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==Bethe晶格上的渗流==
==Bethe晶格上的渗流==
==公式，图表==
===1===
$$\begin{array}{|l|r|r|r|}
\hline \text { Lattice } & \# \mathrm{nn} & \text { Site percolation } & \text { Bond percolation } \\
\hline \hline 1 \mathrm{d} & 2 & 1 & 1 \\
\hline \hline 2 \mathrm{d} \text { Honeycomb } & 3 & 0.6962 & 1-2 \sin (\pi / 18) \approx 0.65271 \\
\hline 2 \mathrm{d} \text { Square } & 4 & 0.592746 & 1 / 2 \\
\hline 2 \mathrm{d} \text { Triangular } & 6 & 1 / 2 & 2 \sin (\pi / 18) \approx 0.34729 \\
\hline 3 \mathrm{d} \text { Diamond } & 4 & 0.43 & 0.388 \\
\hline 3 \mathrm{d} \text { Simple cubic } & 6 & 0.3116 & 0.2488 \\
\hline 3 \mathrm{d} \mathrm{BCC} & 8 & 0.246 & 0.1803 \\
\hline 3 \mathrm{d} \mathrm{FCC} & 12 & 0.198 & 0.119 \\
\hline \hline 4 \mathrm{d} \text { Hypercubic } & 8 & 0.197 & 0.1601 \\
\hline \hline 5 \mathrm{d} \text { Hypercubic } & 10 & 0.141 & 0.1182 \\
\hline 6 \mathrm{d} \text { Hypercubic } & 12 & 0.107 & 0.0942 \\
\hline \hline 7 \mathrm{d} \text { Hypercubic } & 14 & 0.089 & 0.0787 \\
\hline \hline \text { Bethe lattice } & \mathrm{z} & \mathbf{1} /(\mathrm{z}-1) & \mathbf{1} /(\mathrm{z}-1) \\
\hline
\end{array}$$
===2===
$$n_{s}(p)=(1-p) p^{s}(1-p)=(1-p)^{2} p^{s}$$
$$\begin{aligned}
n_{s}(p) &=(1-p)^{2} p^{s} \\
&=(1-p)^{2} \exp \left(\ln \left(p^{s}\right)\right) \\
&=(1-p)^{2} \exp (s \ln (p)) \\
&=\left(p_{c}-p\right)^{2} \exp \left(-\frac{s}{s_{\xi}}\right)
\end{aligned}$$
$$s_{\xi}=\frac{-1}{\ln (p)}=\frac{-1}{\ln \left(p_{c}-\left(p_{c}-p\right)\right)} \rightarrow \frac{1}{p_{c}-p}=\left(p_{c}-p\right)^{-1} \quad \text { for } p \rightarrow p_{c}$$
$$s_{\xi} \propto\left|p_{c}-p\right|^{-\frac{1}{\sigma}} \quad \text { for } p \rightarrow p_{c}$$
$$\sum_{s=1}^{\infty} s n_{s}(p)=p \quad \text { for } p<p_{c}$$
$$\begin{aligned}
\sum_{s=1}^{\infty} s n_{s}(p) &=\sum_{s=1}^{\infty} s(1-p)^{2} p^{s} \\
&=(1-p)^{2} \sum_{s=1}^{\infty} p \frac{d\left(p^{s}\right)}{d p} \\
&=(1-p)^{2} p \frac{d}{d p}\left(\sum_{s=1}^{\infty} p^{s}\right) \\
&=(1-p)^{2} p \frac{d}{d p}\left(\frac{p}{1-p}\right) \\
&=p
\end{aligned}$$
$$w_{s}=\frac{s n_{s}(p)}{p}=\frac{s n_{s}(p)}{\sum_{s=1}^{\infty} s n_{s}(p)}$$
$$\begin{aligned}
S(p) &=\sum_{s=1}^{\infty} s w_{s} \\
&=\sum_{s=1}^{\infty} \frac{s^{2} n_{s}(p)}{\sum_{s=1}^{\infty} n_{s}(p) s} \\
&=\frac{1}{p}(1-p)^{2} \sum_{s=1}^{\infty} s^{2} p^{s} \\
&=\frac{1}{p}(1-p)^{2}\left(p \frac{d}{d p}\right)^{2}\left(\sum_{s=1}^{\infty} p^{s}\right)
\end{aligned}$$
