# Do $\rho$ and $\vec j$ change in a moving frame?

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?

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

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.