星状高分子刷的强拉伸理论

为文献 Macromolecules 2012, 45, 7260-7273 中星状高分子刷的强拉伸理论补充一些细节。

对Zhulina的强拉伸理论不熟悉者可参阅博客文章:Zhulina 的高分子刷理论

接枝星状高分子共有$f$ 条臂,其中一臂接枝在平面上,各臂链长均为$n$,总链长为$N=fn$。接枝臂局域拉伸量为$E_0(z_0,z)$,其中$z_0$为星状高分子中心所在位置。自由臂局域拉伸量为$E_1(z_1,z)$,其中$z_1$为自由臂末端所在位置。链拉伸量满足如下归一化条件:

\begin{equation}
\int_0^{z_0} \frac{1}{E_0(z_0,z)}\mathrm dz =n
\label{E0norm}
\end{equation}

\begin{equation}
\int_{z_0}^{z_1} \frac{1}{E_1(z_1,z)}\mathrm dz =n
\label{E1norm}
\end{equation}

由\eqref{E1norm}知,$z_0$是$z_1$的函数,$z_0=z_0(z_1)$,$z_1=z_1(z_0)$。在星状高分子中心,由力的平衡,有:

\begin{equation}
E_0(z_0,z_0)=(f-1)E_1(z_1,z_0)
\label{mechequi}
\end{equation}

设自由臂末端分布为$g(z_1)$,满足归一化条件

\begin{equation}
\int_0^H g(z_1)\mathrm dz_1 =1
\label{gnorm}
\end{equation}

高分子密度分布为

\begin{equation}
\varphi(z)=\frac{1}{s}\int_{z_0(z_1)}^H \frac{g(z_1)\mathrm dz_1}{E_0(z_0,z)}+\frac{f-1}{s}\int_{z}^H \frac{g(z_1)\mathrm dz_1}{E_1(z_1,z)}
\label{dens}
\end{equation}

其中$s$为单个星状高分子所占据面积。密度分布满足如下关系

\begin{equation}
\int_0^H\varphi(z)\mathrm dz=fn=N
\label{densnorm}
\end{equation}

自由能

\begin{equation} \begin{split} F=& \frac{3}{2a^2}\int_{z_0(z_1)}^H g(z_1)\mathrm dz_1 \int_0^{z_0}E_0(z_0,z)\mathrm dz + \\ & \frac{3}{2a^2}\int_{z}^H g(z_1)\mathrm dz_1\int_{z_0}^{z_1} (f-1)E_1(z_1,z)\mathrm dz + \\ \\ & s\int_0^H f[\varphi(z)]\mathrm dz \end{split} \label{Fstar} \end{equation}

其中$f[\varphi(z)]$描述体系中相互作用。

考虑到约束条件\eqref{E0norm}、\eqref{E1norm}、\eqref{densnorm},我们需要对下式变分:

\begin{equation} \begin{split} F'=& F+\lambda\int_0^H \varphi(z)\mathrm dz + \\ &\int_{z_0(z_1)}^H \lambda_0(z_1)\mathrm dz_1\int_0^{z_0(z_1)} \frac{1}{E_0(z_0(z_1),z)}\mathrm dz \\ & + \int_{z_0(z_1)}^H \lambda_1(z_1)\mathrm dz_1\int_{z_0}^{z_1} \frac{1}{E_1(z_1,z)}\mathrm dz \end{split} \label{Fconstraint} \end{equation}

变分,有

\begin{equation} \begin{split} \delta F' = & \frac{3}{2a^2}\int_{z_0(z_1)}^H \delta g(z_1)\mathrm dz_1 \int_0^{z_0}E_0(z_0,z)\mathrm dz + \\ & \frac{3}{2a^2}\int_{z_0(z_1)}^H g(z_1)\mathrm dz_1 \int_0^{z_0}\delta E_0(z_0,z)\mathrm dz + \\ & \frac{3}{2a^2}\int_{z}^H \delta g(z_1)\mathrm dz_1\int_{z_0}^{z_1} (f-1)E_1(z_1,z)\mathrm dz + \\ & \frac{3}{2a^2}\int_{z}^H g(z_1)\mathrm dz_1\int_{z_0}^{z_1} (f-1)\delta E_1(z_1,z)\mathrm dz + \\ & s\int_0^H \frac{\delta f[\varphi(z)]}{\delta \varphi(z)}\delta \varphi(z)\mathrm dz + \lambda\int_0^H \delta \varphi(z)\mathrm dz \\ &- \int_{z_0(z_1)}^H \lambda_0(z_1)\mathrm dz_1\int_0^{z_0(z_1)} \frac{\delta E_0(z_0(z_1),z)}{E_0^2(z_0(z_1),z)}\mathrm dz \\ & - \int_{z_0(z_1)}^H \lambda_1(z_1)\mathrm dz_1\int_{z_0}^{z_1} \frac{\delta E_1(z_1,z)}{E_1^2(z_1,z)}\mathrm dz \end{split} \label{F'var} \end{equation}

