I Euclidean geometry and gravity

[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.

 

[PeterDonis]

The forces required to hold all parts of object static relative to the big mass can all be applied externally

Really? You can externally apply a force to an atom in the center of the object?

You can't apply an external force to a part of an object that's not exposed to the external environment.

there are no internal forces required for this anymore.

Yes, there are. See above.

What is true is that, in special cases where you already know the worldlines of the parts of the object--which is true for the case we're discussing here (and also for the case of an accelerated object in flat spacetime, that I mention below)--the worldlines already dictate the proper accelerations, and hence the forces. But that doesn't mean there aren't internal forces required; it just means you have a simpler way of determining what they are. They're still internal for any part of the object that's not exposed to the external environment.

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

Please show your work: show the math that demonstrates this. Or give a reference that does so.

Before you even try to do that, you might want to consider this: an object being accelerated in flat spacetime, in a flat spatial geometry (spacelike slices of constant time in Rindler coordinates are flat), still requires proper acceleration (and hence force applied) to vary within the object.

 

[A.T.]

You can't apply an external force to a part of an object that's not exposed to the external environment.

For the point I make you don't need to go down to every atom. You can also assume the object is a truss structure with easy access to apply the support forces externally to each small part of the structure

If you build that structure in the curved space region near the big mass, then you can make the stresses arbitrarily small, by applying the right external support forces.

But if you build it in flat space, and then place it in curved space near the big mass, then no matter what external support forces you apply, you will never get rid of the stresses due to the mismatch between the objects flat intrinsic geometry and the curved local spatial geometry.

 

[PeterDonis]

You can also assume the object is a truss structure with easy access to apply the support forces externally to each small part of the structure

The OP specified a piece of paper, not a truss structure.

Also, your claims, if they are correct, should apply to any object, not just a truss structure--i.e., they should work just as well for an object where you do have to have internal forces to hold it in place because much of it isn't exposed to the external environment.

If you build that structure in the curved space region near the big mass, then you can make the stresses arbitrarily small, by applying the right external support forces.

No, you can't, because the proper acceleration of each part of the structure, which determines the net force that has to be applied to it, is determined by the path curvature of its worldlines, which is not "arbitrarily small".

the stresses due to the mismatch between the objects flat intrinsic geometry and the curved local spatial geometry.

Still waiting for you to show the math for this. Note that, as I said above, the actual proper acceleration of each element of the structure, which determines the net force applied to it, is determined by the path curvature of its worldline. That is not the same as the curvature of the local spatial geometry.

 

[A.T.]

No, you can't, because the proper acceleration of each part of the structure, which determines the net force that has to be applied to it, is determined by the path curvature of its worldlines, which is not "arbitrarily small".

Consider two separate objects held static against gravity by external forces. Now you connect them with a beam of negligible mass. The stress in that beam can obviously be zero, because it doesn't make a difference if it's there or not. You can continue this with many such objects to build a truss structure where all the beams are stress free.

But if you build it in flat space, and bring it into curved space, you will never get the beams stress free, no matter what forces you apply to the nodes.

 

[PeterDonis]

Consider two separate objects held static against gravity by external forces. Now you connect them with a beam of negligible mass. The stress in that beam can obviously be zero

A beam of "negligible mass" might as well not be there; you're only getting the result that the stress in it is zero by saying it has no substance.

A beam made of anything that actually exists will have nonzero stress in it, because the proper acceleration of each of its parts will be nonzero. At least, that will be the case in the scenario we're discussing in this thread. But that's not the scenario you appear to be implicitly assuming. See below.

But if you build it in flat space, and bring it into curved space, you will never get the beams stress free, no matter what forces you apply to the nodes.

Here you're implicitly talking about a different scenario from the one we're discussing in this thread. You're assuming we build the object "in flat space" (which should be "flat spacetime") in free fall. That's the only way to have zero internal stress everywhere in it.

Then you're assuming that we "bring it into curved space" (which should be curved spacetime) and still have its center of mass in free fall. That's the only way to have the only internal stresses in the object be due to the curvature of the spacetime. The proper acceleration of the center of mass will still be zero. But for the object to hold itself together against tidal gravity, there must be nonzero internal stresses elsewhere in the object.

Even in this scenario, you still are incorrectly claiming that "space curvature" is what's determining the internal stresses. What actually determines them is tidal gravity (i.e., spacetime curvature), plus the properties of the material. Those are the things that determine what worldlines the parts of the object that aren't the center of mass will follow (the center of mass worldline is determined in this scenario by the geodesic equation and the initial conditions). Neither of those things is the same as "space curvature".

But on top of that, in the actual scenario we're discussing in this thread, the object is not in free fall. So the appropriate comparison with flat spacetime is not a freely falling object with zero internal stresses. It's an accelerated object, whose worldlines, in the simple case we're considering, will be Rindler worldlines, and which will have internal stresses due to the proper acceleration of each of the worldlines. Basically we're comparing an object in flat spacetime described by the Rindler congruence, with an object in Schwarzschild spacetime described by the timelike Killing congruence of that spacetime. And "space curvature" is not a correct description of how the internal stresses inside the object will be different in those two scenarios. If you disagree, as I've already asked you more than once, you need to show the math.

