= Solution
The scaling solution has $\rho=3M^2p^2/t^2$ and $\rho+P=\dot\varphi^2=4M^2/(\lambda^2t^2)$. Therefore
$$
\boxed{w=-1+\frac{\lambda^2}{3},\qquad q_0=\frac1p-1=\frac{\lambda^2}{2}-1.}
$$
More generally any real canonical <scalar field> with positive total <energy density> obeys $\rho+P=\dot\varphi^2\ge0$, so \b[it cannot have $w<-1$]. A negative potential does not reverse this conclusion.
The proper <particle horizon> is $a(t)\int_0^t dt'/a(t')$. For $a\propto t^p$ the lower-limit integral diverges exactly when $p\ge1$. Thus, assuming the same power law extends to the initial boundary, there is no finite <particle horizon> for $\boxed{0<\lambda^2\le2}$. For the strict inequality, $p>1$, the proper <cosmological event horizon> and present age are
$$
\boxed{R_{\rm EH,0}=\frac{t_{\rm now}}{p-1}=\frac{2}{(2-\lambda^2)H_0},\qquad
t_{\rm now}=\frac{p}{H_0}=\frac{2}{\lambda^2H_0}.}
$$
At $\lambda^2=2$, $a\propto t$ has an infinite particle-horizon integral but also an infinite future event-horizon integral, so no finite event horizon exists. The age remains $H_0^{-1}$. The $\lambda=0$ constant-potential case is separate: its flat <de Sitter> patch has event-horizon radius $H_0^{-1}$ and no finite-age power-law <Big Bang>.
Back to article page
