Q64P - Introduction to Electrodynamics

Question

In case you're not persuaded that ${\mathbf{\nabla }}^{\mathbf{2}}\left(\frac{1}{r}\right)\mathbf{=}\mathbf{-}\mathbf{4}\mathit{\pi }{\mathit{\delta }}^{\mathbf{3}}\mathbf{\left(}\mathbf{r}\mathbf{\right)}$ (Eq. 1.102) with ${\mathbf{r}}^{\mathbf{\text{'}}}\mathbf{=}\mathbf{0}$ for simplicity), try replacing r  by  $\sqrt{{\mathbf{r}}^{\mathbf{2}}\mathbf{+}{\mathbf{\epsilon }}^{\mathbf{2}}}$ , and watching what happens as $\mathit{\epsilon }\mathbf{\to }{\mathbf{0}}^{\mathbf{16}}$ Specifically, let                                              $\mathbf{D}\mathbf{(}\mathbf{r}\mathbf{,}\mathbf{\epsilon }\mathbf{)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}\mathbf{\pi }}{\mathbf{\nabla }}^{\mathbf{2}}\frac{\mathbf{1}}{\sqrt{{\mathbf{r}}^{\mathbf{2}}\mathbf{+}{\mathbf{\epsilon }}^{\mathbf{2}}}}$

To demonstrate that this goes to ${\mathit{\delta }}^{\mathbf{3}}\mathbf{\left(}\mathit{r}\mathbf{\right)}$ as $\mathit{\epsilon }\mathbf{\to }\mathbf{0}$:

(a) Show that $\mathbf{D}\mathbf{=}\mathbf{(}\mathbf{r}\mathbf{,}\mathbf{\epsilon }\mathbf{)}\mathbf{=}\mathbf{(}\mathbf{3}{\mathbf{\epsilon }}^{\mathbf{2}}\mathbf{/}\mathbf{4}\mathbf{\pi }\mathbf{)}{\mathbf{(}{\mathbf{r}}^{\mathbf{2}}\mathbf{+}{\mathbf{\epsilon }}^{\mathbf{2}}\mathbf{)}}^{\mathbf{-}\mathbf{5}\mathbf{/}\mathbf{2}}$

(b) Check that$\mathbf{D}\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{\epsilon }\mathbf{)}\mathbf{\to }\mathbf{\infty }$ , as $\mathit{\epsilon }\mathbf{\to }\mathbf{0}$

(c)Check that $\mathbf{D}\mathbf{(}\mathbf{r}\mathbf{,}\mathbf{\epsilon }\mathbf{)}\mathbf{\to }\mathbf{0}$ , as $\mathbf{\epsilon }\mathbf{\to }\mathbf{0}$, for all $\mathbf{r}\mathbf{\ne }\mathbf{0}$

(d) Check that the integral of $\mathbf{D}\mathbf{(}\mathbf{r}\mathbf{,}\mathbf{\epsilon }\mathbf{)}$ over all space is 1.

Step-by-Step Solution

Answer

(a)The equation,$\mathrm{D}(\mathrm{r},\mathrm{\epsilon })=\frac{3{\mathrm{\epsilon }}^2}{4\mathrm{\pi }}{\left({\mathrm{r}}^2+{\mathrm{\epsilon }}^2\right)}^{-5/2}$ in part (a) is proved $\nabla .\mathrm{v}=(\mathrm{n}+2){\mathrm{r}}^{\mathrm{n}-1}$

(b)It is proved that as $\mathrm{\epsilon }\to 0, \mathrm{D}(0,\mathrm{\epsilon })\to \infty$

(c)It is proved that as $\mathrm{\epsilon }\to 0. \mathrm{D}(r,\mathrm{\epsilon })\to 0$.

(d)It is proved that the integral of the function $\mathrm{D}(\mathrm{r},\mathrm{\epsilon })$ over all spaces is equal to 1.

