树枝状聚电解质的泊松-玻尔兹曼-弗洛里理论

采用Cell模型，推导树枝状聚电解质体系的泊松-玻尔兹曼-弗洛里理论。本文整理自：PCCP 2018,20,2693 Dendritic polyelectrolytes revisited through the Poisson-Boltzmann-Flory theory and the Debye-Hückel

泊松-玻尔兹曼-弗洛里理论

Cell Model，Cell 半径$L$，树枝状聚电解质半径$R$。

树枝状聚电解质代数为$G$，代段长度$S$，于是总链段数目为$N=2+4S(2^G-1)$，带电分率为$f$。

链弹性自由能：

\begin{equation} F_{el}=k_BT\frac{N}{(GS)^2}R^2b^{-2} \label{Fel} \end{equation}

排除体积相互作用：

\begin{equation} F_{excl}=k_BT\nu b^3\frac{N^2}{R^3} \label{Fexcl} \end{equation}

静电巨势：

\begin{equation} \begin{split} \Omega_{ele} =& \frac{\epsilon}{2}\int_V (\nabla\phi)^2dV + k_BT\sum_j \int_V n_j(\ln n_j -1)dV - \sum_j \mu_j \int_V n_j dV\\ =& -\frac{\epsilon}{2}\int_V (\nabla\phi)^2dV + \int_V \rho \phi dV + \\ & k_BT\sum_j \int_V n_j(\ln n_j -1)dV - \sum_j \mu_j \int_V n_j dV \end{split} \label{Omele} \end{equation}

其中，$\phi$为平均电势，$n_j$为第$j$种离子的数密度，电量为$q_j$，$\mu_j$为第$j$种离子的化学势。上式中第二个等号用到了恒等式：

\begin{equation} \frac{\epsilon}{2}\int_V (\nabla\phi)^2dV = \frac{1}{2} \int_V \rho \phi dV \label{elecident} \end{equation}

其中$\rho$为体系中电荷密度为

\begin{equation} \rho = \sum_j n_jq_j+\rho_{ext} = \sum_j n_jq_j+\frac{3fNe}{4\pi R^3}\Theta(R-r) \label{rho} \end{equation}

将静电巨势变分：

\begin{equation} \frac{\delta \Omega_{ele}}{\delta \phi} = \epsilon \nabla^2\phi + \rho = 0 \label{deltaphi} \end{equation}

这正是泊松方程。

\begin{equation} \frac{\delta \Omega_{ele}}{\delta n_j} = q_j\phi + k_BT\ln n_j - \mu_j = 0 \label{deltanj} \end{equation}

即得离子的玻尔兹曼分布：

\begin{equation} n_j = n_{0j}e^{-q_j\phi/k_BT} = n_{0j}e^{-q_j\beta\phi} \label{nj} \end{equation}

化学势

\begin{equation} \mu_j=k_BT\ln n_{0j} \label{muj} \end{equation}

将\eqref{rho}、\eqref{nj}和\eqref{muj}三式代入\eqref{Omele}，得

\begin{equation} \Omega_{ele} =-\frac{\epsilon}{2}\int_V (\nabla\phi)^2dV + \int_V \rho_{ext} \phi dV - k_BT\sum_j \int_V n_{0j}e^{-q_j\beta\phi}dV \label{Omelen} \end{equation}

树枝状聚电解质内部静电作用能：

\begin{equation} F_{ele,i}=4\pi\int_0^R r^2\rho_{ext}\phi dr - 4\pi\int_0^R r^2 \left [\frac{\epsilon}{2} \left ( \frac{d\phi}{dr}\right)^2+k_BT(n_{0-}e^{\beta \phi}+n_{0+}e^{-\beta \phi}) \right ] dr \label{Felei} \end{equation}

树枝状聚电解质内部静电作用能：

\begin{equation} F_{ele,b}= - 4\pi\int_R^L r^2 \left [\frac{\epsilon}{2} \left ( \frac{d\phi}{dr}\right)^2+k_BT(n_{0-}e^{\beta \phi}+n_{0+}e^{-\beta \phi}) \right ] dr \label{Feleb} \end{equation}

由\eqref{nj}、\eqref{rho}、\eqref{deltaphi}三式得体系泊松-玻尔兹曼方程为

\begin{equation} \begin{split} \frac{d^2\phi}{dr^2}+\frac{2}{r}\frac{d\phi}{dr}=&-\frac{\rho}{\epsilon}=-\frac{1}{\epsilon}\left[ \sum_j n_jq_j+\frac{3fNe}{4\pi R^3}\Theta(R-r) \right ]\\ =&-\frac{e}{\epsilon}\left[ n_{0+}e^{-\beta \phi}-n_{0-}e^{\beta \phi}+\frac{3fN}{4\pi R^3}\Theta(R-r) \right ] \end{split} \label{Poisson} \end{equation}

