Shape of Hubble sphere at relativistic velocity

[PeterDonis]

Isn't that the same as saying, "there are distances over which you cannot at all approximate spacetime as flat"?

Sorry, I misread your earlier post. Yes, your statement here is correct.

Isn't the universe considered to be infinite

Spatially infinite, according to our best current model, yes.

so any finite distance would be infinitesimal in cosmological terms?

No. To approximate spacetime as flat, you must limit your analysis to a small enough patch of spacetime that the effects of spacetime curvature (not space curvature) are not observable. For the universe, "spacetime curvature" means the expansion of the universe, so to approximate the spacetime of the universe as flat, you must limit your analysis to a small enough patch of that spacetime that any effects associated with the expansion of the universe are not observable. But that assumes you are only looking at large scale effects to begin with. See below.

How far, in your understanding, in light-years would the distance over which you meaningfully can be?

It depends on what you are modeling. If you are modeling the orbits of satellites around the Earth, a patch of spacetime large enough to include the Earth cannot be approximated as flat. If you are modeling the orbits of planets in the solar system, a patch of spacetime large enough to include the solar system cannot be approximated as flat; but a patch that just covers the Earth (for a short enough period of time--remember we are talking about a patch of spacetime, not just space) might be if you ignore all effects of the Earth's gravity and just approximate the Earth as a point particle. And so on up the scale.

 

[Chris Miller]

@PeterDonis Again, very much appreciate your explaining in ways I believe I can (sort of) understand. Particularly,

For the universe, "spacetime curvature" means the expansion of the universe...

since my thread question here concerned the effect of Hubble expansion on near-c time dilation and length contraction. I see now that, as I sort of suspected, they're incompatible. I thought they'd eventually cancel each other out, but it seems more complicated than that.

@Ibix I've been thinking about your tile analogy:

But if you keep on tiling a larger and larger area you will, eventually, find that you can't continue to lay the tiles in a square grid because a square grid won't fit on the surface of a sphere. And that's analogous to what happens if you continue trying to use SR concepts on larger and larger regions of spacetime.

I could tile the Earth (some equivalent perfect sphere) no problem. Every tile 2x2' tile would fit flat and snugly with those around it, seemingly level. The minuscule errors would not be apparent or measurable in any local region. Analogizing to to spacetime's "hypercubal" grid, I imagine traveling at very near c from star to star, perhaps only seconds (by my clock) separating, and think, where am I in a year? But then, if I understand you and PeterDonis, their proper distances are affected by Hubble expansion (spacetime curvature) which messes with the placement of my tiles (Minkowski coordinates).

Given the seconds in a year (31,536), light-years in a megaparsec (3,261,564), kilometers in a megaparsec (30,856,775,714,409,000,000) and Hubble expansion (73 km/sec) , I calculate that in the one second it takes me (by my clock) to cross a megaparsec, it will have expanded by 1/4109 (0.024 percent) which seems still pretty flat and wouldn't seem to have had that much bearing on my trip? At c, am I staying equidistant from my original, proper, (at rest relative to CMB) Hubble horizon?

 

[jbriggs444]

I could tile the Earth (some equivalent perfect sphere) no problem. Every tile 2x2' tile would fit flat and snugly with those around it, seemingly level. The minuscule errors would not be apparent or measurable in any local region.

You can tile any small region to a decent fit, yes. You cannot tile the whole thing with one square tesselation.

Try it.

Start the layout of your square grid with the 2x2' resolution at the north pole. You lay a strip of some 30 million tiles south along the prime meridian until you hit the equator. You lay another strip of some 30 million tiles south at 90 degrees to the first. You lay another strip of some 30 million tiles along the equator from the first strip to the second. Those can all mesh cleanly together if you are willing to adjust your 2 foot tile size by a micron or two.

But now you have a triangular region which you have to tile in a square pattern. That will be an interesting challenge. [Actually, you have no freedom here. Once you place the first tile, the rest of the pattern is fixed. And contradictory].

 

[PeterDonis]

in the one second it takes me (by my clock) to cross a megaparsec, it will have expanded by 1/4109 (0.024 percent)

More precisely, if we assume that the two objects you are traveling between are both moving with the Hubble flow (i.e., they are not gravitationally bound to each other), it will take you, not one second, but slightly longer--1 plus 0.024 percent seconds. Actually, it will be an average value of the amount of expansion over the length of your trip; if we assume that the expansion is linear to a good enough approximation during your trip, the elapsed time by your clock would be 1 plus 0.012 percent seconds.

