Question

Check that the potentials of a point charge moving at constant velocity (Eqs. 10.49 and 10.50) satisfy the Lorenz gauge condition (Eq. 10.12).

Short answer

The Lorentz gauge conditions satisfied.

$\nabla \cdot A=-{\mu }_0{\epsilon }_0\frac{\partial V}{\partial t}$

Step-by-step solution

Write the given data from the question.

The expression for the scaler potential from equation 10.49,

$V(r,t)=\frac{1}{4\pi {\epsilon }_0}\frac{qc}{\sqrt{{(c^{2}t-r\cdot v)}^2+(c^2-v^2)(r^2-c^2{t}^2)}}$

The expression for the vector potential from equation 10.50,

$A(r,t)=\frac{{\mu }_0}{4\pi }\frac{qcv}{\sqrt{{(c^{2}t-r\cdot v)}^2+(c^2-v^2)(r^2-c^2{t}^2)}}$

The Lorentz conditions,

$\nabla \cdot A=-{\mu }_0{\epsilon }_0\frac{\partial V}{\partial t}$

Determine the formulas to satisfy the Lorentz gauge condition.

The expression for the scaler potential for moving charge is given as follows.

$\mathit{V}\mathbf{(}\mathbf{r}\mathbf{,}\mathbf{t}\mathbf{)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}\mathbf{\pi }{\mathbf{\epsilon }}_{\mathbf{0}}}\frac{\mathbf{q}\mathbf{c}}{\mathbf{(}\mathbf{r}\mathbf{c}\mathbf{-}\mathbf{r}\mathbf{\times }\mathbf{v}\mathbf{)}}$

The expression for the vector potential for moving charge is given as follows.

$\mathit{A}\mathbf{(}\mathbf{r}\mathbf{,}\mathbf{t}\mathbf{)}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}}{\mathbf{4}\mathbf{\pi }}\frac{\mathbf{q}\mathbf{c}\mathbf{v}}{\mathbf{(}\mathbf{r}\mathbf{c}\mathbf{-}\mathbf{r}\mathbf{\times }\mathbf{v}\mathbf{)}}$

Satisfy the Lorentz gauge condition.

Calculate the expression of$\nabla \cdot A$ .

$\begin{array}{l}\nabla \cdot A=\nabla \cdot \left[\frac{{\mu }_0}{4\pi }\frac{qcv}{\sqrt{{(c^{2}t-r\cdot v)}^2+(c^2-v^2)(r^2-c^2{t}^2)}}\right]\\ \nabla \cdot A=\frac{{\mu }_0}{4\pi }qcv\cdot \nabla {[{(c^{2}t-r\cdot v)}^2+(c^2-v^2)(r^2-c^2{t}^2)]}^{-\frac{1}{2}}\\ \nabla \cdot A=\frac{{\mu }_{0}qc}{4\pi }v\cdot \left\{\begin{array}{l}-\frac{1}{2}{[{(c^{2}t-r\cdot v)}^2+(c^2-v^2)(r^2-c^2{t}^2)]}^{-\frac{3}{2}}\\ +\nabla [{(c^{2}t-r\cdot v)}^2+(c^2-v^2)(r^2-c^2{t}^2)]\end{array}\right\}\\ \nabla \cdot A=-\frac{{\mu }_{0}qc}{8\pi }\left\{\begin{array}{l}{[{(c^{2}t-r\cdot v)}^2+(c^2-v^2)(r^2-c^2{t}^2)]}^{-\frac{3}{2}}\\ +v\cdot [-2(c^{2}t-r\cdot v)\nabla (\widehat{r}\cdot \widehat{v})+(c^2-v^2)\nabla \left(r^2\right)]\end{array}\right\}\end{array}$

Solve further as,

$\begin{array}{l}\nabla \cdot A=-\frac{{\mu }_{0}qc}{8\pi }\left\{\begin{array}{l}{[{(c^{2}t-r\cdot v)}^2+(c^2-v^2)(r^2-c^2{t}^2)]}^{-\frac{3}{2}}\\ +v\cdot [-2(c^{2}t-r\cdot v)v+(c^2-v^2)2r]\end{array}\right\}\\ \nabla \cdot A=-\frac{{\mu }_{0}qc}{4\pi }\left\{\begin{array}{l}{[{(c^{2}t-r\cdot v)}^2+(c^2-v^2)(r^2-c^2{t}^2)]}^{-\frac{3}{2}}\\ +[(c^{2}t-r\cdot v)v^2+(c^2-v^2)r\cdot v]\end{array}\right\}\\ \nabla \cdot A=-\frac{{\mu }_{0}qc}{4\pi }\left\{\begin{array}{l}{[{(c^{2}t-r\cdot v)}^2+(c^2-v^2)(r^2-c^2{t}^2)]}^{-\frac{3}{2}}\\ +[c^{2}t{v}^2-(r\cdot v)v^2-c^2(r\cdot v)+v^2(r\cdot v)]\end{array}\right\}\end{array}$

