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First of all, the famous Einstein's elevator experiment is quite clear in my head, both of versions.

But now, consider the following:

Suppose then you wake up inside a car that is traveling in perfect straight path in a autoban (but you don't know that). The car have a constant velocity $v$ and is a self-driving car with totally dark-glass windows. You don't have any information about the outside world. After a time $t$ travelling in the straight path, the car enters in a curve. You then fells an acceleration (exactly with $9,8 m/s^2$) accelerating you.

Now, in my opinion, the person inside the car cannot say that the centrifugal acceleration is different from artificial constant gravitational field. The equivalence principle states something similar, since a person inside a elevator in a gravitational field is equivalent to a person inside a elevator which is accelerated with $9,8 m/s^2$. Furthermore, we can construct a ring-like structure to produce, via circular motion, a artificial gravitational field.

So, can I say that any accelerated frame, due to equivalence principle, is equivalent to a gravitational field?

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  • $\begingroup$ the car enters in a curve. You then fells an acceleration (exactly with 9,8m/s2) accelerating you. What makes you think that? $\endgroup$
    – Gert
    Mar 17 '21 at 23:15
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    $\begingroup$ I supposed that we can construct a situation that the centrifugal acceleration have this very value. $\endgroup$ Mar 17 '21 at 23:16
  • $\begingroup$ Only locally, as soon as you look through the windows of your elevator, that is taking the global picture, you will see that it isn't gravity you're experiencing but acceleration ... $\endgroup$ Mar 18 '21 at 0:58
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An accelerated frame is only locally equivalent to a gravitational field. Globally, you will not be able to "fake" the gravitation of a planet by just accelerating. Only if the passenger ignores tidal forces will he be unable to distinguish an accelerated frame from a static gravitional field.

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If the acceleration of the frame is due to rotation, there is, in addition to the centrifugal force which locally looks like gravity, also Coriolis force, which acts perpendicular to the motion of the test particle, and makes it possible to distinguish the frame from one which doesn’t rotate.

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Locally, yes. The geodesic equation of GR and differential geometry contains $\Gamma$. When you move to rotating frame these $\Gamma$s are nonzero. The geodesic equation doesn't care if the terms in $\Gamma$ come from a real gravitational field or a non-inertial reference frame.

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