Using Lorentz Invariance of Charge To Calculate Current Density

Question

I'm attempting a problem from Zwiebach: A First Course in String Theory and am completely stuck. Could anyone give me a hint? The problem is as follows.

Consider $S$, $S'$ two Lorentz frames with $S'$ boosted along the $+x$ axis. In frame $S$ we have a cubic box with sides of length $L$ at rest. The box is filled by a material, also at rest, of uniform charge density $\rho$. In $S$ we assume that the charge density $\underline{j}=0$. Use the Lorentz invariance of charge to calculate the charge density $\rho'$ and current density $\underline{j}'$ in $S'$. Verify that $(c\rho,\underline{j})$ a 4-vector.

The charge density is easy. Indeed $L^3\rho = Q = Q' = L'^3\rho'=\frac{L^3}{\gamma}\rho'$ so $\rho' = \gamma \rho$. I know I'm right here because this agrees with what we'd expect from a 4-vector under Lorentz boost.

To do the current density I tried to use $0=\frac{\textrm{d}Q}{\textrm{d}t}=\frac{\textrm{d}Q'}{\textrm{d}t'}=\int_S\underline{j}'.\textrm{d}\underline{a}=j'_xL^2$ so $\underline{j}'=0$ since $j'_y=j'_z=0$ clearly must be zero.

I know this is wrong though, because it doesn't agree with what I'd expect from a 4-vector! What am I doing wrong? And is this the right way to go about this question?

Many thanks in advance!

Answer 1

In $S'$, there is a flux of material across any surface of constant $x'$ as the material has the velocity of the $S$ frame in $S'$. The material flux is the product of the material density in S' and $v_{Sx}$. The invariance of charge means that the current density goes as the material flux$^1$.

$j'_x = \rho' v_{Sx} = \gamma \rho v_{Sx}$

[1] Extended answer to address OPs comment.

Let $n$ be the number density, the number of particles per unit volume, in the frame of reference $S$ in which all of the particles are at rest.

Let $q$ be the electric charge, in $S$, carried by each particle.

In $S$, the charge density is $\rho = qn$.

Now, we know that $n' = \gamma n$ due to length contraction.

However, only if q is Lorentz invariant will $\rho' = qn' = \gamma q n = \gamma \rho$

In $S'$, there is a number flux across surfaces of constant $x'$:

$N'_x = n'v_{Sx} = \gamma n v_{Sx}$.

But, since each particle carries charge, there is a charge flux, a current density $j_x = \rho' v_{Sx} = \gamma \rho v_{Sx}$.

Note that if $q$ were not Lorentz invariant, e.g., $q' = \gamma q$ , then the charge and current density would not be components of a four-vector but rather a four-tensor since we would get a factor of $\gamma^2$ under a Lorentz transformation.

Answer 2

There are two sides to integrate on. The fact that $\int \vec{j}'\cdot\mathrm{d}\vec{S}'=0$ simply implies a current density is the same everywhere.