Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 349 (Evolution of Galaxies)

These solutions refer to the official examination paper above. Assumptions about the gas, velocity distribution, angular sampling, and stellar recycling are made explicit below.

Write $Q={\dot{N}}_{\mathrm{ion}}$ for the constant hydrogen-ionizing photon production rate. Assume an initially neutral gas of hydrogen atoms, a sharp spherical ionization front, complete ionization inside it, and a fixed gas density. The electron and proton number densities inside are then both $n$. Take the recombination coefficient $\alpha$ to be constant, with the usual Case B recombination interpretation: photons from recombination directly to the ground state are absorbed locally. Neglect absorption by dust, pressure-driven gas motion, and light-travel delays.

The total rate of Case B recombination within radius $R$ isAt photoionization equilibrium, $Q=\mathcal{R}(R_S)$, giving the Strömgren radiusThe square of the density appears because recombination requires encounters between an electron and a proton.

During Ionization-front growth of a Strömgren sphere, each photon either replaces an ion lost through Case B recombination or ionizes an additional atom. There are $N_i=(4\pi /3)n{R}^3$ ions inside the ionization front, so photon counting givesThe recombination time is $t_{\mathrm{rec}}=(\alpha n{)}^{-1}$. Solving this first-order differential equation with $R(0)=0$ givesFor an initial radius $R_0$, the corresponding result is $R^3(t)=R_S^3+(R_0^3-R_S^3)e^{-t/t_{\mathrm{rec}}}$. Initially, the idealized photon-counting solution has $R\simeq [3Qt/(4\pi n){]}^{1/3}$; at late times, $R/R_S\simeq 1-\frac{1}{3}{e}^{-t/t_{\mathrm{rec}}}$. The initially divergent front speed in this approximation is a consequence of neglecting light-travel delays; the formula describes the growth once that approximation is applicable.

For a power law density, specify a reference radius $r_0$ so that $n(r)=n_0(r/r_0{)}^{-\beta }$. With an inner cutoff $r_{\mathrm{in}}>0$, the recombination integral for a power-law nebula isEquivalently, a logarithmic interval in radius contributes $d\mathcal{R}/d\ln r=4\pi \alpha n_0^2{r}_0^{2\beta }{r}^{3-2\beta }$. For $\beta <3/2$, the central integral converges as $r_{\mathrm{in}}\to 0$, and the recombination rate grows as $R^{3-2\beta }$. At $\beta =3/2$, each logarithmic interval contributes equally and the centre gives a logarithmic divergence. For $\beta >3/2$, the unmodified profile gives a power divergence at the centre, while its outer recombination integral converges. The change at $\beta =3/2$ is set by the competition between the spherical volume element and the squared density.

To regularize the centre, choose the following explicit cored power-law density profile, taking $r_0=r_c$ and $n_0$ as the core density:For $R\le r_c$, $\mathcal{R}(R)=(4\pi /3)\alpha n_0^2{R}^3$. For $R>r_c$, put $u=R/r_c$ and integrate the core and envelope separately:This rate is continuous and strictly increasing with $R$. If $\beta >3/2$, it has a finite supremum, the critical ionizing photon rate of a cored power-law nebula:For $0<Q<Q_{\mathrm{crit}}$ there is a unique finite radius at photoionization equilibrium. At $Q=Q_{\mathrm{crit}}$ the equality is approached only as $R\to \infty$; for $Q>Q_{\mathrm{crit}}$ even ionizing the whole profile cannot consume all photons through Case B recombination, so no equilibrium radius exists. For $\beta \le 3/2$, the envelope integral grows without bound and there is no finite critical rate in this infinite-extent model.

For a concrete choice, take any $n_0>0$, $r_c>0$, and $\beta =2$. Then $Q_{\mathrm{crit}}=16\pi \alpha n_0^2{r}_c^3/3$. In the envelope, defining $q=Q/(4\pi \alpha n_0^2{r}_c^3)$ gives $R=r_c/(4/3-q)$ for $1/3<q<4/3$, explicitly showing the divergence as the critical rate is approached. A different core shape changes the numerical coefficient of $Q_{\mathrm{crit}}$, while the outer-slope condition $\beta >3/2$ remains the same.

