# Newtonian derivation of perturbation in density

In Barbara Ryden's Introduction to Cosmology, chapter 12.3, she derives an equation describing the evolution of mass perturbations with time, for small perturbations $|\delta|\ll 1$. Before she starts the derivation, a disclaimer is added stating that

By performing Newtonian analysis of this problem, we are implicitly assuming that the radius $R$ is small compared to the Hubble distance and large compared to the Jeans length.

I have added the derivation below, and was wondering where these implicit assumptions are taken into account.

Suppose a universe with pressure-less matter, with mean mass density $\overline{\rho}(t)$. As the universe expands, the density decreases as $\overline{\rho}(t)\propto a^{-3}(t)$, where $a$ is the scale factor of the universe. Consider a spherical region of radius $R$, to which a small amount of matter is added (or removed) so the density within the sphere is now $\rho(t) = \overline{\rho}(t)(1+\delta(t))$, where $|\delta| \ll1$. The total gravitational acceleration at the surface will be $$\ddot{R} = -\frac{GM}{R^2} = -\frac{4\pi}{3}G\overline{\rho}R - \frac{4\pi}{3}G\overline{\rho}\delta R$$

By mass conservation, we know that $$\frac{4\pi \overline{\rho}(t)}{3}R^3(t)(1+\delta(t)) \equiv \textrm{const}$$

From this, we get that $$R(t)\propto a(t)(1+\delta)^{-\frac{1}{3}}\approx a(t) - \frac{1}{3}\delta a(t)$$

Taking double time derivative, we get $$\ddot{R} \approx \ddot{a}(t) - \frac{1}{3}\ddot{\delta}a-\frac{1}{3}\delta\ddot{a} - \frac{2}{3}\dot{\delta}\dot{a}$$ Dividing by $R$, we get $$\frac{\ddot{R}}{R} \approx \frac{\ddot{a}(t) - \frac{1}{3}\ddot{\delta}a-\frac{1}{3}\delta\ddot{a} - \frac{2}{3}\dot{\delta}\dot{a}}{a(1-\frac{1}{3}\delta)} \approx \frac{\ddot{a}}{a}\left(1+\frac{1}{3}\delta\right) - \frac{1}{3}\ddot{\delta}\left(1+\frac{\delta}{3}\right) - \frac{1}{3}\delta\frac{\ddot{a}}{a}\left(1+\frac{1}{3}\delta\right) - \frac{2}{3}\dot{\delta}\dot{a}\frac{1}{a}\left(1+\frac{1}{3}\delta\right)$$ Taking only linear terms in $\delta$ (we neglect $\delta^2,\ddot{\delta}\delta$ as they are of second order in $\delta$), we get $$\frac{\ddot{R}}{R} \approx \frac{\ddot{a}}{a} - \frac{1}{3}\ddot{\delta}-\frac{2}{3}\frac{\dot{a}}{a}\dot{\delta}$$

Comparing with the gravitation equation, we get that the linear term, in charge of the small perturbation, yields $$\ddot{\delta} + 2H\dot\delta = 4\pi G\overline{\rho}\delta$$ where $H$ is the Hubble parameter.

At which point are the above assumptions are implicitly applied?

## 1 Answer

When you write out the gravitational acceleration as $$\ddot{R} = -\frac{GM}{R^2},$$ you are approximating gravitation as Newtonian. Your source is telling you that this approximation implicitly demands that $R$ is much smaller than the Hubble length, and much greater than the Jeans length. I would wager the latter is so that we can treat the matter content as approximately homogeneous, while the former is so that the expansion speed is small.