# Horizon Length at the Surface of Last Scattering

Horizon length is the distance travelled by light photons from ‘The Big Bang’ to ‘The Recombination Era’. The 1^{st} peak of the angular spectrum is at θ = 1◦ (l = 180), which is a very special length scale.

The proper distance between two points is given by −

$$r_p = \int_{0}^{t}cdt$$

When we take the time frame of t = 0 to t = t_{rec}, then

$$r_H = \int_{0}^{t_{rec}}cdt$$

Where $r_H$ is the proper horizon distance.

Now, we know that −

$$\dot{a} = \frac{\mathrm{d} a}{\mathrm{d} t}$$

$$dt = \frac{da}{\dot{a}}$$

When t = 0, a = 0.

Then $t = t_{rec}, a = a_0 / (1 + z_{rec})$.

Hence, we can write,

$$r_H(z_{rec})=\int_{0}^{a_{rec}} c\frac{da}{aH}$$

$$H(a_{rec}) = H(z_{rec}) = H_0\sqrt{\Omega_{m,0}}a^{-3/2}$$

During the **Recombination period universe** was matter dominated. i.e., **Ωrad << Ωmatter**. Therefore, the term radiation is dropped.

$$r_H(z_{rec}) = \frac{c}{H_0\sqrt{\Omega_{m,0}}}\int_{0}^{a_{rec}} \frac{da}{a^{-1/2}}$$

$$r_H(z_{rec}) = \frac{2c}{3H_0\sqrt{\Omega_{m,0}}}\frac{1}{(1+z_{rec})^{3/2}}$$

$$\theta_H(rec) = \frac{r_H(z_{rec})}{d_A(z_{rec})}$$

Which is equal to 0.5 degrees, if we put all the known values in the equation.

The **Electromagnetic radiation** is opaque from the surface of last scattering. Any two points ‘not’ lying within the horizon of each other need not have the same properties. So, it will give different temperature values.

We can get two points on this surface which did not intersect with each other, which means at one point the universe expanded faster than the speed of light which is the inflationary model for expansion.

## Points to Remember

The horizon length is the distance travelled by light photons from ‘The Big Bang’ to ‘The Recombination Era’.

During the Recombination period, the universe was matter dominated.

Electromagnetic radiation is opaque from the surface of last scattering.