Question

Show that the differential equations for V and A (Eqs. 10.4 and 10.5) can be written in the more symmetrical form

$\begin{array}{r}\begin{array}{cc}{𝆏}^{2}V+\begin{array}{r}\partial L\\ \partial t\end{array}=-\frac{1}{{\epsilon }_{\circ }}p& \\ {𝆏}^{2}A-\nabla L=-{\mu }_{\circ }J& \end{array}\\ \end{array}\}$

Where

${\mathbf{𝆏}}^{\mathbf{2}}\mathbf{\equiv }{\mathbf{\nabla }}^{\mathbf{2}}\mathbf{-}{\mathit{\mu }}_{\mathbf{\circ }}{\mathit{\epsilon }}_{\mathbf{\circ }}\frac{{\mathbf{\partial }}^{\mathbf{2}}}{\mathbf{\partial }{\mathbf{t}}^{\mathbf{2}}}\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{}\mathit{L}\mathbf{\equiv }\mathbf{\nabla }\mathbf{.}\mathit{A}\mathbf{+}{\mathit{\mu }}_{\mathbf{\circ }}{\mathit{\epsilon }}_{\mathbf{\circ }}\frac{\mathbf{\partial }\mathbf{V}}{\mathbf{\partial }\mathbf{t}}$

Short answer

The differential equations for V and Ain the symmetrical form are derived as

${\nabla }^{2}V+\frac{\partial }{\partial t}\left(\nabla .\mathit{A}\right)=-\frac{1}{{\epsilon }_{\circ }}pand\left({\nabla }^{2}A-{\mu }_{\circ }{\epsilon }_{\circ }\frac{{\partial }^2\mathit{A}}{\partial t}\right)=-{\mu }_{\circ }\mathit{J}$

Step-by-step solution

Expression for the differential equations for V and A:

Using equation 10.6, write the differential equation for V.

$𝆏{\nabla }^{2}V+\frac{\partial }{\partial t}=-\frac{1}{{\epsilon }_{\circ }}p$ …… (1)

Here.$𝆏$ is d’ Alembertian.

Similarly, write the differential equation for A.

${𝆏}^2\mathit{A}\mathbf{-}\nabla L=-{\mu }_{\circ }\mathit{J}$

Determine the differential equations for V and A in the symmetric form:

Substitute ${𝆏}^2={\nabla }^2-{\mu }_{\circ }{\epsilon }_{\circ }\frac{{\partial }^2}{\partial t^2}\mathbf{and}\mathbf{}\mathrm{L}=\nabla .\mathbf{A}+{\mathrm{\mu }}_{\circ }{\mathrm{\epsilon }}_{\circ }\frac{\partial \mathrm{V}}{\partial \mathrm{t}}$and in equation (1).

$$\begin{gathered} \left({𝆏}^2-{\mu }_{\circ }{\epsilon }_{\circ }\frac{{\partial }^2}{\partial t^2}\right)V+\frac{\partial }{\partial t}\left(\mathbf{\nabla }\mathbf{.}\mathbf{A}+{\mathrm{\mu }}_{\circ }{\mathrm{\epsilon }}_{\circ }\frac{\partial \mathrm{V}}{\partial \mathrm{t}}\right)=-\frac{1}{{\mathrm{\epsilon }}_{\circ }}p \\ {\nabla }^{2}V-{\mu }_{\circ }{\epsilon }_{\circ }\frac{{\partial }^{2}V}{\partial t^2}+\frac{\partial }{\partial t}\left(\nabla .\mathbf{A}\right)+{\mathrm{\mu }}_{\circ }{\mathrm{\epsilon }}_{\circ }\frac{{\partial }^2\mathrm{V}}{\partial {\mathrm{t}}^2}=-\frac{1}{{\mathrm{\epsilon }}_{\circ }}\mathrm{p} \\ {\nabla }^2\mathrm{V}+\frac{\partial }{\partial \mathrm{t}}\left(\nabla .\mathbf{A}\right)=-\frac{1}{{\mathrm{\epsilon }}_{\circ }}\mathrm{p} \end{gathered}$$

Which is equal to the equation 10.4 as${\nabla }^2\mathrm{V}+\frac{\partial }{\partial \mathrm{t}}\left(\nabla .\mathbf{A}\right)=-\frac{1}{{\mathrm{\epsilon }}_{\circ }}\mathrm{p}$.

Substitute${𝆏}^2={\nabla }^2-{\mu }_{\circ }{\epsilon }_{\circ }\frac{{\partial }^2}{\partial t^2}\mathbf{and}\mathbf{}\mathrm{L}=\nabla .\mathbf{A}+{\mathrm{\mu }}_{\circ }{\mathrm{\epsilon }}_{\circ }\frac{\partial \mathrm{V}}{\partial \mathrm{t}}$in equation (2).

$$\begin{gathered} \left({\nabla }^2-{\mu }_{\circ }{\epsilon }_{\circ }\frac{{\partial }^2}{\partial t^2}\right)\mathit{A}-\nabla \left(\mathbf{\nabla }\mathbf{.}\mathbf{A}+{\mathrm{\mu }}_{\circ }{\mathrm{\epsilon }}_{\circ }\frac{\partial \mathrm{V}}{\partial \mathrm{t}}\right)=-{\mu }_{\circ }\mathit{J} \\ \left({\nabla }^2\mathit{A}-{\mu }_{\circ }{\epsilon }_{\circ }\frac{{\partial }^2\mathit{A}}{\partial t^2}\right)-\nabla \left(\nabla .\mathit{A}+{\mathrm{\mu }}_{\circ }{\mathrm{\epsilon }}_{\circ }\frac{\partial \mathrm{V}}{\partial \mathrm{t}}\right)=-{\mu }_{\circ }\mathit{J} \end{gathered}$$

Which is equal to the equation 10.5 as .

$\left({\nabla }^2\mathit{A}-{\mu }_{\circ }{\epsilon }_{\circ }\frac{{\partial }^2\mathit{A}}{\partial t^2}\right)-\nabla \left(\nabla .\mathit{A}+{\mathrm{\mu }}_{\circ }{\mathrm{\epsilon }}_{\circ }\frac{\partial \mathrm{V}}{\partial \mathrm{t}}\right)=-{\mu }_{\circ }\mathit{J}$

Therefore, the differential equations for V and Ain the symmetrical form are derived as ${\nabla }^{2}V+\frac{\partial }{\partial t}\left(\nabla .\mathit{A}\right)=-\frac{1}{{\epsilon }_{\circ }}p$and .

$\left({\nabla }^2\mathit{A}-{\mu }_{\circ }{\epsilon }_{\circ }\frac{{\partial }^2\mathit{A}}{\partial t^2}\right)-\nabla \left(\nabla .\mathit{A}+{\mathrm{\mu }}_{\circ }{\mathrm{\epsilon }}_{\circ }\frac{\partial \mathrm{V}}{\partial \mathrm{t}}\right)=-{\mu }_{\circ }\mathit{J}$