Electron and Proton Charges: A Fundamental Mystery or a Natural Phenomenon?

[mathman]

The charge of an electron is exactly equal in magnitude to that of a proton (2 up quarks plus down quark). What is the theoretical basis for this, or is essentially a fact of nature that is accepted?

 

[Vanadium 50]

It's known to be very close to equal experimentally. If this weren't the case, matter would blast itself apart by electrostatic repulsion. Theoretically, you need the sum of the electric charges of all (types of) particles to be zero if you want your theory to be free of anomalies - i.e. to be predictive.

 

[Dead Boss]

So the answer is anthropic principle?

 

[bcrowell]

It's known to be very close to equal experimentally. If this weren't the case, matter would blast itself apart by electrostatic repulsion.

I don't think this is correct. This requires the additional assumption that the numbers of protons and electrons in a macroscopic object are nearly equal, but that assumption can't be verified to high precision.

The actual experimental upper limits are based on neutrality of individual free atoms and molecules. The classic experiment was J.G. King, PRL 5 (1960) 562, which showed that the hydrogen molecule was electrically neutral to about 10^-20e. According to this http://arxiv.org/abs/hep-ph/9209259 more recent review, King's upper limit had only been improved on by about one order of magnitude by 1992.

So the answer is anthropic principle?

No, there is nothing in King's experiment that requires the invocation of the anthropic principle.

 

[tom.stoer]

There are theoretical reasons based on the standard model why charges of the electron (-e) and the quarks (e/3, 2e/3) fulfil certain relations. This depends both on the fraction of charges and on the number and types of particles you have in one family of the standad model. A violation of such a relation would imply that the quantum theory of the standard model becomes anomalous and therefore mathematicaly inconsistent [this was one reason why physicists had to find the top quark; w/o the top quarks the third quark familiy would have been incomplete and the consistency of the SM would have been spoiled].

But afaik we do not know any principle from which we can derive uniquely the set of particles with its charges and relations they must obey. There could be different particles with different charges plus different relations between them.

 

[Phrak]

It's known to be very close to equal experimentally. If this weren't the case, matter would blast itself apart by electrostatic repulsion. Theoretically, you need the sum of the electric charges of all (types of) particles to be zero if you want your theory to be free of anomalies - i.e. to be predictive.

This can't be right. Assume an inequality. Matter would arrange itself so that total charge was neutral.

 

[Dadface]

It's known to be very close to equal experimentally. If this weren't the case, matter would blast itself apart by electrostatic repulsion. Theoretically, you need the sum of the electric charges of all (types of) particles to be zero if you want your theory to be free of anomalies - i.e. to be predictive.

This can't be right. Assume an inequality. Matter would arrange itself so that total charge was neutral.

Phrak,aren't you just paraphrasing Vanadiums comment?You both seem to be describing the tendency there will be towards neutrality but Vanadium has an event where matter is "blasting itself apart" whereas in your case things seem to be going more gently with matter arranging itself.

 

[Phrak]

Phrak,aren't you just paraphrasing Vanadiums comment?You both seem to be describing the tendency there will be towards neutrality but Vanadium has an event where matter is "blasting itself apart" whereas in your case things seem to be going more gently with matter arranging itself.

You'd have to ask Vanadium. If, say, the charge of a quark were 0.34 instead of 0.33..., matter would still clink together, but with so many free electrons everything would be a conductor....On the other hand, if the Universe had a net charge imbalance maybe they'd call it dark energy...

 

[Vanadium 50]

Dadface has put his finger on it - if you want matter to be neutral because of an imbalance of non-equally charged electrons, you end up needing a lot of ions in unionized matter. (that is, not ionized, not matter that hasn't joined a labor union)

 

[mathman]

My own wild guess - it is related to the question of what happened to the anti-matter after the big bang.

 

[inflector]

My own wild guess - it is related to the question of what happened to the anti-matter after the big bang.

Yeah, it sure seems related in some way. After all, an antimatter particle is of opposite charge as it's matter equivalent.

It seems like the charges must have been created together somehow from nothing, sort of like you can create angular momentum in one direction only by creating it in the opposite direction at the same time in order to preserve the zero momentum starting state because of conservation of angular momentum.

So whatever caused the anti-matter imbalance seems like a reasonable candidate for creating the charge balance in stable electrons and protons. Opposite charge comes in two forms, the matter/antimatter form like with electrons and positrons, and the pure matter or pure antimatter form with electrons and protons or positrons and antiprotons. The fact that protons and electrons don't annihilate meant they could stick around as pairs with neutral charge.

The hypothesis that the universe is neutrally charged is analogous to the idea that the total energy content of the universe is zero, as well.

