So, with some assumptions, it seems like we can put some lower bounds on the size of the unobservable universe.

Suppose the the universe is unbounded, closed, finite, and flat (is that right?). Can we think of it like a gigantic 4D sphere? If so, does that give us a way to compute its minimum size?

As I understand it we know the size of the “observable universe,” and across nearly this entire part of the universe space looks flat. This means its curvature is indistinguishable from zero. But we only know this to some number of significant digits (how many and in what units?). So, thinking of it as a 4D sphere, and assuming that there is a tiny postive curvature, the maximum possible within measurement error, we should be able to compute the total surface (a volume in this case) of the whole universe (observable and unobservable). Perhaps we could report it as multiples of the observable universe.

Is this a sensible way to estimate the minimum size of the unobservable universe (given those starting assumptions)? I wonder if the @physicists know…

The universe is only ~flat to experimental confirmation when projected to three dimensions. It is decidedly not flat in 4D.

Last I checked, the naive “lower bound” of the spatial size given the experimental error bars is something in the ~trillions of light years, though it’s been a while since I checked, and that number might be outdated now. However, the universe can easily be smaller if we allow for example, that the universe repeats (e.g., like a torus), or there is some sudden change of curvature beyond the observable universe.

You can take their cosmological parameters and their error bars, and plug it to a cosmology calculator like this one: Ned Wright's Javascript Cosmology Calculator to compute the spatial size.

Note also the Hubble tension, which you can read about in the wikipedia page Hubble's law - Wikipedia, so you might want to repeat the exercise with the cosmological parameters from the other experiments listed on that wikipedia page in addition to Planck’s.

This paper might be outdated by now, but they put a lower bound in the size of the universe of approximately 250 times the size of the observable universe, by volume:

Ethan Seigel has a 2018 article on this, it seems:

Observations from the Sloan Digital Sky Survey and the Planck satellite are where we get the best data. They tell us that if the Universe does curve back in on itself and close, the part we can see is so indistinguishable from “uncurved” that it must be at least 250 times the radius of the observable part.

This means the unobservable Universe, assuming there’s no topological weirdness, must be at least 23 trillion light years in diameter, and contain a volume of space that’s over 15 million times as large as the volume we can observe. If we’re willing to speculate, however, we can argue quite compellingly that the unobservable Universe should be significantly even bigger than that.

That makes a little more sense. It was 250 times the length (minimum), but the volume is different, 15625000 times the size of the observable universe. It still seems to be lower than I would have expected.

For example, we observe that the Universe is spatially flat on the largest scales: it’s neither positively nor negatively curved, to a precision of 0.25%. If we assume that our current laws of physics are correct, we can set limits on how large, at least, the Universe must be before it curves back on itself.

That may be the issue for me. I thought we knew to far greater precision that the universe seemed flat…

Depends on what you mean by “size”. Because of the expansion of the universe, the distance between two objects continuously changes.

The “comoving distance” gives the distance to an object with the expansion of the universe “factored out”. This does not change with time.

The “proper distance” is just the “comoving distance” divided by (1+z), and gives the distance between point A and point B at ONE particular moment in time. This changes with time.

There is the “light travel distance”, which is just the “light travel time” multiplied by c. This gives the distance that a beam of light travels to get from point A to point B. This is in line with the idea of measuring a “physical distance” by sending a beam of light and measuring how long it takes for the light to travel that distance.

Although now that I paid attention, that website doesn’t actually have the option for closed universes, only open and flat. My mistake, I used that website a lot for quick calculations, but never had a use for computing anything for closed universes.

To do the closed universe calculation:
Note that the calculator website is just computing the equations found here: Distance measures (cosmology) - Wikipedia which you can do yourself with your favorite calculator/coding language. Note that the comoving distance for closed universes is slightly modified from that of flat universes. You can read more about them here: Comoving and proper distances - Wikipedia

It seems this just covers the size of the observable universe. I’m interested in the size of the unobservable universe. Though it is possible I’m misreading wiki.

As I mentioned in my previous post, you can modify the equations in this page to that for the closed, bounded universe:

Once you do that, you can use the modified equations to compute the size of the unobservable universe, given that you plugged in the right parameters to make the universe finite (otherwise you’ll get infinity).

@swamidass for your pleasure I did the computation with the Planck 2018 values:

y-axis: Ratio of volume of universe vs volume of observable universe
x-axis: How curved the spatial universe is; higher number = more curved, the Planck 2018 values gives a limit of ±0.002 for this parameter, the high end of this errorbar is the point at the right end of the plot (1 gives a flat universe and anything less gives an open universe, both have infinite volume for this kind of topology)

Thanks! Unfortunately, your descriptions are terse and I’m not knowledgeable enough on cosmology to follow you. Can you write out the formula? Note: you can use \LaTeX on this forum…

Sure, the volume of any space is given by an integral over it:

\int_\mathrm{space} \sqrt{g} \; \mathrm{d}(\textrm{all the coordinates})

where g is the determinant of the “metric”, i.e., a mathematical object that determines how to measure how close or how far something is in that space. You don’t need to think too hard on what g is at the moment; it is just something that comes out of your model of cosmology. The \mathrm{d}(\textrm{all the coordinates}) is just \mathrm{d}x \mathrm{d}y \mathrm{d}z \ldots etc (or any other ways you would like to coordinatize the space). For 3D space, we usually write this term as \mathrm{d}^3x.

So, the ratio of the spatial volume of the universe vs the spatial volume of the observable universe is:
\frac{V}{V_{\mathrm{obs}}} = \frac{\int_\mathrm{the \; entire \; universe} \sqrt{g} \; \mathrm{d}^3x}{\int_\mathrm{observable \; universe} \sqrt{g} \; \mathrm{d}^3x}

The experimental results from cosmology enters in two places (how to compute these you can read from the wikipedia pages on cosmology):

determining what g is

determining which part of the universe is observable

It turns out that the most important parameter measured by Planck for V/V_\mathrm{obs} is what is called the “Curvature energy density”, which is written as \Omega_k, which has been measured to be 0.001 \pm 0.002. The more “curved” the universe is, the higher the \Omega_k.

What I plotted is V/V_{\mathrm{obs}} in the y axis versus this parameter \Omega_k in the x-axis where I write \Omega_k = 0.001 - x and plot x in the x-axis. x ranges from 0.001 to 0.002, i.e, it includes all of the range of curvature allowed by Planck 2018 errorbars. Note that the x-axis is written in units of 10^{-3} and when x=0.001, \Omega_k=0 and the universe is flat, hence the notation \Omega_k-\Omega_{k_\textrm{flat}} in the x-axis label.

Other cosmological parameters measured by Planck also affect this calculation, but not by much, so I did not plot them.