How does GR handle metric transition for a spherical mass shell?

[PAllen]

Taking a large, hollow sphere, and counting the number of smaller spheres you can pack into it, to measure it's volume, would (at least in principle) give you a measure of spatial curvature. But it wouldn't give a measure of space-time curvature, it would measure the spatial curvature of some particular spatial slice.

I think that's what was wanted, though I haven't been following in detail and the thread is too long to try and catch up.

Another minor issue is that the Riemann of a plane only has 1 component, but the Riemann of a three-space should have 3. So the circle-packing tells us as much as we can know about the curvature of a plane, but sphere-packing doesn't tell us everything about the curvature of some particular spatial slice.

Well I was interested in something intrinsic. I see no reason you can't construct a non-euclidean spacelike 3-surface in Minkowski flat spacetime. What would be the physical significance of that? Whereas, with spacetime curvature present, while you can generally find a flat 2-surface, you cannot find a flat 3-surface (a while back I opened a thread on embedding like this, and determined this based on number of coordinate conditions that can be imposed on a metric). So, to have real meaning, the condition to look for isn't ability to find a curved spatial slice; instead, it is inability to find a flat one.

Separately, I don't know if every 3-surface with non-vanishing Riemann tensor must deviate from the Euclidean sphere area/volume ratio. Have you determined that this is true?

 

[pervect]

Kinda sorta. You bring up a good point, we've been using the word "curvature" without always being clear about what kind.

When I said that tidal gravity is the same as curvature, I meant specifically *spacetime* curvature. (I said so explicitly at least once.)

There is also, as you say, the curvature of a spatial slice. That, of course, depends on how you cut the slice, so to speak, out of spacetime. Also, as you note, the sphere packing, which measures what I've been calling the K factor, is not a complete measure even of the spatial curvature. (Also, as I've noted, the measurement you describe samples the K factor over a range of radial coordinates, or sphere areas, so it's more complicated than just measuring the K factor between two spheres that are very close together. I'm trying to stick to the "local" case, where K is effectively constant, until we get that sorted out, before bringing in variation in K.)

(Another minor point is that what you described is the *intrinsic* curvature of the spatial slice; there is also the extrinsic curvature of the slice, which is something else again.)

I agree with everything you wrote, though I'm still not sure which one of the various aspects of curvature is of interest. I suspect that the idea is just to overall describe curvature.

I don't regard extrinsic curvature as being physically very interesting, because we'd have to stand outside of space-time to do define it. So I'm mostly interested in intrinsic curvature. I suppose that the extrinsic curvature might be of some use if you're doing ADM stuff, but it's outside the scope of my current interests.

As regards intrinsic space-time curvature, I'd go with the perhaps overly mathematical approach that says that it's completely defined by the Riemann tensor, and that tidal forces are described by one part of the Riemann tensor, the part that's sometimes called the electro-gravitic part in the Bel decomposition.

There are two other parts of the Bel decomposition in the 4d spacetime of GR. One of them is the topo-gravitic part. This describes the purely spatial part of the curvature.

The remaining part is the magneto-gravitic part, that describes frame dragging effects.

So my take is that tidal gravity is part of the mathematical entity (the Riemann) that completely describes all the aspects of space-time, curvature, but it's not the complete story.

Though I think that if you have the tidal forces for observers in all state of motion (rather than just the tidal forces for one observer), you can recover the Riemann, just as you can do it from a set of multiple sectional curvatures of planar slices, though I couldn't write down exactly how to perform either operation.

If we focus on the deviation between a reference geodesic and nearby geodesics (via the geodesic deviation equation), we can neatly categorize the various parts of the Bel decompositon as follows:

The geodesic deviation (the relative acceleration between nearby geodesics) will depend on both the spatial separation (and be proportional to it), and will also depend on the relative velocity (said velocity being measured in the fermi-normal frame of the reference geodesic).

The deviation turns out to be quadratic with respect to the velocity. The terms independent of velocity, presnet at zero velocity, will give rise to the electro-gravitic part of the tensor, and are described by the tidal forces.

The parts that are proportional to velocity describe the magnetic part.

The parts that are proportional to velocity squared are due to the spatial curvature (i.e. the topogravitic part of the tensor). They're rather analogous to the v^2/R type forces that an object moving in a circular path of radius R with velocity v experiences.

 

[PeterDonis]

Whereas, with spacetime curvature present, while you can generally find a flat 2-surface, you cannot find a flat 3-surface (a while back I opened a thread on embedding like this, and determined this based on number of coordinate conditions that can be imposed on a metric).

You can in certain special cases. One has been mentioned in this thread: Painleve coordinates in Schwarzschild spacetime; the slices of constant Painleve time are flat. Another is FRW spacetime with k = 0; spacetime as a whole is curved but the spatial slices of constant "comoving" time are flat. I don't know of any other such cases off the top of my head.

So, to have real meaning, the condition to look for isn't ability to find a curved spatial slice; instead, it is inability to find a flat one.

Provided it isn't one of the special cases.

 

[PeterDonis]

I don't regard extrinsic curvature as being physically very interesting, because we'd have to stand outside of space-time to do define it.

Not the extrinsic 3-curvature of a spatial slice taken out of a 4-d spacetime. But I agree extrinsic curvature isn't very interesting compared to intrinsic; certainly not for the topic of this thread.

As regards intrinsic space-time curvature, I'd go with the perhaps overly mathematical approach that says that it's completely defined by the Riemann tensor,

Agreed.

and that tidal forces are described by one part of the Riemann tensor, the part that's sometimes called the electro-gravitic part in the Bel decomposition.

There are two other parts of the Bel decomposition in the 4d spacetime of GR. One of them is the topo-gravitic part. This describes the purely spatial part of the curvature.

The remaining part is the magneto-gravitic part, that describes frame dragging effects.

So my take is that tidal gravity is part of the mathematical entity (the Riemann) that completely describes all the aspects of space-time, curvature, but it's not the complete story.

Well, the term "tidal gravity" may be a bit ambiguous. I posted earlier the definition I was using: *anything* that causes geodesic deviation. All parts of the Riemann tensor describe this in some form; you describe how later in your post.

I recognize now that my terminology may be non-standard. I first saw it in Kip Thorne's Black Holes and Time Warps, and it made sense to me. But my arguments in this thread are essentially unchanged if "tidal gravity" is taken to describe just the electrogravitic part of the full curvature tensor, as you say. It's still true that what I am calling the K factor is not the same as tidal gravity, and is not equivalent to it, nor is it equivalent to curvature in full.

Though I think that if you have the tidal forces for observers in all state of motion (rather than just the tidal forces for one observer), you can recover the Riemann, just as you can do it from a set of multiple sectional curvatures of planar slices, though I couldn't write down exactly how to perform either operation.

I think this is true, yes.

 

[PAllen]

You can in certain special cases. One has been mentioned in this thread: Painleve coordinates in Schwarzschild spacetime; the slices of constant Painleve time are flat. Another is FRW spacetime with k = 0; spacetime as a whole is curved but the spatial slices of constant "comoving" time are flat. I don't know of any other such cases off the top of my head.
Provided it isn't one of the special cases.

Ok, interesting. I only looked at the general case. I see you are right. So I would interpret that as saying that your K can be considered as an artifact of the 'natural' way stationary (not inertial) observers set up simultaneity.

Also interesting, is that G-P coordinates have the feature (just like SC) that (t,t) metric component becomes spacelike inside the horizon. It looks to me, then, that inside the horizon, where the metric is not static, you do not have flat spatial slices. I might guess that you can't, in this region.

 

[Passionflower]

You can in certain special cases. One has been mentioned in this thread: Painleve coordinates in Schwarzschild spacetime; the slices of constant Painleve time are flat.

Yes an observer traveling radially at escape velocity observes his space as flat in that the distance between two r-coordinate values is exactly the difference. Other observers, who are also on a radial path or stationary, observe this distance differently. One can obtain this distance by applying the Lorentz factor (local speed wrt a free falling at escape velocity observer) before integration between these two r-coordinates.

 

[PeterDonis]

Ok, interesting. I only looked at the general case. I see you are right. So I would interpret that as saying that your K can be considered as an artifact of the 'natural' way stationary (not inertial) observers set up simultaneity.

Yes, I agree.

Also interesting, is that G-P coordinates have the feature (just like SC) that (t,t) metric component becomes spacelike inside the horizon. It looks to me, then, that inside the horizon, where the metric is not static, you do not have flat spatial slices. I might guess that you can't, in this region.

Actually, you still can. G-P coordinates are very weird inside the horizon; *all four* coordinates are spacelike! That means that, even though the time coordinate T is spacelike, the surfaces of constant T are as well. I agree this is extremely odd, but it's true.

