In Chapter 5, I showed that it is always possible to pick a vector potential whose divergence is zero (the Coulomb gauge). Show that it is always possible to choose$\mathbf{\nabla }\mathbf{.}\mathit{A}\mathbf{=}\mathbf{-}{\mathit{\mu }}_{\mathbf{0}}{\mathit{\epsilon }}_{\mathbf{0}}\left(\partial V/\partial t\right)$, as required for the Lorenz gauge, assuming you know how to solve the inhomogeneous wave equation (Eq. 10.16). Is it always possible to pick$\mathbf{V}\mathbf{=}\mathbf{0}$ ? How about$\mathbf{A}\mathbf{}\mathbf{=}\mathbf{0}$ ?

Write the expression for the potential V and the constant A.

$V=0,A=\left\{\begin{array}{l}\frac{{\mu }_{0}k}{4c}\left(ct-{\left|X\right|}^2\right)\hat{Z}for\left|X\right|<ct\\ 0,for\left|X\right|>ct\end{array}\right.$

Take the gradient of A.

$$\begin{gathered} \nabla .\mathit{A}=\left(\hat{x}\frac{\partial }{\partial x}+\hat{y}\frac{\partial }{\partial y}+\frac{\partial }{\partial z}\right)‐\frac{{\mu }_{0}k}{4c}\left(ct-{\left|x\right|}^2\right)\hat{z} \\ \nabla .\mathit{A}=\frac{\partial }{\partial z}\left(\frac{{\mu }_{0}k}{4c}\right)\left(ct-{\left|x\right|}^2\right) \\ \nabla .\mathit{A}=0 \end{gathered}$$

Take the derivative of V.

$\frac{\partial V}{\partial t}=0$

Hence,$\nabla .\mathit{A}={\mu }_0{\epsilon }_0\frac{\partial V}{\partial t}$

So, it can be observed that both the scalar and potentials are in the coulomb and in the Lorentz gauge.

Suppose$V=0,\mathit{A}=-\frac{1}{4\pi {\epsilon }_0{r}^2}{r}^2$

Take the gradient of A.

localid="1655873040623" role="math" $$\begin{gathered} \nabla .\mathit{A}=-\frac{qt}{4\pi {\epsilon }_0}\nabla .\left(\frac{\hat{r}}{r^2}\right) \\ \nabla .\mathit{A}=-\frac{qt}{{\epsilon }_0}{\sigma }^3\left(r\right)=0 \end{gathered}$$

Take the derivative of V.

$\frac{\partial V}{\partial t}=0$

Hence,$\nabla .\mathit{A}={\mu }_0{\epsilon }_0\frac{\partial V}{\partial t}$

So, it can be observed that both the scalar and vector potentials are not in coulomb and in Lorentz gauge.

Similarly, consider$V=,\mathit{A}=A_0\sin \left(kx-\omega t\right)y$

Take the gradient of A.

$$\begin{gathered} \nabla .\mathit{A}=\left(\hat{x}\frac{\partial }{\partial x}+\hat{y}\frac{\partial }{\partial y}+\hat{z}\frac{\partial }{\partial z}\right)‐A_0\sin \left(kx-\omega t\right)\hat{y} \\ \nabla .\mathit{A}=\frac{\partial }{\partial y}{A}_0\sin \left(kx-\omega t\right)\hat{y} \\ \nabla .\mathit{A}=0 \end{gathered}$$

Take the derivative of V.

$\frac{\partial V}{\partial t}=0$

Hence,$\nabla .\mathit{A}={\mu }_0{\epsilon }_0\frac{\partial V}{\partial t}$

So, it can be observed that both the scalar and vector potentials are not in coulomb and in Lorentz gauge.

Yes, it is always possible to pick V=0 but cannot pick A=0 because that would make B=0.

Therefore, it is not always possible to choose$\nabla .\mathit{A}={\mu }_0{\epsilon }_0\frac{\partial V}{\partial t}$for the Lorentz gauge. For the case of the scalar potential, it is always possible to pick$V=0$, and for the vector potential, it is not possible to pick $A=0$.