受限壁附近高分子链的构象统计

计算无限大平面附近的柔性链和硬棒——柔线双嵌段高分子链的构象统计。

柔性链


无限大平面附件的柔性高分子链

如上图所示,高分子链一端位于距离平面$l$处,另一端坐标为$z$,设高分子链链节大小$b$,高分子链链长为$N_c$,则高分子链构象配分函数为

\begin{equation} \begin{split} G(l,z)=&\sqrt{\frac{3}{2\pi N_cb^2}}\left (\exp\left [-\frac{3(z-l)^2}{2N_cb^2} \right ] - \exp\left [-\frac{3(z+l)^2}{2N_cb^2} \right ] \right )\\\\ =&\frac{1}{2\sqrt{\pi }R_g}\left (\exp\left [-\frac{(z-l)^2}{4R_g^2} \right ] - \exp\left [-\frac{(z+l)^2}{4R_g^2} \right ] \right ) \end{split} \label{Gcoil} \end{equation}

其中,$R_g=\sqrt{\frac{N_c}{6}}b$,为柔性链的回转半径。

配分函数为

\begin{equation} \begin{split} \mathcal Z=&\int_0^\infty G(l,z)\mathrm dz\\ =&\frac{1}{2\sqrt{\pi }R_g}\int_0^\infty \exp\left [-\frac{(z-l)^2}{4R_g^2} \right ] \mathrm dz\\ &-\frac{1}{2\sqrt{\pi }R_g}\int_0^\infty \exp\left [-\frac{(z+l)^2}{4R_g^2} \right ] \mathrm dz \\ =&\frac{1}{\sqrt{\pi }}\int_{-l/2R_g}^\infty \exp\left (-t^2\right ) \mathrm dt \quad \left (\mathrm{where} \quad t=\frac{z-l}{2R_g}\right )\\ &-\frac{1}{\sqrt{\pi }}\int_{l/2R_g}^\infty \exp\left (-t^2\right ) \mathrm dt \quad \left (\mathrm{where} \quad t=\frac{z+l}{2R_g}\right )\\ =&\frac{1}{\sqrt{\pi }}\int_{-l/2R_g}^0 \exp\left (-t^2\right ) \mathrm dt + \frac{1}{\sqrt{\pi }}\int_0^\infty \exp\left (-t^2\right ) \mathrm dt \\ &- \left ( \frac{1}{\sqrt{\pi }}\int_0^\infty \exp\left (-t^2\right ) \mathrm dt -\frac{1}{\sqrt{\pi }}\int_0^{l/2R_g} \exp\left (-t^2\right ) \mathrm dt \right )\\ =&\frac{1}{\sqrt{\pi }}\int_{-l/2R_g}^0 \exp\left (-t^2\right ) \mathrm dt + \frac{1}{\sqrt{\pi }}\int_0^{l/2R_g} \exp\left (-t^2\right ) \mathrm dt\\ =&\frac{2}{\sqrt{\pi }}\int_0^{l/2R_g} \exp\left (-t^2\right ) \mathrm dt\\ =&\mathrm {erf}\left (\frac{l}{2R_g}\right ) \end{split} \label{Zcoil} \end{equation}

上式用到了误差函数:

\begin{equation} \mathrm {erf}(x) = \frac{2}{\sqrt {\pi}} \int_0^x \exp(-t^2) \mathrm dt = \frac{2}{\sqrt {\pi}} \int_{-x}^0 \exp(-t^2) \mathrm dt \label{error} \end{equation}

自由能为

\begin{equation} \frac{F_c(l,N_c)}{k_BT} = -\ln \mathcal{Z} = -\ln \mathrm {erf}\left (\frac{l}{2R_g}\right )= -\ln \mathrm {erf}\left (\sqrt{\frac{3}{2N_c}}\frac{l}{b}\right ) \label{Fcoil} \end{equation}

rod-coil高分子链


无限大平面附近的硬棒——柔线(rod-coil)rod-coil双嵌段高分子链

如上图所示,柔线一端位于距离受限壁$l$处,硬棒可以自由旋转,并忽略硬棒与柔线的相互作用。设硬棒和柔线的链长分别为$N_d$和$N_c$,两嵌段库恩长度相等,均为$b$。构象分布为:

\begin{equation} \begin{split} G(l,z,\theta)=&\sqrt{\frac{3}{2\pi N_cb^2}}\left (\exp\left [-\frac{3(z-l)^2}{2N_cb^2} \right ] - \exp\left [-\frac{3(z+l)^2}{2N_cb^2} \right ] \right )\\ &\frac{1}{2}\exp\left [ -U(z,\theta) \right ]\\ =&\frac{1}{2}\exp\left [ -U(z,\theta) \right ]G(l,z) \end{split} \label{Grodcoil} \end{equation}

其中$\theta$为硬棒与$z$轴夹角,硬棒与受限壁相互作用为

\begin{equation} U(z,\theta)= \begin{cases} 0,&\theta \lt \pi-\arccos{\frac{z}{N_db}}, or \quad z \ge N_d b \\\\ \infty, & \theta \ge \pi-\arccos{\frac{z}{N_db}} ,and \quad z \le N_d b \end{cases} \label{U} \end{equation}

配分函数为:

\begin{equation} \begin{split} \mathcal Z&=\int_0^\infty \mathrm dz G(l,z,\theta)\int_0^\pi \mathrm d\theta \frac{1}{2}\exp\left [ -U(z,\theta) \right ]\sin \theta \\\\ &=\int_0^{N_db} \mathrm dz G(l,z,\theta)\int_0^\pi \mathrm d\theta \frac{1}{2}\exp\left [ -U(z,\theta) \right ]\sin \theta + \int_{N_db}^\infty \mathrm dz G(l,z)\\\\ &=\int_0^{N_db} \mathrm dz G(l,z) \frac{N_db+z}{2N_db} + \int_{N_db}^\infty \mathrm dz G(l,z)\\\\ &=\int_0^\infty \mathrm dz G(l,z)-\frac{1}{2}\int_0^{N_db} \mathrm dz G(l,z) +\frac{1}{2N_db}\int_0^{N_db} z G(l,z) \mathrm dz \\\\ &=\mathrm {erf}\left (\sqrt{\frac{3}{2N_cb^2}}l\right )-\frac{1}{2}I_1+\frac{1}{2N_db}I_2 \end{split} \label{Zrodcoil} \end{equation}

下面计算$I_1$:

\begin{equation} \begin{split} \int_0^{N_db} \mathrm dz G(l,z)= &\frac{1}{2\sqrt{\pi }R_g}\int_0^{N_db} \mathrm dz \exp\left [-\frac{(z-l)^2}{4R_g^2} \right ]\\ &-\frac{1}{2\sqrt{\pi }R_g}\int_0^{N_db} \mathrm dz \exp\left [-\frac{(z+l)^2}{4R_g^2} \right ] \\ =& \frac{1}{\sqrt{\pi}}\int_{-l/2R_g}^{(N_db-l)/2R_g} \exp(-t^2)\mathrm dt \\ &-\frac{1}{\sqrt{\pi }}\int_{l/2R_g}^{(N_db+l)/2R_g} \exp(-t^2)\mathrm dt\\ =& \frac{1}{\sqrt{\pi}}\int_{-l/2R_g}^0 e^{-t^2}\mathrm dt +\frac{1}{\sqrt{\pi}}\int_0^{(N_db-l)/2R_g} e^{-t^2}\mathrm dt \\ &-\frac{1}{\sqrt{\pi }}\int_0^{(N_db+l)/2R_g} e^{-t^2}\mathrm dt + \frac{1}{\sqrt{\pi }}\int_0^{l/2R_g} e^{-t^2}\mathrm dt\\ =&\mathrm {erf}\left (\frac{l}{2R_g}\right )+\frac{1}{2}\mathrm {erf}\left (\frac{N_db-l}{2R_g}\right )-\frac{1}{2}\mathrm {erf}\left (\frac{N_db+l}{2R_g}\right )\\ =&\mathrm {erf}\left (\sqrt{\frac{3}{2N_c}}\frac{l}{b}\right )+\frac{1}{2}\mathrm {erf}\left (\sqrt{\frac{3}{2N_c}}\frac{N_db-l}{b}\right )-\frac{1}{2}\mathrm {erf}\left (\sqrt{\frac{3}{2N_c}}\frac{N_db+l}{b}\right ) \end{split} \label{I1} \end{equation}