 

[bcrowell]

Dadface has put his finger on it - if you want matter to be neutral because of an imbalance of non-equally charged electrons, you end up needing a lot of ions in unionized matter. (that is, not ionized, not matter that hasn't joined a labor union)

This is less incorrect than your #2, but still incorrect. As I pointed out in #4, the best upper bound does not come from from observations of this type. Please take a look at the paper on arxiv that I linked to in #4.

[EDIT] The relevant part of the Foot paper is on p. 12 (p. 13 of the PDF file), at "Direct experimental measurements on atomic neutrality ...," which clearly shows that these values are based on atomic neutrality, not bulk neutrality. There is also a good description of this type of experiment in Purcell, Electricity and Magnetism, McGraw-Hill, 1965, pp. 5-6. Although the Purcell book is quite old, it describes the result by King, which is only one order of magnitude worse than the best current limit.

 

[tom.stoer]

It is a simple exercise to derive based on the Gauss law that physical states must be charge-neutral. This applies to QED and e.g. to QCD as well.

The idea is as follows:

The Gauss law must not be interpreted as an operator equation as this would violate the operator algebra / commutation relations. Therefore it is translated into a constraint equation for the physical sector of the theory:

G(x) = \nabla E(x) + \rho(x)

G(x)|\text{phys}\rangle = 0

Now one can integrate these equations

\int_V d^3x G(x) = \oint_{\partial V} dA E(x) + Q

[Q - \text{surface charge}]|\text{phys}\rangle = 0

For universe with closed topology = vanishing surface the boundary must be equally zero, that means for S³, T³ etc. the total charge must be exactly zero. For an open universe one could introduce "surface charges" but this seems tobe rather unnatural. These surface charges would then cancel the volume charges.

That means that the Gauss law is equivalent to vanishing total charge.

 

[Phrak]

It is a simple exercise to derive based on the Gauss law that physical states must be charge-neutral. This applies to QED and e.g. to QCD as well.

The idea is as follows:

The Gauss law must not be interpreted as an operator equation as this would violate the operator algebra / commutation relations. Therefore it is translated into a constraint equation for the physical sector of the theory:

G(x) = \nabla E(x) + \rho(x)

G(x)|\text{phys}\rangle = 0

Now one can integrate these equations

\int_V d^3x G(x) = \oint_{\partial V} dA E(x) + Q

[Q - \text{surface charge}]|\text{phys}\rangle = 0

For universe with closed topology = vanishing surface the boundary must be equally zero, that means for S³, T³ etc. the total charge must be exactly zero. For an open universe one could introduce "surface charges" but this seems tobe rather unnatural. These surface charges would then cancel the volume charges.

That means that the Gauss law is equivalent to vanishing total charge.

In classical electromagnetism we could say that Gauss's law is the directly mathematical result of imposing a twice differentiable vector field on a pseudo Riemann manifold. Where A is this vector field, then J=-*d*dA, where J is the current and charge density, all expressed in differential k-forms. The result is Gauss’s law and Ampere’s law combined—they’re really one and the same law, each expressing a part of this law in a different subspace of space and time. J, current and charge density is a particular second derivative of A, the combined electric and magnetic potential, so that J is nothing new, but only second derivative of A.

To oversimplify a bit, charge is defined as the divergence of the electric field rather than a distinct physical quantity that just so happens to be exactly equal, everywhere and at all times, to the divergence of E.

What you have is certainly different than this. I don’t question your conclusion. It’s just that I don’t understand it. I think you have something very interesting and insightful to say.

I’m sure I’m not the only one who doesn’t know what you mean but would wish someone would ask for clarification. Could you possibly explain it in simpler language the rest of us would understand?

 

[tom.stoer]

To oversimplify a bit, charge is defined as the divergence of the electric field rather than a distinct physical quantity ...

What you have is certainly different than this. ...

... Could you possibly explain it in simpler language the rest of us would understand?

Of course.

We have to look at QED, QCD etc.,

There we have the physical gauge fields (only two polarizations out of four A-components after gauge fixing; in QED the photons, in QCD the gluons with an additional color index) and the physical fermion spinor fields \psi (in electromagnetism the electrons, in QCD the quarks with an additional color index). Now the currents are neither defined in terms of a vector potential nor are they introduced by hand as classical sources. Instead the are defined via the fermionic degrees of freedom and are therefore independent physical objects

j^\mu = \bar{\psi} \gamma^\mu \psi

The Gauss law relates the 0-component of the current (charge) living in the fermionic sector of the Hilbert space with the divergence of the E-field living in the bosonic (gauge field) sector of the Hilbert space. This is one of the reasons why they can't be identical and that's why the equation "divergence electric field = fermionic charge density" does no longer hold as operator equation; it remains valid as an equation acting on physical states which are defined as the states on which the Gauss law operator vanishes (eigenstates of the Gauss law with eigenvalue 0 = the kernel of the Gauss law).

