首先考虑最简单的 1D 模型. 设步长为 $a$, 总步数为 $N = n_{l} + n_{r}$, 则最后位置为 $R = (n_{r}-n_{l})a$ (假定向右为正方向). 向左和向右的概率均为 $1/2$, 因此处于位置 $R$ 的概率为

$$ \begin{aligned} p(R,N) &= \frac{N!}{n_{l}!n_{r}!}\frac{1}{2^{N}} = \frac{N!}{(\frac{Na+R}{2})!(\frac{Na-R}{2})!}\frac{1}{2^{N}}\\ \ln{p} &\approx \ln{\sqrt{\frac{2}{\pi N}}} - \frac{R^{2}}{2Na^{2}}\\ p(R,N) &\approx \sqrt{\frac{2}{\pi N}}\exp{\left(-\frac{R^{2}}{2Na^{2}}\right)} \end{aligned} $$

计算 $\int_{-Na}^{Na}p\mathrm{d}R = 2a$, 因此概率密度需要再归一化:

$$ p(R,N) = \frac{1}{\sqrt{2\pi Na^{2}}}\exp{\left(-\frac{R^{2}}{2Na^{2}}\right)} = \frac{1}{\sqrt{2\pi\sigma_{R}^{2}}}\exp{\left(-\frac{R^{2}}{2\sigma_{R}^{2}}\right)} $$

这是一个高斯分布, 均值为 $\bar{R} = 0$, 方差为 $\sigma_{R}^{2} = Na^{2}$.

若将 $N$ 视为时间变量, 即 $t = N\tau$ (其中 $\tau$ 是每一 step 的耗时), 则

$$ \begin{aligned} p(R,t) &= \frac{1}{\sqrt{2\pi ta^{2}/\tau}}\exp{\left(-\frac{R^{2}}{2ta^{2}/\tau}\right)} = \frac{1}{\sqrt{4\pi\frac{a^{2}}{2\tau}t}}\exp{\left(-\frac{R^{2}}{4\frac{a^{2}}{2\tau}t}\right)}\\ &= \frac{1}{\sqrt{4\pi Dt}}\exp{\left(-\frac{R^{2}}{4Dt}\right)}, \quad D = \frac{a^{2}}{2\tau} \end{aligned} $$

连续性方程: $\frac{\partial n(\vec{x},t)}{\partial t} = -\nabla\cdot\vec{j}(\vec{x},t)$; Fick's law: $\vec{j}(\vec{x},t) = -D\nabla n(\vec{x}, t)$

扩散方程: $\frac{\partial n(\vec{x},t)}{\partial t} = D\nabla^{2}n(\vec{x},t)$. $d$ 维解 $n(\vec{x},t) = \frac{N}{(\sqrt{4\pi Dt})^{d}}\exp{\left(-\frac{|\vec{x}|^{2}}{4Dt}\right)}$

$\langle x\rangle = 0$, $\langle x^{2}\rangle = \frac{1}{N}\int_{-\infty}^{+\infty}x^{2}n(\vec{x},t)\mathrm{d}^{d}\vec{x} = 2dDt$.

方差 $\sigma_{x}^{2} = \langle x^{2}\rangle - \langle x\rangle^{2} = 2dDt\sim t$.

一维 Ito SDE(stochastic differential equation, 随机微分方程)为

$$ \mathrm{d}x = A(x)\mathrm{d}t + \sqrt{2D}\mathrm{d}W_{t} $$

$\langle\mathrm{d}W_{t}\rangle = 0$, $\langle(\mathrm{d}W_{t})^{2}\rangle = \mathrm{d}t$.

设概率密度为 $f(x,t)$, 且引入辅助函数 $\phi(x)$, 其期望为

$$ \langle \phi(x)\rangle = \int \phi(x)f(x,t)\mathrm{d}x $$

微元展开:

$$ \mathrm{d}\phi = \phi^{\prime}(x)\mathrm{d}x + \frac{\phi^{\prime\prime}(x)}{2}(\mathrm{d}x)^{2} $$

代入 $\mathrm{d}x = A(x)\mathrm{d}t + \sqrt{2D}\mathrm{d}W$

需要注意到性质 $$ W_{t+\Delta t} - W_{t}\sim \mathcal{N}(0,\Delta t)\Rightarrow \Delta W\sim\sqrt{\Delta t} $$

$$ (\mathrm{d}x)^{2} = (A(x)\mathrm{d}t + \sqrt{2D}\mathrm{d}W)^{2} = 2D\mathrm{d}t $$

