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?

  • $\begingroup$ Well you're not really accelerating right now with your feet touching the ground. If there wasn't a big rock under us preventing us from moving, we would be free-falling and then we would not be able to tell whether we are in an accelerating frame or not. So unless you were writing this post on a laptop, free falling in the world's largest vaccum chamber, you are not accelerating right now. :) $\endgroup$ Jul 8, 2020 at 20:53
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    $\begingroup$ There are two problems: the first one is that you are not suppose to answer a question in the comments. For me is fine but eventually a moderator will remove your comment. And the second problem is that what you are saying is in contrast with the definition of inertial frame in GR. You are answering based on the classical definition of non inertial frame. $\endgroup$
    – Noumeno
    Jul 8, 2020 at 21:09
  • $\begingroup$ Point taken on the first problem. But i don't think i'm wrong by saying that the answer to your first question is no, you are not accelerating. But on the larger interpretation of the equivalencd principal, i agree that i am not really answering the question. $\endgroup$ Jul 8, 2020 at 21:27

4 Answers 4


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.

  • $\begingroup$ Just a small clarification. Referring to where you answer the third highlighted text- is the definition of an inertial frame $g_{\mu\nu}=\eta_{\mu\nu}$ and $\partial_\alpha(g_{\mu\nu}=0$ point. I don’t think so this is correct because we can set up polar coordinates in an inertial frame and have non vanishing Christoffel symbol. Is that right. 2)you say “ means is that all of the curvature tensors on spacetime are 0 over the local region”- but curvature is given by 2nd derivatives of the metric or derivatives of Christoffel symbols, both of which aren’t zero at the pointCould you pleaseclarify $\endgroup$
    – Shashaank
    Jun 21, 2021 at 19:03
  • $\begingroup$ @Shashaank with your point 2 you are absolutely correct. What I stated is wrong. I will revise to fix that. For your first point, there is actually a lot of disagreement in the literature on what defines a reference frame. Some people associate the reference frame with the coordinate system, in which case polar coordinates would not be inertial. Others associate the reference frame with a tetrad, in which case the coordinates don’t matter. I was speaking in the first sense, although on occasion I use the second sense. I will leave that part as is. $\endgroup$
    – Dale
    Jun 21, 2021 at 23:18
  • $\begingroup$ Thanks for the clarification. Still a related thing. Referring to the part you have edited- “where the metric is the Minkowski metric”- does locally spacetime looks like $M^4$ also mean that at each point we can reduce the Metric to any metric (not just Minkowskian in Cartesian coordinates but the metrics obtained by arbitrary coordinate transformations of Minkowski metric; like transformation to polar coordinates or accelerated frames or whatever) with the only condition that at the given point where this new metric is valid, the new metric will give 0 Reimann tensor components……. $\endgroup$
    – Shashaank
    Jun 22, 2021 at 9:11
  • $\begingroup$ Please not while considering the above that I am not saying that the derivatives of the Christoffel symbols for old ( valid all throughout the Manifold) metric are zero ( like I pointed in the 1st comment) but just that the metric you get at each point should not just necessarily be Minkowski in Cartesian but Minkowski in ANY coordinate with the condition that the Reimann tensor that you get from this metric will be 0. Is that correct. I think so. Because “looks locally like $M^4$ means you can have Minkowski metric in any coordinates; Minkowski metric is flat in all coordinates. $\endgroup$
    – Shashaank
    Jun 22, 2021 at 9:16
  • $\begingroup$ Yes, although the “to first order” part will be different for unusual coordinates $\endgroup$
    – Dale
    Jun 22, 2021 at 10:51

Standing on the surface of the Earth, the reference frame at rest relative to yourself is certainly not a frame with Minkowski metric. Here is the proof: release an object, so that it is in free fall. There is relative acceleration between the object and the chosen frame. Hence the frame is not inertial and its metric is not Minkowskian.

To define a tangent space in general relativity it is not sufficient that the metric be Minkowskian just at one event. It must be Minkowskian AND have no first-order dependence on distance or time near that event. In other words the Christoffel symbols must all vanish. But since the released object is accelerating relative to the frame at rest on Earth, at least one of the Christoffel symbols is not zero.

  • $\begingroup$ This answer is correct in spirit, but I respectfully believe you got some of the details wrong in your second paragraph. In fact, for any point $p$ on any manifold (including curved), there exists a Riemann normal coordinate system about $p$ in which the Christoffel symbols all vanish at $p$. So you can still have gravitational effects at a point even if all the Christoffel symbols vanish. To be truly "gravity-free" at a point (zero curvature), you need the second-order dependence of the metric on spacetime to vanish, or equivalently the Christoffel symbols must not only vanish at $p$ ... $\endgroup$
    – tparker
    Jul 11, 2020 at 18:20
  • $\begingroup$ ... but also have their first-order variation vanish as well. See my (far too long and rambling) answer for further discussion. $\endgroup$
    – tparker
    Jul 11, 2020 at 18:20
  • $\begingroup$ Relatedly: you can always define a tangent space at any point on a (smooth) manifold, whether or not it's flat at that point. The point is that if the base manifold is curved at a point $p$, then the tangent space only "agrees with" the manifold to first order (the "flat order"), and the two start to deviate at second order. $\endgroup$
    – tparker
    Jul 11, 2020 at 18:24
  • $\begingroup$ @tparker Thanks for the comment; I know Riemann geometry stuff so this is a matter of terminology and what we think the equivalence principle is saying. One can always find a frame where the curvature effects only come in at second order not first order in the metric. But in a frame not moving rel. to surface of Earth they come in at first order, so that's not a local inertial frame. You seem to want to say that when the curvature is not zero there is no local inertial frame but that is not the standard terminology. $\endgroup$ Jul 11, 2020 at 21:32
  • $\begingroup$ Okay I see what you're saying, I think I may have misinterpreted (or at least differently interpreted) the OP's question but we agree on the substance. $\endgroup$
    – tparker
    Jul 11, 2020 at 23:02

