Is 'Local Flatness' the Right Term for Describing Spacetime?

[Ibix]

In other words, it's a "simple explanation."

I think the problem is that it's also wrong - as discussed on this thread, curvature is an invariant and not zero in a small region. Something like "the effects of curvature are negligible over a small region" isn't really any more complicated, and is rather more accurate.

 

[Raymond Beljan]

I typically see "local flatness" or similar terms used when representing space-time as a 2D grid as is often the case to show the analogy of gravity's effect on space-time to placing a heavy ball on a sheet. In other words, it's a "simple explanation."

That "ball placed on a rubber sheet" analogy is one I wish would not be used. Why? Well even if the curved rubber sheet is analogous to curved spacetime, the analogy is often accompanied by rolling a small ball on the sheet to show the effect of curvature. The small balls motion then depends on the Earth's actual gravity in the space in which the rubber sheet is embedded. Is the path of the ball really a geodesic? Does the ball really follow a line defined by parallel transport of a vector ? I don't know. But even if it did, a person walks away thinking they understand GR without any idea of a geodesic or parallel transport. Further the path should be realizable if the sheet were on its side or upside down. Arggghhh. I say stop rolling balls on a rubber sheet. And if something moversd it should be a disk which is in the sheet, not on it.

 

[Povel]

One claim in this article seems questionable to me: that you can have, in the interior of some spacetime and bounded by curved regions separating it from a standard flat Minkowski spacetime region, a spacetime region which is flat but has "homogeneous acceleration" relative to the exterior flat region. I have never seen such a solution in the GR literature. Does anyone know what this refers to?

At least in the case of Newtonian gravity, such a region with homogeneous acceleration can be found inside a non-concentric spherical cave in a spherical body with uniform density. There are no tidal forces inside the cave.

I don't know if the same is also true in GR (the demonstration involves the use of the superposition principle), but given how in this situation the gravitational field can be weak and the mass involved small, I don't see how it could give a completely different result.

 

[Nugatory]

At least in the case of Newtonian gravity, such a region with homogeneous acceleration can be found inside a non-concentric spherical cave in a spherical body with uniform density. There are no tidal forces inside the cave.

I'm asking not arguing here, but I'd be interested in seeing the Newtonian calculation showing this result. It's equivalent to showing that the equipotential surfaces are exactly parallel throughout the entire interior of the cavity, is it not?

 

[PeterDonis]

I'd be interested in seeing the Newtonian calculation showing this result.

A full Newtonian calculation seems somewhat hairy, but it's at least easy to check along the radial line from the center of the body to the center of the cavity. If the cavity goes from $r = a$ to $r = b$ along that radial line, with $a < b < R$ ($R$ is the overall radius of the body), then the center of the cavity is at $r = (b - a)/2$, and it is easily checked (by superposition, just take the acceleration that would be due to the body if it were solid, and subtract the acceleration that would have been caused by the cavity if it were solid) that the acceleration of a test object anywhere inside the cavity along that radial line (i.e., from $r = a$ on the opposite side of the body's center from the cavity center, to $r = b$ on the same side of the body's center as the cavity center) is $2 \pi \rho \left( b - a \right) / 3$, where $\rho$ is the constant density of the body, in the direction from the $r = b$ edge of the cavity to the $r = a$ edge.

 

[Nugatory]

but it's at least easy to check along the radial line from the center of the body to the center of the cavity.

Yes, I got that far... but we have to compare the acceleration on that line with the acceleration on a nearby line not quite through the center of the cavity to see if tidal effects vanish... still calculating.

 

[PeterDonis]

we have to compare the acceleration on that line with the acceleration on a nearby line not quite through the center of the cavity

I think the following argument is sufficient to show that the acceleration doesn't change with a displacement perpendicular to the line.

Suppose we are at some point along the line (and inside the cavity), which is a distance $r$ from the center of the body and a distance $s$ from the center of the cavity. The acceleration is what I gave before.

Now we displace a distance $d$ perpendicular to the line. The two accelerations (due to the body, and minus due to the cavity) will now each have two components, one along the line and one perpendicular to the line. The components along the line are the same as before. The components perpendicular to the line cancel, because they will point in opposite directions (due to the opposite signs) and will be of the same magnitude (because the perpendicular component of each force is given by the same ratio to each total force as the ratio of the distance $d$ to the corresponding total distance, $\sqrt{r^2 + d^2}$ and $\sqrt{s^2 + d^2}$ respectively, and the forces are linear in the distances so the reduction of each force by the corresponding ratio ends up giving the same magnitude).

 

[Orodruin]

The argument is quite simple. Inside a homogeneous sphere the potential is proportional to $\rho(x^2 + y^2 + z^2)$. This quadratic term does not change with translations, only introduces a linear term in addition. Thus, superposing the sphere of negative density imside the cavity, the quadratic terms cancel out, leaving only a linear potential, ie, a homogeneous field.

 

[Ibix]

At least in the case of Newtonian gravity, such a region with homogeneous acceleration can be found inside a non-concentric spherical cave in a spherical body with uniform density. There are no tidal forces inside the cave.

I don't know if the same is also true in GR (the demonstration involves the use of the superposition principle), but given how in this situation the gravitational field can be weak and the mass involved small, I don't see how it could give a completely different result.

Note that you need two hollow regions to match Peter's original comment, which was:

in the interior of some spacetime and bounded by curved regions separating it from a standard flat Minkowski spacetime region, a spacetime region which is flat but has "homogeneous acceleration" relative to the exterior flat region.

An off-center hollow sphere has a uniform gravitational field, but you also need a hollow concentric with the sphere in order to have a zero-field region corresponding to a flat Minkowski spacetime.

The obvious difference with full GR is that the non-spherically symmetric internal stresses in the matter contribute as sources of gravity. I don't know if they'll cancel out quite so elegantly.

 

[PeterDonis]

Note that you need two hollow regions to match Peter's original comment

No, you don't. The "standard flat Minkowski spacetime region" I was referring to would be outside the body altogether. The original reference is this article that @atyy linked to:

https://www.mathpages.com/home/kmath622/kmath622.htm

The spacetime illustrated in the image in that article is, of course, not the same as one containing a hollow sphere, since the exterior region in the latter spacetime is not flat, only asymptotically flat. However, I think "asymptotically flat" for the exterior region is actually enough to investigate the question.