I Euclidean geometry and gravity

[PeterDonis]

I note you haven’t commented on the second part of that post

I hadn't looked at it in detail because I was focusing on the first part and wanting to understand it.

Having now looked at it, I'm not sure about this:

it is certainly not the result of either tidal gravity or hovering force. [In some sense, you can think of it as resulting from changing tidal gravity, but it can't be accounted for just by considering the static tidal gravity of final state.]

Does the bolded part just mean that of course you can't just look at the final state, you have to look at the difference between the initial state and the final state?

 

[Jaime Rudas]

Also, there is the possibility of non-smooth embeddings, which are much easier.

Where could I find some introductory text that would help me delve deeper into this?

 

[PAllen]

I hadn't looked at it in detail because I was focusing on the first part and wanting to understand it.

Having now looked at it, I'm not sure about this:


Does the bolded part just mean that of course you can't just look at the final state, you have to look at the difference between the initial state and the final state?

Yes, since you are looking for a cause of changed pressure. The final state by itself has no memory of how it originated.

 

[PAllen]

Where could I find some introductory text that would help me delve deeper into this?

You can start here to look for references:

https://en.wikipedia.org/wiki/Nash_embedding_theorems

Also look at the Whitney embedding theorem and also

https://en.wikipedia.org/wiki/Embedding

 

[PAllen]

You don't seem to understand that the proper acceleration is determined by the beam's worldline. You can't adjust it by applying external forces. For an object that's held static in Schwarzschild spacetime, the proper acceleration is radially outward, and its magnitude (in geometric units where $G = c = 1$) is

$$
\frac{M}{r^2 \sqrt{1 - 2M / r}}
$$

where $r$ is the radial coordinate at which it is held static. There is simply no way to change that. The resultant of all the forces applied to the object must be such as to produce that proper acceleration.

But I think the claim is simply that you can’t deduce balance opposing forces from proper acceleration; that doesn’t mean they don’t exist, or the whole field of statics would vacuous.

 

[PeterDonis]

I think the claim is simply that you can’t deduce balance opposing forces from proper acceleration; that doesn’t mean they don’t exist

No, that's not the claim that was made. Here's the claim that was made:

A beam (as part of a larger structure) is under longitudinal compressive stress, but its proper acceleration is purely transversal due to transversal external forces applied uniformly along the beam.

The beam is supposed to be held static against gravity in Schwarzschild spacetime. That means its proper acceleration is purely radially outward, with the magnitude I gave. The longitudinal compressive stress is fine, since that means the longitudinal direction is the radial direction. But there is no way to have any transverse proper acceleration at all if the beam is static, much less to have the proper acceleration "purely transversal".

 

[PAllen]

No, that's not the claim that was made. Here's the claim that was made:


The beam is supposed to be held static against gravity in Schwarzschild spacetime. That means its proper acceleration is purely radially outward, with the magnitude I gave. The longitudinal compressive stress is fine, since that means the longitudinal direction is the radial direction. But there is no way to have any transverse proper acceleration at all if the beam is static, much less to have the proper acceleration "purely transversal".

I assumed the beam was perpendicular to r, supported against gravity by external transverse force all along the bottom of the beam. Meanwhile, the rest of the structure presses the beam longitudinally, producing balanced longitudinal stresses. These exist without being reflected in proper acceleration of the world lines.

 

[PeterDonis]

I assumed the beam was perpendicular to r, supported against gravity by external transverse force all along the bottom of the beam.

The structure would have to be narrow enough for the variation in $r$ for a transverse beam to be negligible. Otherwise its proper acceleration will no longer be purely transversal, because different parts of the beam will be at different $r$ and different angular coordinates, so their proper accelerations will differ both in magnitude and direction, with the difference being enough to be measurable.

 

[PAllen]

The structure would have to be narrow enough for the variation in $r$ for a transverse beam to be negligible. Otherwise its proper acceleration will no longer be purely transversal, because different parts of the beam will be at different $r$ and different angular coordinates, so their proper accelerations will differ both in magnitude and direction, with the difference being enough to be measurable.

