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Consider a planar circular aperture of diameter $D$ in constant refractive index $n$. Suppose every position in its area $A=\pi D^2/4$ is filled with directions inside the same rotationally symmetric cone about the unit normal, with half-angle ${\theta }_{\max }$. Using spherical coordinates, the exact etendue iswhere $\mathrm{NA}=n\sin {\theta }_{\max }$ is the appropriate side's numerical aperture. The ordinary cone solid angle is $2\pi (1-\cos {\theta }_{\max })$; its cosine-weighted value is $\pi {\sin }^2{\theta }_{\max }$. A full forward hemisphere therefore gives $G=n^{2}A\pi$, rather than $2\pi n^{2}A$. ^[ref]

For an original model example, choose $n=1$, $D=10\mathrm{mm}$, and ${\theta }_{\max }=30^{\circ }$. ThenIf the radiance is uniform across this filled bundle and equals $L=100\mathrm{W}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}$, its radiant flux is $\Phi =LG/n^2\approx 6.1685\mathrm{mW}$. Neither the geometry nor the chosen radiance is an experimental measurement.

A detector collects a range of solid angles, not a single direction. If its nonnegative angular response is $w(\Omega )$, defineIntegrating the polarized beta-decay angular distribution gives a count proportional to $G(1+ak)$. For a uniform forward cone with half-angle ${\theta }_0$, put $c_0=\cos {\theta }_0$. Direct integration givesThe opposite cone has mean cosine $-k$. A hemisphere gives $k=1/2$; a narrow cone tends to $k=1$. Hence a spin-reversal cross-ratio for beta decay measures $ak$, and an acceptance correction is needed to obtain $a$.

With unequal angular responses, let the two laboratory mean cosines be $k_{+}$ and $k_{-}$, including their signs. Then the actual expectation iswhich reduces to the simple square only when $k_{-}=-k_{+}$. The cancellation of stable multiplicative detection efficiencies survives, but their angular shapes still matter. An energy-integrated response likewise averages $A(E)P\beta (E)\cos \theta$ with the accepted spectrum; a mean speed cannot in general be substituted independently of acceptance.

Additive backgrounds, unequal polarization magnitudes or detection efficiencies that change on reversal do not have the assumed factorization. For example, a reversal-dependent efficiency cross-ratio multiplies $R$ by ${\eta }_{+}^{+}{\eta }_{-}^{-}/({\eta }_{-}^{+}{\eta }_{+}^{-})$. Background subtraction also creates statistical complications: positive expected signal counts are assumed in this derivation, whereas finite samples with Poisson distributions can contain zero counts. Taking square roots of negative background-subtracted counts is not a valid estimator. These limits explain why reversal and independent preparation checks belong together.

The focal ratio, or f-number, of a converging optical instrument focused at infinity is the dimensionless quantitywhere $f>0$ is effective focal length and $D$ is the on-axis entrance pupil diameter. The notation f/4 means $N=4$, not $f=4$ in a particular unit. A $100\mathrm{mm}$ focal length with $D=25\mathrm{mm}$ gives f/4. Closing the aperture stop until $D=12.5\mathrm{mm}$ gives f/8 while retaining the same focal length. The mechanical opening need not equal $D$. ^[ref].

For a thin lens in air with a circular aperture stop at the thin lens, the converging bundle from a distant on-axis point has half-angle $\alpha$ satisfyingThe last step uses the small-angle approximation. For a compound optical lens, the cone is set by the image-side optical pupil and its distance from the receiving plane; the simple triangle is not an exact high-angle rule. At finite object distance a thin lens instead gives $v=f(1+|m|)$, by its imaging equation and transverse magnification, so its working cone corresponds to $v/D=N(1+|m|)$. Nominal focal ratio therefore does not alone specify close-range image irradiance. ^[ref].

Two useful scalings follow under stated conditions. For a fixed uniform incident beam and unchanged transmission, collected radiant flux scales as $D^2$. For an extended source with fixed radiance, the central image irradiance scales approximately as $N^{-2}$ in the paraxial approximation: the solid angle of the cone scales as $(D/f{)}^2$. For an ideal uniformly illuminated unobscured circular entrance pupil, the first dark ring of the point spread function has angular radius and focal plane radiusHere $\lambda$ is wavelength. These are first-zero radii, not diameters or a universal rule for resolving arbitrary objects. A proof outline evaluates the circular aperture's Fraunhofer diffraction using its Fourier transform, then uses the first zero of the resulting Bessel function. ^[ref].

