Is the equivalence principle strictly fulfilled by general relativity?

Question

The equivalence principle states

The outcome of any local experiment in a freely falling laboratory is independent of the velocity of the laboratory and its location in spacetime.

Any real local experiment needs some finite space, since nobody is a point particle himself. In principle I will always feel some deviation.

Since in (pseudo-)Riemannian geometry I can only change to flat coordnates in a single point $p$ of the spacetime manifold (and not in a finite neighborhood), is the deduction right, that the equivalence princpil is only fulfilled by general relativity in an approximative sense?

Answer 1

Yes. If your local experiment finds a violation of the equivalence principle, you would have to shrink the size of your experiment. Alternatively you could find an equivalent place to perform the experiment where the curvature is lower.

For example, consider the case where you want to do an experiment at an event horizon. Normally we think that the gravitational "force" at the event horizon of a black hole is extremely large and thus there is a very large curvature that will limit the size of the experiment. But if you find a black hole with a much larger mass then the gravitational force at the event horizon can be decreased as much as you want. For example with a sufficiently massive black hole, the gravitational force at the event horizon could be less than the force on the surface of the earth. (This does depend on a suitable definition of the meaning of force near a black hole: See this and this.)

Answer 2

General Relativity obeys the equivalence principle exactly.

The equivalence principle as stated by Einstein is a little bit different from your statement in terms of coordinates. Any observer in a non-inertial reference frame can account for the local effects by postulating a distribution of masses whose gravitational effects are the cause. I.e., the effects of an accelerated reference frame can always be accounted for by some imaginary, postulated distribution of masses. But it does not state the converse, it is not true that the observable effects of a gravitational field produced by masses can be accounted for by an accelerated frame of reference which produces those effects without any masses.

In terms of mathematics, this means that any local coordinate system is allowed, and the user will attribute the observed motion of objects along geodesics, which in that coordinate system do not appear to be straight lines, to the influence of gravity, i.e., the metric tensor. The failure of the converse to be true means that you cannot find a local coordinate system in which the metric is flat unless there actually is no matter in the Universe.

The fact that Special Relativity is an approximation to General Relativity means that the converse is almost true if the neighbourhood of the point is sufficiently small: you can find coordinates in which the Christoffel symbols vanish at that point and are negligible in your neighbourhood of the point, and hence your coordinate system is approximately inertial.

Consider an elevator which is not in free fall exactly but is in a state of uniform acceleration. (To make things perfectly clear assume there is no matter in this universe...) An observer inside the elevator can exactly say that their system of coordinates is at rest, and is inertial in the sense of Special Relativity, but the observed departures from inertial motion are due to the presence of an infinite wall of matter which is producing a gravitational field. It has to be an infinite wall, not a point source, or the observed effects will not be exactly reproduced.

The Earth is an actual mass that produces a gravitational field. An observer in a freely falling elevator which is sufficiently far away from the centre of the Earth will not quite be able to establish an inertial frame for the inside of their elevator in such a was as to claim that there are no masses. I.e., you cannot completely remove the effect of masses by passing to a freely falling reference frame.

General Relativity fulfils the equivalence principle exactly. No observer can tell whether their local coordinate system is accelerated or not: all coordinate systems obey the same laws of gravity as any other coordinate system.

Answer 3

Since General Relativity is a classical theory (which does not describe the quantum world and Planck constant $\hbar$ is zero), it seems senseless to worry about the actual spatial size of elementary particles. Hence, let us in the following grant ourselves the right to discuss probes of zero spatial extend like mathematical points in space.

Also we shall here assume that spacetime in principle could be any pseudo-Riemannian manifold $(M,g)$, and not wonder how it got its curvature.

OP then raises the question about the existence of a coordinate system in a turbular neighborhood $U$ of the worldline $\gamma$ of a point probe such that the metric $g_{\mu\nu}$ becomes on Minkowski-form with vanishing Christoffel symbols $\Gamma^{\lambda}_{\mu\nu}$ on the worldline $\gamma$ (but not necessarily in the rest of the tubular neighborhood $U\backslash\gamma$).

This a valid concern since Riemann normal coordinates only guarantee that the metric $g_{\mu\nu}$ becomes on Minkowski-form with vanishing Christoffel symbols $\Gamma^{\lambda}_{\mu\nu}$ in a single spacetime event $p\in\gamma$.

However, since the worldline $\gamma$ is a geodesic, there actually exists Fermi normal coordinates that guarantee that the metric $g_{\mu\nu}$ becomes on Minkowski-form with vanishing Christoffel symbols $\Gamma^{\lambda}_{\mu\nu}$ along $\gamma$, such that we can claim that the probe is truly experiencing a free fall.