Expansion of the universe, a poor choice of words?

[Flatland]

How did you leave Earth if you never accelerated?

Yeah that's what I meant by the kinetic energy of the spaceship blasting off. I've always thought that it was the symmetry breaking (ie. one party having to turn around) that caused the time dilation.

 

[bahamagreen]

PeterDonis
I must admit that 3-sphere, gluing, and quotient space are things I've not encountered before (I have never looked at topology at all), but I'm somewhat encouraged that my intuition might be leading toward at least the right mathematical direction. Can you recommend an introductory level topology reference (perhaps something online) that emphasizes basic foundation?

 

[PeterDonis]

Can you recommend an introductory level topology reference (perhaps something online) that emphasizes basic foundation?

Unfortunately I have learned what I know of topology piecemeal as I encountered particular concepts in the course of studying physics. So the only references I'm familiar with are the parts of physics textbooks that discuss topology. Carroll's online lecture notes on GR have a fairly good, if brief, discussion early on.

 

[Orodruin]

How did you leave Earth if you never accelerated?

This has nothing to do with the resolution of the paradox. You could just as well take a spaceship passing by (eg, the time difference would be the same the second lap).

The point is that the situation is not symmetric and you have to consider the geometry of the space-time properly.

That acceleration is the cause of the age difference is a bit of a misconception.

 

[George Jones]

I'm not trained in the jargon so my words probably did hurt your ears. I looked up 3-sphere and think I recognized the idea I was referring to... in topological construction - gluing:

"A 3-sphere can be constructed topologically by "gluing" together the boundaries of a pair of 3-balls. The boundary of a 3-ball is a 2-sphere, and these two 2-spheres are to be identified. That is, imagine a pair of 3-balls of the same size, then superpose them so that their 2-spherical boundaries match, and let matching pairs of points on the pair of 2-spheres be identically equivalent to each other. In analogy with the case of the 2-sphere (see below), the gluing surface is called an equatorial sphere.
Note that the interiors of the 3-balls are not glued to each other."

Yes, I see, you were talking about how the 3-sphere can be defined using a quotient space (but see below for limitations of that). This is mathematically true, but the fact plays no role at all in the physical models used in cosmology.

Also, note that the construction you refer to does not make the entire 3-sphere a quotient space; as the quote you gave shows, only the "equatorial sphere" is a "gluing surface" (i.e., a quotient space). This is similar to the way that a 2-sphere can be constructed by "gluing" the boundaries of two 2-disks together; the "gluing curve" is a circle which corresponds to the "equator" of the 2-sphere. But the rest of the 2-sphere is not a quotient space; only the "equator" is.

Actually, the passage that bahamagreen quotes from Wikipedia does construct the entire 3-sphere as a quotient space. Let ##S_3## be a 3-sphere and ##B_3## be a 3-ball. Then,

$$S_3 \approx \left( B_3 \times B_3 \right)/ \sim ,$$

where "##\approx##" means "is homeomorphic to" and ##\sim## is the equivalence relation on all of ##B_3 \times B_3## generated by the Wikipedia quote.

 

[PeterDonis]

the equivalence relation on all of ##B_3 \times B_3##

As I'm reading the Wikipedia quote, the equivalence relation is only on the boundaries of the two 3-balls--it identifies "corresponding" points on the two boundaries, which are each 2-spheres.

Since I have trouble visualizing 3-balls embedded in a higher dimensional space , I am understanding this by analogy with the 2-sphere case. The 2-sphere construction described in the Wikipedia article, as I understand it, is: take two 2-balls, and identify points on their boundaries--each boundary is a 1-sphere ##S^1##, aka a circle, and we identify "corresponding" points on the two circles. The resulting circle, composed of "points" each of which is actually an equivalence class (a set of two points, one on the boundary of each of the two 2-balls), becomes the "equator" of the 2-sphere. The "north pole" of the 2-sphere is then the point at the "center" (heuristically speaking, I know we're only talking topological spaces so there is no "center" point in terms of distance, but hopefully my meaning is clear) of one of the 2-balls, and the "south pole" of the 2-sphere is the point at the "center" of the other 2-ball. Other points in the northern and southern hemispheres correspond to interior points in one or the other of the 2-balls. But I don't see where there is any equivalence relation between points in the two hemispheres, i.e., between points not on the boundaries of the two 2-balls.

 

[PAllen]

As I'm reading the Wikipedia quote, the equivalence relation is only on the boundaries of the two 3-balls--it identifies "corresponding" points on the two boundaries, which are each 2-spheres.

Since I have trouble visualizing 3-balls embedded in a higher dimensional space , I am understanding this by analogy with the 2-sphere case. The 2-sphere construction described in the Wikipedia article, as I understand it, is: take two 2-balls, and identify points on their boundaries--each boundary is a 1-sphere ##S^1##, aka a circle, and we identify "corresponding" points on the two circles. The resulting circle, composed of "points" each of which is actually an equivalence class (a set of two points, one on the boundary of each of the two 2-balls), becomes the "equator" of the 2-sphere. The "north pole" of the 2-sphere is then the point at the "center" (heuristically speaking, I know we're only talking topological spaces so there is no "center" point in terms of distance, but hopefully my meaning is clear) of one of the 2-balls, and the "south pole" of the 2-sphere is the point at the "center" of the other 2-ball. Other points in the northern and southern hemispheres correspond to interior points in one or the other of the 2-balls. But I don't see where there is any equivalence relation between points in the two hemispheres, i.e., between points not on the boundaries of the two 2-balls.

My understanding matches yours. I found it quite intriguing to try to imagine a 3-sphere as two balls with their surfaces identified, and each ball interior being 'hemis-3-sphere' of the total 3-sphere.

 

[George Jones]

Yes, only the boundaries are identified, Not all of the equivalence classes need to have more than one element.

 

[PeterDonis]

Not all of the equivalence classes need to have more than one element.

I still don't see how this helps. To use the construction of ##S^2## again as an example, the "points" on the equator are really pairs of points, one on each ##S^1## boundary of a 2-ball ##B^2##. These can be thought of as elements of the set ##S^1 \times S^1##. But if we think of these as equivalence classes of elements in ##S^1 \times S^1##, each one only has one element--the element that pairs the specific "matching" points on each 2-ball boundary. In other words, the set of equator "points" on the 2-sphere already picks out a small proper subset of all the elements of ##S^1 \times S^1##.

Now, I suppose, since the set ##S^1 \times S^1## is a subset of the set ##B^2 \times B^2##, the equator "points" can also be thought of as equivalence classes of elements of ##B^2 \times B^2##--each one with only one element. But ISTM that the rest of the points in ##S^2## are not mapped to pairs of points in the two 2-balls; they are just points in one or the other 2-ball. For example, the "north pole" of the 2-sphere is the "center" point in one 2-ball; I don't see how it corresponds to a pair of points, one in each 2-ball, which is what would be necessary, ISTM, to treat it as an element of ##B^2 \times B^2## (or an equivalence class of such elements with one member).

 

[nikkkom]

one obvious logical thought that springs to my mind is:
if the density of all the matter in the observable universe was very high 13 billion years ago,

Yes

then that means that all the matter in the early observable universe was very close together,

No.
Every two selected bits of matter were closer, yes.
*All* matter was very close? If there is infinitely much of it, then not.

Analogy: if you take a set of all real numbers and divide each by one billion, the resulting numbers "become closer to each other". But you can still easily find two numbers which are arbitrarily far away from each other. Shrinking infinity by a finite amount does not make it finite.

and that implies to me that there was a center.

No. Even in a finite Universe, how does that follow?