Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 346 / 1

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.