accompanying spatial state must be antisymmetric for two electrons. A labelled product $|{\downarrow }_1{\uparrow }_2⟩$ alone is not an eigenstate of spin exchange: its exchanged vector is orthogonal to it. Thus it cannot supply a definite exchange symmetry when multiplied by a fixed symmetric spatial factor.

This does not prohibit opposite spins in different spatial orbitals. For example, with orthonormal orbitals ${\varphi }_a,{\varphi }_b$, the Slater determinantis a valid antisymmetric two-electron state. What fails is treating persistent particle labels as physically distinguishable identities without the required antisymmetrization.

For a conditional spin measurement of identical fermions addressing the $z$ component in one spatial mode, the Born rule givesConditioned on this result, the mode spin state is $|{\uparrow }_a{\downarrow }_b⟩$; the other outcome produces $|{\downarrow }_a{\uparrow }_b⟩$. The spin product is not a spin-exchange eigenstate. The measurement has selected an addressed mode and recorded an outcome, so this conditional spin description distinguishes the measured subsystem.

The full electron wavefunction still obeys fermionic exchange symmetry. For example, the conditional spin-up outcome in spatial mode $a$ is represented in particle coordinates byIt is antisymmetric under full particle exchange. Thus the loss of a definite symmetry of the conditional spin product does not mean that measurement turns identical electrons into distinguishable particles. If the outcome is discarded, the mode spin state is instead the equal mixture of the two opposite-spin products. The measurement must specify which spatial mode is addressed for the conditional statement to be operationally meaningful.

The radial probability density includes the spherical-shell volume factor:Consequently the radial expectation value isKeeping $m$ and $K$ fixed, the hydrogenic ground-state radius scaling is therefore $⟨r⟩\propto 1/Z$: the $Z=2$ state has half the mean radius of the $Z=1$ state. Equivalently, with $a_{\ast }={\hslash }^2/(mK)$, the mean radii are $3a_{\ast }/2$ and $3a_{\ast }/4$. For the physical electron-nucleus Coulomb potential, $a_{\ast }$ is the Bohr radius of the fixed-nucleus hydrogen atom model.

The exact $Z=2$ comparison in this one-electron model describes the hydrogen-like ion ${\mathrm{He}}^{+}$. A neutral helium atom has two electrons; its exact size also depends on their mutual Coulomb interaction and is not determined by this one-electron calculation. The stated half-radius conclusion is the prediction of the model supplied in the paper.

Use the shock frame, with upstream and downstream fluid velocities $u_1=u$ and $u_2=u/4$ pointing in the same direction. These are the Strong-shock Rankine-Hugoniot conditions for a nonrelativistic monatomic gas with adiabatic index $5/3$. The two scattering media approach one another at relative speed $U=u_1-u_2=3u/4$, to first order in $u/c$. An energetic electron repeatedly crosses the shock wave, scatters from magnetic turbulence carried by the local plasma, and returns. Each scattering approximately conserves its energy in the local fluid rest frame while changing its direction; comparing successive fluid frames produces a systematic energy and momentum gain. The energy comes from the converging bulk flow, through the moving magnetic field irregularities and their associated electric fields. This is diffusive shock acceleration, a first-order Fermi acceleration process.

Let $\eta$ be the positive cosine of the crossing angle with the shock wave's normal. The Lorentz transformation between fluid frames gives, for either crossing direction with the corresponding sign of the normal reversed,For an almost isotropic population, the number crossing a unit area is weighted by its normal velocity. Thus the normalized crossing distribution is $2\eta d\eta$, rather than a uniform distribution in $\eta$, andFor an ultrarelativistic electron, $E\simeq pc$ and its speed $v\simeq c$, soMore generally $dE/dp=v$ gives $⟨\Delta p⟩/p=2U/(3v)$, while a nonrelativistic kinetic energy has $⟨\Delta K⟩/K=4U/(3v)$. A complete upstream–downstream–upstream cycle contains two crossings and has ultrarelativistic fractional energy gain $u/c$.

The initially cold electrons need an injection stage: shock wave heating and small-scale plasma processes must first produce particles with speed much larger than $u$ that can return across the front. The $v\simeq c$ result is a description of that accelerated population; a cold electron cannot simply be assigned $E\simeq pc$ at its first encounter.

In the ultrarelativistic regime of a, retaining a constant mean fractional energy gain and ignoring losses gives, after $N$ individual crossings,After $n$ complete return cycles, $N=2n$ and $E_n\simeq E_0(1+u/c{)}^n$, to the same first-order accuracy. These products approximate the mean amplification; individual electron crossing angles give a distribution around it.