Use a stationary, non-streaming tracer population with spherical symmetry in its full velocity distribution. In particular, the mean velocities vanish, the mixed covariances vanish, and the two tangential components have equal velocity dispersions. WriteHere ${\sigma }_t^2$ is the variance of one tangential component. In the alternative convention using the sum of both tangential variances, the same velocity-anisotropy parameter is $1-({\sigma }_{\theta }^2+{\sigma }_{\varphi }^2)/(2{\sigma }_r^2)$. These definitions agree; they must not be mixed by inserting an extra factor of two.

The time derivative and mean-velocity terms in the supplied Jeans equation vanish, as do its angular mixed-moment terms. Its remaining geometric term is $2\nu ({\sigma }_r^2-{\sigma }_t^2)/r$. Thus the Spherical Jeans equation becomesThe no-streaming assumption excludes a steady radial flow; stationarity of the density alone would not remove its advective terms.

By the shell theorem, the gravitational potential of a spherical mass distribution satisfies $d\Phi /dr=GM(<r)/r^2$. Expanding the derivative above, dividing by $\nu$, and using logarithmic derivatives givesThe tracer density $\nu$ need not be the density of the gravitating matter. The circular speed is $V_c^2=rd\Phi /dr=GM(<r)/r$, whether or not the tracers themselves follow circular orbits.

Now let $\beta$ be constant, $\nu ={\nu }_0(r/r_0{)}^{-\gamma }$, and $V_c^2=V_0^2(r/r_0{)}^{2\alpha }$. With $y={\sigma }_r^2$, the Spherical Jeans equation readsFor a power-law ansatz $y=K{V}_c^2$, $y^{\prime }=2\alpha y/r$, so $(\gamma -2\beta -2\alpha )y=V_c^2$. Therefore the scale-free spherical Jeans solution isThe positivity condition is necessary for this particular solution to be a physical velocity dispersion.

There is a boundary condition implicit in selecting this solution. The integrating factor $r^{2\beta -\gamma }$ gives, when $D=\gamma -2\beta -2\alpha \ne 0$,The homogeneous term represents an additional radial-pressure boundary contribution. It is removed, for example, by requiring $r^{2\beta }\nu {\sigma }_r^2\to 0$ at infinity. Under that condition,and convergence requires $D>0$, giving the boxed result. If $D=0$, the solution instead has the logarithmic formso the quoted constant proportionality does not apply. A finite outer boundary can retain a homogeneous term even when $D\ne 0$.

To derive the heliocentric projection of a spherical velocity dispersion, put the Sun at vector ${\mathbf{R}}_{\odot }$ of length $R_{\odot }$, and a tracer at $\mathbf{r}=r\hat{\mathbf{r}}$. Its heliocentric line of sight isIf $\chi$ is the angle between $\hat{\mathbf{\ell }}$ and the Galactocentric radial direction, thenAfter correction for the Sun's motion, the intrinsic covariance matrix has eigenvalues ${\sigma }_r^2,{\sigma }_t^2,{\sigma }_t^2$ in the local spherical basis. Projection onto the line of sight therefore givesAt fixed $r$, ${\sigma }_r$ and $\beta$ are constant across the spherical shell. Averaging its squared line of sight velocity, with zero mean, givesFor a uniformly sampled Galactocentric shell, the solid angle measure makes $u$ uniform on $[-1,1]$, giving the geometric correction for heliocentric velocity dispersionTo evaluate it, set $a=r^2+R_{\odot }^2$ and $b=2r{R}_{\odot }$, and divide the integrand:The odd term integrates to zero, while ${\int }_{-1}^{1}du/(a-bu)=b^{-1}\ln [(a+b)/(a-b)]$. For $r>0$, $R_{\odot }>0$, and $r\ne R_{\odot }$, this givesThe removable limit at $r=R_{\odot }$ is $A=1/2$. As $r/R_{\odot }\to \infty$, $A\simeq 2R_{\odot }^2/(3r^2)$, while as $r/R_{\odot }\to 0$, $A\to 2/3$. If $R_{\odot }=0$, $A=0$ and the observed velocity dispersion is exactly radial. For isotropic velocities, $\beta =0$, the correction vanishes at every radius.

