Linearisation of continuity equation (cosmology)

[ergospherical]

Homework Statement

Show linearisation of
$\frac{\partial \rho}{\partial t} + 3H \rho + \frac{1}{a} \nabla \cdot (\rho \mathbf{v}) = 0$
is
$\frac{\partial \delta}{\partial t} + \frac{1}{a} \nabla \cdot (\delta \mathbf{v}) = 0$
where
$\delta \equiv \delta \rho / \bar{\rho}$, $\rho = \bar{\rho} + \epsilon \delta \rho$, $\mathbf{v} = \epsilon \delta \mathbf{v}$, and $\epsilon \ll 1$.

Relevant Equations

N/A

After expanding to first order in $\epsilon$ and subtracting off the unperturbed equation, I get\begin{align*}
\frac{\partial \delta \rho}{\partial t} + 3H \delta \rho + \frac{\bar{\rho}}{a} \nabla \cdot \delta \mathbf{v}=0
\end{align*}I'm not sure how to deal with the $3H \delta \rho$ term. Where does $H$ enter? ($H = \dot{a}/a$ is the Hubble parameter).

 

[strangerep]

The 0'th order eqn is $$ \partial_t \bar\rho ~+~ 3 H \bar\rho ~=~ 0 ~.$$ You must use this in the 1st order eqn.

Additional hint: before writing out the 1st order eqn, compute $\,\partial_t \left( \frac{\delta\rho}{\bar\rho} \right)$ carefully, separately, using the 0'th order eqn.

[Question for other HW helpers: does the above give away too much of the solution in one go? I'm never really sure where the balance lies.]