# If like charges repel, why doesn't a charge break itself apart?

How can it be that, if like charges repel, they don't repel themselves? In other words, why don't charges break apart?

About the possible duplicate: I want to know about charges in general, not just that of an electron.

My response to Lawrence B. Crowell's answer:

Thank you very much for the extensive explanation. Unfortunately it's a bit above my level (I'm a first year electrical engineering bachelor student). This is how I understand it: If the charge is the sum of multiple separate charges, there has to be an external force that keeps this charges together.

However we most often think of electrons and protons as point charges. In other words, we don't think of them as if they are made up from different (smaller) parts.

But the idea of electrons and protons as point charges has its own problems. I didn't know that the electric field has a mass. From the formula, I understand that $$m\propto \frac{1}{r}$$. This would mean that $$m\rightarrow \infty$$ as $$r \rightarrow 0$$ (or $$r = 0$$), which of course is physically not possible.

But there is a way to calculate the radius of an electron based on the mass, the speed of light and the constant of Planck. (Unfortunately I don't know why.)

You can get around this by a technique called called renormalization, which causes the integral to converge.

Unfortunately, I don't understand the rest of the answer due to my level in physics. Nevertheless I am grateful for you answer. It would be awesome if you could confirm or correct my understanding.

• Possible duplicate: physics.stackexchange.com/q/231495/2451 – Qmechanic Jul 19 '16 at 13:58
• @Qmechanic : Here's a conundrum : This Qn is a duplicate of the one you cite. However, this has 5 Answers (1 accepted, with 9 upvotes) whereas the other has 1 brief Answer (3 upvotes, not accepted). Maybe that Qn should be closed as a duplicate of this? – sammy gerbil Jul 19 '16 at 17:07
• @sammy gerbil: Well, you have the power to vote to close as a dupe either way. – Qmechanic Jul 19 '16 at 17:19
• @Qmechanic : I have voted to Leave Open this Qn. I shall propose closing the other. – sammy gerbil Jul 19 '16 at 17:30
• Kevin : Please could you clarify that, when you say you are asking about "charges in general", you mean sub-atomic particles (as interpreted by valerio, LBC, etc) and not macroscopic charged objects (as addressed at first by GNA)? The distinction ought to be clear in your title also, IMO. – sammy gerbil Jul 19 '16 at 18:17

This was one of those big questions in the 19th century. It still causes some consternation. If you have a composite system, such as the nucleus of an atom, some other force is necessary. This force of course is the nuclear interaction. This keeps the protons from flying apart, though for some unstable nuclei there are transitions that eject charged particles, electrons or positrons, due to weak interactions. In the case of the proton it is composed of three quarks and these are bound to each other by the QCD (quantum chromodynamics) interaction. The gauge bosons called gluons interact most strongly at low energy and these keep the quarks, with charges $$2/3,2/3,-1/3$$ in a bound state.

Things are a bit more mysterious with point-like particles, such as the electron and other leptons and quarks. We generally do not regard such particles as composite, though this has not stopped people from proposing constituents called preons or rishons that make them up.

There is a problem with defining the mass of the electron or any point-like electrically charged particle. The mass of the electric field is $$m_\textrm{em}~=~\frac{1}{2}\int E^2~\mathrm d^3r~=~\frac{1}{2}\int_r^\infty\left(\frac{e}{4\pi r^2}\right)^24\pi r^2~\mathrm dr~=~\frac{e^2}{8\pi r}.$$ if the electron has zero radius this is divergent. There is the classical radius of the electron $$r~=~\alpha\lambda_c$$ $$=~2.8\times10^{-13}~\mathrm{cm}$$ for $$\lambda_c~=~\hbar/mc$$ the Compton wavelength. This raises some questions, for the classical radius suggests "structure," and it also has a relationship to something called Zitterbewegung.

