Soft Matter笔记

Soft Matter Solutions

Gibbs free energy: $G=G(N_p,N_s,T,p)$$G=G(N_p,N_s,T,p)$, where $N_p$$N_p$ is the amount of solute molecules and the $N_s$$N_s$ is the amount of solvant molecules.

Use volumn fraction to characterize the concentration of the solution: $\varphi =\frac{V_p}{V_p+V_s}$$\phi={V_p\over V_p+V_s}$. Given that total volumn $V=V(N_s,N_p,T,p)$$V=V(N_s,N_p,T,p)$, define $v_s=\frac{\partial V}{\partial N_s},v_p=\frac{\partial V}{\partial N_p}$$v_s={\part V\over \part N_s}, v_p={\part V\over \part N_p}$, we have $v_s{N}_s+v_p{N}_p=V$$v_sN_s+v_pN_p=V$. (Using the fact that $V$$V$ is an extensive quantity.) Then $\varphi =\frac{v_p{N}_p}{v_s{N}_s+v_p{N}_p}$$\phi={v_pN_p\over v_sN_s+v_pN_p}$. Here $v_{s,p}$$v_{s,p}$ is just the effective volumn of one solute/solvant molecule.

Assuming $v_{s,p}$$v_{s,p}$ is fixed, thus $V$$V$ is fixed. Using $\mathrm{d}G=-S\mathrm{d}T+V\mathrm{d}p$$\newcommand{\dd}{{\rm d}} \dd G=-S\dd T+V\dd p$ and $\mathrm{d}F=-S\mathrm{d}T-p\mathrm{d}V$$\dd F=-S\dd T-p\dd V$ we have $G=F(N_p,N_s,T)+pV$$G=F(N_p,N_s,T)+pV$. We can let $F=Vf(\varphi ,T)$$F=Vf(\phi,T)$.

If two solutions ($V_1,{\varphi }_1\mathrm{\&}{V}_2,{\varphi }_2$$V_1,\phi_1 ~\&~ V_2,\phi_2$) can always mix to get a new homogeneous solution, we have:

for all $V_1,V_2$$V_1,V_2$. Then $f^{″}(\varphi )>0$$f''(\phi)>0$ for all $\varphi$$\phi$.

Brownian Motion

Displacement correlation:

Velocity correlation:

attention: the average is the ensemble average.

One Dimentional Brownian Motion

Langevin equation:

where $-\zeta v$$-\zeta v$ is the friction, and $F_r(t)$$F_r(t)$ is a stochastic force.

For instantaneous force $F_r(t)$$F_r(t)$ we have:

The first term corresponds to $⟨v|v⟩=0$$\braket v=0$, and the last term corresponds to the fact that $F_r$$F_r$ is instantaneous. To get the factor $A$$A$, we calculate $⟨v^2|v^2⟩$$\braket{v^2}$ with the conditions of $F_r$$F_r$, and let $⟨v^2|v^2⟩=k_{B}T/m$$\braket{v^2}=k_BT/m$. Let ${\tau }_v=\zeta /m$$\tau_v=\zeta/m$:

Thus

we have $A=2m{k}_{B}T/{\tau }_v=2\zeta k_{B}T$$A=2mk_BT/\tau_v=2\zeta k_BT$, i.e.

The escape rate over a high barrier

Assume that there is a 1-d potential well $U(x)$$U(x)$, see the following figure.

It diverges on the left side and right side:

which means that you do not need to consider

1. the particles in the well escapes on the left side,

2. the particles comes back after crossing the barrier.

Assume that the well is deep and the barrier is high. SDE and FPE of the system read

Now what we want is the escape rate over the high barrier.

Since the barrier is high, we assume that the probability distribution in the well soon reaches the steady state, namely

Here we use $P_i,U_i$$P_i, U_i$ to denote the probability density and potential at $X_i$$X_i$. Thus the total probability of the particle being inside the well is

Of course here we assume that $U_a^{″}$$U''_a$ is large enough.

