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+### Statement
+
+$14.3.19.$ [Insert the problem statement]
+
+### 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.
+
+#### Answer
+
+[Insert a concise answer or boxed result]