The Gauss law acts as a generator of gauge transformations (in A°=0 gauge with time-independent gauge transformations left). Therefore vanishing of the Gauss law on physical states is equivalent to gauge invariance in the physical subspace.

So in contrast to ED the Gauss law in QED tells us something about the relation of the dynamics of the bosonic and the fermionic degrees of freedom.

 

[Vanadium 50]

This is less incorrect than your #2, but still incorrect. As I pointed out in #4, the best upper bound does not come from from observations of this type.

Did I say this was the best limit? I don't think so - it is, however, a very graphic way to show that the equality is very close.

 

[Bob S]

It is a simple exercise to derive based on the Gauss law that physical states must be charge-neutral. This applies to QED and e.g. to QCD as well..

Suppose the electron and proton charge differed by one or two parts in 10²³. Then the universe may have a few extra protons or electrons (about 1 per gram). Can Gauss's law show that this is not possible? Do measurements of neutral atoms show that this is not true? Does Gauss's Law plus neutron radioactive decay into a proton and electron prove that the magnitudes of electron and proton charges are exactly equal?

Bob S

 

[tom.stoer]

Suppose the electron and proton charge differed by one or two parts in 10²³. Then the universe may have a few extra protons or electrons (about 1 per gram).

Can Gauss's law show that this is not possible?

The Gauss law does not say anything regarding the individual portions of the total charge; it only talks about the total charge. The following charge-neutrality condition would be compatible with the Gauss law:

N_Proton q_Proton + N_Elektron q_Elektron = 0

(of course this is oversimplified as one would have to use quarks instead of of protons and as one would have to include all other charged particles)

Do measurements of neutral atoms show that this is not true?

The measurements show that the total charge of one single atom vanishes (with certain experimental error bars). The Gauss law applies not to single atoms; there could very well be slightly charged atoms with a corresponding anti-charge located at spatial infinity. But these anti-charges would create electic fields which would show up in measurements - which is not the case. Therefore I would say that the total charge is zero already within a very smal region of space.

Does Gauss's Law plus neutron radioactive decay into a proton and electron prove that the magnitudes of electron and proton charges are exactly equal?

If applied to the single neutron - yes. I see no mechanism how a neutral neutron could decay into a charged electron-positron pair (plus neutral neutrinos) plus corresponding anti-charge located at spatial infinity.
Mathematically one would prepare a state |neutron> and describe its decay channel |proton, electron, neutrino> which again has vanishing total charge. As this process is local there is no way to create an anti-charge at spatialinfinity in order to cancel total charge in the decay channel.

I think this argument becomes even more convincing if one uses the non-abelian Gauss law of the electro-weak theory across all fermion families in combination with the condition of vanishing gauge anomalies.

 

[DrDu]

Dear tom,

I feel uneasy with your argument in #13. The Gauss law is only observable in static situations when the time of observation is much larger than the distance of the objects. If you apply it on a cosmological scale, I think it is necessary to take the expansion of space into account.
The argument is in fact very similar to the argument that because the sky at night isn't bright universe has to be expanding.
There is always an open horizon and the range of integration should not be extended further than that. Then the difference between open and closed topologies vanishes.

 

[Phrak]

Thanks, tom. I get the gist of it. Certainly different than I'm used to running into.

 

[tom.stoer]

I feel uneasy with your argument in #13. The Gauss law is only observable in static situations when the time of observation is much larger than the distance of the objects. If you apply it on a cosmological scale, I think it is necessary to take the expansion of space into account.

I don't think that this is true.

First of all the Gauss law is an exact equation of constraint which follows directly from the el.-mag. Lagrangian plus gauge condition A°=0 (which is a good choice as A° is not a dynamical degree of freedom b/c there is no canonical conjugate momentum).

I do not understand what you mean by "static situation" and "observation time". There are no such restrictions. The Gauss law (as a local equation) remains valid even if spacetime becomes dynamical (of course the equation itself becomes more complicated).

 

[DrDu]

I am not sure whether the temporal gauge is a good choice.
It amounts to splitting a retarded field E into two components E_{||} and E_\perp in a non-local way which hence both obey equations which have instantaneous character. In a closed topology these fields fold back infinitely often over the boundaries of the universe and the sum diverges. With a constraint respecting causality probably nothing the like would happen.

 

[tom.stoer]

Of course the A-, E-and B-field must respect the topology.

