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Of course it is not exactly known how big the universe is, but I thougt that the universe is about 100 billion lightyears in diameter. But if that is true can you also make an estimation of the volume of the universe?

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    $\begingroup$ Just use your good old $V = \frac{4\pi}{3}R^3$ (and remember that we're talking about the observable Universe; the rest may be infinite). $\endgroup$
    – pela
    Apr 20 '16 at 8:54
  • $\begingroup$ What you give is the volume of a round sphere but in this picture en.wikipedia.org/wiki/File:Ilc_9yr_moll4096.png it is always not shown as as perfect globe. Why is that? $\endgroup$
    – Marijn
    Apr 20 '16 at 10:10
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    $\begingroup$ You can thank Karl Mollweide for that :) $\endgroup$
    – pela
    Apr 20 '16 at 10:46
  • $\begingroup$ Fun exercise: Once you have calculated the volume, you can then calculate the average approximate density, too. (Estimated mass of an average star (in sols)) * (Estimated number of stars in an average galaxy) * (Total estimated number of galaxies in observable universe) / Volume of observable universe. $\endgroup$ Apr 20 '16 at 12:22
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    $\begingroup$ @TexasTubbs: That will give you the density of stars only, which is only $\sim0.5$% of the total mass of the Universe. $\endgroup$
    – pela
    Apr 21 '16 at 10:13
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Based on your comments, I think your confusion comes from having seen the classic rugby ball-shaped image of the CMB. The CMB we observe is not from the whole cosmos, but only from a thin and completely spherical shell centered on us. This light is redshifted by an amount $z=1100$.

The radius of this shell can be calculated integrating the Friedmann equation from us (at $z=0$) to $z=1100$: $$ R_\mathrm{CMB} = \frac{c}{H_0}\int_0^{1100} \frac{dz}{\sqrt{ \Omega_r(1+z)^4 + \Omega_m(1+z)^3 + \Omega_k(1+z)^2 + \Omega_\Lambda }}, $$ where $c$ and $H_0 \sim 70\,\mathrm{km}\,\mathrm{s}^{-1}\,\mathrm{Mpc}^{-1}$ are the speed of light and the Hubble parameter, respectively, and $\{\Omega_r,\Omega_m,\Omega_k,\Omega_\Lambda\}\sim\{\lt 10^{-4},0.3,0,0.7\}$ are the radiation, matter, curvature, and dark energy density parameters, respectively.

With these approximate numbers, I get $R_\mathrm{CMB} = 44.63\,\mathrm{Glyr}$, while using a Planck 2015 cosmology for the various parameters yields $R_\mathrm{CMB} = 45.36\,\mathrm{Glyr}$.

Just as the shell of Earth can be projected onto a rugby-shaped figure using a Mollweide projection, so can the shell of the CMB. Here's a figure from Universe Adventure that can help visualize:

projection

The volume of the observable Universe

Although we (still) can't see beyond the CMB, the observable Universe is all the way out to redshift $z=\infty$. The difference is not big, though; integrating the equation above from $0$ to $\infty$ yields $R_\mathrm{obs.Uni.} = 46.27\,\mathrm{Glyr}$. Thus, the volume of the observable Universe is $$ V = \frac{4\pi}{3}R_\mathrm{obs.Uni.}^3 = 415,\!065\,\mathrm{Glyr}^3. $$

The volume of the whole Universe is probably much larger, and may in fact easily be infinite.

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  • $\begingroup$ I guess I should read $\mathrm{Gly^3}$ as "cubic giga-light-year," not "billion cubic light-year." $\endgroup$
    – Mike G
    Apr 20 '16 at 19:04
  • $\begingroup$ But still it is unclear why they mostly illustrate the CMG rugby shaped and not just spherical. As I said, for the earth I can imagine why, that is to show all continents in one view. But there isn't such a reason for the CMB? $\endgroup$
    – Marijn
    Apr 20 '16 at 19:11
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    $\begingroup$ @Marijn: The reason is exactly the same as for the Earth: We want to show the entire sky in one view. That's not possible with a spherical projection, unless you invent some crazy projection algorithm that will not only heavily distort the peripheral regions (which to a certain degree will be the case for all projections), but also display some regions multiple times. $\endgroup$
    – pela
    Apr 21 '16 at 6:59
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    $\begingroup$ NB Only true for a universe with no curvature. $\endgroup$
    – ProfRob
    Apr 24 '17 at 11:23
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    $\begingroup$ @Acccumulation: Yes, the observable Universe is extremely (if not completely) flat, with $|\Omega_k| < 0.005$ (Planck Collaboration et al. 2016). $\endgroup$
    – pela
    Oct 11 '17 at 7:08
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Cosmologists estimate the age of the universe at 13.8 billion years. The most distant objects we could theoretically observe emitted light that long ago; light from more distant objects hasn't reached us yet. Due to expansion of the universe, that horizon is now about 46 billion light-years away. If we assume uniform expansion in all directions, we can approximate the observable universe as a sphere of that radius, consistent with the diameter you'd heard. I figure its volume as roughly $4 \times 10^{32}$ cubic light-years. The shape and size of the unobservable universe are anyone's guess.

