# 我们的宇宙可能由闭宇宙演化为开宇宙么？[第二节]

\begin{eqnarray}
\mathrm d l^2&=& K^{-1}[\mathrm d^2 \arcsin r + r^2 (\mathrm d\theta^2+\sin^2\theta\mathrm d^2\phi)], K>0 \\
\mathrm d l^2&=& \mathrm d r ^2 + r^2 (\mathrm d\theta^2+\sin^2\theta\mathrm d^2\phi), K=0 \\
\mathrm d l^2&=&- K^{-1}[\mathrm d ^2\mathrm {sh}^{-1} r + r^2 (\mathrm d\theta^2+\sin^2\theta\mathrm d^2\phi)], K<0
\end{eqnarray}

\mathrm ds^2=-\mathrm dt^2+a^2(t)[\frac{\mathrm dr^2}{1-kr^2}+r^2(\mathrm d\theta^2+\sin\theta\mathrm d\phi^2)],

\left(
\begin{array}{cccc}
-1 & 0 & 0 & 0 \\
0 & \frac{A[t]}{1-k r^2} & 0 & 0 \\
0 & 0 & r^2 A[t] & 0 \\
0 & 0 & 0 & r^2 A[t] \text{Sin}[\theta ]^2
\end{array}
\right)

\left(
\begin{array}{cccc}
\frac{3 \left(4 k A[t]+A'[t]^2\right)}{4 A[t]^2} & 0 & 0 & 0 \\
0 & -\frac{A'[t]^2-4 A[t] \left(k+A”[t]\right)}{4 \left(-1+k r^2\right) A[t]} & 0 & 0 \\
0 & 0 & \frac{r^2 \left(A'[t]^2-4 A[t] \left(k+A”[t]\right)\right)}{4 A[t]} & 0 \\
0 & 0 & 0 & \frac{r^2 \text{Sin}[\theta ]^2 \left(A'[t]^2-4 A[t] \left(k+A”[t]\right)\right)}{4 A[t]}
\end{array}
\right)

\begin{eqnarray}
\frac{3 \left(4 k A[t]+A'[t]^2\right)}{4 A[t]^2}&=&-8 G \pi \rho [t] \\
\frac{A'[t]^2-4 A[t] \left(k+A”[t]\right)}{4 A[t]}&=&8 G \pi A[t] p[t]
\end{eqnarray}

\begin{eqnarray}
\frac{3 \left(-32 G \pi A[t]^2 p[t]+2 A'[t]^2-4 A[t] A”[t]\right)}{4 A[t]^2}&=&-8 G \pi \rho [t] \\
(32 G \pi \rho [t]+16 G \pi (3 p[t]+\rho [t])) A'[t]&=&-32 G \pi A[t] \rho ‘[t] .
\end{eqnarray}

\begin{eqnarray}
6A”[t]-3\frac{A'[t]^2}{A[t]}-16\pi G A[t](\rho[t]+3p[t])&=&0 \\
\frac{3A'[t]}{2A[t]}[\rho[t]+p[t]]+\rho ‘[t]&=&0
\end{eqnarray}

$$A[t]>0$$时，

\rho A^{3(1+w)/2}=\text{Const}

$$A[t]<0$$时，

\rho (-A)^{3(1+w)/2}=\text{Const}

Mathematica文件：OpenOrClose[2]MathematicaFile.