<p>The de Sitter spacetime is maximally symmetric, and so having a full set of Killing vector fields, it must have a static form produced by a timelike Killing vector field. 
One way to get without worrying about coordinate transformations is to take the $M\to 0$ limit of the general spherically symmetric lambdavacuum solution, which is the the Schwarzschild–de Sitter spacetime:
$$\mathrm{d}s^2 = -\left(1-\frac{1}{3}\Lambda r^2\right)\mathrm{d}t^2 + \left(1-\frac{1}{3}\Lambda r^2\right)^{-1}\mathrm{d}r^2 + r^2\,\mathrm{d}\Omega^2\text{,}$$
where $\mathrm{d}\Omega^2 = \mathrm{d}\theta^2 + \sin^2\theta\,\mathrm{d}\phi^2$ is the usual unit $2$-sphere. Except for minor notational differences, this is also identical to the static de Sitter slicing presented <a href="https://en.wikipedia.org/wiki/De_Sitter_space" rel="nofollow">here</a> in terms of different slices of a hyperboloid in Minwkoski $\mathrm{E}^{1,4}$, with $\alpha\equiv\sqrt{3/\Lambda}$.</p>

<p>In these coordinates, the timelike Killing field is simply $\partial_t$, which corresponds to a family of static observers with time dilation relative to the static observer at the origin given by
$$\dot{t}\equiv\frac{\mathrm{d}t}{\mathrm{d}\tau} = \frac{1}{1-r^2/\alpha^2}\text{,}$$
which diverges as $r\to\alpha$.</p>

<p>For <em>geodesics</em>, the Killing field also generates a conserved specific energy $\epsilon = \left(1-r^2/\alpha^2\right)\dot{t}$, which makes solving for radial geodesics not difficult. What's interesting about them is they take an infinite amount of coordinate time $t$ to reach $r=\alpha$, but only a finite amount of proper time $\tau$, and they have no trouble at all continuing beyond it. Additionally, time dilation for static observers becomes infinite there. This behavior is analogous to the event horizon of a black hole, except that the location of the de Sitter horizon is dependent on the choice of origin.</p>

<blockquote>
  <p>Can Hubble red shift be interpreted as time dilation?</p>
</blockquote>

<p>Yes. If we wish, we can think of the redshift due to a particle (e.g., galaxy) at some radial coordinate as a combination of two things: the special-relativistic time dilation of a galaxy with respect to a local static observer and the gravitational time dilation between static observers given above. In fact, the $\epsilon$ parameter above is exactly the (per-mass) energy of the particle as measured by a local static observer, i.e., the relative Lorentz gamma.</p>

CC BY-SA 3.0
2014-10-11T00:42:52.573Z
"What's interesting about them is they take an infinite amount of coordinate time t to reach r=α, but only a finite amount of proper time τ" - so from the point of view of any observer nothing can escape the sphere of finite radius centered at himself? "except that the location of the de Sitter horizon is dependent on the choice of origin." - looks like it is always centered at the observer, seems one can safely drop the Copernicus principle. "Yes. If we wish, we can think" so time and space stops at 5 Gpc from here. Well.
Anixx
CC BY-SA 3.0
2014-10-11T00:51:43.123Z
Does from your formula also follow that dispite finite volume, the internal (comoving) capacity of the sphere will be infinite due to length contraction on the edges? This by the way will also mean that the comoving volume capacity around a black hole should also be infinite (am I wrong?)
Anixx
CC BY-SA 3.0
2014-10-11T01:09:34.013Z
@Anixx that's a property of a particular time coordinate. Just like for a Schwarzschild black hole, things take an infinite amount of Schwarzschild time to cross, but a finite amount of proper time, or Kruskal–Szekeres time, or Gullstrand–Painlevé time, or Lemaître time, etc. The spacetime is quite well-defined beyond the horizon; it doesn't just end. ... Also, in this frame the Hubble volume is finite and static, $pi^2\alpha^3$, while in a comoving frame it's finite and expanding.
Stan Liou
CC BY-SA 3.0
2014-10-11T01:14:18.610Z
hmm it is difficult to see how the co-moving volume is finite. Is not length contraction becomes infinite as r approaches alpha?
Anixx
CC BY-SA 3.0
2014-10-11T01:16:24.277Z
Also I wonder what are historical circumstances that led to adoption of the interpretation that universe has infinite volume but finite age and not vice versa (for instance)? It seems to me that all such interpretations are equal.
Anixx
CC BY-SA 3.0
2014-10-11T01:19:16.233Z
@Anixx diverging functions can have finite integrals, such as $\int_0^1 \mathrm{d}x/\sqrt{x}$. In both static coordinates and comoving coordinates, the result is finite. ... As for the rest, again, this Hubble sphere is not the entirety of the de Sitter universe. Not by a long shot. For analogous reasons as to why the external Schwarzschild metric is not the whole Schwarzschild spacetime.
Stan Liou
CC BY-SA 3.0
2014-10-11T00:28:43.957Z
2014-10-11T00:28:43.957Z
Stan Liou
<p>Can we interpret the de Sitter universe as a spherical cosmic horizon null surface of finite radius, centered at Earth, and containing the <a href="https://en.wikipedia.org/wiki/Hubble_volume" rel="noreferrer">Hubble volume</a> 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?</p>