I don’t see that as relevant to the example. It was just to point out that there can be balanced forces unrelated to, and in different directions, from proper acceleration of mass points. Therefore, that stress is not, in general, derivable from proper acceleration. And that if you supply the proper acceleration by ‘atomic rockets’, you have not necessarily eliminated stress.

 

[PeterDonis]

stress is not, in general, derivable from proper acceleration.

In the general sense of "stress", which in relativity means "the components of the stress-energy tensor", yes, I agree. I should have used the term "unbalanced forces" instead of "stress" to be more specific about what I was concerned about.

 

[PAllen]

You can start here to look for references:

https://en.wikipedia.org/wiki/Nash_embedding_theorems

Also look at the Whitney embedding theorem and also

https://en.wikipedia.org/wiki/Embedding

I found something right on point, but it is a thesis, so I don’t swear to its accuracy (I have not studied it in detail). Its results imply that a portion of a Euclidean plane can be smoothly isometrically embedded in any curved 3 space. It discusses use of the gauss-codazzi equations, as I suggested.


http://davidbrander.org/penn.pdf

 

[A.T.]

The beam is supposed to be held static against gravity in Schwarzschild spacetime. ...

This restriction wasn't part of your original statement:

It's not an argument, it's a definition of what proper acceleration is and what it means physically. You can't say that all of the proper acceleration of a worldline is accounted for by external forces, and also say there are internal stresses left over. That's simply a contradiction.

You claimed above that this is the very definition of proper acceleration, so it should apply in any spacetime, including flat spacetime.

But even assuming Schwarzschild spacetime, I don't get this:

The longitudinal compressive stress is fine, since that means the longitudinal direction is the radial direction.

How does this follow? The longitudinal compressive stress can come from the rest of the structure, that the beam is part of. So what does it have to do with the beam's orientation in Schwarzschild spacetime? Why can't the beam's orientation be perpendicular to purely radial?

 

[PeterDonis]

This restriction wasn't part of your original statement

No, but it was part of the specific scenario we were discussing as I understood it.

You claimed above that this is the very definition of proper acceleration, so it should apply in any spacetime, including flat spacetime.

Yes, that's correct.

even assuming Schwarzschild spacetime, I don't get this

See my discussion with @PAllen, who pointed out what you've pointed out.

 

[A.T.]

Its results imply that a portion of a Euclidean plane can be smoothly isometrically embedded in any curved 3 space.

So a zero thicknes flat sheet of paper could be somehow placed in Flamm's space without distorting / tearing it? I wonder if it's possible to visualize such an embedding. Would it be a relatively simple configuration, or one of those endlessly crumpled shapes?

 

[PAllen]

So a zero thicknes flat sheet of paper could be somehow placed in Flamm's space without distorting / tearing it? I wonder if it's possible to visualize such an embedding. Would it be a relatively simple configuration, or one of those endlessly crumpled shapes?

A smooth embedding is possible. Just try to think about a 2-sphere embedded in Euclidean 3-space. There is no curvature anywhere in the Euclidean space, but the sphere embeds just fine. Even a small part of hyperbolic 2-space is smoothly embeddable in Euclidean 3-space. So just imagine the 'reverse' as it were.

Note, for smooth local embeddings (a global embedding is where you run into all the topological constraints, requiring a large number of dimensions of the target space) the relevant formula is 2n-1. So a small volume of Euclidean 3-space may require a 5-dimensional curved space to embed.

 

[A.T.]

A smooth embedding is possible. Just try to think about a 2-sphere embedded in Euclidean 3-space. There is no curvature anywhere in the Euclidean space, but the sphere embeds just fine. Even a small part of hyperbolic 2-space is smoothly embeddable in Euclidean 3-space. So just imagine the 'reverse' as it were.

Curved 2D space embeded in flat 3D space is easy to visualize, but the reverse is tricky, because curved 3D space itself is hard to visualize.

I guess for an illustration one could project the 3D Flamm embedding space onto flat 3D space, but that is not isometric, so the embedded 2D flat sheet will then look curved. Just as if you had embedded curved 2D in flat 3D.

