6
$\begingroup$

Imagine empty infinite universe with just a single resting electron - let's ask the question about configuration of electric field in such empty universe.

The standard answer would be $E\propto 1/r^2$.

However, if calculating energy $(\propto E^2)$ of such electric field, due to singularity in $r=0$ we get

$$ \int_{0}^\infty 4\pi r^2 r^{-4} dr=\infty $$

In contrast, we know well that in reality this energy should be at most 511keV: released while annihilating with positron.

We would get 511keVs if integrating from $r_0 \approx 1.4$fm instead of zero - we need deformation of electric field in scale of femtometers not to exceed electron's mass with energy of electric field alone.

It is vaguely said that this issue is repaired by QED (how exactly?), but there still remains kind of basic question: what objectively would be electric field for single resting electron in empty universe?

I have met with two trials to solve this fundamental problem:

  1. That vacuum polarization reduces electric field near the singularity (is it satisfactory?).

  2. In soliton particle models (slides) we have $E\propto q(r)/r^2$, where effective charge $q(r)$ is practically constant for large $r$, but $q(r)\to 0$ for $r\to 0$ to prevent infinite energy. It is made by activating Higgs potential - kind of deforming electromagnetism into weak/strong interaction to regularize infinite energy. This kind of effect is observed as running coupling.

Can that vacuum polarization reduce energy of point charge below 511keV? Or maybe there is some other reasonable solutions to this problem?


Clarification: I see nobody defends vacuum polarization explanation, but there are lots of "impossibility claims" and avoiding answers, so let me briefly elaborate on solution to this problem suggested by topological solitons.

Let's look at the simplest vector field in 2D with Higgs-like potential to prefer unitary vectors: enter image description here

Requiring unitary vectors, $u(x)=x/|x|$ configuration would also have infinite energy due to discontinuity in the center. As in the diagram, it is regularized by getting out of minimum of Higgs potential (unitary vectors) - up to zero vector in the center, allowing to realize such topological charge using only finite energy.

To recreate electromagnetism in 3D for topological charges as electric charges, we can use Gauss-Bonnet theorem in place of Gauss law: it says that integrating curvature over a closed surface, we get topological charge inside this surface.

So interpreting curvature of a deeper field as electric field (analogously B), and using standard Lagrangian for it, we can recreate electromagnetism with two corrected issues: Gauss law allowing only integer (topological) charge (included charge quantization), and with charges containing only finite energy - some article.

Is there a problem with such explanation of finite energy of a charge, or maybe there are some better explanations?

$\endgroup$
  • 4
    $\begingroup$ There is currently no solution to this problem. $\endgroup$ – lalala Feb 16 '18 at 16:24
  • $\begingroup$ If the universe around the electron is empty, why would you invoke CED or QED? The results of these two theories are obtained in the presence of a (quantized) electromagnetic field, which would make the universe around the electron no longer empty. $\endgroup$ – DanielC Feb 16 '18 at 16:44
  • $\begingroup$ How could a universe be empty? It has a diameter for some reason. $\endgroup$ – Bill Alsept Feb 16 '18 at 16:48
  • $\begingroup$ Here you can find a basic discussion of this problem by Richard Feynman. $\endgroup$ – valerio Feb 16 '18 at 17:20
  • 1
    $\begingroup$ You are assuming that 1) the rest energy (511 keV) is entirely due to electromagnetic energy of a charge distribution 2) that this energy is proportional to electric field squared. But regarding 1), this is not necessary; the mass could be entirely non-electromagnetic, which is the case for point particles; regarding 2) the EM energy may not be given by the simple Poynting formula in the microscopic domain; again for point particle, the Poynting theorem cannot be interpreted as work-energy theorem. $\endgroup$ – Ján Lalinský Apr 2 '18 at 9:54
4
$\begingroup$

This problem is highly related to the fact that the electron mass requires renormalization in QED. They both arrise from the same basic physical idea: our theories don't hold to arbitrarily small length scales. When calculating the self-energy, you assume that the concept of an electromagnetic field holds to all scales, which is quite possibly not the case.

One way to remedy this situation is to give the electron a finite, but small, radius. Then, the self-energy is given by (setting $\epsilon_0=1$)

$$U=\frac{e^2}{2}\int_{r_e}^{\infty}\frac{\mathrm{d}r}{r^2}=\frac{e^2}{2r_e}.$$

Now, the mass of the electron we measure in the lab will be given by (setting $c=\mu_0=1$)

$$m(r_e)=m_0+\frac{e^2}{2r_e},$$

where $m_0$ is known as the "bare mass." The thing to do now is to think about what happens when we take $r_e\to 0$. Clearly, if $m_0$ is just a number, this would cause the measured mass to become infinite. However, if $m_0$ is formally infinite and negative, then $m(r_e)$ becomes positive and finite, if the value is tuned properly.