At $\lambda =0.55\mu \mathrm{m}$, the example's f/4 and f/8 first-ring radii are $2.684$ and $5.368\mu \mathrm{m}$. Closing the aperture stop halves $D$, quarters collected radiant flux and doubles this diffraction radius. It can also reduce optical aberrations by excluding outer light rays; actual sharpness reflects both effects. This explains why a smaller aperture need not produce a sharper image. ^[ref].

A proposed optical bench check puts a fixed collimator before a converging thin lens and a photodetector large enough to collect the entire image behind it. Compare two centred aperture stops of known diameters while fixing source output, detector gain and integration time. The integrated signal tests the $D^2$ collection law; a spatially resolved optical camera at the focal plane can separately measure the point spread function. Repeat a dark reading and the first aperture stop, and check that neither the incoming beam nor detector clips either bundle. A resolved pinhole source, imperfect collimation, optical aberrations and detector pixel size can dominate the apparent spot width; a single peak reading does not measure total collected radiant flux.

For a narrow angular power pattern $\exp [-(u^2+w^2)/(2{\alpha }^2)]$ about an equatorial beam centre, the solid angle measure is locally $dudw$. Extending the small lobe to the plane gives its beam solid angle ${\Omega }_A\simeq 2\pi {\alpha }^2$. Both sides of the azimuthal lobe must be included. A lossless antenna then has peak antenna gain $G_0\simeq 2/{\alpha }^2$ and peak antenna effective area $A_0\simeq {\lambda }^2/(2\pi {\alpha }^2)$.

For tracers sampled uniformly on a Galactocentric shell of radius $r$, with the Sun at radius $R_{\odot }$, the correction in the heliocentric projection of a spherical velocity dispersion isThe limiting value at $r=R_{\odot }$ is $1/2$; it tends to $2/3$ as $r/R_{\odot }\to 0$ and to $2R_{\odot }^2/(3r^2)$ at large radius. Uniform heliocentric solid angle at fixed $r>R_{\odot }$ instead gives $A=2R_{\odot }^2/(3r^2)$ exactly. Specify the selection measure when applying the correction.

For box-normalized plane waves in volume $\mathcal{V}$, a potential has matrix element $\tilde{W}({\mathbf{k}}_f-{\mathbf{k}}_i)/\mathcal{V}$. The final-state density per energy and solid angle is $\mathcal{V}m{k}_f/[(2\pi {)}^3{\hslash }^2]$. Dividing the golden-rule rate by incident flux $\hslash k_i/(m\mathcal{V})$ gives $(m^2/4{\pi }^2{\hslash }^4)(k_f/k_i)|\tilde{W}{|}^2$. Elastic scattering has $k_f=k_i$; the arbitrary box volume cancels.

The lumen is the International System of Units unit of luminous flux: $1\mathrm{lm}=1\mathrm{cd}\mathrm{sr}$. A source with constant luminous intensity $1\mathrm{cd}$ within a cone of solid angle $1\mathrm{sr}$ supplies $1\mathrm{lm}$ into that cone. A source with the same luminous intensity in every direction supplies $4\pi \mathrm{lm}$ over a full sphere. ^[ref]

Luminous intensity is luminous flux per outgoing solid angle, measured in candelas. It is the visually weighted counterpart of radiant intensity. The same total luminous flux concentrated into a narrower cone can give a greater luminous intensity within that cone.^[ref]

How much does a small directional uncertainty help? Consider an original idealized model: a source direction fixed in advance, uniform background directions over the full sphere, and circular acceptance radius $\psi$. The accepted solid angle and single-event background probability areThe derivation integrates the area element on a unit sphere. For small $\psi$ in radians, $q\simeq {\psi }^2/4$: halving the radius quarters the accidental acceptance, to leading order. At $psi=0$ acceptance vanishes; at $\psi =\pi$ it covers the full sphere. The solid angle has units of steradians; $q$ is dimensionless.

