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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.

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Moonraker
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2 Answers2

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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
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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
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