I'm confused about a detail in cosmology. Consider a static closed universe, of the following metric (consider $a$ as a simple constant with units of length) : \begin{equation}\tag{1} ds^2 = dt^2 - a^2 \big( d\chi^2 + \sin^2 {\chi} \; (d\vartheta^2 + \sin^2 {\vartheta} \; d\varphi^2) \big). \end{equation} Here, the radial coordinate takes values on a bounded domain : $0 \le \chi \le \pi$. The proper radial lenght is defined by this line element ($d\vartheta = d\varphi = 0$) : \begin{equation}\tag{2} d\ell^2 = a^2 \, d\chi^2. \end{equation} Integrating gives trivially $\ell = a \, \chi$.

The proper volume of the whole space is easily found to be $\mathcal{V} = 2 \pi^2 a^3$, and the area of a sphere of coordinate radius $\chi$ is given by $\mathcal{A}(\chi) = 4 \pi a^2 \sin^2 {\chi}$. Thus $\mathcal{A}(0) = \mathcal{A}(\pi) = 0$ and $\mathcal{A}_{\text{max}} = \mathcal{A}(\frac{\pi}{2}) = 4 \pi a^2$.

In a closed universe, it is important to not confuse length and distance.

The question is this :

What is the maximal proper distance from a given stationary observer in this space : $\mathcal{D}_{\text{max}} = \pi \, a$, or $\mathcal{D}_{\text{max}} = \frac{\pi}{2} \; a$ ?

I'm confused because of the area behavior, and I believed that the maximal distance is $\mathcal{D}_{\text{max}} = \pi \, a$ and not $\mathcal{D}_{\text{max}} = \frac{\pi}{2} \; a$, despite the fact that $\mathcal{A}(\pi) = 0$. I'm not sure anymore that it's making sense ! I may have confused distance with length and I need a confirmation.

If you built a linear structure in that space, its maximal length should be $2 \pi \, a$, and the distance between both extremities should be 0, right ? Or is the lenght actually $\pi \, a$ ??

  • $\begingroup$ If you're considering only the spatial part of your metric, which happens to be a Riemannian manifold, your question doesn't make sense. In Riemannian manifolds there is no notion of "maximal" distance or length since one can always make ripples around such a path, ending up with a path which is longer than the original one. Now if you mean the maximal distance along a geodesic, it's just a matter of integrating along a great circle of the 3-sphere over the entire range of the parameter. $\endgroup$ – Mr. K Oct 5 '16 at 14:23
  • $\begingroup$ I don't understand why it doesn't make sense. The radial distance element is defined as $d\ell = a \, d\chi$ (equation (1)). Then the radial distance between observer $\mathcal{O}$ located at $\chi = 0$ and an arbitrary point of coordinate $\chi > 0$ is just $\mathcal{D} = a \, \chi$. The problem is to define the fartest point from $\mathcal{O}$. Since $0 \le \chi \le \pi$, the maximal distance is apparently $\mathcal{D} = \pi \, a$ (not to be confused with the maximal length, which is arbitrary, depending on the curve). What you described is the length of a curve, not a distance. $\endgroup$ – Cham Oct 5 '16 at 14:51
  • $\begingroup$ The radial distance is a geodesic, so it agrees with my comment above. We're just using different nomenclature. For me (and for geometers I guess), distance and length should mean the same thing. $\endgroup$ – Mr. K Oct 5 '16 at 15:00
  • $\begingroup$ I don't agree with your last sentence. length and distance are not the same (or why would we have two words for the same thing ?). However, it is clear that an arbitrary curve (not a geodesic) could have an arbitrary length. This is obvious. The word "distance" is usually used to define the minimal curve length between two points (i.e. the length of a geodesic, by definition). $\endgroup$ – Cham Oct 5 '16 at 15:09
  • $\begingroup$ @Mr.K, so dou you agree that the maximal distance between two points in the closed space is $\mathcal{D}_{\text{max}} = \pi \, a$ ? And that the maximal length of a linear structure is $\ell_{\text{max}} = 2 \pi \, a$ while the distance between both of its extremities is 0 ? $\endgroup$ – Cham Oct 5 '16 at 15:18

The spatial component of the spacetime you describe is a 3-sphere. The largest distance is obtained when $\Delta t=0$ so we can ignore the time direction of the spacetime.

A 3-sphere is natural extension of a 2-sphere. Where a 2-sphere consists of a circle with radius $\sin(\theta)$ for every value of $\theta$, a 3-sphere consists of 2-sphere with radius $\sin(\chi)$ for every value of $\chi$. The largest distance on a 2-sphere is the distance from the north to the south-pole. This corresponds to varying $\theta$ by $\pi$. When the radius is $a$, this gives a distance of $\pi a$.

A similar reasoning lets you travel from one side of the 3-sphere to the other by varying $\chi$ by $\pi$. Here the distance is again $\pi a$.

For the second part of your question. If by maximal length you mean the distance traveled along a space-like path, then there is no maximal length. A path can curl around the 3-sphere and get an infinite length. Even when you demand the path to be straight, i.e., a geodesic, there is no maximal length as the path could travel around the 3-sphere with a small time-like component, thereby forever circling the 3-sphere and obtaining infinite length.

The length of a straight path around the 3-sphere without a time-like component in this coordinate system is $2\pi a$, just as for the 2-sphere.

| cite | improve this answer | |
  • $\begingroup$ Thanks, you then confirm what I was thinking : the longest linear structure in that space would have a length of $2 \pi \, a$ while both extremities are at a distance of 0. So the maximal distance (not length) is $\pi \, a$, and not $\frac{\pi}{2} \; a$. The behavior of the spherical area $\mathcal{A}(\chi)$ is weird, though. $\endgroup$ – Cham Oct 5 '16 at 14:58
  • $\begingroup$ Ok, about the area, I now have a clearer picture from the reduction of a simple 2-sphere. In this simple case, the equivalent of $\mathcal{A}(\chi)$ would be the circumference of the circle drawn on the surface of the sphere : $\ell(\chi) = 2 \pi \sin{\chi}$, which gives $\ell(0) = \ell(\pi) = 0$ while the limiting circles are on opposite sides of the sphere. But it is still weird to "visualize" the area for the 3-sphere. $\endgroup$ – Cham Oct 5 '16 at 15:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.