Ok I thought it was because it were a sound politically incorrect word.
Anyhow, this question lead me to the following case : if an elongated object of length L, moving with a parallel speed v along, is observed perpendicularly to this direction, the gamma factor would be [math] \gamma(v)=\left(1-\frac{v^2}{c^2}\right)^{-1/2}[/math].
If observed perpendicularly, could this intuitively give [math]\gamma(v_x)[/math] ?
How to get this from the 2D Lorentz transformation, knowing that [math]\vec{e}'_x=(-\beta_x\gamma(v), 1+(\gamma(v)-1)\frac{\beta_x^2}{\beta^2},(\gamma-1)\frac{\beta_x\beta_y}{\beta^2})[/math] ?