Then we select a point d between a and b, which is

1. high enough in potential, thus probability density on it is small and the segment on its right nearly contribute nothing to the probability flux across the barrier,

2. far away from b, thus $P_d=P_a\exp \{-(U_d-U_a)/k_{B}T\}$$P_d=P_a\exp\{-(U_d-U_a)/k_BT\}$ is still valid.

Rewrite FPE into the following form

Because of the first condition of d, we can make an approximation that from d to c, the flux is a constant --- it's just the flux of particles escaping from the well. Thus

Use the fact $U_c\to -\mathrm{\infty }$$U_c\to -\infty$ and ${\int }_d^c\mathrm{d}x{e}^{U/k_{B}T}\approx {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\mathrm{d}x{e}^{(U_b+U_b^{″}{x}^2/2)/k_{B}T}=e^{U_b/k_{B}T}\sqrt{\frac{2\pi k_{B}T}{-U_b^{″}}}$$\int_d^c\dd x e^{U/k_BT}\approx \int _{-\infty }^\infty \dd x e^{(U_b+U_b''x^2/2)/k_BT}=e^{U_b/k_BT}\sqrt{ 2\pi k_B T\over- U_b'' }$, we derives

$J$$J$ should be proportional to $P_{\mathrm{i}\mathrm{n}\mathrm{w}\mathrm{e}\mathrm{l}\mathrm{l}}$$P_{\rm in~well }$, their ratio is what we need

Flory's argument of self-avoiding walking

This is an approximation of the exponent of self-avoiding walking. We assume that the characteristic length of a path of self-avoiding walking satisfies the following formula

where $L$$L$ denotes the characteristic length, which is the absolute length from the start point to the end, and $N$$N$ denotes the number of steps.

The space the path occupies can be estimated by

where $d$$d$ is the number of dimensions. To characterize the self-avoiding condition, we assume that there is an energy penalty when the distance between two footholds is less than a certain distance $R$$R$. On average, for each particle, there is

particles being too close to it. Thus the total energy penalty is

Next step, we consider a normal random walking. Since we know the probability density distribution in the space, we can obtain the relative number of path corresponding to a certain $L$$L$, which is

where $A,B$$A,B$ are some coefficients. Since to calculate entropy, we are to take the logarithm of it, both of them are unimportant.

Finally, we obtain the free energy

where $C$$C$ is an unimportant coefficient again. At the real characteristic length, $F$$F$ should be minimized, thus

It's a really good approximation.

第一类不稳定的振幅方程

　　第一类不稳定在线性稳定性分析中只有在特定的波矢 $q_c$$q_c$ 附近有正实部的增长率。一个最简单的例子为Swift-Hohenberg方程：

显然，上述方程有 $u=0$$u=0$ 的基态解。当 $ϵ>0$$\epsilon >0$，对上式做线性稳定性分析，易知在波矢 $q_c$$q_c$ 附近是不稳定的。（这里的非线性项 $g{u}^3$$gu^3$ 的存在是为了防止不稳定模态的无限增长，要求 $g>0$$g>0$。）

　　根据“偏离 $u=0$$u=0$ 的解应该近似于 $u=A{e}^{\mathrm{i}{q}_{c}x}+\mathrm{c}.\mathrm{c}.$$u=Ae^{\ii q_c x}+{\rm c.c. }$” 的事实，我们可以将解记为：

这里的 $A(x,t)$$A(x,t)$ 应该是随 $x,t$$x,t$ 的慢变量，调制基底波的振幅。将上式带入运动方程，得到：

由于认为 $A$$A$ 是慢变量，这里可以认为 ${\partial }_x$$\partial _x$ 是一个小量，只保留到最低阶。同时，可以认为在 $x$$x$ 附近的几个周期 $\frac{2\pi }{q_c}$${2\pi \over q_c}$ 内 $A$$A$ 不变，将上式乘上 $e^{-\mathrm{i}{q}_{c}x}$$e^{-\ii q_c x}$ 并在局域积分，只有含有 $e^{\mathrm{i}{q}_{c}x}$$e^{\ii q_c x}$ 的项保留：

