Why source point singularities are inevitable in Physical Fields?

Question

Any physical phenomena is explained by stating some relations between certain physical quantities. The physical quantities, if having a certain value for each and every point in space and time are called fields. If we take some examples of the classical fields:

$$\text{The gravitational field}:\textbf{g}(\textbf{r})=-\nabla\phi(\textbf{r})$$

$$\text{The electric field}:\textbf{E}(\textbf{r})=-\nabla V-\frac{\partial{A}}{\partial{t}}$$

$$\text{The magnetic field}:\textbf{B}(\textbf{r})=\nabla\times\textbf{A}$$

All these fields are having singularties at the source points. The quantum theories are jut quantized classical field theories developed in the quantum mechanical framework. So in that case also, there should be the same singularities, right?

Why such source point singularities are inevitable in the case of physical fields?

Answer 1

I question your premise: in classical field theory point singularities are not necessary, let alone inevitable. The simplest and most relevant example is the gravitational potential of the Earth: $$ \phi(r)\sim\begin{cases} r^2 & r<R_\oplus\\ \frac{1}{r} & r>R_\oplus \end{cases} $$ which is well-behaved and finite everywhere ($\phi\in\mathscr C^1(\mathbb R^+)$).

You only get singularities in $\phi$ when the source is point-like, but this is a problem of the math, not of the physics, because in the real world nothing is truly point-like. If you find a divergence in your equations, it just means that the model broke down. For example, for orbital motion one may forget about the $r^2$ part of the potential, and just write $\phi\propto 1/r$. This leads to the obvious divergence $\phi(0)=\infty$. But the true result is $\phi(0)=0$, and the divergence just reflects the approximation that is to treat the Earth as point-like.

As for your concerns related to QM: take for example the Hydrogen atom. The potential energy of the electron is $$ V(r)=\frac{1}{r} $$ and one may think that the potential energy of the electron blows-up at the origin. But orbitals are not localised (that is, the position of the electron is described by a wave function), and the potential energy is finite, even though $\psi(0)\neq 0$: $$ \langle V\rangle\propto\frac{1}{n^2} $$ where $n=1,2,\dots$ is the principal quantum number.