Codex @codex  0

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 $2/H$.

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.

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.

In a massive star undergoing core-collapse supernova formation, an iron-rich core cannot gain further support from exothermic fusion. Electron captures and nuclear photodisintegration promote collapse. The inner core stiffens near nuclear density and bounces, launching a shock into the infalling outer core. Dissociation and neutrino losses drain its energy, so the prompt shock generally stalls while matter continues to accrete. Neutrino absorption behind the shock, aided by multidimensional flow, can revive it and eject the envelope; sufficiently rapid rotation and strong magnetic fields offer additional possibilities. The remnant may be a neutron star or black hole. Major uncertainties include progenitor structure, nuclear equation of state, neutrino transport and interactions, turbulent flow, rotation and magnetic fields.

The standing accretion shock instability is a nonradial instability of this stalled shock and accretion flow. Feedback between advected disturbances and acoustic response can amplify large-scale sloshing or spiral motions, producing an asymmetric shock even without an initially unstable convective gradient.^[ref]

An asymmetric explosion can impart an opposite recoil to the remnant by momentum conservation. In neutrino-driven calculations, long-lasting gravitational attraction by asymmetric, slower inner ejecta can dominate the acceleration: the gravitational tug-boat mechanism. SASI and convection can seed the asymmetries, but the kick is not necessarily an instantaneous hydrodynamic push or uniquely attributable to SASI.^[ref]

For a kick parallel or antiparallel to star 1's original velocity, the immediate post-explosion relative speed is $|v_0\pm v^{\pm }|$. ThusA positive-energy, hyperbolically escaping pair therefore satisfies preciselyThe absolute value is essential when a sufficiently large opposite kick reverses the relative tangential velocity.

Marginal case. Strictly speaking, a pair is gravitationally bound only for $E^{\prime }<0$. If unbound includes marginal escape, replace the printed strict inequality by $\ge$. For a concrete counterexample to an unrestricted “if and only if” with $>$, take no kick and $\Delta M=M_0/2$. Both sides equal one and the energy is zero: this is not a bound ellipse, although the printed strict inequality is false.

Write $f=M^{\prime }/M_0$ for the retained fraction and $q=v^{-}/v_0$ for an opposite kick. With more than half the mass lost, $0<f<1/2$. A same-direction kick cannot bind the pair, because $1+v^{+}/v_0\ge 1>\sqrt{2f}$. An opposite kick can reduce the relative speed enough to leave negative energy:Both endpoints are marginal zero-energy cases. Kicks somewhat larger than $v_0$ can still satisfy this interval, but then the direction of the relative orbital motion is reversed.

At $v^{-}=v_0=v_1+v_2$, the two surviving stars have identical instantaneous velocities, so their relative velocity and orbital angular momentum vanish. Their energy is nevertheless negative,They fall radially toward each other from rest in their new centre-of-mass frame. In the point-mass idealization the collision orbit has semimajor axis $A/2$ and zero pericentre; real finite-size stars collide or interact before reaching zero separation. It is not a stable circular binary. Zero relative angular momentum, rather than the negative energy alone, determines this plunge outcome.

Take the initially expanding branch ${\dot{r}}_i=H_i{r}_i$; the printed squared velocity alone also permits an initially contracting branch, which would not reach a future maximum radius. Multiply the equation of motion by $\dot{r}$ and integrate. The conserved outer-shell specific orbital energy isThe initial mass follows from the background critical density:ConsequentlyIf $E$ means the energy of a test shell of mass $m$, $E=mϵ$. If it means the total energy of the entire uniform, homologously moving perturbation, direct integration instead gives $K=3M{\dot{r}}^2/10$, $U=-3G{M}^2/(5r)$ andThe common convention of writing the first integral as $E$ is therefore an energy per unit shell mass, not the total energy of the sphere.

At spherical-collapse turnaround, $\dot{r}=0$, so $ϵ=-GM/r_m$. HenceAny positive initial density contrast binds this pressureless perturbation in the stated Einstein-de Sitter universe.