To be more precise in stating what this means: inside the horizon, the vector $\frac{\partial}{\partial T}$ corresponding to Painleve time T *and* the vector $\frac{\partial}{\partial r}$ corresponding to the radial coordinate r (which is defined the same for Painleve as for Schwarzschild coordinates) are spacelike. Since the angular vectors are also spacelike, you can form a spacelike 3-surface using the three "spatial" coordinates inside the horizon just like you can outside, and these 3-surfaces will be orthogonal to the worldline of an infalling "Painleve" observer. (Note that they are *not* orthogonal to the vector $\frac{\partial}{\partial T}$, obviously, because that vector is spacelike inside the horizon. But the 3-surfaces I've just defined aren't orthogonal to $\frac{\partial}{\partial T}$ *outside* the horizon either, because of the dTdr term in the Painleve metric.)

 

[PAllen]

Yes, I agree.



Actually, you still can. G-P coordinates are very weird inside the horizon; *all four* coordinates are spacelike! That means that, even though the time coordinate T is spacelike, the surfaces of constant T are as well. I agree this is extremely odd, but it's true.

To be more precise in stating what this means: inside the horizon, the vector $\frac{\partial}{\partial T}$ corresponding to Painleve time T *and* the vector $\frac{\partial}{\partial r}$ corresponding to the radial coordinate r (which is defined the same for Painleve as for Schwarzschild coordinates) are spacelike. Since the angular vectors are also spacelike, you can form a spacelike 3-surface using the three "spatial" coordinates inside the horizon just like you can outside, and these 3-surfaces will be orthogonal to the worldline of an infalling "Painleve" observer. (Note that they are *not* orthogonal to the vector $\frac{\partial}{\partial T}$, obviously, because that vector is spacelike inside the horizon. But the 3-surfaces I've just defined aren't orthogonal to $\frac{\partial}{\partial T}$ *outside* the horizon either, because of the dTdr term in the Painleve metric.)

Fascinating. After a little thought I agree. So I guess the SC geometry everywhere has too many symmetries to preclude construction of flat spatial slices.

 

[Q-reeus]

Originally Posted by Q-reeus:
"Unless you can prove that the marble filled hoop will *not* experience changed packing density (restraint = fixed marble count) in heading towards pointy end, you have to face the fact that locally observed effects are present. And one possible *interpretation* by a local flat-land observer, who can't discern curvature directly, is varying hoop size, or alternately, shrinking marbles. Stresses can't explain it, but effects normally put down to changing container size and/or marble size are there. All you have to do to end that argument, is what I asked above - can the hoop packing density/number be independent of surface curvature?"

Forget hoops, filled or otherwise. You cannot detect curvature in a 4-manifold with anything restricted to a 2-surface, in any orientation (anything you think you might detect this way will be a an embedding feature, similar to embedding a curved 2-sphere in flat 3-space). You need lots of measurements of a substantial spatial region, as in the examples Peter and I have been discussing (his volume examples, and Synge's many point, many measurement examples).

I'm putting it down to this: both you and Peter 'know' I probably have everything wrong, so what I actually argue tends to go in one ear and out the other, to be replaced with an image of what 'I probably meant', and then proceed to tear down that straw man. Point to anywhere, not just above but any previous entry, where I claimed what you think I did. I'm not so stupid as to imagine that running a hoop over an 'egg' allows one to detect 3-curvature at all. Have endlessly now referred to this as 2D analogue of 3D situation. Having said that, there *is* I think an in principle 2D manifestation of 3-curvature, which will be discussed in another posting.

 

[Q-reeus]

You could, I suppose, also replace the hoops with spheres, which seems more in spirit with the discussion.

And in #139:

"Taking a large, hollow sphere, and counting the number of smaller spheres you can pack into it, to measure it's volume, would (at least in principle) give you a measure of spatial curvature. But it wouldn't give a measure of space-time curvature, it would measure the spatial curvature of some particular spatial slice.
I think that's what was wanted, though I haven't been following in detail and the thread is too long to try and catch up."

Full marks from me for carefully reading what I have been arguing - yes got the drift admirably.

 

[Q-reeus]

Rather than spend time arguing over every matter of who said and meant what in recent posts, may I propose to look at this from a slightly different angle - literally. Back in #113 angles of triangles in positively curved spacetime was mentioned. Let's take it a bit further. In flat spacetime we have a flat equilateral triangular surface formed from a fine tiling of much smaller uniform equilateral triangles. There are no gaps - and all internal angles = 60⁰. Now move this composite triangle to a region of 'uniformly' positively curved spacetime. Do we all accept as a given that, no matter the orientation, included angles now add to more than 60⁰, -provided that is the sides of all triangles are geodesically 'straight' in that curved spacetime? And with that proviso that the angular departure is larger the larger the triangle? I will take it there is unanimously a yes and yes to the above. So here's the thing. If one constrains the outer triangle to have straight sides and therefore vertex angles significantly > 60⁰, it follows gaps must appear in the tiling, since the much smaller triangular tiles will have vertex angles insignificantly > 60⁰. This straight sides constraint necessarily means lowered surface packing density, or alternately filling with micro tiles between the mini tiles to maintain surface density.

Alternately, imposing the constraint that tiles pack uniformly, we must have that sides are no longer geodesically staight. In particular, the outer triangle must become puckered - inwardly bowing sides. 2D manifestation of 3-curvature. Now go the next step to 3D. Instead of flat tiles, build an outer tetrahedron from much smaller ones. Anyone doubt the same issues will manifest as for 2D, only proportionately greater in effect? Now head back to the s-sided cube arrangement between concentric shells. It didn't fit. But then it was supposed to be a perfect cube. Still wouldn't exactly fit given the above, but depending on assumed constraints, that certified in flat spacetime perfectly cubic cube has either puckered like a pin-cushion, or one notices a 'strange' addition to the vertex angles that 'shouldn't' be there. And 'strangely', one can fit more counting spheres inside said cube given the latter constraint (geodesically straight edges and flat faces).

Now all my yokel arguing here is apparently imposible, but seems inevitable to me. So where is this all falling apart?

 

[pervect]

Rather than spend time arguing over every matter of who said and meant what in recent posts, may I propose to look at this from a slightly different angle - literally. Back in #113 angles of triangles in curved spacetime was mentioned. Let's take it a bit further. In flat spacetime we have a flat equilateral triangular surface formed from a fine tiling of much smaller uniform equilateral triangles. There are no gaps - and all internal angles = 60⁰. Now move this composite triangle to a region of 'uniformly' curved spacetime. Do we all accept as a given that, no matter the orientation, included angles now add to more than 60⁰, -provided that is the sides of all triangles are geodesically 'straight' in that curved spacetime?

Well, to use one of the examples already mentioned, if you consider a FRW space-time is uniformly curved (offhand, I'm not sure of the definition of uniformly curved), the spatial part of the curvature, which is what you're measuring (assuming also that you use the cosmological time-slice) could be positive, negative, or zero, and hence the sum of the angles could be greater, equal, or less than 360.

 

[Q-reeus]

Well, to use one of the examples already mentioned, if you consider a FRW space-time is uniformly curved (offhand, I'm not sure of the definition of uniformly curved), the spatial part of the curvature, which is what you're measuring (assuming also that you use the cosmological time-slice) could be positive, negative, or zero, and hence the sum of the angles could be greater, equal, or less than 360.

Fair comment. I'm assuming we are relating this to the case of something similar to the original setup - stationary spherical mass and stationary observers. That would be positively curved spacetime I presume? In which angles add to more than 180⁰. [have since edited and added word 'positively' to curved - thanks]

 

[PAllen]

I'm putting it down to this: both you and Peter 'know' I probably have everything wrong, so what I actually argue tends to go in one ear and out the other, to be replaced with an image of what 'I probably meant', and then proceed to tear down that straw man. Point to anywhere, not just above but any previous entry, where I claimed what you think I did. I'm not so stupid as to imagine that running a hoop over an 'egg' allows one to detect 3-curvature at all. Have endlessly now referred to this as 2D analogue of 3D situation. Having said that, there *is* I think an in principle 2D manifestation of 3-curvature, which will be discussed in another posting.

Then it is not a straw men. Your last sentence is a mathematical falsehood, with no possible valid interpretation. That is, anything you can detect restricted to a 2-surface (or thin 3-analog of it) will tell you nothing about presence of absence of spatial curvature.

