随机行走

将分子链的构型看成是随机行走（Random walk，也称醉汉行走，Drunkard's walk）产生的结果：Macromolecules are regarded as rigid segments connected by hinges. 要求：分子链长度大于弛豫长度（\(l \gg \xi_p\)）

考虑一维随机行走，每(steps)长为 \(a\)，向左向右的概率都是 50%（not biased），即 \(P(k_n = 1) = P(k_n = -1) = 1/2\). 由此得到递推公式

\[ x_n = x_{n - 1} + k_n a \]

可以证明，\(\braket{x_n} = 0\)，

\[ \begin{aligned} \braket{k_n} &= 1 \times \frac{1}{2} + (-1) \times \frac{1}{2} = 0 \\ \implies \braket{x_n} &= \braket{x_{n - 1} + k_n a} = \braket{x_{n - 1}} + a \braket{k_n} = \braket{x_{n - 1}} = \cdots = \braket{x_0} = 0. \end{aligned} \]

平方统计平均 \(\braket{x_n^2} = N a^2\)，

\[ \begin{aligned} \braket{x_n^2} &= \braket{(x_{n - 1} + k_n a)^2} = \braket{x_{n - 1}^2} + 2 a \braket{x_{n - 1} k_n} + a^2 \braket{k_n^2} \\ \text{given that } k_n^2 &= 1, \quad \braket{x_{n - 1} k_n} = (+1) \times \frac{1}{2} \braket{x_{n - 1}} + (-1) \times \frac{1}{2} \braket{x_{n - 1}} = 0, \\ \implies \braket{x_n^2} &= \braket{x_{n - 1}^2} + a^2 = \cdots = N a^2. \end{aligned} \]

Persistence Length

The length scale over which the tangent-tangent correlation function decays along the chain

\[ \braket{\mathbf{t}(s) \cdot \mathbf{t}(u)} = \mathrm{e}^{-|s - u| / \xi_p} \]

相关函数

\[ \begin{aligned} \mathbf{R} = \int_0^L \mathbf{t}(s) \, \mathrm{d} s \implies \braket{\mathbf{R}^2} &= \left\langle \int_0^L \mathrm{d} s \, \mathbf{t}(s) \cdot \int_0^L \mathrm{d} u \, \mathbf{t}(u) \right\rangle = \int_0^L \mathrm{d} s \int_0^L \mathrm{d} u \, \mathrm{e}^{-|s - u| / \xi_p} \\ &= 2 \int_0^L \mathrm{d} s \int_0^s \mathrm{d} u \, \mathrm{e}^{-(s - u) / \xi_p} = 2 \int_0^L \mathrm{d} s \, \xi_p (1 - \mathrm{e}^{-s / \xi_p}) \\ &= 2 \xi_p L - 2 \xi_p^2 (1 - \mathrm{e}^{-L / \xi_p}) \\ \end{aligned} \]

由于 \(L \gg \xi_p\)，所以

\[ \braket{R^2} = 2 \xi_p L \]