= Solution
Neglect the decaying mode from part a. A mode that enters during matter domination grows in amplitude by
$$
\frac{a(t_0)}{a(t_H)}
=\left(\frac{t_0}{t_H}\right)^{2/3},
$$
so its variance grows by $(t_0/t_H)^{4/3}$. For $k<k_{\rm eq}$, part b gives
$$
P(k)=\frac{C}{k^3}
\left(\frac{k}{k_0}\right)^4
=\boxed{\frac{Ck}{k_0^4}}.
$$
For $k>k_{\rm eq}$, the mode enters during radiation domination and is assumed not to grow significantly until equality. It then grows by $a(t_0)/a_{\rm eq}$. Since the equality mode satisfies
$$
\frac{t_{\rm eq}}{t_0}
=\left(\frac{k_0}{k_{\rm eq}}\right)^3,
$$
the post-equality variance growth is
$$
\left(\frac{a(t_0)}{a_{\rm eq}}\right)^2
=\left(\frac{t_0}{t_{\rm eq}}\right)^{4/3}
=\left(\frac{k_{\rm eq}}{k_0}\right)^4.
$$
Therefore the <broken matter power spectrum from horizon entry> is
$$
\boxed{
P(k)=V\langle|\delta_k(t_0)|^2\rangle
=
\begin{cases}
\dfrac{Ck}{k_0^4},&k<k_{\rm eq},\\[6pt]
\dfrac{Ck_{\rm eq}^4}{k^3k_0^4},&k>k_{\rm eq}.
\end{cases}}
$$
Back to article page
