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By equivalence principle, one can find a local inertial frame at every point of spacetime. Then this is usually used to introduce the general spacetime metric, as a back-transform of the Minkowski metric.

The transformation from Minkowski to general metric depends only on the first derivatives of the coordinate transformation. This means that locally we can get an inertial frame by applying a linear coordinate transform. It is puzzling because in motivating examples we always get accelerating transforms, such as a falling elevator. However, it seems that the acceleration does not play role in the final formula. What is happening?

Is this because linear transform is enough to diagonalize the metric, but the quadratic part (i.e., the acceleration) is needed to make the Christoffel symbols vanish?

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  • $\begingroup$ no upvotes? i will delete my question if no upvote in 10 min $\endgroup$
    – timur
    Commented Jun 19, 2023 at 12:36

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Yes you were on the right track. If you want to recover the Minkowski metric at an event, then yes a simple linear coordinate change is enough starting from a general metric.

However, when you talk about an inertial frame and the equivalence principle, you need to do more than that. You need to kill the first spatial derivatives of the metric, or equivalently the Christoffel symbols. Physically, this would mean that you don't have inertial forces precisely at this event. In this case, a simple linear transformation won't do. Mathematically, you usually introduce normal coordinates and the exponential map.

Btw, this is the best you can do, since second order derivatives can be built to construct gauge invariant quantities like the curvature.

Hope this helps.

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  • $\begingroup$ thank you, you are the best! life is brighter because of people like you $\endgroup$
    – timur
    Commented Jun 19, 2023 at 22:25

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