Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 346 (Formation of Galaxies)

For an adiabatic cosmological perturbation, the baryon and photon fractional density perturbations obey ${\delta }_b=3{\delta }_{\gamma }/4$. Since the pressure of the photon-baryon fluid is supplied by the photons,The photon-baryon sound speed is therefore

Use ${\bar{\rho }}_b\propto a^{-3}$ and ${\bar{\rho }}_{\gamma }\propto a^{-4}$ in the Comoving Jeans lengthWell before matter-radiation equality, photon inertia dominates, $c_s\simeq c/\sqrt{3}$, andOnce baryon loading dominates while tight coupling still holds, $c_s\propto a^{-1/2}$, so ${\lambda }_{J,\mathrm{com}}$ is approximately constant. At cosmological recombination, photon pressure support disappears and the baryonic Jeans scale drops sharply. For a subsequently adiabatic monatomic gas, $T_b\propto a^{-2}$ and $c_s\propto a^{-1}$, givingin an Einstein-de Sitter universe. The requested graph therefore rises as $t^{1/4}$, flattens before recombination, jumps downward there, and then decreases as $t^{-1/3}$.

For collisionless matter, the same instantaneous estimate uses its one-dimensional velocity dispersion ${\sigma }_v$ instead of $c_s$; more precisely, suppression is described by collisionless free streaming. ThusWhile the particles are relativistic, ${\sigma }_v\simeq c$ and ${\lambda }_{J,\mathrm{com}}\propto a^{1/2}$. Once nonrelativistic but thermally coupled to radiation, $T_{\mathrm{cdm}}\propto a^{-1}$, so ${\sigma }_v\propto a^{-1/2}$ and the scale is constant. After kinetic decoupling, momentum redshifts as $a^{-1}$, so ${\sigma }_v\propto a^{-1}$ and ${\lambda }_{J,\mathrm{com}}\propto a^{-1/2}$. The second graph joins these three power laws at $t_{\mathrm{NR}}$ and $t_{\mathrm{dec}}$; unlike the baryonic graph, its final decline begins at dark-matter kinetic decoupling rather than recombination.

The shell feels only radial gravity and the radial force due to the cosmological constant, so its torque vanishes and its specific angular momentum $L=r^2\dot{\theta }$ is conserved. Multiplyingby $\dot{r}$ and integrating gives the conserved specific orbital energy

For a uniform sphere, assembling concentric shells gives its gravitational potential energyThe cosmological-constant potential per unit mass is $-\Lambda r^2/6$. Since $⟨r^2⟩=3r_{\mathrm{ta}}^2/5$ in a uniform sphere,

The scalar virial theorem weights a potential homogeneous of degree $n$ by $-nW$. Gravity has degree $-1$ and the $\Lambda$ potential degree $2$, so the final state obeysAt turnaround $T=0$, while the virial relation gives $E_f=W_{G,f}/2+2W_{\Lambda ,f}$. Conservation of energy, together with $M=4\pi {\rho }_{\mathrm{ta}}{r}_{\mathrm{ta}}^3/3$, then gives, for $x=r_f/r_{\mathrm{ta}}$ and $\eta =\Lambda /(4\pi G{\rho }_{\mathrm{ta}})$,The root connected continuously to the $\Lambda =0$ solution has $x=1/2-\eta /8+O({\eta }^2)$, equivalentlyto first order in $\eta$. With $\Lambda =0$, virialization occurs at half the turnaround radius. At fixed turnaround state, positive $\Lambda$ makes this equilibrium root slightly smaller because its repulsive quadratic potential enters both energy conservation and the virial relation; sufficiently strong repulsion instead prevents a bound virialized state. Negative $\Lambda$ shifts the root in the opposite direction.

The two halo centres orbit their centre of mass with separation $R$. Their relative coordinate obeys $\ddot{\mathbf{R}}=-G(M_1+M_2)\mathbf{R}/R^3$, hence circular motion requiresReflection symmetry about the orbital plane makes the vertical force point back toward $z=0$, while the centrifugal force has no vertical component. An equilibrium away from that plane is therefore impossible.

