6
$\begingroup$

I've heard compelling reasons to think that it is one although why do we assert this in light of the calculation which shows a point particle creates an electrostatic field of infinite energy (see e.g. Griffiths ED 2.4.4 or Feynman Lectures vol. 2 section 28)?

Why do people not consider this calculation a definitive reason to abandon the notion that the electron is a point particle?

$\endgroup$
15
  • 5
    $\begingroup$ In modern physics an electron is a quantum. Quanta are not objects but amounts of energy, momentum, angular momentum and charges (electric charge but also other quantum numbers like lepton number). There are, unfortunately, a lot of really poor explanations in circulation, both among lay people and physicists. $\endgroup$ Commented Apr 12, 2023 at 4:21
  • 2
    $\begingroup$ Does this answer your question? The problem of infinite energy of electron as point charge? $\endgroup$ Commented Apr 12, 2023 at 4:46
  • 2
    $\begingroup$ You seem to have a factual misconception. Namely, that an electron 'is a point particle'. According to modern knowledge, it's not; that's just not how quantum field theory (or even quantum mechanics) works. Also, if you allow a point particle, aren't you worried about the gravitational singularity, too? $\endgroup$
    – Myridium
    Commented Apr 12, 2023 at 4:49
  • 2
    $\begingroup$ @Myridium What localizes an electron in an experiment is always a piece of matter. We can, of course, make statements about the minimal spatial cutoff that we need to explain scattering between electrons or electrons and photons and we know that there is no internal structure in any of the experiments we have done, so far. But there is a difference between field interactions without a spatial scale and a "point particle". It is esoteric for most applications, which is why most physicists don't (have to) care about it. Those who do often seem to end up on the wrong side of the issue, though. $\endgroup$ Commented Apr 12, 2023 at 4:56
  • 1
    $\begingroup$ @Myridium If it's not a point particle, in what sense do people mean it is one when asserting that it is in QFT (e.g. the response from Anna V in link provided in original question)? $\endgroup$ Commented Apr 12, 2023 at 6:11

6 Answers 6

10
$\begingroup$

Why do we insist that the electron be a point particle [...]?

We do not. For example, measurements of the electron validate that it's charge distribution has spherical symmetry to within an error of $10^{-15}$. And point can't have any geometrical symmetries, hence electron is not a point particle.

Point particle approximations are used in some models for useful first order insights.

It's like when you analyze the Sun's trajectory in our galaxy: you can imagine the Sun being a "point particle", because ${d_\odot}/{r} \approx 10^{-12}\ll 1$, where $r$ is our distance to Milky Way galaxy center (about $24~\textrm{kly}$). Hence, the Sun's diameter can be ignored in these types of problems. However, in other types of problems it is important, like evaluating black body radiation, measuring the exact barycenter of solar system rotation etc.

It is the same for particles: if the model at hand doesn't require exact shape of particle, it doesn't use it.

$\endgroup$
16
  • 13
    $\begingroup$ The news page you cite in fact states that the electron is a sphere. The articles cited by that news page, as far as I understand, only state that the eletron's charge is spherically symmetric. Seems to be a strange simplification (nature-style). doi.org/10.1103/PhysRevLett.88.071805 and doi.org/10.1038/nature10104 $\endgroup$ Commented Apr 12, 2023 at 12:07
  • 3
    $\begingroup$ @reciprocallettuce - in essence, those groups are probing for indications that the electron is a composite, not an elemental, particle. $\endgroup$
    – Jon Custer
    Commented Apr 12, 2023 at 12:30
  • 1
    $\begingroup$ @reciprocallettuce In fact if to be precise,- mentioned researches measures electron electric dipole moment. If it has some, then electric fields should change electron spin orientation even a bit. And the fact that they have found none, shows that electron's negative charge is distributed spherically without any "hills" or "pits". This is equivalent of saying that shape of electron is an "ideal" sphere. I don't see the difference between the two. What you say - is more "a word play". $\endgroup$ Commented Apr 12, 2023 at 13:15
  • 9
    $\begingroup$ The difference between an object exhibiting spherical symmetry and an actual sphere is much more than just "word play". The atmosphere of the earth, for example, exhibits spherical symmetry (to a certain approximation), but it is not an actual sphere. Even better, the electric field generated by a point charge (such as an electron) exhibits spherical symmetry, but it is not an actual sphere. $\endgroup$
    – Lee Mosher
    Commented Apr 12, 2023 at 23:51
  • 2
    $\begingroup$ The reason that it can't be non-sharp delimited is simply that this is not what a sphere is – it's a matter of definition, and yours is wrong (or nonstandard if you prefer). A sphere has/is a surface that is a submanifold, i.e. the extent in normal direction is vanishing or at least substantially smaller than the extent in the tangential directions. A radial gradient doesn't have that property. Of course it has level sets that have the property, but that's a different story. $\endgroup$ Commented Apr 13, 2023 at 12:05
5
$\begingroup$

why do we assert this in light of the calculation which shows a point particle creates an electrostatic field of infinite energy

