Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 346 (Formation of Galaxies)

DefineUsing mass conservation to differentiate a material moment givesMultiply the Jeans equation for component $j$ by $x_k$, integrate over space, discard the surface term of an isolated system, and symmetrize in $j,k$. The result isDecompose $⟨v_j{v}_k⟩={\bar{v}}_j{\bar{v}}_k+{\sigma }_{jk}^2$ and defineThis proves the tensor virial theorem

For a steady oblate system rotating about $z$, $T_{zz}=0$, while symmetry gives $T_{xx}=T_{yy}$. The $zz$ and $xx$ equations areUsing ${\Pi }_{zz}=(1-\delta ){\Pi }_{xx}$, $2T_{xx}=M{v}_0^2$, and ${\Pi }_{xx}=M{\sigma }_0^2$ givesWith $W_{xx}/W_{zz}\simeq (1-ϵ{)}^{-0.9}$, the edge-on curves areAt fixed ellipticity, increasing anisotropy lowers $v_0/{\sigma }_0$; the curve starts only once $(1-ϵ{)}^{-0.9}\ge (1-\delta {)}^{-1}$. The $\delta =0$ line is the oblate isotropic rotator reference.

At inclination $i<90^{\circ }$, line-of-sight rotation is reduced approximately by $\sin i$, and the projected ellipticity also decreases as the spheroid is viewed closer to its symmetry axis. A galaxy therefore moves down and left from its edge-on location, with the exact track set by intrinsic thickness and anisotropy.

Low-mass elliptical galaxies mostly occupy the fast-rotating, flattened, nearly isotropic region close to the oblate-rotator line. High-mass systems mostly occupy the slow-rotating region below it and require anisotropy or triaxiality. Gas-rich dissipative evolution retains angular momentum, forms a compact rotating stellar component, and produces low-mass fast rotators. Repeated dry major mergers randomize orbits, lower specific angular momentum, scour central cores through black-hole binaries, and build massive slow rotators.

Four characteristic contrasts are:

These are population trends rather than sharp boundaries in the fast and slow rotator galaxy classification.

A galaxy contains so many stars that its two-body relaxation time is generally much longer than its age. Individual encounters can therefore be neglected and each star moves in the smooth collective potential. Liouville conservation along these Hamiltonian trajectories gives the Collisionless Boltzmann equationIn spherical phase-space coordinates this is

The spherical line element is $d{s}^2=d{r}^2+r^{2}d{\theta }^2+r^2{\sin }^2\theta d{\phi }^2$, soFor a unit-mass star in a spherical potential,The Euler--Lagrange equations, followed by differentiating $v_{\theta }=r\dot{\theta }$ and $v_{\phi }=r\sin \theta \dot{\phi }$, give

Integrating the Boltzmann equation over velocity space, with vanishing velocity-space boundary terms, gives spherical mass conservation:Multiplication by $v_r$ and integration gives the radial Jeans equation. With isotropic dispersionthe geometric dispersion terms cancel, and use of continuity yieldsThe factor $\rho$ on the right is required dimensionally.

In a Lambda-CDM background, the local excess mass contributes $GM(r)/r^2$, homogeneous matter inside radius $r$ contributes $4\pi G{\rho }_{b}r/3$, and the cosmological constant contributes outward acceleration $\Lambda r/3$. HenceSince ${\Omega }_m=8\pi G{\rho }_b/(3H^2)$, ${\Omega }_{\Lambda }=\Lambda /(3H^2)$, and $q={\Omega }_m/2-{\Omega }_{\Lambda }$,

Write $⟨v_r⟩=Hr+v_p$, where $v_p$ is the peculiar velocity. ThenThe acceleration equation gives $\dot{H}+H^2=\ddot{a}/a=-q{H}^2$, which cancels the background term in ${\Phi }^{\prime }$. Therefore

Before recombination, tight coupling makes photons and baryons share an adiabatic perturbation. Since $\delta {\rho }_b/{\rho }_b=(3/4)\delta {\rho }_r/{\rho }_r$ and only radiation supplies appreciable pressure,Thus the photon-baryon sound speed isBecause ${\rho }_r\propto a^{-4}$ and ${\rho }_b\propto a^{-3}$,After cosmological recombination, baryons decouple from radiation; for adiabatically cooling nonrelativistic gas, $T_b\propto a^{-2}$ and $c_s\propto a^{-1}$.

