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To calculate the size of the observable universe, one has to calculate the current distance to the particle horizon (t0: today, c = 1):
[tex]\int_{0}^{t_0} \ dt / a(t)[/tex] (1)
To be able to calculate the integral one has to find an expression for a(t). With some assumptions one can take
[tex]a(t) = (t/t_0)^\frac{2}{3}[/tex] (2)
and get an reasonable result for the integral: 3 c t0.
(see e.g. http://www.astro.ucla.edu/~wright/cosmology_faq.html#DN)
Now, my question: we know that there is an inflationary period during from t = 10^-35 sec. to t = 10^-30 sec. or whatever. During this period the dependence of a(t) is different than equation (2) above and thus the integral (1) should be calculated in two steps.
Although this period is short, it is indeed relevant because there is a huge expansion (an exponential dependence of a)
I have never seen the particle horizon to be calculated in this way. I assume I am missing something, since, in such a case, the observable universe would be far bigger than 3 c t0. So, what is wrong?
Regards.
[tex]\int_{0}^{t_0} \ dt / a(t)[/tex] (1)
To be able to calculate the integral one has to find an expression for a(t). With some assumptions one can take
[tex]a(t) = (t/t_0)^\frac{2}{3}[/tex] (2)
and get an reasonable result for the integral: 3 c t0.
(see e.g. http://www.astro.ucla.edu/~wright/cosmology_faq.html#DN)
Now, my question: we know that there is an inflationary period during from t = 10^-35 sec. to t = 10^-30 sec. or whatever. During this period the dependence of a(t) is different than equation (2) above and thus the integral (1) should be calculated in two steps.
Although this period is short, it is indeed relevant because there is a huge expansion (an exponential dependence of a)
I have never seen the particle horizon to be calculated in this way. I assume I am missing something, since, in such a case, the observable universe would be far bigger than 3 c t0. So, what is wrong?
Regards.
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