= Solution
The normalized <cycloid> has
$$
\frac{R_s}{R_m}=\frac{1-\cos\eta}{2},\qquad
\frac{t}{t_m}=\frac{\eta-\sin\eta}{\pi}.
$$
The initially expanding perturbation reaches maximum radius at $\eta=\pi$, $t=t_m$, then formally collapses to zero radius at $\eta=2\pi$, $t=2t_m$. A finite virialized object is assigned the conventional formation epoch \b[$t_f=2t_m$]; random motions replace the pressureless point-collapse singularity. The <cycloid> alone does not describe that relaxation.
\Image[/past-exam-of-the-formation-of-structure-in-the-universe-course-of-the-university-of-cambridge/2012/paper-1-collapse-history.png]
{title=Spherical-collapse history}
For the standard top-hat <virialization> estimate, write $U=-C_GGM^2/R$ and retain the same gravitational structure factor $C_G$ at turnaround and final equilibrium. The coherent <kinetic energy> vanishes at turnaround, whereas the <virial theorem> gives $2K_v+U_v=0$ in the final isolated equilibrium. <conservation of energy> gives
$$
U_m=K_v+U_v=\frac{U_v}{2},
\qquad\boxed{R_v=R_m/2}.
$$
The final <density> is eight times the turnaround <density>. Meanwhile $\bar\rho\propto a^{-3}\propto t^{-2}$, so $\bar\rho(t_f)=\bar\rho(t_m)/4$. The <Einstein-de Sitter virial overdensity> follows by combining these factors with the supplied turnaround contrast:
$$
\boxed{\Delta_f=8\times4\left(\frac{3\pi}{4}\right)^2
=18\pi^2\simeq177.65\approx200,}
\qquad
\boxed{\rho_v=\Delta_f\rho_0(1+z_f)^3.}
$$
The value 200 is a convenient rounded overdensity convention, not exactly the Einstein-de Sitter result.
The <virial radius of a dark-matter halo> follows from $M_v=4\pi R_v^3\rho_v/3$:
$$
\boxed{R_v=
\left(\frac{3M_v}{4\pi\Delta_f\rho_0}\right)^{1/3}
\frac1{1+z_f}.}
$$
The <hydrostatic temperature of an inverse-square halo> follows from <hydrostatic equilibrium> and the <ideal gas law>:
$$
\frac{dP}{dr}=-\rho_g\frac{GM(r)}{r^2},\qquad
P=\frac{\rho_gk_BT_v}{\mu}.
$$
Since $\rho_g\propto r^{-2}$ and $M(r)=M_vr/R_v$, cancellation of $\rho_g/r$ gives $2k_BT_v/\mu=GM_v/R_v$. Therefore
$$
\boxed{T_v=\frac{GM_v\mu}{2k_BR_v}
=\frac{GM_v^{2/3}\mu}{2k_B}
\left(\frac{4\pi\Delta_f\rho_0}{3}\right)^{1/3}(1+z_f).}
$$
Here $\mu$ is a dimensionful <mean particle mass>, as specified in the paper; a <dimensionless> <mean molecular weight> would need multiplication by a particle-mass unit. The isothermal interior is a further halo idealization, not an exact continuation of the uniform top-hat profile used in the virial-overdensity estimate. Changing its gravitational structure factor in a literal <energy> comparison would change the turnaround-radius relation.
The <formation redshift from halo mean density> uses the cluster data, which imply $\rho_v=3M_v/(4\pi R_v^3)=5.37148\times10^{13}M_\odot\,\mathrm{Mpc}^{-3}$. Thus, using the requested rounded convention,
$$
\boxed{1+z_f=\left(\frac{3M_v}{4\pi R_v^3\Delta_f\rho_0}\right)^{1/3}
=1.91908,\qquad z_f\simeq0.92\quad(\Delta_f=200).}
$$
Using $18\pi^2$ instead gives $z_f\simeq1.00$. In the Einstein-de Sitter time scaling the rounded-convention result is $t_f/t_0=(1+z_f)^{-3/2}\simeq0.376$. Directly from the <mass> and radius,
$$
\boxed{T_v=2.34478\times10^8
\left(\frac{\mu}{m_p}\right)\mathrm K.}
$$
For the estimate, take <mean particle mass> $\mu=0.6m_p$, appropriate approximately to a <hydrogen>-<helium> mixture with complete <ionization>. This gives \b[$T_v\simeq1.41\times10^8\,\mathrm K$], or $k_BT_v\simeq12.1\,\mathrm{keV}$. The composition choice is additional input, not a value supplied by the examination; the boxed expression retains the result for any $\mu$.
Back to article page