　　事实上，利用对称性可以推广结果。一维空间中的 $n$$n$ 维场发生第一类不稳定变化，记其控制参数为（无纲量）$ϵ$$\epsilon$，当 $ϵ>0$$\epsilon >0$ 时发生基态失稳，设最早失稳的线性模式为 $e^{\pm \mathrm{i}{q}_{c}x}\overrightarrow{V}$$e^{\pm \ii q_c x}\vec V$，这里 $\overrightarrow{V}$$\vec V$ 是模式对应的特征矢。在最低阶近似下，失稳后的场可以记为：

考虑到 $A$$A$ 本身是小量（$ϵ\ll 1$$\epsilon \ll 1$）且是慢变量。则根据对称性分析，$A$$A$ 在相位变换下不改变。这是因为 $A$$A$ 的相位变换对应着 $e^{\mathrm{i}{q}_{c}x}$$e^{\ii q_c x}$ 的平移，调制函数的行为与基底波的行为应无关，否则无法独立写出调制函数的行为。这有些一厢情愿，但是出于实用性考虑必须有这条要求。$A$$A$ 满足的最低阶方程为：

最低阶非线性项必须为 $|A{|}^{2}A$$|A|^2A$，二阶项无法满足条件。以上就是振幅方程。通过比较振幅方程和场的运动方程的结果，可以得到 $\tau ,\xi ,\mu$$\tau ,\xi ,\mu$。当 $A\to 0$$A\to 0$ 时，忽略非线性项，$A$$A$ 的增长率为：

记场的运动方程的线性稳定性分析得到的结果为 $\sigma (q)=\sigma (q_c+k)$$\sigma (q)=\sigma (q_c+k)$，应有

而 $g$$g$ 需要通过解 $A{e}^{\mathrm{i}{q}_{c}x}$$Ae^{\ii q_c x}$ 的（近似）稳定 $A$$A$​ 值来得到。

　　当 $\sigma (q)\in \mathbb{R}$$\sigma (q)\in\R$ 时，振幅方程的右边总是能够写成如下的梯度形式：

即总是朝着能量最低的方向运动。

从运动方程到流体方程

Dean方法

　　设运动方程为：

这里的 ${\overrightarrow{x}}^{\alpha }$$\vec x^\alpha$ 是对应个体 $\alpha$$\alpha$ 的运动变量数组，$\mathrm{d}{\overrightarrow{W}}^{\alpha }$$\dd\vec W^\alpha$ 是各分量各指标独立的Wienner过程。（令 $\overrightarrow{B}(\overrightarrow{x},\overrightarrow{x})=0$$\vec B(\vec x,\vec x)=0$。）则在运动变量数组空间中的密度函数为：

　　利用伊藤引理能够直接写出 $\rho$$\rho$ 的运动方程：

利用等式：

可将上式化简为：

再利用等式：

可得：

上式除了随机部分，都是 $\rho (\overrightarrow{x},t)$$\rho (\vec x,t)$ 的闭表达式。

　　下面将随机部分改写成 $\rho$$\rho$ 的形式。有平均值：

又随机变量

满足：

故可用 $\xi$$\xi$ 代替上面推导得到的表达式。最终结果如下：

BBGKY方法

　　依然设运动方程为：

在相空间中的概率分布的运动方程为：

　　单粒子分布函数为：

故对 $n$$n$ 粒子分布函数做积分，得

利用 $f(\overrightarrow{x}{}^{\prime }\mathrm{s},t)$$f(\vec x{\rm 's },t)$ 对不同 ${\overrightarrow{x}}^{\beta }$$\vec x^\beta$ 是轮换对称的这一事实，可将上式简化为：

定义：

则上式即为：

通过一定的近似方法处理 $f^{(2)}$$f^{(2)}$，即可得到 $f^{(1)}$$f^{(1)}$ 的闭表达式。

二者的关系

　　通过定义，易知：

故对Dean方程取平均，即可得到BBGKY结果。

　　一般在涨落可忽略的情况下，会简单将Dean方程中的随机项舍弃；或利用平均场近似，在BBGKY结果中用

两者能得到同一个结果：