Missing Terms in Weinberg's treatment of perturbations on Newtonian Cosmology I was reading Appendix F of Steven Weingberg's book "Cosmology". In this Appendix he works out the perturbations to a cosmological fluid described by non-relativistic hydrodynamics and Newtonian gravity. 

It turns out that the first order perturbations satisfy,

$$
\frac{\partial \delta \rho }{\partial t } + 3 H \delta \rho + H \vec{X} \cdot \nabla \delta \rho + \bar{\rho} \nabla \cdot \vec{v} = 0, \qquad \tag{1}
$$

$$
\frac{\partial \delta \vec{v}}{\partial t } +  H \vec{X} \cdot \nabla \delta \vec{v} + H \delta \vec{v} = - \nabla \delta \phi, \qquad \tag{2}
$$

$$
\nabla^2 \delta \phi = 4\pi G \delta \rho. \qquad \tag{3} 
$$

Weinberg applies the following Fourier transform to these equations,

$$ f(\vec{X},t)  = \int \exp \left(  \frac{i \vec{q} \cdot \vec{X}}{a} \right)  f_{\vec{q}}(t) \ \mathrm{d}^3\vec{q} $$,

where $f(\vec{X},t)$ is a place holder for $\delta \vec{v}, \delta \rho, $ and $\delta \phi$. 

The resulting equations he gets are,

$$
\frac{\mathrm d \delta \rho_{\vec{q}}}{\mathrm d t } + 3 H \delta \rho_{\vec{q}} + \frac{i\bar{\rho}}{a}\ \vec{q} \cdot \delta \vec{v}_{\vec{q}} = 0 \qquad \tag{1'}$$

$$
\frac{\mathrm d \delta \vec{v}_{\vec{q}}}{\mathrm d t } +  H \delta \vec{v}_{\vec{q}} = -\frac{i}{a}\ \vec{q}  \delta \phi_{\vec{q}} \qquad \tag{2'}$$

$$
\vec{q}^2 \delta \phi_{\vec{q}} = -4\pi G a^2 \delta \rho_{\vec{q}} \qquad \tag{3'}$$.

For the most part these new equations can be obtained by making the substitution $\nabla \rightarrow i \vec{q}/a$. 



My question : There doesn't seem to be any terms in the transformed equations which correspond to the terms $ H \vec{X} \cdot \nabla \delta \rho$ and $H \vec{X} \cdot \nabla \delta \vec{v}$. Weinberg makes no comment about their absence. Is anyone aware of a legitimate mathematical reason for these terms to disappear in the transformed equations?
The answer turns out to be embarrassingly simple, and I suspect nobody came up with it because of my poor communication in the question. 

The $a$ is the occurring in the Fourier transform is the scale factor which has a time dependence. So if we look at the term $\frac{\partial}{\partial t} \rho$ we will get,

$$
\frac{\partial}{\partial t} \rho = \frac{\partial}{\partial t} \int  \exp \left( \frac{ i \vec{q} \cdot \vec{X}}{a(t)}\right) \rho_{\vec{q}}(t) \ \mathrm{d}^3 \vec{q} 
$$

$$
 = \int \frac{\partial}{\partial t}\left( \frac{ i \vec{q} \cdot \vec{X}}{a(t)} \right) \exp \left( \frac{ i \vec{q} \cdot \vec{X}}{a(t)}\right) \rho_{\vec{q}}(t) +  \exp \left( \frac{ i \vec{q} \cdot \vec{X}}{a(t)}\right) \frac{ \mathrm{d} \rho_{\vec{q}}}{\mathrm{d} t }  \ \mathrm{d}^3 \vec{q} 
$$

$$
 = \int \left(-\frac{\dot{a}}{a} \frac{ i \vec{q} \cdot \vec{X}}{a(t)} \right) \exp \left( \frac{ i \vec{q} \cdot \vec{X}}{a(t)}\right) \rho_{\vec{q}}(t) +  \exp \left( \frac{ i \vec{q} \cdot \vec{X}}{a(t)}\right) \frac{ \mathrm{d} \rho_{\vec{q}}}{\mathrm{d} t }  \ \mathrm{d}^3 \vec{q} 
$$

$$
 = \int \left(-H\frac{ i \vec{q} \cdot \vec{X}}{a(t)} \right) \exp \left( \frac{ i \vec{q} \cdot \vec{X}}{a(t)}\right) \rho_{\vec{q}}(t) +  \exp \left( \frac{ i \vec{q} \cdot \vec{X}}{a(t)}\right) \frac{ \mathrm{d} \rho_{\vec{q}}}{\mathrm{d} t }  \ \mathrm{d}^3 \vec{q} 
$$

The left hand term in the integrand matches the Fourier transform of $H (\vec{X} \cdot \nabla) \rho$. Which is why the terms cancel.