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Starting with the continuity equation

$$\frac{\partial \rho}{\partial t} + \nabla_r(\rho \vec{u}) = 0$$

making those substitutions. $$\nabla_r = \frac{\nabla_x}{a}$$ $$\rho(\vec{x},t) = \bar \rho(t)[1 + \delta(\vec{x},t)]$$

And then compute the time derivative I get: $$\frac{\partial \bar \rho}{\partial t} \frac{[1 + \delta]}{\rho} + \dot \delta + \frac{1}{a} \nabla_x [1 + \delta]\vec{u} = 0$$

However, based on this document the term $\frac{\partial \bar \rho}{\partial t}$ vanishes.

Is it because for small perturbations the mean density is constant? If so, why exactly?

Thank you

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    $\begingroup$ It's neglected because the transformation is is a linear approximation and the first term is non-linear. See slide 10 of the document you linked. $\endgroup$
    – ScienceSnake
    Commented Mar 22 at 10:54

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