I Euclidean geometry and gravity

[raagamuffin]

I asked a question here, probably over 15 years ago on entanglement and I appreciated the thoughtful answers I received back then. The intervening years haven't made me any more knowledgeable in physics, so forgive my naïveté !

If a have a piece of paper in an area of high gravity, lets say near a black hole, and I draw a triangle on this paper and 'measure' the angles of the triangle, will they add to 180 degrees?
How about if I'm looking at this paper outside of the (reasonable) gravitational sphere of influence ? Does it matter where it's being observed from?

Or is this question nonsensical? For example, presumably the measuring apparatus (protractor) would also be warped by an equal amount?

 

[PeterDonis]

If a have a piece of paper in an area of high gravity, lets say near a black hole, and I draw a triangle on this paper and 'measure' the angles of the triangle, will they add to 180 degrees?

It depends on how the paper is oriented and what its state of motion is. But for a piece of paper of ordinary size and a black hole of stellar mass or larger, any effects of non-Euclidean geometry will be too small to observe.

A better way of probing the spatial geometry around a black hole is this: suppose I construct two spherical shells around the hole, one inside the other. I carefully measure the surface area of each shell. Then I measure the radial distance between the shells. How will that distance compare to the radial distance between two concentric spherical shells as predicted by Euclidean geometry?

The answer to that is that the measured radial distance will be larger than the Euclidean prediction.

How about if I'm looking at this paper outside of the (reasonable) gravitational sphere of influence ? Does it matter where it's being observed from?

The observations I described above, for the two concentric spherical shells, depend on local measurements. Trying to make those observations from a distance brings in other effects, such as the behavior of light in curved spacetime, that make it much more difficult to interpret the results.

presumably the measuring apparatus (protractor) would also be warped by an equal amount?

No, it wouldn't. It's true that objects being held at rest in a gravitational field, so they feel weight, don't behave exactly the same as objects in free fall; internal stresses in the objects can change their shape. But for measurements like the ones I described above with the spherical shells, you can make the measurements using, for example, very small rulers, so such effects are negligible.

 

[raagamuffin]

Thanks, the sphere example is a nice illustration.
Please allow me to probe into your answers a bit.
(If you wouldn't mind sticking to the 2D world, it just helps me visualize things a bit better.)

Let's extend the example, to say that I have a large sheet of paper, say earth sized, bumping up next to a black hole.. would the triangle angles still not add up to 180?

Trying to make those observations from a distance brings in other effects, such as the behavior of light in curved spacetime, that make it much more difficult to interpret the results.

I understand this issue with observation.Theoretically speaking then, should the observer expect a different answer every time, or is there a predictable answer (or set of answers) ?

 

[PeterDonis]

say that I have a large sheet of paper, say earth sized, bumping up next to a black hole.. would the triangle angles still not add up to 180?

If the paper were oriented radially, then yes, you would be able to draw a triangle on it whose angles would not add up to 180 degrees. In effect, it would be as if you were drawing the triangle on a Flamm paraboloid:

https://en.wikipedia.org/wiki/Schwarzschild_metric#Flamm's_paraboloid

Please note that the Flamm paraboloid visualization is very limited, and should be used with caution. But it does capture the specific case we're discussing.

 

[PeterDonis]

Theoretically speaking then, should the observer expect a different answer every time

Not if the conditions are the same. But what someone very distant observes won't be easily interpreted as "I'm seeing non-Euclidean geometry".

 

[raagamuffin]

thanks, this is good information. I'll work through the Flamm paraboloid writeup.

Not if the conditions are the same. But what someone very distant observes won't be easily interpreted as "I'm seeing non-Euclidean geometry".

One last question, if we can't interpret the data well enough that comes from curved space time, would it also be the case that we can't get usable data (about spacetime curvature) from light that comes as a result of gravitational lensing ? or infer the geometry of space from the light that comes from black hole boundaries (instead relying on theoretical data)?
(Perhaps this is a cosmology thread question.)

 

[Ibix]

we can't get usable data (about spacetime curvature) from light that comes as a result of gravitational lensing ?

