Suppose the universe is spherical and its density ratio is

$\Omega \leq 1.00125$

$\Omega = 1.00125$ is approximately the maximum possible value of the density ratio according to the Planck Mission measurements.

$H_0=67.6$ (km/s)/Mpc

$c=300000$ km/s

With these values, according to the Friedmann equations, the minimum radius of the whole universe ($\Omega = 1.00125$) would be:

$$R=\dfrac{c}{H_0 \sqrt{\Omega -1}}$$

$R=409.5$ Gly

And the diameter $D = 809$ Gly

We know that the radius of the observable universe (particle horizon PH) is $PH=46$ Gly, therefore

$R / PH = 9.1$

The ratio of volumes between the whole universe and the observable universe will be the cube of the radii ratio

$V / V_o = 9.1^3=753$

However, today I was surprised when I read Ethan Siegel's article entitled: Ask Ethan: How Large Is The Entire, Unobservable Universe? where he says:

"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".

How did he make that calculation and why is there such a difference with the figures I calculated above?

My theory: Let's assume that Ethan used the same values for his calculation as we did above.

$\Omega = 1.00125$

$H_0=67.6$ (km/s)/Mpc

$c=300000$ km/s

$PH = 46$ Gly

But in doing the operations, would Ethan have made a mistake and forgotten the square root?

$R=\dfrac{c}{H_0 ( \Omega -1 )}$ ??

Without the square root, the following results are obtained:

$R = 11585$ Gly

$R/PH = 252$ ~ 250 times

$Diameter = 2R = 23162 \ Gly$ ~ 23 trillion light-years

$V / Vo = 250^3$ ~ 15 million

What do you think?


EDITED: On the size of the universe assuming it were spherical, it is advisable to see this later thread: Size of the Unobservable Universe

  • $\begingroup$ Maybe I am misunderstanding something here, but when Siegel says if the Universe does curve back in on itself and close that makes me question if your parameters really are the same. I’m no cosmologist but my understanding is that the closer $\Omega$ is to 1, the flatter the universe is, and based off of his description it sounds like he is assuming a closed universe which would have a value greater than 1 (granted your value is greater than 1 by a little, but maybe his is yet larger) $\endgroup$
    – Justin T
    Mar 2, 2023 at 16:38
  • $\begingroup$ Also, again might be missing something, but your equation doesn’t seem to have a curvature constant? $\endgroup$
    – Justin T
    Mar 2, 2023 at 16:50
  • 1
    $\begingroup$ FWIW, I mentioned that Ask Ethan article here: astronomy.stackexchange.com/q/31793/16685 $\endgroup$
    – PM 2Ring
    Mar 3, 2023 at 11:36

2 Answers 2


Different data sets, analysis methods, and statistical interpretations will give different limits on the ratio $\Omega$ between the density and the critical density. However, let's adopt $\Omega=1.00125$, as you say.

$R / PH = 9.1$

This looks to be about right. Keep in mind, though, that $R$ is the radius of a 3-sphere while $PH$ is the radius of a 2-sphere. In particular, $R$ is not a distance in our universe, it's really just a measure of the curvature. The circumference of the universe -- which is a distance within the universe -- would be $2\pi R\sim 55PH$.

$V / V_o = 9.1^3=753$

Not quite because the correct comparison is between the volume $(4\pi/3)PH^3$ enclosed within the 2-sphere and the surface area $2\pi^2 R^3$ of the 3-sphere. So the ratio of volumes is $\sim 3500$. Actually we should also account for how the 2-sphere volume is modified by the curvature, but I'll neglect that.


The Planck (2018) results combined with BAO constraints give a curvature density $\Omega_k = 0.0007 \pm 0.0019$, where the error bar is a 68% confidence limit. That means a reasonable 3-sigma upper limit to any positive curvature is $\Omega_k \geq -0.005$.

Then, using $$ R = \frac{c}{H_0 \sqrt{-\Omega_k}}$$ we get a radius of curvature of at least 205 Gly.

Even using a 1-sigma upper limit gives 418 Gly.


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