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

[Orodruin]

But isn't this still wrong, since Minkowski space is flat, but spacetime in a small region is not necessarily flat?

This is taken care of by the "small enough" and that the "looks like" is informal. The small enough depends on your accuracy in measuring curvature. The "looks like" does not sound as formal as "locally flat" as both "local" and "flat" have precise mathematical definitions. I would also be fine with "looks flat in a small enough region" as "look" is informal.

 

[atyy]

This is taken care of by the "small enough" and that the "looks like" is informal. The small enough depends on your accuracy in measuring curvature. The "looks like" does not sound as formal as "locally flat" as both "local" and "flat" have precise mathematical definitions. I would also be fine with "looks flat in a small enough region" as "look" is informal.

And with infinite precision, would you say that the informal statement is not true, since there would be no region that is small enough?

 

[PAllen]

And with infinite precision, would you say that the informal statement is not true, since there would be no region that is small enough?

Of course.

In fact, I think the quantitative behavior can be made accessible in a B level description. One can say that as the dimensions of a spacetime region (note, including time) are cut in half, the precision needed to detect a deviation from SR using any particular method goes up by 4 (not by 2 or 8 for example).

 

[Ibix]

I hadn't thought of "locally flat" being problematic, but I see the point. What we're trying to convey here is analogous to the Earth being spherical but my kitchen floor being flat. The latter claim is incorrect, but the errors that follow from making the claim aren't even on my list of worries when I'm tiling my floor. Not even if I need to re-pave the town square. Maybe I need to worry if I'm planning on making maps of a mid-sized country. "Flat" is a good enough description of the floor for a small area.

The same applies to spacetime, with caveats that "small area" is strongly context dependent and that the "small" applies to the extent in the time-like direction as well. So I agree with the consensus above - happy to say some variant on "spacetime is near enough flat over a small region that SR applies near enough that no one cares about the errors". And that the formal statement of this is that we can always find a coordinate system where, at a chosen point, the second first derivatives of the metric are zero and are small in a surrounding volume.

 

[PAllen]

... And that the formal statement of this is that we can always find a coordinate system where, at a chosen point, the second derivatives of the metric are zero and are small in a surrounding volume.

Well, you can make the metric diag [-1,1,1,1], and the first derivatives zero, but not the second derivatives.

 

[Ibix]

Well, you can make the metric diag [-1,1,1,1], and the first derivatives zero, but not the second derivatives.

Thanks - corrected above (at least, I hope I got it right this time...).

 

[Dale]

say that spacetime locally looks like Minkowski space as long as you stay in a small enough region

How about “locally Minkowski to first order”. I think that makes it clear that second order properties like curvature may differ from Minkowski spacetime which is flat. It also gives an immediate clue about how small the locally needs to be.

 

[Orodruin]

How about “locally Minkowski to first order”. I think that makes it clear that second order properties like curvature may differ from Minkowski spacetime which is flat. It also gives an immediate clue about how small the locally needs to be.

I would say it depends on the audience. I am not sure I would use that with someone asking relatively basic questions about GR, but in more advanced cases I would be fine with that.

 

[Orodruin]

And with infinite precision, would you say that the informal statement is not true, since there would be no region that is small enough?

There is no such thing as infinite precision and if there was then no region (with a finite extension) would be "small enough". Either way, the informal expression is for conveying a more informal image. If I wanted to convey the mathematical structure with precision I would just say Riemannian/Lorentzian manifold.

 

[Paul Colby]

As such, to speak of the principle of equivalence, you must restrict time to a small fraction of the frequency of change.

Well, space-time is in fact curved in this case. LIGO measures the change in the proper length of 4 km interferometer arms. If one were unlimited by noise etc, the time scale of this length measurement is limited by the transit time of the laser over 8 km which is quite small relative to a light second over which the phenomena of interest occurs. Of course in the real world of LIGO the noise dominates in the extreme. In many ways, LIGO is typical of all experimental work independent of GR. So I don't see your point exactly.

 

[PAllen]

Note that there is

Well, space-time is in fact curved in this case. LIGO measures the change in the proper length of 4 km interferometer arms. If one were unlimited by noise etc, the time scale of this length measurement is limited by the transit time of the laser over 8 km which is quite small relative to a light second over which the phenomena of interest occurs. Of course in the real world of LIGO the noise dominates in the extreme. In many ways, LIGO is typical of all experimental work independent of GR. So I don't see your point exactly.

i see nothing unclear in my point. Tidal gravity is easily measurable, but this is never taken to refute the principle of equivalence. Instead, the POE is taken to apply only to a spacetime region such that tidal gravity is undetectable. Similarly, if curvature is changing in time, the spacetime region must be limited to a time period over which curvature doesn’t change within given measurement precision. This is just the definition of the POE. If the GW has a frequency of e.g. a kilohertz, this means POE only applies to a time period of say .00001 seconds. I don’t see your objection to the basic definition of the POE.

 

[Paul Colby]

i see nothing unclear in my point.

Well, okay. I understand your point I just don't see it as addressing the question. In dream land if distances and times can be reduced to zero in theory, why can't experimental methods in principle be refined indefinitely? In the limit of an infinite precision 0 noise LIGO of arbitrary small size isn't one making a measurement at "a point" at "a time"? I see this as at the heart of OPs pet peeve but could well be wrong.

