How does a weak lense perturb the Minkowski metric? From 'Introduction to Gravitational Lensing' by Massimo Meneghetti:

The metric of unperturbed space-time is the Minkowski metric, $$\eta_{\mu\nu}=\begin{array}{lcr}
\mbox{} & 1 & 0 & 0 & 0  \\
\mbox{} & 0 & -1 & 0 & 0 \\
\mbox{} & 0 & 0 & -1 & 0 \\
\mbox{} & 0 & 0 & 0 & -1 \end{array}$$ whose line element is $$ds^2=\eta_{\mu\nu}dx^\mu dx^\nu.$$ A weak lens perturbs this metric such that $$\eta_{\mu\nu}\rightarrow g_{\mu\nu}=\begin{array}{lcr}
\mbox{} & (1+\frac{2\Phi}{c^2}) & 0 & 0 & 0  \\
\mbox{} & 0 & -(1-\frac{2\Phi}{c^2}) & 0 & 0 \\
\mbox{} & 0 & 0 & -(1-\frac{2\Phi}{c^2}) & 0 \\
\mbox{} & 0 & 0 & 0 & -(1-\frac{2\Phi}{c^2}) \end{array}$$ for which the line element becomes $$ds^2=g_{\mu\nu}dx^\mu dx^\nu.$$

Here, $\Phi$ is the Newtonian gravitational potential. I understand what the Minkowski metric is and its line element, but I don't understand how the metric following the perturbation from the lens is determined. Could someone explain?
 A: If you plug this form of line element into the Einstein's equation and linearize with respect to $\Phi$ you will find that whole set of Einstein's equations (which is tensorial) simplifies to nonrelativistic Poisson equation. Moreover if you look at geodesic equation of this metric, you will find that (again, after some approximations) usual nonrelativistic equation of motion of particle in Newtonian potential $\Phi$. These two observations justify interpretation of this line element as low-energy limit of GR, with $\Phi$ usual Newtonian potential. This form of right element is correct one to use for this, but it is not unique - of course you can change coordinates and get different, equally valid coordinate expression. So there is no way to show that line element needs to be in this form (because it doesn't). As for the question how to find this expression: The only way I know is to make some ansatz, not much more general than the final solution itself and look at Einstein's equations.
