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Ideally I like a single formula or multiple formulas for different time ranges that would cover the time from the end of inflation through 100+ billion years after the big bang using the $\Lambda CDM$ Model. I know that from the end of inflation back to the time of the big bang would be much more speculative, but some wild estimate would be appreciated for that time range also!

The Friedmann–Lemaître–Robertson–Walker metric defines the scale factor, a(t), from the metric: $$-c^2d\tau^2 = -c^2dt^2+a(t)^2d\Sigma^2$$ where $d\Sigma^2$ ranges over the 3 dimensional space of the universe and does not depend on time. Usually the scale of the scale factor is set by defining $a(t_{now}) = a(13.78 B yr) \equiv 1$

Pulsar
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FrankH
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2 Answers2

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The scale factor can be derived from the Hubble parameter $$ H(a) = \frac{\dot{a}}{a} = H_0\sqrt{\Omega_{R,0}\,a^{-4} + \Omega_{M,0}\,a^{-3} + \Omega_{K,0}\,a^{-2} + \Omega_{\Lambda,0}}. $$ The latest values of the parameters, obtained from the Planck mission, are $$ H_0 = 67.3\;\text{km}\,\text{s}^{-1}\text{Mpc}^{-1},\\ \Omega_{R,0} = 9.24\times 10^{-5},\qquad\Omega_{M,0} = 0.315,\\ \Omega_{\Lambda,0} = 0.685,\qquad\Omega_{K,0} = 0. $$ From $$ \dot{a} = \frac{\text{d}a}{\text{d}t} $$ we get $$ \text{d}t = \frac{\text{d}a}{\dot{a}} = \frac{\text{d}a}{aH(a)} = \frac{a\,\text{d}a}{a^2H(a)}, $$ so that $$ \begin{align} t(a) &= \int_0^a \frac{a'\,\text{d}a'}{a'^2H(a')}\\ &= \frac{1}{H_0}\int_0^a \frac{a'\,\text{d}a'}{\sqrt{\Omega_{R,0} + \Omega_{M,0}\,a' + \Omega_{K,0}\,a'^2 + \Omega_{\Lambda,0}\,a'^4}}, \end{align} $$ which is the age of the universe as a function of $a$. By numerically inverting this relation, we get $a(t)$. For more information, see these posts:

Equation for Hubble Value as a function of time

Can space expand with unlimited speed?

Community
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Pulsar
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For the case $\Omega_M + \Omega_\Lambda = 1 = \Omega_\text{total}$, which is a good approximation for $t \gtrsim 10^8 \; \text{yr}$, an explicit formula for a(t) is

$$a(t) = \left[ \frac{\Omega_{M,0}}{\Omega_{\Lambda,0}} \sinh^2 \left( \frac32 \sqrt{\Omega_{\Lambda,0}} \, H_0 \, t \right) \right]^{\frac13}$$

or, plugging in numbers from Pulsar's answer,

$$a(t) \approx 0.772 \, \sinh^{2/3} \, (t \, / \, 11.7 \; \text{Gyr}).$$

You can easily calculate $H = \dot a/a$ from this, but for completeness, it's $H(t) = H_\infty \coth (\tfrac32 H_\infty t)$ where $H_\infty = \sqrt{\Omega_{\Lambda,0}} H_0 \approx 55.7 \; \text{km/s/Mpc}$.

benrg
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