A more standard approach to this is renormalization. A screenshot of this is to look at this integral with the variables $$p~=~1/r$$ so in this integral above $$\mathrm dr/r~\rightarrow~-\mathrm dp/p$$. Here we are thinking of momentum and wavelength or position as reciprocally related. This integral is then evaluated for a finite $$r$$ as equivalent to being evaluated for a finite momentum cut off $$\Lambda$$ $$I(\Lambda)~=~\int_0^\Lambda\frac{\mathrm dp}{p}~\simeq~1~+~2^{-1}~+~3^{-1}~\dots$$ which is equal to $$\lim_{\Lambda\rightarrow\infty}I(\Lambda)~=~-\zeta(1)$$ In some ways this is a removal of infinities. Another curious way to look at this is with $$p$$-adic number theory. This is a topic that could consume a lot of bandwidth.

We have another way to look at this. This comes down to the question of what do we mean by "composite." It also forces us to think about what we mean by the locality of field operators. The Dirac magnetic monopole is a solenoid with an opening to an infinite coil. The condition for the Dirac monopole is that the Aharonov-Bohm phase of a quantum system is zero as it passes the "tube" of the solenoid $$\psi~\rightarrow~\exp\left(ie/\hbar\displaystyle\oint{\vec A}\cdot ~\mathrm d{\vec r}\right)\psi$$. This might be compared to "cutting off the tail" on the magnetic monopole charge. The vanishing of this is equivalent to saying $$2\pi N~=~\frac{e}{\hbar}\displaystyle\oint{\vec A}\cdot ~\mathrm d{\vec r}~=~\frac{e}{\hbar}\iint\nabla\times{\vec A}\cdot{\vec a},$$ for the integral evaluated over units of area of the opening. This is of course the magnetic field $${\vec B}~=~-\nabla\times{\vec A}$$ evaluated in a Gauss' law that gives the magnetic monopole charge $$g~=~\displaystyle\iint\nabla\times{\vec A}\cdot{\vec a}$$ and we use this expression to see the S-duality relationship between the electric and magnetic monopole charge $$eg~=~2\pi N\hbar,$$ sometimes called the Montonen-Olive relationship.

This means that if we have an electric charge we can use the renormalization machinery to illustrate how the vacuum around it is polarized with virtual particles according to $$\alpha~=~\frac{e^2}{4\pi\epsilon\hbar c}$$. The electric charge is comparatively weak in strength with a modest polarization of the vacuum expanded in orders of $$\alpha$$ for $$N$$ internal lines or loops. This S-dual relationship tells us that while this is modest, the magnetic monopole is very strong and the vacuum is a "bee's nest" of lots of particles. This then means the dual of the electric field is a magnetic monopole field that in some ways appears composite.

This means in some ways we have questions needed to be asked about the locality of field operators. Something that appears local, point-like and "nice" may be dual to something that appears not so local, more composite-like and not renormalizable. As a result there are still open questions on this, and even Feynman agreed with Dirac that the situation with QED was not perfectly satisfactory.

• The first two paragraphs are a fine answer to the original question, but the rest about mass and infinities is only tangential and not very illuminating. I would edit down so the vital part does not get lost in long text. – Ján Lalinský Jul 19 '16 at 13:16
• I'd second Ján's point - this is likely to attract a broad audience (probably on the HNQ sidebar soon enough) and we want nice, correct answers which also have easily understood tl;dr's. On a related vein, having unexplained acronyms like QCD is also not great. – Emilio Pisanty Jul 19 '16 at 13:37
• I always found that a strong point of this site was the range of answers it gave, ranging from answers similiar in depth to the above to, ok I admit it, ones that I could understand. The answers are not just for the OP. – user108787 Jul 19 '16 at 13:56
• I can understand some consternation about this, but this question is very subtle and maybe just as important today as in the 19th century. It has some bearing on nonperturbative QFT, locality or nonlocality of fields. Dirac and Feynman admitted that the answers given by regularization can only be approximations or effective. – Lawrence B. Crowell Jul 19 '16 at 17:00
• Fun fact: it's actually "Zitterbewegung" (i.e. "trembling motion"). "Zwitterbewegung" would translate to "hermaphrodite motion"... and I don't think that has something to do with the topic. ;-) – tipavi Jul 20 '16 at 0:41

Composite particles like protons don't break apart because of the strong interaction which holds their constituents (the quarks) together. Elementary particles like electrons don't break apart because they are point-like particles, i.e. they are not made of “parts” (if the Standard Model is correct).

