I wonder is there a relation between the size of the universe and the scale factor calculated by solving Friedmann equations.

I mean if the volume of the universe nowadays is a round $V= 10^{78} m^3$, does this mean the current value of the cosmological scale factor is around $10^{26} m $ ? can we say $V=a^3 ~m^3$?

When solving Friedmann equation:

$$\left( \frac{\dot{a}}{a} \right) = H_0 \sqrt{\Omega_ra(t)^{-4}+\Omega_ma(t)^{-3}+\Omega_{\Lambda}}$$

According to this thread: The scale factor of ΛCDM as a function of time Or according to this code: The scale factor of ΛCDM

It gives the normalized dimensionless scale factor with $a(t_0)=1$, where $t_0$ is the current age of the universe $\sim 13$ Gyr .

Now I think if we wish to get a dimensionful scale factor with units of length, we should use an alternative formula for Friedmann equations. I tried

$$\dot{a}(\eta) = \frac{H_0}{c} \left(\Omega_m a_0^3 a + \Omega_r a_0^4 + \Omega_\Lambda a^4\right)^{1/2}$$

Where $\eta$ is a dimensionless conformal time . This formula is from Notes equation (28). But when using NDSolve in Mathematica in this Thread the equation has not been solved.

So any help to understand that? I thought when the equation is solved it gives $a(\eta_0) = a(13) = 10^{26}$ meter ?

  • $\begingroup$ Where are you getting the volume? Is it just the volume of the observable universe? Because if so, that grows faster than $a^3$ because light from more and more of the universe is able to reach us as time passes. We don't know the universe's total volume (or whether it is even finite), just our observable patch of it. $\endgroup$
    – Sten
    Sep 27, 2022 at 22:26
  • $\begingroup$ You should probably write $a'(η)$ since the dot normally indicates a derivative wrt $t$, not $η$. $\endgroup$
    – benrg
    Sep 27, 2022 at 22:34

1 Answer 1


The size of the universe and the size of the observable universe are different things. The radius of the observable universe is equal to the conformal time times the scale factor if they're appropriately normalized.

If $k=\pm 1$, the scale factor is the radius of curvature of the spatial slices. Commonly it's called $R$ instead of $a$ in that case. If $k=1$ then $R$ is the size of the universe (or its reduced circumference). If $k=-1$, it at least sets a characteristic scale, though the total volume is of course infinite.

If $k=0$, as it is in ΛCDM, then there is no geometrical basis for assigning units to the scale factor. Your formula for $a'(η)$, although correct, doesn't help since it's invariant under a rescaling of $a$ (keeping in mind that the unitless conformal time also depends on $a_0$).

  • $\begingroup$ Hello @benrg . Thanks for the answer. “ If 𝑘=0, as it is in ΛCDM, then there is no geometrical basis for assigning units to the scale factor” what do you mean? still how to calculate the size of the universe from a ? $\endgroup$
    – Dr. phy
    Sep 28, 2022 at 5:25
  • $\begingroup$ @Dr.phy You can't calculate the size from $a$ when $k=0$. A spherical space has a radius, and a hyperbolic space has a radius (of curvature), but a Euclidean space doesn't. And its area is infinite regardless of $a$. $\endgroup$
    – benrg
    Sep 29, 2022 at 3:57

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