I Euclidean geometry and gravity

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    Euclidean geometry
  • #51
Bosko said:
Are you claiming that is not?
I'm wondering why you said this:

Bosko said:
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
 
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  • #52
Jaime Rudas said:
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.
 
  • #53
PeterDonis said:
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.

PeterDonis said:
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.
 
  • #54
Bosko said:
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.
 
  • #55
A.T. said:
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.

A.T. said:
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.
 
  • #56
PeterDonis said:
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.
PeterDonis said:
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.
PeterDonis said:
... 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.
 
  • #57
A.T. said:
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.

A.T. said:
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:

A.T. said:
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.
 
  • #58
A.T. said:
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.
 
  • #59
PAllen said:
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".)
 
  • #60
PeterDonis said:
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.
PeterDonis said:
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.

PeterDonis said:
(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.
 
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  • #61
I think it is important to recall points made in some other threads about the limitations imposed by Herglotz-Noether : they apply to a 3 parameter congruence, i.e. one with nonzero volume in a spatial slice. For a strictly 2-d surface, you can find all sorts of motion plus rotation that preserve rigid positioning within the 2-surface, because it can freely change shape (embedding) within the containing spacetime, while maintaining intrinsic metric rigidity. In effect, it can behave like a non-stretchable fabric, allowing many 'rigid' motions not possible for an object with thickness. Once you add nominal thickness, any bending will necessarily produce expansion and compression.
 
  • #62
What I know and don't know as of this point in the discussion:

1) You can have rigid body (congruence) of parametrized by varying r, ##\theta##, and a small ##\phi## thickness. This can exist at any place outside the horizon of an SC BH, and be of any size whatsoever (as long as it is all outside the horizon).

2) If you inquire as to physical plausibility, the limiting factors are simply the ability to resist the compressive stress of being held in position radially, as well as the tidal gravity stress. Both of these are encoded in the proper acceleration of the congruence world lines.

What I still don't know is what happens if you consider the motion of one such 'body' very far from the BH, radially, slowly, until it is relatively close to the BH horizon. Is it possible for this to be a rigid motion?

The obvious congruence to test would be the Fermi-Normal congruence built from a world line stationary, far away until t1, then slowly (very small rate of change of proper acceleration) decreases in r to e.g. 2 times the SC radius, then remains stationary at the new radius.

A first question to ask is whether the 'stationary' congruence in (1) above is Fermi-Normal. If not, then there are major parts of SR rigid motion theory that don't carry over to GR>
 
  • #63
PAllen said:
the real content of the theorem is what cases are possible. I have not been able to access discussions of this.
I'll see if I can find any references in the previous threads we had here. At least one of them, IIRC, dealt with a scenario in Schwarzschild spacetime.
 
  • #64
PAllen said:
The GP constant position congruence is the same as the SC congruence.
Yes, agreed. But...

PAllen said:
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
They're not orthogonal to the static congruence (the Killing congruence), yes. They are orthogonal to the free-falling Painleve congruence. I was using the latter orthogonality to illustrate how a congruence can be hypersurface orthogonal, with the orthogonal hypersurfaces being flat, but still not satisfy the H-N theorem (because the congruence has nonzero expansion).

PAllen said:
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.
Yes, agreed.
 
  • #65
A.T. said:
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.

My intuition leads to a similar conclusion, but I think we would need to clarify what a "beam" is in this context. I am not sure I have a solid enough definition of what mathematical properties a beam has to make a mathematical argument about why this should be so. The beam being elastic (so that a strain causes a stress) is a start, but the idea of going from a curved space to a flat space is sort of fuzzy. My thinking is that the beam consists of atoms, which attract and repel each other based on their distances apart, but that's not a formal mathematical definition, and it's still rather fuzzy about what we are actually doing to this arrangment of presumed atoms to make it transition from a curved space to a flat one. Are we moving the colleciton of atoms / beam? If so, does it matter how we move it? Can we say for sure that if we move it "slowly" the details of exactly what path we move it and where we apply the force to make it move don't matter?

I suppose my best attempt at justification would be to say that there is probably no born-rigid (in the restrictive first sense) that moves the beam, which implies that the distance between atoms is disturbed. But this still seems far from rigorous.
 
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  • #66
PAllen said:
What I still don't know is what happens if you consider the motion of one such 'body' very far from the BH, radially, slowly, until it is relatively close to the BH horizon. Is it possible for this to be a rigid motion?

pervect said:
I suppose my best attempt at justification would be to say that there is probably no born-rigid (in the restrictive first sense) that moves the beam, which implies that the distance between atoms is disturbed. But this still seems far from rigorous.
I agree this is the fundamental question. I lean in the same direction as you (@pervect ). [This would mean that Herglotz-Noether extended to GR is substantively different in that a whole class of possible rigid motions in SR don't typically exist in GR].

