Why doesn't an electron rip itself apart?

Question

A proton is stable because of the strong force between quarks, which is not there in electron. So what's the reason for electron's stability?

There is still the issue of the self-force of the electron, which is nontrivial: https://physics.stackexchange.com/q/99285/ — Anders Sandberg, Jun 14 '21 at 14:26
How can you tell the electron is not a bound state of an ultra strong force which we have not assayed/seen yet? To upend your argument, if the proton is stable by dint of confinement of its constituents, how can you check an analogous option is not available to the electron? What do you make of all those "Rishon" models? — Cosmas Zachos, Jun 14 '21 at 15:50
Even if negative charge repelling negative charge stopped an electron being a point charge as per classical reasoning, it wouldn't be ripped apart, just end up at a very small (but empirically refuted) size. — J.G., Jun 14 '21 at 18:14
What would it become when torn apart on its own? I think there's nothing it could become, which is why it cannot be torn apart. — Daniel Darabos, Jun 16 '21 at 08:19
You tell me what you want the electron to rip itself apart INTO, and I'll tell you why that does not happen. But first you need to tell me what constituent parts you believe the electron is made out of. — PcMan, Jun 16 '21 at 18:18

score 47 · Answer 1 · answered Jun 14 '21 at 14:35

47

As far as we know, electrons are fundamental particles and have no internal structure or components. Also, an electron cannot decay into other particles (unless it has a very high kinetic energy) because there is no lighter charged lepton for it to decay into. It can, however, annihilate with a positron to produce gamma rays.

answered Jun 14 '21 at 14:35

gandalf61

52,505

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Actually, even at "very high kinetic energy" (which means it's travelling very fast in our local frame) can't decay into anything unless it interacts with another particle, and I wouldn't call that interaction with something else a 'decay' – poncho Jun 15 '21 at 14:03
@poncho many things that involve a particle interaction are refereed to as "decay". For example, K-40 -> Ar-40 – OrangeDog Jun 17 '21 at 14:00

anna v · Answer 2 · 2021-06-14T18:01:43.663

35

An electron is an elementary particle in the standard model of particle physics. . The table axiomatically assumes that elementary particles are point particles in the QFT of the model, i.e. have no constituent parts.

Depending on the quantum number conservation rules and if there exist consistent lower mass particles to decay to, elementary particles can decay, even though they have no constituents.

The electron has the electron quantum number , and the only lower mass particle is the electron neutrino, and the photon with zero mass is available, ( at least two for momentum conservation in the center of mass) but both are neutral so charge would not be conserved. Thus the electron is point like and stable, as far as our data and the theory that fits these data are concerned.

edited Jun 14 '21 at 18:01

answered Jun 14 '21 at 15:13

anna v

233,453

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This does not answer the question. It just postulates the stability of the electron because we have no explanation for it. – my2cts Jun 14 '21 at 17:05
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@my2cts Isnt this the way with all of physics? the why's end up on the axiomatic statements which were chosen in order to model the data, because the theory successfully does so with these axioms? The question touches on an axiom. – anna v Jun 14 '21 at 17:58
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@my2cts Actually it's the reverse: You question presupposes/implies that there actually is "something to rip apart", whereas the electron not ripping itself apart is more like a prime number not being factorizable into other integers. – David Tonhofer Jun 16 '21 at 07:00
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elementary particles can decay, even though they have no constituents thats an incredible point, well made – lineage Jun 17 '21 at 03:07
@annav I agree. However, an axiom is not an answer. On the contrary, it is the statement that we don't have an answer. – my2cts Jun 17 '21 at 08:44
@Will Fortunately, I did not say this. – my2cts Jun 17 '21 at 11:06
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There is this great word for something that can't break into pieces because it isn't made up of multiple constituent parts: atom. Unfortunately physical chemistry has already used it -- for a group of particles which we now know it does not describe. Of course, element also is in existing use, so "elementary particle" is not a satisfying solution either. – Ben Voigt Jun 17 '21 at 16:32

score 20 · Answer 3 · answered Jun 14 '21 at 18:10

A proton is stable because of the strong force between quarks, which is not there in electron

So you suggest a proton must rip itself apart, or has the ability to, as it is made up of quarks. But do the quarks also need to rip themselves apart? Its the same for the electron. We consider quarks and electron, both to be elementary - experimentally and theoretically. There is nothing in them to make them rip themselves apart.

Besides there are other deeper reasons, which other authors have remarked upon.

fraxinus · Answer 4 · 2021-06-16T21:12:25.700

"Why" is more of a philosophical question instead of physics one.

From our observations and experiments in the particle world it looks like two types of particles don't decay:

massless particles (photon, gluon) - they simply don't feel "time".
particles that can't decay without breaking some known conservation law (like electric charge or mass).

All others are known to decay into lighter particles until one gets into one of the two cases above.

