I have the following conceptual doubt.

These are my assumptions:

1) The geometry of spacetime is the same for all observers, regardless their motion

2) All motion is relative (both uniform and not uniform)

Now follow this reasoning:

  • If we neglect the effects of gravity (away from masses and energy), the spacetime is flat in good approximation

  • If in this flat spacetime one particle is submit to a force, it will accelerate (no more geodesic-path)

  • From the particle perspective, there is a local gravitational field (equivalence principle: acceleration <--> gravity)

  • The particle will deduce that the spacetime is locally curved.

  • But for a comoving free falling particle, the spacetime clearly appears flat!

So we have two particles, in the same spacetime local region, who disagree about the effective geometry of spacetime. They can't both be right, because this would violate assumption 1). And it can't be that one particle is "indeed moving", while the other is "indeed at rest", because this would violate assumption 2).

So... where is the wayout?

  • 1
    $\begingroup$ all motion is not relative. There is distinction between inertial frames and non inertial even in general relativity. $\endgroup$ – Umaxo Sep 10 at 10:50

Assumption 2 is false, not all motion is relative. For example, inside a closed box, with no access to anything external, it is possible to use accelerometers to establish the difference between free fall, proper acceleration, and rotation. The reading of an accelerometer is an invariant, and therefore those motions are invariants and thus not relative.

Also, step 4 is incorrect. A local gravitational field in this sense does not imply spacetime curvature. Spacetime curvature is related to tidal gravity, not gravitational acceleration (which is related to the Christoffel symbols). Since there is no tidal gravity in this scenario the spacetime would remain flat even for the particle in the “gravitational” field.

So what can we say about curvature? It is a rank-4 tensor, so like any tensor it is a geometric object which is the same in all frames. However, all of its components are relative to the given reference frame. There are also several invariants of the curvature tensor, including the Ricci scalar.

  • $\begingroup$ How to realize whether what the accelerometers "feels" is due to the rotation of the box or due to a strange, probably time variable mass distribution outside the box? $\endgroup$ – J. Manuel Sep 10 at 13:42
  • $\begingroup$ It doesn't matter. Regardless of the cause of the rotation, the rotation itself is an invariant motion which is measured by the accelerometer (proper acceleration). $\endgroup$ – Dale Sep 10 at 14:22
  • $\begingroup$ It does matter. The principle of relativity affirms that “one cannot determine one’s own state of motion from within his own lab”. In other words, to check whether one his moving or at rest, one must look outside the box. In your case the acceleration measured by the accelerometer is ambiguous. It doesn’t tell you if you are moving or at rest. You must check outside the box to realize whether you are in empty space and the acceleration is due to rotation, or stuck in earth within a strange mass-energy field. Meaning any reference frame can claim to be inertial. $\endgroup$ – J. Manuel Sep 11 at 11:37
  • $\begingroup$ You have a severe misunderstanding. The principle of relativity does not affirm that and not any reference frame can claim to be inertial. If an accelerometer at rest in a reference frame detects a non-zero proper acceleration then the reference frame is non-inertial. Note, that is a sufficient but not a necessary condition to identify a reference frame as non-inertial. Comments are not for extended discussion, I recommend posting an actual question here or looking for a discussion forum. $\endgroup$ – Dale Sep 11 at 14:03
  • $\begingroup$ An accelerometer is a gauge. "The reading of an accelerometer is an invariant" only if the reading of any gauge is an invariant. Is the reading of any gauge an invariant? $\endgroup$ – safesphere Sep 13 at 16:13

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