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The principle of equivalence states that it is possible to choose a locally "inertial" coordinate at every space-time point, in presence of an arbitrary gravitational field, where the explicit effects of gravity disappear.

If we take the simple example of a body ($m$) falling freely under gravity ($mg$)- let $x^{\mu}$ be the lab frame, which is of course inertia, and let $\xi^{\mu}$ be the frame locally chosen using priciple of equivalence. The z-components of the two frames can be connected as $\xi_z= x_z - gt^2$. Using this transformation, the effects of gravitations ( the $mg$ term) disappears from the equation of motion.

Clearly, the frame $\xi^{\mu}$ is accelerated in comparison to $x^{\mu}$. My question is, how can the frame $\xi^{\mu}$ be inertial when it is in acceleration wrt another inertial frame $x^{\mu}$?

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In the context of General Relativity, the "lab frame" is not inertial, because it is not freely falling. This is the point of Einstein's "elevator" thought experiments: a freely falling frame in a gravitational field is equivalent to an inertial frame in deep space, and a stationary frame in a gravitational field is equivalent to an accelerated frame in deep space. Only the first category of frames is seen as "inertial" in the context of GR.

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  • $\begingroup$ This makes sense and it clears up some of my doubts. While reading the calculations in GTR , I found that the proper time is taken to be equal in both stationery and freely falling frame. But I thought proper time remains same only while transforming from one inertial frame to another. So why is the proper time remaining same here while going from freely falling frame to stationery frame? $\endgroup$
    – D.K.
    Commented Apr 23, 2022 at 17:29
  • $\begingroup$ @D.K.: I don't know what calculation you're referring to, to be honest; it might be worth asking that as a separate question. But in general the proper time between two fixed events A and B will not be the same for two observers if they follow different spacetime trajectories. $\endgroup$ Commented Apr 23, 2022 at 19:31
  • $\begingroup$ (including the case where one observer is stationary and the other is freely falling between the two events; for example, consider a clock sitting on Earth's suface and another one that is tossed straight up into the air) $\endgroup$ Commented Apr 23, 2022 at 19:53

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