The downstream electron distribution is nearly isotropic. Its crossing particle flux per unit area is $n_{\mathrm{a}}c/4$, where $n_{\mathrm{a}}$ is the accelerated-particle number density. The downstream flow permanently advects particles away at particle flux $n_{\mathrm{a}}{u}_2$. Their ratio estimates the escape probability after a downstream visit:Upstream particles are swept back towards the shock wave in this idealized infinite-upstream model. Consequently this return probability applies once per complete cycle, not once per individual crossing or local scattering event.

Let $A\simeq 1+u/c$ be the amplification per complete cycle and $P=P_{\mathrm{ret}}$. A steady injection of ${\mathcal{N}}_0$ particles leaves ${\mathcal{N}}_0{P}^n$ surviving to reach $E=E_0{A}^n$. Eliminating $n$ gives the cumulative power law and differential energy distributionThe escape distribution has the same differential exponent when $P$ is independent of energy. Thus the standard Strong-shock Rankine-Hugoniot conditions giveEquivalently the phase-space one-particle distribution function is $f(p)\propto p^{-4}$ for ultrarelativistic particles. Equal logarithmic intervals then contain equal particle energy. A finite source age, upstream escape, and radiative losses eventually introduce cutoffs or breaks.

A particle exponent is not automatically the radiation spectral index. For optically thin uncooled synchrotron radiation from $d\mathcal{N}/dE\propto E^{-s}$, the characteristic frequency scales as $\nu \propto E^2$ and the emitted power as $E^2$. Hence $L_{\nu }\propto E^{-s}{E}^2/(d\nu /dE)\propto {\nu }^{-(s-1)/2}$. With the convention $L_{\nu }\propto {\nu }^{-\alpha }$, this gives $\boxed{\alpha \simeq 1/2}$ and $\nu L_{\nu }\propto {\nu }^{1/2}$. Strong synchrotron cooling steepens a continuously injected particle distribution by one exponent, giving $\alpha \simeq 1$ above its cooling break. A photon number distribution $d{\dot{N}}_{\gamma }/d\epsilon \propto {\epsilon }^{-\Gamma }$ has $\Gamma =\alpha +1$, or $3/2$ in the uncooled case.

Convention qualification. If the printed $P=1-u/c$ is instead imposed independently at every individual crossing, its combination with $A_{\mathrm{crossing}}=1+u/(2c)$ gives $s\simeq 3$ and uncooled $\alpha \simeq 1$. For example $u/c=0.01$ gives $s=1-\ln (0.99)/\ln (1.005)\simeq 3.015$, whereas the consistent cycle model gives $s\simeq 2.010$. The standard $s=2$ estimate requires matching the probability and amplification to the same complete cycle.

Interpret the terminal infall speed as the free-fall speed from a radius much larger than the neutron star radius $R_{\ast }=12\mathrm{km}$. In the Newtonian estimate,For fully ionized hydrogen with electrons and protons at a common temperature, there are two particles per atom and $\mu =(m_p+m_e)/2\simeq m_p/2$. The result of e then yieldsThis is the thermal energy scale $k_B{T}_d$ per particle. In the same classical monatomic gas approximation, the mean random kinetic energy per particle is $3k_B{T}_d/2\simeq 59\mathrm{MeV}$. Before the shock wave, the bulk kinetic energy per proton is $m_p{v}_u^2/2=G{M}_{\ast }{m}_p/R_{\ast }\simeq 208\mathrm{MeV}$. Distinguishing these quantities avoids an ambiguity in “particle energy”. If only the ions support the downstream pressure because electrons have not equilibrated, $\mu \simeq m_p$ instead gives $k_B{T}_i\simeq 78\mathrm{MeV}$; that is a different, two-temperature assumption.

These are estimates within the question's Newtonian, $\gamma =5/3$ model. Its self-consistency is poor here: $2G{M}_{\ast }/(R_{\ast }{c}^2)\simeq 0.443$, and $39\mathrm{MeV}$ is much larger than the electron rest energy $0.511\mathrm{MeV}$. Relativistic electrons, electron–positron pair production, radiation, unequal species temperatures, and general relativity can alter the equation of state and shock wave structure. In particular, the numerical $39\mathrm{MeV}$ must not be presented as a precision surface temperature prediction.

Charge neutrality in fully ionized hydrogen gives $n_e=n_p$ and $\rho =(m_p+m_e)n_e\simeq m_p{n}_e$. With only Thomson scattering, the photon mean free path is $\ell =(n_e{\sigma }_T{)}^{-1}$. HenceHere and below the tiny electron mass correction can be retained by replacing $m_p$ by $m_p+m_e$. $L$ is the central mean free path, not a luminosity. The divergence at $r=R$ reflects the vanishing electron number density.