Actually this is one of the reasons quantum mechanics had to be invented. If there were a singularity at the center of charged particles then when they were attracted to each other they would fall and neutralize so no atoms and molecules as we have measured them would exist, or us either. Bohr with his model tried to solve this by using a planetary model for bound states that has a quantized angular momentum. This eventually led to the theory of quantum mechanics and its postulates where rigorous mathematics can be used to calculate and predict the behavior of particles in the phase space region where classical electrodynamics would predict singularities.

The wave function postulate is important in this, second page in the link. The wave nature of the solution for interactions between particles, controls their behavior in interactions. Only the probability distributions of a particle being at (x,y,z,t) with four vector $(p_x,p_y,p_z,E)$ can be calculated.

See this answer of mine to get a feeling about the probability build up of the wave nature.

$\endgroup$
7
  • 1
    $\begingroup$ But quantum mechanics doesn't resolve the calculation. Quantum mechanics includes the 1/r Coulomb singularity in the H atom and gets quantization in spite of it. Meanwhile, a free electron can still be constructed to have arbitrarily large electrostatic energy via wavepackets. $\endgroup$ Commented Apr 12, 2023 at 6:13
  • $\begingroup$ @greatscissors you can do anything with mathematics , The point is to fit the data, and QFT fits the data and is predictive. The 1/r in the quantum mechanical equation defines the wavefunction solutions, the probabilities, not the location of the particle. $\endgroup$
    – anna v
    Commented Apr 12, 2023 at 6:24
  • 2
    $\begingroup$ @greatscissors Unless I've misunderstood your meaning, the fact that an electron gets arbitrarily large energy (expectation value) when it is arbitrarily localized has nothing to do with electrostatic energy. The same would happen to a neutron (even though it has a radius unlike the electron!) $\endgroup$
    – AXensen
    Commented Apr 12, 2023 at 8:46
  • $\begingroup$ I think this has nothing to do with the question. The reason for introducing "quantum" in the "pre-Bohr classical planetary model of the atom" is that the accelerated charge moving around looses energy through radiation. You don't need the singularity of the electric potential for that, just replace it by $\frac{1}{r+\varepsilon}$, $\varepsilon>0$, and the argument by Bohr should still work, right? $\endgroup$
    – kricheli
    Commented Apr 13, 2023 at 7:11
  • 1
    $\begingroup$ Of course. But if $\varepsilon$ is sufficiently small or you have an electron form factor with a sufficiently small radius, you don't see the difference/the data fits as well. But the singularity is gone. $\endgroup$
    – kricheli
    Commented Apr 13, 2023 at 7:20
5
$\begingroup$

When we write down the equations of the Standard Model of Elementary particles, the electron appears as a field that is represented as an irreducible representation of the Lorentz group, the Dirac spinor. This is an object which has no substructure. Note that this is subtly different from a point particle, in that this is not a statement about a spatial extent. Instead, it is a statement about the field's response to Lorentz transformations (or more generally, Poincaré transformations, this includes rotations and changes in uniform velocity).

In particular, the fields are arranged in such a way that the equations of motions stay independent of a choice of inertial frame (technically, the Lagrangian is invariant under Lorentz transformations, for the free Dirac field this looks like $\mathcal{L}_\textrm{free} = \bar \psi ( i \hbar c \partial\llap{\unicode{x2215}} - m_ec^2) \psi$). This is something that in a classical setting cannot be done for an extended object. This is easy to understand if you are familiar with Lorentz contraction: e.g. a sphere will become an ellipsoid in any frame where it is moving, and thus the rotational symmetry is broken. So in classical theory, only a point-like object can be Lorentz-invariant. I believe this is why we call objects without substructure, i.e. those that transform according to irreducible representations, "point-like".