于是

$$ \begin{aligned} \mathrm{d}\phi &= \phi^{\prime}(x)A\mathrm{d}t + \phi^{\prime}\sqrt{2D}\mathrm{d}W + D\phi^{\prime\prime}\mathrm{d}t\\ &= \left[A\phi^{\prime} + D\phi^{\prime\prime}\right]\mathrm{d}t + \sqrt{2D}\phi^{\prime}\mathrm{d}W \end{aligned} $$

对两侧取期望, 且注意到 $\langle \mathrm{d}W\rangle = 0$, 因此

$$ \mathrm{d}\langle\phi\rangle = \langle A\phi^{\prime} + D\phi^{\prime\prime}\rangle \mathrm{d}t \Rightarrow \frac{\mathrm{d}\langle\phi\rangle}{\mathrm{d}t} = \langle A\phi^{\prime} + D\phi^{\prime\prime}\rangle $$

将 $\langle \cdot\rangle$ 展开为积分形式 $\int \cdot f\mathrm{d}x$:

$$ \frac{\mathrm{d}}{\mathrm{d}t} \int\phi f\mathrm{d}x = \int (A\phi^{\prime} + D\phi^{\prime\prime})f\mathrm{d}x $$

使用分部积分:

$$ \begin{aligned} \int A\phi^{\prime}f\mathrm{d}x &= \int Af \mathrm{d}\phi = Af\phi - \int \phi \frac{\partial(Af)}{\partial x}\mathrm{d}x\\ \int D\phi^{\prime\prime}f\mathrm{d}x &= \int Df\mathrm{d}\phi^{\prime} = Df\phi^{\prime} - \int \phi^{\prime}\frac{\partial Df}{\partial x}\mathrm{d}x\\ &= Df\phi^{\prime} - \int \frac{\partial Df}{\partial x}\mathrm{d}\phi = Df\phi^{\prime} - \frac{\partial D f}{\partial x}\phi + \int \phi \frac{\partial^{2}Df}{\partial x^{2}}\mathrm{d}x \end{aligned} $$

假定边界条件为 $0$, 因此化为形式

$$ \int A\phi^{\prime}f\mathrm{d}x = -\int \phi \frac{\partial }{\partial x}(Af)\mathrm{d}x\\ \int D\phi^{\prime\prime}f\mathrm{d}x = \int \phi D\frac{\partial^{2}f}{\partial x^{2}}\mathrm{d}x $$

代回原等式中:

$$ \frac{\mathrm{d}}{\mathrm{d}t}\int\phi f\mathrm{d}x = \int \phi \left[ -\frac{\partial }{\partial x}(Af) + D\frac{\partial^{2}f}{\partial x^{2}} \right]\mathrm{d}x $$

因为对任意 $\phi$ 都成立, 因此

$$ \frac{\mathrm{d}}{\mathrm{d}t}\int\phi f\mathrm{d}x = \int \phi \frac{\partial f}{\partial t}\mathrm{d}x $$

得到 Fokker-Planck 方程

$$ \boxed{\frac{\partial f}{\partial t} = -\frac{\partial }{\partial x}(Af) + D\frac{\partial^{2}f}{\partial x^{2}}} $$

已得 1D random walk 形式为

$$ p(R_{i},N_{i}) = \frac{1}{\sqrt{2\pi N_{i}a^{2}}}\exp{\left(-\frac{R_{i}^{2}}{2N_{i}a^{2}}\right)} $$

研究 Ideal chain 的熵弹性. 熵变被定义为 Boltzmann 熵差

$$ \Delta S = k_{B}\ln{\Omega(N,x)} - k_{B}\ln{\Omega(N,0)} = k_{B}\ln{\frac{p(N,x)}{p(N,0)}} = -\frac{1}{2}\frac{k_{B}}{Na^{2}}x^{2} $$

自由能变

$$ \Delta G = \Delta U - T\Delta S = \frac{1}{2}\frac{k_{B}T}{Na^{2}}x^{2} $$

因此熵弹性力为

$$ f = \frac{\partial G(N,x)}{\partial x} = \frac{k_{B}T}{Na^{2}}x $$

热力学快速回顾:

焓 $H \equiv U + pV$, Helmholtz 自由能 $F \equiv U - TS$, Gibbs 自由能 $G\equiv H-TS = U - TS + pV$

设两种聚合物混合, 分子链数分别为 $n_{1}$ 和 $n_{2}$, 聚合度分别为 $N_{1}$ 和 $N_{2}$. 混合物总格点数 $M = n_{1}N_{1} + n_{2}N_{2}$.

