91Ó°ÊÓ

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

It is given that $\mathbf{D}\mathbf{(}\mathbf{r}\mathbf{,}\mathbf{Î\mu }\mathbf{)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}\mathbf{Ï€}}{\mathbf{∇}}^{\mathbf{2}}\frac{\mathbf{1}}{\sqrt{{\mathbf{r}}^{\mathbf{2}}\mathbf{+}{\mathbf{Î\mu }}^{\mathbf{2}}}}$. The Laplacian operator with respect to r is simplified as

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

Apply the expression role="math" localid="1654687221342" $\frac{1}{\sqrt{{\mathrm{r}}^2+{\mathrm{Î\mu }}^2}}$to the above simplified Laplacian operator as,

role="math" localid="1654689997951" $$\begin{gathered} {∇}^2\left(\frac{1}{\sqrt{{\mathrm{r}}^2+{\mathrm{Î\mu }}^2}}\right)=\left[\frac{{∂}^2}{∂{\mathrm{r}}^2}+\frac{2}{\mathrm{r}}\frac{∂}{∂\mathrm{r}}\right]\left(\frac{1}{\sqrt{{\mathrm{r}}^2+{\mathrm{Î\mu }}^2}}\right) \\ =\frac{∂}{∂{\mathrm{r}}^2}\left(\frac{1}{\sqrt{{\mathrm{r}}^2+{\mathrm{Î\mu }}^2}}\right)+\frac{2}{\mathrm{r}}\frac{∂}{∂\mathrm{r}}\left(\frac{1}{\sqrt{{\mathrm{r}}^2+{\mathrm{Î\mu }}^2}}\right) \\ =\left\{-{\left({\mathrm{r}}^2+{\mathrm{Î\mu }}^2\right)}^{-3/2}+\frac{3\mathrm{r}}{2}{\left({\mathrm{r}}^2+{\mathrm{Î\mu }}^2\right)}^{-5/2}.\left(2\mathrm{r}\right)+\frac{2}{\mathrm{r}}\left(\begin{array}{c}\left(-\frac{1}{2}{\left({\mathrm{r}}^2+{\mathrm{Î\mu }}^2\right)}^{-3/2}.\left(2\mathrm{r}\right)\right)\\ -\mathrm{r}{({\mathrm{r}}^2+{\mathrm{Î\mu }}^2)}^{-3/2}\end{array}\right)\right\} \\ =-{\left({\mathrm{r}}^2+{\mathrm{Î\mu }}^2\right)}^{-3/2}+3{\mathrm{r}}^2{\left({\mathrm{r}}^2+{\mathrm{Î\mu }}^2\right)}^{-5/2}+\frac{2}{\mathrm{r}}\left(-\mathrm{r}{({\mathrm{r}}^2+{\mathrm{Î\mu }}^2)}^{-3/2}-\mathrm{r}{({\mathrm{r}}^2+{\mathrm{Î\mu }}^2)}^{-3/2}\right) \end{gathered}$$

Simplify further as

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

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

role="math" localid="1654690616425" $$\begin{gathered} \mathrm{D}(\mathrm{r},\mathrm{Î\mu })=\frac{1}{4\mathrm{Ï€}}{∇}^2\frac{1}{\sqrt{{\mathrm{r}}^2+{\mathrm{Î\mu }}^2}} \\ \mathrm{D}(\mathrm{r},\mathrm{Î\mu })=-\frac{1}{4\mathrm{Ï€}}\left(-5{\left({\mathrm{r}}^2+{\mathrm{Î\mu }}^2\right)}^{-3/2}+3{\mathrm{r}}^2{\left({\mathrm{r}}^2+{\mathrm{Î\mu }}^2\right)}^{-5/2}\right) \\ =\frac{3{\mathrm{Î\mu }}^2}{4\mathrm{Ï€}}{\left({\mathrm{r}}^2+{\mathrm{Î\mu }}^2\right)}^{-5/2} \end{gathered}$$

Thus, the equation role="math" localid="1654690728467" $\mathrm{D}(\mathrm{r},\mathrm{Î\mu })=\frac{3{\mathrm{Î\mu }}^2}{4\mathrm{Ï€}}{\left({\mathrm{r}}^2+{\mathrm{Î\mu }}^2\right)}^{-5/2}$, in part (a) is proved.

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

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

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

Simply further as,

localid="1654691126290" $D(0,\mathrm{Î\mu })=\frac{3}{4\mathrm{Ï€}{\mathrm{Î\mu }}^3}$

Substitute 0 for $\mathrm{Î\mu }$, into equation $D(0,\mathrm{Î\mu })=\frac{3}{4\mathrm{Ï€}{\mathrm{Î\mu }}^3}$

localid="1654691262144" $$\begin{gathered} D(0,\mathrm{Î\mu }→0)=\frac{3}{4\mathrm{Ï€}{\left(0\right)}^3} \\ =∞ \end{gathered}$$

Thus, it is proved that as localid="1654692695242" $\mathrm{Î\mu }→0.D(0,\mathrm{Î\mu })→∞$.

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

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

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

Thus, it is proved that as role="math" localid="1654691695727" $\mathrm{Î\mu }→0.D(r,\mathrm{Î\mu })→0$.

It is given that $\mathrm{D}(\mathrm{r},\mathrm{Î\mu })≈{\mathrm{Î\P }}^3\left(\mathrm{r}\right)$as$\mathrm{Î\mu }→0$ . For all spaces the value of r ranges from $∞$to$-∞$ . Integrate the function$\mathrm{D}(\mathrm{r},\mathrm{Î\mu })≈{\mathrm{Î\P }}^3\left(\mathrm{r}\right)$ over all values of r as,

$$\begin{gathered} ∫\mathrm{D}(\mathrm{r},\mathrm{Î\mu })\mathrm{dr}=\underset{-∞}{\overset{∞}{∫}}{\mathrm{Î\P }}^3\left(r\right)dr \\ =\underset{-∞}{\overset{∞}{∫}}\mathrm{Î\P }(\mathrm{r}).\mathrm{Î\P }\left(\mathrm{r}\right).\mathrm{Î\P }\left(\mathrm{r}\right).\mathrm{dr} \end{gathered}$$

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

role="math" localid="1654692146367" $$\begin{gathered} ∫\mathrm{D}(\mathrm{r},\mathrm{Î\mu })\mathrm{dr}=\underset{-∞}{\overset{∞}{∫}}\mathrm{Î\P }\left(\mathrm{r}\right).\mathrm{Î\P }\left(\mathrm{r}\right).\mathrm{Î\P }\left(\mathrm{r}\right).\mathrm{dr} \\ =\underset{-∞}{\overset{∞}{∫}}\mathrm{Î\P }(\mathrm{r})\mathrm{dr} \\ =1 \end{gathered}$$

Thus, it is proved that the integral of the function $D(r,\mathrm{Î\mu })$over all spaces is equal to 1.