My old answer gives some insight into how the spatial "size" of such an object can be usefully defined (a radius would be the square root of the cross-sections mentioned there give or take a factor $\pi$). Now, the only reaction-independent length scale you can construct in a theory of only electrons and photons is $\frac{e^2}{m_ec^2} \approx 10^{-15}\textrm{m}$ which is known as the classical electron radius. Why is it the only one? Because there are only two inputs characterizing the theory: the electron charge $e$ and the electron mass $m_e$, and this is the only combination that has dimensions of a length. Since this is the only length scale, it is bound to appear all over the cross-sections related to electron scattering, and in that sense it can usefully be considered a measure of the size of the electron. Let me emphasize though that the actual cross-section in any reaction will also depend on the other available length scales in the reaction (derived from masses of other particles, center-of-mass energies, energy transfers ...), so one shouldn't think of the classical electron radius as "the" size of the electron.

$\endgroup$
4
$\begingroup$

Why do people not consider this calculation a definitive reason to abandon the notion that the electron is a point particle?

Because that calculation is based on a dubious assumption that energy associated with point charged particle is given by the standard formula for Coulomb energy of a sphere, or extended charged object with spherical symmetry. When we try radius consistent with experiments on electrons ($r<$1e-18 m), we get energy higher than rest energy of the electron. Thus the Coulomb interaction energy formula is extrapolated somewhere where it clearly does not work well.

$\endgroup$
1
  • $\begingroup$ Yes, thank you. This discussion went a bit off-topic from what I was originally looking for but this is really the sort of reasoning I was after. $\endgroup$ Commented Apr 14, 2023 at 1:38
1
$\begingroup$

If you allow me to paraphrase: You are making the case that since the energy of a Coulomb electric field is infinite at the origin, the electron must not really be a point particle. First I will summarize some evidence in favor of point-like particles, without coming to a final conclusion. Then I will address the Coulomb potential and our ability to make a conclusion from that infinity.

On the former: Well, really every experiment that has been done shows particle-like nature on each individual run. Let me give you some example for context.

Check out this image of the trajectory of an electron from a cloud chamber: enter image description here

In addition, in a screen-like setup like a multi-channel plate, which is basically a bunch of cavities that an electron can enter, we only ever get a signal in one channel at once.

What else? Consider the Milikan's oil drop experiment. There's a good video of it here. The charge of the electron was found in this experiment, and every charge seen in the experiment is a multiple of that base charge. That suggests a final granularity of electrons.

There are pretty much endless other examples.

However, as we know, there is a wave nature of the electron as well. This wave-particle "duality" has confused mankind for more than 100 years and we still don't have it all quite cleanly sorted out today. Thus many of the arguments around measurement in Quantum Mechanics.

Finally I would like to add that while some experiments are suggestive of pointlike particles, this does not have to be the final say in nature. I show these experiments only to present some relevant evidence that can be weighed in.

So, how are these compatible with the infinite energy of a Coulomb potential?

The infinite energy of a Coulomb potential is an issue for sure. But we already know that our electrodynamics is not the final story, so an issue there is likely just indication of the limitations of our theory. Quantum Electrodynamics is a more fundamental theory, however even here the singularity still causes issues. We are able to tame these issues with a process called renormalization.

But regardless of whether one considers this an issue in QED... even QED is not the end of the story. Since we don't have a final theory, we don't know what an electron actually IS. So we can't rule out the possibility that there is a way for it to have a point-like nature while still having a total energy that doesn't cause our theories to blow up. The way of calculating energy (if there even is one in a hypothetical final theory) may be completely different than $\int\vec{E}^2 dV$. As such, we have no way of ruling out a pointlike nature of the electron.

$\endgroup$
1
  • $\begingroup$ Comments have been moved to chat; please do not continue the discussion here. Before posting a comment below this one, please review the purposes of comments. Comments that do not request clarification or suggest improvements usually belong as an answer, on Physics Meta, or in Physics Chat. Comments continuing discussion may be removed. $\endgroup$
    – Buzz
    Commented Apr 13, 2023 at 18:34
0
$\begingroup$

An interesting point. I’ve had similar problems with the origins and properties of the electron. And they also indicate that the concept of the electron as a point particle with only neg charge should be questioned. My research shows that in 1897 Zeeman noticed spectral splitting of atomic emission lines.Lorentz explained this as proof of existence of the electron as a point particle ,proof of Lorentz force and validation of his model of electromagnetism. Yet in fact in his theory, his ion ( That’s what he called the electron initially) emits polarised radiation perpendicular to its plane of Rotation. Clearly violating all observations of induction and emission of emr up till that point. ( emr waves can only be emitted parallel to the rotation plane of a dipole). What seems to happen is Lorentz, ignored all empirical observation current to 1897 and erroneously assumes his ion /electron is a point particle with a neg charge. So my answer to,your question is Yes, people should abandon this concept of the electron as a point particle. And the reason is that its properties contradict all known observations of emr.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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