1
$\begingroup$

From Wikipedia, Observable Universe page https://en.wikipedia.org/wiki/Observable_universe

Some parts of the Universe are too far away for the light emitted since the Big Bang to have had enough time to reach Earth, so these portions of the Universe lie outside the observable universe.

I have problems in understanding this sentence. (Assume by observable we mean observable by photon detection - which the reference to "light emitted" in the above source seems to imply too). As from my current understanding of the Big Bang cosmological model, the farthest we can see is up to the last scattering surface of the epoch of recombination. This we observe today as the cosmic microwave background (CMB) - and thus we can indeed "see" it. Mustn't all other parts be less far away than that surface of last scattering? So how can there be other parts we aren't (in principle) able to see?

EDIT:

I think i realize my error of thought. I thought that, given that the epoch of recombination occured relatively early in the age of the univers, i.e. around only after 380 kyears after BB, the space beyond the SoLS would be tiny in comparison to the space before. What i didn't realize is that as is currently assumed in inflationary models, space had already risen to enormous dimensions before this epoch and has been considerably expanding since. Which means, the inside volume of our local SoLS might only span a tiny volume of all of the entire universe.

Improve this question
$\endgroup$

2 Answers 2

Reset to default
5
$\begingroup$

You're right that the surface of last scattering (SoLS) is the farthest we can see in practice. This light is seen as the cosmic microwave background (CMB), observed e.g. with the Planck spacecraft.

The term "observable Universe" refers to the farthest we can see in theory, and is defined as the distance a photon is able to travel in the time from the Big Bang (BB) to now, in the hypothetical event that is does not interact with any other particles. But since the Universe was filled with free electrons — which scatter photons of all wavelengths effectively — until the CMB was emitted 380,000 years after BB, this does not happen in practice.

If at some point we will be able to measure the cosmic neutrino background, which decoupled from matter 1 second after BB (e.g. Fässler et al. 2016) and interacts extremely weakly with matter along its journey, or primordial gravitational waves which are thought to have been emitted during inflation, $10^{-32}$ s after BB, and which doesn't interact with matter at all, this will come from the "edge of the observable Universe"$^\dagger$ (the so-called "particle horizon").

Due to the expansion of the Universe, light from the SoLS is redshifted by a factor of 1100, while (hypothetical) light from the particle horizon is infinitely redshifted. The distance to the SoLS is roughly 45.6 Gly (billion lightyears), while the distance to the particle horizon is slightly larger, 47.1 Gly. The reason that the difference between the two radii is not only 380 kly (the distance that light travels in the 380 kyr that went before the CMB was emitted) is that the expansion rate at that time was much larger (at $t=380\,\mathrm{kyr}$, it was $H = 1.4\times10^6\,\mathrm{km}\,\mathrm{s}^{-1}\,\mathrm{Mpc}^{-1}$, compared to today where it is only $H_0 = 67.8\,\mathrm{km}\,\mathrm{s}^{-1}\,\mathrm{Mpc}^{-1}$).

$\dagger$Although neutrinos probably have mass and thus don't travel quite at the speed of light.

Improve this answer
$\endgroup$
10
  • 1
    $\begingroup$ Neutrinos aside, Do you agree that the only part of the universe which is not observable through photons is beyond the surface of last scattering (SoLS)? Because the Wikipedia article goes on to say that (in inflationary models) "the entire universe's size is at least 3x10^23 times larger than the size of the observable universe". Where is all that additional volume? Behind the SoLS? I've always thought that'll be only a tiny volume, maybe that's where i'm mislead? $\endgroup$
    – Scrontch
    Commented Nov 17, 2016 at 14:40
  • 1
    $\begingroup$ @Scrontch: The observable Universe is the part which is, in principle, observable to us. The whole Universe is probably much, much bigger, perhaps infinitely large but at least 3e23 × larger than the observable Universe (according to your statement). As times goes, photons from farther away will have had the time to reach us, so the observable Universe always increases in size. Like if you're standing on an extremely larger field full of people, and at some point, everybody claps. After 10 seconds, you will only have heard the ones within a circle of radius 3 km, centered on you, while… $\endgroup$
    – pela
    Commented Nov 17, 2016 at 16:11
  • 1
    $\begingroup$ …after 20 seconds you'll have heard everybody within 6 km radius, etc. Wrt. to your first statement, yes, for practical reasons the part of the Universe outside the SoLS (which has a somewhat smaller radius than the observable Universe), is not observable. $\endgroup$
    – pela
    Commented Nov 17, 2016 at 16:15
  • $\begingroup$ Note though that apart from the SoLS, there is a large part that lies closer than the SoLS that cannot be seen, because stars have not yet reionized the Universe. The CMB travels more or less freely because it is quickly redshifted to more or less non-interacting IR, sub-mm, and microwaves. But when the first luminous objects (stars) are born, their light cannot reach us because most of its light is absorbed by the neutral intergalactic medium. After a while these stars reionize the Universe, and it becomes transparent so that we're able to see the newborn stars and galaxies. $\endgroup$
    – pela
    Commented Nov 17, 2016 at 16:19
  • 2
    $\begingroup$ This question relies on a misconception, and has been asked and answered several times on this site, so I won't go into detail, but briefly: The Universe was only arbitrarily small if it's finite. If it's infinite, it was born infinite, and of arbitrarily high density. But the observable Universe — which is finite — was arbitrarily small at BB; not (necessarily) the entire Universe. $\endgroup$
    – pela
    Commented Nov 18, 2016 at 14:17
3
$\begingroup$

One has to be pretty careful to discuss these different horizons and what is outside them. Apart from practical considerations the limit of what we can see now is the particle horizon. Now about 46 Bly from us. You can not take age of the universe and multiply by c to get it because of the expansion. You already know and discussed some of it.

But the universe can be much bigger than that. You know also. But the best way that I've seen to understand it is in this arXiv paper at https://arxiv.org/pdf/astro-ph/0310808v2.pdf

In the first figure with 3 panels it shows the particle horizon, and the other 'horizons' (the Hubble bubble which is not a horizon, the event horizon, the surface of last scattering for light also not a horizon) in spacetime diagrams. I can't paste the figure here, some people might, but if you look at the figure it is easier to see. You can also see the space outside the horizons, and in the last panel it goes to infinite time in a conformal time diagram.

Other figures depict the speed of the expansion for the different cosmological models,

The paper also has the math for the different 'horizons' it plots.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Good paper! I'll read through it. $\endgroup$
    – Scrontch
    Commented Nov 18, 2016 at 13:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.