 

[PAllen]

Well, okay. I understand your point I just don't see it as addressing the question. In dream land if distances and times can be reduced to zero in theory, why can't experimental methods in principle be refined indefinitely? In the limit of an infinite precision 0 noise LIGO of arbitrary small size isn't one making a measurement at "a point" at "a time"? I see this as at the heart of OPs pet peeve but could well be wrong.

No, he doesn’t dispute my point at all. Please read his last several posts in this thread responding to @atyy . Nowhere is @Orodruin questioning POE as physical principle. He disputes calling its phenomenology local flatness.

 

[Paul Colby]

I don’t see your objection to the basic definition of the POE.

Yeah, nor do I and perhaps that's my problem. In your reply in #81 is the claim that based on the POE there is always a limit in which tidal gravity becomes undetectable. This is an explicit statement that any dreamland race between theory and experimental observation will always be lost by the experimentalist. This I can accept. The POE is used to structure theory of which I'm quite content.

 

[atyy]

There is no such thing as infinite precision and if there was then no region (with a finite extension) would be "small enough". Either way, the informal expression is for conveying a more informal image. If I wanted to convey the mathematical structure with precision I would just say Riemannian/Lorentzian manifold.

OK, thanks for clarifying the physics. So it's clear the remainder is just terminology and a matter of taste, like "work", "relativistic mass", "collapse". I still think local flatness is the best, otherwise how can one say that a semi-Riemannian manifold that is nowhere flat is everywhere locally flat :)

 

[Orodruin]

I still think local flatness is the best, otherwise how can one say that a semi-Riemannian manifold that is nowhere flat is everywhere locally flat :)

One could just avoid saying that

 

[vanhees71]

And with infinite precision, would you say that the informal statement is not true, since there would be no region that is small enough?

I think it's simply a matter of accuracy. In a free-falling non-rotating frame of reference you have approximately in a "not too large spatio-temporal extension" a "local inertial reference frame" "up to tidal forces". Whenever you meausure the forces accurately enough there's some deviations from an idealized inertial refeference frame due to these tidal forces.

 

[Raymond Beljan]

I see many posts by several different people referring to spacetime being "locally flat" with the intended meaning of being locally indistinguishable from Minkowski space, i.e., being able to rewrite the metric on orthonormal form and not being able to measure curvature on some local scale. I do not think this is an appropriate nomenclature and the more appropriate nomenclature would be to refer to a local inertial frame. I am aware that some textbook authors, such as Schutz, use the term in this way as well. These are (some of) my issues with the terminology:

• "Local flatness" is typically defined in a different manner in topology, where it is a property of a submanifold. The entire point of using differential geometry is that spacetime can be described without reference to it being a submanifold of some higher-dimensional space.
• Not withstanding the previous point, we otherwise use "local" to describe a property that is only true in a point or in a neighbourhood of that point. "Flat" refers to the curvature being zero. Putting those two together as "locally flat" would therefore typically mean that the curvature at the given event (or neighbourhood) would be zero. This is not generally true as curvature invariants can be computed to be non-zero even though there are local inertial frames at all events.
• There exists other alternative terminology to describe precisely the ideas that "locally flat" intends to convey. The existence of a "local inertial frame" or similar comes to mind.

Any thoughts? Am I just being picky?

I don't think you are being picky.

Leonard Susskind in his Stanford U. lectures discuses this around 03:18 in this lecture:

https://cosmolearning.org/video-lectures/geodesics-gravitational-fields-special-relativity/
From the lectures, I think there can be curved coordinates at a point but the important thing is weather or not another coordinate system can be found, such that, locally ( that is in the limit of differential change ) the coordinates are those of a flat space.

Oh, I should add that for a locally flat space the metric is the kronecker delta and the first derivitives of the metric are zero.

 

[Raymond Beljan]

Well actually the kronecker delta but one minus sign. As noted by Ibix (I think) metric with diagonal [-1,1,1,1]

 

[TonyP0927]

I see many posts by several different people referring to spacetime being "locally flat" with the intended meaning of being locally indistinguishable from Minkowski space, i.e., being able to rewrite the metric on orthonormal form and not being able to measure curvature on some local scale. I do not think this is an appropriate nomenclature and the more appropriate nomenclature would be to refer to a local inertial frame. I am aware that some textbook authors, such as Schutz, use the term in this way as well. These are (some of) my issues with the terminology:

• "Local flatness" is typically defined in a different manner in topology, where it is a property of a submanifold. The entire point of using differential geometry is that spacetime can be described without reference to it being a submanifold of some higher-dimensional space.
• Not withstanding the previous point, we otherwise use "local" to describe a property that is only true in a point or in a neighbourhood of that point. "Flat" refers to the curvature being zero. Putting those two together as "locally flat" would therefore typically mean that the curvature at the given event (or neighbourhood) would be zero. This is not generally true as curvature invariants can be computed to be non-zero even though there are local inertial frames at all events.
• There exists other alternative terminology to describe precisely the ideas that "locally flat" intends to convey. The existence of a "local inertial frame" or similar comes to mind.

Any thoughts? Am I just being picky?

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."

 

[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.