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In the weak-field thin-lens limit, circular symmetry makes the physical deflection depend on the projected enclosed mass. With the mass calculated in b,For ${\sigma }_{\mathrm{DM}}=155\mathrm{km}{\mathrm{s}}^{-1}$, this is $3.36\times 10^{-6}\mathrm{rad}$, orIt is independent of impact radius in this ideal profile.

Enclosed-mass qualification. The source's appeal to Birkhoff's theorem must not be read as permission to use the mass of a three-dimensional sphere of radius $R$ here. Deflection accumulates along the light path. A concrete counterexample is a thin spherical shell of radius $r_s>R$: its spherical enclosed mass at $R$ is zero, but its projected mass is $M[1-\sqrt{1-R^2/r_s^2}]>0$, and it produces a nonzero deflection. The formula above is the axisymmetric projected-mass lens result; it is consistent with b and is not that incorrect spherical-mass interpretation.

Use $R$ for the physical projected radius, or impact parameter, in the lens plane. Although the source calls it angular, a physical length is required in the mass and deflection equations; its relation to an angle is $R=D_L|\theta |$. Integrate along physical line-of-sight coordinate $z$:Consequently the projected cylindrical mass inside the image radius isSubstituting a givesIf an angular radius $\vartheta$ is used instead, write $\Sigma (D_L\vartheta )={\sigma }_{\mathrm{DM}}^2/(2G{D}_L\vartheta )$ and $M_{2D}(<\vartheta )=\pi {\sigma }_{\mathrm{DM}}^2{D}_L\vartheta /G$. The projected mass differs from the spherical mass $2{\sigma }_{\mathrm{DM}}^{2}R/G$; material at three-dimensional radii larger than $R$ still projects into the cylinder.

Interpret ${\sigma }_{\mathrm{DM}}$ as a constant one-dimensional, isotropic velocity dispersion. The effective pressure is $p=\rho {\sigma }_{\mathrm{DM}}^2$, corresponding to the stated gas-temperature analogy. For $\rho =A/r^2$, the spherical enclosed mass isThe two sides of the hydrostatic equilibrium equation are thenEquating them fixes the singular isothermal sphere:For collisionless dark matter this is an isotropic Jeans-equation construction, not a claim that the particles thermalize by gas collisions. The central singularity and infinite total mass show that the untruncated profile is an idealization.

The infrared and optical transitions share the same upper level 3. Using the supplied infrared photon energy givesWith the same $T_e,n_e$ and three-level fraction $f_3$ used in the abundance calculation of b,This exceeds the optical abundance by a factor $5.53$. Because the upper level is common, its population cancels in the intrinsic emissivity ratio:The observed ratio is $1.5\times 10^{-5}$. Thus changing only the electron temperature or a common level-population correction cannot reconcile these two lines from the same gas. Dust extinction suppressing the optical line is a natural explanation. If the far-infrared attenuation is negligible, the required optical attenuation isDifferent apertures, filling factors, optical-depth assumptions or emissivity calibrations could also cause a discrepancy. A total sulfur abundance would still need an ionization correction beyond this ionic comparison.

Use the complete supplied three-level model, including collisions between levels 2 and 3. Put $g_1=4$, $g_2=4$, $g_3=6$, and denote a collisional rate coefficient by $q_{ij}$. Detailed balance givesAt $T_e=10^4\mathrm{K}$, the supplied collision strengths give, in ${\mathrm{m}}^3{\mathrm{s}}^{-1}$,The rounded optical wavelengths set the excitation energies; treating the fine splitting as very small compared with $k_{\mathrm{B}}{T}_e$ changes none of the main conclusions.

Let $D_2=A_{21}+n_e(q_{21}+q_{23})$, $D_3=A_{31}+A_{32}+n_e(q_{31}+q_{32})$, and $R_{32}=A_{32}+n_e{q}_{32}$. Statistical equilibrium for the upper levels isEliminating $n_1$ yields the line-ratio formulaThe same ratio holds for integrated intensities when the two lines sample the same optically thin volume and differential extinction is negligible. Inserting the data gives the particularly simple numerical formThe low- and high-density limits are approximately $1.49$ and $0.442$. The collisional loss channels giveThus the useful density range is around $10^9$–$10^{10}{\mathrm{m}}^{-3}$. Maximizing $|d\log \mathcal{R}/d\log n_e|$ for the rational expression gives its strongest fractional sensitivity at $n_e=7.8607\times 10^8\sqrt{1.49325/0.442056}=1.44\times 10^9{\mathrm{m}}^{-3}$. Omitting the supplied 3–2 collisions would change the critical densities and inferred density appreciably.