However, you were not talking about this in your OP, but about how your Hubble sphere would "appear" to you if you were in this state of motion (one second by your clock to travel 1 megaparsec according to comoving observers--observers moving with the Hubble flow). Those are different questions. The elapsed time question answered above can be answered by just looking at the patch of spacetime that contains your worldline for the one second that elapses by your clock. The question about how the Hubble sphere "appears" to you cannot; to answer that question, you need to look at a much larger patch of spacetime, that includes your entire Hubble sphere for the time it took light to reach you from objects that are on your Hubble sphere "now". The latter patch is nowhere near flat even if the former patch can be approximated reasonably well as flat over the time period you used.

 

[PeterDonis]

The minuscule errors would not be apparent or measurable in any local region.

But the errors build up; they don't cancel out. So your tiling would stop working outside the local region in which you started. And if you started tilings in two different local regions, they would not match up when they met up with each other.

 

[Ibix]

I could tile the Earth (some equivalent perfect sphere) no problem

Not on a square grid, though. And that's the point. You can start a grid anywhere and cover part of the sphere, but you can't extend anyone grid to cover the whole of the sphere.

Consider your two inch tiles. Make a row of them circling the equator. Now make another row next to it, touching on the north side of the first one. That can't quite work because this row is slightly shorter - at latitude $\theta$ the distance around the world (a sphere of radius $R_E$) is $2\pi R_E\cos\theta$. For the first row off the equator $\theta\approx 10^{-8}$ radians and you can easily absorb the error - it's about a nanometer. But by the time you get to 60° north, there is only half the distance around the world that there is at the equator. So you've only got space for half the tiles and you can't possibly line them up with the equatorial ones.

 

[pervect]

I've been thinking about your tile analogy: I could tile the Earth (some equivalent perfect sphere) no problem. Every tile 2x2' tile would fit flat and snugly with those around it, seemingly level. The minuscule errors would not be apparent or measurable in any local region.

I don't think you can. Here's an equivalent problem. Take a single flat piece of graph paper, that isn't stretchy. Real paper will be a bit stretchy - if you tug on the edges of the paper, something drawn on it will distort a tiny bit as the fibers stretch - but it's not very pronounced. That's what I mean by stretchy. A certain amount of stretchiness is necessary, though, for the paper to bend at all if it has finite thickness, by the way. If the fibers making up the paper weren't a little stretchy, they'd break

Now, try and wrap this flat sheet of graph paper around a ball. You'll find that you always have excess material.

This is easy enough to try if you're skeptical. You don't need graph paper - some reasonably rigid (in the above sense) wrapping paper will do. In this version, it becomes "gift wrapping a baseball".

 

[PeterDonis]

Now, try and wrap this flat sheet of graph paper around a ball. You'll find that you always have excess material.

A good contrasting exercise is to wrap the flat sheet of graph paper around a cylinder--which can be done without any excess material. This is a good illustration of what "flat" means in geometry (and of the difference between extrinsic and intrinsic curvature).

 

[pervect]

Spherical triangles gives a slightly more rigorous illustration of the difficulties of tiling the sphere, with the concept of "excess angle". It's a bit different from the issue of tiling the sphere with squared, but related.

The key result is that the sum of the angles of a spherical triangle is always greater than 180 degrees. See for instance https://en.wikipedia.org/wiki/Spherical_geometry#Relation_to_Euclid's_postulates

A statement that is equivalent to the parallel postulate is that there exists a triangle whose angles add up to 180°. Since spherical geometry violates the parallel postulate, there exists no such triangle on the surface of a sphere. The sum of the angles of a triangle on a sphere is 180°(1 + 4f), where f is the fraction of the sphere's surface that is enclosed by the triangle. For any positive value of f, this exceeds 180°.

The angular excess is proportional to the area of the triangle, so halving the length of a side causes the excess angle to drop by a factor of 4.

In the limit, one has 360 degrees in a circle at any point, and greater than 360 degrees for the sum of six finite sized equilateral spherical triangles. So they simply can't fit.

Angular excess can be directly related to the concept of the Riemann curvature in two dimensions. For higher dimensions, one needs something more sophisticated. Riemann curvature reduces to a single number in two dimensions, but not for higher dimensions.

 

[Chris Miller]

Thank you all. You've made it abundantly clear, and I understand, one cannot perfectly tile a perfect sphere with perfectly square tiles of any size. And not just squares, but any polygon? Or collection of polygons? But if locally immeasurable curvatures and distortions were permitted (say in the grouting), then?

@PeterDonis

However, you were not talking about this in your OP, but about how your Hubble sphere would "appear" to you if you were in this state of motion (one second by your clock to travel 1 megaparsec according to comoving observers--observers moving with the Hubble flow).

