Past exam of the general relativity II course of the University of Oxford / 2015 / 3 / d

For $a>0$, the assumed inequality and the Friedmann acceleration equation imply $\ddot{a}\le 0$. Thus $a$ is a concave function. Let $a_0=a(t_0)>0$ and $v_0=\dot{a}(t_0)>0$. At every earlier time in the same regular expanding interval, $\dot{a}(t)\ge v_0$. Integration gives the concavity bound on past FLRW evolution:The right-hand side reaches zero at $t=t_0-a_0/v_0$. Hence an everywhere regular positive scale factor cannot extend arbitrarily far into the past. With the usual implicit continuation assumption that the cosmological solution remains regular while $a>0$, it reaches $a(t_B)=0$ at a finite past boundary satisfyingPast expansion lasts at most the present inverse Hubble parameter. The inequality is the essential argument, and equality is attained by a linearly growing scale factor.

There is a literal qualification to the claim that $a$ must reach zero: the printed hypothesis alone does not exclude an earlier curvature singularity with positive limiting $a$. In arbitrary fixed units, take $k=0$ and $a(t)=1+\sqrt{t}$ for $t>0$, and define $\varrho ,p$ using the two Friedmann equations and the Friedmann acceleration equation. Then $\dot{a}>0$, $\ddot{a}<0$, and $\varrho +3p=-3\ddot{a}/(4\pi Ga)>0$, but $a\to 1$ while the energy density and pressure diverge as $t\downarrow 0$. Thus the rigorous unconditional conclusion is the finite bound on a regular past interval; reaching zero additionally requires the stated continuation assumption. Nor does a zero of $a$ alone prove a curvature singularity, as the Milne model illustrates.