For $0<\psi \le \pi$, take a deliberately crude point spread function: stipulate that every signal direction lies uniformly inside that cap. The signal and background probability density functions per steradian are $S=1/\Omega$ inside the cap and zero outside, and $B=1/(4\pi )$ everywhere. Both integrate to one. For an event inside the cap their likelihood ratio isThis ratio compares two specified directional models; it is not the probability that the event came from the source. Such a conditional probability would additionally require prior source odds and the other selection information. For fixed source fraction $f$ among selected events, a normalized mixture distribution has density $(1-f)B+fS$. For independent events the likelihood is the product of those densities; its ratio to background alone isOutside-cap events contribute factors $1-f$ rather than being silently discarded. At $f=0$, $\Lambda =1$; for a full-sky cap, $S=B$ and every $f$ gives $\Lambda =1$. A directional model that cannot distinguish source from background carries no directional evidence.

As an original numerical example, radii $1^{\circ }$ and $10^{\circ }$ give $q\simeq 7.615\times 10^{-5}$ and $7.596\times 10^{-3}$, respectively: nearly a hundredfold difference. For a fixed number $N$ of independent background events, the binomial distribution givesWith chosen $N=100$, those radii give approximately $0.00759$ and $0.5335$. A small single-event acceptance can become a common coincidence after many opportunities. This probability concerns a fixed cap, rather than a scan for the most favorable cap anywhere. For $m$ independent searches with common local tail probability $p$, the corresponding global tail probability is $1-(1-p{)}^m$; overlapping sky searches generally violate that independence. Actual source tests must include directional response, nonuniform exposure, energy information and the look-elsewhere effect in their null calibration.

For an ultrarelativistic electron scattered elastically by a static, spherical, spin-zero atomic nucleus, a plane-wave Born approximation retains a single electromagnetic interaction. The Fourier transform of its Coulomb potential is proportional to $ZeF(q)/q^2$. Relative to the same point charge, the scattering amplitude is therefore multiplied by $F(q)$ and the differential scattering cross-section by $|F(q){|}^2$:Here $E\gg m_e{c}^2$ is the incident total electron energy, $\theta$ the scattering angle, $d\Omega$ a solid angle element, and $\alpha$ the fine-structure constant. The cosine factor already accounts for the projectile's spin; a spinless projectile would give a different point-charge factor. The formula neglects nuclear recoil, electron rest-mass corrections, radiative corrections and Coulomb distortion in electron scattering. The latter can be substantial at large $Z\alpha$: electron waves then require treatment in the target's Coulomb potential, rather than undistorted plane waves.

Nonzero target spin can introduce magnetic and additional electric contributions, so division by this point-charge expression does not generally isolate one charge form factor for a proton or deuteron. Even for spin zero, finite energy resolution, recoil and Coulomb distortion in electron scattering can change a measured minimum. A zero of the uniform-sphere model is a clear illustration of amplitude cancellation, not a universal prediction of a zero experimental counting rate.

For a uniformly emitting spherical body of radius $R$ at Euclidean distance $d\gg R$, integration of its specific intensity over the apparent solid angle gives $f_{\lambda }=\pi I_{\lambda }(R/d{)}^2$. For a blackbody, $I_{\lambda }=B_{\lambda }(T)$ and the bolometric result is $f={\sigma }_{\mathrm{SB}}{T}^4(R/d{)}^2=L/(4\pi d^2)$. A grey emissivity factor multiplies this expression; isothermality alone does not imply blackbody emission.

Fermi's golden rule converts a weak transition matrix element and the available density of final states into a transition rate. For a static perturbation $W$ and a continuum,When the matrix element is nearly constant on the energy shell, this becomes $(2\pi /\hslash )|W_{fi}{|}^2\rho (E_i)$. The energy delta function is to be integrated, rather than interpreted as a finite rate to one exactly resonant discrete level. The observation time must resolve a smooth final-state continuum while remaining short enough that first-order depletion is small. A periodically driven perturbation instead gives the corresponding absorption and emission energy shells.