Solve further as,

$\nabla \cdot A=-\frac{{\mu }_{0}qc}{4\pi }\left\{\begin{array}{l}{[{(c^{2}t-r\cdot v)}^2+(c^2-v^2)(r^2-c^2{t}^2)]}^{-\frac{3}{2}}\\ +\left[c^2(t{v}^2-r\cdot v)\right]\end{array}\right\}$

$\nabla \cdot A=-\frac{{\mu }_{0}qc}{4\pi }\frac{v^{2}t-r\cdot v}{{[{(c^{2}t-r\cdot v)}^2+(c^2-v^2)(r^2-c^2{t}^2)]}^{\frac{3}{2}}}$ ……. (1)

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

$\begin{array}{l}\frac{\partial V}{\partial t}=\frac{\partial }{\partial t}\left[\frac{1}{4\pi {\epsilon }_0}\frac{qc}{\sqrt{{(c^{2}t-r\cdot v)}^2+(c^2-v^2)(r^2-c^2{t}^2)}}\right]\\ \frac{\partial V}{\partial t}=\frac{qc}{4\pi {\epsilon }_0}\frac{\partial }{\partial t}\left[\frac{1}{\sqrt{{(c^{2}t-r\cdot v)}^2+(c^2-v^2)(r^2-c^2{t}^2)}}\right]\\ \frac{\partial V}{\partial t}=\frac{qc}{4\pi {\epsilon }_0}(-\frac{1}{2}){[{(c^{2}t-r\cdot v)}^2+(c^2-v^2)(r^2-c^2{t}^2)]}^{-\frac{3}{2}}+\frac{\partial }{\partial t}[{(c^{2}t-r\cdot v)}^2+(c^2-v^2)(r^2-c^2{t}^2)]\\ \frac{\partial V}{\partial t}=\frac{qc}{8\pi {\epsilon }_0}{[{(c^{2}t-r\cdot v)}^2+(c^2-v^2)(r^2-c^2{t}^2)]}^{-\frac{3}{2}}+[2(c^{2}t-r\cdot v)c^2+(c^2-v^2)(-2c^{2}t)]\end{array}$

Solve further as,

$\begin{array}{l}\frac{\partial V}{\partial t}=\frac{-q{c}^3}{8\pi {\epsilon }_0}\frac{(-r\cdot v+v^{2}t)2}{{[{(c^{2}t-r\cdot v)}^2+(c^2-v^2)(r^2-c^2{t}^2)]}^{\frac{3}{2}}}\\ \frac{\partial V}{\partial t}=\frac{-q{c}^3}{4\pi {\epsilon }_0}\frac{(v^{2}t-r\cdot v)}{{[{(c^{2}t-r\cdot v)}^2+(c^2-v^2)(r^2-c^2{t}^2)]}^{\frac{3}{2}}}\end{array}$

Multiply the above equation with $-{\mu }_0{\epsilon }_0$.

$-{\mu }_0{\epsilon }_0\frac{\partial V}{\partial t}=(-{\mu }_0{\epsilon }_0)\frac{-q{c}^3}{4\pi {\epsilon }_0}\frac{(v^{2}t-r\cdot v)}{{[{(c^{2}t-r\cdot v)}^2+(c^2-v^2)(r^2-c^2{t}^2)]}^{\frac{3}{2}}}$

$-{\mu }_0{\epsilon }_0\frac{\partial V}{\partial t}=\frac{{\mu }_{o}q{c}^3}{4\pi }\frac{(v^{2}t-r\cdot v)}{{[{(c^{2}t-r\cdot v)}^2+(c^2-v^2)(r^2-c^2{t}^2)]}^{\frac{3}{2}}}$ ……. (2)

From the equation (1) and (2).

$\nabla \cdot A=-{\mu }_0{\epsilon }_0\frac{\partial V}{\partial t}$

Hence, the Lorentz gauge conditions satisfied.