How to find a curvature of the space-time by having $g^{\alpha \beta}$ in the following case without cumbersome calculations? The metric tensor for Fock-Lorentz space-time,
$$
\mathbf r_{||}{'} = \frac{\gamma (u)(\mathbf r_{||} - \mathbf u t)}{\lambda \gamma (u) (\mathbf u \cdot \mathbf r) + \lambda c^{2} (1 - \gamma (u))t + 1},
$$
$$\mathbf r_{\perp}{'} = \frac{\mathbf r_{\perp}}{\lambda \gamma (u) (\mathbf u \cdot \mathbf r) + \lambda c^{2} (1 - \gamma (u))t + 1},
$$
$$
t' = \frac{\gamma (u)(t - \frac{(\mathbf u \cdot \mathbf r)}{c^{2}})}{\lambda \gamma (u) (\mathbf u \cdot \mathbf r ) + \lambda c^{2} (1 - \gamma (u))t + 1},
$$
is given by
$$
g^{\alpha \beta} = \begin{bmatrix} \frac{1}{(1 + c^{2}\lambda t)^{4}} & \frac{c\lambda x}{(1 + c^{2}\lambda t)^{3}} & \frac{c\lambda y}{(1 + c^{2}\lambda t)^{3}} & \frac{c\lambda z}{(1 + c^{2}\lambda t)^{3}} \\ \frac{c\lambda x}{(1 + c^{2}\lambda t)^{3}} & -\frac{1}{(1 + c^{2}\lambda t)^{2}} & 0 & 0 \\ \frac{c\lambda y}{(1 + c^{2}\lambda t)^{3}} & 0 & -\frac{1}{(1 + c^{2}\lambda t)^{2}} & 0 \\ \frac{c\lambda z}{(1 + c^{2}\lambda t)^{3}} & 0 & 0 & -\frac{1}{(1 + c^{2}\lambda t)^{2}} \end{bmatrix},
$$
where $c, \lambda $ are constant.
Is there a slick way to find the Ricci scalar without cumbersome calculations with Christoffel symbols (if at all possible)?
 A: Cleaning up the notation a bit by rescaling coordinates to get rid of $c$ and $\lambda$ and pulling out a common factor gives:
$$g^{\alpha \beta} = \frac{1}{(1 + t)^{2}}
\begin{bmatrix} \frac{1}{(1 + t)^{2}} & \frac{x}{1 + t} & \frac{y}{1 + t} & \frac{z}{1 + t} \\
\frac{x}{1 + t} & -1 & 0 & 0 \\
\frac{y}{1 + t} & 0 & -1 & 0 \\
\frac{z}{1 + t} & 0 & 0 & -1
\end{bmatrix}.
$$
You can use the properties of the Ricci scalar under conformal transformations (google them) to forget about the overall factor by performing a conformal (Weyl) rescaling. You can change time coordinates $t\to\tau$ by integrating
$$ \mathrm{d}\tau = (1+t) \mathrm{d}t,\ \partial_\tau = \frac{1}{1+t} \partial_t.$$
This removes $t$ completely from the conformally rescaled metric:
$$\tilde{g}^{\alpha \beta} = 
\begin{bmatrix} 1 & x & y & z \\
x & -1 & 0 & 0 \\
y & 0 & -1 & 0 \\
z & 0 & 0 & -1
\end{bmatrix}.
$$
Then going to spherical coordinates (scroll to the bottom of the page for the relevant formulae) simplifies the off diagonal part (check this, I haven't been careful!):
$$\tilde{g}^{\alpha \beta} = 
\begin{bmatrix} 1 & r & 0 & 0 \\
r & -1 & 0 & 0 \\
0 & 0 & -\frac{1}{r^2} & 0 \\
0 & 0 & 0 & -\frac{1}{r^2 \sin^2 \theta}
\end{bmatrix}.
$$
You can do some more coordinate transformation mixing $r$ and $\tau$ to diagonalise the metric if you want but I'm getting tired of this. It is straightforward now to compute the Ricci scalar for this metric (and significantly simpler than the original form of the metric). You can probably look up formulae for the curvature tensors of metrics in this form.
A: That metric is diagonal in the spatial coordinates, and its constant-time sections are conformally flat.  It is easy enough to derive what the curvature of a conformally flat metric is in terms of the conformal factor.
Once you have this, plug this solution into the ADM equations.  You can find references that let you reconstruct the Einstein equations from the ADM equations, which obviously then gives you the Ricci scalar.  In particular, $\rho + \gamma^{ab}S_{ab}$, where $\rho$ is the matter density, $\gamma_{ab}$ is your 3-metric, and $S_{ab}$ is the 3-pressure defined by $\gamma_{ac}\gamma_{bd}T^{cd}$ will give you a quantity proportional to the Ricci scalar.