For elastic scattering from a weak localized potential $W(\mathbf{r})$, use box-normalized free states $⟨\mathbf{r}|\mathbf{k}⟩={\mathcal{V}}^{-1/2}{e}^{i\mathbf{k}\cdot \mathbf{r}}$. ThenThe number of final states is $\mathcal{V}{k}_f^2\mathrm{d}{k}_f\mathrm{d}\Omega /(2\pi {)}^3$. With $E_f={\hslash }^2{k}_f^2/(2m)$, the density per energy and solid angle is $\mathcal{V}m{k}_f/[(2\pi {)}^3{\hslash }^2]$. Dividing the resulting golden-rule rate by the incident single-particle flux $j_i=\hslash k_i/(m\mathcal{V})$ givesThis is the golden-rule derivation of the Born cross-section. For the elastic problem $k_f=k_i$, so the first Born scattering amplitude and cross-section areThe box volume cancels, as a physical cross-section must. The minus sign is fixed by the outgoing-wave scattering convention; the golden-rule rate alone determines only the modulus.

For example, $W(r)=g{e}^{-\kappa r}/r$ has $\tilde{W}=4\pi g/(q^2+{\kappa }^2)$, hence $f_B=-2mg/[{\hslash }^2(q^2+{\kappa }^2)]$, with $q=2k\sin (\theta /2)$. Replacing the exact scattering state by the incident plane wave is the essential Born approximation. It can fail for strong scattering, low-energy resonances and situations requiring repeated scattering; a small perturbation and a suitable continuum are separate conditions.^[ref]

For elastic scattering by the Yukawa potential, take $\kappa >0$ and use the first Born approximation in the plane-wave normalizationChoose the polar axis along the momentum transfer $\hslash \mathbf{q}$. The angular integral isThe radial volume element cancels the remaining powers of $r$, leavingFor real $A$, the differential scattering cross-section is thereforeThis also has a finite forward limit when $\kappa >0$. The Born amplitude for a Yukawa potential provides a screened route to the Coulomb potential by taking $\kappa \to 0^{+}$ at fixed $q>0$.

For an alpha particle of charge $2e$ scattering from a heavy nucleus of charge $Ze$, set $A=2Z{e}^2/(4\pi {ϵ}_0)$. Neglect recoil, so that the reduced mass is approximately the projectile mass, $E={\hslash }^2{k}^2/(2m)$, and elastic geometry givesThe Coulomb-limit scattering amplitude becomeswhich gives the Rutherford differential cross-sectionThe unscreened Coulomb potential has long range, so taking this limit does not justify the Born approximation for its full scattering state, particularly for a strongly charged target. The exact Rutherford scattering amplitude differs by a phase and has the same modulus. Thus the resulting differential scattering cross-section remains applicable to the ideal nonrelativistic Coulomb problem beyond the weak-potential regime.

To relate the detector counts to the atomic number, first compute the number of target atoms per illuminated area. With molar mass $M=0.197\mathrm{kg}\mathrm{mo}{\mathrm{l}}^{-1}$, the number density integrated through the foil isThe detector's transverse area is already its area normal to the scattered ray. Its small solid angle isThere is no extra $\cos 45^{\circ }$ factor: the foil is traversed normally and the detector area is transverse to its line of sight.

For a thin foil with negligible multiple scattering and energy loss, the detected fraction is $f=\mathcal{N}(d\sigma /d\Omega )\Delta \Omega$. Thus the measurements implyUsing $E=7.68\times 10^{6}e\mathrm{J}=1.23047\times 10^{-12}\mathrm{J}$ in the Rutherford differential cross-section gives

The stated measurements do not reproduce gold's atomic number 79 in this model. The values and units above agree with the original PDF. Substituting $Z=79$ instead would predict $f\simeq 3.17\times 10^{-7}$ with the other inputs unchanged, rather than $3.7\times 10^{-7}$. No stated correction accounts for that discrepancy; forcing the answer to 79 would require changing the data or adding an experimental correction not specified in the problem.

Choose the baseline vector from the first antenna to the second. Apart from a sign convention for receiver delay, the geometric delay is ${\tau }_g=\mathbf{b}\cdot \mathbf{s}/c$, and the phase difference is $\varphi =2\pi \nu {\tau }_g$. For equal received source powers $P_a$, adding the voltages and averaging over many wave periods givesUncorrelated receiver noise adds a mean level but not the source's coherent cross term. For an east-west baseline and one choice of hour-angle sign, ${\tau }_g=(b/c)\cos \delta \sin H(t)$. Earth rotation therefore sweeps the phase; locally the fringe frequency is $(b/\lambda )\cos \delta \cos H\dot{H}$. Tracking the antenna pointing keeps the source in the primary beam. Electronic fringe stopping is a separate operation: it can remove this phase evolution, making a source at the phase centre constant in the corrected complex visibility. The oscillating power above is the geometrical fringe before that correction.

