Which of the potentials in Ex. 10.1, Prob. 10.3, and Prob. 10.4 are in the Coulomb gauge? Which are in the Lorenz gauge? (Notice that these gauges are not mutually exclusive.)

Write the condition for the potential in the Coulomb gauge.

$\mathbf{\nabla }\mathbf{·}\mathit{A}\mathbf{=}\mathbf{0}$ …… (1)

Here, A is the potential corresponding to the Coulomb gauge.

Write the condition for the potential in the Lorentz gauge.

$\mathbf{\nabla }\mathbf{·}\mathit{A}\mathbf{=}\mathbf{-}{\mathit{\mu }}_{\mathbf{0}}{\mathit{\epsilon }}_{\mathbf{0}}\frac{\mathbf{d}\mathbf{V}}{\mathbf{d}\mathbf{t}}$ …… (2)

Here, V is the potential corresponding to the Lorentz gauge, ${\mu }_0$ is the magnetic permeability and ${\epsilon }_0$is the electrical permittivity of free space.

(a)

Refer to problem 10.1 to obtain the potential as,

$V=0$

Write the general expression for the potential corresponding to the Coulomb gauge.

$A=\frac{{\mu }_{0}k}{4c}\left(ct-x^2\right)$ …… (3)

Here, k is a constant, c is the speed of light, t is the time and x is a point in a space along the x-axis.

The above potential is valid for,$x<ct$.

And for$x>ct$.

Rewrite the equation (3) as,

$A=0$

Write the expression for the divergence of potential along the x-axis.

$\nabla ·A=\left(\hat{x}\frac{d}{dx}+\hat{y}\frac{d}{dy}+\hat{z}\frac{d}{dz}\right)\frac{{\mu }_{0}k}{4c}\left(ct-x^2\right)$ …… (4)

Solve equation (4) to find the value of $\nabla A$for $x<ct$.

$\nabla ·A=\left(\hat{x}\frac{d}{dx}+\hat{y}\frac{d}{dy}+\hat{z}\frac{d}{dz}\right)\frac{{\mu }_{0}k}{4c}\left(ct-x^2\right)$

Substitute $\hat{x}\frac{d}{dx}=0$and $\hat{y}\frac{d}{dx}=0$in equation (4).

$$\begin{gathered} \nabla ·A=\left(0+0+\hat{z}\frac{d}{dz}\right)\frac{{\mu }_{0}k}{4c}\left(ct-x^2\right) \\ \nabla ·A=\hat{z}\frac{d}{dz}\left(\frac{{\mu }_{0}k}{4c}\right)\left(ct-x^2\right) \\ \nabla ·A=0\times \left(ct-x^2\right) \\ \nabla ·A=0 \end{gathered}$$

Substitute role="math" localid="1653969420708" $\nabla ·A=0$in equation (2).

$$\begin{gathered} 0=-{\mu }_0{\epsilon }_0\frac{dV}{dt} \\ \frac{dV}{dt}=0 \end{gathered}$$

Therefore, the two potentials are in the coulomb gauge and Lorentz gauge.

(b)

Write the expression for the potential for the problem 10.3.

$$\begin{gathered} V=0 \\ A=-\frac{1}{4\pi {\epsilon }_0}\left(\frac{qt}{r^2}\right).......\left(5\right) \end{gathered}$$

Here, q is the charge, t is the time, and r is the radial distance.

Substitute the value of $\nabla ·A$from equation (5).

$$\begin{gathered} \nabla ·A=-\frac{qt}{4\pi {\epsilon }_0}\times \nabla \left(\frac{1}{r^2}\right) \\ \nabla ·A=-\frac{qt}{4\pi {\epsilon }_0}\times \left(4\pi {\mathcal{S}}^{3}r\right) \\ \nabla ·A=-\frac{qt}{{\epsilon }_0}\times \left({\mathcal{S}}^{3}r\right) \\ \nabla ·A\ne 0 \end{gathered}$$

Clearly, the potential corresponding to the Coulomb gauge has a non-zero value.

Substitute $-\frac{qt}{{\epsilon }_0}\times \left({\mathcal{S}}^{3}r\right)$for $\nabla ·A$in equation (2) as,

$$\begin{gathered} -\frac{qt}{{\epsilon }_0}\times \left({\mathcal{S}}^{3}r\right)=-{\mu }_0{\epsilon }_0\frac{dV}{dt} \\ \frac{dV}{dt}=\frac{qt}{{\epsilon }_0}\times \frac{\left({\mathcal{S}}^{3}r\right)}{{\mu }_0{\epsilon }_0} \\ \frac{dV}{dt}\ne 0 \end{gathered}$$

Clearly, the potential corresponding to the Lorentz gauge is also having a non-zero value.

Therefore, the potentials are not in the coulomb gauge and Lorentz gauge.

(a)

Write the expression for the potential for the problem 10.4.

$$\begin{gathered} V=0 \\ A=A_0\sin \left(kx-\omega t\right).........\left(6\right) \end{gathered}$$

Here, $A_0,\omega$ and k are the constants.

The equation (5) is the wave function in the Cartesian coordinates. So, the value of $\nabla ·A$is,

$$\begin{gathered} \nabla ·A=\left(\hat{x}\frac{d}{dx}+\hat{y}\frac{d}{dy}+\hat{z}\frac{d}{dz}\right)\left(A_0\sin \left(kx-\omega t\right)\right) \\ \nabla ·A=\frac{\mathcal{S}}{\mathcal{S}y}\left(A_0\sin \left(kx-\omega t\right)\right) \\ \nabla ·A=0 \end{gathered}$$

Clearly, the potential corresponding to the Coulomb gauge is zero.

Substitute 0 for $\nabla ·A$in equation (2).

$$\begin{gathered} 0=-{\mu }_0{\epsilon }_0\frac{dV}{dt} \\ \frac{dV}{dt}=0 \end{gathered}$$

Clearly, the potential corresponding to the Lorentz gauge is also equal to zero.

Therefore, the potentials are in the Coulomb gauge and Lorentz gauge.