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 isnβt 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 π4
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 π4. All π4 means is that you cannot have any tidal effects.

π4 is a flat spacetime manifold and can be equipped with an endless number of coordinate systems, including non-inertial ones. What βlocally spacetime looks like π4β 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.

## Best Answer

So, in general, a coordinate transformation involves giving each coordinate $x$ as a function of the new coordinates $x^{\prime}$:

$$x^{\prime\mu} = f^{\mu}(x^{\nu})$$

This gives a general relationship:

$$ds^{2} = g_{\mu\nu}dx^{\mu}dx^{\nu} = g_{\mu \nu}\left(\frac{dx^{\mu}}{dx^{\prime\alpha}}dx^{\prime \alpha}\right)\left(\frac{dx^{\nu}}{dx^{\prime\beta}}dx^{\prime \beta}\right) = g_{\alpha \beta}^{\prime}dx^{\prime \alpha} dx^{\prime \beta}$$

Where we can transform the old inner product defined by $g$ into a new inner product defined by $g^{\prime}$. So, if $ds^{2}=0$ in our old reference frame, then it is necessarily zero in our new reference frame. You can argue that the formalism is cooked to make this true, and you'd be right, but that's what the underlying mathematics of general and special relativity describe.