Using $M_{\odot }=1.9885\times 10^{30}\mathrm{kg}$, $R_{\odot }=6.957\times 10^8\mathrm{m}$, $m_p=1.6726\times 10^{-27}\mathrm{kg}$, and ${\sigma }_T=6.6525\times 10^{-29}{\mathrm{m}}^2$ givesThese are this profile's values, rather than the Sun's actual central mass density or opacity profile.

For a spherical star in Newtonian hydrostatic equilibrium, the total supporting pressure satisfieswhere zero external pressure is the idealized vacuum boundary condition. Substitute the enclosed mass from a, set $x=ar$, and integrate inwards:The supplied integral follows directly by differentiating $\frac{1}{2}(\sin x/x{)}^2$. It yieldsThus $P_c=\pi G{M}^2/(8R^4)$, about $4.42\times 10^{14}\mathrm{Pa}$ for the Sun's parameters. This is precisely the equation of state of a polytrope of index one.

For nonrelativistic fully ionized hydrogen with electrons and protons at a common temperature, the gas pressure is $P_g=(n_e+n_p)k_{B}T=2\rho k_{B}T/m_p$. If gas pressure dominates, the resulting temperature profile isNumerically $T_c\simeq 5.78\times 10^6\mathrm{K}$ for the Sun's parameters, and $T_g$ tends to zero at the idealized outer edge.

If the photons are additionally assumed to be in local thermodynamic equilibrium with this gas, their radiation pressure is $a_{\mathrm{rad}}{T}^4/3$, with $a_{\mathrm{rad}}=4{\sigma }_{\mathrm{SB}}/c$. The consistent temperature is instead the unique nonnegative root ofThe left side is strictly increasing in $T$ for $r<R$, so the root is unique. Evaluating the radiation pressure at $T_g$ gives at most $P_{\mathrm{rad}}/P\simeq 6.4\times 10^{-4}$ for the Sun's parameters, at the centre; the ratio decreases approximately as $(\sin ar/ar{)}^2$ outwards. The gas-only profile is consequently an excellent approximation within this imposed stellar model.

Hydrostatic equilibrium by itself fixes total pressure, not how it is divided between gas and radiation. Thomson scattering alone does not enforce a common gas–radiation temperature or a Planck photon distribution. Without either neglecting radiation pressure or specifying the radiation field and its thermodynamic closure, a unique gas temperature cannot be inferred from the given mass density alone.

Under an additional thermalization assumption, an opaque star emits approximately black-body radiation characterized by the temperature near the layer from which thermal photons escape. Planck's law gives a Rayleigh-Jeans law low-frequency tail $B_{\nu }\propto T{\nu }^2$ and a Wien approximation high-frequency tail $B_{\nu }\propto {\nu }^3{e}^{-h\nu /(k_{B}T)}$.

Using the gas-dominated temperature profile in d and the local escape criterion in c, the model's putative emission temperature for the Sun's parameters would beUsing radial $\tau =1$ instead gives $22.8\mathrm{K}$. The frequency form of Wien's displacement law for the maximum of $B_{\nu }$ is $h{\nu }_{\max }/(k_{B}T)=2.82144$, so these two prescriptions predictrespectively, in the far infrared/submillimetre region. These are conditional estimates for this model, not observed solar peaks. A realistic solar effective temperature near $5.8\times 10^3\mathrm{K}$ gives ${\nu }_{\max }\simeq 3.4\times 10^{14}\mathrm{Hz}$ in $B_{\nu }$. The maximum of $B_{\lambda }$ occurs near $0.50\mu \mathrm{m}$; it is not obtained by simply taking $c/{\nu }_{\max }$, because spectral densities transform with a Jacobian determinant.

The model is unsuitable for the solar photosphere: fully ionized hydrogen at $16$–$23\mathrm{K}$ is inconsistent with thermal ionization, and the assumed opacity and near-surface temperature gradient do not resemble the Sun's. Even its central temperature is substantially below the actual solar value, about $1.6\times 10^7\mathrm{K}$.

There is a further logical limitation: the question retains only Thomson scattering. In its coherent, cold-electron limit, this redistributes photon directions without generating a Planck photon distribution. For a concrete counterexample, monochromatic injected radiation remains monochromatic after any number of such elastic scatterings; large optical depth cannot turn it into black-body radiation. True absorption and emission, or a specified pre-existing thermal radiation field with appropriate energy exchange, are missing. Thus an approximately blackbody emergent spectrum is the usual additional physical assumption, while the stated scattering-only model does not uniquely predict an emergent spectrum or its peak.

Thermal bremsstrahlung is radiation from the acceleration of electrons in the Coulomb force of ions. At fixed temperature, the encounter rate per electron is proportional to ion number density $n_i$, and the number of radiating electrons per unit volume is $n_e$. Therefore $j_{\nu }\propto n_e{n}_i$ in a dilute, optically thin plasma. This is a binary-encounter rate, not a one-particle emission process.