树枝状聚电解质半径由$\frac{\partial }{\partial R}(F_{el}+F_{excl}+\Omega_{ele})=0$得到：

\begin{equation} \frac{2N}{(GS)^2}Rb^{-2}-3\nu b^3\frac{N^2}{R^4}-\frac{9fNe}{k_BTR^4}\int_0^R r^2 \phi dr+\frac{3fNe}{k_BTR}\phi(R)=0 \label{R} \end{equation}

将方程无量纲化。令$l_B=\frac{e^2}{4\pi \epsilon k_BT}$，$\Phi=\frac{e\phi}{k_BT}$，$y=r/l_B$，$Y=R/l_B$，$m_{0\pm}=4\pi l_B^3n_{0\pm}$，$l=l_B/b$，$D=L/l_B$。

泊松-玻尔兹曼方程\eqref{Poisson}化为：

\begin{equation} \frac{d^2\Phi}{dy^2}+\frac{2}{y}\frac{d\Phi}{dy}=- m_{0+}e^{- \Phi}+m_{0-}e^{\Phi}-\frac{3fN}{Y^3}\Theta(Y-y) \label{DLPoisson} \end{equation}

树枝状聚电解质半径满足的方程\eqref{R}化为

\begin{equation} \frac{2Nl^2}{(GS)^2}Y-\frac{3\nu N^2}{l^3}Y^{-4}-9fNY^{-4}\int_0^Y y^2 \Phi dy+3fNY^{-1}\Phi(Y)=0 \label{DLR} \end{equation}

设离子库中正负离子数密度为$m$，唐南势为$\Phi_D$，则正负离子数密度为

\begin{equation} \begin{split} &m_+=m_{0+}e^{- \Phi}=me^{- (\Phi-\Phi_D)}\\ &m_-=m_{0-}e^{\Phi}=me^{\Phi-\Phi_D} \end{split} \label{m-+} \end{equation}

令 $m_{0-}e^{\Phi}=me^{\Phi-\Phi_D}=e^{\Psi}$，即

\begin{equation} \Phi=\Psi+\Phi_D-\ln m \label{Psi} \end{equation}

\begin{equation} \begin{split} &m_+=m_{0+}e^{- \Phi}=me^{- (\Phi-\Phi_D)}=m^2e^{-\Psi}$\\ &m_-=m_{0-}e^{\Phi}=me^{\Phi-\Phi_D}=e^{\Psi}$ \end{split} \label{m-+n} \end{equation}

将\eqref{m-+n}代入\eqref{DLPoisson}式，得

\begin{equation} \frac{d^2\Psi}{dy^2}+\frac{2}{y}\frac{d\Psi}{dy}=e^{\Psi}-m^2e^{-\Psi}-\frac{3fN}{Y^3}\Theta(Y-y) \label{DLPoissonn} \end{equation}

将\eqref{Psi}式代入\eqref{DLR}式，得

\begin{equation} -9fNY^{-4}\int_0^Y y^2 \Psi_i dy+3fNY^{-1}\Psi_i(Y)+\frac{2Nl^2}{(GS)^2}Y-\frac{3\nu N^2}{l^3}Y^{-4}=0 \label{DLRf} \end{equation}

边界条件：

\begin{equation} \begin{split} & \frac{d\Psi_i}{dy}\bigg |_{y=0}=0 \\ & \frac{d\Psi_b}{dy}\bigg |_{y=D}=0 \\ & \Psi_i(Y)=\Psi_b(Y) \\ & \frac{d\Psi_i}{dy}\bigg |_{y=Y}=\frac{d\Psi_b}{dy}\bigg |_{y=Y} \end{split} \label{bc} \end{equation}

德拜-休克尔近似

如果$|\Psi|\ll 1$，方程\eqref{DLRf}化为

\begin{equation} \frac{d^2\Psi}{dy^2}+\frac{2}{y}\frac{d\Psi}{dy}=1-m^2+(1+m^2)\Psi-\frac{3fN}{Y^3}\Theta(Y-y) \label{DH} \end{equation}

解此方程，得

\begin{equation} \Psi_i[y]=\frac{m^2-1+\frac{3fN}{Y^3}}{m^2+1}+C_1\frac{e^{-\sqrt{m^2+1}y}}{y}++\frac{C_2}{\sqrt{m^2+1}}\frac{e^{\sqrt{m^2+1}y}}{y} \label{Psii}\end{equation}

\begin{equation} \Psi_b[y]=C_3\frac{e^{-\sqrt{m^2+1}y}}{y}++\frac{C_4}{\sqrt{m^2+1}}\frac{e^{\sqrt{m^2+1}y}}{y} \label{Psib}\end{equation}

结合边界条件\eqref{bc}，可定出常数。

将$\Psi_i(y)$代入\eqref{DLRf}，可得树枝状聚电解质大小$Y$。