Is there any proof of why a charge is scalar?

Question

I tried to prove it by contradiction as, if it was a vector then the isotropic space property that nature currently has now would not hold; But that proof eventually broke down. So, I was wondering if a proof actually exists or it is just experimentally told as a property?

Answer 1

The point-like sources of electromagnetic field don't have to be scalar like. We can have dipole sources, quadrupole sources and so on. In general, any feld generated by a point-like source can be decomposed using so-called multipole expansion.

$$ V(r,\theta,\varphi) = \sum_{l=0}^\infty \sum_{m=-l}^l C^m_l(r) Y^m_l(\theta, \varphi)$$

where functions $Y^m_l$ are specific functions called spherical harmonics.

The electric charge is the property of the source that generates the spherically isotropic field (l=0,m=0), and that means it is a scalar by definition.

That is, if we consider only the transformations of the space. If we consider space-time transformation, then charge density is not a scalar density, but just the time-component of a four-vector called the four-current: $(\rho, \vec j)$.

If the question is whether the elementary particles possess only scalar charges, then it depends on how deep do you go. The truly fundamental particles like electrons or quarks seem to have only scalar charges, but composite particles may have for example the dipole moment, which is a vector. This is the case of a neutron, which despite having no net electric charge, still possesses a small electric dipole moment. Even fundamental particles may theoretically have a dipole moment, and there are theoretical predictions to detect the electron's dipole moment, but the predicted value is too small to be observed experimentally. The same for other fundamental particles. For all practical uses, the fundamental particles can be treated as possessing only scalar charge.

Answer 2

Yes, the proof that charge is a scalar is quite simple. Start with Gauss' law: $$\nabla \cdot \vec E = \frac{\rho}{\epsilon_0}$$ The divergence of a vector is a scalar, so the quantity on the right is a scalar. Then define $$Q = \iiint \rho \ dx \ dy \ dz$$ since $\rho$ is a scalar the integral of $\rho$ over a volume is also a scalar, so $Q$ is a scalar.

Answer 3

The properties of a particles cannot be a vector quantity. I believe physically there's no need to define the charge as a vector quantity. Because the important thing is its magnitude. The vector is a mathematical concept that can be used in physics to describe nature. The charge of a particle does not need this kind of description.