For the observed ratio $0.50$, inversion gives $n_e=1.35\times 10^{10}{\mathrm{m}}^{-3}$. This is on the high-density side of the transition, but is not the infinite-density limit. Solving the statistical-equilibrium equations and imposing $f_1+f_2+f_3=1$ gives the level fractionsA rough high-density estimate would instead use $f_i\propto g_i{e}^{-E_i/(k_{\mathrm{B}}{T}_e)}$; it is useful as a limiting check, not an exact solution for $\mathcal{R}=0.50$.

For the separate abundance calculation, use the expressly prescribed $n_e=10^{12}{\mathrm{m}}^{-3}$. The same model now gives $(f_1,f_2,f_3)=(0.77426,0.09045,0.13528)$, nearly the collisional Boltzmann limit. The measured angle-integrated volume emissivity givesThereforeThis is an abundance per free electron; it is approximately $n({\mathrm{S}}^{+})/n({\mathrm{H}}^{+})$ for predominantly ionized hydrogen with $n_e\simeq n({\mathrm{H}}^{+})$. No $4\pi$ factor is inserted because the supplied emissivity is power per volume, not power per volume per steradian. The imposed $10^{12}{\mathrm{m}}^{-3}$ predicts a ratio about $0.443$, rather than exactly $0.50$; the two numerical instructions cannot both describe one uniform zone exactly within this model.

The source's $2p^3$ configuration label is a typographical error for the sulfur valence configuration $3s^{2}3p^3$. This does not alter the stated $J$ values or the calculation above.^[ref]

A spontaneous emission transition has probability per unit time $A_{ji}$ and emits a photon of energy $h{\nu }_{ji}$. For an optically thin gas, its angle-integrated emissivity is $j_{ji}=n_j{A}_{ji}h{\nu }_{ji}$. Electron collisions populate excited levels at rates $n_e{n}_i{q}_{ij}$; they also remove excitation collisionally at rates $n_e{n}_j{q}_{ji}$, returning the energy to the gas rather than producing the line photon. A collisionally excited line is usually emitted by a subsequent spontaneous radiative decay; collisional de-excitation instead quenches it.

The critical density of an atomic transition compares radiative loss from an upper level with its collisional loss. For a two-level model, $n_{\mathrm{crit}}=A_{ji}/q_{ji}$; for several lower or connected levels use the relevant sums. Below this density, excitation is commonly followed by a photon and the line cools the gas. Above it, collisions increasingly suppress radiative emission per excitation and drive the level populations toward Boltzmann ratios.

Optical forbidden emission lines have small $A$ because an electric-dipole transition is forbidden by a selection rule. Metastable levels survive long enough to radiate in dilute ionized nebulae, although collisions quench them in dense gas. Ratios from closely spaced levels with different critical densities are electron-density diagnostics; ratios involving different excitation energies probe the electron temperature through Boltzmann factors. Combining line emissivities with electron density, temperature and an ionization model estimates ionic abundances.

Infrared fine-structure emission lines connect closely spaced sublevels. Their small excitation energies make them useful over temperature ranges where optical excitation is exponentially suppressed, and dust extinction is usually weaker in the far infrared. They are important coolants and density diagnostics in ionized gas, with the choice of ion specifying the ionization zone. Neither an optical nor an infrared ionic abundance automatically equals the total elemental abundance; unobserved ionization stages require an ionization correction.

For the Double-degenerate Type Ia supernova scenario, measure the Type Ia supernova rate per formed stellar mass in old galaxies, grouped by the time since their main star-formation episode. Correct the event counts for observing time and detection efficiency. In an ideal burst population, the normalized rate measures the delay-time distribution directly. If gravitational waves inspiral dominates the delay and the initial white-dwarf separations are uniformly distributed in separation, c predicts a slope $-3/4$.

More generally, if $dN/dA\propto A^{\beta }$ and $t_m\propto A^4$, the predicted rate scales as $t_m^{(\beta -3)/4}$. A distribution uniform in $\log A$ has $\beta =-1$ and gives $t_m^{-1}$. Thus a slope differing from $-3/4$ does not alone rule out mergers: the separation distribution must also be tested.