 

[A.T.]

A beam of "negligible mass" might as well not be there; you're only getting the result that the stress in it is zero by saying it has no substance.

Yes it's an idealization. You can say the stresses are negligible instead of zero.

Then you're assuming that we "bring it into curved space" (which should be curved spacetime) and still have its center of mass in free fall.

Not free fall. You bring the object built far away from the big mass close to the big mass, and keep it hovering there by supporting the nodes against gravity.

Just like you did with the other object, that was built close to the mass. But this time you cannot make the stresses in the beams negligible, even if their masses are negligible.

 

[PeterDonis]

Not free fall. You bring the object built far away from the big mass close to the big mass, and keep it hovering there by supporting the nodes against gravity.

Meaning, you build the object in an accelerating spaceship or something like it in flat spacetime, and then bring it into curved spacetime and have it hover?

Or you build the object in free fall in flat spacetime, and then bring it into curved spacetime and have it hover?

The first comparison at least allows you to try to separate the effects of the proper acceleration required to make the object hover, from the effects of going from flat to curved spacetime. (Though doing so is not as simple as it might look.)

The second does not, and is therefore useless if what you're trying to analyze is the effects of going from flat to curved spacetime.

 

[PeterDonis]

Just like you did with the other object, that was built close to the mass. But this time you cannot make the stresses in the beams negligible, even if their masses are negligible.

If you build the object in curved spacetime, hovering against gravity, you can't make the stresses in it negligible, because it's under proper acceleration to hover. You keep ignoring this crucial point.

 

[PeterDonis]

Yes it's an idealization. You can say the stresses are negligible instead of zero.

Zero or negligible, it's still pointless. You're making it negligible by making the beam itself negligible. Which means we can just neglect it and talk about things that actually matter.

What you can't do, no matter how negligible you make the beam's mass, is make the beam's proper acceleration negligible if it's hovering against gravity. Again, you keep ignoring this crucial point.

 

[Bosko]

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

You pick three points and connect them with ( the segments of ) three "straight lines".
What is the ( segment of a ) "straight line" between two points in this static space?
- a ray of light?
- the shortest path?
- something else?

and 'measure' the angles of the triangle, will they add to 180 degrees?

I think a good and fun way to one better understand your question is to try to find a problem, a mistake, a counterexample ... in the following statement :

If the black hole is inside the triangle then the sum of the angles is greater than 180 and if it is outside then it is less than 180 degrees.

 

[PeterDonis]

What is the ( segment of a ) "straight line" between two points in this static space?

A spacelike geodesic in the Flamm paraboloid.

Note that such a curve is not one that either an ordinary object or a light ray will travel in spacetime. That's one of the reasons why the "space curvature" of the Flamm paraboloid is not what determines the internal stresses in a hovering object.

 

[PeterDonis]

If the black hole is inside the triangle then the sum of the angles is greater than 180 and if it is outside then it is less than 180 degrees.

Are you claiming that this statement is true?

 

[Bosko]

Are you claiming that this statement is true?

Are you claiming that is not?
I claiming that is funny :-)

 

[Jaime Rudas]

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?

Regarding this, I have the following question: Is it possible to embed a flat (Euclidean) two-dimensional surface in a curved, three-dimensional space?

 

[PeterDonis]

Are you claiming that is not?

I'm wondering why you said this:

I think a good and fun way to one better understand your question is to try to find a problem, a mistake, a counterexample ... in the following statement

 

[PAllen]

Regarding this, I have the following question: Is it possible to embed a flat (Euclidean) two-dimensional surface in a curved, three-dimensional space?

Yes.

 

[A.T.]

Zero or negligible, it's still pointless. You're making it negligible by making the beam itself negligible. Which means we can just neglect it and talk about things that actually matter.

I don't follow. For example, we often assume that our test object has negligible mass compared to the big spherical mass, so its effect on spacetime geometry is negligible.

We also often use idealizations like a rope of negligible mass, which still can be under tension or not. The idealized beams have a similar purpose here.

no matter how negligible you make the beam's mass, is make the beam's proper acceleration negligible if it's hovering against gravity.

But it makes the force required for that proper acceleration negligible.

 

[A.T.]

If the black hole is inside the triangle then the sum of the angles is greater than 180 and if it is outside then it is less than 180 degrees.

Yes, it's similar to triangles on a cone. If they include the cone apex, the angle sum is greater than 180°. If not it's just 180°, because a cone, sans apex, is flat. But Flamm's Paroboloid has negative curvature, so if the mass is outside the triangle it's less than 180° there.

 

[PeterDonis]

we often assume that our test object has negligible mass compared to the big spherical mass, so its effect on spacetime geometry is negligible.

We're not talking about having the object affect the spacetime geometry; we're assuming the spacetime geometry is fixed.