It is rather simple on a 3-torus where periodic boundary conditions are sufficient.

Question: what do you mean by "non-local"?

The instantaneous character is no problem as one can show that the Poincare algebra remains valid (I think that even quantization / regularization anomalies have been studied and found to vanish).

Do you know the work regarding canonical quantization of QED and QCD in the temporal gauge from 1990-05? A lot of people studied the theory in this framework; it seems to be a consistent approach.

http://adsabs.harvard.edu/abs/1994AnPhy.233...17L
http://www.springerlink.com/content/dba1dl6p6rxc01v8/
http://cdsweb.cern.ch/record/292166/files/9511450.pdf

 

[DrDu]

No, I don't know these articles as I am not from the high energy community. But the 3-torus is analogous to a crystal of charge distributions with total charge Q and as a solid state guy, it is clear that the gaussian field has to diverge.
With non-local I mean that the splitting of a localized function into transverse and longitudinal components leads to functions which fall of as 1/r^3 at best, so they are not localized.
See e.g.
http://omnis.if.ufrj.br/~reinaldo/Rohrlich.pdf

 

[tom.stoer]

Nonvanishing total charge in the 3-torus is strcctly forbidden by the Gauss law constraint!

 

[DrDu]

Then prove it in the Lorentz gauge!

 

[FizzyWizzy]

This is a fun thread.

Would it be of interest to consider beta decay? That is when a neutron loses an electron it becomes a proton. The neutrino involved -- does it carry any charge? Perhaps it is on the order of the empirical difference between e and p?

Is there really "positive" charge (as defined by Franklin), or merely the lack of negative charge as in this case? That is, the neutron itself lacks any net charge (presumably). When it loses an electron's worth of charge (and mass), a proton results. My single course on particle physics from ...some year long ago ... escapes me at the moment.

Moreover, I suppose I am looking for a deeper answer: an electron has both mass and charge. It has at least two properties. In other words, it has "structure". Do we consider it a fundamental particle as a result? or because we have yet to break an electron into smaller bits?

What is the "best" existing model of what an electron actually is? Or have we completely bowed to quantum theory and now accept that we have no further understanding of Einstein's "Old One"?

Does anyone want to simply think of it as a "string" of one flavor or another? Or a ball of "mass" encased in a "sheet" of charge? Or something else?

 

[tom.stoer]

Then prove it in the Lorentz gauge!

Why? Tell me where my proof is wrong. Or even better tell me where Lenz et al. (see the "Annals of physics" QED paper) made a mistake.

Remark 1: the Lorentz gauge is well-known in high energy physics / physics of el.-mag waves, but for quantization is has several draw backs (not mentioned in standard textbooks). One has to introduce negative norm states in the Hilbert space (or even Fadeev-Popov ghosts and BRST symmetry in non-abelion gauge theories). The temproral gauge eliminates the Lagrange multiplier A°=0 before quantization and provides a theory involving only physical degress of freedom and a standard Hilbert space.

Remark 2: I need the quantization of all physical degrees of freedom wo make my argument really work. Introducing classical / static charges w/o dynamics may spoil the argment b/c you can manipulate them as you like. But if you allow the quantum dynamics to act on all charges the theory immediately tells you that the total charge must vanish (provided that the manifold has no boundary - which is obviously true for the torus; bzw.: the reason to study the 3-torues is that it is compact and flat, a 3-sphere does not allow for a flat geometry).

 

[DrDu]

I had a look at the paper of Lenz et al.
This whole operator G becomes ill-defined on a torus for Q ne 0.
Q is the K=0 Fourier component of the charge density rho(x).
In Fourier space, the Gauss law reads: i\mathbf{K}\cdot \mathbf{E}=\rho(\mathbf{K}),
so \mathbf{E}=-i \mathbf{K} \rho(\mathbf{K})/K^2+\mathbf{E}_\perp with \mathbf{K}\cdot \mathbf{E}_\perp. Obviously is ill defined for K=0 and Q ne 0. Especially Gauss theorem leaves you with an integral of an infinite field over a vanishing surface.

 

[FizzyWizzy]

the Lorentz gauge is well-known in high energy physics / physics of el.-mag waves, but for quantization is has several draw backs (not mentioned in standard textbooks).

This is an interesting comment on something in which I am not well-versed. While defending my dissertation a few years ago, concerning a classical electrostatic correspondence of point charges in a dielectric sphere with the first ionization energies of neutral atoms, one professor jokingly asked if I could demonstrate that my work was Lorentz invariant. He said he saw plainly that my work was correct (though I have since made some revisions -- and have observed many new features), but also said that I should simply reply with something along the lines of "does it matter?"