接下来求解两聚合物在分离和混合下的状态数.

${\color{red}{错误的推导过程}}$

$$ \begin{aligned} \Omega_{\text{sep}} &= \Omega_{1}\Omega_{2} = \frac{(n_{1}N_{1})!}{(N_{1}!)^{n_{1}}n_{1}!}\cdot \frac{(n_{2}N_{2})!}{(N_{2}!)^{n_{2}}n_{2}!}\\ \Omega_{\text{mix}} &= \frac{M!}{(N_{1}!)^{n_{1}}(N_{2}!)^{n_{2}}n_{1}!n_{2}!} \end{aligned} $$

• I-聚合物占据总格点为 $n_{1}N_{1}$, 第一个单体有 $n_{1}N_{1}$ 个选择, 第二个单体有 $(n_{1}N_{1}-1)$ 个选择...因此在可区分条件下共有 $(n_{1}N_{1})!$ 种状态;
• 对于单独一条 I-链, 占据格点数为 $N_{1}$, 而在链内单体间是不可区分的, 因此需要除掉 $N_{1}!$; 又因为一共有 $n_{1}$ 条链, 因此由链内单体不可区分引起的因子为 $(N_{1}!)^{n_{1}}$;
• 由于链与链间不可区分, 因此还需除去其因子 $(n_{1}!)$.

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上面这些说得很有道理, 但是错在忽略了这是聚合物的混合, 同一分子的单体需要彼此相连, 因此单体的位置不能是任意的! 聚合物分子这个整体才是可以自由换位置的.

设该混合晶格的配位数为 $z$. 现在将聚合度为 $N$ 的链放入已经有 $x$ 个格点被占据的晶格中.

• 第 1 单体: 可以放在任意一个空位上, 因此选择数为 $M-x$;
• 第 2 单体: 只能放在第 1 单体的配位格点, 此时空间中空格点的平均概率为 $\begin{aligned}p = \frac{M-x}{M}\end{aligned}$. 因此选择数为 $\begin{aligned}zp = z\frac{M-x}{M}\end{aligned}$;
• 第 $3-N$ 单体: 由于自体积排斥, 后续可选的格点数为 $\begin{aligned}(z-1)p = (z-1)\frac{M-x}{M}\end{aligned}$;

因此, 放入该链的选择数 $\nu$ 为

$$ \nu = (M-x) \cdot \left[z\frac{M-x}{M}\right] \cdot \left[(z-1)\frac{M-x}{M}\right]^{N-2}\overset{z\approx z-1}{\simeq} \left[\frac{z-1}{M}\right]^{N-1}(M-x)^{N} $$

证明近似 $\begin{aligned}(M-x)^{N}\approx\frac{(M-x)!}{(M-x-N)!}\end{aligned}$. 展开等式右边:

$$ \begin{aligned} \frac{(M-x)!}{(M-x-N)!} = (M-x)(M-x-1)(M-x-2)\cdots [M-x-(N-1)]\\ = (M-x)^{N}\left[ 1\cdot \left(1 - \frac{1}{M-x}\right)\left(1 - \frac{2}{M-x}\right) \cdots\left(1 - \frac{N-1}{M-x}\right) \right] \end{aligned} $$

由于 $M\gg N$, 因此 $\begin{aligned}1 - \frac{k}{M-x}\to 1\end{aligned}$

若将 $M-x$ 视为整体 $M$, 该近似公式也可写作 $\begin{aligned}\frac{M!}{(M-N)!} \approx M^{N}\end{aligned}$

采用近似 $\begin{aligned}(M-x)^{N}\approx\frac{(M-x)!}{(M-x-N)!}\end{aligned}$, 第 $i$ 条放入聚合度为 $N$ 链的方法数为

$$ \nu_{i} = \left[\frac{z-1}{M}\right]^{N-1}\frac{(M-x_{i})!}{(M-x_{i}-N)!} $$

• 分离状态 ( I-链和 II-链各自占据自己纯物质的晶格)