Step 1: Describe the given information

It is given that $\mathrm{D}(\mathrm{r},\mathrm{\epsilon })=\frac{1}{4\mathrm{\pi }}{\nabla }^2\frac{1}{\sqrt{{\mathrm{r}}^2+{\mathrm{\epsilon }}^2}}$ and the equation $\mathrm{D}(\mathrm{r},\mathrm{\epsilon })=\frac{3{\mathrm{\epsilon }}^2}{4\mathrm{\pi }}{\left({\mathrm{r}}^2+{\mathrm{\epsilon }}^2\right)}^{-5/2}$  have to be verified. It has to be proved that as$\mathrm{\epsilon }\to 0, \mathrm{D}(0,\mathrm{\epsilon })\to \infty$as $\mathrm{\epsilon }\to 0. \mathrm{D}(r,\mathrm{\epsilon })\to 0$  .and the integral of the function $\mathrm{D}(\mathrm{r},\mathrm{\epsilon })$  over all spaces is equal to 1.

Step 2: Define the Laplacian operator in spherical coordinates

It is given that $\mathbf{D}\mathbf{(}\mathbf{r}\mathbf{,}\mathbf{\epsilon }\mathbf{)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}\mathbf{\pi }}{\mathbf{\nabla }}^{\mathbf{2}}\frac{\mathbf{1}}{\sqrt{{\mathbf{r}}^{\mathbf{2}}\mathbf{+}{\mathbf{\epsilon }}^{\mathbf{2}}}}$ . The Laplacian operator with respect to r is simplified as

 $$\begin{gathered} {{\mathbf{\nabla }}_{\mathbf{2}}}^{\mathbf{r}}\mathbf{=}\frac{\mathbf{1}}{{\mathbf{r}}^{\mathbf{2}}}\frac{\mathbf{\partial }}{\mathbf{\partial }\mathbf{r}}\left(r\frac{\partial }{\partial r}\right) \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\frac{{\mathbf{\partial }}^{\mathbf{2}}}{\mathbf{\partial }{\mathbf{r}}^{\mathbf{2}}}\mathbf{+}\frac{\mathbf{2}}{\mathbf{r}}\frac{\mathbf{\partial }}{\mathbf{\partial }\mathbf{r}} \end{gathered}$$ 

Apply the expression $\frac{1}{\sqrt{{\mathrm{r}}^2+{\mathrm{\epsilon }}^2}}$ to the above simplified Laplacian operator as,

$$\begin{gathered} {\nabla }^2\left(\frac{1}{\sqrt{{\mathrm{r}}^2+{\mathrm{\epsilon }}^2}}\right)=\left[\frac{{\partial }^2}{\partial {\mathrm{r}}^2}+\frac{2}{\mathrm{r}}\frac{\partial }{\partial \mathrm{r}}\right]\left(\frac{1}{\sqrt{{\mathrm{r}}^2+{\mathrm{\epsilon }}^2}}\right) \\ =\frac{\partial }{\partial {\mathrm{r}}^2}\left(\frac{1}{\sqrt{{\mathrm{r}}^2+{\mathrm{\epsilon }}^2}}\right)+\frac{2}{\mathrm{r}}\frac{\partial }{\partial \mathrm{r}}\left(\frac{1}{\sqrt{{\mathrm{r}}^2+{\mathrm{\epsilon }}^2}}\right) \\ =\left\{-{\left({\mathrm{r}}^2+{\mathrm{\epsilon }}^2\right)}^{-3/2}+\frac{3\mathrm{r}}{2}{\left({\mathrm{r}}^2+{\mathrm{\epsilon }}^2\right)}^{-5/2}.\left(2\mathrm{r}\right)+\frac{2}{\mathrm{r}}\left(\begin{array}{c}\left(-\frac{1}{2}{\left({\mathrm{r}}^2+{\mathrm{\epsilon }}^2\right)}^{-3/2}.\left(2\mathrm{r}\right)\right)\\ -\mathrm{r} {({\mathrm{r}}^2+{\mathrm{\epsilon }}^2)}^{-3/2}\end{array}\right)\right\} \\ =-{\left({\mathrm{r}}^2+{\mathrm{\epsilon }}^2\right)}^{-3/2}+3{\mathrm{r}}^2{\left({\mathrm{r}}^2+{\mathrm{\epsilon }}^2\right)}^{-5/2}+\frac{2}{\mathrm{r}}\left(-\mathrm{r} {({\mathrm{r}}^2+{\mathrm{\epsilon }}^2)}^{-3/2}-\mathrm{r} {({\mathrm{r}}^2+{\mathrm{\epsilon }}^2)}^{-3/2}\right) \end{gathered}$$