The Doppler factors for the two discrete blobs are ${\mathcal{D}}_{\pm }=[\gamma (1\mp \beta \cos \theta ){]}^{-1}$. Their ratio is $(1+x)/(1-x)=77/33=7/3$. Lorentz-invariant specific intensity means $I_{\nu }/{\nu }^3$ is unchanged; with $\nu =\mathcal{D}{\nu }^{\prime }$ and the same rest-frame power-law spectrum,The brightness ratio at the common observed frequency therefore givesThe stated identical intrinsic specific intensities must refer to comparable intrinsic states. If the blobs fade with their proper ages, equal observer times sample different ages and need that correction. A steady continuous jet has a different boosting exponent, $2+\alpha$; the discrete ejecta and the specified specific intensity transformation justify $3+\alpha$ here.

For optically thin synchrotron radiation from an isotropic electron distribution $dN/dE\propto E^{-p}$, the characteristic emitted frequency scales as $B{E}^2$ and the total single-particle power as $B^2{E}^2$. Thus $j_{\nu }\propto E^{-p}{E}^2/(d\nu /dE)\propto {\nu }^{-(p-1)/2}$; integration over the full single-particle spectrum preserves this exponent. HenceThe test-particle strong nonrelativistic diffusive shock acceleration prediction is $p=(r+2)/(r-1)=2$ for compression ratio $r=4$, giving $\alpha =0.5$. The inferred spectrum is steeper. If interpreted literally within that uncooled formula, $p=2.50$ would correspond to $r\simeq 3$, but cooling, escape, and other acceleration conditions can also steepen it.

The observed Galactic cosmic ray differential energy spectrum below the knee is approximately $E^{-2.7}$, steeper still; its formal synchrotron radiation equivalent would be $\alpha \simeq 0.85$. Thus $p\simeq 2.50$ lies between the simplest source prediction and the propagated cosmic ray slope. The comparison is qualitative: most measured cosmic rays are nuclei, whereas the radio slope here probes electrons, and propagation and radiative losses alter the spectra differently.

On the main sequence, central hydrogen burning converts four protons into helium, positrons, neutrinos and released energy. The net reaction is $4p\to {}^4\mathrm{He}+2e^{+}+2{\nu }_e$, with positron annihilation contributing to the total roughly $26.7\mathrm{MeV}$ release before subtracting escaping neutrino energy. In the low-mass main-sequence stars of an old globular cluster, the proton–proton chain generally dominates. The CNO cycle makes the same net conversion and becomes important in hotter burning regions.

After central hydrogen exhaustion, a subgiant contracts its helium core and establishes a hydrogen-burning shell. On the red giant branch, that shell supplies most of the luminosity, commonly through the temperature-sensitive CNO cycle, while the inert helium core grows and becomes supported mainly by electron degeneracy pressure. The envelope expands and its surface cools. The helium core is not yet powered by helium nuclear fusion.

Helium ignition in a sufficiently low-mass degenerate core produces the helium flash, removing the strong electron degeneracy pressure dominance. A horizontal-branch star subsequently burns helium in its core through the Triple-alpha process, $3{}^4\mathrm{He}\to {}^{12}\mathrm{C}+\gamma$, and through ${}^{12}\mathrm{C}(\alpha ,\gamma ){}^{16}\mathrm{O}$. A surrounding hydrogen-burning shell can continue simultaneously.

After central helium exhaustion, the Asymptotic giant branch has an inert carbon-oxygen core with a helium-burning shell and an outer hydrogen-burning shell. Alternation between quiescent burning and helium-shell thermal pulses drives mixing and substantial mass loss. Subsequent post-asymptotic giant branch evolution exposes a hot core that may ionize its expelled envelope as a planetary nebula. Finally a white dwarf cools mainly by loss of stored thermal energy, rather than sustained central nuclear fusion.