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I'm assuming you're talking about the visible universe, not the observable universe. If so, the comoving distance we could see is about 45.7 Billion light years. Even though the universe is only 13.8 Billion years old, due to expansion it took longer than that to reach us. It'd be like walking up an escalator thats going down vs walking up stairs, I'll get there but it'll just take much longer. A COMOVING distance would relate better to the person on the stairs, it factors out expansion, thus making calculations much easier. Using the comoving distance, figuring out it's volume is simple math, unless of course you plan on figuring it's volume without mass, just the space. Then of course you'd have to find each masses volume, subtract and blah, blah, and impossible, since we don't know every object in the visible universe. Using V=4/3πr^3, V = approximately 3.99795E+32ly^3. Or, 399 nonillion Ly^3.

Now imagine the escalator you're on starts running just as fast as you, eventually faster; You'll never reach the top, in fact you'll move away from the top despite running towards it. This is what is happening to our visible universe, it's slowly receding faster than light due to expansion. So the light will continue to travel towards us but will never reach us once it's emitted beyond about 4500 megaparsecs away, it's event horizon, sometimes called particle horizon. Point being, the visible universe volume is constantly shrinking.

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  • $\begingroup$ I think your distinguishing between the observable and the visible Universe is not what is normally used. As I understand your answer, you use the term "visible Universe" to refer to the region where light emitted today will be able to reach us. But this is rarely of any interest in a scientific context. Also, note that the 4500 Mpc you're referring to is the comoving distance to the Hubble sphere, outside which stuff recedes faster than the speed of light. But actually light emitted from outside this region right now, up to ~5000 Mpc, will still be able to reach us. $\endgroup$
    – pela
    Apr 24 '17 at 7:18
  • $\begingroup$ As I read your answer, you use the term "visible Universe" for the region bound by the event horizon, which is the horizon outside which light emitted today will never reach us. It is true that it is always shrinking in comoving coordinates, but in physical coordinates it actually doesn't. It asymptotically increases toward a finite size (~17 Gly). $\endgroup$
    – pela
    Apr 24 '17 at 9:07
  • $\begingroup$ How does this differ from Pela's answer? $\endgroup$
    – ProfRob
    Apr 24 '17 at 11:22
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The Universe is infinite and always has been. Yes, it is expanding, but what does that mean when it exists everywhere? Yes, it can appear to have exploded from a point, but that "point" is a universe that is infinite in size and has no edge. There is no way to look at it or describe it from "outside".

The diagrams you often see of the history of the Big Bang show a poor concept of expansion and are just misleading. They imply an outside view is possible.

See the excellent answer to a similar question here: https://physics.stackexchange.com/questions/136860/did-the-big-bang-happen-at-a-point

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  • $\begingroup$ This answer has a number of highly contentious statements (eg the universe "always has been" and it "exploded"). At the very least you need to back them up with references. $\endgroup$ Oct 13 '17 at 7:42
  • $\begingroup$ I you can show that time existed before the Big Bang, then I will retract "always has been". $\endgroup$ Oct 14 '17 at 1:54
  • $\begingroup$ No, that's not how science works. It's up to you to explain how your hypothesis resolves all possible scenarios (eg a cyclic universe). $\endgroup$ Oct 14 '17 at 22:40
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    $\begingroup$ Are you two perhaps actually agreeing? It seems @Chappo objects to the "always has been" statement because this sounds like the Universe has existed forever. But maybe C.T. Springer just means "has been infinite since time started at BB" (which nevertheless is a somewhat bold statement, since this is not really known, although it's true if the rest of the Universe is like the obsevable Universe). $\endgroup$
    – pela
    Oct 16 '17 at 16:10
  • $\begingroup$ "No, that's not how science works." Every measurement, experiment, and observation indicates an infinite universe and the existence of time. The current theory of course has to fit these conditions. A singularity at a Big Bang is an inference. Your objection to "always has been" places the burden on you to show why you claim something outside the evidence is true. $\endgroup$ Oct 17 '17 at 4:20

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