# 数学代写|微积分代写Calculus代写|MTH125

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## 数学代写|微积分代写Calculus代写|Applications of the Dirac Delta Function

Taking the divergence of the Coulomb integral for the electric field of a volume distribution of charge leads to
\begin{aligned} \boldsymbol{\nabla} \cdot \mathbf{E}(\mathbf{r}) & =\boldsymbol{\nabla} \cdot \int \frac{\left(\mathbf{r}-\mathbf{r}^{\prime}\right) \rho\left(\mathbf{r}^{\prime}\right) d^3 r^{\prime}}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|^3} \ & =\int \rho\left(\mathbf{r}^{\prime}\right) d^3 r^{\prime} \boldsymbol{\nabla} \cdot\left[\frac{\left(\mathbf{r}-\mathbf{r}^{\prime}\right)}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|^3 \mid}\right] \ & =4 \pi \int \rho\left(\mathbf{r}^{\prime}\right) d^3 r^{\prime} \delta\left(\mathbf{r}-\mathbf{r}^{\prime}\right) . \end{aligned}
Doing the delta function integral gives
$$\boldsymbol{\nabla} \cdot \mathbf{E}=4 \pi \rho$$
for any continuous charge distribution.
Gauss’s law can be derived by applying the divergence theorem to Eq. (4.9)
$$\oint_S \mathbf{E} \cdot \mathbf{d A}=\int_V \boldsymbol{\nabla} \cdot \mathbf{E} d^3 r=4 \pi \int_V \rho(\mathbf{r}) d^3 r=4 \pi Q_{\text {enclosed }} .$$
Gauss’s law holds for any vector field, including the gravitational field as well as the electric field, that is given by the first integral in Eq. (4.8).

Equation (4.9) can be put in terms of the potential $\phi$, leading to Poisson’s equation $^1$
$$-\boldsymbol{\nabla} \cdot \mathbf{E}=\boldsymbol{\nabla} \cdot(\boldsymbol{\nabla} \phi)=\nabla^2 \phi=-4 \pi \rho .$$
The homogeneous form of Poisson’s equation, with the source function $\rho=0$,
$$\nabla^2 \phi=0,$$
is called Laplace’s equation.

## 数学代写|微积分代写Calculus代写|Singularities of Dipole Fields

Another occurrence of the Dirac delta function is in the singular behavior at the origin of the electromagnetic field of a point electric or magnetic dipole. The potential of a point electric dipole is given by
$$\phi_{\mathbf{p}}(\mathbf{r})=\frac{\mathbf{r} \cdot \mathbf{p}}{r^3} .$$

The electric field of a point electric dipole is given by the negative gradient of the potential:
\begin{aligned} \mathbf{E}{\mathbf{p}}(\mathbf{r}) & =-\boldsymbol{\nabla} \phi{\mathbf{p}}(\mathbf{r})=-\boldsymbol{\nabla}\left(\frac{\mathbf{p} \cdot \mathbf{r}}{r^3}\right) \ & =-\frac{\mathbf{p}}{r^3}-(\mathbf{p} \cdot \mathbf{r}) \boldsymbol{\nabla}\left(\frac{1}{r^3}\right) . \end{aligned}
The term $\boldsymbol{\nabla}\left(1 / r^3\right)$ has to be treated carefully because of its singular nature. We do this by relating it to the known result
$$\nabla \cdot\left(\frac{\mathbf{r}}{r^3}\right)=4 \pi \delta(\mathbf{r}) .$$
Since the gradient in $\boldsymbol{\nabla}\left(1 / r^3\right)$ acts on a function of $r$, it can be written as
\begin{aligned} \boldsymbol{\nabla}\left(\frac{1}{r^3}\right) & =\hat{\mathbf{r}}\left[\hat{\mathbf{r}} \cdot \boldsymbol{\nabla}\left(\frac{1}{r^3}\right)\right]=\frac{\hat{\mathbf{r}}}{r}\left[(\mathbf{r} \cdot \boldsymbol{\nabla})\left(\frac{1}{r^3}\right)\right] \ & =\frac{\hat{\mathbf{r}}}{r}\left[\boldsymbol{\nabla} \cdot\left(\frac{\mathbf{r}}{r^3}\right)-\frac{\boldsymbol{\nabla} \cdot \mathbf{r}}{r^3}\right] \ & =\frac{\hat{\mathbf{r}}}{r}\left[4 \pi \delta(\mathbf{r})-\frac{3}{r^3}\right] . \end{aligned}
Now we can continue Eq. (4.14) as
\begin{aligned} \mathbf{E}_{\mathbf{p}}(\mathbf{r}) & =-\frac{\mathbf{p}}{r^3}-(\mathbf{p} \cdot \mathbf{r}) \nabla\left(\frac{1}{r^3}\right) \ & =\frac{3(\mathbf{p} \cdot \hat{\mathbf{r}}) \hat{\mathbf{r}}-\mathbf{p}}{r^3}-4 \pi \hat{\mathbf{r}}(\hat{\mathbf{r}} \cdot \mathbf{p}) \delta(\mathbf{r}) . \end{aligned}
Equation (4.17) shows the dipole electric field with its singular behavior at the origin. The singular behavior at the origin adds a contact term to the potential energy of two dipoles, $\mathbf{p}$ and $\mathbf{p}^{\prime}$.