The Comoving Jeans length isDuring radiation domination $\bar{\rho }\propto a^{-4}$, so ${\lambda }_{\mathrm{com}}\propto a$. During matter domination before recombination, $\bar{\rho }\propto a^{-3}$ and $c_s\propto a^{-1/2}$, so ${\lambda }_{\mathrm{com}}$ is approximately constant. Recombination causes a sharp downward jump in sound speed, after which ${\lambda }_{\mathrm{com}}\propto a^{-1/2}$. The requested log-log sketch therefore rises with slope one, reaches a plateau after equality, drops at recombination, and then declines with slope $-1/2$.

For collisionless particles, replace $c_s$ by their one-dimensional velocity dispersion $v_{\mathrm{rms}}$:For a thermally produced cold relic, identify four epochs:

Its plot rises through the two relativistic epochs, turns onto a plateau at the nonrelativistic transition, and falls after equality. Recombination does not dynamically affect collisionless dark matter.

Treat the Milky Way and M31 as a radial Kepler orbit of total mass $M$. At maximum separation $r=2a$ the radial speed vanishes, so conservation of energy givesPut $r=a(1-\cos 2\eta )=2a{\sin }^2\eta$. Substitution into the energy equation and integration from the Big Bang gives the cycloidal orbitSince $r^3=8a^3{\sin }^6\eta$,Differentiating the parametric solution givesEliminating $\eta$ numerically gives the stated accurate fit

A remote dwarf sees the Local Group primarily through its total monopole mass, so the same radial timing relation applies with its own $r_i,v_i$. Dimensional consistency requiresrather than the cube-root exponent printed in the converted statement. Let $C=2^{-3/2}\pi$. At the zero-velocity radius $r_0$, ${\omega }_0=0$ and ${\Omega }_{0}t=C$. For the dwarf, ${\Omega }_{i}t=C-0.85{\omega }_{i}t$. Since $\Omega \propto r^{-3/2}$,Therefore

An Einstein-de Sitter universe has $a\propto t^{2/3}$, $H=2/(3t)$, and homogeneous density $\rho =1/(6\pi G{t}^2)$. HenceBefore the perturbation appreciably changes the motion, every shell follows the Hubble flow,

By Newton's shell theorem, only matter currently inside a shell contributes to its radial acceleration. Labelling matter by initial radius givesBefore shell crossing, the enclosed mass of a given shell is $M_i$. Its initial specific energy isAt the turnaround radius, $E_i=-G{M}_i/r_{\ast }$, soThe radial Kepler solution from the Big Bang to apocenter givesAfter turnaround a shell falls inward. Shell crossing makes its enclosed mass time dependent; shells then oscillate through the center, form caustics near successive apocenters, and phase mix into a halo.

For self-similar secondary infall, writeSubstitution in the shell equation givesUsing $G{t}_{\ast }^2/r_{\ast }^3={\pi }^2/(8M_i)$,

Now imposeThe case $ϵ=0$ is a scale-independent fractional overdensity, for which all shells turn around together in the ideal model. The case $ϵ=1$ has constant excess mass $\delta M_i=M_0$, corresponding to a central point-mass seed outside its core.

Since $r_{\ast }/r_i=(M_i/M_0{)}^{ϵ}$, while $r_i^3=(9/2)G{t}_i^2{M}_i$, the turnaround formulas giveThe shell turning at time $t$ therefore hasFor a shell with $\tau =t/t_{\ast }(M_i)$,Consequently its autonomous similarity equation isThe exponent $2/(3ϵ)$ follows directly from the preceding mass ratio; the printed $2ϵ/3$ in the final displayed equation is inconsistent with the supplied initial condition and preceding requested result. The corrected nonlinear equation is integrated numerically after turnaround.