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So in this thread, Can space expand with unlimited speed?, the author Pulsar made amazing diagrams of different horizons and paths for a benchmark model that describes our current universe, and gave a great explanation of how to interpret the diagram. Does someone have a version of the diagram for an empty and negatively curved universe? I'm trying to visualize what happens in that universe, and it seems the particle horizon at our age today 13.7 Gyr would be infinite, and I'm not sure how to interpret this result.

Qmechanic
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ABC
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1 Answers1

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I'll assume that empty and negatively curved means $k<0$ and $ρ=p=Λ=0$.

In that case the Friedmann equations reduce to $\dot a = \pm\sqrt{-k}$ and $\ddot a=0$. The solutions are $a(t)=\pm\sqrt{-k}\,(t-t_0)$. I'll take $t_0=0$, and I'll assume you picked the positive-slope solution, $a(t)=t\sqrt{-k}$, since that has a "big bang" at a finite time in the past.

I put "big bang" in scare quotes because it is actually just a coordinate singularity. You can analytically extend the spacetime beyond it, and when you do, you get Minkowski space.

This diagram, which I adapted from this blog post, shows the relationship between FLRW and polar inertial coordinates.

$ξ$ and $τ$ are the inertial coordinates, and the metric is $ds^2=dτ^2-dξ^2-ξ^2dΩ^2$, where $dΩ^2=dθ^2+\sin^2 θ\,d\phi^2$ – just standard polar coordinates for Minkowski space. The $θ$ and $\phi$ coordinates are shared between the Minkowski and FLRW coordinate systems, and are suppressed as usual. The Minkowski $(ξ,τ)$ and FLRW $(r,t)$ are related by

$$t = \sqrt{τ^2-ξ^2}, \qquad r = \frac{1}{\sqrt{-k}} \tanh^{-1}\,\frac{ξ}{τ}$$

The black curves are surfaces of constant FLRW time, and the blue lines are lines of constant FLRW position, or the worldlines of test particles moving with the Hubble flow (though there is really no Hubble flow in this cosmology, since the universe is empty).

The brown lines show the past light cone of a point at cosmological time $12$. You can see that it crosses all of the blue lines, and even if you added infinitely many more blue lines at evenly spaced rapidities, it would still cross all of them before hitting cosmological time $0$. That's why the particle horizon works out to be infinitely large.

benrg
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  • I apologize for the delayed response and thank you for the work you put into it. A few clarification questions. a) Is the y axis "tau" the same as the t in the metric? b) It looks like the horizontal axis is the comoving coordinate, then how come the "comoving observer" lines not be vertical? c) What is the equation for "constant cosmological time"? Thank you. – ABC Sep 25 '22 at 04:38
  • @ABC See the edit. $t$ for $τ$ was just a typo. The horizontal axis is the Minkowski radial coordinate $ξ$, not the FLRW radial coordinate $r$. I added equations giving the FLRW coordinates in terms of the Minkowski coordinates. Constant cosmological time is constant $t$. – benrg Sep 25 '22 at 05:38
  • Great thank you. a) I probably have a misunderstanding about the Minkowski metric --- are the coordinates and time in Minkowski not comoving and proper time like in the FLRW metric? b) I get that Minkowski is for empty flat and static universe, but we have FLRW and we make graphs like this using FLRW for other universes, so why for empty and negatively curved universe we have to bring Minkowski metric into the picture? – ABC Oct 11 '22 at 14:52
  • c) The "past light cone" transverses all speed, but it does intersect with the x-axis at a finite value at t=0. Doesn't that mean only objects within that /finite/ distance will be visible to us? The particles can have different speed but they need to be within the "past light cone" distance. Thus we're not seeing "infinitely" faraway? – ABC Oct 11 '22 at 14:52
  • @ABC (a) You can put FLRW coordinates on Minkowski space with $a=\text{const}$, and the inertial coordinates would be the comoving and cosmological time of that FLRW coordinate system, but it's a different one. It has $k=0$ so it doesn't fit your criterion of negative curvature. (b) You don't have to use the inertial coordinates; I just thought it was the easiest way to see the causal structure of this cosmology. If you plot comoving position vs conformal time, then the big bang is at $η=-\infty$, so the past light cone reaches arbitrary comoving distances, but that seems less enlightening. – benrg Oct 13 '22 at 18:11
  • @ABC (c) No, all the blue lines are within the future light cone of the origin, so the past light cone crosses all of them at a positive cosmological time. That means it reaches arbitrarily large comoving distances. It doesn't reach arbitrary coordinate distance, but the comoving distance is normalized to a "present day" that is far in the future (arbitrarily large $τ$ as you get farther from the center). – benrg Oct 13 '22 at 18:13
  • Great thank you. About (c) Is "coordinate distance" the proper distance or something else? I thought "comoving" coordinate is the coordinate used in the metric, and multiplied by scale factor will be the "proper distance", thus if cmoving is infinite, then proper distance (frozen in time) is also infinite. But from what you're saying it seems "coordiante distance" is something else. – ABC Oct 13 '22 at 19:57
  • @ABC Sorry, that was a braino, I meant metric distance, i.e. $\int ds$ along a path of constant time (probably $t$, but could be $τ$). There is a finite upper bound on the metric distance from the center to the points of intersection of the past light cone with the blue lines, but no upper bound on the metric distance from the center to the "now" points of the blue lines. – benrg Oct 13 '22 at 20:01
  • " but no upper bound on the metric distance from the center to the "now" points of the blue lines." So it $\bf {does}$ reach $\bf{infinite}$ metric distance? Does this contradict "That means it reaches arbitrarily large comoving distances. It doesn't reach arbitrary coordinate distance"? Thank you. – ABC Oct 13 '22 at 20:16
  • @ABC I meant to say "That means it reaches arbitrarily large comoving distances. It doesn't reach arbitrary metric distance." And what I meant by that is that the metric distance to the actual points of intersection with the past light cone is bounded, but the comoving distance (which can be defined as another metric distance, but to "now" points that are far in the future) does go to infinity. That was meant to resolve your confusion about the apparent infinite size of the past light cone: it's only infinite in comoving distance, which is a somewhat artificial distance measure. – benrg Oct 13 '22 at 20:32
  • The relation between the Minkowski and FLWR time $t = \sqrt {\tau^2 - \xi^2}$, does this come from special relativity Lorentz transformation? If so, how could SR be applied in a curved space? – ABC Oct 13 '22 at 21:53
  • @ABC SR can be applied here because it's flat spacetime. The "space" that is curved is a subspace of it. It's like a sphere in Euclidean 3-space, which is positively curved even though Euclidean space is flat. You can use Euclidean geometry to describe it. If you had a bunch of concentric spheres in Euclidean 3-space, you could identify them by radius: $r=\sqrt{x^2+y^2}$. That's analogous to $t$, which is the "radius" of the "spheres" in this cosmology. – benrg Oct 14 '22 at 16:17