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The EEP says that freely falling frames are inertial. But suppose the freely falling frame is at some spacetime point. At that point you can by a suitable coordinate transformation reduce the metric to $\eta$. But after sometime the freely falling frame is at another location and you can't reduce the metric to $\eta$ at this new point using the same previous coordinate transformation. So why is a freely falling frame inertial on the whole of its trajectory when using the same coordinate system, the metric will be $\eta$ at only one point. In the same coordinate system ( and it's only logical to see the whole trajectory in one coordinate system )the metric won't be $\eta$ at every point of the trajectory.

Secondly consider a frame attached on the Earth ( neglect rotation). See a particle at rest. It remains at rest ( or uniform motion). So why is the frame attached on the earth non inertial. What is accelerating here.. Is the motion in time accelerated...

I am confused with the link of Geodesics with the above facts. Books say, the trajectory of inertial frame is a geodesic. But with respect to what frame is that trajectory a geodesic..Shouldn't we ask that with respect to what is the trajectory of freely falling frames geodesic. And shouldn't we be bothered with what the trajectory of some particles is with respect to those freely falling inertial frames rather than the trajectory of those inertial frames itself. After all we should ve concerned with the trajectory of the particles in those inertial frames not the trajectory of the inertial frames itself ( that too without telling that with respect to what is the trajectory of inertial frames a geodesic).

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In Newtonian mechanics, a frame of reference is always global, global Cartesian coordinate systems always exist, and there is a one-to-one association between frames and coordinate systems. None of these things hold true in general relativity.

If there exists some neighborhood of a point in spacetime on which coordinates exist such that the metric is exactly $\eta$, then spacetime is flat in that whole neighborhood. The equivalence principle doesn't say that spacetime is flat.

Secondly consider a frame attached on the Earth ( neglect rotation). See a particle at rest. It remains at rest ( or uniform motion). So why is the frame attached on the earth non inertial.

I have a bottle of beer sitting on my desk right now. The desk is applying an upward normal force to it, and this is the only force that acts on it. (Gravity is not a force.) In a frame of reference fixed to the desk, the bottle's acceleration is zero. This is clearly not an inertial frame -- in an inertial frame, an object with a nonzero net force acting on it should accelerate. The experimental results are the same results you'd see aboard a spaceship accelerating at 1 g, which is clearly not an inertial frame.

I am confused with the link of Geodesics with the above facts. Books say, the trajectory of inertial frame is a geodesic. But with respect to what frame is that trajectory a geodesic

A curve is either a geodesic or not. That's a primary notion. Frames are a secondary notion. If a test mass has zero force acting on it, then its world-line is a geodesic. We could then ask whether the test mass appears to accelerate in a certain frame. If so, then that frame is not inertial.

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  • $\begingroup$ But wrt what is the trajectory of the freely falling frame inertial. Also, do we define those frames whose trajectory is a geodesic as inertial or vice versa. Also what will be the trajectory of a free particle with respect to a inertial frame. $\endgroup$
    – Shashaank
    Nov 25, 2020 at 8:38
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For the first part, a free falling non rotating frame is always locally a Minkowski spacetime.

At any small interval, all free objects inside or nearby the ISS moves in straight lines at constant speed.

But for longer intervals they start to deviate more and more from that approximation. That discrepancy is a indication that the spacetime is not flat, except for that locally approximation.

For the second part, any free falling frame follows a geodesic in spacetime. If its Minkowski spacetime is only locally valid, instead of the velocities of the celestial bodies around be constant, their covariant velocities (calculated for the valid metric) are constant.

About object at rest on the surface of the earth, they are of couse not following a geodesics, but the meaning of its acceleration is that the covariant derivative of its velocity for the Schwarzschild metric is not zero. Instead, it gives $g$ as result.

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