Finally, the equations of motion for different $l$ decouple and we can therefore pick up a single value for $l$. In summary, for this particular gauge choice, the odd-parity metric perturbations assume the simple form $$h_{\mu \nu}=  -
\begin{pmatrix}
    0 & 0 & 0 & h_0\\
    0 & 0 & 0 & h_1\\
    0 & 0 & 0 & 0\\
    h_0 & h_1 & 0 & 0\\
  \end{pmatrix}  \sqrt{\frac{(2l+1)}{4\pi}}\sin\theta P'_l(\cos \theta)\,,$$ where we simplified the notation $h_0=h_{0,(l,m)}$ and $h_1=h_{1,(l,m)}$ since $(l,m)$ are fixed.

In addition to the metric perturbations, we also need to perturb the scalar field, $\phi=\bar\phi(r)+\delta\phi(t,r,\theta,\phi)$. Depending on the context, we assume that there will be no confusion between the scalar field and the azimuthal angle. This perturbation can also be expanded into spherical harmonics $$\begin{aligned}
\delta\phi=\sum_{l,m}\Phi(t,r)Y_l^m(\theta,\phi)\,.\end{aligned}$$ Scalar field perturbation will contribute only to even-parity modes and therefore we will consider $\delta\phi=0$.

Second-order action
-------------------

To obtain a linear equation for $h_0$ and $h_1$, we expand the action (\[action1\]) to second order in the perturbations. We derive the following action: $$\begin{aligned}
S^{(2)}=\int {\rm d} t\,{\rm d} r\,{\cal L}^{(2)}\,,\label{action-init}\end{aligned}$$ where we performed an integration over the angles $(\theta,\phi)$ by using the standard relations of the Legendre Polynomials and multiple 