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

<p><img src="https://i.sstatic.net/xgtGt.png" alt="enter image description here"></p>

<p>P.S. I know real universe is not de Sitter but it is usually asserted that it approaches de Sitter state asymptotically.</p>

CC BY-SA 3.0
2014-10-10T21:45:26.760Z
Are you asking from the standpoint of further off in time, when our universe is just about (okay, asymptotically) a de Sitter universe, or from the present time?
HDE 226868
CC BY-SA 3.0
2014-10-10T21:46:21.990Z
@HDE 226868 when it's just about de Sitter universe so that the difference is negligible.
Anixx
CC BY-SA 3.0
2014-10-10T21:57:51.417Z
@HDE 226868 Look. My main point is as follows: it is logically plausble to consider the Universe as a sphere of finite volume and Earth as its center. Similarly it is plausible to count the Universe as appearing at the Big Bang (finite in time) or just count the past as infinite but time-diliated due to large mass density. So one can logically arrive at 4 variants, all consistent: universe is finite in time and in volume, infinite in time but finite in volume, infinite in time and volume, finite in time, infinite in volume (seems the dominating interpretation today)
Anixx
CC BY-SA 3.0
2014-10-10T21:59:10.310Z
How are they all consistent? Observations seem to indicate a finite age of the universe.
HDE 226868
CC BY-SA 3.0
2014-10-10T22:05:28.660Z
@HDE 226868 yes it is finite if to look at it from inside because the universe has no observer that could measure its age without being affected by time diliation at far past epoch due to greater density. But measuring relic radiation we can, say find hydrogen lines in it and conclude that hydrogen had many times less frequency than now, so time was going slower those times than today. The closer to "big bang" the slower time was, and it turns out that on non-diliated scale the Big Bang possibly appears at infinite past.
Anixx
<cosmology><redshift><cosmological-horizon>
Can Hubble red shift be interpreted as time dilation?
Anixx
# Can Hubble red shift be interpreted as time dilation?

(Asked by: username_0 on 2014-10-10)

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](https://en.wikipedia.org/wiki/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.

![enter image description here](https://i.sstatic.net/xgtGt.png)

P.S. I know real universe is not de Sitter but it is usually asserted that it approaches de Sitter state asymptotically.

**Tags:** cosmology, redshift, cosmological-horizon
> **Comments:**
> **username_1**: Are you asking from the standpoint of further off in time, when our universe is just about (okay, asymptotically) a de Sitter universe, or from the present time?  
> **username_0**: @HDE 226868 when it's just about de Sitter universe so that the difference is negligible.  
> **username_0**: @HDE 226868 Look. My main point is as follows: it is logically plausble to consider the Universe as a sphere of finite volume and Earth as its center. Similarly it is plausible to count the Universe as appearing at the Big Bang (finite in time) or just count the past as infinite but time-diliated due to large mass density. So one can logically arrive at 4 variants, all consistent: universe is finite in time and in volume, infinite in time but finite in volume, infinite in time and volume, finite in time, infinite in volume (seems the dominating interpretation today)  
> **username_1**: How are they all consistent? Observations seem to indicate a finite age of the universe.  
> **username_0**: @HDE 226868 yes it is finite if to look at it from inside because the universe has no observer that could measure its age without being affected by time diliation at far past epoch due to greater density. But measuring relic radiation we can, say find hydrogen lines in it and conclude that hydrogen had many times less frequency than now, so time was going slower those times than today. The closer to "big bang" the slower time was, and it turns out that on non-diliated scale the Big Bang possibly appears at infinite past.  

