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.]