# How to prove Poisson equation for the peculiar potential in Cosmology

I am self-studying Cosmology using a set of class notes, where I came across a claim that I cannot understand. First, it is stated that the equation of motion for a particle (located at physical position $$\vec{r}$$) in the cosmological fluid that fills the universe is the folowing:

$$\ddot{\vec{r}}=-\vec{\nabla}_r\Phi$$

where $$\vec{x}$$ is the vector of comoving coordinates, $$\vec{r}=a\vec{x}$$ the vector of physical coordinates, $$a$$ the scale factor that measures the expansion of the universe, and $$\vec{\nabla}_r$$ the gradient with respect to physical coordinates. I am not sure of what this potential is or why this equation is so, but the notes continue by defining the so-called \emph{peculiar potential} $$\phi$$ as:

$$\Phi=-\dfrac{1}{2}\mathcal{H}x^2+\phi$$

It is important to note that $$H=\dot{a}/a$$ and $$\mathcal{H}=a'/a$$, where dots denote the derivative $$d/dt$$ with respect to coordinate time and dashes denote the derivative $$d/d\eta$$ with respect to conformal time, and where $$d\eta=dt/a$$.

Now comes the mystery:

The peculiar potential is solely seeded by the energy density fluctuations in the universe. Since we are assuming that dark energy is homogeneous, at late times these energy density fluctuations are dominated by the fluctuations in the matter density. Hence, the Poisson equation for the peculiar potential can be expressed as

$$\nabla_x^2\phi=\dfrac{3}{2}\mathcal{H}^2\Omega_m(a)\delta=\dfrac{3}{2}\Omega_{m,0}H_0^2\dfrac{\delta}{a}$$

How? The expression that I know for the Poisson equation is $$\nabla_x^2\phi=4\pi G\rho$$, but I don't know which potential is in the left hand side or how to get to this expression.

Any help would be greatly appreciated!

• I don't know which potential is in the left hand side. en.wikipedia.org/wiki/Gravitational_potential Aug 13, 2023 at 20:02
• I am self-studying Cosmology. Did you first study classical mechanics? That’s normally where one first learns the concepts of “conservative” forces, leading to Newtonian gravitational potential energy and Newtonian gravitational potential. Taking their negative gradients gives, respectively, Newtonian gravitational force and Newtonian gravitational acceleration. Aug 13, 2023 at 20:09
• @Ghoster Thank you, I wasn't sure if it was the gravitational potential or something else. I have a physics degree but sadly they didn't teach much about fluid mechanics or theoretical cosmology at my university. Thank you for clarifying that! Any clue about how to prove that expression for the Poisson equation? I do know what the involved magnitudes are, but I have no clue how to get what the notes state. Aug 13, 2023 at 21:46
• Do you mean how to get the $\frac32\Omega_{m,0}H_0^2\frac{\delta}{a}$? No, I don’t know how to do that. Aug 13, 2023 at 23:32
• I suspect you need to use the Friedman equation for a matter dominated universe, $H^2=\frac{8\pi G}{3}\bar{\rho}_m$ (where $\bar{\rho}_m$ is the background matter density), plus the definition $\Omega_{m,0} = \frac{8\pi G \bar{\rho}_m}{3H_0^2}$, plus the definition $\delta = \frac{\rho_m - \bar{\rho}_m}{\bar{\rho}_m}$, where $\rho_m$ is the full (background plus perturbation) matter density. If you substitute in all those relations into the Newtonian Poisson equation I suspect you'll get the right answer. Aug 14, 2023 at 2:24

Retaining your definitions of $$\vec x$$ as the comoving coordinate and $$\vec r=a\vec x$$ as the physical coordinate, the Poisson equation is $$\nabla_{\vec r}^2\Phi=4\pi G\rho.$$ Here $$\Phi$$ is the total potential. In a homogeneous universe of density $$\bar\rho$$, the solution would be $$\Phi = \frac{2\pi}{3}G\bar\rho r^2$$, so we define the peculiar potential $$\phi$$ as the deviation from that, such that $$\Phi\equiv \frac{2\pi}{3}G\bar\rho r^2+\phi = \left(\frac{Hr}{2}\right)^2+\phi,$$ where the second equality follows from the first Friedmann equation.
Substituting this into the Poisson equation yields $$\nabla_{\vec r}^2\phi=4\pi G(\rho-\bar\rho).$$ In comoving coordinates, this is $$\nabla_{\vec x}^2\phi=4\pi G a^2(\rho-\bar\rho).$$ Assuming matter is the only component whose density can differ from the average, $$\rho-\bar\rho$$ is equal to the matter density $$\Omega_m\bar\rho$$ times the density contrast $$\delta$$; that is, $$\rho-\bar\rho=\Omega_m\bar\rho\delta.$$ But by the first Friedmann equation, $$\bar\rho=\frac{3H^2}{8\pi G}$$, so that $$\nabla_{\vec x}^2\phi=\frac{3}{2} a^2 H^2 \Omega_m\delta=\frac{3}{2} \mathcal{H}^2 \Omega_m\delta.$$ In the last equality I used that $$aH=\mathcal{H}$$.
Also, since $$H^2\Omega_m$$ is proportional to the matter density, it scales as $$a^{-3}$$. This means that $$H^2\Omega_m=H_0^2\Omega_{m,0}a^{-3}$$ at the scale factor $$a$$, where $$H_0$$ and $$\Omega_{m,0}$$ are the Hubble rate and matter density parameter, respectively, at $$a=1$$. Then $$\nabla_{\vec x}^2\phi=\frac{3}{2} a^2 H^2 \Omega_m\delta=\frac{3}{2} H_0^2 \Omega_{m,0}\frac{\delta}{a}.$$
• Thank you very much, this is a wonderful answer. I just have two quick questions. First, isn't the Friedmann equation $H^2=\dfrac{8\pi G}{3}\rho$, with the total density $\rho$ instead of the background density $\bar{\rho}$? And second, why do you say that $H^2\Omega_m$ is proportional to the matter density? Thanks in advance! Aug 22, 2023 at 17:29
• @WildFeather The Friedmann equations are for a homogeneous universe, i.e. they are derived under the assumption that $\rho=\bar\rho$. The first Friedmann equation (for a flat universe) has $H^2=(8\pi/3)G\rho\propto\rho$, so $H^2\Omega_m\propto\rho\Omega_m=\rho_m$.