An <Einstein-de Sitter universe> has $a\propto t^{2/3}$, $H=2/(3t)$, and homogeneous density $\rho=1/(6\pi Gt^2)$. Hence
$$
\boxed{M_i=\frac{4\pi}{3}\rho(t_i)r_i^3=\frac{2r_i^3}{9Gt_i^2}}.
$$
Before the perturbation appreciably changes the motion, every shell follows the <Hubble flow>,
$$
\boxed{v_i=H_ir_i=\frac{2r_i}{3t_i}}.
$$

By Newton's shell theorem, only matter currently inside a shell contributes to its radial acceleration. Labelling matter by initial radius gives
$$
\boxed{\ddot r(r_i,t)=-\frac{GM(r,t)}{r^2}},
$$
$$
\boxed{M(r,t)=\int_0^\infty
\frac{dM_i}{dr_i}\,
H\!\left(r-r(r_i,t)\right)\,dr_i}.
$$
Before shell crossing, the enclosed mass of a given shell is $M_i$. Its initial specific energy is
$$
E_i=\frac12\left(\frac{2r_i}{3t_i}\right)^2-\frac{GM_i}{r_i}
=-\frac{G\,\delta M_i}{r_i}.
$$
At the <turnaround radius>, $E_i=-GM_i/r_*$, so
$$
\boxed{r_*=r_i\frac{M_i}{\delta M_i}}.
$$
The radial Kepler solution from the Big Bang to apocenter gives
$$
\boxed{t_*=\frac\pi2\sqrt{\frac{r_*^3}{2GM_i}}}.
$$
After turnaround a shell falls inward. Shell crossing makes its enclosed mass time dependent; shells then oscillate through the center, form caustics near successive apocenters, and phase mix into a halo.

For <self-similar secondary infall>, write
$$
M(r,t)=M(t)\mathcal M(r/R(t)),\qquad
r(r_i,t)=r_*(r_i)\Lambda(\tau),\qquad
\tau=t/t_*(r_i).
$$
Substitution in the shell equation gives
$$
\frac{r_*}{t_*^2}\Lambda''
=-\frac{GM(t)}{r_*^2\Lambda^2}
\mathcal M\!\left(\frac{r_*\Lambda}{R(t)}\right).
$$
Using $Gt_*^2/r_*^3=\pi^2/(8M_i)$,
$$
\boxed{
\Lambda''(\tau)=
-\frac{\pi^2}{8\Lambda^2}\frac{M(t)}{M_i}
\mathcal M\!\left(\frac{r_*\Lambda}{R(t)}\right)}.
$$

Now impose
$$
\frac{\delta M_i}{M_i}=\left(\frac{M_0}{M_i}\right)^\epsilon.
$$
The case $\epsilon=0$ is a scale-independent fractional overdensity, for which all shells turn around together in the ideal model. The case $\epsilon=1$ has constant excess mass $\delta M_i=M_0$, corresponding to a central point-mass seed outside its core.

Since $r_*/r_i=(M_i/M_0)^\epsilon$, while $r_i^3=(9/2)Gt_i^2M_i$, the turnaround formulas give
$$
\boxed{t_*=\frac{3\pi}{4}t_i
\left(\frac{M_i}{M_0}\right)^{3\epsilon/2}},
$$
$$
\boxed{r_*(M_i)=
\left[\frac8{\pi^2}t_*^2GM_i\right]^{1/3}}.
$$
The shell turning at time $t$ therefore has
$$
\boxed{M(t)=M_0
\left(\frac{4t}{3\pi t_i}\right)^{2/(3\epsilon)}},
$$
$$
\boxed{R(t)=
\left[\frac{8t^2G}{\pi^2}M(t)\right]^{1/3}}.
$$
For a shell with $\tau=t/t_*(M_i)$,
$$
\boxed{\frac{M(t)}{M_i}=\tau^{2/(3\epsilon)}},
\qquad
\boxed{\frac{r_*(M_i)}{R(t)}
=\tau^{-2/3-2/(9\epsilon)}}.
$$
Consequently its autonomous similarity equation is
$$
\boxed{
\Lambda''=
-\frac{\pi^2}{8}\frac{\tau^{2/(3\epsilon)}}{\Lambda^2}
\mathcal M\!\left(
\frac{\Lambda}{\tau^{2/3+2/(9\epsilon)}}
\right)}.
$$
The exponent $2/(3\epsilon)$ follows directly from the preceding mass ratio; the printed $2\epsilon/3$ in the final displayed equation is inconsistent with the supplied initial condition and preceding requested result. The corrected nonlinear equation is integrated numerically after turnaround.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 346 / 4 / Solution

{title2=$r_*$}
{wiki}

The turnaround radius of a shell is its maximum physical radius before collapse. At a given cosmic time, the currently turning shell defines a natural similarity length.


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 Incoming links: Turnaround radius