For a small angular displacement $d\theta$ along the projected baseline, $d\varphi =2\pi b_{p}d\theta /\lambda$. One full fringe therefore spans $\Delta \theta \sim \lambda /b_p$. Here $b_p=|\mathbf{b}-(\mathbf{b}\cdot \mathbf{s})\mathbf{s}|$. This is the angular period of the measured Fourier component and the characteristic resolution scale when such long baselines are synthesized; precise beam width and image ambiguity also depend on baseline coverage and weighting. A single pair by itself does not give an unambiguous complete image.

Writing $x=h\nu /(k_{B}T)$, the expansion $e^x-1=x+O(x^2)$ immediately gives the Rayleigh-Jeans law,The factor two counts the two independent modes of polarization of waves. A small brightness-temperature perturbation thus has $\Delta I_{\nu }=2k_B{\nu }^2\Delta T_{\mathrm{RJ}}/c^2$ in this limit.

Hot intracluster electrons inverse-Compton scatter the microwave-background photons. The scattering optical depth is $d\tau =n_e{\sigma }_{T}dl$ and the characteristic thermal fractional energy transfer is of order $k_B{T}_e/(m_e{c}^2)$. The integrated distortion is therefore controlled byScattering conserves photon number while shifting photons toward higher frequencies, giving a low-frequency decrement and a high-frequency increment. Thus the printed estimate correctly gives a negative change of order $y{T}_{\mathrm{CMB}}$. Its numerical coefficient, however, is not the standard Rayleigh-Jeans coefficient. For nonrelativistic electrons and small $y$, the Kompaneets equation gives $\Delta n=y{x}^{-2}{\partial }_x(x^4{\partial }_x{n}_0)$. At small $x$, $n_0\simeq 1/x$, so $\Delta n\simeq -2y/x$ andEquivalently the thermodynamic-temperature spectral factor is $x\coth (x/2)-4\to -2$. The original PDF really prints the estimate with $-y$, not $-2y$; both numerical conventions are retained below. Relativistic corrections at the supplied $T_e$ are small compared with this factor-of-two difference.^[ref]

The uniform cube has a constant line-of-sight depth $L=6\times 10^{22}\mathrm{m}$. Using the supplied cross-section rather than replacing it with a more precise value givesAdopt $T_{\mathrm{CMB}}=2.72548\mathrm{K}$, explicitly an external measured constant, not a temperature supplied in the question. The values $k_B=1.380649\times 10^{-23}\mathrm{J}{\mathrm{K}}^{-1}$, $c=299792458\mathrm{m}{\mathrm{s}}^{-1}$ and $m_e\simeq 9.10938\times 10^{-31}\mathrm{kg}$ set the physical units. Hence the printed relation gives $\Delta T\simeq -22\mu \mathrm{K}$; the standard Rayleigh-Jeans result gives $\Delta T\simeq -44\mu \mathrm{K}$. Keeping the answer as $-8.14\times 10^{-6}{T}_{\mathrm{CMB}}$, or twice that, avoids needing a numerical background temperature.^[ref]

The face is a square, so its solid angle is $\Omega =(6\pi /10800{)}^2=3.046\times 10^{-6}\mathrm{sr}$, not the area of a circular six-arcminute aperture. At $15\mathrm{GHz}$,Thus $\Delta S_{\nu }\simeq -4.67\times 10^{-4}\mathrm{Jy}=-0.47\mathrm{mJy}$ using the printed estimate, or $-0.93\mathrm{mJy}$ using the standard factor two. The sign denotes missing brightness relative to the surrounding microwave background. At this frequency $h\nu /(k_B{T}_{\mathrm{CMB}})\simeq 0.264$; the exact nonrelativistic spectral intensity differs from the strict Rayleigh-Jeans approximation by about one percent. This precision does not affect the estimates.

