Consider a cylindrical symmetrical infinite wire with radius $R$. The wire has constant charge density $\rho$ and current density $\vec j$ (pointing to the right). This is measured from the stationary frame $S$.

We let the wire move at a uniform speed $v$ to the right; label the moving frame $\bar S$. The observations are made from the stationary frame $S$.

a) Do $\rho$ and $\vec j$ change? (wrt our stationary reference frame S)? How is this consistent with the conservation of charge?

In $S$ we have:

$$\vec j = \rho u$$

Where $u$ is the speed of the charges (which is quite small, so we'd expect a small current density indeed.

My first thought was that $\rho$ wouldn't change because charge is invariant. But then I thought that (considering $v$ to be a relativistic speed) moving objects get shortened (only along the direction of its motion). Meaning that the volume will be shortened. Thus, based on the definition $\rho = \frac{Q}{V}$

$$\bar \rho > \rho$$

What about $\vec j$? As $v >> u$ let me neglect $u$. Then $\bar {\vec j}$ is:

$$\bar {\vec j} = \bar \rho v$$

Which means:

$$\bar {\vec j} >> \vec j$$

The above is perfectly consistent with local conservation of charge.

In $S$ we have:

$$\frac{\partial \rho}{\partial t} = -\nabla \vec j$$

While in $\bar S$ we have:

$$\frac{\partial \bar \rho}{\partial t} = -\nabla \bar {\vec j}$$

How do you see my reasoning?

  • 2
    $\begingroup$ Yes, $\rho$ and $\vec{j}$ form a four-vector $j^\mu = (\rho, \vec{j})$ and transform like any other four-vector. $\endgroup$
    – knzhou
    Commented Jun 13, 2019 at 9:25
  • $\begingroup$ @knzhou Oh I see thanks! Actually I should work with proper velocity because proper time is invariant (thus we have less work to do). $\endgroup$
    – JD_PM
    Commented Jun 13, 2019 at 9:36

1 Answer 1


As noted by @knzhou, charge and current density form a four-vector. There are several ways of prooving it. One way is to postulate that the electromagnetic fields (as an anti-symmetric tensor) are co-variant, and then properties of current density follow from Maxwell's equations.

It can also be done based solely on Maxwell's equations without asking for co-variance of the fields. Indeed, as you noted, the charge conservation implies ($\rho$ - charge density, $\mathbf{J}$ - current density)

$\partial_t \rho + \boldsymbol{\nabla}.\mathbf{J}=0$

in any reference frame. It is convinient to build a proof for a point-charge. The charge density of a point-particle (with charge $q$) at $\mathbf{\bar{r}}$ is


Where $\delta^{(3)}$ is the 3d delta function. This can be also written as:

$\rho\left(\mathbf{r},t\right)=\int cdt' q\,\delta\left(c\left(t-t'\right)\right) \delta^{(3)}\left(\mathbf{r}-\mathbf{\bar{r}}\left(t'\right)\right)$

i.e. we are taking into account all the motion of the charge (by integrating with respect to time $t'$), but only its position at time $t$ matters for $\rho\left(\mathbf{r},t\right)$

This can be further simplified to:

$\rho\left(\mathbf{r},t\right)=\int dt' \,cq\,\delta^{(4)}\left(\left(ct,\mathbf{r}\right)-\bar{x}\left(t'\right)\right)$

Where $\bar{x}\left(t'\right)^\mu=\left(ct',\mathbf{\bar{r}}\left(t'\right)\right)^\mu$ is the four-position at time $t'$ (in the lab-frame), and $\delta^{(4)}$ is the 4d delta function.

Now the current density, that satisfies the charge conservation equation is:

$\mathbf{J}\left(\mathbf{r},t\right)=\int cdt' q\,\mathbf{\dot{\bar{r}}}\left(t'\right)\delta\left(c\left(t-t'\right)\right) \delta^{(3)}\left(\mathbf{r}-\mathbf{\bar{r}}\left(t'\right)\right)=\int cdt' \,q\,\mathbf{\dot{\bar{r}}}\left(t'\right)\,\delta^{(4)}\left(\left(ct,\mathbf{r}\right)-\bar{x}\left(t'\right)\right)$

Finally we can change integration variable from lab-frame time to proper time of the charged particle ($t'\to \tau$). This does not affect the $\delta^{(4)}$ because Lorentz tranformations do not change 'four-volumes', i.e. the Jacobian corresponding to such coordinate change is unity. But $dt'\to \gamma d\tau$, where $\gamma$ is the Lorentz factor.


$c\rho\left(\mathbf{r},t\right)=\int cd\tau \gamma\,cq\,\delta^{(4)}\left(\left(ct,\mathbf{r}\right)-\bar{x}\left(\tau\right)\right)$

$\mathbf{J}\left(\mathbf{r},t\right)=\int cd\tau \gamma\,q\,\mathbf{\dot{\bar{r}}}\left(\tau\right)\,\delta^{(4)}\left(\left(ct,\mathbf{r}\right)-\bar{x}\left(\tau\right)\right)$

Now the four-velocity of a particle is $u^\mu=\gamma\left(c,\mathbf{\dot{\bar{r}}}\right)^\mu$, thus if we create an object $J^\mu = \left(c\rho,\mathbf{J}\right)^\mu$ we can see that for point-particle it will be (in any frame!):

$J^\mu = \int cd\tau \,u^\mu\, \delta^{(4)}\left(\left(ct,\mathbf{r}\right)-\bar{x}\left(\tau\right)\right)$

and therefore a four-vector (since four-velocity is a four vector). Assuming that all charge densities are made from point charges (even weaker, purely mathematical, assumption is that all charge densities are sufficiently well-behaved to be decomposable into delta-functions) we then establish the four-vector nature of $\left(c\rho,\mathbf{J}\right)^\mu$ in general.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.