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What is the explanation between equality of proton and electron charges (up to a sign)? This is connected to the gauge invariance and renormalization of charge is connected to the renormalization of photon field, but is this explanation enough? Do we have some experimental evidence that quarks have 2/3 and -1/3 charges? By the way I think about bare charge of electron and proton. And I am also wondering if this can be explained by Standard Model.

Qmechanic
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Newman
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5 Answers5

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Because a proton can decay to a positron. It is an experimental fact that the proton and positron charges are very close. To conclude that they are exactly equal requires an argument. If a proton could theoretically decay to a positron and neutral stuff, this is enough.

In QED, charge quantization is equivalent to the statement that the gauge group is compact. This means that there is a gauge transformation by a full $2\pi$ rotation of the fields which is equivalent to nothing at all. Under these circumstances you have the following:

  • Charge is quantized
  • There are Dirac string solutions which have a magnetic flux indistinguishable from no flux (the magnetic flux is the phase around a loop).

If you have any sort of ultraviolet regulator, either a GUT or gravity, the existence of Dirac strings leads to monopoles. If you don't have an ultraviolet regulator, it is consistent to make all the monopoles infinitely massive.

So the question is why is the U(1) of electromagnetism compact. There are two avenues for answering this:

  • A compact U(1) emerges from a higher gauge group, because all higher gauge groups must be compact for the kinetic terms to have the right sign. Breaking a compact group produces a subgroup, which is necessarily compact.

It is also true that in any GUT theory producing electromagnetism, you get monopoles, so you automatically get charge quantization by Dirac's argument.

But even if you have a U(1) which is not part of a GUT, there are constraints from gravity. If you have particle with charge q and a particle with charge q', and they aren't rational multiples of each other, you can produce a particle with charge $nq - m q'$ by throwing n q particles into a black hole, waiting for m q' particles to come out, and letting the resulting black hole decay, while throwing back any charge particle that comes out.

This means that in a consistent quantum gravity, you need either charge quantization or a spectrum of charges that accumulates near zero. Further, in order for the theory to be consistent, a black hole made from the wee charges must be able to naturally decay to wee charged things, and barring a conspiratorial spectrum of charges and masses, this strongly suggests that the mass of the wee charges must be smaller than the charge, meaning that as the charge gets small they become massless.

So in quantum gravity, the only alternative to charge quantization is a theory with nearly massless particles with extremely tiny charges, and this has clear experimental signatures.

I should point out that if you believe that the standard model matter is complete, then anomaly cancellation requires that the charge of the proton is equal to the charge of the positron, because there is instanton mediated proton decay as discovered by t'Hooft, and this is something we might concievable soon observe in accelerators. So in order to make the charge of the proton slightly different from the electron, you can't modify parameters in the standard model, you need to add a heck of a lot of unobserved nearly massless fermions with tiny U(1) charge.

This is enough conspiratorial implausibility, that together with the experimental bound, you can say with certainty that the proton and electron have exactly the same charge.

Ron Maimon
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    There is zero experimental evidence for proton decay to positron. The current state of the field is that the fact that the proton and positron have the same quantum of charge allows the possibility that a proton could decay to a positron and some neutral detritus. – rob Sep 08 '15 at 13:43
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    @rob I think he's referring to beta decay of a proton inside a nucleus, not a hypothetical decay of a free proton. – Timaeus Mar 10 '16 at 20:02
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    @rob, I agree with you that Ron Maimon was referring to proton->positron, not beta decay inside a nucleus. However, Timaeus is also correct in pointing out that we should consider the significance of the more common-place and observable beta decays. – Hackless Oct 29 '19 at 20:41
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    This answer would be more useful if some sources were given. The best experimental limit on electron-proton charge difference actually comes from $SF_6$ neutrality, not any beta decay. (And certainly not from unseen GUT proton decays.) The black hole argument here has been referred to in Physical Review Letters as a 'folk theorem' that may not be rock solid. Do experiments actually sufficiently exclude the existence of the "heck of a lot" of light fermions with tiny electric charges? People are still looking. – David Bailey Nov 16 '22 at 07:32
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On the level of QED and above, the equality of the charges has no theoretical explanation. But it is extremely well established experimentally, as even small deviations would add up to huge amounts of electricity in bulk matter.

On the level of the standard model, the value of the charges of the up and down quark comes from simple arithmetic from those of the proton and neutron, and hence doesn't give an independent piece of information.

On the other hand, if a unified field thoery with a semisimple gauge group were found to be valid, it would forces charge quantization, as there are only a discrete number of irreducible unitary representations (which define the possible quantum numbers = charges). This would turn an approximate equality in an exact equality, and hence prove the equality of the charges of the proton and the electron (apart from the sign). Thus it would explain this equality.

