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I am trying to derive the equation for a case, where we have a dust(zero-pressure) in an expanding universe.

There are 4 equations but I think exercising on one of them would be helpful for me.

So we have a continuity equation

$$\frac{\partial \rho}{\partial t} + \nabla \cdot (\rho \vec{v})=0$$

In an expanding universe

$\rho = \rho_0a^{-3}$

$\vec{v} = \frac{\dot{a}}{a}\vec{r}$

$\vec{r} = \vec{r_0}a$

where $a=a(t)$

So the problem is I am not sure how to define the perturbation

I thought I can write

$$\rho = \rho_0a^{-3} + \delta \rho$$ $$\vec{v} = \frac{\dot{a}}{a}\vec{r} + \delta \vec{v}$$

If I put it into the equation I get

$$\frac{\partial }{\partial t}(\rho_0a^{-3} + \delta \rho) + \nabla \cdot ((\rho_0a^{-3} + \delta \rho)(\frac{\dot{a}}{a}\vec{r} + \delta \vec{v}))=0$$

$$\frac{\partial }{\partial t}(\rho_0a^{-3}) + \frac{\partial }{\partial t}(\delta \rho) + \nabla \cdot (\rho_0a^{-3}\frac{\dot{a}}{a}\vec{r} + \rho_0a^{-3}\delta \vec{v}+ \delta \rho\frac{\dot{a}}{a}\vec{r})=0$$

since $\rho = \rho_0a^{-3}$

$$\frac{\partial }{\partial t}(\rho_0a^{-3})+\frac{\partial }{\partial t}(\delta \rho) + \nabla \cdot (\rho\frac{\dot{a}}{a}\vec{r} + \rho\delta \vec{v}+ \delta \rho\frac{\dot{a}}{a}\vec{r})=0$$

$$\frac{\partial }{\partial t}(\rho_0a^{-3})+\frac{\partial }{\partial t}(\delta \rho) + \rho\frac{\dot{a}}{a} (\nabla \cdot \vec{r}) + \rho(\nabla \cdot \delta \vec{v})+ \delta \rho\frac{\dot{a}}{a} (\nabla \cdot \vec{r})=0$$

The answer is

$$\frac{\partial }{\partial t}(\delta \rho) + 3\delta \rho\frac{\dot{a}}{a}+ \rho(\nabla \cdot \delta \vec{v})+ \delta \rho\frac{\dot{a}}{a} (\vec{r} \cdot \nabla )=0$$

Also $(\nabla \cdot \vec{r}) = 3 ?$

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  • $\begingroup$ You incorrectly transcribed the answer, and in your own attempt at derivation you lost the term $\frac{\partial }{\partial t} \rho_0 a^{-3}$. $\endgroup$ – A.V.S. Feb 9 at 7:30
  • $\begingroup$ @A.V.S. Which term is wrong ? Yes, thanks for noticing. $\endgroup$ – Reign Feb 9 at 8:09
  • $\begingroup$ $\nabla$ is a differential operator, so order of multipliers containing it matters: $(\mathbf{r} \cdot \nabla) \delta \rho $ is not the same as $\delta \rho (\nabla \cdot \mathbf{r})$. $\endgroup$ – A.V.S. Feb 9 at 16:17
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To derive the Jean instability condition, you need both the continuity and Euler equations and linearise them. You start with $$ \frac{\partial\rho}{\partial t}+\nabla\cdot(\rho\mathbf{v})=0 $$ and $$ \frac{\partial\mathbf{v}}{\partial t}+\left( \mathbf{v}\cdot\nabla \right)\mathbf{v}+\frac{\nabla P}{\rho}=0 $$ There might also be $\nabla \Phi$ term in the Euler equation, for which there needs to be a Poisson equation. You can then use $\rho(\mathbf{x},t)=\bar{\rho}(t)+\delta\rho(\mathbf{x},t)$ and $\mathbf{v}(\mathbf{x},t)=\bar{\mathbf{v}}(t)+\delta\mathbf{v}(\mathbf{x},t)$, where $\bar{\mathbf{v}}(t)=H(t)\mathbf{x}$. Once the two equations are linearised, $\delta\mathbf{v}$ can be plugged back into the equation for $\delta\rho$.

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  • $\begingroup$ I am doing similar things. At least can you point out what am I doing wrong ? I am aware that there are 4 equations to get the Jeans Theory, however I dont think that we need additional equations to derive the expression that I am trying to get. $\endgroup$ – Reign Feb 9 at 6:00

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