• @njzk2 : Hmm... I wonder about an upvoting system that would take predicates for revoking upvotes: "+1 Until meticulously-defined-condition". – Eric Towers Jul 19 '16 at 22:41
• @EricTowers: +1 iff this comment does not have my upvote – TLW Jul 20 '16 at 2:21
• This still not explains why is a point particle does not interact with itself (so that it's quantum wavefunction would expand faster). For that, one need to look into the expression of the interaction in the framework of quantum electrodynamics. See van.physics.illinois.edu/qa/listing.php?id=1162 – Zsolt Szatmari Jul 20 '16 at 10:01
• The fact that the particle is point-like is largely irrelevant. The important thing is that electrons, neutrinos, etc. are quantised/indivisible/atomic - so there does not exist anything for them to break apart into. – OrangeDog Jul 20 '16 at 11:43
• @OrangeDog +1 Which should definitely make one wonder especially knowing etymology of "atom". Also, I don't like the answer for suggesting electron is a point particle when it's clearly far from it, e.g. benzene ring. Calling it point-like seems like a delicate thing, but it's very important, precisely because it is meant to drop the localization part of the the definition. – luk32 Jul 20 '16 at 17:57

When you have a charged object, for example a charged metal sphere, of course the charges on the surface of the sphere interfere with each other. Because of these effect the charge gets distributed equally over the sphere.
However, these effects are not big enough to actually break up the sphere or something like that.

If your object is charged high enough there can be discharges to other objects (like the air) because of the potential difference.

Looking at an electron: It is a subatomic particle and can not be split in "half". It is not like an object that carries the charge. The charge is a fundamental property of the electron itself.

Protons contain two $\frac{2}{3}$-positive charged up-quarks and one $\frac{1}{3}$-negative charged down-quark. The quarks are glued together through gluons. These bring up the necessary force to keep the proton together. Quarks are also elementary particles, as per the Standard Model

Using a particle accelerator you can smash protons and other particles into each other. In this case it is possible to destroy the bindings between the quarks and new particles are created.

• Thanks for the answer. So an electron is a charge, instead of having a charge. If the charge is the sum of multiple other charges, there has to be a force that keeps this multiple charges together. – Kevin Jul 19 '16 at 17:37
• @Kevin. Yes. But let's clarify that the electron is NOT a charge. The charge is more or less a property of the electron. Just like mass and spin. – GNA Jul 19 '16 at 19:12
• @Kevin, electron is not only charge, it has many other properties that are not implied by its electric charge. For example, electron has mass 9E-31 kg. We do not know if this is related to its charge 1.6E-19 C in any way. – Ján Lalinský Jul 23 '16 at 17:25

Like charges repel via the electromagnetic interaction which is mediated by exchange particles (gauge bosons) called the photon. Since the photon is massless, the electromagnetic force has infinite range, and all like charges will attempt to break apart from each other. However, those that don't are being held together by a force that is not electromagnetic in nature, attractive forces like the Strong Nuclear Force or Gravity.

It is well known a proton consists of uud quarks, and the reason why the repelling up quarks who each have a +2/3 elementary charge do not separate from one another is due to an even stronger force holding them together, due to the flux tubes between them clearing the gluon field causing stability. This is also the namesake of the strong nuclear force, without it, matter would not exist.

There is also a unit to a charge, at the scale of a fundamental particle. This indivisibility is consistent with why the electron cannot shatter into smaller charges, it IS the smallest charge. According to modern understanding, the electron is a point particle with a point charge and no spatial extent.Attempts to model the electron as a non-point particle are considered to be inconsistent with reality.

Another point of view is to consider the electron as quantum-mechanically smeared due to its permanent coupling to electromagnetic field oscillators. Such a construction is rather "soft" and is easy to excite - radiate and absorb soft photons. In this sense, this construction is not elementary and point-like. Point-likeness is then an inclusive picture, not the "elastic" one.

I would like to add that if we do not consider the elementary particles but think of those charged spheres made of metal, they can actually break. If you keep on removing electrons from a material block and protect the discharge from the neighboring atmosphere, after a stage the repulsion among the like charges become stronger than their cohesive force of chemical bonds and the material will explode. This phenomena is known as Coulomb explosion. It is mainly observed in nano particles and used to generate energetic ions.