Past exam of the mathematics course of the University of Cambridge / 2009 / ii / Paper 4 / 10D (Cosmology)

Take velocities relative to a comoving observer at $O$. Spatial homogeneity means that changing the observer to $A$ gives the same relative velocity law, soWith $O$ as origin this implies $v(0)=0$ and $v(a+b)=v(a)+v(b)$. A physical velocity field is continuous; additivity first gives rational homogeneity and then, by continuity, real homogeneity. Hencewith coefficients independent of position. Without a regularity assumption, the additive functional equation alone admits non-linear pathological solutions; continuity is the usual implicit velocity-field assumption.

Isotropy requires $v(Rr)=Rv(r)$ for every spatial rotation, or $HR=RH$. Rotations about $e_i$ force $H{e}_i$ to be parallel to $e_i$, and rotations exchanging coordinate directions force those diagonal coefficients to agree. Therefore $H_{ij}=H{\delta }_{ij}$ and $\boxed{v=Hr}$, the Hubble law.

Allow $H=H(t)$. Solving $\dot{r}=H(t)r$ gives $r(t)=a(t)x$ for a constant comoving coordinate $x$, where $a(t)=a(t_0)\exp ({\int }_{t_0}^{t}H(s)ds)$. Thus $\boxed{H=\dot{a}/a}$, with $a$ the scale factor.

The physical particle horizon is the greatest physical distance at time $t$ from which a light signal could have reached the observer since the beginning of the cosmological model. If that beginning is $t=0$, radial null propagation has comoving distance ${\int }_0^{t}cd{t}^{\prime }/a(t^{\prime })$, soFor $a(t)=t^{\alpha }$ and $0<\alpha <1$, the integral converges at zero and equals $t^{1-\alpha }/(1-\alpha )$. ConsequentlyThe horizon problem is that widely separated regions of the observed cosmic microwave background have nearly equal temperatures although their past causal regions did not overlap before recombination in the usual decelerating hot-big-bang history. That history offers no causal equilibration mechanism for their observed uniformity. The particle horizon used here is distinct from the Hubble radius $c/H$.