Relativistic transformation of electrical current If, in frame $S$, we have an electrically neutral wire with some current $I$, modelled as positive charges moving in $x$ direction and negative charges moving in $-x$ direction, then how would one find the current as meausred in a reference frame $S'$ moving with velocity $v$ in the $xx'$ direction?
One way I assumed was to find $\lambda'$, the charge density as measured in $S'$, and then multiply that by the drift velocity of the electrons as seen in $S'$, giving $I'=\lambda ' u'$ where $u'$ is the drift velocity of the electrons as seen in $S'$. However, a lot of texts seem to state that current transforms simply as $I'=\gamma I$, and the two results seem inconsistent. 
 A: It depends on whether the current is carried by a conductor or is in free space (an electron beam). In the case of an electron beam, the current will appear to have reversed in direction if you travel faster than the charge carriers, even without relativistic effects.
This web page does the transformation roughly like you have attempted, using a charge density $\rho$ moving at a velocity $u$. They find an expression for the current density:
$$J' = \gamma (J - v\rho) = \gamma J(1-v/u),$$
where $v$ is the velocity of your coordinate system. This formula will apply to a particle beam. Note that you cannot eliminate the drift velocity; it makes quite a difference whether the current is carried by high-velocity particles (say, electrons in a particle accelerator) or not.
For current carried by a conductor with stationary positive charges (charge density $\rho_+$ and moving negative charges ($\rho_-=-\rho_+$), it works differently. You must consider the current carried by positive charges ($J_+=0$) and the current carried by electrons ($J_-=J$). The transformed current is then
$$J'=J'_+ +J'_-=\gamma(J_+-\rho_+v) + \gamma(J_--\rho_-v)=\gamma J.$$
And there you are.
Update: this is for current density in A/m2 and charge density as C/m3. You could just as well read the charge density as the linear charge density (C/m), in which case you can read $J$ as the current.
