Past exam of the short options paper of the physics Prelims of the University of Oxford / 2022 / S02 / 3

Use ${\rho }_r,{\rho }_m$ as energy per volume, as specified in the prose, and let $R$ have dimensions of length so that the positive curvature magnitude $k$ is dimensionless. The Friedmann equation for spatial curvature $-k$ isFor an energy density the cosmological perfect-fluid continuity equation is $\dot{\rho }+3H(\rho +P)=0$. Nonrelativistic matter has negligible pressure and radiation has $P_r={\rho }_r/3$. Thus the four contributions to $H^2$ scale respectively asFor nonzero positive densities and $\Lambda >0$, the usual possible sequence is radiation domination, matter domination, curvature domination, then cosmological-constant domination. A distinct era for each is not guaranteed: coefficients may cause curvature to be overtaken by the cosmological constant before it ever dominates. If $\Lambda =0$, curvature dominates asymptotically; negative $\Lambda$ does not give a late positive vacuum-dominated expansion.

The printed auxiliary relation $P_r={\rho }_r{c}^2/3$ uses mass-density notation, in conflict with the stated energy-density definition. It is consistent if ${\rho }_r$ there is understood as energy density divided by $c^2$; in that convention the density term in the Friedmann equation is $8\pi G\rho /3$. The two conventions must not be mixed.

In the prescribed matter-dominated approximation, neglect curvature, radiation, and the cosmological constant throughout the calculation. The Friedmann equation and ${\rho }_m\propto R^{-3}$ give $\dot{R}\propto R^{-1/2}$, hence $R\propto t^{2/3}$ and $H=2/(3t)$. For $H_0=70\mathrm{km}{\mathrm{s}}^{-1}\mathrm{Mp}{\mathrm{c}}^{-1}=2.269\times 10^{-18}{\mathrm{s}}^{-1}$,The particle horizon is a proper distance on a constant-time slice, not simply $ct$:ThereforeThese are estimates within a spatially flat, matter-only model, rather than the measured age and particle horizon of the actual late-time universe.

Normalize the scale factor to $a(t_0)=1$ and set $a_d=10^{-3}$. In the same matter-only model $a=(t/t_0{)}^{2/3}$, soIts present proper size after expansion with the scale factor isThe present comoving distance to decoupling isThe angular diameter distance is $D_A=a_d{\chi }_d$, so the angular size of a transverse length equal to one particle horizon radius isIf “the horizon” denotes the diameter of the whole causally connected ball, its angular diameter is twice this, approximately $3.74^{\circ }$. The degree-scale conclusion is the same. Neglecting early radiation is an explicit approximation inherited from the question.

The angle estimates the largest transverse region at decoupling that could have established causal contact after the big bang in the assumed decelerating model. The cosmic microwave background has nearly the same temperature in directions separated by much larger angles. This is the horizon problem: those regions have no shared causal history in that simple model, yet their thermodynamic conditions are strikingly alike. The particle horizon should be distinguished from the smaller sound horizon, which controls acoustic propagation in the primordial fluid.

An early epoch dominated by a sufficiently large positive effective cosmological constant gives $\ddot{R}>0$ and approximately $R\propto e^{Ht}$. During this cosmic inflation, the comoving Hubble length $c/(RH)$ shrinks. A patch that was initially in causal contact can expand to contain the whole presently observable cosmic microwave background surface, while expansion suppresses spatial curvature. A finite period of cosmic inflation followed by reheating and the usual hot expansion therefore explains the uniformity without superluminal local signals. An eternally large unchanged cosmological constant would prevent the later matter-dominated history; an exit from cosmic inflation and reheating are required. Sufficient duration and a suitable pre-inflationary causal patch are part of this explanation.