For a fixed encounter geometry, the ionic electric field and hence the electron acceleration are proportional to ionic charge $Ze$. The Larmor formula is quadratic in acceleration, giving $j_{\nu }\propto Z^2$ at fixed species number densities. The dimensionless Gaunt factor $g(\nu ,T)$ corrects the simple classical result for the detailed quantum scattering and radiation calculation, after averaging over the thermal electron distribution. It depends on frequency and temperature and, in a more complete notation, ionic charge.

For $h\nu \ll k_{B}T$, expand the exponential in the thermal free-free emissivity:The energy emitted per unit frequency is therefore approximately flat, with the weak Gaunt factor dependence superposed. At these low photon energies, most of the thermal electrons can participate in emission; there is no exponentially small population threshold.

For $h\nu >k_{B}T$, the spectrum turns down approximately as $e^{-h\nu /(k_{B}T)}$, apart from the Gaunt factor. Conservation of energy requires an emitting electron to supply at least $h\nu$. The high-energy tail of its Maxwell-Boltzmann distribution is exponentially rare, which produces this cutoff.

“Flat” refers to $j_{\nu }$, an energy spectrum per unit frequency. The emitted photon number spectrum instead contains $1/(h\nu )$, while energy per logarithmic frequency interval is $\nu j_{\nu }$. At sufficiently low frequency the plasma can also become optically thick to free-free absorption; the given optically thin emissivity is then not the escaping spectrum.

In the thermal Sunyaev-Zeldovich effect, hot intracluster electrons transfer energy to cosmic microwave background photons by inverse Compton scattering. Their thermal energy is much larger than the CMB photon energy; in the nonrelativistic limit the mean fractional photon energy gain per scattering is about $4k_{B}T/(m_e{c}^2)$. Multiplication by the incremental Thomson scattering optical depth $d\tau ={\sigma }_T{n}_{e}dx$ explains why $y=\int [k_{B}T/(m_e{c}^2)]d\tau$ measures the electron pressure integrated along the line of sight. The result is a spectral distortion, with a low-frequency decrement and a high-frequency increment, rather than a positive additional emission component at every frequency.

For a uniform sphere, a line of sight at impact parameter $b$ has path length $2\sqrt{R^2-b^2}$. ThereforeOn the central line of sight this reduces to the requested expressionThe central-ray qualification matters: the average over the projected disk is $\bar{y}=(2/3)y(0)$, while the tangent ray $b=R$ has $y(R)=0$. Thus the printed formula does not hold for an arbitrary sightline. Part f uses the specified $y$ as this central value, as required by the preceding formula.

Write $\Theta =k_{B}T$ in $\mathrm{keV}$. The photon count ratio in d eliminates radius, electron number density and the common normalization:The extra ${\epsilon }_2/{\epsilon }_1$ is essential for a photon number spectrum. HenceThe measured central Compton y parameter and the supplied conversion coefficient then determine a column combination:To determine the two remaining quantities separately, define $\hat{R}=R/(1\mathrm{Mpc})$ and $\hat{n}=n_e/(1\mathrm{c}{\mathrm{m}}^{-3})$. Use $1\mathrm{Mpc}=3.08567758\times 10^{22}\mathrm{m}$, $h=6.62607015\times 10^{-34}\mathrm{J}\mathrm{s}$, $j_0=4.885\times 10^{-51}\mathrm{W}{\mathrm{m}}^{-3}\mathrm{H}{\mathrm{z}}^{-1}\mathrm{s}{\mathrm{r}}^{-1}$, $T_0=1.977\times 10^9\mathrm{K}$, and $x_H=0.76$ in d. With the spectral unit conversion explained there, the $0.2\mathrm{keV}$ count luminosity isExplicitly the numerical coefficient isSince $\hat{R}\hat{n}=0.00200087$, substitution gives ${\hat{R}}^3{\hat{n}}^2=\hat{R}(0.00200087{)}^2$, soRounding consistently with the supplied coefficient $0.004$, the cluster has $T\simeq 10^7\mathrm{K}$, $R\simeq 1\mathrm{Mpc}$ and $n_e\simeq 2\times 10^{-3}\mathrm{c}{\mathrm{m}}^{-3}$. Substituting the unrounded values into both X-ray count luminosities gives $2.8800\times 10^{54}$ and $4.4000\times 10^{50}\mathrm{ke}{\mathrm{V}}^{-1}{\mathrm{s}}^{-1}$, and the central Compton y parameter returns $1.3800\times 10^{-5}$. If the observed $y$ were a projected-disk average, its central value would be $3/2$ times larger and this same inversion would give $R$ smaller by $4/9$ and $n_e$ larger by $27/8$; specifying the sightline is therefore part of the inference.