Under the condition of spherical symmetry the metric is reduced to $$\begin{aligned}
ds^2&=&\left[-N^2(t,r)+(N^r(t,r))^2V^2(t,r)\right]dt^2+2N^r(t,r)V^2(t,r)dt dr\nonumber\\
\label{mod.int}
&+&V^2(t,r)dr^2+W^2(t,r)\left(d\theta^2+\sin^2\theta d\varphi^2\right).\end{aligned}$$ where $N^r=N^1$ is the only component of the shift vector. In this model we have two gauge variables $N$ and $N^r$ which are fixed by two gauge conditions $$\begin{aligned}
\label{g.c.1}
N&=&f(V,W);\\
\label{g.c.2}
N^r&=&f^r(V,W).\end{aligned}$$ $f(V,W)$, $f^r(V,W)$ are arbitrary functions. In differential form the gauge conditions will introduce missing velocities into the effective Lagrangian, so ensuring [*an actual extension of phase space*]{}: $$\begin{aligned}
\label{d.c.1}
\dot N&=&\frac{\partial f}{\partial V}\dot V+\frac{\partial f}{\partial W}\dot W;\\
\label{d.c.2}
\dot N^r&=&\frac{\partial f^r}{\partial V}\dot V+\frac{\partial f^r}{\partial W}\dot W.\end{aligned}$$

We shall consider the Faddeev – Popov effective action including gauge and ghost sectors as it appears in the path integral approach to gauge field theories, $$\label{full-act1}
S_{(eff)}=S_{(grav)}+S_{(gauge)}+S_{(ghost)}.$$

The gravitational part of the effective action $$\label{grav-act1}
S_{(grav)}=\int d^4x\sqrt{-g}R$$ is invariant under gauge transformations (\[g\_transf\]), but the gravitational Lagrangian involves second derivatives of metric components. To get field equations it is much easier to make use of the Lagrangian which is quadratic in first derivatives of metric components and can be obtained from the original one by omitting total derivatives. However, we shall have to return to the original gravitational Lagrangian when deriving the BRST 