Sure we can get usable data. But you have to take into account the effect of curvature in the spacetime the light travels through, as well as the curvature at the area you're looking at. So what you actually see doesn't have a straightforward interpretation in terms of the local geometry. You can still deduce what's there.

It's not radically different from looking at something through a bottle. The image is distorted so what you see isn't what's there, but if you know how to do the maths you can correct for the distortion. The maths is a bit harder in GR than normal optics, that's all.

 

[PeterDonis]

if we can't interpret the data well enough that comes from curved space time

That's not what I said. What I said is that, if you're observing from a distance objects that are at rest fairly close to a black hole's horizon, there won't be a simple way to interpret your observations as showing you the non-Euclidean geometry of space in the region you're observing. That's a much narrower statement than the one you made in the quote above.

would it also be the case that we can't get usable data (about spacetime curvature) from light that comes as a result of gravitational lensing ?

Not at all. There are certainly ways of getting useful data about spacetime curvature from a distance.

But the non-Euclidean geometry of space close to a black hole is not the same thing as spacetime curvature. It is a consequence of the geometry of spacetime around a black hole, which is curved, but it has no simple relationship to spacetime curvature.

or infer the geometry of space from the light that comes from black hole boundaries

"The geometry of space" is not even an invariant. Remember that GR is a theory of spacetime, not space. The geometry of spacetime is an invariant. The non-Euclidean "geometry of space" around a black hole that we've been discussing depends on splitting up spacetime into "space" and "time" in a particular way. That way is the one that makes the most sense for an observer at rest relative to the black hole, but it's still a choice that ultimately has nothing to do with the physics; it's just a convenient way of organizing information so we humans can work with it. The geometry of spacetime is the fundamental thing.

 

[PAllen]

If the paper were oriented radially, then yes, you would be able to draw a triangle on it whose angles would not add up to 180 degrees. In effect, it would be as if you were drawing the triangle on a Flamm paraboloid:

https://en.wikipedia.org/wiki/Schwarzschild_metric#Flamm's_paraboloid

Please note that the Flamm paraboloid visualization is very limited, and should be used with caution. But it does capture the specific case we're discussing.

Some additional cautions may be stated:

A paper so oriented is not inertial, so you have to imagine some force holding it in place, that has no effect on the spacetime. The force must be different at different points on the paper.

Also, if I choose a different set of events (using a different possible time coordinate) on each stationary paper world line to be simultaneous, I can discover a strictly Euclidean triangle - the Gullestrand Panlieve foliation.

 

[A.T.]

Let's extend the example, to say that I have a large sheet of paper, say earth sized, bumping up next to a black hole.. would the triangle angles still not add up to 180?

If you manufacture a large enough sheet of paper in flat space, and then move it into curved space, it will be torn apart, no matter if/how you apply forces to try preventing that.

 

[PeterDonis]

If you manufacture a large enough sheet of paper in flat space, and then move it into curved space, it will be torn apart, no matter if/how you apply forces to try preventing that.

First, you should say flat spacetime and curved spacetime. Those are the invariants.

Second, while paper made out of the ordinary materials we usually make paper out of probably would be torn apart, yes, we could imagine making paper out of thin sheets of Kevlar, or carbon nanotubes, or something that would be strong enough to hold itself together against the stresses it would undergo while being moved. But such an object, while it would not be torn apart, would change its shape. That's the crucial point.

 

[Jaime Rudas]

A better way of probing the spatial geometry around a black hole is this: suppose I construct two spherical shells around the hole, one inside the other. I carefully measure the surface area of each shell. Then I measure the radial distance between the shells. How will that distance compare to the radial distance between two concentric spherical shells as predicted by Euclidean geometry?

The answer to that is that the measured radial distance will be larger than the Euclidean prediction.

From this, I understand that for a sphere of area $A$ around a very massive and dense star, its radius $R$ will be greater than $\frac12 \sqrt{\frac A \pi}$. If this is correct, for the case of the sphere around the black hole, can its radius be calculated, or is it undefined?

 

[PeterDonis]

From this, I understand that for a sphere of area $A$ around a very massive and dense star, its radius $R$ will be greater than $\frac12 \sqrt{\frac A \pi}$.

