91影视

Write the condition for the potential in the Coulomb gauge.

$\mathbf{鈭�}\mathbf{路}\mathit{A}\mathbf{=}\mathbf{0}$ 鈥︹�� (1)

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

Write the condition for the potential in the Lorentz gauge.

$\mathbf{鈭�}\mathbf{路}\mathit{A}\mathbf{=}\mathbf{-}{\mathit{渭}}_{\mathbf{0}}{\mathit{蔚}}_{\mathbf{0}}\frac{\mathbf{d}\mathbf{V}}{\mathbf{d}\mathbf{t}}$ 鈥︹�� (2)

Here, V is the potential corresponding to the Lorentz gauge, ${渭}_0$ is the magnetic permeability and ${蔚}_0$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.

$A=\frac{{渭}_{0}k}{4c}\left(ct-x^2\right)$ 鈥︹�� (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.

$鈭�路A=\left(\hat{x}\frac{d}{dx}+\hat{y}\frac{d}{dy}+\hat{z}\frac{d}{dz}\right)\frac{{渭}_{0}k}{4c}\left(ct-x^2\right)$ 鈥︹�� (4)

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

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

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

$$\begin{gathered} 鈭�路A=\left(0+0+\hat{z}\frac{d}{dz}\right)\frac{{渭}_{0}k}{4c}\left(ct-x^2\right) \\ 鈭�路A=\hat{z}\frac{d}{dz}\left(\frac{{渭}_{0}k}{4c}\right)\left(ct-x^2\right) \\ 鈭�路A=0脳\left(ct-x^2\right) \\ 鈭�路A=0 \end{gathered}$$

Substitute role="math" localid="1653969420708" $鈭�路A=0$in equation (2).

$$\begin{gathered} 0=-{渭}_0{蔚}_0\frac{dV}{dt} \\ \frac{dV}{dt}=0 \end{gathered}$$

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

(b)

Write the expression for the potential for the problem 10.3.

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

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

Substitute the value of $鈭�路A$from equation (5).

$$\begin{gathered} 鈭�路A=-\frac{qt}{4蟺{蔚}_0}脳鈭�\left(\frac{1}{r^2}\right) \\ 鈭�路A=-\frac{qt}{4蟺{蔚}_0}脳\left(4蟺{\mathcal{S}}^{3}r\right) \\ 鈭�路A=-\frac{qt}{{蔚}_0}脳\left({\mathcal{S}}^{3}r\right) \\ 鈭�路A鈮�0 \end{gathered}$$

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

Substitute $-\frac{qt}{{蔚}_0}脳\left({\mathcal{S}}^{3}r\right)$for $鈭�路A$in equation (2) as,

$$\begin{gathered} -\frac{qt}{{蔚}_0}脳\left({\mathcal{S}}^{3}r\right)=-{渭}_0{蔚}_0\frac{dV}{dt} \\ \frac{dV}{dt}=\frac{qt}{{蔚}_0}脳\frac{\left({\mathcal{S}}^{3}r\right)}{{渭}_0{蔚}_0} \\ \frac{dV}{dt}鈮�0 \end{gathered}$$

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.

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

Here, $A_0,蝇$ and k are the constants.

The equation (5) is the wave function in the Cartesian coordinates. So, the value of $鈭�路A$is,

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

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

Substitute 0 for $鈭�路A$in equation (2).

$$\begin{gathered} 0=-{渭}_0{蔚}_0\frac{dV}{dt} \\ \frac{dV}{dt}=0 \end{gathered}$$

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

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