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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 物理代写|统计物理代写Statistical Physics of Matter代考|Transition Between Dispersed and Condensed Phases

Now we consider the detail of the critical condition for the transition by focusing on the case with $\epsilon=0$. For $q \beta b$ above the critical value the $\theta-p$ curve yet develops a wiggle. The critical point is the inflection point where $d p / d \theta=0=p(1 / \theta(1-\theta)-$ $q \beta b), d^2 p / d \theta^2=0=p(2 \theta-1) /[\theta(1-\theta)]^2$, i.e., via $(8.20)$ and $(8.21)$

$\theta=1 / 2 ;$ and $q \beta b=4$, leading to $p=\alpha e^{-2}$.
At $T$ lower than the critical temperature $T_c=q b /\left(4 k_B\right)$, (or for an attraction strength $q b$ higher than $\left.q b_c=4 k_B T\right)$, and simultaneously at an ambient pressure lower than $p_c=\alpha e^{-2}$, the condensation to an aggregate occurs with a discontinuous jump in the coverage $\theta$.

P8.2 If the molecules adsorb on surface with the binding energy $\epsilon$, how much is the coverage affected? Consider that the $b$ has a strength of the covalent bonding and $\theta$ is nearly 1 .

The vertical isotherm in the $\theta-p$ diagram is obtained by a semi-empirical scheme called the Maxwell construction as in the $\gamma-\theta$ phase diagram given below. The surface tension $\gamma$ is obtained by
\begin{aligned} \gamma &=\frac{\partial F(N / M, M, T)}{a \partial M}=\frac{f(\theta, T)}{a}+M \frac{\partial f(\theta, T)}{a \partial \theta} \frac{\partial \theta}{\partial M} \ &=\frac{1}{a}{f(\theta, T)-\mu \theta} \ &=\frac{k_B T}{a} \ln (1-\theta)+\frac{q}{2 a} b \theta^2 \end{aligned}
where $f(\theta, T)=F(\theta, M, T) / M=-q b \theta^2 / 2+k_B T{\theta \ln \theta+(1-\theta) \ln (1-\theta)}$ is the Helmholtz free energy per site and the relation $\theta=N / M$ should be noted in taking the derivative with respect to $M$.

## 物理代写|统计物理代写Statistical Physics of Matter代考|Mixing and Phase Separation

The lattice model can be adapted to binary mixtures of liquids, colloids, polymers, as well as lipid mixtures in membranes and non-membrane-bound liquid drops within cells (Anthony et al. 2014). We consider an incompressible mixture in which every cell is occupied by a particle of either species $A$ or species $B$, so that the total number of molecules $N=N_A+N_B=M$ is fixed. The occupation number $n_i$ is 0 when the cell $i$ is occupied by a particle of species $A$ and is 1 when it is occupied by a particle of species $B$. Only particles in the nearest neighborhood interact, with bond energies $b_{A A}, b_{B B}, b_{A B}=b_{B A}$, for $A-A, B-B$, and $A-B$ pairs respectively. The Hamiltonian can be written as $$\mathcal{H}=-{ }2^1 \sum{\langle i j\rangle}\left[b_{A A}\left(1-n_i\right)\left(1-n_j\right)+b_{B B} n_i n_j+b_{A B}\left(1-n_i\right) n_j+b_{A B} n_i\left(1-n_j\right)\right],$$
where the sum is over all $q N / 2$ nearest neighbor pairs, and $q$ is the coordination number. Then the Hamiltonian can be rewritten as
$$\mathcal{H}=\frac{1}{2} \sum_{\langle i j\rangle} b\left(1-n_i\right) n_j+\sum_i h n_i+C,$$
where $b=b_{A A}+b_{B B}-2 b_{A B}, h=q\left(b_{A A}-b_{B B}\right) / 2$ chosen to be negative, and $C$ is the (trivial) constant energy that the mixture would have if the particles were identical.

We use the mean field approximation as in the earlier section. Replacing $n_i$ in the Hamiltonian by the relative coverage of species $B$
$$\theta=\frac{\left\langle\sum_i^M n_i\right\rangle}{N}=\frac{N_B}{N},$$
the internal energy then is approximated as
$$E=N\left[\frac{1}{2} q b(1-\theta) \theta+h \theta\right] .$$

# 统计物理代考

## 物理代写|统计物理代写Statistical Physics of Matter代考|Transition Between Dispersed and Condensed Phases

$\theta=1 / 2 ;$ 和 $q \beta b=4$ ，导致 $p=\alpha e^{-2}$.

$$\gamma=\frac{\partial F(N / M, M, T)}{a \partial M}=\frac{f(\theta, T)}{a}+M \frac{\partial f(\theta, T)}{a \partial \theta} \frac{\partial \theta}{\partial M} \quad=\frac{1}{a} f(\theta, T)-\mu \theta=\frac{k_B T}{a} \ln (1-\theta)+\frac{q}{2 a} b \theta^2$$

## 物理代写|统计物理代写Statistical Physics of Matter代考|Mixing and Phase Separation

$$\theta=\frac{\left\langle\sum_i^M n_i\right\rangle}{N}=\frac{N_B}{N}$$

$$E=N\left[\frac{1}{2} q b(1-\theta) \theta+h \theta\right]$$

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