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This question already has an answer here:

I've been told to calculate the age of the Universe at $T = 1 \, \text{MeV}$, assuming that $a(t=0)=0$.

Now, I've already calculated the value of $H$ at that temperature, which was around $H(1\,\text{MeV}) \approx 0.6 \,\text{s}^{-1}$. I've also shown, that in a radiation dominated Universe, which I assume much be the case at $T = 1 \, \text{MeV}$, that:

$$H = \frac{1}{2}t^{-1}$$

So basically, my idea was just to solve for $t$ in that equation, and use the value for $H$ I calculated, and then I end up with $t = 0.85 \, \text{s}$, which seems okay reasonable to me, but, my main question is the info: "assuming that $a(t=0)=0$". I haven't really used that information here, so either it's just not important, or I have missed something. But what ?

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marked as duplicate by John Rennie, Kyle Kanos, Brandon Enright, JamalS, Danu Nov 28 '14 at 23:04

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ There's a standard way of calculating the age at a specific H that involves integrating over $a$ from $a(t=0)$ until the era of interest. I don't quite remember it off-hand. Try going through some of the related questions on this site for the age of the universe. I don't remember where it is, but I know I've seen it here somewhere $\endgroup$ – Jim Nov 28 '14 at 15:06
  • $\begingroup$ But I don't know the scale factor $a$, at least in my case, at that temperature, do I ? $\endgroup$ – Denver Dang Nov 28 '14 at 17:52
  • $\begingroup$ isn't the scale factor proportional to $\sqrt{t}$ during radiation domination? or am I confused about something else $\endgroup$ – Jim Nov 28 '14 at 17:56
  • $\begingroup$ That is correct. But it's the age of the universe, i.e. time $t$, I'm trying to find. So I know neither of them - at least what I can see. But then I maybe can use the relation between $a$ and $T$, i.e. $a \propto 1/T$, or...? $\endgroup$ – Denver Dang Nov 28 '14 at 18:21
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    $\begingroup$ Dear close voters: This is not a duplicate. The issue is the implicit limit of integration, which is nowhere mentioned in the referenced question -- it, like most sources, never draws attention to the fact that we are tacitly defining $t = 0$ to be when $a = 0$. $\endgroup$ – user10851 Nov 28 '14 at 23:07