# A 1D universe with Hubble expansion

I'm thinking about a 1D universe model to understand the expansion of the universe in terms of special relativity. At the origin there's the planet earth and the sun at $\,x_0=0\,$ at rest. There are an infinite line of other stars moving at constant speeds according to

$$x_n=a_n+\beta_nct,\quad n\in\mathbb{Z},$$

where $\,a_0=\beta_0=0$, $\,a_{\pm1}=\pm a$, $\,\beta_{\pm1}=\pm\beta$. So $a$ is the current distance the two nearest stars are from us at $\,t=0\,$ and $\,\beta c\,$ is their recession speed. Following the cosmological principle, let us now assume that other stars are seeing the same picture in their reference frames. We can use Lorentz transformation to go to the reference frame of the $n$-th star and obtain

$$\begin{pmatrix} ct^{(n)}\\x_m^{(n)} \end{pmatrix}=\begin{bmatrix} \gamma_n & -\beta_n\gamma_n\\ -\beta_n\gamma_n & \gamma_n \end{bmatrix}\begin{pmatrix} ct\\a_m+\beta_mct \end{pmatrix}.$$

And we would like to require for $\,m=n\pm 1\,$ that

$$\frac{dx_{n\pm1}^{(n)}}{dt^{(n)}}=\pm\beta c,\quad \left.x_{n+1}^{(n)}+x_{n-1}^{(n)}-2x_n^{(n)}\right|_{t^{(n)}}=0.$$

After solving this model, I find that

$$\beta_n=\frac{(1+\beta)^n-(1-\beta)^n}{(1+\beta)^n+(1-\beta)^n},\quad a_n=\frac{a}{\beta}\beta_n.$$

So Hubble's law applies to the whole 1D universe with Hubble's constant $\,H=\beta c/a$. The age of the universe is $\,1/H=a/\beta c$. At $\,t=-a/\beta c$, all stars were together, which was the moment of the "big bang". At $t=0$, the universe has a boundary with $a_{\pm\infty}=\pm a/\beta$. The boundary does not violate the cosmological principle because it is expanding at the speed of $c$. As one approaches the speed of $c$, relativistic effects would unfold the contracted length to keep the boundary from being reached.

My questions: How well does this model resemble our current understanding of the universe? What does the accelerated expansion mean? Does it mean the recession speeds of the stars are increasing, or does it mean Hubble's constant $H$ is increasing with time?

• Special relativity doesn't have much bearing on interpreting the expansion of the universe - which is governed by General Relativity. Hubble's law permits expansion at $>c$, which your model does not. Nov 9 '17 at 10:59
• I just realized that in the model, the size of the universe at $\,t=0\,$ is $\,[−a/\beta,a/\beta]$, while it appears like $[−a/2\beta,a/2\beta]$ to the telescopes, due to the looking-back time. All stars are seen but they appear closer than they actually are because what we see is where they were. Nov 11 '17 at 1:00

## 2 Answers

You are correct, except in the Hubble law, the velocity is proportional to the distance (in your case, $$\beta\propto a$$) for non-relativistic velocities. The accelerated expansion means that you should add a small second derivative (acceleration) term to your equations.

The Hubble law is only a part of "our current understanding of the universe" based on the $$\Lambda\text{-CDM}$$ Model. This model is based on Friedmann's solution of the General Relativity equations (with some add-ons like inflation). This is not the only possible model based on the Hubble law. For example, the model of Milne (whole student was Walker as W in FLRW) also is based on the Hubble law. The Milne Model is based on Special Relativity and is ruled out, but it is a great educational tool to study the principles of cosmology and Special Relativity.