In the uniformly rotating frame, an equilibrium is a stationary point of the gravitational plus centrifugal effective potentialSet $R=1$, divide by $G(M_1+M_2)/R$, and write $\alpha =M_2/(M_1+M_2)$. The primary and secondary lie at $x=-\alpha$ and $x=1-\alpha$, so on their line

For the two roots near the secondary, put $x=1-\alpha \mp d$ in $F^{\prime }(x)=0$. Dominant balance gives $3d\simeq \alpha /d^2$, so $d=(\alpha /3{)}^{1/3}$. Expanding the root beyond the primary directly in powers of $\alpha$ givesto the requested orders. The other equilibria are the two Triangular Lagrange points

The distance from $H_2$ to either nearby collinear point is the Hill radius. Since $\alpha \simeq M_2/M_1$,Inside this tidal radius, the subhalo's gravity dominates the host's differential gravitational field; outside it, material can escape through the neighborhoods of $L_1$ and $L_2$. For an extended spherical host, $M_1$ is replaced by enclosed mass and the coefficient becomes $3-d\log M_1/d\log R$, giving the Jacobi tidal radius. An extended subhalo requires the bound mass inside $r_t$ to be found self-consistently. On an eccentric orbit there is no time-independent rotating potential or exact tidal boundary; stripping is strongest near pericentre and the instantaneous radius varies around the orbit.

Because the dark component is more extended, tidal stripping first sends dark matter through both $L_1$ and $L_2$, producing leading and trailing dark-matter tidal tails. The compact stellar component is stripped more deeply and also produces a leading and a trailing stellar tail. The two constituents therefore give four tails distinguished by composition, with the dark tails broader and more extended.

The definition of the virial radius of a dark-matter halo immediately gives the virial mass of a dark-matter haloIn an Einstein-de Sitter universe, spherical collapse gives $v=18{\pi }^2\simeq 178$, conventionally rounded to about $200$.

For the Navarro--Frenk--White profile, put $x=r/r_s=cs$. Direct integration givesUsing $r_s=r_v/c$ and ${\delta }_{\mathrm{char}}=v{c}^{3}g(c)/3$ therefore yieldsAt small radius the bracket is $(cs{)}^2/2+O((cs{)}^3)$, so $M(r)\propto r^2$, consistent with the central $\rho \propto r^{-1}$ cusp. At large radius it is $\log (cs)-1+o(1)$, so the mass diverges logarithmically unless the halo is truncated.

The primordial free streaming of warm dark matter erases small-scale density fluctuations and lowers the central phase-space density, tending to replace the smallest, earliest cold-dark-matter cusps by shallower central profiles or cores.

Since $V_v^2=G{M}_v/r_v$, the spherical potential that vanishes at infinity isIndeed, $rd\Phi /dr=GM(r)/r$, and normalization at $s=1$ requiresThe circular speed is consequentlyIt rises as $s^{1/2}$ near the centre, peaks at $r\simeq 2.16r_s$, and then declines approximately as $\sqrt{\log r/r}$. Increasing the concentration of a dark-matter halo moves the peak inward in units of $r_v$ and raises it relative to $V_v$. Thus an NFW curve can be fairly broad but is not exactly a flat galaxy rotation curve; stellar and gas contributions matter when comparing with an observed galaxy rotation curve.

Lower-mass haloes typically collapse earlier, when the cosmic background density is larger. Their characteristic inner densities are consequently larger relative to the present virial density, producing the mass-concentration relation of dark-matter haloes in which concentration decreases weakly with mass.

For a mass smaller by $1024=2^{10}$,and $c=101024^{0.1}=20$. HenceA halo of roughly $10^9{M}_{\odot }$ and kiloparsec scale naturally hosts a dwarf galaxy, possibly an extremely faint one if star formation is inefficient.

Write proper position as $\mathbf{r}=a(t)\mathbf{x}$ and proper velocity asSubstitute this decomposition into the proper-coordinate Euler equations for an inviscid fluid, use ${\nabla }_r=a^{-1}{\nabla }_x$, and subtract the homogeneous-background acceleration. With $\rho =\bar{\rho }(1+\delta )$, one obtains the comoving peculiar-velocity equationThe peculiar gravitational potential $\Phi$ is the total Newtonian potential with the potential of the exactly homogeneous expanding background subtracted. Its gradient therefore generates only accelerations relative to the Hubble flow.

For a pressureless fluid, linearization discards the quadratic advection term, leavingIn the early Einstein-de Sitter universe, the growing mode has $\Phi (\mathbf{x},t)=[D(a)/a]{\Phi }_i(\mathbf{x})$. Integration gives

The linear growth factor obeysSince $\bar{\rho }={\bar{\rho }}_i/a^3$ after choosing $a_i=1$, this is equivalent toIt follows that $a\mathbf{v}=-a^2\dot{D}\nabla {\Phi }_i/(4\pi G{\bar{\rho }}_i)$. Since $\dot{\mathbf{x}}=\mathbf{v}/a$, one final integration gives the Zel'dovich approximationwhere an additive initial displacement has been absorbed into ${\mathbf{x}}_i$.