An extended star formation history gives a convolution, $R(t)=\int d\tau \mathrm{SFR}(t-\tau )\Psi (\tau )$, rather than the single-burst expression. The time since star formation stopped need not equal each progenitor's age. Stellar lifetimes before formation of the second white dwarf, common-envelope evolution, masses and mass ratios, metallicity, delay-dependent explosion efficiencies, galaxy-age estimates and survey selection all affect the comparison. Some CO-white-dwarf mergers need not make Type Ia supernovae, and other channels can contribute to the observed rate. Agreement in slope and normalization therefore supports the hypothesis under stated assumptions but is not a unique identification of the channel.

Keep both component masses constant during a circular, gravitational-radiation-driven inspiral. Since $J=\mu \sqrt{GMA}$, differentiation gives $\dot{J}/J=\dot{A}/(2A)$. Combining this with the supplied angular-momentum loss yieldsThe expression $A/\dot{A}$ printed in the source is negative during inspiral. A positive characteristic shrinking timescale isThe actual remaining coalescence time follows by integrating $A^{3}dA=-Kdt$:This constant factor does not change the requested separation scaling.

For an old, approximately coeval population, let $dN/d{A}_0=C$ over the relevant separation interval, and write the positive delay as $t_m=B{A}_0^4$. Conservation of event counts under this change of variable givesThereforeThe scaling applies between the minimum and maximum delays of the assumed separation distribution. It is a rate per delay interval, not the number per logarithmic delay interval.

Treat the stated $120\mathrm{Mpc}$ as the radius of a complete, volume-limited search, with uniform galaxy density $n_g$. Its surveyed volume and number of galaxies areThe observing time is $\Delta t=117/365.25=0.3203\mathrm{yr}$. One detection estimatesor one merger per approximately $2.3\times 10^4$ years per Milky-Way-type galaxy. The equivalent volume rate is about $431\mathrm{Gp}{\mathrm{c}}^{-3}\mathrm{y}{\mathrm{r}}^{-1}$. The event's actual $40\mathrm{Mpc}$ distance is not the survey radius and must not replace $120\mathrm{Mpc}$ in the exposure.

A representative Galactic core-collapse supernova rate is about two per century, roughly $4.6\times 10^2$ times larger than this merger-rate estimate. The historical ${}^{26}\mathrm{Al}$ estimate was $1.9\pm 1.1$ per century.^[ref] The one-event Poisson uncertainty is large. Detector orientation, duty cycle, completeness and whether the quoted radius is an optimal horizon or an effective range also change the exposure; the boxed value is the specified simple-volume estimate.

When $k^{(p)}\gg k_J^{(p)}$, omit the gravitational term and define $\omega (t)=c_s(t)k/a(t)$. ThenThis is an acoustic damped harmonic oscillator with time-dependent coefficients. Freezing the background locally at $t_0$, its amplitude damping rate is $H_0$, soThe oscillation-averaged quadratic oscillator energy decays twice as fast in this frozen-frequency model, with time $1/(2H_0)$. Stating which quantity is damped avoids a factor-of-two convention error. Locally the solution isFor evolution over many expansion times, retain the varying frequency. Put ${\delta }_{\mathbf{k}}=\chi /a$. Direct substitution givesWhen $\omega \gg H$ and $|\dot{\omega }|\ll {\omega }^2$, the WKB approximation yieldsThese are rapidly oscillating sound waves in comoving density, diluted by the expanding background. For constant $c_s$ their envelope scales as $a^{-1/2}$, not universally as $a^{-1}$, because $\omega \propto a^{-1}$. In the Einstein-de Sitter universe, introducing conformal time $d\eta =dt/a$ gives the exact constant-$c_s$ pressure-only solution ${\delta }_{\mathbf{k}}=[C\sin (k{c}_s\eta )+D\cos (k{c}_s\eta )]/\eta$, consistent with $a\propto {\eta }^2$. For that background the frozen-coefficient damping time is $3t/2$, while the slowly varying envelope has instantaneous decay time $2/H$.

