the other hand, equation $0-0$ takes the form $$\begin{aligned}
2\left(3{ H}^2\Phi+3{ H}\dot{\Phi}-\frac{1}{a^2}\Delta \Phi\right)=\dot{\bar\phi}(\Phi\dot{\bar\phi}-\delta\dot{\phi})-\bar V_{\phi} \delta\phi-
3H\bar{f}_{\chi\chi}\delta\chi,
$$ and, after a somewhat cumbersome calculation, one can see that it is equivalent to the following one, $$\begin{aligned}
\frac{1}{a^2}\Delta \Phi=\frac{{\dot{\bar{\phi}}}^2}{2H}\frac{d}{dt}\left[\frac{H\delta\phi}{\dot{\bar{\phi}}}+\Phi+
\frac{H}{a^2{\dot{\bar{\phi}}}^3}\bar{f}_{\chi\chi}\Delta\delta\phi  \right]-\frac{{\dot{\bar{\phi}}}^2}{2}\frac{d}{dt}\left( \frac{\bar{f}_{\chi\chi}\Delta\delta\phi}{a^2{\dot{\bar{\phi}}}^3 } \right).\end{aligned}$$

Introducing gauge-invariant variables (recall we are working in the Newtonian gauge) and using the conformal time $$\begin{aligned}
v=a\left(\delta\phi+\frac{{\bar{\phi}'}}{{\mathcal H}}\Phi +  \frac{1}{{{(\bar{\phi}')}}^2}\bar{f}_{\chi\chi}\Delta\delta\phi  \right), \quad z=a\frac{{\bar{\phi}'}}{{\mathcal H}}, \quad
u=\frac{2a\Phi}{{\bar{\phi}'}\sqrt{\Omega}}, \quad \mbox{and} \quad \theta=\frac{1}{z\sqrt{\Omega}},\end{aligned}$$ where ${\mathcal R}\equiv \frac{v}{z}=\Phi-\frac{H}{\dot{H}}(\dot{\Phi}+H\Phi)$ is the [*curvature fluctuation in co-moving coordinates*]{}, one obtains the following M-S equations $$\begin{aligned}
 \sqrt{\Omega}\Delta u=z\left(\frac{v}{z}\right)'-{{{\bar{\phi}}'}}\left( \frac{a\bar{f}_{\chi\chi}\Delta\delta\phi}{{(\bar{\phi}')}^3 } \right)' 
 ,\qquad \theta\left(\frac{u}{\theta}\right)'=\sqrt{\Omega} v,
 \end{aligned}$$ which, after inserting the second into the first one, lead to the equation for the potential (\[lqcperturbed\]) in the simple form $$\begin{aligned}
u''-\Omega\Delta u-\frac{\theta''}{\theta}u=\bar{\phi}'\sqrt{\Omega}\left( \frac{a\bar{f}_{\chi\chi}\Delta\delta\phi}{({\bar{\phi}')}^3 }  \right)',\end{aligned}$$ while the equation for the variable $v$, after taking the Laplacian from the second equation and using the first one, becomes $$\begin{aligned}
\label{v}
v''-\Omega\Delta v-\frac{z''}{z}v=\frac{1}{z}\left(z\bar{\phi}'\left( \frac{a\bar{f}_{\chi\chi}\Delta\delta\phi}{({\bar{\phi}')}^3 }  \right)'\right)'.\end{aligned}$$

Note that in our approach, the right hand side of Eq. (\[v\]) does not vanish, meaning that the variable $v$, which encodes the scalar perturbations (it depends on $\Phi$ and $\delta\phi$), is not independent and one needs another equation in order to calculate the evolution 