对于 I-链, 总晶格数为 $M_{1} = n_{1}M_{1}$, 需往空晶格(初始 $x_{1}=0$) 放入 $n_{1}$ 条 I-链. 被占格点数 $x_{i} = (i-1)N $. 引入链的不可区分性因子 $\begin{aligned}\frac{1}{n_{1}!}\end{aligned}$, 得到总方法数为

$$ \begin{aligned} \Omega_{1} &= \frac{1}{n_{1}!}\prod_{i=1}^{n_{1}}\nu_{1,i} = \frac{1}{n_{1}!}\prod_{i=1}^{n_{1}}\left[\frac{z-1}{M_{1}}\right]^{N_{1}-1}\frac{[M-(i-1)N]!}{(M_{1}-iN_{1})!}\\ &= \frac{1}{n_{1}!}\left[\frac{z-1}{M_{1}}\right]^{n_{1}(N_{1}-1)}\prod_{i=1}^{n_{1}}\frac{[M-(i-1)N]!}{(M_{1}-iN_{1})!}\\ &= \frac{1}{n_{1}!}\left[\frac{z-1}{M_{1}}\right]^{n_{1}(N_{1}-1)} \frac{M_{1}!}{(M_{1}-N_{1})!} \frac{(M_{1}-N_{1})!}{(M_{1}-2N_{1})!}\cdots \frac{N_{1}!}{0!}\\ &= \frac{M_{1}!}{n_{1}!}\left[\frac{z-1}{M_{1}}\right]^{n_{1}(N_{1}-1)} \end{aligned} $$

II-链同理可得 $\begin{aligned}\Omega_{2} = \frac{M_{2}!}{n_{2}!}\left[\frac{z-1}{M_{2}}\right]^{n_{2}(N_{2}-1)}\end{aligned}$, 分离状态下的总状态数为

$$ \begin{aligned} \Omega_{\text{sep}} &= \Omega_{1}\Omega_{2} = \frac{M_{1}!}{n_{1}!}\left[\frac{z-1}{M_{1}}\right]^{n_{1}(N_{1}-1)} \frac{M_{2}!}{n_{2}!}\left[\frac{z-1}{M_{2}}\right]^{n_{2}(N_{2}-1)}\\ &= \frac{M_{1}!M_{2}!}{n_{1}!n_{2}!} \left[\frac{z-1}{M_{1}}\right]^{n_{1}(N_{1}-1)} \left[\frac{z-1}{M_{2}}\right]^{n_{2}(N_{2}-1)} \end{aligned} $$

• 混合状态. 现在晶格总数为 $M = n_{1}N_{1} + n_{2}N_{2}$. 前面已经推导出, $M$ 总格点数中已有 $x$ 格点被占, 放入 $N$ 聚合度链的方法数 $\nu$ 为

$$ \nu(x,N) = \left[\frac{z-1}{M}\right]^{N-1}\frac{(M-x)!}{(M-x-N)!} $$

1. 填入 $n_{1}$ 条 I-链. 对于第 i 条被填入的 I-链, 已被占格点数 $x_{1,i}=(i-1)N_{1}$. 那么填入该 (I,i) 链的方法数为

$$ \nu_{1,i} = \left[\frac{z-1}{M}\right]^{N_{1}-1}\frac{[M-(i-1)N_{1}]!}{(M-iN_{1})!} $$

引入 I-链不可区分性的因子 $\begin{aligned}\frac{1}{n_{1}!}\end{aligned}$, 填入这 $n_{1}$ 条 I-链的总方法数为

$$ \begin{aligned} W_{1} &= \frac{1}{n_{1}!}\prod_{i=1}^{n_{1}}\nu_{1,i} = \frac{1}{n_{1}!}\prod_{i=1}^{n_{1}} \left[\frac{z-1}{M}\right]^{N_{1}-1} \frac{[M-(i-1)N_{1}]!}{(M-iN_{1})!}\\ &= \frac{1}{n_{1}!} \left[\frac{z-1}{M}\right]^{n_{1}(N_{1}-1)} \prod_{i=1}^{n_{1}} \frac{[M-(i-1)N_{1}]!}{(M-iN_{1})!}\\ &= \frac{1}{n_{1}!} \left[\frac{z-1}{M}\right]^{n_{1}(N_{1}-1)} \frac{M!}{(M-N_{1})!} \frac{(M-N_{1})!}{(M-2N_{1})!}\cdots \frac{[M-(n_{1}-1)N_{1}]!}{(M-n_{1}N_{1})!}\\ &= \frac{1}{n_{1}!}\left[\frac{z-1}{M}\right]^{n_{1}(N_{1}-1)}\frac{M!}{(M-n_{1}N_{1})!} \end{aligned} $$