That's correct. Note, though, that when you have a star in the center instead of a black hole, the spacetime geometry inside the star is different from the geometry outside, and that has to be taken into account.

If this is correct, for the case of the sphere around the black hole, can its radius be calculated, or is it undefined?

It's undefined. The black hole has no center in the usual geometric sense. The locus $r = 0$, the singularity, is not a place in space at the center of the hole; it's a moment of time that's to the future of all events inside the hole.

 

[PAllen]

First, you should say flat spacetime and curved spacetime. Those are the invariants.

Second, while paper made out of the ordinary materials we usually make paper out of probably would be torn apart, yes, we could imagine making paper out of thin sheets of Kevlar, or carbon nanotubes, or something that would be strong enough to hold itself together against the stresses it would undergo while being moved. But such an object, while it would not be torn apart, would change its shape. That's the crucial point.

Alternatively, make it out of something like rubber, and it will change shape without great stress or tearing.

 

[PAllen]

Claim: There is no difficulty defining a congruence of world lines representing a piece of paper far from a BH, that then has each worldline accelerate and decelerate such that you have a representation of a sheet of paper stationary near the BH. The kinematic decomposition of this congruence will necessarily show expansion and compression. However whether a material approximation to this math breaks or not depends on material properties. Unless one is talking about getting close to the horizon of stellar ( rather than supermassive) BH, the stress would not be extreme.

 

[pervect]

Thanks, the sphere example is a nice illustration.
Please allow me to probe into your answers a bit.
(If you wouldn't mind sticking to the 2D world, it just helps me visualize things a bit better.)

Let's extend the example, to say that I have a large sheet of paper, say earth sized, bumping up next to a black hole.. would the triangle angles still not add up to 180?


I understand this issue with observation.Theoretically speaking then, should the observer expect a different answer every time, or is there a predictable answer (or set of answers) ?

It's unclear to me what mathematics models a "large sheet of paper" in the current context. So I don't have anything rigorous. But I'll talk about the "small piece of paper" case.

A "small piece of paper" will have a sectional curvature that can be computed from the Riemann tensor, wiki has the formula https://en.wikipedia.org/wiki/Sectional_curvature

The easiest thing to do is to define the "small paper" by two unit vectors that are orthogonal, then the wiki formula for the sectional curvature is

$$R_{abcd} X^a Y^b Y^c X^d = -R_{abcd} X^a Y^b X^c Y^d$$

Sorry for the tensor notion, but I don't know how to avoid it. You may be familiar with matrices. In tensor notation they would be two index tensors, like say $Q_{ab}$. If you imagine a sort of matrix that's 4 dimensional rather than 2 dimensional, you'll have some vague idea of what the Riemann tensor is about. Space time is 4 dimensional, so the riemann tensor has 4x4x4x4=256 components, but most of them are zero or the same. There's only about 20 or so unique values of it in a 4d space-time. I forget how many unique componets are in 3d space, and in 2d there is only one non-zero component.

What we are relying on here is the idea that 2 dimensional surfaces have only one non-vanishing curvature component, which is given by the sectional curvature formula Then the wiki formula above tells you how to get this value at any point on your piece of large paper.

But the point is that the sectional curvature is varying with position. Thus, If you have a large sheet of paper, the sectional curvature will vary with the position on the paper, because parts of the paper will be far away from the black hole and other parts will be close. The "small paper" works when the Riemann tensor components don't change much in the paper.

You will find formula for the sum of the angles of a spherical triangle being greater than 180, and the amount the sum is greater than 180 will depend on the area of the triangle. This pretty much carries over to the case of interest.

Wiki covers this briefly in https://en.wikipedia.org/wiki/Spherical_geometry. Here is a brief summary of some (not all) of the points wiki makes about spherical triangles and the sum of angles of spherical triangles.

The angle sum of a triangle is greater than 180° and less than 540°.

The area of a triangle is proportional to the excess of its angle sum over 180°.

Two triangles with the same angle sum are equal in area.

So this method lets us get the sum of angles of a "small" triangle, where the Riemann curvature tensors components are constant within the triangle. For Schwarzschild geometries, I believe the components are proportional to 1/r^3, so you want r not to vary much.