For the array noise, the system temperature sets the receiver noise scale; let $A_e={\eta }_A\pi d^2/4$ be the effective area of each dish, and let ${\eta }_c$ describe correlation losses. No efficiencies or polarization of waves count are supplied, so first take ${\eta }_A={\eta }_c=1$ and a single independent polarization of waves. An unpolarized point source produces antenna temperature $T_a=S_{\nu }{A}_e/(2k_B)$, and the system-equivalent flux density is $\mathrm{SEFD}=2k_B{T}_{\mathrm{sys}}/A_e$. The interferometric radiometer equation gives one baseline an rms flux error $\mathrm{SEFD}/\sqrt{2\Delta \nu t}$. There are $N(N-1)/2=45$ cross-correlation baselines; averaging their independent thermal noise for a point source givesThe ideal single-dish effective area is $4\pi {\mathrm{m}}^2$, giving $\mathrm{SEFD}=6.59\times 10^3\mathrm{Jy}$ andCombining two independent parallel-hand polarization of waves measurements gives $6.3\mu \mathrm{Jy}$. Nonunit efficiencies increase these estimates by $1/({\eta }_A{\eta }_c)$. Here signal equal to noise means a weak-source signal-to-noise ratio of one after integration; it does not mean equating the source to the entire unaveraged receiver-noise power.^[ref]

At the band centre, $\lambda =0.01999\mathrm{m}$. Taking a representative maximum projected baseline of $20\mathrm{m}$ gives ${\theta }_{\mathrm{res}}\sim 3.4$ arcminutes. A corner-to-corner baseline of $20\sqrt{2}\mathrm{m}$ would give about $2.4$ arcminutes before projection and weighting. The exact antenna locations and source declination are not supplied, so neither a unique synthesized beam shape nor an exact FWHM is determined. The individual-dish primary-beam scale, roughly $\lambda /4\mathrm{m}\sim 17$ arcminutes, is the field of view, not the synthesized resolution.

To extract the cluster signal, retain frequency-resolved complex visibilities and fit an extended SZ template together with contaminating compact sources. For the ideal uniform square centred at the phase centre and with a nearly constant primary beam across it, the Fourier transform isThis is why the integrated decrement must not be treated as a point-source flux on every baseline. The first axial visibility null occurs at $b=\lambda /\Theta \simeq 11.5\mathrm{m}$. Short baselines carry most of the large-scale decrement; the longest baselines are useful for separating unresolved positive radio sources. natural visibility weighting or a suitable taper gives a clearer extended-source image than emphasizing the longest baselines, while a visibility-domain matched filter uses all samples with weights proportional to template amplitude divided by noise variance. Its amplitude estimate is $\hat{a}=\mathrm{Re}(T^{†}{C}^{-1}d)/(T^{†}{C}^{-1}T)$ for a real template amplitude. A joint fit propagates radio-source uncertainty instead of subtracting exact, noise-free source models.

Earth-rotation and multifrequency synthesis improve Fourier coverage, but each channel must use its own baseline in wavelengths, primary beam and SZ spectrum. Use the longer baselines and spectra to identify variable point sources, flag bad data, and test the decrement with time, antenna and frequency splits. Primary-beam and absolute-flux calibration must be included. Large-scale microwave-background fluctuations and residual ground or atmospheric signals can remain even below the thermal-noise estimate. The narrow low-frequency band alone gives limited separation of SZ and primary microwave-background temperature structure. There is no zero-spacing interferometric datum: integrated flux is model dependent unless sufficiently short baselines or additional total-power information are available. Exact cluster detection significance cannot be obtained from the point-source noise alone without the actual projected baseline distribution and weighting.

There are four relevant tree-level Feynman diagrams: $s$-channel photon, Z boson and Higgs boson exchange, and $t$-channel electron-neutrino exchange. The electron and Positron are incoming, and $W^{-}$ and $W^{+}$ outgoing. The Higgs diagram is normally tiny because its electron coupling is proportional to $m_e$; it should nevertheless be included in the requested list.

