Special theory of relativity > The transformation of electric and magnetic fields > Problem 14.3.20

Statement

$14.3.20$ In a straight wire, the current density is $j$. How will this density change when
the wire moves with the speed $\beta_1 c$ in the longitudinal direction? What is the
volume charge that appears in the wire?

Solution

To solve this problem we can use the Lorentz transformation for current density and charge density.

The transformations are:

\begin{equation}
j_{\text{Earth}} = \gamma (j + \rho \beta c) \qquad \text{and} \qquad \rho_{\text{Earth}} = \gamma (\rho + \frac{j \beta_1}{c}) \quad \text{where} \quad \gamma = \frac{1}{\sqrt{1 - \beta_1^2}}
\end{equation}

It is also necessary to understand that in the frame where the wire is at rest, there is no charge density. This idea may seem strange, but
you need to think of the current as two currents: one of positive charge and one of negative charge moving in opposite directions.
The difference between these current densities is $j$. You can choose an arbitrary volume around the wire and the sum of the charges inside it is zero at every moment.

Using this idea in the Lorentz transformations we have:

\begin{equation}
j_{\text{Earth}} = \gamma j = \frac{j}{\sqrt{1-\beta_1^2}}
\end{equation}

And:

\begin{equation}
\rho_{\text{Earth}} = \frac{\gamma j \beta_1}{c} = \frac{j \beta_1}{c \sqrt{1-\beta_1^2}}
\end{equation}

Answer

\begin{equation}
j_{\text{Earth}} = \gamma j = \frac{j}{\sqrt{1-\beta_1^2}}
\end{equation}

\begin{equation}
\rho_{\text{Earth}} = \frac{\gamma j \beta_1}{c} = \frac{j \beta_1}{c \sqrt{1-\beta_1^2}}
\end{equation}