 

[PAllen]

Rather than spend time arguing over every matter of who said and meant what in recent posts, may I propose to look at this from a slightly different angle - literally. Back in #113 angles of triangles in positively curved spacetime was mentioned. Let's take it a bit further. In flat spacetime we have a flat equilateral triangular surface formed from a fine tiling of much smaller uniform equilateral triangles. There are no gaps - and all internal angles = 60⁰. Now move this composite triangle to a region of 'uniformly' positively curved spacetime. Do we all accept as a given that, no matter the orientation, included angles now add to more than 60⁰,

No, this is false. As long as you restrict to a 2-surface, and use physical procedures (which naturally pick out the flattest possible interpretation of local reality), you will detect no deviation. To make this plausible, go right back to my argument that you accused of being strawman - 2-sphere embedded in Euclidean 3-space. The 2-surface is curved, the 3-space is flat. As a result, anything you find restricted to a 2-surface cannot be distinguished from an embedding artifact; further, in seeking straightness, you will pick out a flat 2-surface and find no deviation.

The discussion I had with Peter went further than this. That if spacetime curvature is simple enough, then even fully general measures of 3-space curvature (spanning large regions) can show exact flatness. In each case Peter mentioned it is the 'most natural observer' - the inertial one, or the one who sees CMB radiation as isotropic - that detects no spatial at curvature at all, by any means. His K, in SC geometry, only applies to stationary observers, who are non-inertial. One can detect similar things for non-inertial observers in flat spacetime, so even K is related more to non-inertial motion than intrinsic spatial curvature.

 

[PeterDonis]

Now move this composite triangle to a region of 'uniformly' positively curved spacetime.

What is the triangle made of? The "triangle" is not a physical object; it's an abstraction. To "move" it, you have to realize the abstraction somehow. How?

Do we all accept as a given that, no matter the orientation, included angles now add to more than 60⁰, -provided that is the sides of all triangles are geodesically 'straight' in that curved spacetime?

How are the sides going to be kept geodesically straight? As an abstraction, yes, a "triangle" in a curved spacetime, with geodesically straight sides, will have its angles add to something different from the Euclidean sum--more if the spacetime is positively curved, less if it's negatively curved. That much is simple geometry. But to move from that to the actual physical behavior of a physical object, you have to bring in, well, physics.

Btw, in the examples we've been discussing, both in 2-D and 3-D, the "hoops" and 2-spheres are *not* necessarily composed of geodesics. In the 2-D case, the "hoops" are lines of latitude, which are not geodesics except for the equator. In the 3-D case, the 2-spheres themselves have geodesic tangent vectors, but the static objects sitting between them do not if you include the time dimension; static objects hovering above a gravitating body are accelerated.

If one constrains the outer triangle to have straight sides and therefore vertex angles significantly > 60⁰, it follows gaps must appear in the tiling, since the much smaller triangular tiles will have vertex angles insignificantly > 60⁰. This straight sides constraint necessarily means lowered surface packing density, or alternately filling with micro tiles between the mini tiles to maintain surface density.

How will you constrain the triangle to have straight sides (where I assume "straight" means "geodesic in the curved surface"--again, precise terminology really helps in these cases)? If you work it out, you will see that, if you imagine taking a triangle of actual, physical material and moving it from flat to curved space this way, constraining the sides as you suggest will necessarily cause stresses in the material. As soon as that happens, you can no longer use the material as a standard of distance to compare things with flat space, because in flat space the material was unstressed.

Alternately, imposing the constraint that tiles pack uniformly, we must have that sides are no longer geodesically staight. In particular, the outer triangle must become puckered - inwardly bowing sides. 2D manifestation of 3-curvature.

In the sense that you can't "wrap" a flat surface onto a curved surface without distorting it, yes--so if you don't distort the flat surface, it won't fit properly onto the curved surface. That's a quick way of summarizing what you've been saying. But the distortion is a *physical change* in the flat object, so once that object is distorted you can't use it to compare the flat with the curved surface. That's what you appear to be missing.

I won't bother commenting on the 3D case because it works out exactly the same.

Now all my yokel arguing here is apparently imposible, but seems inevitable to me. So where is this all falling apart?

None of what you say is incorrect. It just doesn't mean what you appear to think it means.

Added in edit, after seeing PAllen's post: Here's another way to say what I just said. I say none of what you have said is incorrect; but at the same time, PAllen is right when he says the comment he quoted is false. That's because I am saying you are correct as a matter of abstract geometry only, while he is saying you are incorrect in how you are trying to relate the geometry to the physics. *Both* of our comments are valid. Can you see why?

 

[Q-reeus]

Originally Posted by Q-reeus: "Now move this composite triangle to a region of 'uniformly' positively curved spacetime."

What is the triangle made of? The "triangle" is not a physical object; it's an abstraction. To "move" it, you have to realize the abstraction somehow. How?

From #151 "In flat spacetime we have a flat equilateral triangular surface formed from a fine tiling of much smaller uniform equilateral triangles."
Miss that bit - tiling, as in tiles? A composite, physical object.

Originally Posted by Q-reeus:
"Do we all accept as a given that, no matter the orientation, included angles now add to more than 60⁰, -provided that is the sides of all triangles are geodesically 'straight' in that curved spacetime?"

How are the sides going to be kept geodesically straight? As an abstraction, yes, a "triangle" in a curved spacetime, with geodesically straight sides, will have its angles add to something different from the Euclidean sum--more if the spacetime is positively curved, less if it's negatively curved. That much is simple geometry. But to move from that to the actual physical behavior of a physical object, you have to bring in, well, physics.

Like gaps appearing for instance. I gave two restraint examples - gaps appear, or sides curve. Not physics? Assuming some kind of elastic glue joining the triangle tiles, naturally, forcing the outer triangle's to be straight requires imposing stresses - evidence of living in curved spacetime. Gaps or curved sides or stresses. Not there, or not needed in flat spacetime. Can't see any physics at work; no makings of an in principle 'detector'?

Btw, in the examples we've been discussing, both in 2-D and 3-D, the "hoops" and 2-spheres are *not* necessarily composed of geodesics. In the 2-D case, the "hoops" are lines of latitude, which are not geodesics except for the equator. In the 3-D case, the 2-spheres themselves have geodesic tangent vectors, but the static objects sitting between them do not if you include the time dimension; static objects hovering above a gravitating body are accelerated.

Fine, but in present scenario, whether triangles have perfectly 'straight' sides is just a convenience - we are only really interested in detecting change. Straight sides just makes the job easier. Point is, there is detectable change - choose a restraint - see the effect.

Originally Posted by Q-reeus:
"If one constrains the outer triangle to have straight sides and therefore vertex angles significantly > 600, it follows gaps must appear in the tiling, since the much smaller triangular tiles will have vertex angles insignificantly > 60⁰. This straight sides constraint necessarily means lowered surface packing density, or alternately filling with micro tiles between the mini tiles to maintain surface density."

How will you constrain the triangle to have straight sides (where I assume "straight" means "geodesic in the curved surface"--again, precise terminology really helps in these cases)? If you work it out, you will see that, if you imagine taking a triangle of actual, physical material and moving it from flat to curved space this way, constraining the sides as you suggest will necessarily cause stresses in the material. As soon as that happens, you can no longer use the material as a standard of distance to compare things with flat space, because in flat space the material was unstressed.

Gone over that above. But another angle on it - what assumptions are you making in saying one must force the outer triangle to have straight sides? That assumes, as I guessed in answering above, that the tiles are glued together somehow. That's one configuration. Another could be tiles just sitting together snuggly but loosely on a plane. Moved en masse to curved space, why would they not simply drift apart slightly owing to curvature? There are any number of possible constraints - everyone I have considered results in detectable change - my idea of physics at work.

Originally Posted by Q-reeus:
"Alternately, imposing the constraint that tiles pack uniformly, we must have that sides are no longer geodesically staight. In particular, the outer triangle must become puckered - inwardly bowing sides. 2D manifestation of 3-curvature."

In the sense that you can't "wrap" a flat surface onto a curved surface without distorting it, yes--so if you don't distort the flat surface, it won't fit properly onto the curved surface. That's a quick way of summarizing what you've been saying. But the distortion is a *physical change* in the flat object, so once that object is distorted you can't use it to compare the flat with the curved surface. That's what you appear to be missing.

Huh!? We *want* physical change - that's the detection of being in curved spacetime. Things change. Please, no endless circling here.

 

[PeterDonis]

Warning: long post!

Huh!? We *want* physical change - that's the detection of being in curved spacetime. Things change. Please, no endless circling here.

Yes, things change. And the change means you can't use the things as standards of distance.

Here's another way of stating what I just said: the things you are saying change as a result of curvature, change because you are specifying that the objects have forces exerted on them in order to match the curvature. Those forces are external forces; they are not due to the curvature itself, at least not in the scenarios you are describing.