2. 继续填入 $n_{2}$ 条 II-链. 此时 I-链已经占据 $n_{1}N_{1}$ 个格点. 所以对于放入的第 $j$ 条 II-链, 已被占据的格点数为 $x_{2,j} = n_{1}N_{1} + (j-1)N_{2}$, 填入 (II,j) 链的方法数为

$$ \nu_{2,j} = \left[\frac{z-1}{M}\right]^{N_{2}-1}\frac{[M-n_{1}N_{1}-(j-1)N_{2}]!}{(M-n_{1}N_{1}-jN_{2})!} $$

引入 II-链的不可区分性因子 $\begin{aligned}\frac{1}{n_{2}!}\end{aligned}$, 那么填入这 $n_{2}$ 条 II-链的总方法数为

$$ \begin{aligned} W_{2} &= \frac{1}{n_{2}!}\prod_{j=1}^{n_{2}}\nu_{2,j} = \frac{1}{n_{2}!}\prod_{j=1}^{n_{2}} \left[\frac{z-1}{M}\right]^{N_{2}-1}\frac{[M-n_{1}N_{1}-(j-1)N_{2}]!}{(M-n_{1}N_{1}-jN_{2})!}\\ &= \frac{1}{n_{2}!} \left[\frac{z-1}{M}\right]^{n_{2}(N_{2}-1)}\prod_{j=1}^{n_{2}}\frac{[M-n_{1}N_{1}-(j-1)N_{2}]!}{(M-n_{1}N_{1}-jN_{2})!}\\ &= \frac{1}{n_{2}!} \left[\frac{z-1}{M}\right]^{n_{2}(N_{2}-1)} \\ &\times \frac{(M-n_{1}N_{1})!}{(M-n_{1}N_{1}-N_{2})!} \frac{(M-n_{1}N_{1}-N_{2})!}{(M-n_{1}N_{1}-2N_{2})!}\cdots \frac{[M-n_{1}N_{1} - (n_{2}-1)N_{2}]!}{(M-n_{1}N_{1}-n_{2}N_{2})!}\\ &= \frac{1}{n_{2}!} \left[\frac{z-1}{M}\right]^{n_{2}(N_{2}-1)} \frac{(M-n_{1}N_{1})!}{(M-n_{1}N_{1}-n_{2}N_{2})!} \\ &= \frac{1}{n_{2}!} \left[\frac{z-1}{M}\right]^{n_{2}(N_{2}-1)} \frac{(M-n_{1}N_{1})!}{0!} \end{aligned} $$

于是混合状态下的方法数为

$$ \begin{aligned} \Omega_{\text{mix}} &= W_{1}W_{2} \\ &= \frac{1}{n_{1}!}\left[\frac{z-1}{M}\right]^{n_{1}(N_{1}-1)}\frac{M!}{(M-n_{1}N_{1})!} \frac{1}{n_{2}!} \left[\frac{z-1}{M}\right]^{n_{2}(N_{2}-1)} \frac{(M-n_{1}N_{1})!}{0!}\\ &= \frac{M!}{n_{1}!n_{2}!}\left[\frac{z-1}{M}\right]^{n_{1}(N_{1}-1) + n_{2}(N_{2}-1)} \end{aligned} $$

混合熵被定义为 Boltzmann 熵变化量 $\begin{aligned}\Delta S = k_{B}\ln{\left(\frac{\Omega_{\text{mix}}}{\Omega_{\text{sep}}}\right)}\end{aligned}$, 而

$$ \frac{\Omega_{\text{mix}}}{\Omega_{\text{sep}}} = \frac{\frac{M!}{n_{1}!n_{2}!}\left[\frac{z-1}{M}\right]^{n_{1}(N_{1}-1) + n_{2}(N_{2}-1)}}{\frac{M_{1}!M_{2}!}{n_{1}!n_{2}!} \left[\frac{z-1}{M_{1}}\right]^{n_{1}(N_{1}-1)} \left[\frac{z-1}{M_{2}}\right]^{n_{2}(N_{2}-1)}} = \frac{M!}{M_{1}!M_{2}!}\frac{M_{1}^{n_{1}(N_{1}-1)}M_{2}^{n_{2}(N_{2}-1)}}{M^{n_{1}(N_{1}-1) + n_{2}(N_{2}-1)}} $$

