# Tidal force in equivalence principle

The inertial frame of reference in a gravitational field is defined locally, but couldn't a sufficiently sensitive instrument detect tidal forces in the gravitational field and thus make the frame in the gravitational field distinguishable from a uniformly moving reference frame? If the answer is to take a frame that is "local enough" for the tidal forces to be indistinguishable, then why can't similar restrictive conditions be applied to other non-inertial frames?

• I am not an expert, so only a comment. The equivalence priciple should not be taken strictly (as the answers pointed out). If I remember correctly, MTW discusses the deviation of free falling particles (in that elevator), (which I think you mean by tidal forces), and this deviation of the paths defines the Riemann tensor. So in a sense the whole point (slight exaggeration) of GR are the tidal 'forces'. Apr 7 at 7:01

The trick is in the definition of the word "local". What physicists typically call inertial coordinates or freely falling coordinates, mathematicians will call normal coordinates. The thing about these coordinates is: "Given a point $$p$$, you can always choose coordinates such that the metric is $$\text{diag}(-1,+1,+1,+1)$$ at $$p$$ and such that the Christoffel symbols vanish at $$p$$". Hence, at $$p$$, you can't distinguish the spacetime you're in from a flat spacetime, but no one claims that that will still hold if you consider what happens in the vicinity of $$p$$.

Let me try to provide some mathematical intuition into that. The Christoffel symbols are the derivatives of the metric, and they do depend on coordinate choices. However, the curvature tensors, which are the second derivatives of the metric, will vanish in some choice of coordinates if, and only if, they vanish in all choices of coordinates. Hence, while picking inertial coordinates will allow us to set the derivatives of the metric to zero at some point $$p$$, they won't allow us to set their second derivatives to zero. So for example, no amount of coordinate changes will ever allow you to get rid of the Kretschmann scalar in the Schwarzschild spacetime, which is sort of a measure of the tidal forces.

In short, your intuition is correct: a sufficiently precise experiment will be able to distinguish gravity from acceleration. The reason behind it is that the two effect are undistinguishable at a point, but the experiment will happen across a bunch of points. Naturally, if you were to perform an experiment at $$p$$ alone, you would never be able to detect tidal forces, since those are precisely due to the difference of effects in two close points.

This is a rather confusing topic. At any event in any spacetime manifold, mathematically, you can calculate various curvature invariants. Mathematically they are defined at that event. So in that sense, if a spacetime is curved then it is not flat even at a single point.

Now, at any event in any spacetime you can define coordinates such that the metric, expressed in those coordinates, is equal to the flat spacetime Minkowski metric to first order. This means that for any given sensitivity and given curvature there is always a length scale below which the deviations from flat spacetime are undetectable. So in that sense even a curved spacetime is “locally flat”.

However, as you correctly point out, if spacetime is curved then for any length scale there is some sensitivity that will detect the deviations from flatness. So in that sense the concept of “local flatness” is rather weak.

In the end the idea of local flatness is simply that curvature is a second-order concept. So to first order a curved manifold and a flat manifold are indistinguishable.

why can't similar restrictive conditions be applied to other non-inertial frames?

The metric expressed non-inertial frames differs from that in local inertial frames to first order.