Let $v=2m_W/g$. The Higgs-electron coupling is $m_e/v$, while the Higgs boson coupling to W bosons is $2m_W^2{g}_{\mu \nu }/v$. Multiplying the vertices and scalar propagator gives, up to an irrelevant overall phase,A scalar propagator has no vector index, the Yukawa interaction supplies the scalar spinor bilinear, and the second vertex contracts the two polarization vectors. We retain the tree propagator and neglect widths, consistently with the stated leading-order calculation.

For unpolarized incoming beams the spin average is $1/4$, not a sum without normalization. The fermion spin sums and gamma-matrix trace identities giveThe terms with one gamma matrix have zero trace; the two-gamma trace gives $4p_1\cdot p_2$. Both final polarization states are specified as longitudinal, so no final polarization average is appropriate.

In the centre-of-momentum frame, take $k_1=(E,k\hat{\mathbf{n}})$ and $k_2=(E,-k\hat{\mathbf{n}})$, where $E=\sqrt{s}/2$ and $k^2=E^2-m_W^2$. A pair of longitudinal polarization vectors isEach obeys ${ϵ}_i\cdot k_i=0$ and ${ϵ}_i^2=-1$. Their contraction is angle independent:Consequently

For the two-body scattering cross section, the outgoing-to-incoming momentum ratio isThe source's final sentence interchanges the particle names attached to these two momenta; physical final-state phase space and initial-state flux fix the ratio as written here. Since the Higgs-only spin-averaged expression is isotropic, integrating the solid angle gives $4\pi$:There is no identical-particle factor: $W^{+}$ and $W^{-}$ are distinguishable. The threshold factor ${\beta }_W$ makes this contribution vanish when the outgoing momentum vanishes. This is the isolated squared Higgs amplitude for a specified longitudinal pair, not the complete physical cross section: the full result squares the coherent sum of all four amplitudes and includes their interference.

The solar corona is a hot, dilute plasma in which collisional ionization mainly proceeds by direct electron impact and by excitation-autoionization. The principal inverse ion-changing processes are radiative recombination and dielectronic recombination. Three-body recombination is negligible at ordinary coronal densities. Photoionization, stimulated recombination and charge exchange are normally secondary in this fully ionized, collision-dominated regime, although they should be reconsidered near irradiated or neutral interfaces.

For collisional ionization equilibrium, the adjacent ion stages satisfytogether with conservation of the element's total abundance. Thus the low-density ion fraction depends primarily on $T_e$. This fixes the amount of Si X available for line emission, but not the populations of its excited levels: a coronal approximation does not impose a Boltzmann level distribution.

Within an ion, electrons collisionally excite, de-excite and redistribute levels. Their collision rate coefficient for an atomic transition isThe rate per ion is $N_e{q}_{ij}$. Maxwellian detailed balance gives, for $E_j>E_i$,Strong allowed transitions undergo rapid spontaneous emission; forbidden magnetic-dipole and electric-quadrupole transitions give much smaller decay rates and can leave a metastable atomic level appreciably populated. Radiative cascades feed levels from higher states. Absorption and stimulated emission add rates proportional to the local radiation field through the Einstein coefficients when significant. Proton collisions can also mix closely separated levels separated by fine structure and should be included when their rates matter.

Write $f_i=N_i/N(\mathrm{Si}\mathrm{X})$. A stationary collisional-radiative model solveswhere, in the simplest electron-plus-radiative model, $P_{ij}=N_e{q}_{ij}+A_{ij}$ with $A_{ij}=0$ for an upward transition. Add proton and radiation rates when appropriate. Include enough levels and radiative branches to capture population feeding. The thermal collision data, radiative probabilities and wavelengths then determine each line's optically thin atomic line intensity:The factor $4\pi$ distributes isotropic emission over solid angle. photon intensities omit $h\nu$. A practical source of the necessary atomic rates and population calculations is the CHIANTI reference guide.

Label the two ground-configuration levels $1,2$ and the two indicated excited levels $3,4$:The common $1s^2$ core is omitted. Si X is silicon with nine electrons removed, hence five bound electrons; the left superscript $2$ on $P$ or $D$ is the spin multiplicity, not another electron-occupancy exponent.