At the risk of piling on scenario after scenario, consider the following:

Take a piece of paper and try to wrap it around a globe. You can't; the paper will buckle. That's a physical realization of one scenario you were describing, the one where the sides of the triangles remain straight in the Euclidean sense; if we imagine the paper tesselated with tiny triangles, the paper behaves in such a way as to keep the triangles as flat Euclidean triangles. The paper may buckle along the edges between triangles, but the triangles themselves maintain their shape. So the paper can't possibly conform to any curved surface.

Now take a piece of flat rubber and wrap it around the same globe. You can do it, but only by exerting force on the rubber to deform it into the shape of the globe. This is a physical realization of another scenario you described, where we allow the triangles to adjust to the curvature of the surface; if we imagine the rubber tesselated by tiny triangles, the deformation of the rubber will deform the triangles too. But the deformation is not somehow magically caused by the globe's curvature; it's caused by us, exerting external force on the rubber to change its shape.

We can use the rubber to measure the K factor as follows: suppose that, when sitting on a flat Euclidean plane, the paper and the rubber have identical shape and area. We cut the paper into little tiny squares, and use them to tile the rubber. They fit exactly. Now we take the rubber and stretch it over the globe, and it deforms; we can see the deformation by watching the little squares and seeing that they no longer exactly tile the rubber; we might, as you say, see little gaps open up between the squares, assuming that the rubber is being deformed appropriately. But in order to know exactly how the squares will tile the rubber once it's wrapped around the globe, we have to specify *how*, exactly, the rubber is being wrapped.

Here's one way of specifying the wrapping. Take a circular disk of rubber instead of a square, and a circular piece of paper that, on a flat Euclidean plane, is exactly the same size. We cut the paper up into tiny shapes (triangles, squares, whatever you like) and verify that on a flat Euclidean plane, they tile the rubber exactly. Now take a sphere and draw a circle on it, centered on the North Pole, such that the circumference of the circle is exactly the same as the circumference of the rubber circle when the latter is sitting on the flat Euclidean plane. If we wrap the rubber circle onto the sphere in such a way that its circumference lies exactly on the circle we drew on the sphere, we will need to stretch the rubber; its area will now be larger than it was on the flat plane, and we can verify this by trying to tile it with the little pieces we cut out of the paper and seeing that there is still area left over. This is a manifestation of the K factor being greater than 1. But again, we had to physically stretch the rubber--exert external force on it--to get it to fit the designated area of the sphere.

(As I've noted before, the K factor will vary over the portion of the sphere being used for this experiment, but the average will be greater than 1; if we wanted to limit ourselves to a single value for K, we could cut a small annular ring of rubber on the flat plane, draw two circles on the sphere that matched its outer and inner circumferences, and stretch the rubber between them, and verify that the area increased by tiling with little pieces cut out of a paper annulus that exactly overlapped the rubber on a flat plane. We'll need this version for the 3-D case, as we'll see in a moment.)

All of the above is consistent with what you've been saying about how curvature of a surface manifests itself, and nobody has been disputing it. Now carry the analogy forward to the 3-D case. Here, since we're going to have a gravitating body in the center, we need to use the "annular" version of the scenario, as I just noted. We go to a region far away from all gravitating bodies, so spacetime in the vicinity is flat, and we cut ourselves a hollow sphere of rubber, with inner surface area A and outer surface area A + dA. We also cut a hollow sphere of metal (we use this instead of paper as our standard for material that will maintain its shape instead of deforming) with the same inner and outer surface area. Then we cut the metal up into little tiny pieces and verify that the pieces exactly fill the volume of the rubber. (This is a thought experiment, so assume that we can do this "tiling" in the 3-D case as we did in the 2-D case.)

Now take the hollow rubber sphere and place it around a gravitating body, in such a way that its inner and outer surface area are exactly the same as they were in the flat spacetime region. (Again, this is a thought experiment, so assume we can actually do this as we did in the 2-D case.) We will find that we need to physically stretch the rubber in order for the inner and outer surface areas to match up; there is "more volume" between the two surfaces than there was in the flat spacetime region. Again, we can verify this by trying to tile the volume of the rubber with the little metal pieces, and finding that we run short; there is volume left over when all the pieces have been used up. This excess volume is a physical measure of the K factor. (If dA is small enough compared to A, K is basically constant in this scenario, as it was in the "annular" 2-D case.)

The key fact, physically, is that we have to apply an external force to the rubber, stretch it, to make it fit when it's wrapped around a gravitating body, just as we had to apply an external force to the 2-D rubber circle to make it fit in the specified way around the sphere. So we *cannot* say that the rubber was stretched "by the K factor"--the K factor wasn't what applied the force. We did. Remember that we are assuming that tidal forces, and any other "forces", are negligible; the *only* effect we are considering is the K factor. (Again, this is easier to do because we are now dealing with a "local" scenario, where K is constant; if we covered a wider area, so K varied, we would also find it impossible, or at least a lot harder, to ignore or factor out the effects of tidal gravity. In the "local" case it's easy to set up the scenario so that tidal gravity is negligible, while the K factor itself is not; we just choose the mass M of the gravitating body and the radial coordinate r at which we evaluate K appropriately.)

You may ask, well, what about if we *don't* exert any force on the rubber? What then? In the 2-D case, the answer is that if we exert no force on the rubber, we can't make it conform to the curved surface at all. We specified one way of fitting the rubber to the sphere--so that its circumference remained the same (we checked that by drawing in advance a circle on the sphere centered on the North Pole, to use as a guide). We could specify another way--keep the total area of the rubber constant. But if we do that, then the rubber's circumference will be smaller, so again the rubber has to deform. There is *no* way to make the flat piece of rubber exactly fit to the curved surface without deforming it somehow.

For the spacetime case, it's a little more complicated, because if we take the hollow sphere of rubber we made in a flat spacetime region, and move it around a gravitating body, *something* will happen to it if we exert no force on it (other than the force needed to move it into place). But the general point still applies: there is no way to "fit" the object into a curved spacetime region without deforming it somehow. Since we can't actually wrap a full hollow sphere around a gravitating body without breaking it, consider a portion of the sphere covering a given solid angle, as PAllen proposed in a previous post. Suppose we have done all the work described earlier in a flat spacetime region: we have the inner surface area A and outer surface area A + dA measured (now not covering the total solid angle 4 pi of the sphere, but something less), and we have a piece of metal of the exact same shape and size in flat spacetime cut up into little pieces, which tile the volume of the rubber exactly in flat spacetime.

Now we slowly lower the rubber into place "hovering" over a gravitating body. We have to specify *some* constraint for where it ends up. Suppose we specify that the inner surface lies in a portion of a 2-sphere such that the solid angle it covers is exactly the same as what it covered in flat spacetime--i.e., the ratio of the inner surface area A of the rubber piece to the total area of the 2-sphere is the same as it was for the 2-sphere that the inner surface was cut out of in flat spacetime. That tells us that the inner surface of the 2-sphere is not deformed; but now we have to decide what to do with the outer surface. We have the same problem as we did in the 2-D case: if we specify that the outer surface area remains the same, relative to the inner surface area (i.e., the outer surface "fits" into a corresponding 2-sphere the same way the inner surface does--the outer surface is not deformed), then the rubber will have to stretch to match up, and there will be extra volume in the rubber, according to the K factor (which we can verify by trying to tile with the little pieces of metal and seeing that there is volume left over). Or, if we specify that we want the volume of the rubber to remain the same (i.e., we want to be able to tile it exactly with the little pieces of metal), then the outer surface area will "fit" to a smaller 2-sphere, so the shape of the outer surface will deform. And, of course, if we relax the constraint on the inner surface, there are even more ways we could deform the rubber; but there is *no* way to fit it without *some* kind of deformation.

As far as how the rubber would deform if we carefully exerted *no* force on it except to lower it into place, I think it would end up sitting with the inner surface not deformed and with the volume the same (meaning the outer surface would be deformed, as above). But even in this "constant volume" case, the shape of the rubber would still not be exactly the same. This is a manifestation of curvature, but I don't think you can attribute it to the K factor, because the K factor is measured by "volume excess" when I impose a particular constraint (having the inner and outer surfaces both the same shape as in flat spacetime, as above), and we're now talking about a *different* constraint, one where we keep the volume the same, since, as I said, that's the "natural" constraint that I think would hold if we carefully exerted no "extra" forces on the rubber.

