If the particle accelerates radially, is the metric the same as in cartesian coordinates?
Article Kinematic parameters in the spherical Rindler frame spacetime
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...
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