A straight wire along thez-axis carries a charge density$\mathbf{\lambda }$traveling in the +z direction at speed v. Construct the field tensor and the dual tensor at the point role="math" localid="1654331549769" $\left(x,0,0\right)$.

Write the expression for the electric field strength.

$\mathbf{E}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}\mathbf{\pi }{\mathbf{\epsilon }}_{\mathbf{0}}}\frac{\mathbf{2}\mathbf{\lambda }}{\mathbf{x}}\overbrace{\mathbf{x}}$ .....(1)

Here, ${\epsilon }_o$ is the permittivity of free space, $\mathrm{\lambda }$ is the charge density, and x is the distance from the wire.

Write the expression for the magnetic field strength.

$$\begin{gathered} \mathbf{B}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}}{\mathbf{4}\mathbf{\pi }}\frac{\mathbf{2}\mathbf{\lambda \nu }}{\mathbf{x}}\hat{\mathbf{y}} \\ \mathbf{B}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}}{\mathbf{2}\mathbf{\pi }}\frac{\mathbf{\lambda \nu }}{\mathbf{x}}\hat{\mathbf{y}} \end{gathered}$$

It is known that:

$\frac{1}{{\mathrm{\epsilon }}_0}={\mathrm{c}}^2{\mathrm{\mu }}_0$

Substitute$\frac{1}{{\mathrm{\epsilon }}_0}={\mathrm{c}}^2{\mathrm{\mu }}_0$ in equation (1).

$$\begin{gathered} \mathrm{E}=\frac{1}{4\mathrm{\pi }\left({\displaystyle \frac{1}{{\mathrm{c}}^2{\mathrm{\mu }}_0}}\right)}\frac{2\mathrm{\lambda }}{\mathrm{x}}\hat{\mathrm{x}} \\ \mathrm{E}=\frac{{\mathrm{\mu }}_0}{2\mathrm{\pi }}\frac{{\mathrm{\lambda c}}^2}{\mathrm{x}}\hat{\mathrm{x}} \end{gathered}$$

Write the equation for filed tensor as an array form.

${\mathrm{F}}^{\mathrm{\mu \nu }}=\left.\left\{\begin{array}{cc}0& {\mathrm{E}}_{\mathrm{x}}/\mathrm{c}\\ -{\mathrm{E}}_{\mathrm{x}}/\mathrm{c}& 0\\ -{\mathrm{E}}_{\mathrm{y}}/\mathrm{c}& -{\mathrm{B}}_{\mathrm{x}}\\ -{\mathrm{E}}_{\mathrm{z}}/\mathrm{c}& {\mathrm{B}}_{\mathrm{y}}\end{array}\right.\begin{array}{cc}{\mathrm{E}}_{\mathrm{x}}/\mathrm{c}& {\mathrm{E}}_{\mathrm{x}}/\mathrm{c}\\ {\mathrm{B}}_{\mathrm{z}}& -{\mathrm{B}}_{\mathrm{y}}\\ 0& {\mathrm{B}}_{\mathrm{x}}\\ -{\mathrm{B}}_{\mathrm{x}}& 0\end{array}\right\}\begin{array}{}\\ \\ \\ \end{array}$

Substitute ${\mathrm{B}}_{\mathrm{x}}=0,{\mathrm{B}}_{\mathrm{z}}=0,{\mathrm{E}}_{\mathrm{y}}=0,{\mathrm{E}}_{\mathrm{z}}=0{\mathrm{E}}_{\mathrm{x}}=\frac{{\mathrm{\mu }}_0}{2\mathrm{\pi }}\frac{{\mathrm{\lambda c}}^2}{\mathrm{x}}\mathrm{and}{\mathrm{B}}_{\mathrm{y}}=\frac{{\mathrm{\mu }}_0}{2\mathrm{\pi }}\frac{\mathrm{\lambda v}}{\mathrm{x}}$in equation (2).

$$\begin{gathered} \mathrm{F\mu \nu }=\left(\begin{array}{cccc}0& \left(\frac{{\mathrm{\mu }}_0{\mathrm{\lambda c}}^2}{{\displaystyle \frac{2\mathrm{\pi }\mathrm{x}}{\mathrm{c}}}}\right)& 0& 0\\ -\left(\frac{{\mathrm{\mu }}_0{\mathrm{\lambda c}}^2}{{\displaystyle \frac{2\mathrm{\pi }\mathrm{x}}{\mathrm{c}}}}\right)& 0& 0& 0\\ 0& 0& 0& 0\\ 0& \left(\frac{{\mathrm{\mu }}_0{\mathrm{\lambda c}}^2}{{\displaystyle 2\mathrm{\pi }\mathrm{x}}}\right)& 0& 0\end{array}\right) \\ {\mathrm{F}}^{\mathrm{\mu \nu }}=\frac{{\mathrm{\mu }}_0\mathrm{\lambda }}{{\displaystyle 2\mathrm{\pi }\mathrm{x}}}\left(\begin{array}{cccc}0& \mathrm{c}& 0& 0\\ -\mathrm{c}& 0& 0& -\mathrm{v}\\ 0& 0& 0& 0\\ 0& \mathrm{v}& 0& 0\end{array}\right) \end{gathered}$$

Write the equation for the dual tensor as an array form.