Sorry for the long post, but this is not a simple question, and I wanted to try to get as much as possible out on the table for discussion. To summarize: the K factor is measured by "area excess" or "volume excess" under a particular physical constraint; with that constraint, K is a physical observable, but you have to be careful about interpreting what it means. More generally, curvature manifests itself as the inability to "fit" an object with a defined shape in a flat region, onto a curved region without changing its shape somehow. That change in shape requires external forces to be exerted on the object, which distort the object and make it unusable as a standard of "size" or "shape"; this is always true in the 2-D case we were discussing, and in the spacetime case, it is "almost always" true; there is *some* way to place the object without exerting any "extra" forces, and even in this case, the object's shape will not be the same as it was in flat spacetime, but its volume will be the same, as long as we can ignore tidal gravity.

 

[Q-reeus]

Warning: long post!
You are I deduce a speed typist - massive effort!

...Here's another way of stating what I just said: the things you are saying change as a result of curvature, change because you are specifying that the objects have forces exerted on them in order to match the curvature. Those forces are external forces; they are not due to the curvature itself, at least not in the scenarios you are describing.

Understand and agree with that almost completely. Sure the curvature is not exerting any direct forces, so in that sense yes is not physics. Had thought though we could use the 'deformity correction' externally applied forces as a gauge of 'geometry'. More on that below.

As far as how the rubber would deform if we carefully exerted *no* force on it except to lower it into place, I think it would end up sitting with the inner surface not deformed and with the volume the same (meaning the outer surface would be deformed, as above). But even in this "constant volume" case, the shape of the rubber would still not be exactly the same. This is a manifestation of curvature, but I don't think you can attribute it to the K factor, because the K factor is measured by "volume excess" when I impose a particular constraint (having the inner and outer surfaces both the same shape as in flat spacetime, as above), and we're now talking about a *different* constraint, one where we keep the volume the same, since, as I said, that's the "natural" constraint that I think would hold if we carefully exerted no "extra" forces on the rubber.

This is where something, perhaps much different than straight K, should still allow measurement of a kind. Take an equalateral triangle composed of rigid tubes joined by free-hinging joints. In this configuration, one should expect vertex angles will exceed 60 degrees as discussed before. If small triangular gussetts were initially glued to the apexes, owing to their reduced susceptibility to angular change, ought to partially restrain angular expansion of the much larger triangle tube joints. In other words, there should be some internal stresses - miniscule but in principle measurable. I think.

Sorry for the long post, but this is not a simple question, and I wanted to try to get as much as possible out on the table for discussion. To summarize: the K factor is measured by "area excess" or "volume excess" under a particular physical constraint; with that constraint, K is a physical observable, but you have to be careful about interpreting what it means. More generally, curvature manifests itself as the inability to "fit" an object with a defined shape in a flat region, onto a curved region without changing its shape somehow.

Many thanks Peter for going to all that fuss - I do now appreciate more just why K factor, relating to a global geometry restraint, requires an implied stretching to make sense. So an isolated, stress-free container (or sheet in 2D case) cannot exhibit that unless forced into an equivalent geometry - like the solid angle segment mentioned. OK then agree that my earlier subdividing of concentric shells argument is not appropriate. And the hoop/marbles thing is presumably analogous. To explain a null result, we must assume there is no effective sampling of curvature, which is still hard for me to fathom re curved vs flat surface. Or is it a fundamental difference betwen 2D vs 3D , which may have been talked about elsewhere.

That change in shape requires external forces to be exerted on the object, which distort the object and make it unusable as a standard of "size" or "shape"; this is always true in the 2-D case we were discussing, and in the spacetime case, it is "almost always" true; there is *some* way to place the object without exerting any "extra" forces, and even in this case, the object's shape will not be the same as it was in flat spacetime, but its volume will be the same, as long as we can ignore tidal gravity.

So how to interpret goings on in the case of say a very rigid tubular frame in the shape of a perfect tetrahedron in flat spacetime. It is the simplest polyhedron that is fully constrained re shape when all sides are connected via freely hinging ball joints. A 3D arrangement of four triangles. If we accept the vertex angles will be greater in +ve curved spacetime, there is a sense that the enclosed volume should be greater than in flat spacetime. Can we say just how that would not be true? S'pose it's down to fierce math proof - yes? Scratching head over that one.
Thanks again, Cheers :zzz:

 

[PeterDonis]

You are I deduce a speed typist

Thanks to my Dad forcing me to learn to type one summer in high school, yes. I learned on the old Olympic typewriter that he had used to type his master's thesis, which was older than I was. So my fingers learned how to squeeze a decent word rate out of those old, half-rusted keys that took about ten pounds of effort each to strike, and now they just laugh at a typical computer keyboard, which takes practically no effort by comparison. (It probably helps that I play keyboards too.)

This is where something, perhaps much different than straight K, should still allow measurement of a kind. Take an equalateral triangle composed of rigid tubes joined by free-hinging joints. In this configuration, one should expect vertex angles will exceed 60 degrees as discussed before.

I think this is right, provided that the tetrahedron is large enough that spatial curvature is measurable over its size, and that we deform the tetrahedron to conform to the curvature. Consider the analogous case on a 2-D surface, where we make a triangle out of rigid tubes joined by free-hinging joints, and then measure the joint angles after carefully wrapping the triangle onto a sphere so it conforms to the sphere's curvature. This case is exactly like that of the circular rubber disk or annulus; to see any change in shape, we need to exert external force on the rubber to make it conform to the shape of the sphere, and we have to decide how it is going to conform. We also have to make the shape we are trying to wrap around the sphere large enough that it "sees" the curvature; if it's too small we will be unable to use it to detect any difference from a flat plane (like the tiny pieces of paper we cut out of the paper disk or annulus in my previous example).

Given that these conditions are met, yes, we can in principle use the triangle as described (or the tetrahedron in the 3-D case) to measure differences between the space we are interested in and flat space; but there is still the question of how we decide to constrain the deformation. In the case quoted above, you are essentially constraining the side lengths of the tetrahedron (or triangle) to be constant, and letting the angles vary. Consider the 2-D case first (the triangle), and note that to physically realize this, we not only need to let the joint angles expand; we also need to bend the sides of the triangle since they are no longer Euclidean straight lines, but geodesics of the sphere, i.e., segments of great circles. (At least, I assume this was your intention in specifying side lengths constant and letting angles vary.) For thought experiment purposes, we can stipulate that this can be done while keeping the length of the sides constant, and similarly, in 3-D, the sides of the tetrahedron will have to bend slightly, but we can stipulate that their lengths are still held constant. In both cases, however, the sides will clearly undergo deformation, and we will have to exert external force on the tetrahedron to effect this deformation.

If we accept the vertex angles will be greater in +ve curved spacetime, there is a sense that the enclosed volume should be greater than in flat spacetime.

With the constraint as described above, yes; I would expect the volume inside the tetrahedron (or the area of the triangle, in the analogous 2-D case) to be larger if the side lengths are stipulated to be held constant. But as I just noted, to realize this case we have to exert external force to deform the tetrahedron (or triangle), and this external force is what causes the volume to expand.

OTOH, we could stipulate a different constraint, that the tetrahedron should be "unstressed"--or more precisely, that we should not *impose* any stress on it by exerting external force. In this case, I'm not sure what would happen, other than that I would expect the tetrahedron's volume to remain constant--always assuming that we can neglect tidal forces (as I noted previously, we can always choose the mass M of the central body and the radius r appropriately to make K measurably different from 1 while still having tidal gravity negligible). I *think* that the angles would still expand, but the sides would shorten, keeping the resulting volume constant but changing the shape.

 

[JDoolin]

Hi, I apologize if this seems off topic. You guys are talking about a stationary shell, but another interesting idea to consider is an expanding shell.

The particles making up the shell need not be attached, but are simply Lorentz-contracted and time-dilated to an extremely high density.

This is a question I've been pondering for some time; precisely what would the effect of gravity be on the internal particles? It's quite possible that you would have large regions where the gravitational potential would be extremely great, but fairly constant, so there is no net force in any direction.

Since everything would be so uniform, would there be any observable effect at all?

 

[PeterDonis]

The particles making up the shell need not be attached, but are simply Lorentz-contracted and time-dilated to an extremely high density.

Lorentz-contrated and time-dilated relative to what?

This is a question I've been pondering for some time; precisely what would the effect of gravity be on the internal particles?

It would depend on what the shell was made of and how you specified the initial conditions of the expansion.

For a shell made of "normal" matter, i.e., with pressure no greater than 1/3 its energy density, the gravity of the shell would cause the expansion to decelerate, similar to a matter-dominated expanding FRW model. Depending on the initial conditions (velocity of expansion vs. shell energy density and pressure), the shell might stop expanding altogether and re-contract, or it might go on expanding forever, continually slowing down but never quite stopping.

It's quite possible that you would have large regions where the gravitational potential would be extremely great, but fairly constant, so there is no net force in any direction.

Since everything would be so uniform, would there be any observable effect at all?