By the way, bare charges of charged elementary particles are infinite and devoid of any physical meaning.

Arnold Neumaier
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    Even in the standard model the charge of the proton must be equal and opposite to the charge of the electron by anomaly cancellation. This follows from the instanton induced proton decay in the SM. – Ron Maimon Aug 12 '12 at 20:09
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    @RonMaimon: I thought the proton is stable in the SM. Please give a reference for its decay. – Arnold Neumaier Aug 13 '12 at 13:54
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    This is a common misconception, not only is B not conserved, it may be violated already at 40TeV collisions for all we know ( http://physics.stackexchange.com/questions/32080/what-is-the-rate-of-b-violation-expected-in-the-standard-model-during-high-energ). The original reference is a very clear classic: G. 't Hooft, Phys. Rev. Lett. 37, 37 (1976). The argument is that each instanton changes the occupation number of zero modes that go from negative to positive energy through the pseudoparticle in such a way to turn anomaly cancellation link between leptons and quarks into a physical process. – Ron Maimon Aug 13 '12 at 19:27
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    The correct reference is G. 't Hooft, Phys. Rev. Lett. 37, 8 (1976). - I need to digest this first. – Arnold Neumaier Aug 14 '12 at 08:41
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The answer is "because". It is an experimental fact.

It is among the first data that were gathered which supported the atomic theory. If they were not the same the atoms would not be neutral, there would always be left over charge and the chemistry and atomic physics data would be different, if there were chemistry and atoms at all.

This fact together with a multitude of facts studied since a century and more, have lead to positing the standard model for particle physics. This model simplifies the information in the data in a similar way as when one knows the symmetries which describe a crystal lattice the crystal is describable by a few parameters and equations. The Standard Model of Particle physics incorporates beautiful symmetries which in the end must arise from any theoretical models about particle physics which aims to be a theory of the whole.

anna v
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    It is amusing that there exist people who think they are interested in physics and cannot accept an experimental fact. They need convoluted theoretical arguments which in the end of course end up on the experimental fact. Data trumps theory every single time. Without data theory is science fiction. – anna v Mar 24 '12 at 04:45
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    The issue is that experiment only establishes it to finite precision, while the result is true to arbitrary precision. To extend a finite precision result to arbitrary precision requires a theoretical principle always. – Ron Maimon Aug 12 '12 at 20:10
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    It's an experimental fact on Earth that the universe revolves around the Earth - why? – Larry Harson Aug 12 '12 at 22:34
  • @LarryHarson The theory and models arising from it answer the how, not the why. Theory just upgrades the problem to another level of "why", more convoluted but still unanswerable. – anna v Aug 13 '12 at 03:13
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    @RonMaimon I agree that a successful theory extends the accuracy and available phase space of which the data is a manifestation. It still only intrinsically replies to "how" by transferring the "why" to the axioms and premises of the theory." Why this theory? Because it fits the data." is a circular argument. – anna v Aug 13 '12 at 03:19
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    @annav: I agree, it's a combination of common sense/ occam's razor/ mathematical simplicity and experimental data, but you can't say charge quantization is pure experiment, because it's exact. – Ron Maimon Aug 13 '12 at 04:00
  • @RonMaimonmy argument with this last is that if it were not exact we would not have had atoms , and atoms are an experimental fact . In any case progress in understanding the hows of nature data and theory go together. – anna v Aug 13 '12 at 04:20
  • @annav: exact to one part in 10^{80} is certainly enough for atoms. Then you can ask why the remaining exactness. – Ron Maimon Aug 13 '12 at 19:30
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    @annav The same you could say about the cosmological constant. Why is it so tiny? Because it is an experimental fact. That is certainly not enough. – Newman Aug 17 '12 at 14:54
  • @Newman, i would say this is exactly enough, actually it is the definition of enough. No matter how good (theoritically) i find other answers, the point of departure of all, is at some point the expreminetal data of proton charge ~ -electron charge, upon which the lorentz force and electromagnetism was built and upon which in turn all $SU(n)$ gauge groups are built and extrapolated – Nikos M. Aug 06 '14 at 03:12
  • @Newman, to be more precise, one can have perfectly valid symmetries and physics even if proton charge had another value, one could still make beautiful symmetric theories, still they would not explain the charge value per se – Nikos M. Aug 06 '14 at 03:14
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    To me, the matched charges of the electron and the proton means there should be some relationship between electrons and quarks. But none has been established. The electron is considered an elementary particle with no relationship to quarks. That really doesn't make any sense. – Jiminion Feb 10 '15 at 13:28
  • @Jim This is addressed in higher symmetries for example the grand unified theory http://en.wikipedia.org/wiki/Grand_Unified_Theory – anna v Feb 10 '15 at 14:06
  • Addressed? Not really. None of the GUTs have much purchase at this point. Plus super symmetry seems to be toast. – Jiminion Feb 10 '15 at 14:25
  • @Jim The future will show , we have to wait and see. The model though that can explain the existence of 1/3 charge as the basic quantum of charge exists. – anna v Feb 10 '15 at 14:39
  • Which one is that? Is it a string GUT? – Jiminion Feb 10 '15 at 15:18
  • @Jim A final theory of everything will have to include the quantum of charge in its group representations.A GUT, in any form, allows for that classification. It describes observed symmetries and therefor it will be embedded in TOE if/when theorists reach there. The standard model is an encapsulation of data, and a GUT theory goes one step further in including the observed trend to unification of couplings at high energies. A TOE has to have them embedded in some form. – anna v Feb 10 '15 at 16:01
  • TY We'll see what happens, maybe. – Jiminion Feb 10 '15 at 19:44
  • Kepler's laws were experimental facts. Was it a fool's errand to seek a deeper explanation? – WillO Nov 15 '22 at 13:24
  • @WillO imo it was not a "deeper explanation" , but a different theory – anna v Nov 15 '22 at 14:12
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It's due to the observed fact that all charges come in common multiples of the electron charge. The electron charge is the minimum charge an isolatable particle can have. Quarks have a charge of $1\over 3$$e$, so that could be considered the minimum charge. However, quarks are never found by themselves, due to a property called quark confinement. Instead, they are always found in either groups of three (baryons), or two (mesons). Any one of these hadrons does however have a charge that is a multiple of the electron charge.