Yes, that's the question. Say, for "simplicity," even a nanosecond (or Planck unit) instead of a second, i.e., c in the sense that .9999... = 1.

The elapsed time question answered above can be answered by just looking at the patch of spacetime that contains your worldline for the one second that elapses by your clock. The question about how the Hubble sphere "appears" to you cannot; to answer that question, you need to look at a much larger patch of spacetime, that includes your entire Hubble sphere for the time it took light to reach you from objects that are on your Hubble sphere "now". The latter patch is nowhere near flat even if the former patch can be approximated reasonably well as flat over the time period you used.

So ahead, does one "now" remain equidistant from the "former" Hubble sphere? Had the Hubble sphere "now" formerly been receding at 2c? How about behind? Is there "now" only empty space? Isn't time dilation still a thing? Do the countless megaparsec-separated (by their measure) observers you pass (almost simultaneously by your clock) watch you come and go for cumulative quadrillions of years by their clocks?

 

[jbriggs444]

one cannot perfectly tile a perfect sphere with perfectly square tiles of any size. And not just squares, but any polygon? Or collection of polygons?

Polygons are probably not the right idea to be chasing here. One can tile a sphere with triangles -- in a way we call a tetrahedron [or an octahedron or an icosahedron]. One can tile it with squares -- in a way we call a cube. One can tile it with pentagons -- in a way we call a a dodecahedron.

But these approaches do not give you a tiling that translates smoothly between a flat piece of paper and the surface of a sphere. A square tiling on a piece of paper would typically have four squares meeting at each vertex. On a cube there are just three.

There is an intrinsic difference between a flat piece of paper and a sphere. That difference is intrinsic curvature.

You might find it both entertaining and educational to read Sphereland.

 

[Ibix]

icosahedron

And you can further subdivide the icosahedron's triangles, which is an excellent way of building "spheres" in 3d modelling. But the triangles are typically different sizes and some of those triangles meet 6 at a corner and some 5 at a corner.

 

[PeterDonis]

does one "now" remain equidistant from the "former" Hubble sphere?

There is no well-defined answer to this question, or to all the other questions you are asking, because there is no well-defined "now" for you. It depends on how you choose your coordinates. If you are a "comoving" observer, there is a "natural" choice of coordinates, which is just the standard FRW coordinates that cosmologists use. But if you are not a comoving observer, and you can't be if you are traveling between galaxies, then there is no "natural" choice of coordinates for you.

 

[PeterDonis]

Do the countless megaparsec-separated (by their measure) observers you pass (almost simultaneously by your clock) watch you come and go for cumulative quadrillions of years by their clocks?

This question, at least, can be answered since it is about direct observables. The answer is yes (assuming you do "come and go", i.e., change direction to pass by the same megaparsec-separated observers multiple times).

 

[Chris Miller]

All your replies to my questions and (misguided) conjectures are informative, and make me wish I knew more, understood more.

@jbriggs444

One can tile a sphere with triangles ...

Doh! Of course! Why didn't I see that? Six equilateral triangles with apexes at either pole meeting at the equator. Not a sphere, but the more you subdivide them into smaller equilateral triangles the closer you can approximate (adhere to) a sphere. So maybe there is a way to tile the hypercube with small regions of locally flat spacetime?

@PeterDonis

This question, at least, can be answered since it is about direct observables. The answer is yes (assuming you do "come and go", i.e., change direction to pass by the same megaparsec-separated observers multiple times).

I appreciate the clarification, but don't fully get why I must bounce back and forth as if between two mirrors to observe for myself a quadrillions of years old universe. I've also given some thought to what exactly "no well defined answer" means, especially since I intend to plagiarize/paraphrase it in my SF. I know this is wrong, but it gives off an undecidedness vibe that I, for some reason, associate with QM.

 

[jbriggs444]

Doh! Of course! Why didn't I see that? Six equilateral triangles with apexes at either pole meeting at the equator. Not a sphere, but the more you subdivide them into smaller equilateral triangles the closer you can approximate (adhere to) a sphere.

A tetrahedron (three equilateral triangles meeting at the north pole) stuck to a second tetrahedron (three equilateral triangles meeting at the south pole) does not make a figure that can be inscribed within a sphere. An octahedron (four triangles meeting at the poles to make two four-sided pyramids joined square face to square face at the equator will work).

But you already realize this.

Dividing each face into smaller equilateral triangles does nothing to unflatten those faces. You still need to unflatten them up against the surface of the sphere.

Those of us who played DnD in our youth have a firm picture of the regular polyhedra. The six sided regular polyhedron has squares for faces, not triangles. It is called a "cube".