The proton cannot decay into anything while still conserving electrical charge, baryon number and mass. All other known particles are either heavier, wrong baryon number, or electrical charge.

Electrons are limited in the same fashion - electrical charge, mass and lepton number are all conserved (as far as we know) properties.

Then again, we are not absolutely sure that electrons and protons don't decay. A lot of effort is made searching for decay modes for both the proton and the electron and their half-life is (as of now) limited to not less than some mind-boggling number of years like 10^35.

Observing a proton decay will invalidate some of the conservation laws as we know them.

edit:

A proton being a bound state of quarks doesn't change the picture. We don't know if the quarks are stable if unbound, they may as well not be, or at least the down quark may be able to decay into up one. We cannot separate them for long enough to see.

But, the bound state being stable while free particles unstable is pretty much known in the atomic nuclei. Neutrons are prone to beta-decay when free and pretty much stable when bound in a stable nucleus. The bound state has lower enough mass than its constituents to make the decay impossible.

score 1 · Answer 5 · answered Jun 14 '21 at 16:17

1

The electron is a point particle as far as physicists know. If you apply the electrostatic self energy formula for a charge distribution to a point particle, you will find infinity. The only conclusion we can draw from this is that we cannot consider an electron as a static charge distribution.

answered Jun 14 '21 at 16:17

my2cts

24,097

Do you mean that its charge sign alternates in time -in which case it would oscillate between a state in which it is, looks like an electron and a state in which it acts like a positron? – Anton Jun 28 '21 at 00:35

Vladimir Kalitvianski · Answer 6 · 2021-06-14T16:41:00.363

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An electron is not point-like, as a matter of fact. Its "size" is state-dependent. Anyway, if you puch a still free electron, you will get an excited final state - a moving electron plus soft radiation. Looks like it was not "free", but coupled permanently within the EMF oscillators, to say the least. This coupling smears the elastic (non destructive) picture (photo) of a "free" electron. Point-like picture is inclusive one - it includes all possible excitations during observation.

edited Jun 14 '21 at 16:41

answered Jun 14 '21 at 15:20

Vladimir Kalitvianski

93

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This answer is flawed. It confuses the wave function of an electron with the particle. – my2cts Jun 14 '21 at 16:11
@my2cts: No confusion is here. Read my popular paper https://arxiv.org/abs/0806.2635 – Vladimir Kalitvianski Jun 14 '21 at 16:37
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You cannot calculate the electron self interaction by considering the charge distribution described by its orbital. This idea, expressed as 'is "size" is state dependent', is fundamentally flawed. – my2cts Jun 14 '21 at 17:03
The idea of "self-interaction" is fundamentally flawed. The idea of interaction is fundamentally correct.I consider the latter. – Vladimir Kalitvianski Jun 14 '21 at 17:19
The question is unmistakably about self-interaction. By the way, I disagree that the concept of self-interaction is flawed. There are problems with it, but it is essential to physics. – my2cts Jun 14 '21 at 17:31
@my2cts: In CED this idea fails miserably. https://vladimirkalitvianski.wordpress.com/2013/11/12/on-the-higgs-fake-book-by-alexander-unzicker/ – Vladimir Kalitvianski Jun 15 '21 at 07:19
Without the self interaction term in the lagrangian a particle has no field at all. – my2cts Jun 15 '21 at 08:06
@my2cts: too self-confident statement. – Vladimir Kalitvianski Jun 15 '21 at 08:13
Why 'self' confident? This is not personal. This is fact. – my2cts Jun 15 '21 at 08:54
@my2cts: Because you do not see the flawness of this concept: on one hand you fix the electron shape (a sphere, for example); on the other hand you calculate the force that may not change the shape by definition. In CED you obtain an infinite self-induction for a point-like charge, which is not a small ratiation reaction at all. (I omit the static self force.) – Vladimir Kalitvianski Jun 15 '21 at 10:30
I am echoing the mainstream QED point of view, which is extremely successful and which I am aligned with. Mainstream physics does not deny the infinities but deals with these successfully. Equating particle shape to $|\psi|^2$ as you do is not a solution but a fatal flaw. – my2cts Jun 15 '21 at 10:38
I am not equating particle shape to $|\psi|^2$; I just distinguish elastic and inelastic channels from an inclusive "picture". – Vladimir Kalitvianski Jun 15 '21 at 11:04
Talking about "self-confident": "please read my popular paper" which has zero citations. (As context hopefully makes clear, that is not to say that citations are a measure of quality.) – tobi_s Jun 16 '21 at 03:19
@tobi_s: my paperhas zero citations because it is a hundred years late, so the mainstream subject has changed a lot. – Vladimir Kalitvianski Jun 17 '21 at 05:38
@VladimirKalitvianski The phrasing 'distinguish elastic and inelastic channels from an inclusive "picture".' has no meaning. Also, you state that your paper is 'popular'. That suggests readers, citations and acceptance which are not there. – my2cts Jun 17 '21 at 08:50
@Yes, everything is here: it has been published in peer review journals and it has been read by many. – Vladimir Kalitvianski Jun 17 '21 at 09:55
@VladimirKalitvianski I read in the abstract 'the "positive charge cloud" size in the Hydrogen ground state is much larger than the proper proton size'. So the paper is not about electrons. Also the electron does not see this cloud as proton and electron are correlated. The electron sees a near point charge, flattened only at the scale of the proton radius. This fact is actually used to determine the proton radius. – my2cts Jun 17 '21 at 14:20
@my2cts: read farther,please. – Vladimir Kalitvianski Jun 17 '21 at 15:44