The observed lines correspond to decays $3\to 1$ and $4\to 2$. For approximately uniform plasma at the indicated $T_e\simeq 1.3\times 10^6\mathrm{K}$, form the theoretical energy-intensity ratioThe common abundance, Si X fraction and path length cancel, leaving a ratio governed by level populations. The frequency factor is ${\nu }_{42}/{\nu }_{31}=347/356$ for the rounded wavelengths. With observed photon counts, first apply the detector calibration and use the corresponding photon-ratio convention.

The electron-density diagnostic from metastable populations can be seen in a reduced rate model. The upper ground level $2$ has only a weak radiative decay to $1$. Ignoring higher-level feeding for this explanation, their balance givesAt low density $r\simeq N_e{q}_{12}/A_{21}$; above the critical density of an atomic transition $A_{21}/q_{21}$ it approaches $q_{12}/q_{21}=(g_2/g_1)e^{-\Delta E_{21}/k{T}_e}$, approximately two when the gap due to fine structure is small compared with $k{T}_e$. This saturation can occur while the strong upper-level transitions still remain radiatively dominated.

Let $A_3,A_4$ be the total radiative loss rates of levels $3,4$, and $b_{31}=A_{31}/A_3$, $b_{42}=A_{42}/A_4$ their line branching fractions. In that upper-level low-density regime,ThusThe two lines receive different relative feeding from the two ground levels, so their ratio changes as collisions populate the metastable level. More explicitly,Different, nonproportional feeding coefficients give a usable density-sensitive interval. With $s=R/K$, this reduced model even permits explicit inversion:Physical solutions require positive density and a ratio within the allowed calibration range. The actual Si X calibration should use the full multilevel population system, not discard cross-excitation, cascades or all competing branches simply to use this illustrative formula.

Operationally, evaluate the model ratio at the known temperature over a density grid and match the calibrated observed ratio to it, propagating line-flux and atomic-data uncertainties. No observed ratio or numerical atomic rate set is supplied here, so the result is this inference procedure rather than a numerical density. At low- or high-density saturation, the ratio constrains density poorly. Deconvolve blends or sum every unresolved transition in the theoretical feature; for example, the neighboring ${}^2{D}_{3/2}\to {}^2{P}_{3/2}$ branch can contribute near the 356-Angstrom feature when unresolved. Atomic line identifications and that nearby branch are tabulated in Del Zanna's coronal-spectroscopy thesis.

For a nonuniform sightline, the ratio is insteadA single-density inversion gives an emission-weighted effective density from a line ratio, generally not the volume-average electron number density. The density estimate is meaningful within its temperature, optical-depth, line-blending and spatial-uniformity assumptions. This is the precise interpretation of an average density obtained from the observed pair.

Use the luminosity convention in the supplied flux relation: $P_{\nu }=L_{\nu }/(4\pi )$, where $L_{\nu }$ is the total spectral luminosity. For a general spectrum $P_{\nu }\propto {\nu }^{\alpha }$, the cosmological spectral flux-density relation givesIn a spatially flat universe, a shell subtending the solid angle $d\Omega$ contains $dN=n_0{r}_e^{2}d{r}_{e}d\Omega$ sources, because $n_0$ is their comoving number density. The cosmological spectral background intensity per unit solid angle is consequentlyThis is the cosmological background intensity from comoving emissivity specialized to identical, nonevolving sources in an Einstein-de Sitter universe. For $\alpha =1$ the integrand reduces to $(1+z{)}^{-3/2}$, or directly $d{I}_{{\nu }_0}=n_0{P}_{{\nu }_0}d{r}_e$. The finite comoving particle horizon therefore givesFor $\alpha =2$, instead,The emitted frequency grows with redshift, and the rising spectral luminosity now defeats the redshift dimming. The Einstein-de Sitter spectral-background convergence criterion is $\alpha <3/2$; equality produces a logarithmic divergence. This divergence describes the unbounded power-law, eternal-population model. A finite formation epoch, a high-frequency spectral break or absorption changes that model and can make its background finite.

An isotropic source's bolometric luminosity is distributed over a wavefront of physical area $4\pi a_0^2{r}^2$. Each photon loses an energy factor $(1+z{)}^{-1}$ through cosmological redshift, and successive photons arrive with a further rate factor $(1+z{)}^{-1}$ through cosmological time dilation. Therefore the radiative flux and luminosity distance areThese are bolometric formulas; a fixed observed frequency would require the shifted emission spectrum as well.