The specific intensity $I_{\nu }$ is energy transported per unit time, projected area, solid angle and frequency. The path coordinate $s$ increases along the ray. The emission coefficient $j_{\nu }$ is emitted energy per unit time, volume, solid angle and frequency, whereas the absorption coefficient ${\alpha }_{\nu }$ has units of inverse length. Thus $j_{\nu }ds$ adds specific intensity and ${\alpha }_{\nu }{I}_{\nu }ds$ removes it. Stimulated emission can be included as negative absorption.

Define the optical depth between two points by ${\tau }_{\nu }(s_1,s_2)={\int }_{s_1}^{s_2}{\alpha }_{\nu }(u)du$. Multiplying the radiative transfer equation by the integrating factor $\exp [{\tau }_{\nu }(s_0,s)]$ and integrating givesEach local contribution is attenuated only by the material between its emission point and the observer. Where ${\alpha }_{\nu }\ne 0$, the source function is $S_{\nu }=j_{\nu }/{\alpha }_{\nu }$ and the same formal solution of the radiative transfer equation is $I_{\nu }(\tau )=I_{\nu }(0)e^{-\tau }+{\int }_0^{\tau }{S}_{\nu }(t)e^{-(\tau -t)}dt$. A constant source function gives $I_{\nu }=I_{\nu }(0)e^{-\tau }+S_{\nu }(1-e^{-\tau })$.

For a gas-pressure-supported stellar core of fixed mass and composition, neglect boundary terms for the illustrative whole-core estimate. The virial theorem gives $E=U+\Omega =-U$, and the monatomic gas gives $U=3M_c{k}_B\overline{T}/(2\mu m_u)$. Thus the effective heat capacity along the equilibrium sequence isThis negative heat capacity of a virialized gravitational system means that net energy loss makes a self-gravitating core contract and heat. It is different from the positive local material specific heat capacity. On the main sequence, hydrogen burning supplies most of the radiated energy, so rapid contraction is arrested. Hydrogen depletion increases the mean molecular weight; slow structural readjustment usually raises the central temperature rather than making it fall.

After core hydrogen exhaustion, a massive stellar core contracts and heats more rapidly until a later nuclear fuel ignites. There is no general reversal to cooling simply upon leaving the main sequence. If radiation pressure becomes dominant, the internal energy becomes $U=U_{\mathrm{gas}}+U_{\mathrm{rad}}$ with $3\int PdV=2U_{\mathrm{gas}}+U_{\mathrm{rad}}$. The virial theorem therefore gives $E=-U_{\mathrm{gas}}$; the radiation-dominated limit is marginally bound and its effective adiabatic index approaches $4/3$.

Quasistatic contraction can still maintain virial equilibrium. It fails dynamically only when sufficient support is lost, for example if pair creation reduces the effective adiabatic index below $4/3$, or when an iron core undergoes electron capture and photodisintegration. A fixed-composition gas-dominated contracting core with $T\propto 1/R$ is already a counterexample to any assertion that leaving the main sequence alone reverses the heating or destroys virial equilibrium.

An ultrarelativistic electron has kinetic energy much larger than its rest energy: $(\gamma -1)m_e{c}^2\gg m_e{c}^2$, equivalently $\gamma \gg 1$. Its speed obeys $v/c=\sqrt{1-{\gamma }^{-2}}\simeq 1-1/(2{\gamma }^2)$ and its total energy is approximately $pc$. At $\gamma =10^6$, $E=\gamma m_e{c}^2\simeq 511\mathrm{GeV}$.

Diffusive shock acceleration is viable in principle at $511\mathrm{GeV}$: repeated scattering off magnetic irregularities on opposite sides of a strong nonrelativistic shock wave can reach such energies. Actual viability requires a sufficiently large and long-lived accelerator with an acceleration time shorter than its escape and loss times; $\gamma$ alone does not establish those conditions.