Indeed, the classical electrostatics model that I developed leads to results that are consistent with Pauli's exclusion (Rather than most other models that simply presume exclusion a priori).

I have come to recognize a slight difference between my model and the electrostatics textbook. It is very much apparent to me that textbooks on the subject treat continuous charges, charge densities, and so forth, and do not treat discrete point charges in a proper manner when dealing with dielectrics. Only recently I've managed to form a substantially complete understanding of why there is a discrepancy between my work and the textbook. While they all set this term to zero, they each do so in a different manner. One author simply sets the term to zero to validate his subsequent discussion of dielectrics in the presence of metal electrodes. He never returns to the subject to discuss the generally nonzero term. Another introduces an internal energy term, insisting that it negates the energy term in question. Yet another author (of a published paper) produced four models that she claims support the textbook result -- though each model, when carefully thought-through, exposes the non-physical nature of discrete point charges. One final author (who coincidentally was my doctoral advisor) simply argued that since the energy term is a "self-interaction energy", the term must be chopped in half (thus, negating the same quantity of energy as the textbooks and other authors for an ad hoc reason -- "just to agree with the standard textbook"). It is clear to me, since the above mentioned authors have several unique approaches that are all based on non-physical assumptions, and that the energy in question is actually "delivered" to, and required by, the system, -- it must be nonzero. So, why is there such an apparent error?

The textbook approach is that of a continuous charge density and/or metallic electrodes -- both representative as an approximation for large N-electron systems. ...my approach was simpler: "What happens when we introduce electrons one-by-one?" Instead of an integral formulation, I obtain a summation that I believe to be general (with respect to N) and exact while the textbook expression is merely an approximation for large N that is quite good, but up to 50% in error for N=1!

So, when I find discussion of Lorentz invariance and Gauss' Law, I am not surprised to find conflict. I will have to spend some time looking over this thread much closer.

Incidentally, I have stumbled upon an energy difference in my model as a function of N that I have not yet fully understood as the data set is fit superbly to

\Delta E \propto \sqrt{N}.

My first guess is that the data set represents the first standard deviation from perfect spherical symmetry. But there is nothing statistical about the data set. It is numerically exact. In relation to this thread, I believe this relationship is intrinsic to the difference between continuous and discrete charge systems.

 

[tom.stoer]

This whole operator G becomes ill-defined on a torus for Q ne 0.
...
so \mathbf{E}=-i \mathbf{K} \rho(\mathbf{K})/K^2+\mathbf{E}_\perp

I think this is not true. E lives the bosonic sector of the Hilbert space whereas the charge density lives in the fermionic sector. That means you can't solve the equation as an operator equation.

 

[tom.stoer]

So, when I find discussion of Lorentz invariance and Gauss' Law, I am not surprised to find conflict.

There is no conflict.

The problem seems tobe that chosing e.g. the temporal gauge (or the Coulomb gauge) the equations are no longer Lorentz invariant explicitly; one has to check Lorentz invariance explicitly. This has to be done on the level of the operator algebra for the Poincare generators H, Pⁱ, Lⁱ and K_i. After very many pages of boring calculations one finds that the algebra still closes w/o anomaly.

So Lorentz invariance still holds.

btw.: I don't know whether there's confision between Lorentz invariance and Lorentz gauge.

 

[DrDu]

I think this is not true. E lives the bosonic sector of the Hilbert space whereas the charge density lives in the fermionic sector. That means you can't solve the equation as an operator equation.

I was arguing classically.
You may use \mathbf{E}_{||}+i \mathbf{K} \rho(\mathbf{K})/K^2=0 as the constraint.
It can be seen that the longitudinal part of the electric field has to be divergent.
My question is whether this really indicates that solutions with Q ne 0 aren't admissible or whether this is an artifact of splitting a non-divergent field E into divergent longitudinal and transverse parts as a consequence of imposing an unsuitable gauge?

 

[FizzyWizzy]

I apologize. I didn't mean to suggest there was conflict between Gauss' law and Lorentz invariance, but that conflict in discussions concerning these topics tends to exist. (Conflict having a more social than mathematical meaning.) In fact, I often refer to the "Gauss model" (terminology adopted by my advisor) in my dissertation work, and my model being in conflict with it. At the time, I could not explain all the differences, but in the past few months I have converged on a substantial understanding and have pinpointed it to a single single term that approaches a negligible fraction of the total (potential) energy of the system as N grows very large -- hence, in agreement with the textbook.

I, therefore, wanted to draw out the notion that Gauss' law and Lorentz invariance may include implicit assumptions about the nature of a given system. Quantum theory, for instance, appears almost exclusively concerned with dynamic (implicitly read as "statistical") properties while electrostatics is purely static (and perhaps somewhat non-physical in its assumption as well -- though my position is that the electrostatic configuration is that toward which a given N-charge system is driven).

