In “The picture of our universe: A view from modern cosmology” (https://arxiv.org/abs/astro-ph/0209504 - eqn 63, page 26) there's an explicit equation for the scale factor for a flat universe with $\Omega_{M}+\Omega_{\Lambda}=1=\Omega_{\text{total}}$:$$a\left(t\right)=A^{1/3}\sinh^{2/3}\left(\frac{t}{t_{\varLambda}}\right),$$where $A=\Omega_{m,0}/\Omega_{\varLambda,0}$ and $t_{\varLambda}=\frac{2}{\sqrt{3\varLambda}}$. For how long in the history of the universe has the $\Omega_{M}+\Omega_{\Lambda}=1=\Omega_{\text{total}}$ condition been true and how is this known?

  • $\begingroup$ So you're asking how long the Universe has attained critical density? $\endgroup$
    – JamalS
    Jul 14, 2018 at 14:38
  • $\begingroup$ @JamalS - To be honest, I'm often not too sure what I'm asking. But yes, I think that's what I'm asking. The derivation of the equation does appear, though, to assume negligible radiation density. $\endgroup$
    – Peter4075
    Jul 14, 2018 at 16:01
  • $\begingroup$ @JamalS - On reflection, no, I'm not asking that. I'm asking how long radiation density has been negligible and therefore how long the explicit equation for the scale factor I give has been valid. $\endgroup$
    – Peter4075
    Jul 16, 2018 at 12:45
  • $\begingroup$ This is a duplicate of How long was the universe radiation dominated? $\endgroup$ Jul 20, 2018 at 16:18
  • $\begingroup$ @JohnRennie - Surely not. My question is how long has the $sinh$ explicit scale factor equation been valid. In other words, how long has radiation density been negligible. That's different to how long was the universe radiation dominated. $\endgroup$
    – Peter4075
    Jul 21, 2018 at 9:24

3 Answers 3


I'll make some rough estimates, doing everything to the nearest order of magnitude. That will be good enough, because it's not like your equation will suddenly stop working at some certain time, it'll just gradually degrade as we go back in time.

We know that the radiation contribution today is $$\Omega_{r,0} \approx 10^{-4}.$$ We also know the matter density falls off as $1/a^3$, while the radiation density falls off as $1/a^4$, so $$\frac{\Omega_m}{\Omega_r} \propto a$$ which indicates matter and radiation were equally important at around $a_{\text{eq}} \sim 10^{-4}$. For smaller scale factors, your result surely doesn't hold.

If you want to convert this into a time since the Big Bang, you can use the heuristic that the Hubble parameter $H = \dot{a}/a$ should obey $$H \sim \frac{1}{t}$$ where $t$ is the current age of the universe. This is true whenever the universe is dominated by a single component which isn't dark energy, essentially by dimensional analysis. The Friedmann equation may be rewritten as $$H(a)^2 = H_0^2 (\Omega_{r,0} a^{-4} + \Omega_{m,0} a^{-3} + \Omega_{k,0} a^{-2} + \Omega_{\Lambda, 0})$$ where $\Omega_{k,0}$ is a term that formally accounts for the curvature of the universe. We know that at the current moment, $$\Omega_{k,0} \approx 0, \quad \Omega_{r,0} \approx 0, \quad \Omega_{m,0} + \Omega_{\Lambda,0} \approx 1.$$ Also note that since the curvature term falls off as $a^{-2}$, it becomes less and less relatively important in the early universe compared to matter and radiation, so we never have to worry about it. (In particular, if the density is almost critical now, it was even closer to critical in the past.) Counting only the radiation at time $a_{\text{eq}}$ for simplicity, $$H(a_{\text{eq}})^2 = H_0^2 \frac{\Omega_{r,0}}{a_{\text{eq}}^4} \approx H_0^2 a_{\text{eq}}^{-3}, \quad H(a_{\text{eq}}) \approx 10^6 H_0.$$ Therefore, the age of the universe at time $t_{\text{eq}}$ is $$t_{\text{eq}} \approx 10^{-6} t_0 \sim 10^4 \text{ years}.$$ Your equation should begin to be fairly accurate after about $10^5$ years.

  • $\begingroup$ Why doesn't $\frac{\Omega_m}{\Omega_r} \propto a$ and how do you get $a_{\text{eq}} \sim 10^{-4}$? Thank you. $\endgroup$
    – Peter4075
    Jul 16, 2018 at 16:04
  • $\begingroup$ @Peter4075 Whoops, I made a typo, thanks! I get the latter result because $\Omega_{m,0} / \Omega_{r,0} \sim 10^4$, where I'm just working to the nearest order of magnitude. You can refine everything I did here to be more precise if you want. $\endgroup$
    – knzhou
    Jul 16, 2018 at 16:13
  • $\begingroup$ Still chugging through it! Shouldn't that be $H_{0}^{2}\frac{\Omega_{r,0}}{a_{\text{eq}}^{4}}\approx H_{0}^{2}a_{\text{eq}}^{-3}$. $\endgroup$
    – Peter4075
    Jul 16, 2018 at 17:56
  • $\begingroup$ @Peter4075 Yup, thanks for spotting that too! $\endgroup$
    – knzhou
    Jul 16, 2018 at 18:02
  • $\begingroup$ Final comment - strange, but I would have thought this explicit equation for the scale factor would be much more frequently quoted than it appears to be. For example, there's no mention in the Wikipedia scale factor page (en.wikipedia.org/wiki/Scale_factor_(cosmology)) nor in most cosmology textbooks I've looked at. $\endgroup$
    – Peter4075
    Jul 16, 2018 at 18:27

