electromagnetism - How can we prove charge invariance under Lorentz Transformation?

We have gravitational force between two massive particles and we have electromagnetic force between two charged particles. When special relativity suggests that mass is not an invariant quantity why we have electric charge as an invariant quantity ?

Answer

Let $j=(\rho,\boldsymbol j)$ be the current density of a system. This four numbers are, by hypothesis, a vector. This means that the charge density $\rho$ transforms just like $t$ does, i.e., it gets "dilated" when changing from reference frame to reference frame:

$$\rho'\to\gamma \rho \tag{1}$$

Charge is, by definition, the volume integral of the charge density:

$$Q\equiv \int\mathrm d\boldsymbol x\ \rho \tag{2}$$

In a different frame of reference the charge is

$$Q'=\int\mathrm d\boldsymbol x'\ \rho'=\int\mathrm d\boldsymbol x'\ \gamma \rho \tag{3}$$ where I used $(1)$.

Next, we need to know what $\mathrm d\boldsymbol x'$ is. The trick to evaluate this is to note that the product $\mathrm dt\;\mathrm d\boldsymbol x$ is invariant (in SR). This means that we can write $\mathrm dt'\;\mathrm d\boldsymbol x'=\mathrm dt\;\mathrm d\boldsymbol x$; solving for $\mathrm d\boldsymbol x'$ we get

$$\mathrm d\boldsymbol x'=\frac{\mathrm dt}{\mathrm dt'}\mathrm d\boldsymbol x=\frac{1}{\gamma}\mathrm d\boldsymbol x \tag{4}$$

If we plug this into $(3)$, we find

$$Q'=\int\mathrm d\boldsymbol x'\ \gamma \rho=\int\mathrm d\boldsymbol x\ \rho=Q \tag{5}$$ that is, $Q=Q'$.