---

## 1 answers

### Answer by username_2 on 2014-10-11. Score: 3 | Accepted answer

The de Sitter spacetime is maximally symmetric, and so having a full set of Killing vector fields, it must have a static form produced by a timelike Killing vector field.
One way to get without worrying about coordinate transformations is to take the $M\to 0$ limit of the general spherically symmetric lambdavacuum solution, which is the the Schwarzschild–de Sitter spacetime:
$$\mathrm{d}s^2 = -\left(1-\frac{1}{3}\Lambda r^2\right)\mathrm{d}t^2 + \left(1-\frac{1}{3}\Lambda r^2\right)^{-1}\mathrm{d}r^2 + r^2\,\mathrm{d}\Omega^2\text{,}$$
where $\mathrm{d}\Omega^2 = \mathrm{d}\theta^2 + \sin^2\theta\,\mathrm{d}\phi^2$ is the usual unit $2$-sphere. Except for minor notational differences, this is also identical to the static de Sitter slicing presented [here](https://en.wikipedia.org/wiki/De_Sitter_space) in terms of different slices of a hyperboloid in Minwkoski $\mathrm{E}^{1,4}$, with $\alpha\equiv\sqrt{3/\Lambda}$.

In these coordinates, the timelike Killing field is simply $\partial\_t$, which corresponds to a family of static observers with time dilation relative to the static observer at the origin given by
$$\dot{t}\equiv\frac{\mathrm{d}t}{\mathrm{d}\tau} = \frac{1}{1-r^2/\alpha^2}\text{,}$$
which diverges as $r\to\alpha$.

For *geodesics*, the Killing field also generates a conserved specific energy $\epsilon = \left(1-r^2/\alpha^2\right)\dot{t}$, which makes solving for radial geodesics not difficult. What's interesting about them is they take an infinite amount of coordinate time $t$ to reach $r=\alpha$, but only a finite amount of proper time $\tau$, and they have no trouble at all continuing beyond it. Additionally, time dilation for static observers becomes infinite there. This behavior is analogous to the event horizon of a black hole, except that the location of the de Sitter horizon is dependent on the choice of origin.

> Can Hubble red shift be interpreted as time dilation?

Yes. If we wish, we can think of the redshift due to a particle (e.g., galaxy) at some radial coordinate as a combination of two things: the special-relativistic time dilation of a galaxy with respect to a local static observer and the gravitational time dilation between static observers given above. In fact, the $\epsilon$ parameter above is exactly the (per-mass) energy of the particle as measured by a local static observer, i.e., the relative Lorentz gamma.

> **Comments:**
> **username_0**: "What's interesting about them is they take an infinite amount of coordinate time t to reach r=α, but only a finite amount of proper time τ" - so from the point of view of any observer nothing can escape the sphere of finite radius centered at himself? "except that the location of the de Sitter horizon is dependent on the choice of origin." - looks like it is always centered at the observer, seems one can safely drop the Copernicus principle. "Yes. If we wish, we can think" so time and space stops at 5 Gpc from here. Well.  
> **username_0**: Does from your formula also follow that dispite finite volume, the internal (comoving) capacity of the sphere will be infinite due to length contraction on the edges? This by the way will also mean that the comoving volume capacity around a black hole should also be infinite (am I wrong?)  
> **username_2**: @Anixx that's a property of a particular time coordinate. Just like for a Schwarzschild black hole, things take an infinite amount of Schwarzschild time to cross, but a finite amount of proper time, or Kruskal–Szekeres time, or Gullstrand–Painlevé time, or Lemaître time, etc. The spacetime is quite well-defined beyond the horizon; it doesn't just end. ... Also, in this frame the Hubble volume is finite and static, $pi^2\alpha^3$, while in a comoving frame it's finite and expanding.  
> **username_0**: hmm it is difficult to see how the co-moving volume is finite. Is not length contraction becomes infinite as r approaches alpha?  
> **username_0**: Also I wonder what are historical circumstances that led to adoption of the interpretation that universe has infinite volume but finite age and not vice versa (for instance)? It seems to me that all such interpretations are equal.  
> **username_2**: @Anixx diverging functions can have finite integrals, such as $\int_0^1 \mathrm{d}x/\sqrt{x}$. In both static coordinates and comoving coordinates, the result is finite. ... As for the rest, again, this Hubble sphere is not the entirety of the de Sitter universe. Not by a long shot. For analogous reasons as to why the external Schwarzschild metric is not the whole Schwarzschild spacetime.  

---

