Dear Cristini

The main and second question are not compatible. In the second question you use

Cartesian coordinate, Do you mean Minkowskian? In Relativity ,Cartesian coordinate

is not a useful tool for accelerated observer.

sorry, wrote too quickly.  I guess my real question would be to see a formal proof of the equivalence of the metric in gravity and in accelerated frames

I think I just figured it out... I Taylor expanded the Schwarzschild metric around a point in space-time and obtained the Rindler metric!

Your calculation may be an innovation, However confirm it by a view on last paragraph of this link:

http://en.wikiversity.org/wiki/Rindler_coordinates#Comparing_to_Schwarzschild

Colleagues! Rindler metric (Moeller) flat. Schwarzschild metric has a non-zero curvature. Probably that the metric should be quite different. For example uniformly accelerated

https://www.researchgate.net/publication/287980615_Solution_of_the_Einstein_equation_and_metric_of_accelerated_system

But I think this is not quite true. Now I hope the next result. Quite unexpected

https://www.researchgate.net/publication/303844514_Gravitational_field_equations_and_relativistic_homogeneous_gravitational_field?ev=prf_pub

Preprint Solution of the Einstein equation and metric of accelerated system

Preprint Gravitational field equations and relativistic homogeneous g...