The averaging measure matters. If directions are instead uniform in heliocentric solid angle at a fixed Galactocentric radius $r>R_{\odot }$, let $\vartheta$ be the angle between the line of sight and ${\mathbf{R}}_{\odot }$. Geometry gives ${\sin }^2\chi =(R_{\odot }/r{)}^2{\sin }^2\vartheta$, so that particular weighting gives $A=2R_{\odot }^2/(3r^2)$ exactly. The logarithmic expression above uses uniform tracer weighting on a Galactocentric shell. For an observational sample, use its actual angular selection weights in $A=⟨{\sin }^2\chi ⟩$; the boxed dispersion relation remains valid.

Use the instantaneous recycling approximation with a well-mixed gas reservoir, constant stellar yield $p$, no gas inflow, and an irreversible galactic outflow. Let $M_s$ be the mass locked into long-lived stars and remnants after prompt recycling, and put $\psi (t)={\dot{M}}_s(t)\ge 0$. The yield $p$ is the newly synthesized mass of heavy elements returned to the gas per unit increase of $M_s$. Thus $M_h$ in $Z=M_h/M_g$ denotes the heavy-element mass in the gas, excluding metals already locked into stars.

Assume $M_s(0)=0$, $M_g(0)=M_0$, and $Z(0)=Z_0$; initially pristine gas means $Z_0=0$. Take the outflow rate to be $\alpha \psi$ with constant $\alpha \ge 0$, and assume its metallicity equals the current gas-phase metallicity $Z$. This is the well-mixed leaky-box model of galactic chemical evolution. The parameter $\alpha$ is a mass-loading factor relative to the net rate of locking mass into stars. If a gross star formation rate $\Psi$ and a prompt returned fraction $\mathcal{R}$ are used instead, $\psi =(1-\mathcal{R})\Psi$ and a wind rate $\eta \Psi$ corresponds to $\alpha =\eta /(1-\mathcal{R})$.

Conservation of mass givesand the baryonic mass remaining in the box is $M_g+M_s=M_0-\alpha M_s$. Newly made metals enter the gas at rate $p\psi$, while pre-existing metals are locked into stars at rate $Z\psi$ and leave in the wind at rate $\alpha Z\psi$. HenceUsing $M_h=Z{M}_g$ and the product rule,The cancellation expresses the fact that a well-mixed wind removes gas and metals in the same proportion, while fresh stellar production raises the abundance in the remaining gas.

Divide this equation by ${\dot{M}}_g=-(1+\alpha )\psi$ along the evolving reservoir, or integrate in time through intervals with no star formation. The resulting gas-phase metallicity isEquivalently, its fully explicit dependence on a prescribed net star formation rate isvalid while $M_g(t)>0$. With pre-existing stars, replace $M_s(t)$ in the gas-mass relation by $M_s(t)-M_s(0)$. Without specifying $\psi (t)$ or a gas-consumption law, the model fixes $Z$ as a function of gas mass but does not determine a unique function of time.

For the closed-box model of galactic chemical evolution, set $\alpha =0$. Then $M_g=M_0-M_s$ andAt the same remaining fraction $g=M_g/M_0$ of the initial gas reservoir, the leaky box has the logarithmic coefficient $p/(1+\alpha )$, lower than the true stellar yield by the wind factor. This coefficient is often called an effective yield, but the definition of the gas fraction must be specified.

In particular, the observable gas fraction of a galaxy is often $\mu =M_g/(M_g+M_s)$, using the mass still present, rather than $g$. In our initially star-free model,ThereforeFor $0<\mu <1$ and $\alpha >0$, the leaky-box enrichment is smaller at the same $\mu$: $\ln [1+\alpha (1-\mu )]\le \alpha (1-\mu )<\alpha \ln (1/\mu )$. If effective yield is defined observationally as $(Z-Z_0)/\ln (1/\mu )$, it is consequently not simply the constant $p/(1+\alpha )$ at all $\mu$.

As an explicit time example, choose a linear gas-consumption law $\psi =ϵM_g$, with constant $ϵ>0$. The mass and metallicity equations giveThe closed box with the same $ϵ$ has the same metallicity at a given time, while the leaky box exhausts its gas faster. Thus a comparison at fixed time depends on the chosen star formation rate, even though the comparison at fixed gas fraction above is unambiguous.

Finally, the assumption about wind metallicity is essential. If the expelled gas has abundance $Z_w$ instead, its metal-loss rate is $\alpha Z_w\psi$, and the same bookkeeping yieldsA wind enriched relative to the ambient gas lowers the enrichment rate; the logarithmic solution derived above applies to $Z_w=Z$.