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# 物理代写|电磁学代写Electromagnetism代考|PHYS404 Dirac’s Delta Function ( δ-Function)

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## 物理代写|电磁学代写Electromagnetism代考|Dirac’s Delta Function ( δ-Function)

The $\delta$-function is particularly useful in the following, which is why we will introduce it here. It shall be noted that our exposition here does not substitute a rigorous mathematical introduction.

A rough, illustrative way to describe the character of the $\delta$-function is to note that it vanishes everywhere except for one particular point of its argument (namely $0)$, where it takes an infinite value, exactly such that its integral equals 1 .
$\delta\left(x-x^{\prime}\right)=\left{\begin{array}{lll}0 & \text { for } & x \neq x^{\prime} \ \infty & \text { for } & x=x^{\prime}\end{array}\right.$
$$\int_{-\infty}^{+\infty} \delta\left(x-x^{\prime}\right) d x=1$$
The $\delta$-function is not a function in the usual sense. It belongs to a more general category of functions, which sometimes are called improper functions, generalized functions, or distributions. Another possibility is to imagine the $\delta$-function as the limit of a series of functions. It can be constructed in various ways, for example,

The limit of a series of rectangular functions as illustrated in Fig. 3.7. Thus

$$g_h(x)= \begin{cases}h & \text { for }\left|x-x^{\prime}\right| \leq \frac{1}{2 h} \ 0 & \text { else }\end{cases}$$
and
$$\delta\left(x-x^{\prime}\right)=\lim _{h \rightarrow \infty} g_h(x)$$

The limit of a series of Gaussian functions as illustrated in Fig. 3.8. Now
$$f_a(x)=\frac{1}{a \sqrt{\pi}} \exp \left[-\frac{\left(x-x^{\prime}\right)^2}{a^2}\right]$$
where
$$\int_{-\infty}^{+\infty} f_a(x) d x=1$$
and

$$\delta\left(x-x^{\prime}\right)=\lim _{a \rightarrow 0} f_a(x) .$$

## 物理代写|电磁学代写Electromagnetism代考|Point Charge and δ-Function

Poisson’s equation applies to the case of a point charge.
$$\nabla^2 \varphi=-\frac{\rho}{\varepsilon_0},$$
where
$$\rho(\mathbf{r})=Q \delta\left(\mathbf{r}-\mathbf{r}^{\prime}\right),$$
i.e.,
$$\nabla^2 \varphi=-\frac{Q \delta\left(\mathbf{r}-\mathbf{r}^{\prime}\right)}{\varepsilon_0},$$
$$\varphi=-\frac{Q}{4 \pi \varepsilon_0} \cdot \frac{1}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|},$$
i.e., we may now write for all locations, including the location of the point charge, (which was previously impossible)
$$\nabla^2 \frac{Q}{4 \pi \varepsilon_0} \cdot \frac{1}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|}=-\frac{Q}{\varepsilon_0} \cdot \delta\left(\mathbf{r}-\mathbf{r}^{\prime}\right),$$
or
$$\nabla^2 \frac{1}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|}=-4 \pi \delta\left(\mathbf{r}-\mathbf{r}^{\prime}\right) .$$
In Section 2.3, we had to exclude the locations of the point charges, because we were unable to differentiate there. With the use of the $\delta$-function, those difficulties or restrictions are now removed.

\$\delta$\backslash$left(xx^{\prime$} \backslash$right$)=\backslash$left { 0 for$x \neq x^{\prime} \infty$for$x=x^{\prime}$泟确的。$\int_{-\infty}^{+\infty} \delta\left(x-x^{\prime}\right) d x=1 T h e \backslash$三角洲 functionisnotafunctionintheusualsense. Itbelongstoamoregeneralcategoryof functions, whichsometimesarecalledimproper functions, generalized fun 一系列矩形函数的极限如图$3.7$肵示。 因此 $$g_h(x)=\left{h \quad \text { for }\left|x-x^{\prime}\right| \leq \frac{1}{2 h} 0\right. \text { else }$$ $$\delta\left(x-x^{\prime}\right)=\lim {h \rightarrow \infty} g_h(x)$$ 一系列高斯函数的极限如图$3.8$所示。, 现在 $$f_a(x)=\frac{1}{a \sqrt{\pi}} \exp \left[-\frac{\left(x-x^{\prime}\right)^2}{a^2}\right]$$ 在邭里 $$\int{-\infty}^{+\infty} f_a(x) d x=1$$ $$\delta\left(x-x^{\prime}\right)=\lim _{a \rightarrow 0} f_a(x) .$$ ## 物理代写|电磁学代写Electromagnetism代考|Point Charge and$\delta\$-Function

$$\nabla^2 \varphi=-\frac{\rho}{\varepsilon_0},$$

$$\rho(\mathbf{r})=Q \delta\left(\mathbf{r}-\mathbf{r}^{\prime}\right),$$
IE。
$$\nabla^2 \varphi=-\frac{Q \delta\left(\mathbf{r}-\mathbf{r}^{\prime}\right)}{\varepsilon_0},$$

$$\varphi=-\frac{Q}{4 \pi \varepsilon_0} \cdot \frac{1}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|},$$
$$\nabla^2 \frac{Q}{4 \pi \varepsilon_0} \cdot \frac{1}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|}=-\frac{Q}{\varepsilon_0} \cdot \delta\left(\mathbf{r}-\mathbf{r}^{\prime}\right),$$

$$\nabla^2 \frac{1}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|}=-4 \pi \delta\left(\mathbf{r}-\mathbf{r}^{\prime}\right) .$$

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