Check that the retarded potentials of an oscillating dipole (Eqs. 11.12 and 11.17) satisfy the Lorenz gauge condition. Do not use approximation 3.

Write the expression for the Lorenz gauge condition.

$\mathbf{\nabla }\mathbf{·}\mathit{A}\mathbf{=}\mathbf{-}{\mathit{\mu }}_{\mathbf{0}}{\mathit{\epsilon }}_{\mathbf{0}}\left(\frac{\partial V}{\partial t}\right)$ …… (1)

Here,${\mu }_0$ is the magnetic permeability${\epsilon }_0$ is the magnetic permittivity, A is the vector potential, and V is the scalar potential.

Write the expression for the vector potential (using equation 11.17 ).

$$\begin{gathered} A=-\frac{{\mu }_0{p}_0\omega }{4\pi }\frac{1}{r}\sin \left(\omega \left(t-\frac{r}{c}\right)\right)\hat{z} \\ A=-\frac{{\mu }_0{p}_0\omega }{4\pi }\frac{1}{r}\sin \left(\omega \left(t-\frac{r}{c}\right)\right)\left(\cos \theta \hat{r}-\sin \theta \hat{\theta }\right) \end{gathered}$$

Calculate the value of $\nabla ·A$.

$$\begin{gathered} \nabla ·A=\frac{1}{r^2}\frac{\partial }{\partial r}\left(r^2{A}_r\right)+\frac{1}{r\sin \theta }\frac{\partial }{\partial \theta }\left(\sin \theta A_{\theta }\right)+\frac{1}{r\sin \theta }\frac{\partial \varphi }{\partial \varphi } \\ \nabla ·A=\frac{1}{r^2}\frac{\partial }{\partial r}\left[\begin{array}{c}{r}^2\left[\left(-\frac{{\mu }_0{p}_0\omega }{4\pi }\right)\frac{1}{r}\sin \left(\omega \left(t-\frac{r}{c}\right)\right)\cos \theta \right]\\ \frac{1}{r\sin \theta }\frac{\partial }{\partial \theta }-\left[\left(\frac{{\mu }_0{p}_0\omega }{4\pi }\right)\frac{1}{r}\sin \left(\omega \left(t-\frac{r}{c}\right)\right)\left(-{\sin }^2\theta \right)\right]\end{array}+\right] \\ \nabla ·A=-\frac{{\mu }_0{p}_0\omega }{4\pi }\left[\begin{array}{c}\frac{1}{r^2}\frac{\partial }{\partial r}\frac{1}{r}{r}^2\sin \left(\omega \left(t-\frac{r}{c}\right)\right)\cos \theta -\left[\frac{\omega }{rc}\cos \left(\omega t-\frac{\omega r}{c}\right)\cos \theta \right]-\\ \frac{2\sin \theta \cos \theta }{r^3\sin \theta }\sin \left(\omega t-\frac{\omega r}{c}\right)\end{array}\right] \\ \nabla ·A=-\frac{{\mu }_0{p}_0\omega }{4\pi }\left[\frac{1}{r^2}\sin \left(\omega t-\frac{\omega r}{c}\right)\frac{\omega }{rc}\cos \left(\omega t-\frac{\omega r}{c}\right)-\frac{2}{r^2}\sin \left(\omega t-\frac{\omega r}{c}\right)\right]\cos \theta \end{gathered}$$

On further solving, the above equation becomes,

localid="1653907297258" $$\begin{gathered} \nabla ·A=-\frac{{\mu }_0{p}_0\omega }{4\pi }\left[\frac{\left(2-1\right)}{r^2}\sin \left(\omega t-\frac{\omega r}{c}\right)+\frac{\omega }{rc}\cos \left(\omega t-\frac{\omega r}{c}\right)\right]\cos \theta \\ \nabla ·A=-{\mu }_0\omega \left[\frac{p_0\omega }{4\pi {\epsilon }_0}\frac{1}{r^2}\sin \omega \left(t-\frac{r}{c}\right)+\frac{\omega }{rc}\cos \omega \left(\omega -\frac{r}{c}\right)\right]\cos \theta ....\left(1\right) \end{gathered}$$

Write the expression for the scalar potential for an oscillating dipole potential (using equation 11.12 ).

$V=\frac{p_0\cos \theta }{4\pi {\epsilon }_{0}r}\left[-\frac{\omega }{c}\sin \left[\omega \left(t-\frac{r}{c}\right)\right]+\frac{1}{r}\cos \omega \left(t-\frac{r}{c}\right)\right]$

Calculate the value of $\frac{\partial V}{\partial t}$.

$$\begin{gathered} \frac{\partial V}{\partial t}=\frac{p_0\cos \theta }{4\pi {\epsilon }_{0}r}\left[-\frac{{\omega }^2}{c}\cos \omega \left(t-\frac{r}{c}\right)-\frac{\omega }{r}\sin \omega \left(t-\frac{r}{c}\right)\right] \\ \frac{\partial V}{\partial t}=\frac{p_0\omega }{4\pi {\epsilon }_0}\left[\frac{1}{r^2}\sin \omega \left(t-\frac{r}{c}\right)+\frac{\omega }{rc}\cos \omega \left(t-\frac{r}{c}\right)\right]\cos \theta .......\left(2\right) \end{gathered}$$

From equations (1) and (2),

$\nabla ·A=-{\mu }_0\omega \frac{\partial V}{\partial t}$

Therefore, the retarded potentials of an oscillating dipole satisfy the Lorenz gauge condition.