Suppose you are in radial free fall at some point outside the event horizon of a Schwarzchild metric. The strong equivalence principle implies that locally you would be unable to discern whether you are in fact in free fall near a gravitating body or simply at rest in flat spacetime. The operative word being "locally." If you had, say, two bowling balls freely falling with you nearby, then (due to differences in the gravitational field) you would observe them gradual approach one another - an effect called geodesic deviation.

This phenomenon is usually cited as an example of the inexact nature of the equivalence between gravity and acceleration or of the local nature of the equivalence principle, but it seems to me that (at least under certain circumstances) this equivalence can be made truly exact by appealing to other explanations for the observations. For example, in the situation described above, the two bowling balls would move together even in nominally flat spacetime due to their mutual gravitational attraction. Granted, they may come together more rapidly in the freely falling situation due to the addition of tidal forces, but without any other data, one might simply conclude that the bowling balls have more mass than they actually do - just enough to account for the extra attraction. Or, alternatively, one might do some experiment to determine the inertial mass of the bowling balls and then use their mutual attraction in the freely falling frame to determine the gravitational constant.

Even if these mental maneuvers work as explanations, it is fairly easy to come up with examples that would seem to undermine an attempt to explain away tidal effects completely. For example, suppose you are in an unstable circular orbit with one bowling ball at a slightly smaller radius and the other at a slightly larger radius. Over time, the three of you would diverge - one spiraling inward, one gradually going out to infinity, and you staying in your circular orbit. Without some profound changes in one's theory of gravity, this observation seems impossible to square with flat spacetime.

On the other hand, both of these examples would require very careful preparation and very delicate measurements if they were to actually be performed, so it doesn't seem entirely implausible that the subtlety of the effects being measured would leave room for apparently minor details that I'm ignoring to change the outcome of the experiments - either leaving the door open to plausible explanations consistent with flat spacetime or providing an additional piece of information to distinguish between them.

Are there any situations in which the equivalence principle can be strengthened such that a freely falling observer under these specific conditions is entirely unable to determine whether or not they are in the presence of a gravitating body? If so, what would the necessary conditions be?


The equivalence principle has a precise mathematical statement, and when written down this way its meaning is immediately obvious to general relativists, but of course not to everyone else. Hence the various forms of it you see written down in somewhat vague terms. You simply cannot convert the mathematical statement to natural language without losing the precision.

It is certainly true that people criticise the equivalence principle on the grounds of its vagueness, and indeed we have had such statements expressed here. These comments betray a lack of understanding of the real meaning, but it's hard to know how to tackle this without directing the complainant to a book on differential geometry.

The mathematical statement is that the four acceleration is the sum of two terms:

$$ A^\alpha = \frac{\mathrm d^2x^\alpha}{\mathrm d\tau^2} + \Gamma^\alpha{}_{\mu\nu}U^\mu U^\nu $$

On the right hand side the first term is a coordinate acceleration, what we think of as acceleration in Newtonian mechanics, while the second term is a gravitational acceleration that arises due to the curvature of spacetime.

The reason that the equation embodies the equivalence principle is that by simply changing our coordinates we can make the first term zero or the second term zero. A coordinate change is not a physical change and the magnitude of the four-acceleration is unaffected by a coordinate change, so by a choice of coordinates the same four-acceleration can be made to appear entirely coordinate acceleration, entirely gravitational acceleration, or indeed a combination of the two. This means it is a fundamental principle in GR that coordinate and gravitational accelerations are indistinguishable since they can be interchanged simply by an appropriate choice of coordinates.

The local nature of the equivalence is because at any point in spacetime I can choose coordinates that make the gravitational acceleration zero, these are called the Fermi normal coordinates, but this equivalence applies only at the point I have chosen. If I move away from this point in either a spatial or time direction then the second term is no longer zero - these are the tidal accelerations. The fact the tidal accelerations exist does not disprove the equivalence principle, because that isn't what the principle is concerned with.