If the shell's density is uniform (i.e., uniform throughout the shell at any particular "time" in the shell's comoving frame--the density could still change with time, as long as it remained uniform spatially within the shell), then I think you are right to guess that there would be no observable effect from the "potential" within the shell itself. There might still be an effect relative to the potential in the spacetime exterior to the shell. (The potential interior to the shell would be the same as the potential on the shell's inner surface, just as for a static shell.)

 

[Q-reeus]

Originally Posted by Q-reeus:
"This is where something, perhaps much different than straight K, should still allow measurement of a kind. Take an equalateral triangle composed of rigid tubes joined by free-hinging joints. In this configuration, one should expect vertex angles will exceed 60 degrees as discussed before."

I think this is right, provided that the tetrahedron is large enough that spatial curvature is measurable over its size, and that we deform the tetrahedron to conform to the curvature. Consider the analogous case on a 2-D surface, where we make a triangle out of rigid tubes joined by free-hinging joints, and then measure the joint angles after carefully wrapping the triangle onto a sphere so it conforms to the sphere's curvature. This case is exactly like that of the circular rubber disk or annulus; to see any change in shape, we need to exert external force on the rubber to make it conform to the shape of the sphere, and we have to decide how it is going to conform.

Having accepted your explanation of why K cannot apply to a simple container that does not enclose the source of curvature, there remains a point of disagreement here. My understanding of triangles adding to more than 180 degrees in +ve curved 3-space is that the weirdness here is precisely due to that, as measured by say laser theodolite, the tubular sides of said triangle will be *exactly* straight and thus entirely unstressed (assuming 'gussetts' are absent). It is understood here measurements are taken in the curved environment - not some distant coordinate reference. Hence the specification of free-hinging pinned joints, and rigid tubes that are not 'floppy'. Otherwise, what is implied is surely an intrinsic, locally measurable curvature of just one straight rod, in going from flat spacetime to curved. But in 3-curvature, how will the 'straight' rod/tube 'know' which way to bend?

I think the proper analogy here in going from flat to curved is not trying to wrap a flat object onto a curved surface. Rather, think of drawing an equalateral triangle on the surface of a large balloon (low surface curvature). 2D flat-landers living on the balloon surface cannot directly detect the surface curvature, but with their 2D confined 'laser theodolites' will confirm the triangle sides are straight, and the vertex angles are 'near enough' to 60 degrees. Now deflate the balloon to a much smaller radius. Flat-landers now attempt to construct another equalateral triangle of the same side lengths as before (meaning triangle occupies a much larger portion of the balloon surface than before). Their theodolites continue to say the sides are perfectly straight, but are puzzled to find the vertex angles now significantly exceed 60 degrees. That's how I got what curvature does here - there's a faint whiff of sanity to Dr Who's 'Tardis' if you like.

We also have to make the shape we are trying to wrap around the sphere large enough that it "sees" the curvature; if it's too small we will be unable to use it to detect any difference from a flat plane (like the tiny pieces of paper we cut out of the paper disk or annulus in my previous example).

Yes, and is it not just this size related differential that allows flat-landers to detect curvature induced angular changes by means of a small, 'standard' protractor that minimally 'feels' curvature. Thoughts?

 

[PAllen]

Having accepted your explanation of why K cannot apply to a simple container that does not enclose the source of curvature, there remains a point of disagreement here. My understanding of triangles adding to more than 180 degrees in +ve curved 3-space is that the weirdness here is precisely due to that, as measured by say laser theodolite, the tubular sides of said triangle will be *exactly* straight and thus entirely unstressed (assuming 'gussetts' are absent). It is understood here measurements are taken in the curved environment - not some distant coordinate reference. Hence the specification of free-hinging pinned joints, and rigid tubes that are not 'floppy'. Otherwise, what is implied is surely an intrinsic, locally measurable curvature of just one straight rod, in going from flat spacetime to curved. But in 3-curvature, how will the 'straight' rod/tube 'know' which way to bend?

I think the proper analogy here in going from flat to curved is not trying to wrap a flat object onto a curved surface. Rather, think of drawing an equalateral triangle on the surface of a large balloon (low surface curvature). 2D flat-landers living on the balloon surface cannot directly detect the surface curvature, but with their 2D confined 'laser theodolites' will confirm the triangle sides are straight, and the vertex angles are 'near enough' to 60 degrees. Now deflate the balloon to a much smaller radius. Flat-landers now attempt to construct another equalateral triangle of the same side lengths as before (meaning triangle occupies a much larger portion of the balloon surface than before). Their theodolites continue to say the sides are perfectly straight, but are puzzled to find the vertex angles now significantly exceed 60 degrees. That's how I got what curvature does here - there's a faint whiff of sanity to Dr Who's 'Tardis' if you like.

Yes, and is it not just this size related differential that allows flat-landers to detect curvature induced angular changes by means of a small, 'standard' protractor that minimally 'feels' curvature. Thoughts?

One part you don't get is the issue of embedding. Please think about how a curved spherical surface is embedded in flat 3-space without telling you anything about the curvature of the 3-space. Similarly, in curved spacetime you can embed flat planes, and any procedure looking for straight lines will pick out this embedded flat plane. Thus no procedure limited to a plane can detect curvature of a 4-manifold.

I have explained this multiple times and you have continued to ignore it.

 

[Q-reeus]

I have explained this multiple times and you have continued to ignore it.

Ignore is not perhaps the right word, as all I could pick up were assertions of how it is - may well be true, but to me they were just assertions.

One part you don't get is the issue of embedding. Please think about how a curved spherical surface is embedded in flat 3-space without telling you anything about the curvature of the 3-space. Similarly, in curved spacetime you can embed flat planes, and any procedure looking for straight lines will pick out this embedded flat plane. Thus no procedure limited to a plane can detect curvature of a 4-manifold.

I have quoted a proof by J.L. Synge that five vertices are the minimum to detect curvature of spacetime (that is, even a tetrahedron can always be constructed to conform to Euclidean expectations).

Allright, given that is so, what sense does one make of the 'popular' statement that the internal angles of a triangle do not generally add to 180 degrees in curved spacetime? Having not studied the subject, I took my que from such as:
"In other words, in non-Euclidean geometry, the relation between the sides of a triangle must necessarily take a non-Pythagorean form." at http://en.wikipedia.org/wiki/Pythagorean_theorem#Non-Euclidean_geometry
[Another grab, from: http://en.wikipedia.org/wiki/Curved_space#Open.2C_flat.2C_closed
"Triangles which lie on the surface of an open space will have a sum of angles which is less than 180°. Triangles which lie on the surface of a closed space will have a sum of angles which is greater than 180°. The volume, however, is not (4 / 3)πr³" In context this appears to me to be a generalized statement applicable to higher than 2D curvature. As I say , haven't studied this subject at all.]

 

[PAllen]

Ignore is not perhaps the right word, as all I could pick up were assertions of how it is - may well be true, but to me they were just assertions.

Allright, given that is so, what sense does one make of the 'popular' statement that the internal angles of a triangle do not generally add to 180 degrees in curved spacetime? Having not studied the subject, I took my que from such as:
"In other words, in non-Euclidean geometry, the relation between the sides of a triangle must necessarily take a non-Pythagorean form." at http://en.wikipedia.org/wiki/Pythagorean_theorem#Non-Euclidean_geometry

If you look at that link, the context is geometry on a 2-surface. Popular statements don't get into the issue of embedding - they are over-simplifed. I am not just making assertions, I am asking you to think, as I'll do again. Apply your idea to 3-space "if a triangle is non-pythagorean, the *space* is non-euclidean' to flat 3-space. On a 2-sphere in 3-space, you conclude, correctly, that the 2-sphere is curved (triangle is non-pythagorean). What does that tell you about the 3-space: nothing. The 3-space is still flat.

One specific argument that flat planes are embeddable in a 4-manifold is simply to note that for an arbitrary symmetric metric with 10 components, you can apply 4 coordinate conditions. This is sufficient to make e.g. the x,y components of the metric [[1,0],[0,1]], that is a Euclidean plane. As a result, any non-Euclidean behavior of a plane is just a function of coordinate choice, and is not telling you anything intrinsic about the manifold.

 

[Q-reeus]

If you look at that link, the context is geometry on a 2-surface.

More inclusive quote from that same passage:
"The Pythagorean theorem is derived from the axioms of Euclidean geometry, and in fact, the Pythagorean theorem given above does not hold in a non-Euclidean geometry.[51] (The Pythagorean theorem has been shown, in fact, to be equivalent to Euclid's Parallel (Fifth) Postulate.[52][53]) In other words, in non-Euclidean geometry, the relation between the sides of a triangle must necessarily take a non-Pythagorean form. For example, in spherical geometry, all three sides of the right triangle (say a, b, and c) bounding an octant of the unit sphere have length equal to π/2, and all its angles are right angles, which violates the Pythagorean theorem because a2 + b2 ≠ c2."

