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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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  • Simply from the form your solution takes one can tell something is wrong here. According to this, there is only one value of $t$ for each $a$, which is not necessarily true for the properly chosen values of the various $\Omega_i$. Also, care to justify this a bit more : $dt = \frac{da}{\dot{a}}$ ? – ticster Aug 05 '14 at 22:26
  • @ticster: if the universe doesn't recontract, then $a(t)$ is a one-to-one function. Current observation has ruled out the recollapsing models of the universe. So, it is completely valid that there is only one value of $t$ for each value of $a$ – Zo the Relativist Aug 05 '14 at 22:40
  • @JerrySchirmer No, it isn't. His solution is a general one and is clearly wrong for the values of the $\Omega$ where the universe recollapses. Those scenarios are implicitly part of his solution. It's a sanity check that his solution clearly fails. – ticster Aug 05 '14 at 23:02
  • @JerrySchirmer It'd be like someone calculating the maximal height of a cannonball as a function of initial velocity, and got an answer that wasn't $0$ for initial velocity $0$, then saying "that's fine, the particular case I want to apply it to has $v_0 \neq 0$". Again, failing a very specific sanity check doesn't mean your solution might be correct for other, more general situations. That's why it's called a sanity check. – ticster Aug 05 '14 at 23:07
  • @ticster: it's still true over the piecewise invertible part of the function. You'd get a singularity in the integral at the recollapse point ($\frac{da}{dt} =0 \rightarrow \frac{dt}{da} = \infty$), and then you'd know that you'd have to assemble a piecewise solution. It's no fancier than inverting a parabola to get a square root function in the recollapsing case. – Zo the Relativist Aug 06 '14 at 14:53
  • @JerrySchirmer I understand what you're saying, but it's valid in the sense that you point out when you say "It's no fancier than inverting a parabola to get a square root function in the recollapsing case.", but this is still not the right way to go about solving this particular problem. If a freshman physics student tried solving a pendulum problem by finding time as function of pendulum angle, you would conclude that he hasn't understood the role that time plays in the problem, and is looking at things backwards. – ticster Aug 07 '14 at 20:58
  • @ticster: it's a standard approach for solving ODEs. The solution is valid for the entire range of parameters. There is a caustic at the turnaround point, but who cares? It's written in a way that is convenient for measurement and for easy integration, and the integral is not analytically evaluable as is. If you hate it so much, invert it. – Zo the Relativist Aug 07 '14 at 21:04
  • @JerrySchirmer How is it more convenient than the more intuitive $a(t)$ ? – ticster Aug 07 '14 at 21:12
  • Because $H(t)$ is a known function of $a$, but not of $t$, as shown above. – Zo the Relativist Aug 07 '14 at 21:21
  • You should probably give a reference to Planck papers and mention the year when they were obtained – Andrii Magalich Jun 20 '16 at 18:35
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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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  • do you have a source for the $t\gtrsim10^{8};\text{yr}$ figure and is there a name for the $\Omega_{M}+\Omega_{\Lambda}=1=\Omega_{\text{total}}$ era? Thanks. – Peter4075 Jul 13 '18 at 14:47
  • @Peter4075, I don't think I had a source for $t \gtrsim 10^8 ; \text{yr}$, I just eyeballed it. $\Omega_\text{total} = 1$ is a critical density universe. $\Omega_M + \Omega_\Lambda = \Omega_\text{total}$ is the matter and/or dark energy dominated epoch. In ΛCDM, it could also be called the non-radiation-dominated epoch, since $\Omega_M + \Omega_\Lambda + \Omega_r = \Omega_\text{total}$ in ΛCDM. – benrg Jul 15 '18 at 22:45