Can we interpret the de Sitter universe as a spherical cosmic horizon null surface of finite radius, centered at Earth, and containing the Hubble volume of space where time is dilated and radial dimensions contract closer to the edge in such a way that objects closer to the edge do not recognize that they are radially contracted?

Everything is attracted to the edge, but the total radius remains more or less constant and emits de Sitter radiation at finite temperature.

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What you are describing is the static coordinate system for the de Sitter space, in which the metric could be written as $$ ds^2 = -\left(1-\frac{r^2}{\alpha^2}\right)dt^2 + \left(1-\frac{r^2}{\alpha^2}\right)^{-1}dr^2 + r^2 d\Omega_{2}^2. $$ This is a static universe (not just 'more or less'). We see that at $r=\alpha$ the metric has a cosmological horizon. The region $r<\alpha$ contains the operationally meaningful portion of the de Sitter space, which can be probed by a single observer located at origin.

The factor at $dt^2$ could indeed be interpreted as defining position-dependent time dilation. An object held at a fixed distance from the observer appears redshifted. If released, such object will be accelerating toward the horizon, so your description is accurate in this respect.

One point I object to in your description is the 'contraction' of radial dimension. The $dr^2$ part of the metric tells us how we actually measure the radial distance (from origin): $$ R = \int\limits _0^r\frac{dr}{\sqrt{1-\frac{r^2}{\alpha^2}}}.$$

At the same time your illustration is qualitatively correct in displaying objects in the $(r,theta)$ coordinates.

  • $\begingroup$ By length contraction I meant that anything moving fast enough from the observer will be relativistically contracted. $\endgroup$
    – Anixx
    Jan 12 '14 at 21:53
  • $\begingroup$ The R in your formula appears to be the "co-moving" distance, the sum of co-moving distances between objects, that is the distance, perceived by the traveler. While the distance from the static observer to any point will not exceed the radius of the horizon. $\endgroup$
    – Anixx
    Jan 12 '14 at 21:57
  • $\begingroup$ Can you also clarify whether the time dilation prevents any object leaving the cosmological horizon in finite time. $\endgroup$
    – Anixx
    Jan 12 '14 at 21:58
  • $\begingroup$ You say this universe is static, but what about de Sitter radiation? does it increase or decrease the cosmic horizon? $\endgroup$
    – Anixx
    Jan 13 '14 at 2:02
  • $\begingroup$ The dS horizon is observer dependent. If there is an event which occurred on horizon, the signal from it cannot ever reach observer at origin (for which this horizon is defined). But this horizon is just an ordinary null surface. From the point of view of an object crossing this horizon is not special in any way. $\endgroup$
    – user23660
    Jan 13 '14 at 17:15

Events are always undergoing acceleration as they evolve forward in the time dilated continuum. Therefore, when we look out into space beyond the solar system, and back in time, we are also looking down a time dilation gradient into slower time. The observer’s invariant relative rate of time is always faster than that in frames in the perceived past, and we find that as D → ~13.9 Gly, difference in the rate of time, denoted here as "dRt", → 1 s/s, recessional V → c, and lateral V → 0, just as it does near the event horizon of a black hole. Slower time results in lower frequency and the Hubble shift.

Assuming a Hubble constant of 70 $\frac{\frac{km}{s}}{Mpc}$, we find the apparent recessional velocity reaches c at 4282.7494 Mpc = 13.968062372 Gly.

For a 1s/s dRt at this distance the rate of change is: 1/13968062372 = 7.1592*10^-11 s/s/ly = 2.3349516024*10^-4 s/s/Mpc.

So for each Mpc the $dRt = 2.3349516024 \cdot 10^{-4}$ s/s and: $c \cdot (1 + dRt) = (299792.458) m/s \cdot ((1+(2.3349516024 \cdot 10^{-4})) s = 299862.458 m$ and: 299862.458 - 299792.458 = 70 km/s/Mpc = the Hubble constant

This indicates that the forward evolution of time includes a universal constant of acceleration.

Because we are always being accelerated forward in the rate of time, and therefore apparently space, events in the past must appear to accelerate away from us in the opposite direction.

This also creates the impression we are at the center of the universe and leading it in its evolution.

Please also note that the solution works for a difference in the rates of time of exactly 1 s/s. Does any other theory you know of account for a 1 s/s difference in the rates of time between us and 13.9 Gly?

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    $\begingroup$ Hi Cass, welcome to PSE! First of all, thank you for taking the time to type this. Please note that this site supports mathjax (which is similar to Latex). You can use it to have professional looking equations, which will make your post look much better. Cheers! $\endgroup$ Feb 14 '18 at 23:36
  • $\begingroup$ I've made some example latexification. The others are going similarly. Welcome! $\endgroup$
    – peterh
    Feb 15 '18 at 0:36
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    $\begingroup$ This is difficult to understand because it's written in the language of arithmetic rather than algebra. The treatment of significant figures also shows a lack of basic scientific literacy. Events are always undergoing acceleration This doesn't make sense. An event is a point in spacetime. Events don't accelerate. as they evolve forward in the time dilated continuum. This is just gobbledygook. Time dilation doesn't apply to cosmological spacetimes. $\endgroup$
    – user4552
    Feb 15 '18 at 2:22
  • $\begingroup$ Thanks for the hint on the math. Events do evolve forward. They are "happenings". GR and QM describe the evolution of events within the continuum. Time dilation applies everywhere or there would be no gravity. $\endgroup$
    – CaptCass
    Feb 16 '18 at 1:29
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    $\begingroup$ Author's Note: This is a simple derivation of the acceleration constant IF time accelerates. It is not meant to be a proof of the acceleration. There seems to be some confusion over this. $\endgroup$
    – CaptCass
    Feb 17 '18 at 16:58

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