Simplify further as

$$\begin{gathered} {\nabla }^2=\left(\frac{1}{\sqrt{r^2+{\epsilon }^2}}\right)=-{\left({\mathrm{r}}^2+{\mathrm{\epsilon }}^2\right)}^{-3/2}+3{\mathrm{r}}^2{\left({\mathrm{r}}^2+{\mathrm{\epsilon }}^2\right)}^{-5/2}+\frac{2}{\mathrm{r}}\left(-2\mathrm{r} {({\mathrm{r}}^2+{\mathrm{\epsilon }}^2)}^{-3/2}\right) \\ =-{\left({\mathrm{r}}^2+{\mathrm{\epsilon }}^2\right)}^{-3/2}+3{\mathrm{r}}^2{\left({\mathrm{r}}^2+{\mathrm{\epsilon }}^2\right)}^{-5/2}-4\mathrm{r} {({\mathrm{r}}^2+{\mathrm{\epsilon }}^2)}^{-3/2} \\ =-5{\left({\mathrm{r}}^2+{\mathrm{\epsilon }}^2\right)}^{-3/2}+3{\mathrm{r}}^2{\left({\mathrm{r}}^2+{\mathrm{\epsilon }}^2\right)}^{-5/2} \end{gathered}$$

Step: 3 Verify the equation in part (a)

Substitute $-5{\left({\mathrm{r}}^2+{\mathrm{\epsilon }}^2\right)}^{-3/2}+3{\mathrm{r}}^2{\left({\mathrm{r}}^2+{\mathrm{\epsilon }}^2\right)}^{-5/2}$  for ${\nabla }^2=\left(\frac{1}{\sqrt{r^2+{\epsilon }^2}}\right)$into.

$$\begin{gathered} \mathrm{D}(\mathrm{r},\mathrm{\epsilon })=\frac{1}{4\mathrm{\pi }}{\nabla }^2\frac{1}{\sqrt{{\mathrm{r}}^2+{\mathrm{\epsilon }}^2}} \\ \mathrm{D}(\mathrm{r},\mathrm{\epsilon })=-\frac{1}{4\pi }\left(-5{\left({\mathrm{r}}^2+{\mathrm{\epsilon }}^2\right)}^{-3/2}+3{\mathrm{r}}^2{\left({\mathrm{r}}^2+{\mathrm{\epsilon }}^2\right)}^{-5/2}\right) \\ =\frac{3{\mathrm{\epsilon }}^2}{4\mathrm{\pi }}{\left({\mathrm{r}}^2+{\mathrm{\epsilon }}^2\right)}^{-5/2} \end{gathered}$$

 

Thus, the equation $\mathrm{D}(\mathrm{r},\mathrm{\epsilon })=\frac{3{\mathrm{\epsilon }}^2}{4\mathrm{\pi }}{\left({\mathrm{r}}^2+{\mathrm{\epsilon }}^2\right)}^{-5/2}$, in part (a) is proved.

Step 4: Verify the equation in part (b)

From the result of part (a)$\mathrm{D}(\mathrm{r},\mathrm{\epsilon })=\frac{3{\mathrm{\epsilon }}^2}{4\mathrm{\pi }}{\left({\mathrm{r}}^2+{\mathrm{\epsilon }}^2\right)}^{-5/2}$, .

