Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 45 / 3 / f / Solution

Use the physical root of the cubic which tends to $1/2$ as $\eta \to 0$. For $0<\eta <1$, the cubic has value $1$ at $x=0$ and value $-\eta /4$ at $x=1/2$, and its derivative $6\eta x^2-(2+\eta )$ is negative throughout that interval. Thus it has exactly one root there. Implicit differentiation givesThe smallest limiting value occurs as $\eta \to 1^{-}$, since larger $\eta$ cannot have a finite turnaround. At $\eta =1$ the equation factors:The continuation of the collapsing root is thereforeThe lower endpoint is an infimum, not an attained minimum for a shell that turns around in finite time. This is the minimum virial-radius ratio with positive cosmological constant. The separate root $x=1$ at $\eta =1$ describes the limiting static situation and is not the collapse branch.

The smaller ratio does not mean vacuum repulsion makes turnaround easier. At fixed turnaround data, $U_m=-A-B$ is more negative than for $\Lambda =0$. If one tried $x=1/2$, the virialized energy would be $-A-B/2$, which is not negative enough to equal $U_m$. Extra contraction is necessary to increase the magnitude of the gravitational binding energy. Thus vacuum repulsion inhibits formation of a bound shell, while the shell that does form has a smaller idealized final-to-turnaround ratio under the stated energy and profile assumptions.