I don't have anything rigorous for "large sheets of paper", as I said. But a model comes to mind.

Imagine have a piece of paper that's flat at infinity with a "bump" in the middle. The "bump" represents the curved sections, where the geometry isn't euclidean, and in the envision model it's basically spherical. (You could also have a saddle surface which needs hyperbolic geometry, but I'm not imaginig that case).

Using this model, we can see that probably a large triangle can be oriented so most of its edges are in the "flat" portion, and the angles of the large triangle will sum to 180, but you can change this by having parts of the triangle pass through the section with a high curvature. You can break the large triangle up into a lot of small triangles and use the small triangle formula to get the sum-of-angles of the small triangles, but I'm not sure how to get anything useful out of that at the moment.

 

[A.T.]

Second, while paper made out of the ordinary materials we usually make paper out of probably would be torn apart, yes, we could imagine making paper out of thin sheets of Kevlar, or carbon nanotubes, or something that would be strong enough to hold itself together against the stresses it would undergo while being moved.

To clarify: I wasn't referring to stresses due to "moving" the object around by applying "badly" distributed external forces.

Even if you where able to apply external forces, to hold the object at rest w.r.t the big mass, with arbitrarily smooth force distribution which minimizes the stresses, those stresses still cannot be zero, because the intrinsic spatial geometry of the object is flat, while space near the mass, in the rest frame of the mass, is intrinsically curved over a large enough region.

The limitation, under the above conditions, is not how strong the material is, but rather how far it can be deformed without breaking. Under the above conditions, a sheet of kevlar will tear faster than a sheet of rubber.

 

[PAllen]

I am not quite sure how much of embedding theory corresponds to physics in pseudo-riemannian geometry, but I think it is worth pointing out a few possibly relevant points. T

All real materials have some small amount of rubbery behavior, the amount varying greatly. Mathematically, this means they can change their smooth embedding (bend smoothly with no stretch or compression, e.g. a sheet of paper bent into a cylinder), change their intrinsic geometry (stretch/compress). Thin sheets may also tolerate non-smooth embedding (creasing/crinkling). Thus an idealized non-stretchable fabric that lays flat with no crinkling, can be deformed without stretching in Euclidean 3-space into a flat torus - but it will have crinkling representing non-smooth embedding. Note that a constant positive curvature 2-surface is smoothly embeddable in Euclidean 3-space, while a constant negative curvature surface can only be non-smoothly embedded (e.g. using a crinkly fabric that cannot lay flat).

For the case at hand, the interesting point is that not only can flat Euclidean 3-space be embedded in Kruskal geometry, but any two quadrants (representing one exterior and 'half' of interior) can be fully foliated by flat Euclidean 3-space hypersurfaces. This is proven by the existence of Gullestrand-Panlieve coordinates. Further, this is a smooth embedding. This would seem to say there is no obstacle to the existence of intrinsically flat paper (i.e. smoothly embedded with no stretching or crinkling) of any size anywhere in a BH spacetime.

 

[PeterDonis]

those stresses still cannot be zero

Yes, agreed.

The limitation, under the above conditions, is not how strong the material is, but rather how far it can be deformed without breaking.

But how far it is deformed, under a given stress, depends on how strong the material is. A stronger material will deform less under a given applied stress. Kevlar will tear under a smaller deformation than rubber, but it also will deform less under a given stress than rubber will.

 

[PeterDonis]

not only can flat Euclidean 3-space be embedded in Kruskal geometry, but any two quadrants (representing one exterior and 'half' of interior) can be fully foliated by flat Euclidean 3-space hypersurfaces.

But those hypersurfaces are not geodesic hypersurfaces, so it's not clear that an "intrinsically flat" object can be embedded in them without stresses.

 

[A.T.]

But how far it is deformed, under a given stress ...

Under the above conditions, it's not the stress that is given, but the deformation. Once you distribute the support forces to minimize the deformations, all that is left is the deformation determined by the given spatial geometry.

If an object cannot deform to that given geometry, the stresses will become extremely large, and it will break, no matter how strong the material is.

On the other hand, a material that deforms easily will reshape without producing high stresses, so it doesn't need to be strong.