两边同取对数, 且使用 Stirling 近似 $\ln{X!}\approx X\ln{X} - X$, 第一项

$$ \begin{aligned} \ln{\frac{M!}{M_{1}!M_{2}!}} &\approx M\ln{M} - M - M_{1}\ln{M_{1}} + M_{1} - M_{2}\ln{M_{2}} + M_{2} \\ &= M\ln{M} - M_{1}\ln{M_{1}} - M_{2}\ln{M_{2}}\\ &= (M_{1}+M_{2})\ln{M} - M_{1}\ln{M_{1}} - M_{2}\ln{M_{2}}\\ &= -M_{1}\ln{\left(\frac{M_{1}}{M}\right)} - M_{2}\ln{\left(\frac{M_{2}}{M}\right)} \end{aligned} $$

第二项

$$ \begin{aligned} \ln{\left(\frac{M_{1}^{n_{1}(N_{1}-1)}}{M^{n_{1}(N_{1}-1)}}\frac{M_{2}^{n_{2}(N_{2}-1)}}{M^{n_{2}(N_{2}-1)}}\right)} &= n_{1}(N_{1}-1)\ln{\left(\frac{M_{1}}{M}\right)} + n_{2}(N_{2}-1)\ln{\left(\frac{M_{2}}{M}\right)}\\ &= (M_{1}-n_{1}) \ln{\left(\frac{M_{1}}{M}\right)} + (M_{2}-n_{2})\ln{\left(\frac{M_{2}}{M}\right)} \end{aligned} $$

约定体积分数为 $\phi_{i}$ 聚合物占据格点数 $N_{i}$ 与总格点数 $M$ 之比, 即 $\begin{aligned}\phi_{1} = \frac{M_{1}}{M} = \frac{n_{1}N_{1}}{M}\end{aligned}$, $\begin{aligned}\phi_{2} = \frac{M_{2}}{M} = \frac{n_{2}N_{2}}{M} = 1- \phi_{1}\end{aligned}$. 于是混合熵为

$$ \begin{aligned} \frac{\Delta S_{\text{mix}}}{k_{B}} &= -M_{1}\ln{\phi_{1}} - M_{2}\ln{\phi_{2}} + (M_{1}-n_{1}) \ln{\phi_{1}} + (M_{2}-n_{2})\ln{\phi_{2}}\\ &= - n_{1}\ln{\phi_{1}} - n_{2}\ln{\phi_{2}}\\ &= M\left[ - \frac{\phi_{1}}{N_{1}}\ln{\phi_{1}} - \frac{\phi_{2}}{N_{2}}\ln{\phi_{2}} \right] \end{aligned} $$

单位格点的混合熵为

$$ \begin{aligned} \frac{1}{M}\frac{\Delta S_{\text{mix}}}{k_{B}} = \boxed{\frac{\Delta\bar{S}_{\text{mix}}}{k_{B}} = -\frac{\phi_{1}}{N_{1}}\ln{\phi_{1}} - \frac{\phi_{2}}{N_{2}}\ln{\phi_{2}}} \end{aligned} $$

首先需要推导出 Flory-Huggins 自由能, 核心假设是溶剂和 polymer 的单体都会占据一个格点.

因此设总格点数为 $M$, 各格点体积为 $a^{3}$, polymer 长度均为 $N$, 且一共有 $n_{p}$ 条聚合物链.

每个溶剂分子占据 1 个格点, 且共有 $n_{s}$ 个溶剂分子. 因此存在关系式

$$ M = Nn_{p} + n_{s} $$

polymer 的体积分数为 $\begin{aligned}\phi = \frac{Nn_{p}}{M}\end{aligned}$, 溶剂的体积分数为 $\begin{aligned}1 - \phi = \frac{n_{s}}{M}\end{aligned}$. 可以把溶剂分子视为聚合度为 1 的聚合物.

