Solution

$14.3.19$ Electron velocity in a parallel beam $\beta c$. How will the electron density change
when moving relative to the beam with the velocity $\beta_1 c$ in the longitudinal
direction?

To solve this problem we can use the Lorentz transformation for current density and charge density.
Also, we need to keep in mind that the current density in the Earth frame is related to the charge density in the Earth frame as follows:

\begin{equation} j = -\rho \beta c \end{equation}
The negative sign is due to the sign of the charge, because it is the current density of the electrons.
Now, the Lorentz transformations for current density and charge density are:

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

Using the equation for the charge density we can calculate it in the moving frame:

\begin{equation} \rho' = \gamma (\rho - \frac{j \beta_1}{c}) = \gamma (\rho + \frac{\rho \beta c \beta_1}{c}) =\rho \frac{(1 + \beta \beta_1)}{\sqrt{1-\beta_1^2}} \end{equation}

So, the charge density becomes $\frac{(1 + \beta \beta_1)}{\sqrt{1-\beta_1^2}}$ times the original.