91影视

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

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

Take the gradient of A.

$$\begin{gathered} 鈭�.\mathit{A}=\left(\hat{x}\frac{鈭�}{鈭�x}+\hat{y}\frac{鈭�}{鈭�y}+\frac{鈭�}{鈭�z}\right)鈥�\frac{{渭}_{0}k}{4c}\left(ct-{\left|x\right|}^2\right)\hat{z} \\ 鈭�.\mathit{A}=\frac{鈭�}{鈭�z}\left(\frac{{渭}_{0}k}{4c}\right)\left(ct-{\left|x\right|}^2\right) \\ 鈭�.\mathit{A}=0 \end{gathered}$$

Take the derivative of V.

$\frac{鈭�V}{鈭�t}=0$

Hence,$鈭�.\mathit{A}={渭}_0{蔚}_0\frac{鈭�V}{鈭�t}$

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

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

Take the gradient of A.

localid="1655873040623" role="math" $$\begin{gathered} 鈭�.\mathit{A}=-\frac{qt}{4蟺{蔚}_0}鈭�.\left(\frac{\hat{r}}{r^2}\right) \\ 鈭�.\mathit{A}=-\frac{qt}{{蔚}_0}{蟽}^3\left(r\right)=0 \end{gathered}$$

Take the derivative of V.

$\frac{鈭�V}{鈭�t}=0$

Hence,$鈭�.\mathit{A}={渭}_0{蔚}_0\frac{鈭�V}{鈭�t}$

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

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

Take the gradient of A.

$$\begin{gathered} 鈭�.\mathit{A}=\left(\hat{x}\frac{鈭�}{鈭�x}+\hat{y}\frac{鈭�}{鈭�y}+\hat{z}\frac{鈭�}{鈭�z}\right)鈥�A_0\sin \left(kx-蝇t\right)\hat{y} \\ 鈭�.\mathit{A}=\frac{鈭�}{鈭�y}{A}_0\sin \left(kx-蝇t\right)\hat{y} \\ 鈭�.\mathit{A}=0 \end{gathered}$$

Take the derivative of V.

$\frac{鈭�V}{鈭�t}=0$

Hence,$鈭�.\mathit{A}={渭}_0{蔚}_0\frac{鈭�V}{鈭�t}$

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$鈭�.\mathit{A}={渭}_0{蔚}_0\frac{鈭�V}{鈭�t}$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 $A=0$.