Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 62 / 1

Use units with $c=1$. For the flat FRW metric, the Friedmann equation is $H^2=8\pi G\rho /3+\Lambda /3$, where $H=\dot{a}/a$ is the Hubble parameter. The cosmological continuity equation for a separately conserved component with constant equation of state $P=w\rho$ gives $\rho \propto a^{-3(1+w)}$. Consequently radiation in cosmology scales as $a^{-4}$, pressureless matter as $a^{-3}$, and cosmological constant energy is constant. With $a_0=1$, define the present critical density and cosmological density parameters byThe Hubble constant $H_0$ measures the expansion rate today. Here ${\Omega }_r$ includes photons and relativistic neutrinos, ${\Omega }_m$ includes baryons and cold dark matter, and ${\Omega }_{\Lambda }$ measures the vacuum contribution. Spatial flatness implies ${\Omega }_r+{\Omega }_m+{\Omega }_{\Lambda }=1$. ThusSince conformal time satisfies $d\tau =dt/a$, we have $a^{\prime }=a\dot{a}=a^{2}H$. Multiplication by $a^4$ provesFor an exam-level estimate, $H_0=100h\mathrm{km}{\mathrm{s}}^{-1}\mathrm{Mp}{\mathrm{c}}^{-1}$ with $h\simeq 0.7$, ${\Omega }_m\simeq 0.3$, ${\Omega }_{\Lambda }\simeq 0.7$, and the specified radiation in cosmology gives ${\Omega }_r=4.2\times 10^{-5}{h}^{-2}\simeq 8.6\times 10^{-5}$. These are rounded illustrative flat-universe values; precision cosmological density parameters depend on the fitted model. For comparison, the base flat-model fit in Planck 2018 cosmological parameters gives $H_0\simeq 67.4\mathrm{km}{\mathrm{s}}^{-1}\mathrm{Mp}{\mathrm{c}}^{-1}$ and ${\Omega }_m\simeq 0.315$.

On the expanding branch of the Friedmann equation, dropping the vacuum term gives the conformal timeThe lower limit sets $\tau =0$ at the Big Bang. The approximation requires ${\Omega }_{\Lambda }{a}^4\ll {\Omega }_{m}a$, and remains safe earlier when radiation in cosmology dominates. At matter-radiation equality, ${\rho }_{m,0}{a}^{-3}={\rho }_{r,0}{a}^{-4}$, so $a_{\mathrm{eq}}={\Omega }_r/{\Omega }_m$. Substitution yieldsIn conformal coordinates the radial FRW metric is $d{s}^2=a^2(-d{\tau }^2+d{r}^2)$. A radial light ray therefore travels a comoving distance $dr=d\tau$. The maximum distance traversed since the Big Bang is ${\int }_0^{t_{\mathrm{eq}}}dt/a={\tau }_{\mathrm{eq}}$: this is the comoving particle horizon, rather than the instantaneous comoving Hubble radius or the acoustic horizon. Restoring $c$ in the numerical conversion givesFor ${\Omega }_m=0.3$ and $h=0.7$, the comoving particle horizon at matter-radiation equality is about $110\mathrm{Mpc}$. The corresponding scale factor is $a_{\mathrm{eq}}\simeq 2.9\times 10^{-4}$.

Dropping radiation in cosmology in the Friedmann equation givesTo evaluate this matter-vacuum conformal age integral, put $a=({\Omega }_m/{\Omega }_{\Lambda }{)}^{1/3}y$. The square root in the denominator becomes ${\Omega }_m^{2/3}{\Omega }_{\Lambda }^{-1/6}\sqrt{y(1+y^3)}$, whereas $da={\Omega }_m^{1/3}{\Omega }_{\Lambda }^{-1/3}dy$. ThereforeThe radiation-free approximation is not valid arbitrarily close to the lower limit. Its extension to zero nevertheless gives a good leading conformal time estimate: the early-time correction has size $O(H_0^{-1}\sqrt{{\Omega }_r}/{\Omega }_m)$, small compared with the total age when matter-radiation equality occurs sufficiently early. The apparent $y^{-1/2}$ endpoint singularity is integrable.

An object at matter-radiation equality lies at comoving radial distance $\chi ={\tau }_0-{\tau }_{\mathrm{eq}}$ in the flat FRW metric. Its angular diameter distance is $D_A=a_{\mathrm{eq}}\chi$, while one particle horizon radius has physical size $a_{\mathrm{eq}}{\tau }_{\mathrm{eq}}$. Its small angular radius is consequentlyThe last step uses ${\tau }_{\mathrm{eq}}\ll {\tau }_0$. If ${\Omega }_{\Lambda }/{\Omega }_m\gg 1$, the matter-vacuum conformal age integral tends to $I$: its integrand behaves as $y^{-1/2}$ at zero and $y^{-2}$ at infinity. Inserting the two conformal times givesThe negative matter exponent follows directly from ${\Omega }_m^{-1}/{\Omega }_m^{-1/3}$. This agrees with the original PDF; the TeX transcription drops the minus sign. If desired, $z=y^3$ evaluates the constant through the beta function and gamma function:The infinite-upper-limit approximation requires vacuum dominance much stronger than in the illustrative parameters of part (a), so it is an asymptotic formula rather than a precision angular prediction for those values.