# Is there any proof of why a charge is scalar?

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?

• Hard to say, but it has something to do that elementary charge is space and time invariant – Agnius Vasiliauskas Mar 17 '20 at 9:19
• Does this answer your question? How can we prove charge invariance under Lorentz Transformation? – AccidentalFourierTransform Mar 17 '20 at 14:25
• You cannot mathematically prove something about the physical universe. All the mathematics does is rearrange where you are making assumptions. But in the end, you are still making assumptions. The actual "proof" is the experimental evidence accrued supporting those assumptions. (And the reason I quoted "proof" is because despite using the same word, the mathematical and scientific concepts of "proof" are significantly different.) – Paul Sinclair Mar 17 '20 at 16:43
• Can you clarify what you intend by “scalar”? Do you just mean something that is not a three dimensional vector or do you mean the technical meaning in terms of four-vectors and tensors? – Dale Mar 17 '20 at 19:19

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.

• I've understood that OP is asking about elementary charges, not multipoles – Agnius Vasiliauskas Mar 17 '20 at 9:15
• @AgniusVasiliauskas Introducing multipole expansion however lets you define the charge as the monopole moment, and then it's scalar by definition. – Adam Latosiński Mar 17 '20 at 9:22
• @AdamLatosiński . So BY DEFINATION means there’s no proof? – Ashish Kumar Mar 17 '20 at 16:33
• Small nitpick: technically we don't know that the neutron electric dipole moment is nonzero. We certainly expect it to be nonzero, but we have only measured null results at current sensitivity. – probably_someone Mar 17 '20 at 17:21
• @AshishKumar I'm confused. If you define "two" as the next natural number after "one", do you require a proof that "two" is the next natural number after "one"? If you define electric charge as the scalar (i.e. monopole or isotropic) source of electric field, do you need to prove that the electric charge is a scalar? I mean, if you use different definition of electric charge, then maybe you do need a proff. But then please tell us what definition you use. – Adam Latosiński Mar 17 '20 at 20:45

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.

• $\nabla \cdot \vec E$ is not a scalar but the time component of a fourvector density. – my2cts Mar 17 '20 at 12:56
• I understood that the OP was asking about “scalars” in the vernacular meaning simply not a (three-)vector”. I do not believe that they were asking about scalars in the technical meaning of the term as a pseudo-Riemannian scalar such as the contraction of two four-vectors. – Dale Mar 17 '20 at 18:49

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.

• The properties of a substance cannot be a vector quantity Wrong. Take a look at complex refractive index of material, which is defined in complex plane as : $$\begin{bmatrix}n\\i\kappa\end{bmatrix}$$. Or take a look into second order Cauchy stress tensor $\sigma$ of material in continuum mechanics: $$\left[{\begin{matrix}\sigma _{11}\,\sigma _{12}\,\sigma _{13}\\\sigma _{21}\,\sigma _{22}\,\sigma _{23}\\\sigma _{31}\,\sigma _{32}\,\sigma _{33}\\\end{matrix}}\right]$$ .So in principle, material property can be described by tensor of any order – Agnius Vasiliauskas Mar 17 '20 at 19:06
• @AgniusVasiliauskas By substance I mean leptons and quarks. Not normal material. – Layla Mar 17 '20 at 22:37
• Spin angular momentum or magnetic dipole moment of particle IS vector quantity – Agnius Vasiliauskas Mar 18 '20 at 10:02
• @AgniusVasiliauskas "In some ways, spin is like a vector quantity; it has a definite magnitude, and it has a "direction" (but quantization makes this "direction" different from the direction of an ordinary vector)". So its not a vector quantity. – Layla Mar 18 '20 at 10:34
• No, it is. Quantized vector is still a vector. – Agnius Vasiliauskas Mar 18 '20 at 11:54