Chapter 1: Dirac Delta Function

Section 5.1/2: Divergence of \(\frac{\hat{r}}{r^2}\) and One Dimensional Dirac

Divergence: \(\nabla \cdot v = \frac{1}{r^2} \frac{\partial}{\partial{r}} (r^2 * \frac{1}{r^2}) = \frac{1}{r^2} * \frac{\partial}{\partial{r}} (1) = 0 \)
Apply Divergence Theorem: \( \oint_S v \cdot \partial{a} = \int \frac{1}{r^2} \boldsymbol{\hat{r}} \cdot \partial{l_{\theta}} \partial{l_{\phi}\boldsymbol{\hat{r}}} = \int \frac{1}{r^2} * r^2 \sin{\theta} \partial{\theta} \partial{\phi} = \int_0^{\pi} \sin{\theta} \partial{\theta} \int_0^{2\pi} \partial{\phi} = 2* 2\pi = 4\pi\)
BUT \(\int \nabla \cdot v \partial{\tau} = 0?\)
We need a special mathematical object to represent situations such as mass of point particle. Mass of a point particle is finite, zero everywhere except at the location of the particle. This mathematical object is called Dirac Delta Function

Definition: \(\delta(x) = \left\{ \begin{array}{c} 0, \text{if}\ x \neq 0 \\ \infty , \text{if}\ x = 0 \end{array} \right. ,\ \int_{-\infty}^{\infty} \delta(x) \partial{x} = 1\)

For \(f(x)\) continuous \(\int_{-\infty}^{\infty} f(x) \delta(x) \partial{x} = f(0) \int_{-\infty}^{\infty} \delta(x) \partial{x} = f(0) \)

Shifted spike: \(\delta(x - a) = \left\{ \begin{array}{c} 0, \text{if}\ x \neq a \\ \infty , \text{if}\ x = a \end{array} \right. ,\ \text{with} \int_{-\infty}^{\infty} \delta(x - a) \partial{x} = 1\)

For \(f(x)\) continuous \(\int_{-\infty}^{\infty} f(x) \delta(x - a) \partial{x} = f(a) \)

Section 5.3: Three-Dimensional Dirac

Generalization: \( \delta^3(\boldsymbol{r}) = \delta(x)\delta(y)\delta(z)\), where \(\int \delta^3(\boldsymbol{r}) \partial{\tau} = 1\) and \(\int_V f(\boldsymbol{r}) \delta^3(\boldsymbol{r - a}) \partial{\tau} = f(\boldsymbol{a}) \)

Thus behavior of divergence of \(\frac{\hat{r}}{r^2}\) can be modeled by Dirac delta:
\(\nabla \cdot (\frac{\boldsymbol{\hat{r}}}{r^2}) = 4\pi * \delta^3(\boldsymbol{\overrightarrow{r}})\), more generally \(\nabla \cdot (\frac{\hat{\boldsymbol{r}}_{sep}}{\overrightarrow{r^2}_{sep}}) = 4\pi * \delta^3(\overrightarrow{\boldsymbol{r}}_{sep})\), where \(\overrightarrow{\boldsymbol{r}}_{sep}\ = \boldsymbol{r} - \boldsymbol{r'} \)

From Prob 1.13 \(\nabla(\frac{1}{\overrightarrow{r}_{sep}}) = \frac{\hat{\boldsymbol{r}}_{sep}}{\overrightarrow{r^2}_{sep}}\)
Thus \(\nabla^2(\frac{1}{\overrightarrow{r}_{sep}}) = 4\pi * \delta^3(\overrightarrow{\boldsymbol{r}}_{sep})\)

Relevant problems:

1.49: Evalute the volume integral \(J = \int_v e^{-r} (\nabla \cdot \frac{\boldsymbol{\hat{r}}}{r^2}) \partial{\tau}\) by two different methods:

Evalute the integral directly utilizing \(\nabla \cdot \frac{\boldsymbol{\hat{r}}}{r^2} \) result and using divergence theorem

Direct evaluation: \(J = \int_v e^{-r} (\nabla \cdot \frac{\boldsymbol{\hat{r}}}{r^2}) \partial{\tau} = \int_v e^{-r} * 4\pi \delta^3(\boldsymbol{\overrightarrow{r}}) \partial{\tau} = 4\pi * e^0 = \require{enclose}\enclose{box}[mathcolor="black"]{4\pi} \)
Using one of the product rules from Section 2: \(\nabla \cdot (f\boldsymbol{A}) = f(\nabla \cdot \boldsymbol{A}) + \boldsymbol{A} \cdot (\nabla{f})\)
In our case: \(\nabla \cdot (e^{-r}\frac{\boldsymbol{\hat{r}}}{r^2}) = e^{-r}(\nabla \cdot \frac{\boldsymbol{\hat{r}}}{r^2}) + \frac{\boldsymbol{\hat{r}}}{r^2} \cdot (\nabla{e^{-r}})\)
Rearrange: \(e^{-r}(\nabla \cdot \frac{\boldsymbol{\hat{r}}}{r^2}) = \nabla \cdot (e^{-r}\frac{\boldsymbol{\hat{r}}}{r^2}) - \frac{\boldsymbol{\hat{r}}}{r^2} \cdot (\nabla{e^{-r}})\)
\( \int_v e^{-r}(\nabla \cdot \frac{\boldsymbol{\hat{r}}}{r^2})\partial{\tau} = \int_v \nabla \cdot (e^{-r}\frac{\boldsymbol{\hat{r}}}{r^2}) \partial{\tau} - \int_v \frac{\boldsymbol{\hat{r}}}{r^2} \cdot (\nabla{e^{-r}}) \partial{\tau}\)
Simplify using divergence theorem: \( \int_v e^{-r}(\nabla \cdot \frac{\boldsymbol{\hat{r}}}{r^2})\partial{\tau} = \int_S e^{-r}\frac{\boldsymbol{\hat{r}}}{r^2} \partial{a} - \int_v \frac{\boldsymbol{\hat{r}}}{r^2} \cdot (\nabla{e^{-r}}) \partial{\tau}\)
Solve each integral separately:
Radius of an infinite shell is constant (r=R) \(\int_S e^{-R}\frac{\boldsymbol{\hat{r}}}{R^2} \cdot R^2 \sin{\theta}\partial{\theta}\partial{\phi}\boldsymbol{\hat{r}} = \int_S e^{-R} \sin{\theta}\partial{\theta}\partial{\phi} = 4\pi e^{-R} \)
To find \(\nabla{e^{-r}}\), we look at del operator in spherical coordinates. Our function is a function of r only, so we just take the derivative: \(\frac{\partial}{\partial{r}} e^{-r} = -e^{-r} \boldsymbol{\hat{r}}\)
Our integral has a negative in front, which "cancels" negative in front of the derivative. Now integral looks: \( \int_v \frac{\boldsymbol{\hat{r}}}{r^2} \cdot -e^{-r} \boldsymbol{\hat{r}} r^2 \sin{\theta} \partial{r}\partial{\theta}\partial{\phi} \)
Simplify further: \( \int_v -e^{-r} \sin{\theta} \partial{r}\partial{\theta}\partial{\phi} = 4\pi \int_0^R -e^{-r}\partial{r} = 4\pi (1 - e^{-R})\)
Combine results together: \( \int_v e^{-r}(\nabla \cdot \frac{\boldsymbol{\hat{r}}}{r^2})\partial{\tau} = 4\pi e^{-R} + 4\pi (1 - e^{-R}) = \enclose{box}[mathcolor="black"]{4\pi}\)
Results match.