B Shape of Hubble sphere at relativistic velocity

[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.