The exact relationship between the scale factor “a” and cosmic time “t” for our universe, considered flat, is given by the following expression:

$$\displaystyle t=\dfrac 1{H_0} \int_0^{a} \dfrac{dx}{\sqrt{\Omega_{\Lambda_0} x^2+\Omega_{M_0} x^{-1}+\Omega_{R_0} x^{-2}}}$$

The best values currently available (given for Planck Mission 2018), for the parameters are:

$H_0=67.66 \ (km/s)/Mpc$

$\Omega_{R_0}=9.15 \cdot 10^{-5}$



An approximation of the relationship between the scale factor and time for our universe, (considered flat), when the radiation density ratio is negligible, is given by the following expression:

$$\displaystyle a \approx \left (\frac{\Omega_{M_0}}{\Omega_{\Lambda_0}}\right )^{1/3} \sinh^{2/3}\left ( \frac{3 H_0 \sqrt{\Omega_{\Lambda_0}}}{2} \ t \right )$$

This expression is the same as in equation 63 page 26 of the document "The picture of our universe: A view from modern cosmology", taking into account that:

$\displaystyle t_{\Lambda}=\left (\dfrac 4{3\Lambda c^2} \right )^{1/2}=\dfrac{2}{3 H_0 \sqrt{\Omega_{\Lambda_0}}}$

For the density ratios in the approximate expression, we will use:

$\Omega_{M_0} \approx 0.3111$

$\Omega_{\Lambda_0} \approx 0.6889$

Now, we make a table in which for each cosmic time from the beginning until now, we tabulate the scale factor calculated with the exact expression (by numerical methods), the scale factor calculated with the approximate expression, and the relative error we make if we keep the approximate expression.

enter image description here

The value of the error we accept will determine from what cosmic time the approximate expression is acceptable

  • If we do not accept an error greater than 20% the approximation is valid from t=180 thousand years until now.

  • If we do not accept an error greater than 10% the approximation is valid from t=1 million years until now.

  • If we do not accept an error greater than 2% the approximation is valid from t=22 million years until now.

  • And if we only accept an error of less than 1% the approximation is valid from t=71 million years until now, ...

Best regards.

  • $\begingroup$ You've quantified the error. Nice one. Why didn't I think of that? $\endgroup$
    – Peter4075
    Jul 21, 2018 at 9:28
  • $\begingroup$ This thread reminds me of when someone asks me: At what speed should I stop using Newtonian Mechanics and must I use Special Relativity? I usually reply, that the question asked in this way is incomplete. And then I ask him, what precision do you need? What is the maximum error you are willing to admit? :) Best regards $\endgroup$
    – Albert
    Jul 22, 2018 at 9:46

Yes, $\Omega_{M}+\Omega_{\Lambda}=1=\Omega_{\text{total}}$ implies that the universe is flat. A spatially flat universe requires $K = 0$, with $K$ the curvature constant. As you may be aware of this constant can also have the values $-1$ or $+1$ in case the spatial curvature is hyperbolic or spherical respectively.

Now for a given universe $K$ can't change. A change from e.g. flat to spherical can't happen and moreover would imply a change of the topology. What changes in the flat case is the ratio $\Omega_{\Lambda}/\Omega_{M}$. It increases, but the sum of the dimensionless densities remains $constant = 1$.

  • $\begingroup$ But at one stage our flat universe's radiation density $\Omega_{R}$ would not have been negligible and $\Omega_{M}+\Omega_{\Lambda}+\Omega_{R}=1.$ The explicit equation for the scale factor in my question would not then have been valid (I think). Maybe I should have simply asked for how long has the radiation density been negligible? $\endgroup$
    – Peter4075
    Jul 15, 2018 at 16:47
  • $\begingroup$ In this answer here (physics.stackexchange.com/questions/92805/…), for example, @benrg says $\Omega_{M}+\Omega_{\Lambda}=1=\Omega_{\text{total}}$ has been true for $t\gtrsim10^{8}\;\text{yr}.$ But where does that figure come from? $\endgroup$
    – Peter4075
    Jul 15, 2018 at 16:48
  • $\begingroup$ This case is discussed also here, 4.5. Indeed radiation played a significant role in the early universe. But this doesn't mean that the universe wasn't spatially flat then. The criterion for flatness, total density equals the critical density, holds at that time. Sorry I'm off-line now for about 1 week. I think I can clarify that figure you mentioned after being back. $\endgroup$
    – timm
    Jul 15, 2018 at 20:23
  • $\begingroup$ Hi Peter, in case you are still interested in this: I think $10^8$ yr means the time of last scattering, 380000 years after the big bang. However according to the The Big Bang article the radiation dominated era ended about 50000 years after the big bang. $\endgroup$
    – timm
    Jul 22, 2018 at 7:53
  • $\begingroup$ No sorry, I was mistaken. This figure fits to the first appearance of stars, but that doesn't make sense either in my opinion. $\endgroup$
    – timm
    Jul 22, 2018 at 12:50

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