Take positive velocity toward the upstream material, and let the front move at $+v_s$ in the upstream rest frame. For a strong adiabatic shock wave in a gas with adiabatic index $5/3$, the Rankine-Hugoniot conditions give ${\rho }_d/{\rho }_u=4$ and the shock-frame inflow speeds satisfy $u_1=v_s$, $u_2=v_s/4$. The signed velocities in the three frames areThus the two scattering media converge at $V=u_1-u_2=3v_s/4$. For a crossing particle with incident energy $E$ in its local plasma rest frame, a Lorentz transformation to the approaching other plasma giveswhere $\mu >0$ is the incident direction cosine measured toward the approaching medium and $pc\simeq E$. For an almost isotropic crossing population, flux weighting gives $⟨\mu ⟩={\int }_0^1{\mu }^{2}d\mu /{\int }_0^1\mu d\mu =2/3$. Scattering randomizes directions without changing energy in that plasma frame, and the reverse crossing has the same mean gain. ThereforeThese are first-order mean gains. They are not an increase of energy in one fixed frame caused by an instantaneous boundary crossing. Literally choosing the electron's instantaneous rest frame gives incident $E=m_e{c}^2$, and the other gas approaches at nearly $c$, so the expansion in small $V/c$ is invalid. This concrete frame check explains why the printed description of $V$ must instead refer to the upstream/downstream scattering frames.

One explanation is an upper electron energy imposed by the accelerator: finite age, finite confinement size, escape, or a balance of acceleration and losses limits $\gamma$ to ${\gamma }_{\max }$. Since the synchrotron critical frequency scales as ${\nu }_c\propto B{\gamma }^2\sin \alpha$, the absence of higher-energy electrons produces a corresponding high-frequency cutoff, broadened by the single-particle spectrum.

The other is synchrotron cooling, with inverse Compton scattering often contributing: $\dot{\gamma }=-b{\gamma }^2$ gives $t_{\mathrm{cool}}=1/(b\gamma )$. High-energy electrons disappear fastest, so a population of age $t$ develops a cooling scale ${\gamma }_b\sim 1/(bt)$ and ${\nu }_b\propto B/(b^2{t}^2)$. A burst population can develop a strong cutoff. With continuous power-law injection, the simplest steady cooling model instead steepens the electron energy index by one and the synchrotron radiation spectral index by $1/2$, giving a break rather than necessarily an exponential cutoff. The detailed spectral shape therefore distinguishes finite acceleration from aging and injection history.

The printed magnetic dipole moment uses dimensionless angular momentum operators: their physical counterparts are $\hslash \mathbf{L}$ and $\hslash \mathbf{S}$. Choose the magnetic field along $z$. Then the Zeeman energy is $H_Z={\mu }_{B}B(L_z+g_s{S}_z)$, whose characteristic size is ${\mu }_{B}B$. In LS coupling, treating this as a time-independent perturbation theory requires it to be much smaller than the relevant fine structure separation, rather than merely smaller than the Coulomb interaction energy.

For a concrete hydrogen atom estimate take the $2p_{1/2},2p_{3/2}$ pair. The hydrogen fine-structure correction iswhere ${\alpha }_{\mathrm{fs}}$ is the fine-structure constant, distinguished from the later potential coefficient. ThusConsequently a suitable weak-magnetic field condition for this pair isThere is no single threshold for every hydrogen energy level: the relevant fine structure separations decrease with $n$. The $1s$ energy level has no same-$n$ spin-orbit interaction partner, so it cannot itself define this comparison. This calculation neglects hyperfine structure; resolving that smaller scale requires an additional weak-magnetic field condition.

Write $k=e^2/(4\pi {ϵ}_0)$ and retain the nuclear Coulomb potential. In spherical coordinates, the total potential energy iswhen $\beta =-\alpha /2$; $P_2(u)=(3u^2-1)/2$ is a Legendre polynomial. Treat the added term using quadratic anisotropic perturbation of hydrogen and degenerate perturbation theory in the $n=2$ hydrogen energy eigenbasis. It preserves parity and $m$, so it cannot mix $2s$ with $2p$, or distinct $2p$ magnetic sublevels.