自由能 $F \equiv U - TS$, 混合(视为等温过程)后产生变化 $\Delta F_{\text{mix}} = \Delta U_{\text{mix}} - T\Delta S_{\text{mix}}$. 也可通过同除以格点数 $M$ 和 能量量纲 $k_{B}T$ 化为形式 $\begin{aligned}\frac{\Delta\bar{F}_{\text{mix}}}{k_{B}T} = \frac{\Delta\bar{U}_{\text{mix}}}{k_{B}T} - \frac{\Delta\bar{S}_{\text{mix}}}{k_{B}}\end{aligned}$.

设晶格配位数为 $z$. 那么将存在 $\varepsilon_{pp}$ (溶质-溶质), $\varepsilon_{ss}$ 溶剂-溶剂, $\varepsilon_{ps}$ 溶质-溶剂三种相互作用的能量.

对于溶质-溶剂(polymer-solvent), 采取平均场假设, 即溶质和溶剂配位的概率为 $P = \phi(1-\phi)$. 则 polymer-solvent 的总配位数为 $\begin{aligned}N_{ps} = \frac{1}{2}zM\phi(1-\phi)\end{aligned}$, 其中 $\begin{aligned}\frac{1}{2}\end{aligned}$ 是为了消除重复计数.

交换能被定义为 $\begin{aligned}\varepsilon = \varepsilon_{ps} - \frac{1}{2}(\varepsilon_{pp} + \varepsilon_{ss})\end{aligned}$. 那么由混合引起的内能变化则为

$$ \Delta U_{\text{mix}} &= N_{ps}\varepsilon = \frac{z\varepsilon}{2}M\phi(1-\phi) \\ \Rightarrow \frac{\Delta\bar{U}_{\text{mix}}}{k_{B}T} &= \frac{\Delta U_{\text{mix}}}{Mk_{B}T} = \frac{z\varepsilon}{2k_{B}T} \phi (1-\phi) = \chi\phi(1-\phi) $$

其中 $\begin{aligned}\frac{z\varepsilon}{2k_{B}T}\end{aligned}$ 被定义为 Flory-Huggins 参数.

代入前面推导的混合熵公式, 即得到各格点自由能变化量(Flory-Huggins 自由能密度):

$$ \frac{\Delta\bar{F}_{\text{mix}}}{k_{B}T} = \frac{\phi}{N}\ln{\phi} + (1-\phi)\ln{(1-\phi)} + \chi\phi(1-\phi) $$

在稀溶液极限 ($\phi\to 0$) 下, 有 $\begin{aligned}\ln{(1-x)} = - x - \frac{x^{2}}{2}\end{aligned}$, 则

$$ (1-\phi)\ln{(1-\phi)} = -\phi + \frac{\phi^{2}}{2} + \mathcal{O}(\phi^{3}) $$

自由能可写作

$$ \frac{\Delta\bar{F}_{\text{mix}}}{k_{B}T} = \overset{\text{ideal mix entropy}}{\frac{\phi}{N}\ln{\phi}} + \underset{A_{1}}{(\chi-1)}\phi + \underset{A_{2}}{\left(\frac{1}{2}-\chi\right)}\phi^{2} + \mathcal{O}(\phi^{3}) $$

$$ \begin{aligned} \frac{f}{k_{B}T} = \frac{\phi}{N}\ln{\phi} + A_{1}\phi + A_{2}\phi^{2} + A_{3}\phi^{3} + \cdots \end{aligned} $$

其中 $A_{1}$ 代表的是溶质和溶剂的平均能量贡献, $A_{2}$ 是溶质与溶质的两体相互作用.

$$ A_{2}\sim \frac{v}{a^{3}}\Rightarrow v\propto a^{3}\left(\frac{1}{2}-\chi\right) \Rightarrow v = a^{3}(1-2\chi) $$

这样就回到了熟悉的通过体积排斥参数 $v$ 来讨论溶剂性能的领域.

