0

What is the spatial distance of two stars, if each of them is outside of the observable universe of the other? Even by the means of a thought experiment, we could not synchronize the two clocks of both stars, the synchronization being required for the indication of the spatial distance with respect to the reference frame of an observer, taking into account Lorentz contraction. By consequence, no distance could be defined, as it seems to be confirmed by the answer to this question. Distances beyond the observable universe would become meaningless.

Edit: In his answer below, John Rennie shows a current way how distance is calculated:

The distance to the star is the spatial distance between our current position ($t_0$,0,0,0) and the star's current position ($t_0$,x,y,z). To calculate this distance we have to construct a spacelike hypersurface with constant time t0. For the flat FLRW spacetime this is dead easy because the sapcetime naturally foliates into Euclidean submanifolds (x,y,z) of constant comoving time, and the distance is just: (…) where a(t) is the scale factor.

The problem is how to define a common $t_0$ for both our position and the star's position. This seems impossible because relativity of simultaneity is always limited to the observer's observed universe. An observer can define simultaneous points only within the limits of his observed universe. Even in a thought experiment this is not possible.

Community
  • 1
Moonraker
  • 3,135
  • We will tell you as soon as we get a sample of stars from outside of the observable universe. :-) – CuriousOne Jul 10 '16 at 08:13
  • Re your edit: see Doesn't dating the universe violate the concept of spacetime's inseparability?. For an FLRW universe like (approximately) ours there is a natural choice of time called comoving time. The distance I describe is between points of the ssame comoving time. To a good approximation all galaxies, stars, etc agree on the comoving time. – John Rennie Jul 10 '16 at 14:04
  • @John Rennie: The reference to a comoving frame for questions such like the universe is acceptable because it is obvious that every particle has its own proper time and we must define one frame among several options. But your application of this technique to distance measurements is not consistent with short-distance measurements which are based on a clock synchronization process according to the principles of relativity. Short-distance measurements and long-distance measurements should follow the same rules. – Moonraker Jul 10 '16 at 14:23

2 Answers2

0

You are mixing up two different ideas.

Let's assume that the spacetime geometry of our universe can be approximately described by the flat FLRW metric. That means there is a coordinate system, that we call the comoving coordinates, $(t,x,y,z)$, that we can use to label all points in the universe. In this coordinate system our location is $(t_0, 0, 0, 0)$, where $t_0$ is the age of the universe as observed by us i.e. 13.8 billion years or so.

For a star to be in our observable universe just means that the star must have been at some past spacetime point $(t \lt t_0,x,y,z)$ such that a light ray can get from that past location to us. More precisely, there must be a null geodesic that connects us and that position of the star.

But the distance to the star isn't simply the spatial separation of these two points, because you're comparing positions at different times. Admittedly we tend to talk about, for example. the most distant galaxy being $13.4$ billion light years away, but of course that isn't true. It was $13.4$ billion light years away when the light we can currently see left it - it's actually around 46 billion light years away now.

The distance to the star is the spatial distance between our current position $(t_0,0,0,0)$ and the star's current position $(t_0,x,y,z)$. To calculate this distance we have to construct a spacelike hypersurface with constant time $t_0$. For the flat FLRW spacetime this is dead easy because the sapcetime naturally foliates into Euclidean submanifolds $(x,y,z)$ of constant comoving time, and the distance is just:

$$ s = a(t)\sqrt{x^2 + y^2 + z^2} $$

where $a(t)$ is the scale factor.

To return to where we came in: the distance between the stars is defined no matter how far apart the stars are, and regardless of whether they share the same observable universe. If you define distance as the light the light travelled to reach us then you are correct that this isn't defined for an object farther away than the cosmological horizon. However this isn't what we normally mean by distance, and it isn't what Sam means by distance in his answer to the question you linked.

John Rennie
  • 355,118
  • Thank you for these indications, but I doubt that distances beyond the observed universe may be calculated in the way you have shown, this does not seem to comply with the synchronization principles of relativity, see my EDIT. – Moonraker Jul 10 '16 at 12:58
-3

This is a comment and a suggestion. I think what you mean to find is the length of the geodesic connecting the two stars. You would need the metric in universe 1, universe 2 and the effective metric at the boundary of the two universes obtained by imposing the boundary conditions on the metric. Once you have these quantities you can calculate the length of the geodesic connecting the two stars. This is possible only if you have the information about the other universe which as pointed out by CuriousOne in the third comment is not possible as for that you will have to see beyond the event horizon.

Abhishek Pal
  • 511
  • 1
    How, exactly, are you going to get the metric in "the other universe" and how are you going to pass the border between the two? It's not the kind or border where one can get trough with a passport and a visa... – CuriousOne Jul 10 '16 at 08:18
  • 1
    i gave a solution . Obviously metric of the other universe cannot be obtained unless we discover some interaction between the two universes. Once the metric is obtained then the above procedure holds. Its true that inter-universe passports have not been made available but once interactions between the two universes(if present) are found then it wont be surprising to find a flux of refugees and tourists to and fro the other universe. – Abhishek Pal Jul 10 '16 at 08:24
  • I didn't see a solution, merely a suggestion that you know (how?) what's on the other side of an event horizon. The very definition of "observable universe" is that we don't cross said event horizon and that whatever lies beyond... if anything lies beyond, which we also can not know, is in principle unknowable. – CuriousOne Jul 10 '16 at 08:29
  • what you are saying is true and if you ask me then in my opinion the question itself does not make sense at present day. Maybe i should have written the "answer" in the comment box. Nevertheless it would be exciting to find a prove of multiverses and maybe interactions between them in which case my above "comment" is a heurestic algorithm. – Abhishek Pal Jul 10 '16 at 08:34
  • 1
    @AbhishekPal Do you really consider this a comment, and not an answer to the question? In that case, you shouldn't post it as an answer. That being said, maybe it does constitute an answer. If so you should remove the first sentence. – David Z Jul 10 '16 at 16:47