= Solution
A superhorizon <density contrast> is gauge dependent. The stated growth laws describe the growing <adiabatic mode> in the <comoving-gauge density contrast> (and the corresponding usual synchronous growing mode), not an arbitrary coordinate density perturbation. Work with a flat, single-fluid background with constant $w=p/\rho$, negligible <anisotropic stress>, $c=1$, and a comoving Fourier wavenumber $K$.
Here is a derivation from the linear gravitational constraints and evolution equation. For an <adiabatic mode>, the scalar gravitational potential $\Phi$ obeys
$$
\Phi''+3(1+w)\mathcal H\Phi'+wK^2\Phi+
[2\mathcal H'+(1+3w)\mathcal H^2]\Phi=0,
\qquad \mathcal H=\frac{a'}a.
$$
Primes denote <conformal time> derivatives. The <Friedmann equation> gives $a\propto\tau^{2/(1+3w)}$ and $\mathcal H=2/[(1+3w)\tau]$, so the bracket vanishes. On <superhorizon scales>, $K\tau\ll1$, the leading solutions are a constant potential and a decaying solution proportional to $\tau^{-(5+3w)/(1+3w)}$. Keeping the growing adiabatic branch therefore makes $\Phi$ time independent to leading order.
Combining the time-time and time-space linear Einstein constraints eliminates the coordinate velocity and gives the comoving density constraint
$$
-K^2\Phi=4\pi Ga^2\rho\,\Delta,
\qquad
\Delta=-\frac23\left(\frac{K}{aH}\right)^2\Phi.
$$
This is the relativistic comoving constraint, not a subhorizon Newtonian approximation. Since $\rho\propto a^{-3(1+w)}$, it implies $\Delta\propto(a^2H^2)^{-1}\propto a^{1+3w}$. Also $a\propto t^{2/[3(1+w)]}$ in <cosmic time>, whence
$$
\boxed{\frac{\delta(t)}{\delta_i}=
\begin{cases}
t/t_i,&w=1/3\quad\text{(radiation domination)},\\
(t/t_i)^{2/3},&w=0\quad\text{(matter domination)}.
\end{cases}}
$$
Both epochs here mean their respective leading growing solution; matching through <matter-radiation equality> requires the full two-component evolution. The constant superhorizon curvature or potential is consistent with this growth because the comoving density perturbation carries the additional factor $(K/aH)^2$.
Back to article page
