When I take the FLRW equation

$$H_{(t)}=H_0\cdot\sqrt{\Omega_R\cdot a_{(t)}^{-4}+\Omega_M\cdot a_{(t)}^{-3}+\Omega_K\cdot a_{(t)}^{-2}+\Omega_{\Lambda}}$$

and calculate the value of the Hubbleparameter with time it goes to Limit Infinity when I go to the Limit of $t\rightarrow 0$.

In reality the Hubbleparameter did of course not start infinitely high. There is said to be a short time of inflation where $H_{(t)}$ had a high, but constant value, which then quickly fell of when the radiation dominated era took over.

Does anyone know which numerical value $H_{(t)}$ had during inflation, and with which value the radiation dominated era began?

The value of $H_{(t)}$ today is $67150 \text{m}/\text{Mpc}/\text{sec} =2.2\cdot 10^{-18} \, \text{sec}^{-1}$. But how high did it start?

My actual progress is plotted here and here which brings me near to the solutions published by the Planck Team, but the inflation phase is still missing.

(Of course I could try to extrapolate this data from the Log-Plot in Link 1, where we see that the scalefactor was somewhere near $10^{-30}$, but there must be a better and exacter reference than counting the Pixels between $10^{-30}$ and $10^{-20}$ to estimate the rough order of magnitude. Also, since this sketch is older and propably drawn from WMAP data instead of Planck 2013 not very quotable)

  • $\begingroup$ I think the doubling time during inflation is normally taken to be around $10^{-34}$ to $10^{-35}$ seconds, so the Hubble constant would be the inverse of this i.e. $10^{34}$ to $10^{35}$ s$^{-1}$. $\endgroup$ – John Rennie Feb 14 '15 at 12:33
  • $\begingroup$ And that would also be the value where the radiation phase blends in, so I would first have to solve for the time/scalefactor when $H_{(t)}$ was $10^{-34} \text{s}^{-1}$ and then use this value constant from there to $a$ and $t$ = $0$? $\endgroup$ – Yukterez Feb 14 '15 at 12:40
  • $\begingroup$ You could just do the integration taking $\Lambda$ to be a function of time/scale instead of a constant. The trouble is that we know too little about how inflation worked for the results to be very reliable. $\endgroup$ – John Rennie Feb 14 '15 at 16:48
  • $\begingroup$ I would program the function using the Minimum-Function reference.wolfram.com/language/ref/Min.html which would look like $H_{(t)}=Min[ H_{Max} \text{ , } \, H_0\cdot\sqrt{\Omega_R\cdot a^{-4}+...}]$ so when the Hubbleparameter gets higher than the specificied value $H_{Max}$ it gets flattened into a constant. $\endgroup$ – Yukterez Feb 15 '15 at 3:33

Until radiation-matter equality at redshift $z_\mathrm{eq} \simeq 3370$ corresponding to a scale factor $a_\mathrm{eq} \simeq 0.0003$ (Planck Collaboration et al. 2016), the Universe was radiation-dominated, meaning that the scale factor scales with time $t$ like $a \propto t^{1/2}$. That is $$ a_\mathrm{rad}(t) = a_\mathrm{eq} \left( \frac{t}{t_\mathrm{eq}} \right)^{1/2}, $$ where $t_\mathrm{eq}$ is the age of the Universe at radiation-matter equality. This age can be calculated to $t_\mathrm{eq} \simeq 52\,\mathrm{kyr}$ by integrating the Friedmann equation with a Planck 2016 cosmology$^\dagger$.

If you assume that inflation ended at $t=10^{-32}\,\mathrm{s}$, then the equation above yields $$ a_\mathrm{just\,after\,inflation} = 2.3\times10^{-26}, $$ or $z_\mathrm{just\,after\,inflation} = 4.3\times10^{25}$. Again from the Friedmann equation you then have$^\ddagger$ $$ H(t_\mathrm{just\,after\,inflation}) \sim 10^{51}\,\mathrm{km}\,\mathrm{s}^{-1}\,\mathrm{Mpc}^{-1}. $$

$^\dagger$ In Python: astropy.cosmology.Planck15.age(3370)

$^\ddagger$ In Python: astropy.cosmology.Planck15.H(4.3e25)


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