• 当 $T$ 较高, 使得 $\begin{aligned}\chi = \frac{z\varepsilon}{2k_{B}T}<\frac{1}{2}\end{aligned}$, 则 $v = a^{3}(1-2\chi)>0$, 链膨胀;
• 当 $\begin{aligned}T = T_{\theta} = \frac{z\varepsilon}{k_{B}}\end{aligned}$, 使得 $\begin{aligned}\chi = \frac{z\varepsilon}{2k_{B}T} = \frac{1}{2}\end{aligned}$, 则 $v = a^{3}(1-2\chi)=0$, 理想链;
• 当 $T$ 较低, 使得 $\begin{aligned}\chi = \frac{z\varepsilon}{2k_{B}T}>\frac{1}{2}\end{aligned}$, 则 $v = a^{3}(1-2\chi)<0$, 链塌缩;

根据 $T_{\theta}$ 的定义, 可将其代入至 $\chi$ 式中替换系数:

$$ v = a^{3} \left(1 - 2 \frac{T_{\theta}}{2T}\right) = a^{3} \left(1 - \frac{T_{\theta}}{T}\right)\propto \boxed{1 - \frac{T_{\theta}}{T}} $$

假设液晶分子长轴取向为单位向量 $\vec{u}$. 于是平均取向可以被定义为 $\vec{n} = \left\langle \vec{u} \right\rangle$. $\vec{u}$ 和 $\vec{n}$ 之间的夹角 $\theta$ 可以被视为描述液晶分子的参数. 接下来需要定义描述液晶分子的序参量, 即其大小和液晶整体的有序性相关.

Recall:

在 Ising Model 中, 序参量被定义为 $\begin{aligned}L = \frac{1}{N}\sum_{i}\sigma_{i} = \frac{N_{+}-N_{-}}{N}\end{aligned}$, 即自旋的平均值.

$\langle\cdots\rangle$ 表示对所有分子做系综平均

• 如果借用 Ising Model 的定义, 我们有 $S_{1} = \langle\theta\rangle$. 然而, 液晶分子通常自身具有对称性(即长轴取向 $\vec{n}$ 和 $-\vec{n}$ 等价), 偏离平均取向 $\vec{n}$ 的分子 $\theta$ 和 $\theta+\pi$ 实际上是等价的, $S_{1}$ 却会错误的计算出 $\langle\theta\rangle = \theta+\pi/2$ 的结果.

• 序参量 $S_{2} = \left\langle\cos{\theta}\right\rangle$ 同样没好到哪里去, 虽然 $\cos{\theta}$ 具有 $\cos{(2\pi-\theta)} = \cos{\theta}$ 的对称性, 但是仍然具有 $\cos{(\theta+\pi)} = -\cos{\theta}$ 的问题: 明明取向相同, 却为 $S_{2}$ 添加了 $0$ 的成分.

• 最简单的消去符号的方法是对其做平方, 即拟定序参量为 $S_{3} = \langle\cos^{2}{\theta}\rangle$. 我们注意到:

• • 完全有序即 $\theta = 0$ 时, $S_{3} = \cos^{2}{0} = 1$;
• • 完全无序时, $\theta$ 均匀分布在整个球壳上, 即

$$ f(\theta,\phi) = \frac{1}{4\pi} $$

计算拟定序参量

$$ \begin{aligned} \langle\cos^{2}(\theta)\rangle &= \int_{0}^{2\pi}\mathrm{d}\phi \int_{0}^{\pi}\cos^{2}{\theta}\cdot \frac{1}{4\pi}\sin{\theta}\mathrm{d}\theta = -\frac{1}{2}\int_{0}^{\pi}\cos^{2}{\theta}\mathrm{d}(\cos{\theta}) \\ &= -\frac{1}{2}\int_{1}^{-1}u^{2}\mathrm{d}u = \frac{1}{2}\frac{1}{3}u^{3}\bigg|_{-1}^{1} = \frac{1}{3} \end{aligned} $$

我们希望在完全无序时, 序参量为 $0$. 因此对该单调函数进行线性变换即可得到最后结果.

从更深层次的角度而言, 任何方向分布函数都可以展开为 Legendre 多项式级数:

$$ f(\theta) = \sum_{l=0}^{\infty} \frac{2l+1}{2}\langle P_{l}(\cos{\theta})\rangle P_{l}(\cos{\theta}) $$

• $l = 1$ 对应 $\langle\cos{\theta}\rangle$, 对应极性有序;
• $l = 2$ 对应 $\left\langle\frac{3\cos^{2}{\theta}-1}{2}\right\rangle$, 对应取向有序.

可以借助电荷的电场线来理解 winding number. 对于点电荷 $q$, 在二维平面内研究其围线上电场方向的旋转. 如果以顺时针方向沿着围线旋转一周, 会发现电场方向也是以顺时针方向旋转一周 ($2\pi$). 因此其拓扑荷为 $m = 2\pi / (2\pi) = 1$, 这一点对于正负电荷都一样.