# Horizon distance at last scattering surface

1. Sep 22, 2014

### Orodruin

Staff Emeritus
I am reading the textbook by Ryden and I just ran into a statement that puzzles me (on page 201 if you have it at hand):

"For instance, we found that, in the absence of inflation, the horizon size at the time of last scattering was $d_{\rm hor}(t_{\rm ls}) \approx 0.4\,\rm Mpc$. Given a hundred e-foldings of inflation in the early universe, however, the horizon size at last scattering would have been $\sim 10^{43}\,\rm Mpc$, obviously gargantuan enough for the entire last scattering surface to be in causal contact."

Now, 100 e-foldings is around $10^{43}$ which is essentially the ratio of those two numbers. However, it would seem to me that this just assumes that the horizon will always be this factor larger than the value without inflation and does not take into account how the horizon grows. For example, I would expect the following behaviour for the proper distance to the horizon in a radiation dominated universe:

$$d(t) = a(t) \int_0^t \frac{dt'}{a(t')} = \frac{a(t)}{a(t_i)} a(t_i)\int_0^{t_i}\frac{dt'}{a(t')} + a(t)\int_{t_i}^t\frac{dt'}{a(t')} \simeq \sqrt{\frac{t}{t_i}} e^N d_i + 2 t,$$

where $d_i$ is the size of the universe before inflation and $N$ is the number of e-foldings. Without inflation, this would reduce to $2t$. Similar arguments could be made for a matter dominated universe. The problem here is that it is $d_i$, which would essentially be proportional to $t_i$, is the distance that is blown up by $e^N$, not the $2t$ which Ryden seems to assume. Now, this of course still is not enough to bring the horizon problem back, it is just for my own peace of mind. Could anyone shed some light on this?

2. Sep 24, 2014

### George Jones

Staff Emeritus
Ryden applies exponential expansion to the size of the universe generated by radiation-dominated expansion prior to inflation, not to radiation-dominated expansion after inflation. See the last two sentences on page 200.

Instead of splitting the integral $\int_0^t$ into $\int_0^{t_i} + \int_{t_i}^t$, Ryden considers $\int_0^{t_i} + \int_{t_i}^{t_f} + \int_{t_i}^t$, where $t_i$ is the start of inflation and $t_f$ is the end of inflation. Before inflation, Ryden assumes that the universe in radiation-dominated, so the first term gives $d_i = 2 t_i$ as the input to the inflationary period.

3. Sep 24, 2014

### Orodruin

Staff Emeritus
Yes, this is exactly my problem. The integral $\int_{0}^{t_f}$ gives a contribution
$$a(t)\int_{0}^{t_f} \frac{dt'}{a(t')} = \frac{a(t)}{a_f} e^N (d_i + H^{-1}),$$
where $a(t)/a_f \sim \sqrt{t/t_i}$ during radiation dominated expansion and so the contribution from this term is $\sim\sqrt{t/t_i} e^N (d_i + H^{-1})$. The remaining term gives essentially $2t$, which would be the size of the horizon without inflation, so this is not really relevant. The ratio of the horizon size with and without inflation is thus
$$\sim \sqrt{t_i/t} e^N \sim$$
assuming that $d_i \sim t_i$. What Ryden claims is that the horizon size at last scattering (ok the universe is not radiation dominated to that point, but just to get an idea) is a factor $e^N$ larger than it would be without inflation (1e43 Mpc rather than 0.4 Mpc). My question is related to the additional ratio of scale factors that seem to appear in what I did above which would suggest that (with $t_i \sim 10^{-36}$s and $t_{CMB} \sim 300000$yr) something closer to 1e18 Mpc. Still enough to solve the horizon problem, but significantly smaller than 1e43 Mpc ...

4. Sep 24, 2014

### George Jones

Staff Emeritus
Okay, I will try to think more thoroughly about this, but I doubt this will happen before the weekend.

5. Sep 27, 2014

### George Jones

Staff Emeritus
I haven't checked my calculations, but I am getting something close to this, something like $2 \times 10^{20} \mathrm{Mpc}$.

6. Sep 27, 2014

### Orodruin

Staff Emeritus
Thanks, I was being very hand-waving and the 1e18 Mpc essentially assumed radiation domination from inflation to last scattering (I might also have ignored a factor of 100 which I considered order 1 in the comparison ;) 2e20 Mpc is close enough to 1e18 Mpc for me to believe) I was mainly after the qualitative behaviour and wanted to double check that I was not way off when I thought it was not 1e43 Mpc.

7. Sep 28, 2014

### Chronos

I've seen the e43 value mentioned before, but, I think that was mega light years, not Mpc's. That would make sense.