So maybe there is a way to tile the hypercube with small regions of locally flat spacetime?

For any local region, you can pretend that the region is flat. But pretending does not make it so. The global topology does not change just because you can flatten out any particular region by distorting it a little. The little distortions build up. Eventually the paper tears. Or wrinkles. Or both.

gives off an undecidedness vibe

It is not that there is any ambiguity about the answer to a properly posed question. It is that the question, as posed, is ambiguous -- and pretends that it is not.

 

[Chris Miller]

@jbriggs444

Dividing each face into smaller equilateral triangles does nothing to unflatten those faces. You still need to unflatten them up against the surface of the sphere.

Been looking at the icosahedron, which approximates a sphere tiled with equilateral triangles. Are there not polyhedrons composed of infinitely more uniform polygons that approach spherical?

The little distortions build up. Eventually the paper tears. Or wrinkles. Or both.

I was under the impression, from your clarification and examples, distortions could be disseminated, localized.

Which may or may not tie into your metaphor.

It is not that there is any ambiguity about the answer to a properly posed question. It is that the question, as posed, is ambiguous -- and pretends that it is not.

Sorry, any pretense on my part was actually genuine ignorance. I won't try to rephrase my question since I sense my lack of understanding has us in the sort of discourse loop I tend to enjoy when Jehovah Witnesses call. But I might ask, how should it be posed?

 

[jbriggs444]

Been looking at the icosahedron, which approximates a sphere tiled with equilateral triangles. Are there not polyhedrons composed of infinitely more uniform polygons that approach spherical?

What is a "uniform polygon"? Are you thinking of a Uniform Polyhedron?

We've already enumerated all of the regular polyhedra. An icosahedron is as good as you can get that way. A regular polyhedron is one that can be constructed from a set of congruent convex regular polygons. A regular polygon is one that is equilateral and equiangular.

There are also only finitely many uniform polyhedra that are suitable for your purposes. All I know about them is what I can Google up.

I have to repeat myself and say that thinking about polygons is irrelevant to the matters at hand. You seem to be trying to avoid curvature by changing curves into sharp corners and then trying to avoid sharp corners by making enough of them so that you get a curve instead.

I was under the impression, from your clarification and examples, distortions could be disseminated, localized.

I am not sure what "disseminated, localized" means in this context.

You can flatten out any small section of a sphere, for instance (e.g. flattening a beach ball against the deck of a pool). But that does not change the global topology. It is still a [mis-shapen] sphere.

Sorry, any pretense on my part was actually genuine ignorance. I won't try to rephrase my question since I sense my lack of understanding has us in the sort of discourse loop I tend to enjoy when Jehovah Witnesses call. But I might ask, how should it be posed?

As I recall, you are trying to ask about whether the distance to the boundaries of the Hubble sphere shift when we change velocity.

So you first have to clarify what you mean by velocity -- velocity relative to what?

Then you have to clarify what you mean by what distance you are talking about. Distance from an event here and and now to exactly where and when? The when part is important -- does that change with your "velocity"?

 

[Chris Miller]

@jbriggs444

A regular polyhedron is one that can be constructed from a set of congruent convex regular polygons. A regular polygon is one that is equilateral and equiangular.

Thanks for clarifying the terminology (assumed flat, confused by "convex"). Not related to my OP, but surprised there are finite regular polyhedra . What if the constructing polygons only have to be equal but not regular? Still some maximum number of some optimal shape that can be used?

I am not sure what "disseminated, localized" means in this context.

I meant that the minuscule distortions ("sharp corners" approaching 180 degrees) do not accumulate or collect into a big distortion but are spread out as negligible errors in numerous Lorentz calculations. I can lay a 2' level around the Earth and determine its circumference, but only realize it must be curved when I wind up where I began.

So you first have to clarify what you mean by velocity -- velocity relative to what?

Relative to the CMB. Any direction that minimizes its wavelength in front (via Doppler effect) and maximizes it behind.

Then you have to clarify what you mean by what distance you are talking about. Distance from an event here and and now to exactly where and when?

Distance to Hubble horizon (about 14 billion light-years). Fourteen billion years ago.

The when part is important -- does that change with your "velocity"?

That's kind of my question. The spacetime shape/age of the universe in my frame of reference.

 

[PeterDonis]

The spacetime shape/age of the universe in my frame of reference.

There is no unique frame of reference for the state of motion you describe (large dipole anisotropy in the CMB). There would be a unique global inertial frame defined by this state of motion if spacetime were flat, but it isn't. In curved spacetime there are no global inertial frames. So, as I have said already in this thread, your question does not have a well-defined answer.