# Curvature of spacetime and the equivalence principle

Assuming the Einstein equivalence principle, formulated as following:

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

If now the curvature of spacetime is an invariant, shouldn't this violate the EEP (Einstein equivalence principle)?

After all, a ball dropping to the floor due to Earth's gravity and the same ball dropping to the floor in an accelerating spaceship is exactly the same situation – except that one involves curvature of space and the other does not.

If the curvature of space is invariant, one could measure the necessarily quantities, calculate the curvature and conclude that the two situations is in fact not equal. This would certainly contradict the EEP.

So, for the EEP to hold up, my reasoning must obviously be wrong. Can someone clarify for me?

• The catch in that phrase is the word "local". There exist some effects of gravity which "violate" this principle - such as tidal forces - but all the examples I ever saw are non local. To measure the curvature you need to measure the second derivatives of the metric. Assuming you're not near a black hole - it's not something any local experiment can achieve due to resolution required from the measuring apparatus. – Alexander Jul 17 '15 at 23:15
• Comment to question (v2): OP's quote from Wikipedia is truncated. The very next sentence on Wikipedia reads: Here "local" has a very special meaning: not only must the experiment not look outside the laboratory, but it must also be small compared to variations in the gravitational field, tidal forces, so that the entire laboratory is freely falling. It also implies the absence of interactions with "external" fields other than the gravitational field. – Qmechanic Jul 17 '15 at 23:20
• @Qmechanic However, I chose the EEP for this reason instead of the SEP variant. This does not answer my question though. Can't you measure any curvature of spacetime locally? – Madde Anerson Jul 17 '15 at 23:32
• @MaddeAnerson: yes, you can measure the curvature of spacetime locally. But this doesn't violate the very narrow definition of the Einstein Equivalence principle, which at root ssays that youc an always choose a laboratory small enough and an experimental timeframe short enough that the effects go away. – Jerry Schirmer Jul 17 '15 at 23:43
• It's worth mentioning that if you're in a vacuum (or in a low density environment such as air), the Ricci scalar is zero (or nearly zero). Of course, there are other scalars that may not be zero. – Javier Jul 18 '15 at 1:12

You mention a couple of times the curvature being an invariant. I wonder if you're confusing this with being dimensionless. I guess that you're talking about the Ricci scalar, which is a relativistic invariant, but it is not dimensionless; IIRC it has units of inverse distance squared. On distance scales much smaller than scale implied by the curvature, the curvature will be negligible and will not be detectable. For instance, a curvature of 1/(1000 km) will be easy to see on scales of 1000 km, but hard to see on scales of 1 m. That's why you don't see the curvature of the Earth's surface when you stand on the ground and look at things a few meters away.

When the equivalence principle talks about "local" measurements, it means measurements taking place over a small-enough region of spacetime that the curvature is negligible. You can always restrict yourself to a small enough region (formally, an infinitesimal region) so that spacetime looks flat, no matter how strong the curvature or how sensitive your instruments. So, you can't discover the curvature by doing "local" experiments in the sense of the principle.

The content of the equivalence principle is then that freely falling in a gravitational field is (locally) indistinguishable from freely floating in space. In other words, the gravitational field can be made to vanish within an infinitesimal region by finding the appropriate reference frame - a freely falling one. There is no "leftover" part of the gravitational field that cannot be removed by freely falling.

Note that this is not generally the case for other fields - for instance, electromagnetic fields cannot be made to vanish by choosing a particular reference frame. A pure electric field in one frame may appear as a mixture of electric and magnetic fields in another frame, but if there is a nonzero field at a point, there will not be any frame in which both electric and magnetic fields vanish entirely (even locally). So, gravitation is quite special in this regard.

• I can still measure the curvature of the Earth, even locally, and even if I can't do that practically it's still possible with instruments sensitive enough. Thus, I can distinguish between the two situations and the EEP fails. – Madde Anerson Jul 18 '15 at 1:09
• This, ignoring tidal effects or anything like that. The fact that the curvature is an invariant – ALONE – makes the claim "to fall freely and to accelerate with one standard gravitational acceleration is the same thing," false. One involves curvature of a manifold and the other does not. – Madde Anerson Jul 18 '15 at 1:14
• @MaddeAnerson However sensitive your instruments are, you can always restrict yourself to a small enough region that you cannot see the curvature - formally, an infinitesimal region - and in that small region you can't tell the difference. That's the meaning of "local" in the equivalence principle. – Nathan Reed Jul 18 '15 at 4:52
• I see. You essentially restrict yourself to an infinitesimal area and there cannot be any curvature in a such region. I feel though that this makes the EEP much less revolutionary. What Einstein basically said was that "any curvature cannot be noticed in an infinitesimal area." Not to be rue, but "so what?" – Madde Anerson Jul 18 '15 at 11:58
• @MaddeAnerson Yes, it's true that would not be very revolutionary. The content of the EEP is more than just "any curvature cannot be noticed in an infinitesimal area", though. I've added some more to my answer about that. – Nathan Reed Jul 18 '15 at 22:40