score 0 · Answer 7 · answered Jun 15 '21 at 15:25

According to some theories (as Cosmas Zachos rightly commented), the electron can be torn apart. If you consider the electron made up of three more elementary particles (each with a unit charge of -1/3, so they actually are antiparticles) then it's also easy to see how they can change identity in high energy interactions. For example, an electron can interchange its constituents with a quark or neutrino to transform into a quark (while a quark can turn into an electron).
This shifts the question to the more elementary particles though. Why should a charged point particle be stable? First of all, it should be noted that all (electric) charge is ultimately made up from elementary charges. So it doesn't make sense to ask why they can't be subdivided further. They are just elementary charges, not made up of smaller charges. The infinite self-energy question isn't a question. This energy simply is not there.
Secondly, it could well be that the charges are not point-like. In a quantum theory of spacetime, it might well be that the pointy structures are some strange, very small (maybe Planck-sized) distortion of higher-dimensional space. It could be that this curling up of space (within the large, global structure of spacetime as described by general relativity) is all that a particle is. To say that a charge is present in this small curled-up space would be unnecessary, superfluous. Like in string theory charge is represented by the vibration of strings, which themselves contain no charge.

I've never heard some of this before. Is this your own theories? — wizzwizz4, Jun 15 '21 at 17:51
Yeah, and the -⅓ means antiparticle bit, and the “quantum theory of spacetime” bit. — wizzwizz4, Jun 15 '21 at 18:15
As stated by @CosmasZachos above in the comment there exists a rishon model. This states that electrons are made out of three anti-T-rishons. Likewise, the neutrino is three V-rishons. Etc. The spacetime distortion theory of particles can be seen in loop quantum gravity and related braid theory (which also predicts three antiparticles for an electron). https://en.wikipedia.org/wiki/Rishon_model The notion of particles being spacetime distortions is looked for in quantized spacetime, as spacetime itself is quantized and is as such well fitted for these structures. — , Jun 15 '21 at 18:21
@user304539 'spacetime itself is quantized' This is not a generally accepted concept. — my2cts, Jun 17 '21 at 08:26

score 0 · Answer 8 · answered Jun 21 '21 at 00:04

0

Even if one uses the metaphor / classical image of the electron as a spherical body of radius $R$ with a homogenous charge density, the value of the field does not diverge towards the center.

answered Jun 21 '21 at 00:04

Serge Hulne

1,744

score 0 · Answer 9 · answered Nov 17 '23 at 11:01

The simplest answer is: because it's held together by its own gravity. In fact, we could take this one step further: it actually is a gravitational soliton; nothing more, nothing less, and that there is nothing there to fly apart.

Here is a simple exercise that will help focus attention on the issue: write down an axially-symmetric solution to Einstein's field equations that has the same mass, same electric charge and same angular momentum as an elementary fermion - treating its spin as internal angular momentum: specifically the electron. What is that solution?

Answer: a naked Kerr-Newman singularity.

Now, having determined that to be the case, this raises the following question: since electrons (and more generally, fermions) are described by the Dirac equation, then shouldn't there then be some kind of close, and deep, correspondence between solutions to the Dirac equation and Kerr-Newman solutions? If what I said is true, then there must be, and this correspondence will then provide witness to the assertion made.

Well... actually, there is: The Dirac - Kerr-Newman Electron.

It's one of those "well-known" results that's been hanging around in the attic collecting dust, and hardly anyone's gone up to take a close look at it - partly because of the implication - and nobody's really figured out what to do with it. I'll cite the key points in the abstract:

We [...] show that the Dirac equation may naturally be incorporated into Kerr-Schild formalism as a master equation controlling the Kerr-Newman geometry. As a result, the Dirac electron acquires an extended space- time structure of the Kerr-Newman geometry - singular ring of the Compton size [...]

The rest you can read there in the link above. Do a web search on ArXiv for "Kerr-Newman Electron" and you'll find more recent material published on the matter. It's an area of active research.

Why doesn't an electron rip itself apart?

9 Answers9