 

[PAllen]

Curved 2D space embeded in flat 3D space is easy to visualize, but the reverse is tricky, because curved 3D space itself is hard to visualize.

I guess for an illustration one could project the 3D Flamm embedding space onto flat 3D space, but that is not isometric, so the embedded 2D flat sheet will then look curved. Just as if you had embedded curved 2D in flat 3D.

From the thesis I referenced earlier is a notion of what embeddings are 'hard' versus 'easy'. 'Hard' implies you are likely only to get a local smooth isometric embedding. 'Easy' implies that if there are no topological constraints, you are likely to get a global embedding. The determining factor (when curvature is simple for both source and target) is source curvature minus target curvature. If this is negative, the extrinsic curvature of the embedding will be negative; if positive, it will be positive. The 'hard' case is negative extrinsic curvature. So plane into hyperbolic space is 'easy' (0 - negative=positive)), the reverse is 'hard'. Since Flamm space is, I think, mostly negative curvature, plane into this is actually the easy case. Note, sphere into $E^3$ is easy, while plane into 3-sphere is hard.

 

[A.T.]

From the thesis I referenced earlier is a notion of what embeddings are 'hard' versus 'easy'.

Yes, but I was referring to the ease of visualization / illustration.

 

[pervect]

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

I've been assuming the answer to this question is, informally, "the shortest path". The mathematical term is the "Levi-Civita connection". A straight line, more formally a geodesic in a more general treatment is a line that continues "in the same direction".

The general concept of a connection allows geodesics to be defined that are not paths of shortest distance. But we don't use those in General relativity, we only use the Levi-Civita connection. So the path of "shortest distance" is mostly good enough for what we need, though there are some interesting mathematics and other fine points that I'm not discussing having realized they are irrelevant.

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

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

Sorry, this just isn't right. The meaning of the term"space" around a black hole is slightly ambiguous. Using the usual defintions of what this means, though, the equatorial plane of a black hole is the flamm paraboloid as other posters have already mentioned, see the wiki entry in https://en.wikipedia.org/wiki/Schwarzschild_metric, there is a section labelled "flamm paraboloid".

The flamm paraboloid is not flat, and the sum of the angles of the triangles will not be 180. I'm not sure offhand whether it will be greater or less than 180.

The space-time around a black hole is not ambiguous, it's the Schwarzschild space-time geometry. The ambiguity comes in when one makes a map from the 4d space-time to a 3d space. The usual mapping involves the concept of "static observers", then the worldline of every static observer maps to a unique "point in space", the "space" around the black hole can be described mathematically as a "quotient manifold" of the 4d Schwarzschild space-time.

 

[A.T.]

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

Sorry, this just isn't right...

Sorry, I don’t see the exact point of disagreement between the statement above and what you wrote after disagreeing. The assumptions you state, in order to interpret the above statement, seem all reasonable. But based on this interpretation, what is wrong with the statement?

 

[raagamuffin]

I have been following this thread with great interest.

The details that people went into over what I initially thought was a simple question was surprising to me and trying to grasp some of the stuff require some heavy lifting. (Please note my background, I’m a mere layman).

Many of your thoughts around suspending an object in such a state as to (draw a triangle) seem to center around stability due to (gravitational) tidal forces. This made me wonder, what is the smallest unit of space(time?) near a black hole such that an object of that size won’t break apart ? I guess its size would follow the (approximately) classical Newtonian r squared as distance from the BH increases?

Or is this a nonsensical question?

 

[PeterDonis]

what is the smallest unit of space(time?) near a black hole such that an object of that size won’t break apart ? I guess its size would follow the (approximately) classical Newtonian r squared as distance from the BH increases?

I'm not sure what you're asking. What kind of scenario do you have in mind where an object might or might not break apart?

 

[raagamuffin]

I'm not sure what you're asking. What kind of scenario do you have in mind where an object might or might not break apart?

