# Functional dependence of the size of the visible universe

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1. Jun 3, 2015

### A. Neumaier

I am looking for reliable information about the functional dependence of the diameter $d(t)$ of the visible universe on the time $t$ since the big bang singularity, based on the different hypotheses currently deemed competitive.

Last edited by a moderator: Jun 3, 2015
2. Jun 3, 2015

### marcus

Google "LambdaCDM bounce" for the article by Yi-Fu Cai and Edward Wilson-Ewing
http://arxiv.org/abs/1412.2914
A ΛCDM bounce scenario
Yi-Fu Cai, Edward Wilson-Ewing
(Submitted on 9 Dec 2014 (v1), last revised 28 Jan 2015 (this version, v2))
We study a contracting universe composed of cold dark matter and radiation, and with a positive cosmological constant. As is well known from standard cosmological perturbation theory, under the assumption of initial quantum vacuum fluctuations the Fourier modes of the comoving curvature perturbation that exit the (sound) Hubble radius in such a contracting universe at a time of matter-domination will be nearly scale-invariant. Furthermore, the modes that exit the (sound) Hubble radius when the effective equation of state is slightly negative due to the cosmological constant will have a slight red tilt, in agreement with observations. We assume that loop quantum cosmology captures the correct high-curvature dynamics of the space-time, and this ensures that the big-bang singularity is resolved and is replaced by a bounce. We calculate the evolution of the perturbations through the bounce and find that they remain nearly scale-invariant. We also show that the amplitude of the scalar perturbations in this cosmology depends on a combination of the sound speed of cold dark matter, the Hubble rate in the contracting branch at the time of equality of the energy densities of cold dark matter and radiation, and the curvature scale that the loop quantum cosmology bounce occurs at. Importantly, as this scenario predicts a positive running of the scalar index, observations can potentially differentiate between it and inflationary models. Finally, for a small sound speed of cold dark matter, this scenario predicts a small tensor-to-scalar 14 pages, 8 figures. Published JCAP(2015)
http://inspirehep.net/record/1333367?ln=en

Last edited: Jun 3, 2015
3. Jun 3, 2015

### Chalnoth

One caveat: few cosmologists use the diameter of the universe. Sometimes the radius of the observable universe is used. There's no measure for the size of the universe beyond the observable patch.

That said, there is no closed-form expression. It has to be estimated numerically.

The equation of interest for determining this is the first Friedmann equation:
$H^2 = {8\pi G \over 3}\rho - {kc^2 \over a^2}$

Where:
$H = {1 \over a}{da \over dt}$

Here $a(t)$ is the scale factor, usually defined so that $a = 1$ at the current time. If the diameter of the universe at the current time is $d_0$, then the diameter of the universe at any other time will be $d_0 a(t)$. $\rho$ is the energy density of the universe divided by $c^2$. $k$ is the spatial curvature.

What is usually done is to make use of stress-energy conservation to show how the energy density of each component of the universe changes over time. For example, the energy density of matter scales as $1/a^3$, the energy density of radiation scales as $1/a^4$, and the energy density from dark energy is constant (in the simplest case). We then parameterize the equation in terms of the current density fraction, like so:

$H^2 = H_0^2\left({\Omega_r \over a^4} + {\Omega_m \over a^3} + {\Omega_k \over a^2} + \Omega_\Lambda \right)$

Here each $\Omega$ is a dimensionless number, and $H_0$ is the current expansion rate. Since we have defined $a = 1$ at the current time, when $a = 1$, $H^2 = H_0^2$. Thus for the above equation to be valid, $\Omega_r + \Omega_m + \Omega_k + \Omega_\Lambda = 1$. This is why the $\Omega$ terms are called the density fractions for each component. These four density fractions and the current Hubble expansion rate $H_0$ must be measured experimentally. Once you have values for the five parameters, it's possible to use a differential equation solver to get $a(t)$, which you can use to get the diameter as a function of time.

4. Jun 3, 2015

### marcus

Just in case newcomers to the topic could be reading thread, I want to mention the simple answer to this question based on the usual conventional model. This would probably be familiar to both A.N. and Chalnoth.
One can assume that the cosmological curvature constant Lambda is in fact simply that---a small constant curvature that is intrinsic to spacetime (as in Einstein's original treatment) not associated with any imagined "dark energy".
One can assume that the U is spatially flat.
Almost the entire history has radiation energy density negligible compared to matter. So we neglect the small contribution of the very early radiation-dominated era.

