0

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?

UKH
  • 4,881
  • It would help if you could revise your question, to make it more specific. – TLDR May 14 '16 at 04:44
  • 1
    It's not unavoidable, but why would you introduce another length scale (for the size of the source) when the measurements do not show such a scale to exist? We have to go by what nature tells us. One can, of course, always introduce an artificial high energy cutoff and we do that quite a bit, but at the end of the day a cutoff is by no means a natural solution to the singularity problem. Where does it come from? What sets its scale? It doesn't really answer a question but it opens a new one. – CuriousOne May 14 '16 at 06:22

1 Answers1

0

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.

AccidentalFourierTransform
  • 53,248
  • 20
  • 131
  • 253
  • "in the real world nothing is truly point-like" We do not know that. Electron is known to be smaller than $10^{-18}$ m and could be point. – Ján Lalinský May 14 '16 at 11:14
  • Gravitational potential inside Earth is not proportional to $r^2$ but it is $C + kr^2$ with non-zero $C$, so $\phi(0)$ is not zero. – Ján Lalinský May 14 '16 at 11:16
  • @JánLalinský 1) electrons is QFT. Unless you have a well-defined concept of volume in QFT, your comment is pointless (pun slightly intended). 2) yes, thank you (but that doesn't really change the point of my post :P, the potential is still finite at the origin) – AccidentalFourierTransform May 14 '16 at 11:18
  • "Unless you have a well-defined concept of volume in QFT, your comment is pointless" My comment was not limited to QFT. Measurements of $g$-factor of electron give indication of size smaller than $10^{-18}$ m - see http://cerncourier.com/cws/article/cern/29724 and http://iopscience.iop.org/article/10.1088/0031-8949/1988/T22/016/meta In QFT, electrons are successfully described as point particles. – Ján Lalinský May 14 '16 at 11:46
  • @JánLalinský Im pretty familiar with electrons g-2 (Ive been working about that for more than half a year now, as it is my undergraduate final project). The measurement of g-2 indicates that the electron is not composite, but it says nothing about its volume. Volume is undefined in QFT, and the bound $10^{-\color{red}{22}}\ \mathrm{m}$ is simply stating that electrons are fundamental, not that they are point-like. "Fundamental" is well defined; "point-like" is not: its just a meaningless concept. Again: if you can define volume in QFT, please tell me how. Until then, this argument is pointless – AccidentalFourierTransform May 14 '16 at 11:52
  • @JánLalinský In QFT, electrons are "fundamental" or "elementary". This doesnt mean that they have zero volume (they might, they might not: we dont know because we dont even know how to define volume in QFT). For example, see http://physics.stackexchange.com/questions/41676/why-do-physicists-believe-that-particles-are-pointlike?lq=1 – AccidentalFourierTransform May 14 '16 at 12:00
  • I didn't say anything about volume, I only claimed we do not know whether electrons are points or not. The bounds published in the cited works are bounds on the electron (linear) size, they do not answer the question whether electron is a point or a fundamental particle. – Ján Lalinský May 14 '16 at 12:24