OK, I'm making an assumption. So let me start at the beginning. If I have a large sheet of paper, in 'neutral' space, all is fine. As this sheet draws near to a BH something happens to this sheet. What happens to it? (I assumed it would start bending and breaking apart).

 

[PeterDonis]

If I have a large sheet of paper, in 'neutral' space, all is fine. As this sheet draws near to a BH something happens to this sheet. What happens to it?

It depends on how the sheet is getting close to the BH. If it's freely falling, that's one thing. If it's being supported against the hole's gravity by rockets or something like that, so that it remains static or close to static, that's a different thing.

If the sheet is freely falling towards the hole, it will be stretched radially and compressed tangentially. Since paper doesn't have a very large tensile strength, the radial stretching will eventually tear it.

If the sheet is being supported against gravity by a rocket, and slowly lowered so that it remains close to static the whole time, it will be compressed radially (assuming the rocket is at the bottom of the sheet), just as any object that is being held up against its own weight. If it's wide enough tangentially, there could be some bending because of tangential tidal gravity. It's not clear at what point the paper might tear, since most of the forces in this scenario are compression, not tension.

 

[PAllen]

It depends on how the sheet is getting close to the BH. If it's freely falling, that's one thing. If it's being supported against the hole's gravity by rockets or something like that, so that it remains static or close to static, that's a different thing.

If the sheet is freely falling towards the hole, it will be stretched radially and compressed tangentially. Since paper doesn't have a very large tensile strength, the radial stretching will eventually tear it.

If the sheet is being supported against gravity by a rocket, and slowly lowered so that it remains close to static the whole time, it will be compressed radially (assuming the rocket is at the bottom of the sheet), just as any object that is being held up against its own weight. If it's wide enough tangentially, there could be some bending because of tangential tidal gravity. It's not clear at what point the paper might tear, since most of the forces in this scenario are compression, not tension.

Well, arbitrarily close any horizon, irrespective of how mild the tidal forces are near the horizon of a supermassive BH, the compressive forces are from holding an object stationary grow without bound, so any material would become a molecular monolayer, at some point. Of course, that’s assuming you have a rocket that can provide unbounded thrust.

 

[pervect]

This can't be right. By that logic, you can't embed a 2-sphere in Euclidean 3-space since it has positive scalar curvature and every such plane will have zero curvature. Thus, this appears to be irrelevant to the problem.

I would say the obvious place to start would be the Gauss-Codazzi equations.

Also, there is the possibility of non-smooth embeddings, which are much easier. Two examples of surfaces that cannot be smoothly embedded to Euclidean 3-space but can be embedded without this restriction are the flat torus and a surface of constant negative curvature.

Sorry for this very late response. My first thought was that you had a good point, but as I was thinking about this more later, and I realized that this is not a counter-example to my argument.

Specifically, the argument doesn't claim that you can't embed a 2-sphere in a flat 3-space. It does say, though, that you can't embed a flat three-space on a 3-sphere (which is the surface of a 4-ball), or perhaps more relevantly, you can't embed a flat plane on said 3-sphere.

Regardless, I don't have a textbook example, so while I don't see any problems, I'm still cautious about presenting this as a fact, it's more of an argument.

The problem I have (had, in the sense it's probably a necro thread by now) is / was that I'm approaching the issue using the tool of differential geometry, which can be informally thought of as dealing with very small regions of space / space-time, hence the differential part. So it applies well (in some sense) to the limit of small triangles, where you can make statements about parallel transport (with the standard treatment), or the sum-of-angles of a triangle made out of geodesic segments (in the popularized treatment) for small triangles. But it doesn't have much to say about large ones.

I suspect there are different tools to answer more global questions, but - I'm not personally familiar with them, not even enough to say what these tools might be called.

I do think differential geometry has some useful things to say about curvature, though, which I see as the main thrust of the thread. I was attempting to popularize, as much as I could, what differential geometry has to say, as that's the tool I'm familiar with.

I still do not have a textbook reference that says you cannot embed a flat 2-plane in a 3 sphere (again, this is the surface of a 4-ball), but my argument does claims that you can't do it. Probably it's moot at this point., though it's still an interesting question that differential geoemtry an say something about.