• I finally worked out my model and got shocked. Hubble's law applies to the whole universe, which has a boundary expanding at the speed of $c$ but is not reachable, because as you get close to $c$, the contracted length unfolds so the boundary is always that far from you. I'll update this part. Nov 9 '17 at 5:37
• Well, you seem to be describing the Milne model. In this model, galaxies follow the Hubble law by inertia from the Big Bang, but space doesn't expand. In the accepted Friedmann model, space does expand. As a consequence, speeds faster than light are possible. Then the speed-of-light horizon you are describing only limits the observable universe, but not the entire universe. In another yet model, space also expands and speeds faster than light are possible, although appear slower than light to observers and the entire universe is observable. Hope you've found my answer helpful and acceptable :) Nov 9 '17 at 5:51
• I see. I'll read about the links first. Experimentally acceleration should mean the redshift is increasing, and that could be due to either the expansion of space (change of spacetime metric) or the motion of stars. Hopefully I'll understand this better soon. Nov 9 '17 at 6:35
• Hubble's law is that velocity is proportional to (proper) distance. There is no caveat that the velocities be non-relativistic. Nov 9 '17 at 9:41
• @RobJeffries Correct. I did not say the Hubble law was only for non relativistic velocities. I said, the distance was proportional to non-relativistic velocities. If you take a string of galaxies and apply this definition, you would eventually get to a relativistic velocity, but in the original frame it would not be proportional to distance and you'd have to use "proper distance" instead. Nov 9 '17 at 18:11

Your model only approximates to our universe for small, non-relativistic velocities.

Your model is equivalent to saying that the recession velocity of a galaxy is given by $$v = c \left(\frac{(1+z)^2 -1}{(1+z)^2 +1}\right)$$ where $z$ is a measured redshift.

In this model, the space between the galaxies is not expanding, they are simply moving away from each other with a speed proportional to their distance away from us. SR then ensures that the recession velocities cannot exceed $c$.

In the expanding universe, governed by GR, this isn't the case. Hubble's law relating the recession velocity now with the proper distance to a galaxy applies for all distances and hence recession velocities greater than the speed of light are permitted. Indeed, the relationship between recession velcity and redshift is not that provided by SR. It is $$v = c\frac{\dot{R}(t)}{R_(t_0)} \int_0^{z} \frac{dz'}{H(z')},$$ where $H(z)$ is the Hubble parameter at redshift $z$, $R$ is the scale factor of the universe and $t$ is the epoch for which the recession velocity is calculated.

Essential Reading - Davis & Lineweaver (2003).

• Just to confirm my understanding. Is the recession speed always defined as $v=dr/dt$ (all in our reference frame)? If so, to my understanding, $v$ is no longer directly measurable (we deduce it from redshift $z$ and some model). Experimentally what's the difference of the redshift $z$ v.s. distance $r$ relationships given by the two models? I mean, how do we know the special-relativistic view is inconsistent with observations? Nov 9 '17 at 17:30
• @ZhuoranHe Rob's answer makes the same point as my earlier comment that you are describing the Milne model without the space expansion. In reality space does expand making velocities faster then light possible. From this point on however, things depend on the chosen model. The official model is based on the FLRW solution that produces a cosmic horizon meaning that the observable universe is smaller than the entire universe that moves away from us faster than light. However, other models are possible that avoid this limitation and make the entire expanding universe observable. Nov 9 '17 at 18:23
• @ZhuoranHe The issue with your definition of $v=\text{d}r/\text{d}t$ is that the distance $r$ can be defined in different ways, but more importantly, it is not a direct observable, but is derives from other observations like the redshift, supernova brightness, and others. To use this definition, you need to be very clear what exactly you mean by distance. A more convenient way is to use the redshift, as it is a direct observable and also has much less ambiguity than distance. In any case you should start by choosing a model (e.g. expanding, non-expanding, etc.). Nov 9 '17 at 18:32
• @safesphere, we need at least two direct measurements, e.g., redshift $z$ and distance $r$ to confirm Hubble's law. If $z$ is the only thing we measure, then everything else would just be some story we make up. To my knowledge, $r$ is measurable but not very accurately, but to measure $dr/dt$ directly is nearly impossible. We cannot hope to measure any change in $r$ to a distant star even years later. Nov 9 '17 at 18:43
• @ZhuoranHe Yes, we need two measurements to establish the Hubble law, but Hubble already has done it. So I was not talking about establishing the Hubble law, but of using it as already given: "Doppler-shift-measured velocity, of various galaxies receding from the Earth, is approximately proportional to their distance from the Earth for galaxies up to a few hundred megaparsecs away." So we can use the redshift as a measure of the distance: en.wikipedia.org/wiki/Hubble%27s_law Nov 9 '17 at 18:54