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## 物理代写|统计物理代写Statistical Physics of Matter代考|Induced Dipoles and Van der Waals Attraction

Now we consider the electrostatic interaction involving nonpolar molecules. An external field $\boldsymbol{E}(\boldsymbol{r})$ can induce a polarization even in a nonpolar molecule, $\boldsymbol{P}{\text {ind }}=\alpha \boldsymbol{E}$, where $\alpha$ is the polarizability, thereby reducing the electrostatic energy. The energy change induced by the field that polarizes the molecule is \begin{aligned} \varphi{n} &=-\int_{0}^{\boldsymbol{E}} \boldsymbol{P}{\text {ind }} \cdot d \boldsymbol{E} \ &=-\alpha \int{0}^{\boldsymbol{E}} \boldsymbol{E} \cdot d \boldsymbol{E}=-\frac{\alpha \boldsymbol{E}^{2}}{2} \end{aligned}
This is an attractive interaction between the nonpolar molecule and the object that emanates the electric field. For example, the potential energy between the nonpolar molecule and an ion of charge $q$ is given by
$$\varphi_{i n}=-\frac{\alpha E^{2}}{2}=-\frac{\alpha}{2}\left(\frac{q}{4 \pi \varepsilon_{w} r^{2}}\right)^{2}$$
which is comparable to (6.14) of ion-dipole attraction; both are identical if $\alpha$ is replaced by that of a free dipole $\alpha_{d}=p^{2} /\left(3 k_{B} T\right)$.

Consider the interaction between small nonpolar molecules 1 and 2 with their dipole moments instantaneously induced with the polarizabilities, $\alpha_{1}, \alpha_{2}$. A detailed derivation of the interaction is too complicated involving quantum fluctuations as well as thermal fluctuations to be relevant here. To find how it depends on the distance $r$ between the two objects, we give a simple argument following (6.29). The potential energy of the polarized molecule 2 due to the field $\boldsymbol{E}{2}$ emanating from molecule 1 is $\varphi{n n}=-\alpha_{2}\left\langle E_{2}^{2}\right\rangle / 2$, where $\boldsymbol{E}{2}$ is now recognized as a fluctuating field due to an instantaneous dipole of molecule 1 .

## 物理代写|统计物理代写Statistical Physics of Matter代考|The Poisson-Boltzmann Equation

Understanding the behaviors of the ions thermally fluctuating under long-range Coulomb interactions is a many-body problem, for which rigorous use of statistical mechanics (Chap. 4) to solve it is a formidable task. Here we present a simple approximation that can capture the main physical features for the case of dilute ionic (electrolyte) solutions. First, the ionic solution is regarded as a continuum, so that the electric potential $\phi$ at a position $r$ satisfies the basic equation in electrostatics, namely, the Poisson equation
$$\nabla^{2} \phi(\boldsymbol{r})=-\frac{\rho_{e}(\boldsymbol{r})}{\varepsilon},$$
where $\rho_{e}(\boldsymbol{r})$ is the charge density and $\varepsilon$ is its electric permeability, which is nearly that of water, $\varepsilon \cong \varepsilon_{w}$, for the cases of dilute ionic solutions we consider throughout. We further assume that an ion at $\boldsymbol{r}$ is subject to a one-body electric potential $\phi(\boldsymbol{r})$, that is, a mean field, which effectively includes the influence of the other ions. In this mean field theory, for the ions each with charge $q$, the charge density in the solution is given by the Boltzmann factor
$$\rho_{e}(\boldsymbol{r})=\rho_{\infty} e^{-\beta q \phi(\boldsymbol{r})},$$
where $\rho_{\infty}$ is the reference charge density at the point in the bulk where the potential is zero. Equation (6.35) then becomes a nonlinear equation for $\phi(\boldsymbol{r})$, called the Poisson-Boltzmann (PB) equation
$$\nabla^{2} \phi(\boldsymbol{r})=-\frac{\rho_{\infty}}{\varepsilon} e^{-\beta q \phi(\boldsymbol{r})} .$$
As will be shown in the next section, this equation is exactly solved in one dimension, namely, for the potential and charge distribution at a vertical distance from a planar charged surface/membrane.

# 统计物理代考

## 物理代写|统计物理代写Statistical Physics of Matter代考|Induced Dipoles and Van der Waals Attraction

$$\varphi n=-\int_{0}^{E} P_{\text {ind }} \cdot d \boldsymbol{E} \quad=-\alpha \int 0^{\boldsymbol{E}} \boldsymbol{E} \cdot d \boldsymbol{E}=-\frac{\alpha \boldsymbol{E}^{2}}{2}$$

$$\varphi_{i n}=-\frac{\alpha E^{2}}{2}=-\frac{\alpha}{2}\left(\frac{q}{4 \pi \varepsilon_{w} r^{2}}\right)^{2}$$

## 物理代写|统计物理代写Statistical Physics of Matter代考|The Poisson-Boltzmann Equation

$$\nabla^{2} \phi(\boldsymbol{r})=-\frac{\rho_{e}(\boldsymbol{r})}{\varepsilon},$$

$$\rho_{e}(\boldsymbol{r})=\rho_{\infty} e^{-\beta q \phi(r)},$$

$$\nabla^{2} \phi(\boldsymbol{r})=-\frac{\rho_{\infty}}{\varepsilon} e^{-\beta q \phi(\boldsymbol{r})} .$$

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