 

[PeterDonis]

Under the above conditions, it's not the stress that is given, but the deformation.

No, that's not correct. Knowing the spacetime curvature does not tell you the deformation, because spacetime curvature only tells you geodesic deviation. But for an object held together by internal forces, the worldlines of its parts are not geodesics (except for the worldline of its center of mass). To determine what those worldlines actually are, which is what you need to determine the deformation, you have to know the material properties--how the material resists the stress applied by tidal gravity.

 

[PAllen]

But those hypersurfaces are not geodesic hypersurfaces, so it's not clear that an "intrinsically flat" object can be embedded in them without stresses.

I don't know that that is definitive. See my longer post on this, coming soon.

 

[PAllen]

Thinking more about this, I come back to believing that for physics of materials, the only valid approach is in terms of kinematic decomposition of congruences.

Before discussing that, consider a solid cylinder in Euclidean 3-space. You can slice it to produce 2-surfaces of all different geometries. Further, if you extend this in time, in flat spacetime, any of these slices is a possible material 2-surface not under any stress. Further, comparing measurements at different points using different simultaneity conventions, you can 'perceive' different spatial geometries, none of which are related to stresses on the surface. Thus I think all the physics is in the congruence, not is spatial geometry.

So what I am confident of is that you can define a stationary congruence of constant latitude (varying r and $\theta$), anywhere outside of the horizon of an SC BH. This congruence will have different proper acceleration for different world lines, but no expansion, compression, or shear. Assuming one point is externally held fixed (magically), you can also state that if internal forces are supplying the difference in proper acceleration between congruence world lines, the required strength of those forces grows without bound on close approach to a BH horizon. However, up to the point that no such forces are physically plausible, you have a rigid sheet object (with intrinsic sheet geometry subject to choice in how you define simultaneity, i.e. not physically relevant).

So, now the question is, what happens if you imagine such a congruence near, but not too near, a BH after some time t2 (choose any reasonable coordinates you want for this), and much further out before some time t1, connected going from t1 to t2. How close to Born rigidity can you maintain during the transition? If not possible, how large must expansion/compression be? And, how do the required internal forces (as describe above) have to be during the transition as compared to the times before t1 and after t2 (the latter being deemed physically possible)?

My current best guess is that you cannot achieve rigidity during the transition, but that the expansion/compression will not be large. And that the required internal force required for 'closest to rigid' motion will not be much larger than those required within the stationary congruence after t2.

 

[PeterDonis]

for physics of materials, the only valid approach is in terms of kinematic decomposition of congruences.

The kinematic decomposition is useful, yes, but to obtain it, you have to already know the worldlines. But you need to know the material properties in order to determine the worldlines, since the material properties determine the internal forces between the different parts of the material. You also need to know the spacetime curvature, since that tells you the geodesic deviation, which in turn tells you what the motion of the different parts of the material would be in the absence of internal forces.

I think all the physics is in the congruence, not is spatial geometry.

I agree with this, but, as above, you need to know the spacetime curvature and the material properties in order to determine the congruence.

You could take an approach in which you specify the congruence by fiat--for example, you could say that the object is hovering at rest above a black hole, so every worldline in the congruence that describes it is an integral curve of the timelike KVF of the spacetime. In such a special case, yes, you can derive everything else from the congruence. But that's a special case. (It happens to be the one under discussion in this thread, yes. But I think it's still worth pointing out that it's a special case.)

 

[PeterDonis]

you can define a stationary congruence of constant latitude (varying r and ), anywhere outside of the horizon of an SC BH. This congruence will have different proper acceleration for different world lines, but no expansion, compression, or shear.

Yes, this is correct. Since the worldlines in the congruence are integral curves of a timelike KVF, you can derive the result straightforwardly from the properties of that KVF.

 

[PeterDonis]

My current best guess is that you cannot achieve rigidity during the transition

This is also correct. I don't know if we discussed this particular case in previous threads some time back on the Herglotz-Noether theorem, but I'm pretty sure we discussed similar cases. The case you're describing is not one of the special cases that the H-N theorem says can be Born rigid. So it can't be.