$$S(p)=\frac{1+p}{1-p}=\frac{p_{c}+p}{p_{c}-p}$$
$$S(p)=\frac{p_{c}+p}{p_{c}-p} \rightarrow \frac{2 p_{c}}{p_{c}-p} \propto\left(p_{c}-p\right)^{-1} \quad \text { for } p \rightarrow p_{c}^{-}$$
$$S(p) \propto\left|p_{c}-p\right|^{-\gamma} \quad \text { for } p \rightarrow p_{c}$$
$$g(\mathbf{r})=\exp \left(\ln \left(p^{r}\right)\right)=\exp (r \ln (p))=\exp \left(-\frac{r}{\xi}\right)$$
$$\xi=-\frac{1}{\ln (p)}=\frac{-1}{\ln \left(p_{c}-\left(p_{c}-p\right)\right)} \rightarrow \frac{1}{\left(p_{c}-p\right)}=\left(p_{c}-p\right)^{-1} \quad \text { for } p \rightarrow p_{c}=1$$
$$\xi \propto\left|p_{c}-p\right|^{-\nu} \quad \text { for } p \rightarrow p_{c}$$
===3===
$$\begin{aligned}
\text { Total no. sites } &=1+3 \cdot\left(1+2+\cdots+2^{g-1}\right) \\
&=1+3 \cdot \frac{1-2^{g}}{1-2} \\
&=3 \cdot 2^{g}-2
\end{aligned}$$
$$\frac{\text { No. of surface sites }}{\text { Total no. of sites }}=\frac{3 \cdot 2^{g-1}}{3 \cdot 2^{g}-2} \rightarrow \frac{1}{2} \quad \text { for } g \rightarrow \infty$$
$$\frac{\text {No. of surface sites}}{\text {Total no. of sites}} \rightarrow \frac{z-2}{z-1} \text { for } g \rightarrow \infty$$
$$\text { Surface } \propto \text { Volume }^{\frac{d-1}{d}}=\mathrm{Volume}^{1-\frac{1}{d}}$$
$$\frac{\text { No. loops }}{\text { Total no. chains }}=\frac{2 d \cdot(2 d-2) \cdot 1}{2 d \cdot(2 d-1)^{2}}=\frac{(2 d-2)}{(2 d-1)^{2}} \rightarrow 0 \quad \text { for } d \rightarrow \infty$$
$$p_{c}(z-1)=1 \Leftrightarrow p_{c}=\frac{1}{z-1}$$
$P(p)=(\text { Prob. site is occupied }) \times(\text { Prob. at least } O N E \text { branch lead to infinity })$ $=p\left(1-Q^{3}\right)$
$Q=(\text { Prob. site is empty })+(\text { Prob. site is occupied }) \times(\text { Prob. no subbranch leads to infinity })$ $=(1-p)+p Q^{2}$
Q=\frac{1 \pm \sqrt{(2 p-1)^{2}}}{2 p}=\left\{\begin{array}{l}
1 \\
\frac{1-p}{p}
\end{array}\right.
P(p)=\left\{\begin{array}{ll}
0 & \text { for } p<p_{c} \\
p\left(1-\left(\frac{1-p}{p}\right)^{3}\right) & \text { for } p \geq p_{c}
\end{array}\right.
P(p) \propto\left(p-p_{c}\right) \quad \text { for } p \rightarrow p_{c}^{+}
$S(p)=$ average cluster size to which the origin belongs
$=(\text { contribution from origin })+(\text { contributions from the } 3$ branches) $=1+3 T$
T=(1-p) \cdot 0+p \cdot(1+2 T) \Leftrightarrow T=\frac{p}{1-2 p} \quad \text { for } p<p_{c}
S(p)=1+3 T=\frac{1+p}{(1-2 p)}=\frac{1+p}{2\left(\frac{1}{2}-p\right)}=\frac{1+p}{2\left(p_{c}-p\right)}=\Gamma_{2}\left(p_{c}-p\right)^{-1}
P(p)+\sum_{s=1}^{\infty} s n_{s}(p)=p \quad \forall p