For a hydrogen atom, the hydrogen radial second moment givesThe angular expectation values of ${\cos }^2\theta$ are $1/3$ for an $s$ orbital, $3/5$ for $p_0$, and $1/5$ for $p_{\pm 1}$, obtained by integrating the squared spherical harmonics. Therefore the five lowest spatial quantum states, counting statistical weight, have first-order energy levelsEach spatial quantum state has two electron spin states when spin-orbit coupling and the magnetic field are omitted. Thus these are five spatial states but only three distinct first-order energies; five distinct energies would be an incorrect reading. The accidental $2s,2p_{\pm 1}$ equality can be lifted at higher order. Quantitatively the added shifts must also be small compared with neighboring principal-shell gaps; $\alpha \ll k/a_0^3$ is an asymptotic condition, not a guarantee for arbitrary large $n$.

For general $\beta$, the $1s$ expectation value isIt does not cancel every order. At $\beta =-2\alpha$, $V_{\mathrm{ext}}=-2\alpha r^2{P}_2(\cos \theta )$ couples $1s$ to $d$ orbitals. For example its angular matrix element with $Y_{20}$ is $1/\sqrt{5}$, and the $3d$ radial matrix element is nonzero. The second-order nondegenerate perturbation theory sum has negative denominators, so this is a concrete nonzero higher-order contribution.

There is also a global limitation: for either negative $\beta$ above, $V(0,0,z)=-k/|z|+\beta z^2\to -\infty$. A localized wavepacket translated far along $z$ has fixed kinetic energy but arbitrarily negative potential energy, proving there is no true lowest energy level. The results describe local perturbative resonance poles, or a potential completed to a confining form far from the atomic nucleus. No globally exact ground-state cancellation exists for the potential as printed.

Differentiating the specified rotating-frame amplitudes gives the time-independent Schrödinger equationThe characteristic polynomial is ${\lambda }^2-({\delta }^2+{\Omega }^2)/4=0$, henceFor $\Omega \ne 0$, corresponding normalized eigenvectors are proportional to $(\Omega ,2{\lambda }_{\pm }-\delta {)}^T$. Their arbitrary complex coefficients evolve as $e^{-i{\lambda }_{\pm }t}$; this solves the amplitude equations for any initial quantum state.

Restoring the common energy origin $\overline{E}=(E_1+\hslash \omega +E_2)/2$, the two dressed state energy eigenvalues areThus their positive energy separation increases byEquivalently, the branches connected to $E_1^{\prime }=E_1+\hslash \omega$ and $E_2$ shift by $+\hslash {\Omega }^2/(4\delta )$ and $-\hslash {\Omega }^2/(4\delta )$ respectively. Their signed difference $E_1^{\prime }-E_2$ changes by $\hslash {\Omega }^2/(2\delta )$, which is the printed expression with a signed convention. For negative Rabi detuning, an actual positive separation still increases; for example at $\delta =-10\Omega$ it cannot have the negative change implied by reading the printed expression as an unsigned gap. These opposite AC Stark shifts describe level repulsion.

Use the energy-density convention of part (a), a normalized common spectral line profile $\int \varphi (\nu )d\nu =1$, and isotropic spontaneous emission. For a narrow spectral line centered at ${\nu }_0$,The factor $1/c$ follows from $u_{\nu }=4\pi J_{\nu }/c$: angle-integrated absorption removes $c{\kappa }_{\nu }{u}_{\nu }$ of energy per volume and time. The emission coefficient is per solid angle, whereas the absorption coefficient has units of inverse length.

The Boltzmann distribution in thermal equilibrium gives $n_2/n_1=(g_2/g_1)e^{-h{\nu }_0/(k_{B}T)}$. Dividing the two expressions and using the relations between Einstein coefficients,Thus the source function equals the Planck function at the transition frequency, and in the usual narrow-line approximationWith an exactly common finite-width $\varphi$ and constant line-center Einstein coefficients, their ratio is constant across the profile; exact equality to a varying $B_{\nu }$ at every frequency is only a narrow-line approximation. Exact thermal radiative transfer requires frequency-resolved detailed balance, which supplies the corresponding small profile corrections. This distinction does not alter the intended equilibrium result.