Substitute 0 for into equation $\mathrm{D}(\mathrm{r},\mathrm{\epsilon })=\frac{3{\mathrm{\epsilon }}^2}{4\mathrm{\pi }}{\left({\mathrm{r}}^2+{\mathrm{\epsilon }}^2\right)}^{-5/2}$ .

 $$\begin{gathered} \mathrm{D}(\mathrm{r},\mathrm{\epsilon })=\frac{3{\mathrm{\epsilon }}^2}{4\mathrm{\pi }}{\left({\left(0\right)}^2+{\mathrm{\epsilon }}^2\right)}^{-5/2} \\ =\frac{3{\mathrm{\epsilon }}^2}{4\mathrm{\pi }}{\left({\mathrm{\epsilon }}^2\right)}^{-5/2} \\ =\frac{3{\mathrm{\epsilon }}^2{\mathrm{\epsilon }}^{-5}}{4\mathrm{\pi }} \\ =\frac{3{\mathrm{\epsilon }}^{-3}}{4\mathrm{\pi }} \end{gathered}$$

Simply further as, 

 $D(0,\epsilon )=\frac{3}{4\pi {\epsilon }^3}$

Substitute 0 for $\epsilon$, into equation $D(0,\epsilon )=\frac{3}{4\pi {\epsilon }^3}$

$$\begin{gathered} D(0,\epsilon \to 0)=\frac{3}{4\pi {\left(0\right)}^3} \\ =\infty \end{gathered}$$

 Thus, it is proved that as $\epsilon \to 0.D(0,\epsilon )\to \infty$.

Step 5: Verify the equation in part (c)

From the result of part (a),$\mathrm{D}(\mathrm{r},\mathrm{\epsilon })=\frac{3{\mathrm{\epsilon }}^2}{4\mathrm{\pi }}{\left({\mathrm{r}}^2+{\mathrm{\epsilon }}^2\right)}^{-5/2}$ 

 

Substitute 0 for $\epsilon$, into equation$\mathrm{D}(\mathrm{r},\mathrm{\epsilon })=\frac{3{\mathrm{\epsilon }}^2}{4\mathrm{\pi }}{\left({\mathrm{r}}^2+{\mathrm{\epsilon }}^2\right)}^{-5/2}$ 

$$\begin{gathered} \mathrm{D}(\mathrm{r},\mathrm{\epsilon })=\frac{3{\left(0\right)}^2}{4\mathrm{\pi }}{\left({\mathrm{r}}^2+{\left(0\right)}^2\right)}^{-5/2} \\ =\frac{3{\left(0\right)}^2}{4\mathrm{\pi }}{\left({\mathrm{r}}^2\right)}^{-5/2} \\ =0 \end{gathered}$$

 

Thus, it is proved that as $\epsilon \to 0.D(r,\epsilon )\to 0$ .

Step 6: Verify the statement in part (d)

It is given that  $\mathrm{D}(\mathrm{r},\mathrm{\epsilon })\approx {\mathrm{\zeta }}^3\left(\mathrm{r}\right)$ as $\epsilon \to 0$ . For all spaces the value of r ranges from $\infty$to$-\infty$ . Integrate the function $\mathrm{D}(\mathrm{r},\mathrm{\epsilon })\approx {\mathrm{\zeta }}^3\left(\mathrm{r}\right)$ over all values of r as, 

$$\begin{gathered} \int \mathrm{D}(\mathrm{r},\mathrm{\epsilon })\mathrm{dr}=\underset{-\infty }{\overset{\infty }{\int }} {\zeta }^3 \left(r\right)dr \\ =\underset{-\infty }{\overset{\infty }{\int }} \mathrm{\zeta }( \mathrm{r}) . \mathrm{\zeta }\left(\mathrm{r}\right).\mathrm{\zeta }\left(\mathrm{r}\right).\mathrm{dr} \end{gathered}$$

The multiplication of  $\mathrm{\zeta }( \mathrm{r})$with itself, any number of times, gives the delta function, $\mathrm{\zeta }( \mathrm{r})$ . Thus above integral becomes,

$$\begin{gathered} \int \mathrm{D}(\mathrm{r},\mathrm{\epsilon })\mathrm{dr}=\underset{-\infty }{\overset{\infty }{\int }} \mathrm{\zeta }\left(\mathrm{r}\right).\mathrm{\zeta }\left(\mathrm{r}\right).\mathrm{\zeta }\left(\mathrm{r}\right).\mathrm{dr} \\ =\underset{-\infty }{\overset{\infty }{\int }} \mathrm{\zeta }( \mathrm{r})\mathrm{dr} \\ =1 \end{gathered}$$

 

Thus, it is proved that the integral of the function $D(r,\epsilon )$ over all spaces is equal to 1.