 

[PAllen]

This is also correct. I don't know if we discussed this particular case in previous threads some time back on the Herglotz-Noether theorem, but I'm pretty sure we discussed similar cases. The case you're describing is not one of the special cases that the H-N theorem says can be Born rigid. So it can't be.

So far as I know, Herglotz-Noether is purely a theorem of SR. The case described would definitely be a possible rigid motion in SR - you can specify any world line for one point of a rigid body. Then there is at least one rigid congruence for the whole body (but, typically only one), representing irrotational movement. In our case, of changing spacetime curvature, I am guessing there is no rigid motion. But I don’t really know.

I am not familiar with what has been achieved in the case of GR.

 

[PeterDonis]

So far as I know, Herglotz-Noether is purely a theorem of SR.

While the theorem was originally proved before GR was even discovered, AFAIK it does not actually require flat spacetime. The two types of motions that it says can be Born rigid (where "Born rigid" means the congruence must have zero expansion and shear) are both types of motions that can be specified in a curved spacetime, in terms of properties of the congruences:

The first type is an irrotational motion (zero vorticity), which must be everywhere orthogonal to a set of hyperplanes (i.e., flat spacelike 3-surfaces). Note that any irrotational motion must be hypersurface orthogonal by the Frobenius theorem, but the orthogonal hypersurfaces don't generally need to be flat--the flatness criterion is the additional constraint imposed by the H-N theorem (in addition to the constraints of zero expansion and shear imposed by the Born rigidity requirement).

The second type is a motion that can be rotational (nonzero vorticity--though that's not actually required, as will be seen in a moment), which must be a Killing congruence. A congruence describing a static object in Schwarzschild spacetime will be such a congruence. (Note that this congruence is actually irrotational, and therefore hypersurface orthogonal, as above, but the hypersurfaces it is orthogonal to are not flat--so it does not meet the requirements of the first type above.) But a congruence describing an object being moved from static at one altitude to static at a different altitude will not be a Killing congruence, so it won't meet this criterion of the H-N theorem--and while the motion can be irrotational, it won't be orthogonal to a set of flat hyperplanes, so it won't meet the first criterion above either.

(Note that the Painleve congruence--the congruence of worldlines orthogonal to 3-surfaces of constant Painleve coordinate time--is irrotational, and orthogonal to a set of flat hyperplanes, but it has nonzero expansion--that's why that congruence isn't Born rigid.)

 

[PAllen]

(Note that the Painleve congruence--the congruence of worldlines orthogonal to 3-surfaces of constant Painleve coordinate time--is irrotational, and orthogonal to a set of flat hyperplanes, but it has nonzero expansion--that's why that congruence isn't Born rigid.)

I am going to take some time responding to the rest of your post, but on this want to point out that the stationary congruence (constant r, $\theta$,$\phi$) has the same representation in GP coordinates as SC coordinates. The only thing different is the time slicing of the congruence. This time slicing is NOT orthogonal to the congruence (producing the non-diagonal metric element), but also producing an intrinsically flat hypersurface.

 

[A.T.]

No, that's not correct. Knowing the spacetime curvature does not tell you the deformation, ...

To determine what those worldlines actually are, which is what you need to determine the deformation, ...

The object can be held static at rest w.r.t the big mass indefinitely, by using smoothly distributed forces, which minimize internal stresses. Since nothing changes over time, the only thing that determines the constant deformation of the object, is the spatial curvature in the rest frame of the big mass.

 

[PeterDonis]

the stationary congruence (constant r, $\theta$,$\phi$) has the same representation in GP coordinates as SC coordinates.

Yes, that's true, but...

This time slicing is NOT orthogonal to the congruence

Yes, which is why the stationary congruence does not fall into the first type of congruence that the H-N theorem says can be Born rigid--it's irrotational, but is not orthogonal to a set of flat hyperplanes.

 

[PeterDonis]

the only thing that determines the constant deformation of the object, is the spatial curvature in the rest frame of the big mass.

I strongly suggest that you do the math. If you do, you will find that your statement here is false.

First, the "spatial curvature" you are referring to is not the same as the space-space components of the Riemann tensor. It's the curvature of the Flamm paraboloid. That curvature is not involved in determining the internal forces required to hold the various pieces of a static object static. This is one of the reasons I warned the OP in post #4 that the Flamm paraboloid visualization is very limited.

