Question

Recall that the horizon distance at a time $t$ is given by $d_H(t) = R(t) \int_0^t \frac{dt'}{R(t')}$. Suppose that the scale factor $R(t)$ follows $R(t) = R(t_{mr}) \times \begin{cases} \left(\frac{t}{t_{mr}}\right)^{1/2} & 0 < t \le t_{mr},\\ \left(\frac{t}{t_{mr}}\right)^{2/3} & t > t_{mr},\end{cases}$ where $t_{mr} \approx 50\,\,000\,\text{y}$ is the time of matter-radiation equality. Show that the horizon distance at the time of last scattering, $t_{ls} \approx 380\,\,000\,\text{y}$ is $d_H(t_{ls}) = 3t_{ls} - t_{ls}^{2/3} t_{mr}^{1/3}$.

          Recall that the horizon distance at a time $t$ is given by
$d_H(t) = R(t) \int_0^t \frac{dt'}{R(t')}$.
Suppose that the scale factor $R(t)$ follows
$R(t) = R(t_{mr}) \times \begin{cases} \left(\frac{t}{t_{mr}}\right)^{1/2} & 0 < t \le t_{mr},\\ \left(\frac{t}{t_{mr}}\right)^{2/3} & t > t_{mr},\end{cases}$
where $t_{mr} \approx 50\,\,000\,\text{y}$ is the time of matter-radiation equality. Show that the horizon distance at
the time of last scattering, $t_{ls} \approx 380\,\,000\,\text{y}$ is
$d_H(t_{ls}) = 3t_{ls} - t_{ls}^{2/3} t_{mr}^{1/3}$.

Recall that the horizon distance at a time t is given by
dH(t) = R(t) ∫0^t (dt')/(R(t')).
Suppose that the scale factor R(t) follows
R(t) = R(tmr) ×((t)/(tmr))^1/2 0 < t ≤ tmr,
((t)/(tmr))^2/3 t > tmr,
where tmr≈ 50 000 y is the time of matter-radiation equality. Show that the horizon distance at
the time of last scattering, tls≈ 380 000 y is
dH(tls) = 3tls - tls^2/3 tmr^1/3.

Added by Sarah B.

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