下面计算$I_2$:

\begin{equation} \begin{split} \int_0^{N_db} zG(l,z) \mathrm dz = &\frac{1}{2\sqrt{\pi }R_g}\int_0^{N_db} z\exp\left [-\frac{(z-l)^2}{4R_g^2} \right ] \mathrm dz\\ &-\frac{1}{2\sqrt{\pi }R_g}\int_0^{N_db} z\exp\left [-\frac{(z+l)^2}{4R_g^2} \right ] \mathrm dz\\ =& \frac{1}{\sqrt{\pi }}\int_{-l/2R_g}^{(N_db-l)/2R_g} (2R_gt+l)\exp(-t^2)\mathrm dt \\ &-\frac{1}{\sqrt{\pi }}\int_{l/2R_g}^{(N_db+l)/2R_g} (2R_gt-l)\exp(-t^2)\mathrm dt\\ =&\frac{R_g}{\sqrt{\pi }}\left \{\exp\left [-\left ( \frac{l}{2R_g} \right )^2 \right ]-\exp\left [-\left ( \frac{N_db-l}{2R_g} \right )^2 \right ] \right \}\\ &+\frac{l}{\sqrt{\pi }} \int_{-l/2R_g}^{(N_db-l)/2R_g} \exp(-t^2)\mathrm dt \\ &-\frac{R_g}{\sqrt{\pi }}\left \{\exp\left [-\left ( \frac{l}{2R_g} \right )^2 \right ]-\exp\left [-\left ( \frac{N_db+l}{2R_g} \right )^2 \right ] \right \}\\ &+\frac{l}{\sqrt{\pi }} \int_{l/2R_g}^{(N_db+l)/2R_g} \exp(-t^2)\mathrm dt\\ =& \frac{R_g}{\sqrt{\pi }}\left \{\exp\left [-\left ( \frac{N_db+l}{2R_g} \right )^2 \right ] -\exp\left [-\left ( \frac{N_db-l}{2R_g} \right )^2 \right ]\right \}\\ &+\frac{l}{\sqrt{\pi }} \int_{-l/2R_g}^{(N_db-l)/2R_g} \exp(-t^2)\mathrm dt\\ &+ \frac{l}{\sqrt{\pi }} \int_{l/2R_g}^{(N_db+l)/2R_g} \exp(-t^2)\mathrm dt\\ =& \frac{R_g}{\sqrt{\pi }}\left \{\exp\left [-\left ( \frac{N_db+l}{2R_g} \right )^2 \right ] -\exp\left [-\left ( \frac{N_db-l}{2R_g} \right )^2 \right ]\right \} \\ &+\frac{l}{2}\mathrm {erf}\left (\frac{N_db-l}{2R_g}\right )+\frac{l}{2}\mathrm {erf}\left (\frac{N_db+l}{2R_g}\right )\\ =& \sqrt{\frac{N_c}{6\pi}}b\left \{\exp\left [-\frac{3}{N_c}\left ( \frac{N_db+l}{b} \right )^2 \right ] -\exp\left [-\frac{3}{N_c}\left ( \frac{N_db-l}{b} \right )^2 \right ]\right \} \\ &+\frac{l}{2}\mathrm {erf}\left (\sqrt{\frac{3}{2N_c}}\frac{N_db-l}{b}\right )+\frac{l}{2}\mathrm {erf}\left (\sqrt{\frac{3}{2N_c}}\frac{N_db+l}{b}\right ) \end{split} \label{I2} \end{equation}

自由能为:

\begin{equation} \frac{F_{d-c}(l,N_c,N_d)}{k_BT} = -\ln \left [ \mathrm {erf}\left (\sqrt{\frac{3}{2N_cb^2}}l\right )-\frac{1}{2}I_1+\frac{1}{2N_db}I_2 \right ] \label{Frodcoil} \end{equation}

标签: rod-coil, 构象统计, 自由能, 配分函数

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