Size of the Unobservable Universe

Question

If you search that question on Google, the answers will be "250 times the size of the observable universe" or "more than 15.000.000 the size of the observable universe". Both of them come from a different interpretation of the same paper (https://academic.oup.com/mnrasl/article/413/1/L91/1747653#25823893), which estimate the universe size using the Bayesian model. Even if most articles say that the universe is 15.000.000 bigger than the observable, I don't think It's true, since I read the article and from what I understood the minimum size of it is 250 larger than a Hubble sphere (i.e. the observable universe), not 250 time the radius of the Hubble sphere. Am I right or wrong? An answer will be very appreciated, thanks in advance.

Answer 1

The article Applications of Bayesian model averaging to the curvature and size of the Universe cited in the question is from 2011. It is now 2023 and we have much better data, in particular from the Planck satellite measurements: "Planck 2018 results. VI. Cosmological parameters". According to these data $\Omega_k=0.0007 \pm 0.0019$

As the central value is $\Omega_k>0$, the most favoured result is that, (if the topology of the universe is simple), the universe is infinite. But since the tolerance in the value of Omega allows it to be negative, we can calculate the value of the minimum size of the universe if it were spherical-finite.

The radius of curvature of a spherical universe $R$ as a function of the curvature density ratio is: $$R=\dfrac c{H_0} \cdot \dfrac 1{\sqrt{- \Omega_k}}$$ Using the measurements of the Planck Collaboration-2018 $$\Omega_k=0.0007 \pm 0.0019$$ If the universe is spherical, its minimum size corresponds to: $$\Omega_k=-0.0012$$ If the universe is spherical with $H_0=67.66 \ (km/s)/Mpc$, its minimum radius of curvature is: $$R=\dfrac c{H_0} \cdot \dfrac 1{\sqrt{- \Omega_k}}=\dfrac{299792.458}{67.66} \cdot \dfrac 1{\sqrt{0.0012}}=127908 \ Mpc= 417 \ Gly$$ The expression for the 3-volume as a function of the radius of curvature is: $$V=2\pi^2 R^3$$ Therefore, the minimum size of the whole spherical universe is: $$\boxed{V_w=1433486966 \ Gly^3}$$ Let us now calculate the size of the observable universe. The differential volume using spherical coordinates: $$dV=\sqrt{\vert g \vert} \ dr \ d\theta \ d\phi$$ The Friedmann-Lemaitre-Robertson-Walker Metric Tensor: $$g = \begin{bmatrix} -c^2 & 0 & 0 & 0\\ 0 & \dfrac{a^2}{1-Kr^2} & 0 & 0\\ 0 & 0 & a^2 r^2 & 0\\ 0 & 0 & 0 & a^2 r^2\sin^2\theta \end{bmatrix}$$ The square root of the absolute value of the determinant of the spatial part of the metric tensor: $$\sqrt{\vert g \vert}=a^3 \dfrac{r^2 \sin \theta}{\sqrt{1-K r^2}}$$ $$dV=a^3 \ \dfrac{r^2 \sin \theta}{\sqrt{1-K r^2}}dr d\theta d\phi$$ $$\displaystyle V(r)=a^3 \ \int_0^r \dfrac{r^2 dr}{\sqrt{1-K r^2}} \int_0^{\pi} \sin \theta d\theta \int_0^{2\pi} d\phi$$ $$\displaystyle V(r)=a^3 \ 4 \pi \int_0^r \dfrac{r^2 dr}{\sqrt{1-K r^2}}$$ $$V(r)=a^3 \ 2\pi \left ( \dfrac{\arcsin (r \sqrt K)}{K^{3/2}} - \dfrac{r\sqrt{1-Kr^2}}K \right )$$ The relationship between the coordinate $r$ and the distance $h$ is: $$\displaystyle h=\int_0^r \dfrac{dr}{\sqrt{1-Kr^2}}$$ $$h=\dfrac{\arcsin (r \sqrt K)}{\sqrt K}$$ $$r=\dfrac 1 {\sqrt K} \sin (h\sqrt K)$$ If we substitute in the volume expression: $$V(h)=a^3 \ \dfrac{2\pi}{K^{3/2}} \left ( h\sqrt K-\dfrac{\sin (2 h \sqrt K)}2 \right )$$ Now $a=1$ and the relationship between the curvature $K$ and the radius of curvature $R$ is: $$K=\dfrac 1{R^2}$$ Finally: $$\boxed{V(h)=2\pi R^3 \left ( \dfrac h R-\dfrac{\sin \dfrac{2 h}R}2 \right )}$$ If we perform the Taylor-Maclaurin approximation of this last expression: $$V_T(h) \simeq \dfrac 4 3 \pi h^3 \left [ 1-\dfrac 1 5 \left ( \dfrac h R \right )^2 \ \right ]$$ In which it is clear that, when the distance $h$ is small with respect to the radius of curvature $R$, we recover the expression for the volume of the sphere: $$V=\dfrac 4 3 \pi h^3$$ To calculate the size of the observable universe $V_o$ we must put as the value of $h$ the particle horizon $h=46.2 \ Gly$ in the boxed expression and we obtain: $$\boxed{V_o=412048 \ Gly^3}$$ Note that, if we had used the conventional volume of the sphere, the error we make is small, only 0.25% $$V_0\simeq \dfrac 4 3 \pi \ 46.2^3=413061 \ Gly^3$$ The quotient between the size of the minimum possible spherical universe and the observable universe according to the best measurements currently available is. $$\dfrac{V_w}{V_o}=\dfrac{1433486966}{412048}=3479$$ If the universe is spherical, its volume is at least 3479 times larger than the observable universe.

If we prefer to give the result as a function of Hubble spheres, we know that the actual Hubble radius is: $$R_H=14.5 \ Gly$$ Using the boxed expression we calculate the volume of the Hubble sphere: $$V_H=12767 \ Gly^3$$ $$\dfrac{V_w}{V_H}=\dfrac{1433486966}{12767}=112281$$ If the universe is spherical, its volume is at least 112281 times larger than the Hubble Volume.

Best regards.

Answer 2

Yes, you are right. It refers to the Hubble volume rather than the Hubble radius, specifically it is 251 times the Hubble volume, but based on the volume, we can calculate the radius of the whole universe being 88.2 billion light years..