Alexander描述

参考 Langmuir 2003, 19, 4027-4033

胶体粒子为球形，半径为\(a\)，带电为\(-Ze\)，cell半径为\(R\)。

在一个cell里，局域无量纲电势满足PB方程：

\begin{equation}

\nabla^2\phi(r)=\kappa_{res}^2\sinh\phi(r)

\label{eq:PB} \end{equation}

边界条件：

\[\vec{n} \cdot \nabla \phi(r)|\_{r=a}=\frac{Z \lambda_B}{a^2}\]

\[\vec{n}\cdot \nabla\phi(r)|_{r=R}=0\]

把方程(\ref{eq:PB})线性化，即在\(\phi(R)=\phi_R\)处将其展开。

方程(\ref{eq:PB})左边

\[\nabla^2\phi(r)=\nabla^2(\phi(r)-\phi_R)=\nabla^2\widetilde{\phi}(r)\]

方程(\ref{eq:PB})右边

\begin{equation}

\begin{split}

\kappa_{res}^2\sinh\phi(r)&=\kappa_{res}^2[\sinh\phi_R+\cosh\phi_R(\phi(r)-\phi_R)]\

&=\kappa_{res}^2\cosh\phi_R[\tanh\phi_R+\widetilde{\phi}(r)]\

&=\kappa_{res}^2\cosh\phi_R[\gamma_0+\widetilde{\phi}(r)]\

&=\kappa_{PB}^2[\gamma_0+\widetilde{\phi}(r)]

\end{split}

\end{equation}

将以上两式合在一起，得线性化的PB方程：

\begin{equation}

\nabla^2\widetilde{\phi}(r)=\frac{1}{r^2}\frac{d}{dr}\left [r^2\frac{d}{dr}\widetilde{\phi}(r)\right ]=\kappa_{PB}^2[\gamma_0+\widetilde{\phi}(r)]

\label{eq:LPB} \end{equation}

边界条件：

\[\widetilde{\phi}(r)|\_{r=R}=0\]

\[\vec{n}\cdot\nabla\widetilde{\phi}(r)|\_{r=R}=0\]

方程\ref{eq:LPB}的解为

\[\widetilde{\phi}(r)=\gamma_0\left [-1+\frac{\kappa_{PB}+1}{2\kappa\_{PB}}e^{-\kappa_{PB}R}\frac{e^{\kappa_{PB}r}}{r}+\frac{\kappa_{PB}-1}{2\kappa\_{PB}}e^{\kappa_{PB}R}\frac{e^{-\kappa_{PB}r}}{r}\right ]\]

根据下式计算等效电量(Effective charge, Renormalized charge)：

\[\frac{d\widetilde{\phi}(r)}{dr}\Bigg|\_{r=a}=\frac{Z\_{eff}\lambda_B}{a^2}\]

得

\begin{equation}

Z_{eff}=\frac{\gamma_0}{\lambda_B \kappa_{PB}}{(\kappa_{PB}^2aR-1)\sinh[\kappa_{PB}(R-a)]+\kappa_{PB}(R-a)\cosh[\kappa_{PB}(R-a)]}

\label{eq:Zeff}

\end{equation}

计算\(Z_{eff}\)步骤：

1. 解方程(\ref{eq:PB})，得\(\phi_R\)
2. 计算\(\kappa_{PB}^2=\kappa_{res}^2\cosh\phi_R\)
3. 带入方程(\ref{eq:Zeff})，计算\(Z_{eff}\)

Renormalized jellium model

参考：

• PHYSICAL REVIEW E 69, 031403 (2004)
• THE JOURNAL OF CHEMICAL PHYSICS 126, 014702 (2007)
• THE JOURNAL OF CHEMICAL PHYSICS 133, 234105 (2010)

假设胶体离子均匀分布，作为小离子分布的背景。Poisson-Boltzmann方程：

\begin{equation}

\nabla^2\phi(r)=4\pi\lambda_BZ_{back}\rho+\kappa_{res}^2\sinh\phi(r)

\label{eq:RJPB} \end{equation}

其中，\(\rho\)为胶体平均密度，

\[\kappa_{res}^2=8\pi\lambda_B \sinh\phi(\infty) \]

边界条件：

\[\vec{n}\cdot \nabla \phi(r)|\_{r=a}=\frac{Z\lambda_B}{a^2}\]

\[\vec{n}\cdot \nabla \phi(r)|_{r=\infty}=0\]

并有

\[ 4\pi\lambda_B Z_{back}\rho+\kappa_{res}^2\sinh\phi(\infty)=0 \]

等效电荷\(Z_{eff}=Z_{back}\)

需要用迭代法求出\(Z_{eff}\)，见THE JOURNAL OF CHEMICAL PHYSICS 126, 014702 (2007)。