We're talking about the stresses in the object itself, due to the proper acceleration required to hold it static. Assuming "negligible mass" assumes away the very stresses we're supposed to be analyzing.

it makes the force required for that proper acceleration negligible.

By making the object itself not exist for all practical purposes. Which is pointless if you're trying to analyze the stresses required to hold an object static--the object has to exist and not be negligible for such an analysis to be done at all.

 

[A.T.]

We're not talking about having the object affect the spacetime geometry;

Which already makes it an idealization. Yet somehow you don't say that this makes the object non-existent and that therefore its pointless to even talk about it.

We're talking about the stresses in the object itself, due to the proper acceleration required to hold it static.

Who is 'we' here? Only you are obsessing about this. A non-issue, which in an idealization, can be made negligible by assuming smoothly distributed support forces and/or negligible mass of the object itself.

... if you're trying to analyze the stresses required to hold an object static

I'm not trying to do that. I'm talking about the stresses that remain after we have eliminated the stresses required to hold an object static.

 

[PeterDonis]

Which already makes it an idealization. Yet somehow you don't say that this makes the object non-existent and that therefore its pointless to even talk about it.

Of course not. We're not disputing that treating the object as a test object as far as determining the spacetime geometry is concerned is a valid idealization.

What you don't seem to grasp is that treating the object as having negligible mass as far as the spacetime geometry is concerned, is not the same thing as treating the object has having negligible mass as far as its own internal stresses are concerned. The latter is not a valid idealization if the whole point of doing the analysis in the first place is to determine the internal stresses in the object.

Who is 'we' here?

Um, this entire discussion, which is about how to analyze the internal stresses in an object that's being held static in a gravitational field?

If you are discussing something else, I have no idea what it is or how it's relevant to this thread. Unless it's this:

I'm not trying to do that. I'm talking about the stresses that remain after we have eliminated the stresses required to hold an object static.

And how are you going to determine that? The congruence of worldlines describing the object is one congruence, with one kinematic decomposition that determines its internal stresses. There's no way to pick out just part of that and say that's the part that's required to hold the object static, and the rest is what you're interested in, just from that congruence itself.

So the only way to even approach what you say you're interested in here is to do some kind of comparison with a different congruence, in flat spacetime, that "corresponds" in some way to the one describing the object being held static in Schwarzschild spacetime. I said in an earlier post that to me, the Rindler congruence is the obvious one to use for such a comparison. I'm not clear on whether you agree with that, or whether you think an inertial congruence in flat spacetime (i.e., an object in free fall) is the one to use--which I don't, for reasons given in an earlier post.

And in any case, I have no idea why you think idealized massless beams have anything at all to do with any of this. So at this point I'm very confused about what point you're trying to make and why you think it's relevant.

 

[PAllen]

I'm not trying to do that. I'm talking about the stresses that remain after we have eliminated the stresses required to hold an object static.

The only such stresses on a rigid object, whatever its history, are those of tidal gravity.

 

[PeterDonis]

The only such stresses on a rigid object, whatever its history, are those of tidal gravity.

Note that this implies a comparison with the corresponding accelerated object in flat spacetime--in the case under discussion in this thread, it's a comparison between the Killing congruence in Schwarzschild spacetime, and the Rindler congruence.

(Even then there are subtleties associated with what counts as "the corresponding accelerated object".)

 

[PAllen]

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:

But the real content of the theorem is what cases are possible. I have not been able to access discussions of this. The only things I find are similar to the following, from the 1960s, and I can't see the real content. I have not seen any more recent texts or reviews that discuss generalizations of the theorem

https://www.jstor.org/stable/2415272

For example, most relevant for the case at hand, are the following, which are true in SR:

- given an arbitrary enclosed volume (simply connected possibly required) in an arbitrary hypersurface, and an arbitrary world line specified for any one point of this volume, there exists a (typically) unique rigid congruence containing the given world line and also world lines for all other points of the volume. This congruence is irrotational. The only caveat is on 'size' of the body - depending on the proper accelerations of the chosen world line, there are limits on the size of congruence because to maintain rigid configuration, other congruence elements would approach infinite proper acceleration.

- This congruence may be derived by simply building Fermi Normal coordinates from the chosen world line (generalized for arbitrary origin world line, rather than geodesic, but NOT generalized for rotation - that is, the FN basis must be Fermi-Walker transported along the world line). Then, lines of constant FN position define the required congruence, restricted to those that intersect the desired defining volume.

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).

This is an example of what I have not been able to access any discussion of. For example, suppose the feature of an FN congruence being rigid remains true. Is it necessarily true in GR that the hypersurfaces orthogonal to an FN congruence in GR are flat? I suspect this is not true.

(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.)

The GP constant position congruence is the same as the SC congruence. It has no expansion and is rigid. The only thing that differs in GP coordinates is that the congruence is sliced differently for surfaces of constant time. These slices are definitely not orthogonal to the congruence, but they have nothing to do with the properties of the congruence. The congruence remains hypersurface orthogonal (meaning there exists a non-intersecting family of slices orthogonal to the congruence), irrespective of simultaneity choice for coordinate construction.