Then the radius of the observable region is straightforward to calculate as an integral over s = 1/a = the reciprocal of the normalized scale factor.

$$r_{obs} = R_\infty \int_1^\infty \frac{ds}{\sqrt{0.4433s^3 + 1}}$$

where R is the longterm Hubble radius c/H
and 0.4433 is (Hnow/H)2 - 1

5. Jun 3, 2015

### marcus

What that integral gets you is essentially the 46 billion LY which is the particle horizon distance in the a = S = 1 row of this table:
$${\scriptsize\begin{array}{|c|c|c|c|c|c|}\hline R_{0} (Gly) & R_{\infty} (Gly) & S_{eq} & H_{0} & \Omega_\Lambda & \Omega_m\\ \hline 14.4&17.3&3400&67.9&0.693&0.307\\ \hline \end{array}}$$ $${\scriptsize\begin{array}{|r|r|r|r|r|r|r|r|r|r|r|r|r|r|r|r|} \hline a=1/S&S&z&T (Gy)&R (Gly)&D_{par}(Gly) \\ \hline 0.001&1090.000&1089.000&0.0004&0.0006&0.001\\ \hline 0.003&339.773&338.773&0.0025&0.0040&0.006\\ \hline 0.009&105.913&104.913&0.0153&0.0235&0.040\\ \hline 0.030&33.015&32.015&0.0902&0.1363&0.249\\ \hline 0.097&10.291&9.291&0.5223&0.7851&1.491\\ \hline 0.312&3.208&2.208&2.9777&4.3736&8.733\\ \hline 1.000&1.000&0.000&13.7872&14.3999&46.279\\ \hline 3.208&0.312&-0.688&32.8849&17.1849&184.083\\ \hline 7.580&0.132&-0.868&47.7251&17.2911&458.476\\ \hline 17.911&0.056&-0.944&62.5981&17.2993&1106.893\\ \hline 42.321&0.024&-0.976&77.4737&17.2998&2639.026\\ \hline 100.000&0.010&-0.990&92.3494&17.2999&6259.262\\ \hline \end{array}}$$

The particle horizon is conventionally taken to be the current radius of the observable region. It can be defined as follows: Imagine that at the beginning of expansion our matter sent out a particle at speed c which by some miracle was never absorbed or scattered but just continued traveling in a straight line. How far from us would that particle be now? According to standard Friedmann model cosmology the answer is 46 billion LY.

This is also the farthest away some other matter could be now, if we are receiving some kind of signal from it today. You just turn the picture around. So it is the radius of the observable--the farthest any matter could be today, if we could, in principle, be receiving signal from it.

As time goes on this observable region includes more and more distant matter.

6. Jun 4, 2015

### A. Neumaier

Thanks for the answers. Did anyone in the literature draw a plot of the estimated radius $r(t)$ of the observable universe extrapolated into the far past $t$, based on the information given? For the moment, I am mostly interested in such a plot (or table), though I'll ultimately want to understand the explanation and the numerical uncertainty the estimates entail.

7. Jun 4, 2015

### marcus

Based on standard cosmology, yes, for example in Charles Lineweaver's 2003 article "Inflation and the CMB"

==abstract==
http://arxiv.org/abs/astro-ph/0305179
Inflation and the Cosmic Microwave Background
Charles H. Lineweaver (School of Physics, University of New South Wales, Sydney, Australia)
(Submitted on 12 May 2003)
I present a pedagogical review of inflation and the cosmic microwave background. ...
34 pages, 13 figures.
==endquote==
Lineweaver's tutorial is something of a classic. It is also online at the Caltech "Level 5" website.
http://ned.ipac.caltech.edu/level5/March03/Lineweaver/Lineweaver_contents.html
It's interesting to notice that in the bottom diagram which uses a "conformal" timescale (fake time making speed of light constant in comoving distance) the particle horizon, in comoving distance, is the "mirror image" of the event horizon

Last edited: Jun 4, 2015