I can only take that one way - in *any* non-Euclidean geometry. Just how that applies to triangle in 3-curvature is the question.

One specific argument that flat planes are embeddable in a 4-manifold is simply to note that for an arbitrary symmetric metric with 10 components, you can apply 4 coordinate conditions. This is sufficient to make e.g. the x,y components of the metric [[1,0],[0,1]], that is a Euclidean plane. As a result, any non-Euclidean behavior of a plane is just a function of coordinate choice, and is not telling you anything intrinsic about the manifold.

Not familiar with this to argue what you say here, other than to ask you to explain the full passage I quoted above. Put it simply please - are you saying that angles will add to 180 degrees in a generally 3-curved space, or not? :zzz:

 

[PAllen]

More inclusive quote from that same passage:
"The Pythagorean theorem is derived from the axioms of Euclidean geometry, and in fact, the Pythagorean theorem given above does not hold in a non-Euclidean geometry.[51] (The Pythagorean theorem has been shown, in fact, to be equivalent to Euclid's Parallel (Fifth) Postulate.[52][53]) In other words, in non-Euclidean geometry, the relation between the sides of a triangle must necessarily take a non-Pythagorean form. For example, in spherical geometry, all three sides of the right triangle (say a, b, and c) bounding an octant of the unit sphere have length equal to π/2, and all its angles are right angles, which violates the Pythagorean theorem because a2 + b2 ≠ c2."

I can only take that one way - in *any* non-Euclidean geometry. Just how that applies to triangle in 3-curvature is the question.

Not familiar with this to argue what you say here, other than to ask you to explain the full passage I quoted above. Put it simply please - are you saying that angles will add to 180 degrees in a generally 3-curved space, or not? :zzz:

The whole passage refers to geometry of a 2-surface. Euclid's Parrallel postulate is a postulate about plane geometry. Spherical geometry is the geometry of a 2-sphere - the *surface* of a sphere. And I ask you again to think about my simple embedding example. The surface of a globe is non-euclidean. The globe is sitting in a flat euclidean 3-space. Contradiction? No. The embedded space is curved, the space embedded in happens to be flat.

 

[PAllen]

More inclusive quote from that same passage:
"The Pythagorean theorem is derived from the axioms of Euclidean geometry, and in fact, the Pythagorean theorem given above does not hold in a non-Euclidean geometry.[51] (The Pythagorean theorem has been shown, in fact, to be equivalent to Euclid's Parallel (Fifth) Postulate.[52][53]) In other words, in non-Euclidean geometry, the relation between the sides of a triangle must necessarily take a non-Pythagorean form. For example, in spherical geometry, all three sides of the right triangle (say a, b, and c) bounding an octant of the unit sphere have length equal to π/2, and all its angles are right angles, which violates the Pythagorean theorem because a2 + b2 ≠ c2."

I can only take that one way - in *any* non-Euclidean geometry. Just how that applies to triangle in 3-curvature is the question.

Not familiar with this to argue what you say here, other than to ask you to explain the full passage I quoted above. Put it simply please - are you saying that angles will add to 180 degrees in a generally 3-curved space, or not? :zzz:

I am saying if you try to make Euclidean triangles in a general 4-d semi-riemannian manifold, you will succeed. For 3-space, I'm not sure whether you can *always* do it. There are fewer degrees of freedom, and the simple counting argument I used for 4-manifold does not work. It is certainly true that not all 2-manifolds can be embedded in flat 3-space, so it is plausible that some 3-manifolds will not embed a section of flat 2-surface.

Also recall my discussion with Peter: the SC geometry allows embedding of completely flat 3-space regions. The K factor is actually a feature of a particular class of observers (static observers, which are non-inertial observers), not something intrinsic to the geometry. It is analogous to distortions seen in an accelerating rocket in flat spacetime - a feature of the observer, not the intrinsic spacetime geometry. GP observers in the same SC geometry, experience absolutely flat 3-space.

 

[JDoolin]

Lorentz-contrated and time-dilated relative to what?

If you look carefully, there is one particle in the diagram that doesn't move. All motion in this diagram, then, is relative to that stationary particle.

This deserves further animations; I'd rather show you, if I can, rather than tell you. But suffice it to say, the animation above could be Lorentz transformed so as to place any particle in the center of the circle. However, because there are only a finite number of particles in the picture, there would be a gravitational asymmetry for every particle, except for one.

It would depend on what the shell was made of and how you specified the initial conditions of the expansion.

There are a couple different ways I might specify the initial conditions; (1) starting with an equipartition of rapidity (2) starting with an equipartition of momentum.

In this animation I assumed equal masses, and used equipartition of momentum.

Let

$$q\equiv \left \|\frac{\vec p}{m c} \right \| = \left \|\frac{(\vec v/c)}{\sqrt{1-v^2/c^2}} \right \|<3$$$$\frac{v}{c}=\sqrt{\frac{q^2}{q^2+1}}$$

Once I calculated the velocities, I animated, finding position, by simply multiplying the velocities by t.
The animation renders about 79,000 dots of equal mass with random q between 0 and 3 in random directions. The particle at the center of this distribution should experience no net acceleration; however, closer to the edges, there is more and more gravitational force.

There's no limit to what q might be. If I set q=10^100, then all the particles that are in this animation would be so close to the center that they would experience essentially no net force.

For a shell made of "normal" matter, i.e., with pressure no greater than 1/3 its energy density, the gravity of the shell would cause the expansion to decelerate, similar to a matter-dominated expanding FRW model. Depending on the initial conditions (velocity of expansion vs. shell energy density and pressure), the shell might stop expanding altogether and re-contract, or it might go on expanding forever, continually slowing down but never quite stopping.

If the shell's density is uniform (i.e., uniform throughout the shell at any particular "time" in the shell's comoving frame--the density could still change with time, as long as it remained uniform spatially within the shell), then I think you are right to guess that there would be no observable effect from the "potential" within the shell itself. There might still be an effect relative to the potential in the spacetime exterior to the shell. (The potential interior to the shell would be the same as the potential on the shell's inner surface, just as for a static shell.)

I have a plan in mind to present the animation from the reference frame of a particle on the edge of the mass by Lorentz Transforming all of the velocities. Definitely, as long as the total mass of the distribution is finite, then there would be particles on the edge that experience extreme accelerations toward the center. If the mass were infinite (and thus no edge), you could invoke symmetry, and there would be no net force in any direction for any particle.

But then you'd have the problem, being inside a shell of infinite mass, that at any given point inside, you are at an infinite negative gravitational potential

On the other hand, with a finite mass, the particles at the edge could experience, at least for a time, acceleration equivalent to a black hole. I'm not at all sure what theoretical ramifications that would have.

Anyway, I'll work on the other animations, and hope that makes my meaning clearer.

 

[PeterDonis]

Having accepted your explanation of why K cannot apply to a simple container that does not enclose the source of curvature, there remains a point of disagreement here. My understanding of triangles adding to more than 180 degrees in +ve curved 3-space is that the weirdness here is precisely due to that, as measured by say laser theodolite, the tubular sides of said triangle will be *exactly* straight and thus entirely unstressed (assuming 'gussetts' are absent).

If the triangle is *assembled* in the curved space, in the right way, this will be true. But you specified that the triangle (or tetrahedron) was assembled in flat space, and then *moved* to curved space. That means the "natural" configuration of the sides is the flat configuration, Euclidean straight lines, and they must be deformed to get into the curved configuration, geodesics on a curved surface.

If, instead, you assembled the triangle carefully in a curved region, so that the "natural" configuration of its sides was as geodesics on the curved surface, and the "natural" angles at each vertex summed to more than 180 degrees, then the triangle would deform if you tried to make it conform to a flat Euclidean plane. Similarly, if you assembled a tetrahedron in a region of curved space, hovering over a gravitating body, so that it was unstressed in that configuration, and then took it far away from gravitating bodies where space was flat, it would undergo stress and deformation in the course of conforming to the flat space.

I think the proper analogy here in going from flat to curved is not trying to wrap a flat object onto a curved surface. Rather, think of drawing an equalateral triangle on the surface of a large balloon (low surface curvature). 2D flat-landers living on the balloon surface cannot directly detect the surface curvature, but with their 2D confined 'laser theodolites' will confirm the triangle sides are straight, and the vertex angles are 'near enough' to 60 degrees. Now deflate the balloon to a much smaller radius. Flat-landers now attempt to construct another equalateral triangle of the same side lengths as before (meaning triangle occupies a much larger portion of the balloon surface than before). Their theodolites continue to say the sides are perfectly straight, but are puzzled to find the vertex angles now significantly exceed 60 degrees. That's how I got what curvature does here - there's a faint whiff of sanity to Dr Who's 'Tardis' if you like.