Why is electric charge quantized? That is, why do all charges come in multiples of $e$? Paul Dirac attempted to find a solution to this by showing that the existence of magnetic monopoles would require that electric charges come in discrete multiples. In very simple terms, the basic argument is that if there are magnetically charged particles (magnetic monopoles) they too must have well-defined quantum states, and that this requirement places a constraint on electrical fluxes. That constraint leads to the requirement that electrical charge be quantized. For a derivation, see here:

http://bado-shanai.net/map%20of%20physics/mopDiracsMagMonopoles.htm

However, no experiment has found monopoles to date. So, there really isn't a very compelling explanation for the quantization of charge.

Mark M
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I will not give you an answer for the sign difference but I will give you an answer for the same scalar value the elementary charge $e$.

The spin magnetic dipole moment of the electron is for g-factor=2 one Bohr Magneton $μ_{Β}$:

$$ \mu_{\mathrm{B}}=\frac{e \hbar}{2 m_{\mathrm{e}}} $$

$$ μ_{Β}=9.2740100783 \times 10^{-24} \mathrm{~J} \mathrm{~T}^{-1} $$

The spin magnetic moment of the proton (i.e. not the total magnetic moment) when effectively considered as a Dirac particle is one Nuclear Magneton:

$$ \mu_{\mathrm{N}}=\frac{e \hbar}{2 m_{\mathrm{p}}} $$

$$ \mu_{\mathrm{N}}=5.050783699(31) \times 10^{-27} \mathrm{~J} / \mathrm{T} $$

As we observe from the above equations and values, the smaller Dirac spin magnetic moment of the proton compensates for its larger mass (i.e. smaller dimensions) giving the same absolute charge value $e$ for both particles when the equations are solved for $e$.

Why is this happening? It is a fact of nature. Not satisfied? At least I can tell you what fundamentally and physically is happening?

The electric flux of both the electron and the proton are the same (in V⋅m SI units):

$$ \Phi_{E}=\unicode{x222F}_{S} \mathbf{E} \cdot \mathrm{d} \mathbf{S}=\frac{Q}{\varepsilon_{0}} $$

  • $\mathbf{E}$ is the electric field,
  • $\mathbf{S}$ is any closed surface,
  • $Q$ is the value of the elementary charge, absolute vale of e,
  • $\varepsilon_{0}$ is the electric constant (a universal constant, also called the "permittivity of free space") $\left(\varepsilon_{0} \approx 8.854187817 \times 10^{-12} \mathrm{~F} / \mathrm{m}\right)$

or,

$$ \Phi_{E}=\frac{2 \mu_{\mathrm{B}} m_{e}}{\varepsilon_{0} \hbar}=\frac{2 \mu_{N} m_{p}}{\varepsilon_{0} \hbar}=18.09 \times 10^{-9} V⋅m $$

Physical interpretation of the above last equations: Both electron and proton contain the same amount of electric flux $\Phi_{E}$, meaning the flux density increases with smaller size particle (i.e. larger mass). However, so does proportionally decrease also the volume therefore the total amount of the electric flux remains constant for both particles thus the same charge elementary absolute value $e$.

Markoul11
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