Second, even if we correct the above and look at the Riemann tensor, its space-space components are not the ones involved in determining geodesic deviation for timelike worldlines (which is what's involved in determining the internal forces required to hold the various pieces of a static object static). The Riemann tensor components involved in that are time-space components like $R^{r}{}_{t r t}$, $R^{\theta}{}_{t \theta t}$, and $R^{\varphi}{}_{t \varphi t}$. (For a static object, I don't think even the last two of those are relevant since there's no tangential motion.)

 

[PeterDonis]

The Riemann tensor components involved in that are time-space components

Btw, for the particular special case under discussion in this thread, we don't even need the Riemann tensor, because we know that the congruence of worldlines describing a static object is a Killing congruence, and we have a coordinate chart that's adapted to the timelike KVF, so computing the proper accelerations of the worldlines can be done just from knowing the metric in those coordinates. But not all cases are that simple.

 

[pervect]

I'll start out by giving some (I hope) interesting references on the topic of sector models, which describe an approach to learning about curvature based on the simple idea of cutting and gluing.

There's a whole series of papers by Kraus and Zahn, for instance https://arxiv.org/abs/1405.0323, which is part 1 of a series, and some refinements such as https://arxiv.org/abs/2406.02324, which describe a program which does some graphics so one can do the "cutting and gluing" on a computer instead of with actual paper. They've written quite a lot- robphy , another poster at PF, may have more up-to-date info on which of their papers are the "best".

I'll next describe informally my take on the general idea, which won't be as formal or well thought out as the professional papers, but my intent is mainly motivational, and to talk about some of the ideas that they will explore more fully and rigorously.

To learn some basics about curvature, start with learning about 2 dimensional curved surfaces. The altenrative is to take an entirely abstract approach, which also works - if one has the needed mathematical background. But this is rather demanding, and if one does not have the needed background to learn differential geometry, one can get some basic understanding by considering the simple, easy-to-visualize cases.

It might be helpful to learn some spherical trignometry along the way - it may not be necessary, but it will add some insight and ability to make actual numerical predictions. Specifically interesting is the relationship between the sum-of-angles of a triangle and it's area on a sphere.

Going back to my previous point, start by learning about the sphere. Cutting and tearing were mentioned. With a bit of real world experience , it's not hard to observe that if one take a piece of rubber formed in the shape of a hemisphere (say, half a tennis ball), it will stretch and - if it cant stretch enough - tear, if one tries to squish it flat. This comes down to the fact that the sphere is missing material compared to the plane. So, there isn't enough material in the hemisphere to make a plane - the material in the h emisphere h as to stretch to make up the plane when you try to press it flat.

Consider the approximately spherical regular geometric figures, the dodecahedron and the icosahedron. For specificity, we'll use only the icosahedron. One can form six equilateral triangles joining at a vertex to form a flat plane. The icosahedron, however, has five equilateral triangles meeting at each vertex, not six. The missing sixth equilateral triangle illustrates the "material deficit" that underlies curvature in two dimensions.

I have specifically seen Kruas and Zahn make a paper model that can be assembled into 3d pieces that represent pieces of the Schwarzschild geometry, so that they can be 'glued together" at the edges just as paper can. But one can't glue them together in flat space without gaps.

Note that while the idea of missing and extra material can be used to visualize curvature in more than two dimensions, it's only in two dimensions that the curvature is as simple as a single number that can be related to a global "lack of matggerial" or a global "excess of material". The distribution of exactly where there material is missing and where there is excess makes the idea more complex, and perhaps not as useful - but it still can work, and lacking a lot of abstract mathematics "sector models" are one of the best ways I know of of dealing with curvature beyond the curvature of two dimensional surfaces.

 

[A.T.]

That curvature is not involved in determining the internal forces required to hold the various pieces of a static object static.

The forces required to hold all parts of object static relative to the big mass can all be applied externally, so there are no internal forces required for this anymore.

Then, the only internal forces that remain, are those due to the mismatch between the objects flat intrinsic geometry and the curved local spatial geometry.