It depends on what you want to make an analogy to. If we are trying to construct an analogy to what happens when we take an object constructed far away from gravitating bodies and move it into a gravity well, the above is *not* a good analogy for that, because deflating the balloon corresponds to contracting the spacetime as a whole. That would do as an analogy for a collapsing universe, but *not* for moving into a gravity well. A better analogy for that would be to consider a surface like the Flamm paraboloid...

http://en.wikipedia.org/wiki/File:Flamm's_paraboloid.svg

...and think of moving a triangle from the flat region to the curved region.

 

[PeterDonis]

If you look carefully, there is one particle in the diagram that doesn't move. All motion in this diagram, then, is relative to that stationary particle.

Then I'm not sure I would call this a "shell". "Shell" implies a thin region of matter with complete vacuum inside and outside it, at least in the context of this thread.

The animation renders about 79,000 dots of equal mass with random q between 0 and 3 in random directions. The particle at the center of this distribution should experience no net acceleration; however, closer to the edges, there is more and more gravitational force.

How are you calculating the force?

 

[JDoolin]

Then I'm not sure I would call this a "shell". "Shell" implies a thin region of matter with complete vacuum inside and outside it, at least in the context of this thread.
How are you calculating the force?

This is a good question, or more specifically, how should I (or how should we) calculate the force? So far, essentially, I'm not calculating the force. Invoking symmetry, there is no net force. There's a gravitational potential, but no net force on a particle at or near the center of the distribution.

But let's look at the other extreme now, from the very edge of the distribution.

Here, obviously we don't have symmetry. The stationary particle should experience a net force to the right.

Let me give you a couple of premises of how I would go about calculating the force, and see if you agree with these premises, or if I am hopelessly naive.

#1 The net gravitational force on a particle is $$F = G m \sum_i \frac{M_i}{r_i^2} \vec u_i$$
#2 The force on a particle should be calculated based on the reference frame of the particle that is undergoing the force.
#3 The mass of each particle affecting gravitation is the rest-mass of the particle. i.e. the Lorentz factor affects momentum, but not gravitational attraction.
#4 The location of the particle is not the "simultaneous" location, but rather the speed-of-light delayed location of the particle. i.e. the speed of gravity is the same as the speed of light, so we must find an intersection of the past-light-cone of our particle of interest with the world-lines of the particles involved.

Another peculiarity of relativity, (whether using Galilean Transformation or Lorentz Transformation) is that when one accelerates toward a future event, it leans toward him, becoming directly in his future, but when one accelerates toward a past event, it leans away. We may find that even if the particles at the edge accelerate "toward" the center, that in fact, the end result is not at all what common-sense would suggest. By accelerating toward the center, the particle continually enters reference frames where the center is further and further away.

 

[PeterDonis]

Here, obviously we don't have symmetry. The stationary particle should experience a net force to the right.

Well, the problem as a whole has spherical symmetry--or at least, you can impose that condition as a reasonable idealization to make the problem tractable. If the problem has spherical symmetry, then the "force" on any particle (a) must point in the radial direction, and (b) must be a function only of its radial coordinate. If the matter is all ordinary matter, as I described in my last post, then (a) can be further refined: the force on any particle must point radially *inward*, i.e., the expansion of the shell must be decelerating. As I note below, viewing this deceleration as caused by a "force" may not be the simplest way to view this problem.

#1 The net gravitational force on a particle is $$F = G m \sum_i \frac{M_i}{r_i^2} \vec u_i$$

This requires the problem to be non-relativistic. That is not consistent with your #4. In fact, the inconsistency between #1 and #4 was one of the stumbling blocks on the road to General Relativity, back in the early 1900's. See further note below.

#2 The force on a particle should be calculated based on the reference frame of the particle that is undergoing the force.

A better way to state this would be: the 4-force on a particle is a covariant geometric object, it is the particle's rest mass times the particle's 4-acceleration, which is the covariant derivative of its 4-velocity with respect to its proper time.

#3 The mass of each particle affecting gravitation is the rest-mass of the particle. i.e. the Lorentz factor affects momentum, but not gravitational attraction.

A better way to state this would be: the "gravitational field" is determined by the stress-energy tensor of the matter, which is a covariant geometric object. The SET determines the field via the Einstein Field Equation. It also correctly accounts for the fact that "rest mass" is what affects gravity, not momentum due purely to kinematics.

#4 The location of the particle is not the "simultaneous" location, but rather the speed-of-light delayed location of the particle. i.e. the speed of gravity is the same as the speed of light, so we must find an intersection of the past-light-cone of our particle of interest with the world-lines of the particles involved.

If you are assuming this, then you can't use #1 as your force equation. Consider a simple example: the force on the Earth at a given instant does *not* point towards where the Sun was 8 minutes ago by the Earth's clock. It points towards where the Sun is "now" by the Earth's clock. (Technically, there are some small correction factors, but they can be ignored for this discussion.) So if you use your #1, you have to plug in the position of the Earth relative to the Sun "now", not the "retarded" position, or you'll get the wrong answer. (Steve Carlip wrote an excellent paper some time ago that explains all this: it's at http://arxiv.org/abs/gr-qc/9909087.)

As I noted above, using #1 requires the problem to be non-relativistic, and you don't seem to be imposing that limitation. For the relativistic case, you can't really use a "force" equation for this problem, or at least it does not seem to be the easiest way to approach it. A better model would be an expanding matter-dominated FRW-type model, as I mentioned in a previous post, especially since it doesn't seem like your particles are a "shell", since there's no interior vacuum region, as far as I can see. This type of model does not view gravity as a "force"; it just solves for the dynamics of a curved spacetime using the EFE and an expression for the stress-energy tensor of the matter. You can still view individual particles as being subject to a "force", but that force can be more easily calculated *after* you have constructed the model from the EFE.

If the spatial extent of the expanding "shell" is limited, then at the surface of the shell, the FRW-type solution would be matched to an exterior Schwarzschild vacuum solution; this would basically be the time reverse of the Oppenheimer-Snyder solution for the gravitational collapse of a star.

Another peculiarity of relativity, (whether using Galilean Transformation or Lorentz Transformation) is that when one accelerates toward a future event, it leans toward him, becoming directly in his future, but when one accelerates toward a past event, it leans away. We may find that even if the particles at the edge accelerate "toward" the center, that in fact, the end result is not at all what common-sense would suggest. By accelerating toward the center, the particle continually enters reference frames where the center is further and further away.

Huh? I don't understand what you're getting at here at all. How can you accelerate toward a past event? For that matter, how can you accelerate toward a future event? You accelerate toward a point in space, not an event in spacetime. Also, what kind of "reference frames" are you talking about? If you're talking about ordinary Lorentz frames, they are only valid locally; you can't use them to correctly evaluate distances to faraway objects.

It looks like this discussion might be better moved to a new thread, since it appears to be getting further from the topic of this one.

 

[JDoolin]

This requires the problem to be non-relativistic. That is not consistent with your #4. In fact, the inconsistency between #1 and #4 was one of the stumbling blocks on the road to General Relativity, back in the early 1900's. See further note below.

Okay, I see where I had one mistake.

http://en.wikipedia.org/wiki/Gauss's_law#Relation_to_Coulomb.27s_law

Strictly speaking, Coulomb's law cannot be derived from Gauss's law alone, since Gauss's law does not give any information regarding the curl of E (see Helmholtz decomposition and Faraday's law). However, Coulomb's law can be proven from Gauss's law if it is assumed, in addition, that the electric field from a point charge is spherically-symmetric (this assumption, like Coulomb's law itself, is exactly true if the charge is stationary, and approximately true if the charge is in motion).

I was thinking that coulomb's law and gauss law were equivalent. But they are only equivalent when the charge is not moving. I'll try to read more of the article to see what I've missed. May take me a bit of time to catch up so I don't think I'm ready to start another thread at this time.

If indeed, the gravitational field from a receding object has some predictable variation based on its relative velocity, then we should, of course, use that modification. But if you want to claim that there is no predictable variation; no predictable "meaningful" position we can use for distant objects, I beg to differ.

On another topic, if possible, can you support your argument that "ordinary Lorentz frames...are valid only locally." I've heard this time and time again, but no one has ever explained what it means. Are you saying that when I point at a distant galaxy, that that direction that I am pointing only exists locally? Are you saying that the very concept of direction is only a local phenomenon?