In the formal static-background limit, $H=0$ and $\bar{\rho }$ is constant. The gravity-dominated linear cosmological density perturbation equation reduces toThe Static-universe Jeans modes areThe unstable perturbation grows exponentially rather than as $t^{2/3}$. In the expanding case, the $2H\dot{\delta }$ term resists peculiar motion and the falling background density also lengthens the gravitational growth timescale. A globally homogeneous static self-gravitating fluid is not obtained merely by turning off expansion; this calculation is the conventional local Jeans-background idealization, with the uniform background force removed or otherwise supported.

For $k^{(p)}\ll k_J^{(p)}$, neglect pressure support. The growing and decaying Jeans instability modes obeyIn an Einstein-de Sitter universe, $a\propto t^{2/3}$ gives $H=2/(3t)$, and the matter-density relation $8\pi G\bar{\rho }=3H^2$ gives $4\pi G\bar{\rho }=2/(3t^2)$. ThusTry a power law ${\delta }_{\mathbf{k}}\propto t^p$. Its characteristic equation isTherefore the Einstein-de Sitter density-growth modes areThis is a linear result while $|\delta |\ll 1$; a growing mode eventually leaves that regime.

Expand the density contrast in spatial Fourier modes using fixed comoving coordinates:The spatial Laplacian multiplies a mode by $-k^2$, while its comoving wavenumber $k=|\mathbf{k}|$ is time-independent. Writing $H=\dot{a}/a$, the linear cosmological density perturbation equation becomesThe proper wavenumber is $k^{(p)}=k/a$, since the proper wavelength is $2\pi a/k$. The proper Jeans wavenumber balances the pressure and gravity terms:

A massive main-sequence donor loses material in a stellar wind. Gravitational capture by the neutron star can produce a nearly radial, supersonic infall when the captured material has little specific angular momentum. A magnetic field may channel the final infall toward polar accretion columns, where a standing shock decelerates the flow. Large angular momentum instead favours an accretion disk; strictly radial accretion is an idealization.

For mean particle mass $\mu$, the ideal gas law is $p_{\mathrm{d}}={\rho }_{\mathrm{d}}{k}_{\mathrm{B}}{T}_{\mathrm{d}}/\mu$. Combining it with b givesFor Newtonian free fall from a distance large compared with the stellar radius $R_{\ast }$, $v_{\mathrm{u}}^2\simeq 2G{M}_{\ast }/R_{\ast }$. ThereforeTODO: a unique numerical temperature and characteristic photon band require source data not supplied here. The held question gives $M_{\ast }=1.4M_{\odot }$ but gives neither $R_{\ast }$ nor a composition fixing $\mu$. The mass alone does not fix the free-fall speed or the temperature. For example, doubling $R_{\ast }$ at the same mass and composition halves both quantities in the last box.

A normalization, rather than an additional claimed datum, makes the scale explicit:Here $k_{\mathrm{B}}T$ is the characteristic particle energy; the mean translational energy of a nonrelativistic thermal particle is $3k_{\mathrm{B}}T/2$. If one additionally chose $R_{\ast }=10\mathrm{km}$ and $\mu =m_p$, this ideal one-temperature estimate would correspond to gamma-ray-scale thermal photon energies. Those are conditional choices, not information recovered from the source. Real accreting neutron stars commonly emit X-rays: electron-ion disequilibrium, radiative cooling, radiation pressure, reprocessing and relativistic infall invalidate a direct identification of this ideal ion-heating estimate with the escaping photon spectrum.

Use a steady, plane, nonradiative normal shock wave in a perfect gas, and measure the normal speeds $v_{\mathrm{u}},v_{\mathrm{d}}>0$ in the shock frame. The labels denote upstream and downstream. Conservation of mass, normal momentum and total energy givesandThe enthalpy term includes the pressure work. For very large upstream Mach number, neglect $p_{\mathrm{u}}$ compared with ${\rho }_{\mathrm{u}}{v}_{\mathrm{u}}^2$. If $r={\rho }_{\mathrm{d}}/{\rho }_{\mathrm{u}}$ is the shock compression ratio, the first two equations implySubstitution into the energy equation givesDiscard the no-shock solution $r=1$. For a nonrelativistic monatomic gas, $\gamma =5/3$, so the Strong-shock Rankine-Hugoniot conditions becomeThe factor four requires this adiabatic index and negligible radiative, magnetic and escaping-particle contributions to the flux budget. For example, the same strong, nonradiative calculation with $\gamma =4/3$ gives compression seven, not four.