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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?

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    $\begingroup$ Yes, $\rho$ and $\vec{j}$ form a four-vector $j^\mu = (\rho, \vec{j})$ and transform like any other four-vector. $\endgroup$ – knzhou Jun 13 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 Jun 13 at 9:36
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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

$\rho\left(\mathbf{r},t\right)=q\delta^{(3)}\left(\mathbf{r}-\mathbf{\bar{r}}\left(t\right)\right)$

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.

So:

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

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

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.

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