# 微积分代考

## 数学代写|微积分代写Calculus代写|Applications of the Dirac Delta Function

$$\boldsymbol{\nabla} \cdot \mathbf{E}(\mathbf{r})=\boldsymbol{\nabla} \cdot \int \frac{\left(\mathbf{r}-\mathbf{r}^{\prime}\right) \rho\left(\mathbf{r}^{\prime}\right) d^3 r^{\prime}}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|^3} \quad=\int \rho$$

$$\boldsymbol{\nabla} \cdot \mathbf{E}=4 \pi \rho$$

$$\oint_S \mathbf{E} \cdot \mathbf{d A}=\int_V \boldsymbol{\nabla} \cdot \mathbf{E} d^3 r=4 \pi \int_V \rho(\mathbf{r}) d^3 r=4 \pi Q$$

$$-\boldsymbol{\nabla} \cdot \mathbf{E}=\boldsymbol{\nabla} \cdot(\boldsymbol{\nabla} \phi)=\nabla^2 \phi=-4 \pi \rho .$$

$$\nabla^2 \phi=0,$$

## 数学代写|微积分代写Calculus代写|Singularities of Dipole Fields

Dirac delta 函数的另一个出现是在点电偶极子或磁偶极子的电磁场原点处的奇异行为。点电偶极子的电势由下 式给出
$$\phi_{\mathbf{p}}(\mathbf{r})=\frac{\mathbf{r} \cdot \mathbf{p}}{r^3} .$$

$$\mathbf{E} \mathbf{p}(\mathbf{r})=-\boldsymbol{\nabla} \phi \mathbf{p}(\mathbf{r})=-\boldsymbol{\nabla}\left(\frac{\mathbf{p} \cdot \mathbf{r}}{r^3}\right) \quad=-\frac{\mathbf{p}}{r^3}$$

$$\nabla \cdot\left(\frac{\mathbf{r}}{r^3}\right)=4 \pi \delta(\mathbf{r})$$

$$\boldsymbol{\nabla}\left(\frac{1}{r^3}\right)=\hat{\mathbf{r}}\left[\hat{\mathbf{r}} \cdot \boldsymbol{\nabla}\left(\frac{1}{r^3}\right)\right]=\frac{\hat{\mathbf{r}}}{r}\left[(\mathbf{r} \cdot \boldsymbol{\nabla})\left(\frac{1}{r^3}\right)\right]$$

$$\mathbf{E}_{\mathbf{p}}(\mathbf{r})=-\frac{\mathbf{p}}{r^3}-(\mathbf{p} \cdot \mathbf{r}) \nabla\left(\frac{1}{r^3}\right) \quad=\frac{3(\mathbf{p} \cdot \hat{\mathbf{r}}) \hat{\mathbf{r}}}{r^3}$$

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