However, I think looking at this in terms of stresses is not right. Consider that case of spinning up a disc in SR. You can write a plausible congruence representing the motion of 'atoms' from a state of constant rotation, then increasing rotation, then constant rotation at a higher rate. The initial stage and final stage will be rigid, the transition will not be rigid because it can't be. It is not as if there are extreme stresses on the object, or that it somehow 'could' be rigid if strong enough. Each world line will have a smaller proper acceleration during the initial state, increasing during the transition, and then constant and larger at the end. These proper accelerations are what determine the stresses in the material. The change in mutual relative position of points is just something that happens because it must. These geometric changes will affect underlying fields as well, so I don't think they are associated with real stresses. Spinning up a disc is clearly possible and has no dramatic consequences unless you spin it up to the point where it flies apart.

So assuming the rigid nominally thick paper, oriented in an equatorial plane, is moved radially from very far away to near the BH horizon, the position of atoms will change with respect to each other during the the transition (because they must). At the beginning and end, it will be rigid (equilibrium). I claim there will be no unusual stresses during the transition - just those associated with increasing compressive stress from resisting free fall, and additional stress from increasing tidal gravity. At the end, it will be no different from a paper constructed in the higher curvature region.
 
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  • #67
PAllen said:
a whole class of possible rigid motions in SR don't typically exist in GR
In a generic curved spacetime, yes, because the two types of rigid motions allowed by the H-N theorem each depend on the spacetime having some kind of symmetry. In Schwarzschild spacetime, there is at least one such symmetry, the timelike KVF.
 
  • #68
Another thought on this is to emphasize that spatial geometry is simply a choice of simultaneity, with the same conventionality as in any other case. The paper near the BH would be measured as having Flamm paraboloid geometry by a collection of stationary instruments. On the other hand, measured by a 'rain' of instruments at free fall from infinity, it would appear to be Euclidean flat. The invariant statement (assuming post #66 and others that have proposed the same are correct) is that the motion described cannot be rigid.
 
  • #69
PAllen said:
The only such stresses on a rigid object, whatever its history, are those of tidal gravity.
If the external support forces are distributed smoothly throughout the body and locally match the force need to hover, then tidal gravity is handled as well, and the stresses can, in principle, approach zero.

But I disagree with the 'whatever its history' part. I think it does matter if:
a) the object was constructed hovering stress-free near the mass, using the above forces
b) the object was constructed stress-free far away, and then moved to hover near the mass, using the above forces

I think that in case b) it's not possible to make the stresses negligible, even in principle. And the reason is this:

PAllen said:
The paper near the BH would be measured as having Flamm paraboloid geometry by a collection of stationary instrumen

The geometry of the paper, measured by instruments at rest to the paper, is the paper's intrinsic geometry. Changing the paper's intrinsic geometry, from Euclidean to Flamm, will introduce stresses, that cannot be removed by any smoothly applied external forces.
 
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  • #70
A.T. said:
If the external support forces are distributed smoothly throughout the body and locally match the force need to hover, then tidal gravity is handled as well, and the stresses can, in principle, approach zero.
In my view, this is not a sensible approach. What you have is a collection of independent rockets, not a body. In a body, in any normal context, stresses come from body’s resistance to external forces. As soon as you replace your independent rockets with inter atomic forces, you have stress.
A.T. said:
But I disagree with the 'whatever its history' part. I think it does matter if:
a) the object was constructed hovering stress-free near the mass, using the above forces
b) the object was constructed stress-free far away, and then moved to hover near the mass, using the above forces

I think that in case b) it's not possible to make the stresses negligible, even in principle.
I disagree with this. Once the body is in its new position, the only thing needed to hold each element in place is a force to produce the proper acceleration of the element world line. There is no place for a memory effect. You try to get around this by proposing massless struts between the body elements with properties you specify to produce your conclusion, but they are irrelevant. Really, if you add them, they have to part of the same congruence with the same properties as the rest of the body, and then your argument falls apart.
 
  • #71
PAllen said:
There is no place for a memory effect.
I think there is.

Every beam/strut has a relaxed proper length, and the current stress in it depends on how the current proper length deviates from the stress-free proper length. The combination of all current proper beam lengths defines the current proper geometry of a static object.

If the object is built stress-free far away from any mass, then its beams are only all stress-free if the object's proper geometry is Euclidean.

If you then place that object hovering near a big mass, the object's proper geometry cannot stay Euclidean. So the beams cannot all be stress-free, no matter how smoothly you support the structure against gravity.
 
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  • #72
A.T. said:
If the external support forces are distributed smoothly throughout the body and locally match the force need to hover, then tidal gravity is handled as well, and the stresses can, in principle, approach zero.
Yes, but are not the tidal gravity forces just those originating from the difference between flat and curved spacetime ?
 
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