What is the Lorentz transformation of the (scalar) electric current, $I$ ? I got two answers that are not consistent: Consider a lab frame where the charges are moving with a velocity $\vec{\beta}_q$, and consider another (primed) frame that moves with velocity $\vec{\beta}$ (all velocities are relative to the lab frame).

In the case of moving charges (with the same sign) in the same direction of the area unit vector, we use $I = dq/dt$ and obtain \begin{equation} I'= \frac{dq'}{dt'} = \gamma \,\rho'\frac{dV'}{ dt} =\gamma^2 \, \rho'\frac{dV}{dt} =\gamma^2 \,\frac{\rho'}{\rho}\frac{dq}{dt} =\gamma^2 \,\frac{\rho'}{\rho} I = \gamma^3 \,\left(1-\frac{\vec{j}\cdot\vec{\beta}}{c\rho}\right)I\:\:\:\overset{\vec{j}=c\,\rho\,\vec{\beta}_q}{=}\:\:\: \gamma^3\left(1-\vec{\beta}\cdot\vec{\beta}_q\right)I \end{equation} The second answer I got is (note that we break the area vector into two orthogonal components parallel and perpendicular to the velocity of the primed frame) $$dI' = \vec{j}'\cdot d\vec{A}' = c\rho[\gamma^2(\vec{\beta}_q-\vec{\beta})\cdot d\vec{A}_\parallel+\vec{\beta}_q\cdot d\vec{A}_\perp]$$

Let us test the two equations. Consider the primed frame moves with the same velocity as the charges (and the charges move in a cylinder, for example); thus, $\vec{\beta}=\vec{\beta}_q, \:\: d\vec{A} = d\vec{A}_\parallel=dA \,\hat{\beta}$, we get from the first equation $$I' = \gamma \, I\neq 0$$ while the second equation gives $$dI' = 0$$

I think the second equation is more accurate, but I do not know what is wrong with the first equation of $I'$.


1 Answer 1


$I$ isn't actually a scalar, so it doesn't have a well-defined transformation property under the Lorentz transformation. (If $I$ were a scalar, then it would keep its value after a Lorentz transformation.)

When working in a relativistic setting, $I$ isn't the right quantity to use anyway; any formulas that contain it, like the Biot–Savart law, won't be correct. (The Biot–Savart law doesn't work when the wire is moving, and in this case there will also be an electric field in addition to the magnetic one.)

You have to work with the four-current density, $J$, which is a proper Lorentz four-vector.

  • $\begingroup$ Do you mean that the definition of the spacetime function (call it current) $I=\int \vec{J}\cdot d\vec{A}$ is not working? Can you elaborate more? I think any spacetime function can be transformed normally using Lorentz transformation. Here, I used that for the current density (which is the spatial component of the four-current). $\endgroup$ Commented Jan 17 at 9:36
  • $\begingroup$ @AhmadHaitham Knowing the value of $I$ in one reference frame does not give you enough information to determine the value it would have after a Lorentz transformation. In that sense, $I$ does not have a well-defined transformation property under Lorentz transformations. $\endgroup$
    – Brian Bi
    Commented Jan 17 at 15:07

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