Great question. If I could somewhat reframe your question, I believe you confused by the apparent contradiction between these two statement about the equivalent principle:

  1. Any manifold in GR locally looks like Minkowski space.
  2. Even (very small) local frames can demonstrate gravitational effects (e.g. you can feel yourself "accelerating upward" if you are standing on the surface of the Earth).

Your objection is that the statement 1 seems to imply that there can't be any observable gravitational effects over very small regions of spacetime, while statement 2 seems to imply that there can be.

The resolution to this apparent contradiction is that statements 1 and 2 and using different quantitative notions of the word "local", and statement 1 is restricting the word "local" to smaller regions than statement 2 is.

More precisely: statement 1 can be more precisely rephrased as:

For any point $p$ on any pseudo-Riemannian manifold (i.e. spacetime), there exists a local coordinate system around $p$ in which the Taylor expansion of metric tensor agrees with the Minkowski metric $\eta$ to first order about $p$.

In other words, $g(p) = \eta$ and $\partial_\mu g(p) \equiv 0$ in these particular coordinates (which are known as Riemann normal coordinates). So if you define "local" to mean "so small that only first-order variations are non-negligeable," which is the implicit assumption in statement 1, then indeed no gravitational effects can be detected locally.

But it turns out that the effects of curvature (or the acceleration of the "owner" of a local coordinate system) necessarily come in at second order in the metric. More precisely:

A manifold has intrinsic curvature at a point $p$ iff the second-order Taylor expansion of the metric tensor about $p$ deviates from the Minkowski metric.

Or even more precisely:

At any point $p$ on any pseudo-Riemannian manifold, the second-order partial derivatives of the metric tensor $\partial_\mu \partial_\nu g(p)$ are either identically zero in every coordinate system, or have some nonzero components in every coordinate system. Therefore, the proposition $\partial_\mu \partial_\nu g(p) \equiv 0$ is coordinate independent. The Riemann curvature tensor vanishes at $p$ iff $\partial_\mu \partial_\nu g(p) \equiv 0$ in some coordinate system (and therefore in all of them).

Therefore, you can always make the Taylor expansion of the metric about a point agree with the Minkowski metric to first order (by using Riemann normal coordinates), but you can't make it agree to second order if the manifold is curved at point $p$. Since gravitational effects are a physical manifestation of the curvature of the spacetime manifold, you can detect them if your local frame is big enough to capture second-order deviations about the point $p$. This slightly weaker sense of "local" is the sense being used in statement 2. If your region of spacetime is only "first-order big" in the time direction, then you won't have time to measure any relative acceleration of a nearby test particle.

(By the way, you can't actually forces that induce an acceleration, but only forces that induce a tidal acceleration - defined broadly as any spatial variation in the acceleration field. The only reason that you can feel the Earth accelerating you upward is because your body is big enough that the second-order terms in the metric (which are proportional to the acceleration constant $g$) are non-negligeable. You many not be used to thinking of the electrostatic repulsion acceleration of the Earth pushing up on you as a "tidal" acceleration, but it is: the only reason you can feel it is that it's applied at the soles of your feel but not elsewhere on your body, which induces internal compression forces within your body that you feel. If it were somehow distributed in such a way as to induce a uniform acceleration over your entire body, then it would work just like gravity and you wouldn't be able to feel it.)

  • $\begingroup$ Thanks for the answer. Could you also explain the difference between an inertial frame, a Lorentz frame, a locally Minkowskian space, and a Minkowski flat space? $\endgroup$ Jul 14, 2020 at 21:28

General relativity is quite clear. As an observer on the surface of the Earth, you are in an accelerating (non-inertial) frame. Your other formulation is consequently not good enough. Only inertial frames look like Minkowski frames as used in special relativity. This is the essence of the equivalence principle, and it should be clear that only one of your formulations is correct.

The standard test of an inertial frame is to use an accelerometer (you may even have one in and app for your mobile phone). You can therefore tell whether you are in an inertial frame without looking outside of your immediate locality.

  • $\begingroup$ But in GR we also state that spacetime is a Lorentzian Manifold, and a Lorentzian Manifold in every point looks like Minkowski spacetime. So even on the surface of the earth I should get Minkowski spacetime in a local reference frame, and that means that I should be in an inertial SR frame. Do you see my problem? $\endgroup$
    – Noumeno
    Jul 8, 2020 at 22:44
  • $\begingroup$ You are not understanding what this means. Every point has a Minkowski tangent space, but the Minkowski tangent spaces refer to frames in free fall. The surface of the Earth is not in free fall. $\endgroup$ Jul 9, 2020 at 6:30

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