$G^{\mathrm{\mu \nu }}=\left.\left\{\begin{array}{cc}0& {\mathrm{B}}_{\mathrm{x}}\\ -{\mathrm{B}}_{\mathrm{x}}& 0\\ -{\mathrm{B}}_{\mathrm{y}}& {\mathrm{E}}_z/\mathrm{c}\\ -{\mathrm{B}}_{\mathrm{z}}& {\mathrm{E}}_{\mathrm{y}}/\mathrm{c}\end{array}\right.\begin{array}{cc}{\mathrm{B}}_{\mathrm{y}}& {\mathrm{B}}_{\mathrm{z}}\\ -{\mathrm{E}}_{\mathrm{z}}/\mathrm{c}& {\mathrm{E}}_{\mathrm{y}}/\mathrm{c}\\ 0& -{\mathrm{E}}_{\mathrm{x}}/\mathrm{c}\\ {\mathrm{E}}_{\mathrm{x}}/\mathrm{c}& 0\end{array}\right\}\begin{array}{}\\ \\ \\ \end{array}$..........(3)

Substitute $0\mathrm{for}{\mathrm{B}}_{\mathrm{z}},0\mathrm{for}{\mathrm{B}}_{\mathrm{x}},0\mathrm{for}{\mathrm{E}}_{\mathrm{y}},,0\mathrm{for}{\mathrm{E}}_{\mathrm{z}},\frac{{\mathrm{\mu }}_0{\mathrm{\lambda c}}^2}{2\mathrm{\pi }\mathrm{x}}\mathrm{for}{\mathrm{E}}_{\mathrm{x}}\mathrm{and}\frac{{\mathrm{\mu }}_0\mathrm{\lambda c}}{2\mathrm{\pi }\mathrm{x}}\mathrm{for}{\mathrm{B}}_{\mathrm{y}}$in equation (3).

role="math" localid="1654682762200" $$\begin{gathered} {\mathrm{G}}^{\mathrm{\mu \nu }}=\left\{\begin{array}{cc}0& 0\\ 0& 0\\ -\left(\frac{{\mathrm{\mu }}_0\mathrm{\lambda \nu }}{2\mathrm{\pi }\mathrm{x}}\right)& 0\\ 0& 0\end{array}\right.\left.\begin{array}{cc}\left(\frac{{\mathrm{\mu }}_0\mathrm{\lambda \nu }}{2\mathrm{\pi }\mathrm{x}}\right)& 0\\ 0& 0\\ 0& -\left(\frac{{\mathrm{\mu }}_0{\mathrm{\lambda c}}^2}{{\displaystyle \frac{2\mathrm{\pi }\mathrm{x}}{\mathrm{c}}}}\right)\\ \left(\frac{{\mathrm{\mu }}_0{\mathrm{\lambda c}}^2}{{\displaystyle \frac{2\mathrm{\pi }\mathrm{x}}{\mathrm{c}}}}\right)& 0\end{array}\right\} \\ {\mathrm{G}}^{\mathrm{\mu \nu }}=\left\{\begin{array}{cc}0& 0\\ 0& 0\\ -\mathrm{v}& 0\\ 0& \mathrm{v}\end{array}\right.\left.\begin{array}{cc}0& 0\\ 0& 0\\ 0& -\mathrm{c}\\ 0& 0\end{array}\right\} \end{gathered}$$

Therefore, the field tensor and the dual tensor at the point $\left(\mathrm{x},0,0\right)$is

${\mathrm{F}}^{\mathrm{\mu \nu }}=\frac{{\mathrm{\mu }}_0\mathrm{\lambda }}{2\mathrm{\pi x}}\left(\begin{array}{cccc}0& \mathrm{c}& 0& 0\\ -\mathrm{c}& 0& 0& -\mathrm{v}\\ 0& 0& 0& 0\\ 0& \mathrm{v}& 0& 0\end{array}\right)\mathrm{and}{\mathrm{G}}^{\mathrm{\mu \nu }}=\frac{{\mathrm{\mu }}_0\mathrm{\lambda }}{2\mathrm{\pi x}}\left(\begin{array}{cccc}0& 0& \mathrm{v}& 0\\ 0& 0& 0& 0\\ -\mathrm{v}& 0& 0& -\mathrm{c}\\ 0& 0& \mathrm{c}& 0\end{array}\right)\mathrm{respectivety}$