Is gravity fictitious

A.T.
Third, the last metric on the mathpages page, where the acceleration does not vary with position, would seem to be a better one to describe "uniform" acceleration, at least in the sense that Einstein was talking about in his thought experiments on the subject. However, as you can see, the metric in this form is both time-dependent and non-diagonal; after reading that page, I think that feature is what was behind my intuitive guess in an earlier post about "uniformly accelerating coordinates" raising issues.
Yes, that is the one I meant. I get your objection now.

For example, if the curvature is zero everywhere, which it is in both the "Rindler metric" case and the "constant proper acceleration" metric (the last one on the mathpages page), then the geometry is flat Minkowski spacetime.
Yes. It kind of follows from the EP that that a "globally uniform gravitational acceleration" is equivalent to an uniformly accelerating frame in flat space-time.

A final note: some of your examples don't have curvature zero everywhere, just in a particular region; but the geometry is still independent of which particular coordinate system is used to describe it.
Yes, but the point of this example (off center spherical cavity) was merely to show that in an extended region of zero curvature , you still can have "gravitational acceleration" or "mass attraction".

So the general point that the laws of physics should be written in terms of things like curvature, not things like "gravitational acceleration", still holds.)
Yes, which I guess answers the OPs question: Gravitational acceleration is "fictitious" in GR.

Last edited:
PeterDonis
Mentor
2019 Award
Yes. It kind of follows from the EP that that a "globally uniform gravitational acceleration" is equivalent to an uniformly accelerating frame in flat space-time.
I think that's what Einstein was getting at, yes, but as the mathpages page shows, there really isn't a way to actually set this up globally, only locally. Every example on that page fails to meet at least one condition that would seem necessary to really have a "globally uniform acceleration". As I noted before, I would say the last one comes closest, but even if we leave out the time-dependence and non-diagonal metric, observers "at rest" in these coordinates do not remain at a fixed proper distance from each other.

Yes, but the point of this example (off center spherical cavity) was merely to show that in an extended region of zero curvature , you still can have "gravitational acceleration" or "mass attraction".
Agreed.

PeterDonis
Mentor
2019 Award
Am I reading that page correctly that the solution for an infinite plane of mass is the Rindler metric? If so, that's pretty cool!
I read it that way too, but with a couple of qualifications. First, it's only one possible solution, for the special case where all the metric coefficients except g_tt are constant. (The rest of the page covers the other possibilities pretty thoroughly.)

Second, for the "usual" case of the Rindler metric as just an alternate set of coordinates on Minkowski spacetime, that metric does not cover the entire spacetime, only "region I", where x > 0 and |t| < x (in Minkowski coordinates). In this spacetime, with an actual physical wall, unless I'm missing something, the metric is supposed to cover the entire spacetime, or at least all of it on one side of the wall (but by symmetry, the solution would be the same on the other side except for a parity transformation, so that the "acceleration vectors" point in the opposite direction in x). So this is not just flat Minkowski spacetime in disguise.

Another way to put this is: in the usual Rindler metric, the "Rindler horizon" at X = 0 is a null surface, but as I'm reading this, that is not the case for the wall; the infinite wall is a normal timelike surface, for which X = 0 and T, Y, Z can all range from minus infinity to plus infinity.

Still another way to put it: the solution assumes that the Ricci tensor vanishes, but strictly speaking, that only applies for X > 0 (and, by symmetry, for X < 0 with a parity flip). But if the wall is actually physically there, then the Ricci tensor should not vanish on the wall, at X = 0. That means the spacetime is not vacuum everywhere.

All of which makes me wonder if I'm missing something, because the metric does look, formally, exactly like the Rindler metric, which makes me think that if I computed its Weyl tensor, it would vanish (since the Weyl tensor certainly vanishes in Minkowski spacetime). But if the Weyl tensor vanishes, it's hard for me to see how the infinite wall at X = 0 can be a timelike surface instead of a null surface as the normal Rindler horizon is.

PeterDonis
Mentor
2019 Award
it's hard for me to see how the infinite wall at X = 0 can be a timelike surface instead of a null surface as the normal Rindler horizon is.
Actually, after thinking this over, it's even more of an issue than I thought. If the Weyl tensor does vanish for X > 0, then the entire region X > 0 in these coordinates must be isometric to "region I" of Minkowski spacetime. But that means that the apparent 3-surface X = 0 in these coordinates is actually only a 2-surface (i.e., if we leave out the Y and Z coordinates, it is a point instead of a line--the point x = 0, t = 0 in standard Minkowski coordinates). So where is the wall? That is, where is the non-zero Ricci tensor region that is the "source" of the gravitational field? The Ricci tensor can't just be nonzero at that single point in spacetime; that would violate conservation of energy. It can't be nonzero along the entire null "boundary" of region I, because that would cause the Weyl tensor to be nonzero in region I itself (because of curvature "propagating" from the "past horizon" null surface, x = -t). But if the Ricci tensor is zero everywhere, then there is no wall at all; the derivation basically amounts to a proof that an infinite "wall" can only exist if its stress-energy tensor is identically zero. Was that actually the intent, or am I missing something?