胶体电荷重整化

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：
egin{equation}
abla^{2phi(r)=kappa_{res}}2sinhphi(r)
label{eq:PB} end{equation}
边界条件：

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

[vec{n}cdot ablaphi(r)|_{r=R}=0 ]

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

方程( ef{eq:PB})左边

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

方程( ef{eq:PB})右边
egin{equation}
egin{split}
kappa_{res}^{2sinhphi(r)&=kappa_{res}}2[sinhphi_R+coshphi_R(phi(r)-phi_R)]
&=kappa_{res}^2coshphi_R[ anhphi_R+widetilde{phi}(r)]
&=kappa_{res}^2coshphi_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：
egin{equation}
abla^{2widetilde{phi}(r)=frac{1}{r}2}frac{d}{dr}left [r^2frac{d}{dr}widetilde{phi}(r) ight ]=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 ablawidetilde{phi}(r)|\_{r=R}=0 ]

方程 ef{eq:LPB}的解为

[widetilde{phi}(r)=gamma_0left [-1+frac{kappa_{PB}+1}{2kappa\_{PB}}e^{-kappa_{PB}R}frac{e^{kappa_{PB}r}}{r}+frac{kappa_{PB}-1}{2kappa\_{PB}}e^{kappa_{PB}R}frac{e^{-kappa_{PB}r}}{r} ight ] ]

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

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

得
egin{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( ef{eq:PB})，得(phi_R)
2. 计算(kappa_{PB}^2=kappa_{res}^2coshphi_R)
3. 带入方程( ef{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：
egin{equation}
abla^{2phi(r)=4pilambda_BZ_{back} ho+kappa_{res}}2sinhphi(r)
label{eq:RJPB} end{equation}
其中，( ho)为胶体平均密度，

[kappa_{res}^2=8pilambda_B sinhphi(infty) ]

边界条件：

[vec{n}cdot abla phi(r)|\_{r=a}=frac{Zlambda_B}{a^2} ]

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

并有

[4pilambda_B Z_{back} ho+kappa_{res}^2sinhphi(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)。