其中密度分布的变分为:

\begin{equation} \begin{split} \delta \varphi(z) =& \frac{1}{s}\int_{z_0(z_1)}^H \frac{\delta g(z_1)\mathrm dz_1}{E_0(z_0,z)}+\frac{f-1}{s}\int_{z}^H \frac{\delta g(z_1)\mathrm dz_1}{E_1(z_1,z)} \\ & -\frac{1}{s}\int_{z_0(z_1)}^H \frac{g(z_1)\delta E_0(z_0,z)\mathrm dz_1}{E_0^2(z_0,z)} \\ &-\frac{f-1}{s}\int_{z}^H \frac{g(z_1)\delta E_1(z_1,z)\mathrm dz_1}{E_1^2(z_1,z)} \end{split} \label{denvar} \end{equation}

整理以上两式,有

\begin{equation} \begin{split} \frac{\delta F'}{\delta E_1}=&\frac{3}{2a^2}g(z_1)-\frac{\lambda_1(z_1)}{E_1^2(z_1,z)}-\\ &\frac{g(z_1)}{E_1^2(z_1,z)}\left (\lambda + \frac{\delta f[\varphi(z)]}{\delta \varphi(z)} \right )\\ =&0 \end{split} \label{varE1} \end{equation}

由\eqref{varE1}知,$E_1(z_1,z)=\frac{2a^2}{3}\frac{\lambda_1(z_1)}{g(z_1)}-\frac{2a^2}{3}\left (-\lambda-\frac{\delta f[\varphi(z)]}{\delta \varphi(z)} \right ) =\sqrt{U_1(z_1)-U_2(z)}$,$E_1(z_1,z_1)=0$,因此$E_1(z_1,z)=\sqrt{U(z_1)-U(z)}$,再考虑到\eqref{E1norm},得

\begin{equation}
E_1(z_1,z)=k\sqrt{z_1^2-z^2}
\label{E1expr}
\end{equation}

\begin{equation}
z_0=z_1\cos kn
\label{z0z1}
\end{equation}

\begin{equation}
\frac{\delta f[\varphi(z)]}{\delta \varphi(z)}=-\lambda-\frac{3}{2a^2}k^2z^2
\label{kfield}
\end{equation}

同理,由$\frac{\delta F'}{\delta E_0}=0$,得$E_0(z_0,z)=k\sqrt{V(z_1)-z^2}=k\sqrt{V(z_0/\cos kn)-z^2}$。由\eqref{mechequi},得

\begin{equation}
V=(f-1)z_1^2=(f-1)\frac{z_0^2}{\cos^2kn}
\label{V}
\end{equation}

所以

\begin{equation}
\begin{split}
E_0(z_0,z)=&k\sqrt{(f-1)z_1^2-z^2}\
=&k\sqrt{(f-1)\frac{z_0^2}{\cos^2kn}-z^2}
\end{split}
\label{E0expr}
\end{equation}

将此式代入\eqref{E0norm},得

\begin{equation}
k=\frac{1}{n}\arccos\sqrt{\frac{f-1}{f}}
\label{k}
\end{equation}

星状高分子刷的密度分布由\eqref{kfield}、\eqref{k}得到,其中常数$\lambda$由$\varphi(H)=0$得到,$H$由\eqref{densnorm}得到。末端分布解以下积分方程得到:

\begin{equation} \begin{split} \varphi(z)=&\frac{1}{s}\int_{0}^H \frac{\theta(z_1\cos kn-z)g(z_1)\mathrm dz_1}{E_0(z_1,z)}+\\ &\frac{f-1}{s}\int_{z}^H \frac{g(z_1)\mathrm dz_1}{E_1(z_1,z)} \end{split} \label{endinteg} \end{equation}

标签: 高分子刷, 强拉伸理论, sst, 星状高分子

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