Now, this seems pretty philosophical (and it kind of is high level handwaving), but it can be used to actually make predictions. Roughly speaking, the radius we gave the electron is inversely proportional to the highest energy scale in the theory, call it $\Lambda$. In units where $\hbar=1$, we can simply write $r_e=1/\Lambda$. Then we have

$$m(\Lambda)=m_0+\frac{e^2}{2}\Lambda.$$

Now, if we consider that we are able to probe another energy cutoff $\Lambda'$, then

$$m(\Lambda')-m(\Lambda)=\frac{e^2}{2}\left(\Lambda'-\Lambda\right),$$

which allows us to predict how the electron mass changes with energy scale of observation! This is essentially the philosophy of renormalization, and takes on a life of its own in Quantum Field Theories.

I hope this helps!

PS: Feel free to correct me anywhere. I'm sure I made some algebraic/conceptual mistakes somewhere.

$\endgroup$
  • $\begingroup$ So what happens below r_e? We cannot just make a hole in a model, like ignoring the center of sun in modeling due to lack of direct measurement. And physics hates discontinuities due to their infinite energy - maybe we should search for a continuous solution, like in soliton models: E ~ q(r)/r^2 where q=e for large r, but q(r)->0 for r->0? $\endgroup$ – Jarek Duda Feb 17 '18 at 10:37
  • $\begingroup$ That is absolutely a possibility, and I invite you to try it. Generally, you should arrive at the same answer! This is because the low energy physics (trying to measure the electron mass at some energy scale) should always be independent of how you “regularize” the high energy physics. Regularization independence can be proven for a wide variety of problems (see, for instance, Schroeder’s discussion of the Casimir force), and it’s highly related to an important physical concept of universality (one low energy theory can arise (“flow”) from a large class of high energy theories). $\endgroup$ – Bob Knighton Feb 17 '18 at 17:09
3
$\begingroup$

One should separate the models used in describing elementary particles .

The classical model of $\tfrac{1}{r^2}$ singularity, depends on measuring the electric field with a test charge, i.e. it is the force felt by the test charge in the field of the subject charge. Usually the charge is in an extended volume in classical measurements, but it is true that the theoretical formalism leads to an infinity if the particle is a point.

Quantum mechanics solves this problem for the two body interaction by the quantization of the energy levels as the test charge nears the center of the subject charge, and has a ground state which does not allow overlap of charges, if it is an electron measuring a positron. It is how atoms are formed, the electron never lands on the proton in hydrogen.

Positronium exists for some time until the probability of overlap of electron positron leads to two photons, and the lifetimes agree with QED calculations.

It is true that the standard model has the elementary particles in its table as point particles , but it is a quantum field theory model, point particles are not the same as real particles. Real particles are modeled by wavepackets.

I am sure that more complicated models can be devised for specific measurements, but one should remember that the underlying level of classical physics is quantum mechanical. Classical emerges from quantum,, not the other way around, and the classical singularity is just a singularity in a model that fits classical measurements, that's all, in my opinion.

$\endgroup$
  • $\begingroup$ It is interesting to go through the use of wave packets for another puzzle , here physics.stackexchange.com/questions/369902/… $\endgroup$ – anna v Feb 17 '18 at 4:52
  • $\begingroup$ I see you agree that electron is not a perfect point, rather e.g. a wavepacket. The question is how electric field behaves around the center of such wavepacket. Can effective charge be reduced in high energy collisions, for example E ~ q(r)/r^2 where q=e for large r, but q(r)->0 for r->0 to avoid infinite energy of the field? $\endgroup$ – Jarek Duda Feb 17 '18 at 10:45
  • $\begingroup$ It is complicated but you could get a feeling by reading the link I gave in the comment , where Zachos addresses neutrino oscillations at the neutrino's rest frame using wave packets. A wave packet scattering off a wave packet arxiv.org/pdf/physics/9909042.pdf $\endgroup$ – anna v Feb 17 '18 at 11:26
  • $\begingroup$ have a look at this definition to charge density en.wikipedia.org/wiki/… . Probabilities are never infinite, but bounded by 1. $\endgroup$ – anna v Feb 17 '18 at 11:29
  • $\begingroup$ This is about charge density, while my question regards electric field. If electron as wavepacket should be understood as superposition of point charges, each such situation has still infinite energy, so this is not a solution. However, if we assume that 'e' charge is objectively smeared, calculating electric field as average over density, there might be a chance for finite energy of electric field (?) Have you maybe seen a paper with such claims? However, shouldn't smeared charge of electron be seen in high energy collisions? Running coupling suggests that effective charge is reduced there. $\endgroup$ – Jarek Duda Feb 17 '18 at 12:38
1
$\begingroup$

The infinite electrostatic energy of an electron is related to the idea that the necessary work to accumulate its charge by infinitesimal charges in one point against the electrostatic forces they exert on each other is infinite. This is also related to the infinite electrostatic field energy of the point particle. The electron, however, has not been assembled by such a process. As far as we know, it has always existed as a point particle with an elementary charge. Therefore, this supposed infinite potential energy cannot be extracted from the electron. It is usually disregarded or "subtracted" like other infinities in theoretical physics

$\endgroup$
  • $\begingroup$ We are certain that this is not infinite energy as 511keV can be released in annihilation, charge can be constructed from 2*511keV in pair creation. There is a problem with model - requiring a way to regularize the singularity. It can be repaired e.g. if modeling electric charge as topological charge (above). $\endgroup$ – Jarek Duda Feb 17 '18 at 10:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.