When ripples collapse — the spherical collapse model

Think of the dark matter as a collection of roughly spherical overdense clouds. Take one of them, a uniform sphere, sitting in an Einstein–de Sitter background — flat, matter-only, $\Omega = 1$. At first the sphere expands with the universe. But it contains more mass than average, so its own gravity decelerates it more strongly: it lags behind the background, reaches a maximum radius — turnaround — and then falls back in, eventually settling into a virialised halo.

This is the same physics as the closed universe in 8.286: by Birkhoff's theorem (Newtonian version: only the enclosed mass matters), an overdense sphere behaves like a small closed universe embedded in a flat one.

Measure everything relative to turnaround (subscript "ta"). The background obeys the Einstein–de Sitter Friedmann equation $H = H_0a^{-3/2}$. Define

so $x$ is the background scale factor and $y$ the sphere's radius, both equal to 1 at turnaround. Differentiating $x$ with respect to $\tau$, and using $H/H_\text{ta} = x^{-3/2}$:

The sphere's radius obeys Newton's law, with the enclosed mass fixed:

Write the sphere's density at turnaround in units of the background density then, which is the critical density $3H_\text{ta}^2/8\pi G$:

where $\xi > 1$ for an overdensity. (This $\xi$ has nothing to do with the correlation function of the previous page.) In the dimensionless variables the equation of motion becomes

— the sphere is momentarily at rest at turnaround and starts from zero size at the Big Bang. [In eq. (430) the notes repeat the equation with $3$ in place of $\xi$; $\xi$ is meant.]

The background is easy: $x' = x^{-1/2}$ integrates to

the familiar $a \propto t^{2/3}$. [Eq. (432) prints $x = (3/2)\,\tau^{2/3}$; the exponent applies to $3/2$ as well, as the next line of the notes shows; that line also writes $dx/dt$ for $dx/d\tau$.] For the sphere, multiply $y''$ by $y'$ and integrate once — energy conservation. Since $y' = 0$ at $y = 1$:

with $+$ before turnaround (expanding) and $-$ after (collapsing). The notes check by differentiating again that this reproduces eq. (428) (eq. 434). Separating variables and integrating from $y = 0$ gives an implicit solution:

Turnaround is at $x = y = 1$, $\tau = 2/3$. Then $\arcsin(1) = \pi/2$ and

At turnaround the sphere is about 5.6 times denser than the background. Collapse: the motion is symmetric in time about turnaround, so the sphere returns to $y = 0$ at $\tau = 4/3$, when the background has

Substituting $y = \tfrac12(1 - \cos s)$ turns eq. (435) into the cycloid $\tau = \frac{2}{3\pi}(s - \sin s)$ — the same curve as the closed universe in 8.286. (This parametric form is this page's own step; it is used for the plot below.)

What would linear theory have said about this sphere? At early times $y \ll 1$, and expanding eq. (435):

The density inside the sphere relative to the background is the turnaround value $\xi$ scaled by how much each has expanded since — the background density goes as $x^{-3}$, the sphere's as $y^{-3}$:

[The notes print $(x/y)\,\xi$ in eq. (439) but use the cube in eq. (440).] Equating the two expressions for $\tau$ at early times, $(x/y)^3 \approx \xi^{-1}(1 + 3y/5)$, so

Linear perturbations grow like the scale factor, so extrapolating this early $\delta$ to turnaround multiplies it by $a_\text{ta}/a = 1/x$, and to lowest order $1/x \approx (3\pi/4)^{2/3}/y$:

and at collapse, a further factor $x_c = 4^{1/3}$:

[The notes write $\delta_\text{ra}$ for $\delta_\text{ta}$ in the middle step.] This is the key result. The real sphere has collapsed, with formally infinite density, at the moment the linear theory says $\delta = 1.69$. So we can use linear theory as a bookkeeping device: evolve the density field linearly, and wherever it exceeds $\delta_c \approx 1.69$, a halo has formed. The notes illustrate this with a one-dimensional sketch: as the linear field grows, more and more of its peaks poke above the $\delta_c$ line.

A real halo does not collapse to a point: the particles' random motions build up and the halo settles into virial equilibrium, $2T + V = 0$. Energy is conserved, and at turnaround there is no kinetic energy:

[The notes write $V_\text{ra}$ for $V_\text{ta}$.] The potential energy goes as $1/R$, so the halo's radius is half its turnaround radius, $y = 1/2$. Taking virialisation to be complete at the collapse time, when $x = x_c$:

A virialised halo is about 178 times denser than the mean background. That is why halo masses and radii are usually quoted as $M_{200}$ and $R_{200}$: the mass inside the sphere whose mean density is 200 times the mean (or the critical) density of the universe.

• Once $\delta \sim 1$ linear theory fails and halos form. The spherical collapse model follows one uniform overdense sphere in an Einstein–de Sitter background: it obeys $y'' = -\xi/(2y^2)$, expands, turns around and collapses.
• At turnaround the sphere is $\xi = (3\pi/4)^2 \approx 5.55$ times denser than the background; collapse happens at twice the turnaround time, when the background has grown by $4^{1/3}$.
• Linear theory would give $\delta_\text{ta} \approx 1.06$ at turnaround and $\delta_c = \frac35(3\pi/2)^{2/3} \approx 1.69$ at collapse: a region has collapsed when its linear δ reaches 1.69.
• The virial theorem halves the radius, giving a halo $\Delta_V = 18\pi^2 \approx 178$ times denser than the background — hence the common $M_{200}$ halo definitions.