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• Statistical Inference 统计推断
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• Advanced Mathematical Statistics 高等数理统计学
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• 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|The Coulomb Interaction

The water medium affects fundamentally the interaction between two ions. Phenomenologically the interaction between two ions of charges $q_{1}$ and $q_{2}$ separated by a distance $r_{12}$ is just the Coulomb interaction,
$$\varphi_{12}=\frac{q_{1} q_{2}}{4 \pi \varepsilon_{w} r_{12}}$$
where $\varepsilon_{w}$ is the electric permeability of water. As mentioned in Sect. 4.4, this effective interaction is formally obtained by integrating (averaging) over all the degrees of freedom of the water molecules surrounding the two ions, with the distance between the two ions $r_{12}$ given as fixed. The effect of the water medium, treated as a continuum, is incorporated by $\varepsilon_{w}$, which depends on $T$.

Equation (6.4) is based on a coarse continuum picture. While it neglects the microscopic details involving the water molecules at short distance, it often gives a reasonable understanding when the distance $r$ longer than the small molecule scale. At such separation the electrostatic interaction energy can be smaller than the thermal energy. If two elementary charges $e$ are separated by $\sim 1 \mathrm{~nm}$ in a vacmum, they would have the Coulomb energy in the order of $1 \mathrm{eV}$, but in water its high dielectric constant $\varepsilon_{w} / \varepsilon_{0}=78.5$ can reduce the Coulomb energy below the thermal energy which is $k_{B} T \approx 1 / 40 \mathrm{eV}$ at room temperature. A convenient scale to assess dominance of the thermal energy is
$$l_{B}=\frac{e^{2}}{4 \pi \varepsilon k_{B} T}$$
called the Bjerrum length, above which the elementary Coulomb interaction energy is less than $k_{B} T$; in water at $25^{\circ} \mathrm{C}$ it is $7.1 \AA$.

Consider a single ion of charge $q$ in water. Assuming that the ion is a sphere of radius $a$, the work necessary to charge the sphere from zero to $q$ in water continuum is
$$\varphi_{b}=\int_{0}^{q} \frac{q^{\prime} d q^{\prime}}{4 \pi \varepsilon_{w} a}=\frac{q^{2}}{8 \pi \varepsilon_{w} a}$$

## 物理代写|统计物理代写Statistical Physics of Matter代考|Ion-Dipole Interaction

Consider an ion of charge $q$ and a molecule of permanent dipole moment $\boldsymbol{p}$ where the charges $q_{d}$ and $-q_{d}$ are separated by distance $l$ (Fig. 6.5). The energy of electrostatic interaction between the ion and the dipole at a distance $r$ is
\begin{aligned} \varphi_{i d}(r, \Omega) &=\frac{q_{d} q}{4 \pi \varepsilon_{w}|\boldsymbol{r}+\boldsymbol{l} / 2|}-\frac{q_{d} q}{4 \pi \varepsilon_{w}|\boldsymbol{r}-\boldsymbol{l} / 2|} \approx q_{d} \boldsymbol{l} \cdot \boldsymbol{\nabla} \frac{q}{4 \pi \varepsilon_{w} r} \ &=-\boldsymbol{p} \cdot \boldsymbol{E}=-p E \cos \theta \end{aligned}
where $E=q /\left(4 \pi \varepsilon_{\mathrm{w}} r^{2}\right)$ is the electric field from the ion, $\theta$ is the polar angle that the dipole vector $\boldsymbol{p}=q_{d} l$ makes with the field direction (Fig. 6.5).

The dipole is undergoing thermal fluctuations (rotation). The induced polarization along the field observed at equilibrium is the thermal average over the orientation,
\begin{aligned} P_{d}=p\langle\cos \theta\rangle &=p \frac{\int d \Omega \cos \theta e^{-\beta \varphi_{i d}(r, \Omega)}}{\int d \Omega e^{-\beta \varphi_{i d}(r, \Omega)}}=p \frac{\int d \Omega \cos \theta e^{\beta p E \cos \theta}}{\int d \Omega e^{\beta p E \cos \theta}} \ &=p \frac{\partial}{\partial(\beta p E)} \ln Z_{i d}=p\left[\operatorname{coth}(\beta p E)-\frac{1}{\beta p E}\right] \equiv p \mathcal{L}(\beta p E) \end{aligned}
where $d \Omega=d(\cos \theta) d \phi$ is the solid angle element, $Z_{i d}=\int d \Omega e^{\beta p E \cos \theta}=$ $(4 \pi \sinh \beta p E) / \beta p E$, and $\mathcal{L}(x)$ is the Langevin’s function earlier seen in polymer extension problem (3.57).

Now we note that $\beta p E=p q /\left(4 \pi \varepsilon_{w} r^{2} k_{B} T\right) \sim l l_{B} / r^{2}$ where the $l_{B}$ is about $7 \AA$ at room temperature. If, for example, the dipole is due to a small polar molecule like a water molecule, the charge separation length $l$ is also a length in the order of $1 \AA$, so, at a distance $r$ over the nanoscale, $\beta p E(r) \ll 1$. We will consider this case throughout.

# 统计物理代考

## 物理代写|统计物理代写Statistical Physics of Matter代考|The Coulomb Interaction

$$\varphi_{12}=\frac{q_{1} q_{2}}{4 \pi \varepsilon_{w} r_{12}}$$

$$l_{B}=\frac{e^{2}}{4 \pi \varepsilon k_{B} T}$$

$$\varphi_{b}=\int_{0}^{q} \frac{q^{\prime} d q^{\prime}}{4 \pi \varepsilon_{w} a}=\frac{q^{2}}{8 \pi \varepsilon_{w} a}$$

## 物理代写|统计物理代写Statistical Physics of Matter代考|Ion-Dipole Interaction

$$\varphi_{i d}(r, \Omega)=\frac{q_{d} q}{4 \pi \varepsilon_{w}|\boldsymbol{r}+\boldsymbol{l} / 2|}-\frac{q_{d} q}{4 \pi \varepsilon_{w}|\boldsymbol{r}-\boldsymbol{l} / 2|} \approx q_{d} \boldsymbol{l} \cdot \boldsymbol{\nabla} \frac{q}{4 \pi \varepsilon_{w} r} \quad=-\boldsymbol{p} \cdot \boldsymbol{E}=-p E \cos \theta$$

$$P_{d}=p\langle\cos \theta\rangle=p \frac{\int d \Omega \cos \theta e^{-\beta \varphi \text { id }(r, \Omega)}}{\int d \Omega e^{-\beta \varphi \operatorname{id}(r, \Omega)}}=p \frac{\int d \Omega \cos \theta e^{\beta p E \cos \theta}}{\int d \Omega e^{\beta p E \cos \theta}} \quad=p \frac{\partial}{\partial(\beta p E)} \ln Z_{i d}=p\left[\operatorname{coth}(\beta p E)-\frac{1}{\beta p E}\right] \equiv p \mathcal{L}(\beta p E)$$

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