Past exam of the astrophysics course of the University of Oxford / 2022 / 1

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 $I_{\nu }=I_{\nu }(0)e^{-\tau }+S_{\nu }(1-e^{-\tau })$.

Use the Einstein coefficients convention in which $B_{ij}$ multiplies spectral radiation energy density $u_{\nu }$, with $u_{\nu }=c^{-1}\int I_{\nu }d\Omega$. A normalized spectral line profile obeys $\int \phi (\nu )d\nu =1$. The spontaneous emission probability per upper-state atom per unit time is $A_{21}$; the stimulated emission and upward absorption rates per atom are respectively $B_{21}\int u_{\nu }\phi (\nu )d\nu$ and $B_{12}\int u_{\nu }\phi (\nu )d\nu$.

For isotropic spontaneous emission, $n_2{A}_{21}$ transitions occur per unit volume and time, each supplying photon energy $h{\nu }_{12}$. Dividing among $4\pi$ steradians givesFor a small incident beam, its contribution to $u_{\nu }$ is $I_{\nu }d\Omega /c$. Absorption removes and stimulated emission adds photons in the incident direction. Matching the net loss to ${\alpha }_{\nu }{I}_{\nu }d\Omega$ therefore givesReplacing ${\nu }_{12}$ by $\nu$ across the narrow spectral line profile gives the printed expressions. The negative sign is essential: an inverted population can amplify the incident specific intensity. Thermal equilibrium and the Planck function imply $g_1{B}_{12}=g_2{B}_{21}$ and $A_{21}/B_{21}=8\pi h{\nu }_{12}^3/c^3$, where $g_i$ is the state degeneracy. A convention using mean intensity instead changes both $B$ coefficients by $4\pi /c$.

With the spontaneous emission term suppressed, $j_{\nu }=0$. A uniform absorption coefficient over length $\ell$ gives directlyThe normalized specific intensity deficit is unchanged by conversion between wavelength and frequency, since both numerator and denominator acquire the same Jacobian determinant. The Boltzmann distribution and the Einstein coefficients give $n_2{B}_{21}/(n_1{B}_{12})=e^{-h{\nu }_{12}/(k_B{T}_e)}\ll 1$. Consequently, in the optically thin limit,Here $N_1=n_1\ell$ is the column density, and $\nu \simeq {\nu }_{12}$ has been used in slowly varying factors. This linear relation between equivalent width and column density fails once the central optical depth becomes appreciable.

For the unresolved Lyman-alpha line, the lower $1s$ and combined upper $2p$ state degeneracies are $g_1=2$ and $g_2=6$, so the Einstein coefficients giveThe measured equivalent width $W=7.0\times 10^{-12}\mathrm{m}$ then impliesAssume a central transverse line of sight through a uniform filament, so $\ell =d=0.5\mathrm{Mpc}=1.54\times 10^{22}\mathrm{m}$. Nearly all neutral hydrogen is in its ground state at this temperature, henceThe cylinder's length $L$ determines its total volume but does not enter a transverse column density. Geometry must be specified: an axial line of sight has $\ell =L$ and gives $n_{\mathrm{HI}}=4.2\times 10^{-7}{\mathrm{m}}^{-3}$ for the same measured equivalent width. Intermediate chords give intermediate path-dependent estimates. The numbers above treat $W$ as a rest-frame equivalent width; an observed value at redshift $z$ must first be divided by $1+z$.

True thermal equilibrium requires a thermal ionizing radiation field as well as chemical balance; equality of photoionization and recombination rates alone would require a specified external radiation field. Under the stronger equilibrium assumption, apply the Saha equation to pure hydrogen. The electron spin factor cancels the ground-state electronic degeneracy, givingAt $T=10^4\mathrm{K}$, $k_{B}T=0.862\mathrm{eV}$ and $S=3.38\times 10^{20}{\mathrm{m}}^{-3}$. Charge neutrality gives $n_e=n_p$. Importantly, the known quantity from d is the neutral number density, not the total number density; use Saha neutral fraction from a measured neutral density to obtain $n_p=\sqrt{S{n}_{\mathrm{HI}}}=5.31\times 10^7{\mathrm{m}}^{-3}$. ThereforeThis extraordinarily small neutral fraction is the result of applying thermal equilibrium to dilute intergalactic gas; it is not a realistic inference from photoionization equilibrium with a nonthermal cosmic background. For the axial geometry of d, the equilibrium neutral fraction is smaller by $\sqrt{20}$.

Adopt the stipulated neutral fraction $f_{\mathrm{HI}}=10^{-6}$, independently of the equilibrium estimate in e. For pure hydrogen, $n_e=(1-f_{\mathrm{HI}})n_{\mathrm{HI}}/f_{\mathrm{HI}}\simeq 8.3{\mathrm{m}}^{-3}$. The thermal Sunyaev-Zeldovich effect measures integrated electron pressure:Using the same absorbing line of sight as in d,Thus one such filament is below the stated measurement sensitivity. If the number density inferred transversely is instead used to view the same cylinder end-on, its path is twenty times longer and $y=2.9\times 10^{-10}$, still below the stated sensitivity. Changing the assumed path of the original absorption measurement while keeping its column density fixed does not change its predicted $y$. The quoted neutral-fraction floor is a premise of this estimate, not a universal bound derivable from thermal equilibrium.