  • $\begingroup$ I'm a little confused. Your equation obviously reduces to the geodesic equation for the Levi-Civita connection when the 4-acceleration is zero. And my understanding (which may be wrong) is that the 4-acceleration is zero for any body interacting exclusively by way of gravity. Therefore, in all the examples I give, the particles should be undergoing geodesic motion since no interaction aside from gravity is present. The question is: Can one describe the geodesic motion of the balls as entirely due to the local mass distribution (possibly with a small modification to e.g. Einstein's Equation)? $\endgroup$ – Geoffrey Jun 17 '20 at 17:08
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    $\begingroup$ @Geoffrey any freely falling body has a zero four-acceleration, and yes if you set $\mathbf A = 0$ you get the geodesic equation. You cannot describe the geodesic motion of the balls as entirely due to the local mass distribution as the Christoffel symbols depend on all the mass present not just the local mass present. $\endgroup$ – John Rennie Jun 17 '20 at 17:28

The equivalence principle as it is normally stated and understood (if it is stated correctly that is!) is indeed a statement about a limit, as you correctly say. I mean the limit of small regions of spacetime. Most people thinking about this would probably judge that this is already the best way to see this equivalence, because then it connects directly to the mathematical structure of GR. In particular, it connects straightforwardly to the notion that a Riemannian manifold is locally flat, and it allows that the curvature tensor is indeed a tensor---and therefore it can't be made to go to zero in one frame if it is not zero in another. What this means is that if the curvature is non-zero then it is there, ready and waiting to be measured, as one might put it. In particular, it has a value at each event and causes physical effects such as geodesic deviation.

Your suggestion amounts to saying, could we perhaps attribute the tidal effect of gravity (and hence the curvature) to some other phenomenon that did not involve spacetime curvature. One answer is: yes of course. I mean one does not have to adopt the geometric interpretation of GR. One can always just announce that spacetime is flat, with a Minkowski metric, and attribute gravitational phenomena to forces acting in this flat spacetime. Such an approach indeed works quite well when the gravitational effects are weak. Large numbers of useful calculations are done in pretty much this perspective. (Because one can interpret linearized GR as adopting this approach. One does not have to interpret it that way, but one may.) However sooner or later the decision to avoid a geometric interpretation and the tools of differential geometry will look more and more artificial, like doing GR with one hand tied behind your back.

But I see your motivation is to try to make the equivalence principle more precise, or extend it somewhat. There is a problem with doing this along the lines you suggest I think. If you start out by saying you will simply think of some other reason for all the effects you are trying to avoid attributing to gravitation, and you are prepared to introduce new forces and interactions, then I suppose you will always succeed. But the result will not be explanatory.

Now to answer your final bold font paragraph. I think the answer is no. This is because if there is a gravitating body then $T^{ab} \ne 0$ somewhere either at or near to your observer. Hence $R^{ab} \ne 0$ somewhere at or near to your observer. But then the field equation implies $R^a_{\;bcd} \ne 0$ right at your observer (unless something extraordinary happens, see below). So they are located at a spot where the curvature is not zero. They can tell this by measuring the curvature.

But might it happen that something extraordinary happens, namely that all the components of $R^a_{\;bcd}$ just happen to go to zero at some point, even though they are not zero at nearby points? I don't know of such a case apart from the flat spacetime inside a spherical cavity centrally located inside a spherically symmetric body (and no other bodies). But I think this is not the sort of situation you are looking for. A case where one was outside the gravitating body and yet $R^a_{\;bcd}= 0$ would, I think, be such an artificial and limited case that it too would not be the sort of thing you are looking for. But others may know more about this.

Finally, a gravitational situation that mimics acceleration quite well is the situation of a long cylindrical planet. Then the local acceleration due to gravity falls as $1/r$ and this is like the effective gravity in a constantly accelerating reference frame in flat spacetime. So many effects such as time dilation and the motion of light in the plane containing the axis of the cylinder are very closely mimicked. But effects out of that plane are not.

  • $\begingroup$ I take your point, but I'd like to clarify some things. I am not suggesting that there is any interaction other than spacetime curvature here. The two gravitating balls are still interacting by way of spacetime curvature. The question instead is whether the motion they undergo be entirely attributable to the local mass distribution. I take your point that the curvature is measurable, but how can one measure the curvature other than with an experiment like the ones described? If there is no other way, then could such an observer really delineate between effects from local and global curvature? $\endgroup$ – Geoffrey Jun 17 '20 at 17:17

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