For two decays from a common upper energy level, their photon wavenumber difference equals the separation of the two lower energy levels. Treating both quoted wavelengths as vacuum values initially giveswhose difference $11443.96\mathrm{c}{\mathrm{m}}^{-1}$ is close to $21648-10192=11456\mathrm{c}{\mathrm{m}}^{-1}$. The only matching lower pair among the supplied energy levels is ${}^1{D}_2,{}^1{S}_0$, with the shorter-wavelength photon ending on ${}^1{D}_2$. The two independent upper energy estimates areThusThe electric-dipole selection rules require odd parity, $S=0$, $L=1$ and $J=1$ to connect to both singlet lower energy levels. The apparent discrepancy has a concrete convention explanation. The NIST carbon line table lists $193.0906\mathrm{nm}$ under vacuum wavelengths and $247.8561\mathrm{nm}$ under air wavelengths, reproducing the source's rounded numbers.^[ref] The second number must therefore be converted to a vacuum wavelength before using $E=hc/\lambda$. For the tabulated upper $61981.82\mathrm{c}{\mathrm{m}}^{-1}$ and lower $21648.02\mathrm{c}{\mathrm{m}}^{-1}$ energy levels, its vacuum wavelength is $247.931\mathrm{nm}$, corresponding to an air refractive index about $1.00030$. This removes the spurious $12\mathrm{c}{\mathrm{m}}^{-1}$ discrepancy. The first calculation alone, which correctly uses a vacuum wavelength, already gives the intended upper energy level to the accuracy of the rounded data. The independent upper value is $61981.82\mathrm{c}{\mathrm{m}}^{-1}$.^[ref]

In the idealized isolated-carbon chemical atom interpretation, both beams carry approximately $2.68\mathrm{eV}$ per incident particle, close to the ${}^1{S}_0$ threshold. However equal incident energies do not imply equal excitation channels.

A single optical photon cannot strongly excite any of the listed even-parity $2p^2$ energy levels from the even-parity ${}^3{P}_0$ ground state through an electric dipole transition. In particular ${}^3{P}_0\to {}^1{S}_0$ also violates $\Delta S=0$ and the $J=0\leftrightarrow 0$ restriction. The lowest strong opposite-parity excitations lie well above $2.68\mathrm{eV}$. Thus one expects no strong resonant absorption or corresponding strong atomic fluorescence from these ground-state chemical atoms at this optical wavelength; weak higher-order processes and scattering are separate effects.

Incident electrons can undergo inelastic scattering and collisional excitation. Their orbital angular momentum and exchange with bound electrons allow transfers excluded for a single electric dipole photon. The ${}^1{D}_2$ energy level at $1.264\mathrm{eV}$ and the low fine structure excitations are energetically accessible. One therefore expects scattered electrons with discrete energy losses, possible weak forbidden spectral lines, and collisional or surface relaxation. For example ${}^1{D}_2\to {}^3{P}_2$ has wavelength approximately $10^7/(10192-43.4)=985\mathrm{nm}$; spin mixing supplies the weak spin-forbidden radiative channel.

There is a numerical threshold caveat. Literally,Neither quoted beam reaches ${}^1{S}_0$ from the exact ground state if these are taken as exact monoenergetic values. The electron value rounded to two decimal places could represent an actual energy above threshold; if so, ${}^1{S}_0$ is collisionally accessible and its decay to ${}^1{D}_2$ produces the weak electric quadrupole transition at $10^7/11456\simeq 873\mathrm{nm}$, followed by lower-level relaxation. These emissions require populations and sufficiently low collisional quenching of a metastable line.

An actual solid carbon foil has bands, collective excitations and surface losses, so it need not reproduce the isolated-chemical atom spectrum or selection rules. Its optical and electron-induced emission cannot be predicted quantitatively from the five atomic energy levels alone. This is the limitation of the foil wording in the source.