Further still, I see clearly that a distinction must be made between large N and few N systems -- perhaps insofar as all these models/theories are concerned. DFT works well in some cases, while QMech. works better in others. My model (which could potentially develop into a nice, new theory) appears to be a bit more like DFT, but instead of merely knowing (trial) density or wave functions, we might look for a more fundamental spatial symmetry function (of point charges) from which both a wavefunction and density functional may be obtained with greater precision -- if not, dare one say, exactness.

So, when we discuss electron and proton charge -- as is the intent of this thread, do we wish to expound on mathematics concerning continuous charge distributions? or should we constrain ourselves to the discrete nature of charge? If so, to which mathematics and theories are we to resort? I am not convinced that Gauss' Law is explicitly meant to be concerned with discrete, few N systems. I can see how it may apply to N=1, but even then, I personally see a factor of 2 that must be involved. -- though I haven't worked through Gauss' law carefully enough to see that it applies explicitly to N=1. At a glance, I think we must make some further assumptions about the single charge itself in order to fully justify Gauss' law in the discrete regime. One such assumption is that an electron or proton's charge is continuously distributed -- or perhaps not. Does it matter if it is a twisted "string" of charge? or a shell/sheet? or a broken sheet that spins incessantly about an axis and generates a magnetic moment? ...whereupon in free space, if an electron spins in this classical manner, does it really have a magnetic field (if we do not have a reference frame)? Key to this line of thought is interaction. Now one may be considered in reference to the other, and the magnetic moment certainly plays a role.

I just think there is so much more to be learned and understood. Perhaps existing models and ideas need to be carefully revisited, reworked, -- or something new from scratch.

 

[tom.stoer]

You may use \mathbf{E}_{||}+i \mathbf{K} \rho(\mathbf{K})/K^2=0 as the constraint.
It can be seen that the longitudinal part of the electric field has to be divergent.

I still can't see why there should be something divergent.

OK, let's do it that what. First we restrict to compact one-dim. space = to a circle which is T¹ = S¹ instead of T³. You can introduce creation and annihilation operators for the E and the fermion fields. Then can transform to momentum space. The constraint

G(x) = \partial_x E(x) + \rho(x)

is translated as follows

G(x) = \sum_n G_n e^{ip_nx} = \sum_n \left[(ip_n) E_n + \rho_n\right] e^{ip_nx}

with p_n \sim n/L

\rho_n is bilinear in the b_{k+n} and b^\dagger_k

Of course - as you already said -

\rho_0 \sim Q

The constraint is translated as follows

G_n|\text{phys}\rangle = 0 \quad \forall n

But for n=0 the E-field drops out due to the combination (nE_n)_{n=0} and one finds

Q|\text{phys}\rangle = 0

My question is whether this really indicates that solutions with Q ne 0 aren't admissible or whether this is an artifact of splitting a non-divergent field E into divergent longitudinal and transverse parts as a consequence of imposing an unsuitable gauge?

The gauge isn't unsuitable. It's püerfectlywell defined and in the context of canonical quantization it's the gauge that makes most sense! The problem is that most people are not familiar with it as standard QFT textbooks do only talk about Lorentz gauge.

The aim is to eliminate the longitudinal part of the gauge field (which means to move it to the unphysical sector of the Hilbert space). The presence of the constraint G(x) ensures that these unphysical degrees of freedom of the A- and E-field stay within this sector under time evolution (w.r.t. to the physical Hamiltonian); so an unphysical state stays unphysical and doesn't mix with the physical sector.

 

[tom.stoer]

@FizzyWizzy: I hope it becomes clear what the context of my arguments is: it's QED with both electrons and photons being quantized [it my no longer be if one studies quantized electric particles in a classical el.-mag background field; and it my break down if one couples quantized photons to static - infinitly heavy - electric charges]. But QED is the most general context I can think about.

I think one can show that my argument is valid in all cases (1-dim., 3-dim., different topolgies etc.) All what happens is that in non-compact cases one may get surface charges and somekind of background fields. I think this is the only way to escape from the Q=0 conclusion.

 

[tom.stoer]

One can use even simpler (!) arguments in the case of non-abelian gauge theories.

Usually the calculations in QCD become awfully complicated. But I think I can provide a short cut. Again one finds a Gauss law constraint which now lives in color space. It reads

G^a(x) = \partial_x E^a(x) + \rho^a(x)

where a=1..8 is the SU(3) color index and the charge density has a quark and a gluon contribution (the latter one being the special ingredient of the non-abelian gauge group)

The Gauss law operators satisfy a local SU(3) algebra, i.e.

