# Are there local coordinates for Schwarzchild spacetime?

In the Schwarzchild metric, $$t$$ is the time on the clock of an observer at infinity, and $$r$$ is the related to the area of a sphere by $$A=4\pi r^2$$. Are there more physical coordinates one could use, where $$t$$ is the time on the clock of an observer at finite $$r$$, and $$r$$ refers to distances this observer could actually measure with a meter stick? I know the metric will depend on the motion of the observer, but perhaps there are some "preferred frames" that are more relevant than others?

• The metric for a uniformly accelerated observer is the Rindler metric, and this has already been discussed to death on this site, not least by me. For time dependent acceleration the metric will be time dependent but will look locally like the Rindler metric. Also see this question. Can you clarify exactly what it is you are asking that isn't covered by previous questions? – John Rennie Dec 2 '18 at 9:36
• Now you've updated the question isn't it just asking for the transformation between the accelerated and non-accelerated coordinates? If so that's just the relativistic rocket equations which have also been discussed to death on this site. – John Rennie Dec 2 '18 at 11:06
• It now looks as if you're asking about the local geometry for shell observers – John Rennie Dec 2 '18 at 11:27
• Actually I don't get it - in the shell article you just sent me, the metric is $-\sigma_t^2 + \sigma_r^2 + ...$. Doesn't that mean that the metric is flat? I know it's locally flat, but here it looks totally flat. – Eric David Kramer Dec 2 '18 at 14:08
• @EricDavidKramer The quantities $\sigma_\mu$ are one-forms, but they are not exact one-forms. In other words, they don't necessarily satisfy $d\sigma_\mu=0$, like they would if $\sigma_\mu=dx_\mu$. So a metric of the form $-\sigma_t^2+\sigma_x^2+\sigma_y^2+\sigma_z^2$ is not necessarily flat. Case in point: the Schwarzschild metric in the usual spherical coords can be written in this form, just by replacing each term of the form $A\,dx_\mu^2$ with $\sigma_\mu^2$ using $\sigma_\mu=\sqrt{A}\,dx_\mu$. – Chiral Anomaly Dec 2 '18 at 16:08