General Relativity – Definition and Importance of Locally Inertial Frames

Question

The Einstein Equivalence Principle states that in a sufficiently small frame of reference is impossible to know if we are into a gravitational field or not. Equivalently we cannot say if we are in an accelerating frame or not. This is because gravity and inertia are equivalent (hence the name of the principle).

So because in a local frame we indeed can't establish if we are accelerating or not, makes more sense to define a local inertial frame of reference (synonym of not accelerating frame) as a "free falling frame". In fact into a local free falling frame things behave as if in a perfectly inertial special relativity frame. ($\mathbb{M}^4$)

Perfect, but seems to me that this should imply that a local observer standing on the earth (so not free falling at all) should be considered as an accelerating, non inertial frame.

Ok, this seems also fine. But we know that there is another, more geometrical, equivalent formulation of EEP:

Locally spacetime looks like $\mathbb{M}^4$

This is not the precise formulation of the geometrical formulation, but it's good enough. This means that in every sufficiently small region of spacetime it's like being into a inertial special relativity frame, so no accelerating, no gravity, no shenanigans.

But: earlier we said that me, writing this question on the surface of the earth, should be considered as an accelerating frame! But the geometrical formulation states that every sufficiently small reference frame, myself included, should be like an inertial SR frame!

So, in the context of GR, am I accelerating? Or on the contrary am I into a local inertial SR frame? And most important of all: why this two formulation of EEP seems to contradict each other?

Answer 1

seems to me that this should imply that a local observer standing on the earth (so not free falling at all) should be considered as an accelerating, non inertial frame.

Yes, an observer standing on the earth is not inertial in relativity. The definitive test is to have the observer carry a good accelerometer. In this case it will indicate an acceleration of 1 g upwards, conclusively showing that the observer is non-inertial.

Just a nitpick on language: an observer isn’t a reference frame, he or she has a reference frame, or even better there is a reference frame where he or she is at rest.

there is another, more geometrical, equivalent formulation of EEP: Locally spacetime looks like 𝕄4 This is not the precise formulation of the geometrical formulation, but it's good enough.

Agreed, it is good enough for present purposes.

This means that in every sufficiently small region of spacetime it's like being into a inertial special relativity frame, so no accelerating, no gravity, no shenanigans.

It does not mean that at all. You can certainly have accelerating reference frames with pseudo-gravitational forces in 𝕄4. All 𝕄4 means is that you cannot have any tidal effects.

𝕄4 is a flat spacetime manifold and can be equipped with an endless number of coordinate systems, including non-inertial ones. What “locally spacetime looks like 𝕄4” means is that there exist local coordinates where the metric is the Minkowski metric (to first order), but it does not restrict you to using those coordinate systems.

More physically it means that tidal effects become negligible at small scales. The measurable effects from curvature, or tidal effects, are second order so they go away to first order at small enough scales.

But the geometrical formulation states that every sufficiently small reference frame, myself included, should be like an inertial SR frame!

No, the observer is unambiguously non-inertial. The geometrical formulation does not contradict that at all. The geometrical formulation merely says that in a small region spacetime is flat, not that an observer is inertial. It is perfectly consistent to have non-inertial observers and reference frames in flat spacetime. Only tidal effects are forbidden.