[G^a(x), G^b(y)] = if^{abc}G^c(x) \delta(x-y)

Again one can integrate the Gauss law constraint and derive the global SU(3) algebra

[Q^a, Q^b] = if^{abc}Q^c

Now comes the funny thing: As G(x) generates "topologically small" local gauge transformations, Q simply generates "global" gauge transformations, i.e. gauge transformations where the gauge parameter is space-time independent.

Now the requirement is to have gauge-invariant physical states, i.e. every physical state is a color-singulet! But this automatically means that all Q's have zero eigenvalue on physical states! Please note that this shortcut is not possible in QED as the gauge group has an abelian structure which does not immediately single out color singulets.

So in QCD the color-neutrality is an almost algebraic property following directly from the local algebra of the "color-electric" Gauss law.

------

Of course this reasoning remains valid in 3+1 dim. spacetime

 

[DrDu]

Sigh. You don't even try to understand my argument. Did you ever do classical electrodynamics?

Ok, let's try me another way of argumentation: You say that this argumentation applies to all kind of massless gauge bosons. However, if a symmetry gets broken, the symmetry broken state is one of unsharp charge. If only the state with Q=0 is available, how can I end up by symmetry breaking in a state in whicha measurement of Q may yield something different from 0?

 

[tom.stoer]

You don't even try to understand my argument. Did you ever do classical electrodynamics?

I tried to, but I think it's not relevant as soon as you quantize the theory. The crucial point is that you are no longer allowed to "solve" the equation KE = ... as E = (...)/K. Yes, I studied classical electrodynamics, but I don't know whether it says soemthing different. If you run into a contradiction with QED it's the classical reasoning that must be wrong.

Ok, let's try me another way of argumentation: You say that this argumentation applies to all kind of massless gauge bosons. However, if a symmetry gets broken, the symmetry broken state is one of unsharp charge. If only the state with Q=0 is available, how can I end up by symmetry breaking in a state in whicha measurement of Q may yield something different from 0?

I always knew that you wouldcome up with this question :-) I have to admit that I haven't studied this case in detail, so I can't say what happens to the physical states.

Can you explain where you think my argument fails? Is it because the vacuum may be no longer a singulet state?

 

[DrDu]

Actually, the proof of the Goldstone theorem which is intimately related to broken symmetry is very similar to your argumentation why total charge has to vanish, i.e. it also relates the total charge operator (or better to say the limit of a local operator approximating the latter) to an integral over the boundary. The limit is somewhat intricate, that's why I am questioning so hard your argumentation

On the other hand, in a finite system the ground state is unique, hence in a closed topology symmetry can never be broken in the strict sense. So following your argumentation, symmetry breaking is also unnatural in open topologies?

 

[tom.stoer]

I think it depends if you talk about global or local symmetries. Global symmetries can be broken via the Goldstone mechanism. There is no Gauss law associated with global symmetries (the Gauss law is the relict of the local gauge symmetry and reflects the fact that A° is not a dynamical degree of freedom but a Lagrange multiplier).

b/c there is no equation like the Gauss law for global symmetries my argument isn't valid.

For local symmetries it's different as I do not see that the gauge symmetry is really broken. I think this is - strictly speaking - not true. You can derive U(1) and SU(2) Gauss law constraints from the variation with respect to the A° and B° gauge fields. The SU(2) Gauss law constraint has again an non-abelian gauge field current term plus a Higgs term. But nevertheless it must violate the physical states in the same way as the abelian Gauss law.

I didn't check all the details but I am pretty sure that the action of the Gauss law isn't that much different from the SU(3) or QCD case.

 

[enotstrebor]

The charge of an electron is exactly equal in magnitude to that of a proton (2 up quarks plus down quark). What is the theoretical basis for this, or is essentially a fact of nature that is accepted?

I note in many of the answers to this question, there is one significant problem.

There seems to be a consistent errant view, that within the present theory, fundamentals like charge equivalence have a theoretical basis, as if theory defines nature rather than the theory and mathematical models are the result of nature (based on measured/experimental evidence).

If nature did not first present an equivalence of charge experience, then the theory would not either, or the theory would fail to match experience/experiment.

As it turns out, the continued extensions of mathematical model(s) of nature consistently evolved yielding the resulting base CPT symmetry of the present theory (Not to getting into violations of this symmetry in nature) and THUS the manipulation of these equations end in charge equivalence, but they are not the source of nature's behavior.

This experimental equivalence of charge like the experimental equivalence of a particles mass to energy does not have a theoretical base within the present theory.