 

[PAllen]

Sorry for this very late response. My first thought was that you had a good point, but as I was thinking about this more later, and I realized that this is not a counter-example to my argument.

Specifically, the argument doesn't claim that you can't embed a 2-sphere in a flat 3-space. It does say, though, that you can't embed a flat three-space on a 3-sphere (which is the surface of a 4-ball), or perhaps more relevantly, you can't embed a flat plane on said 3-sphere.

But your argument was you can’t embed a plane in e.g. a 3 sphere because the plane has zero curvature and the sectional curvature in every planar direction in the 3 sphere is positive. My argument is that for a 2 sphere embedded in flat 3 space, you have a similar curvature mismatch, but it does not prevent embedding.

Regardless, I don't have a textbook example, so while I don't see any problems, I'm still cautious about presenting this as a fact, it's more of an argument.

The problem I have (had, in the sense it's probably a necro thread by now) is / was that I'm approaching the issue using the tool of differential geometry, which can be informally thought of as dealing with very small regions of space / space-time, hence the differential part. So it applies well (in some sense) to the limit of small triangles, where you can make statements about parallel transport (with the standard treatment), or the sum-of-angles of a triangle made out of geodesic segments (in the popularized treatment) for small triangles. But it doesn't have much to say about large ones.

I suspect there are different tools to answer more global questions, but - I'm not personally familiar with them, not even enough to say what these tools might be called.

It's an embedding problem and the most basic tool for this is the Gauss-Codazzi equations.

I do think differential geometry has some useful things to say about curvature, though, which I see as the main thrust of the thread. I was attempting to popularize, as much as I could, what differential geometry has to say, as that's the tool I'm familiar with.

I still do not have a textbook reference that says you cannot embed a flat 2-plane in a 3 sphere (again, this is the surface of a 4-ball), but my argument does claims that you can't do it. Probably it's moot at this point., though it's still an interesting question that differential geoemtry a say something about.

Regarding your last paragraph, I provided a reference that says you can do exactly this - embed a flat plane in a 3 sphere with a smooth isometric embedding. The largest size piece of plane you can so embed is inversely proportional the the curvature of the 3 sphere.

 

[Jaime Rudas]

Specifically, the argument doesn't claim that you can't embed a 2-sphere in a flat 3-space. It does say, though, that you can't embed a flat three-space on a 3-sphere (which is the surface of a 4-ball), or perhaps more relevantly, you can't embed a flat plane on said 3-sphere.

Is that "flat plane" two-dimensional or three-dimensional?

 

[raagamuffin]

You folks have put a lot of good information and I have been working through some of it. But I can't do the advanced math bits, so you'll hopefully overlook my naivety when it comes to that.
I looked at the Flamm paraboloid , and I think I have rudimentary understanding how the 'paper' will be affected. Then I read a reference to 'frame dragging' in rotational black holes. What effect, if any, will that have on the 'paper'?

 

[PAllen]

In the side discussion of embedding a piece of a plane smoothly, isometrically, in a 3-sphere (following is part of a response to @pervect 's last post:

Regarding your last paragraph, I provided a reference that says you can do exactly this - embed a flat plane in a 3 sphere with a smooth isometric embedding. The largest size piece of plane you can so embed is inversely proportional the the curvature of the 3 sphere.

I thought I would make explicit the construction in that reference (which is presented it very compactly in any number of dimensions). Making it explicit for flat 2-surface section in a 3-sphere, the mechanism is simply to use a properly sized flat torus. As explained in:
https://en.wikipedia.org/wiki/Clifford_torus
a flat torus of just the right size will smoothly fit in a unit 3-sphere, both embedded in R4. But then, the embedding can be just ignored, and you are left with a smooth embedding of a piece of Euclidean plane in a 3-sphere.

The formalization and generalization of this is the general approach to embedding $E^n$ into higher dimesnsional spheres used in the reference I quoted earlier. The general achievement is a local embedding of $E^n$ into an m-sphere, with $m=2n-1$.