In order to have a "theoretical basis" requires that, for the present theory's "point particle" there exists an underlying (theoretical) source model where the underlying source produces an equivalence of positive and negative charge (and the wave behavior and the point behavior) and answers the question what is the underlying reason a massed particle resists a change to velocity and why the energy content (as seen in particle anti-particle annihilation), and the particle's mass (as measured by resistance to a change in velocity) is directly proportional to the resultant photon energy.

But as the present theory denies that an underlying source can exist, no theoretical basis can exist within the present theory.

 

[tom.stoer]

Is I said in one post: there is indeed no direct reason why the charge of electron and proton match, but afair there is some support.

The standard model requires that certain charges of different types of leptons and quarks add up to zero b/c of anomaly cancellation. w/o this perfect match the theory would have triangle anomalies in the chiral el.-weak sector which would spoil its mathematical consistency.

 

[DrDu]

I pondered about our interesting discussion and wanted to propose the following alternative solution to vanishing total charge:
In my view, the problem is due to the long range instantaneous and hence artificial nature of the Coulomb interaction in the constraint. Hence, before disccussing this constraint, it should be regularized.
A simple way would be to start from the Proca equation and then consider the limit m->0.
The constraint becomes:
\nabla \cdot \mathbf{E}-\rho+m^2 \phi =0
For a toroidal topology, a potential problem arises especially for the k=0 Fourier component of the constraint. Reproducing your argument, the k=0 component of \rho -m^2 \phi has to vanish. In the limit m->0, this can either be achieved by \rho(k=0)=0 and finite phi (your proposal) or by an arbitrary total charge and \phi(k=0)->\infty.
I don't see that an infinite constant value of the potential \phi makes any problem as it does not influence the fields which are the only observables in the m=0 limit.

This resembles the resolution of the paradox with the negative energy states in the Dirac equation. If these states are filled, the electrons would provide a (negative) infinite mass. However, leaving gravity aside, this would be unobservable and corresponds only to a shift of the zero point of energy.

 

[tom.stoer]

I still think that the long range instantaneous Coulomb interaction is not artificial but physical. This gauge is widely used in some QED and even in QCD calculations, the latter one making good progress towards the explanation of confinement. I agree that there are scenrios where different gauges are easier to handle, but the gauge itself is not a problem in principle.

Regarding the Proca equation: the potential is still instantaneous and seems to violate causality Lorentz invariance in the same way (of course one can show that it doesn't). So you do not get rid of an instantaneous interaction. All what you get is an exponential decay instead of an 1/r decay, but I do not see the benefit. If you try to regularize the infinite IR contributions due to 1/r for non-compact space, I think compactification (e.g. the 3-torus) is easier to handle.

There is one severe issue with the Proca equation, namely that it explicitly breaks the gauge invariance due to the mass term A²(x). This introduces a third physical polarization and alters the theory completely. Especially the form and the meaning of the of the Gauss law is completely different. I don't think that you can recover QED with massless photons from the Proca theory.

 

[DrDu]

I don't think that you can recover QED with massless photons from the Proca theory.

But that's the way used e.g. by Zee in "QFT in a Nutshell".

 

[tom.stoer]

Interesting; how does he get rid of the longitudinal photon?

anyway - I think Proca theory is irrelevant here; I still do not understand your problem with the 1/r potential and/or the A°=0 & div A = 0 gauge.

 

[DrDu]

Apparently, the longitudinal photon decouples from matter in the limit m->0. See e.g.
http://archive.numdam.org/article/AIHPA_1972__16_1_79_0.pdf
where also different forms of the constraint are discussed.

 

[tom.stoer]

Thanks, I have to check the details. I guess it will not work in non-abelian gauge theories; I haven't seen massive gluons and Pauli-Willars for QCD.

Nevertheless: I do not understand the problem with the 1/r potential and/or the A°=0 & div A = 0 gauge in massless QED.

 

[DrDu]

In contrast to what I have been writing on the beginning of this thread I don't think anymore that my problem has to do with a specific choice of gauge but only with the long range nature of the Coulomb potential. In massaging the constraint you implicitly assume the electric field to be well defined (which indeed it is in the case Q=0). For finite Q it isn't well defined so you also cannot argue that it's divergence will make no contribution using some differential geometric identities. The 3 torus can also be viewed at as an infinite periodic array (a "crystal"). In the case of a Yukawa interaction (or more generally a short range interaction), the electric field at some point in the crystal can be approximated by summing over the fields generated by the charges which are subsequently at further and further distance. For a Coulomb potential, this sum won't converge.

Btw, shouldn't your argument also show that in a closed topology total mass (or better the energy momentum tensor) has to be 0?