- #26

DrDu

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Then prove it in the Lorentz gauge!

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- #26

DrDu

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Then prove it in the Lorentz gauge!

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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?

- #28

tom.stoer

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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.Then prove it in the Lorentz gauge!

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 text books). 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).

- #29

DrDu

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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: [tex] i\mathbf{K}\cdot \mathbf{E}=\rho(\mathbf{K}) [/tex],

so [tex] \mathbf{E}=-i \mathbf{K} \rho(\mathbf{K})/K^2+\mathbf{E}_\perp [/tex] with [tex] \mathbf{K}\cdot \mathbf{E}_\perp [/tex]. 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.

- #30

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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?"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 text books).

Indeed, the

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

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

[tex]\Delta E \propto \sqrt{N}[/tex].

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.

- #31

tom.stoer

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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.This whole operator G becomes ill-defined on a torus for Q ne 0.

...

so [tex] \mathbf{E}=-i \mathbf{K} \rho(\mathbf{K})/K^2+\mathbf{E}_\perp [/tex]

- #32

tom.stoer

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There is no conflict.So, when I find discussion of Lorentz invariance and Gauss' Law, I am not surprised to find 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, P

So Lorentz invariance still holds.

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

- #33

DrDu

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I was arguing classically.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.

You may use [tex] \mathbf{E}_{||}+i \mathbf{K} \rho(\mathbf{K})/K^2=0 [/tex] 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?

- #34

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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

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.

- #35

tom.stoer

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I still can't see why there should be something divergent.You may use [tex] \mathbf{E}_{||}+i \mathbf{K} \rho(\mathbf{K})/K^2=0 [/tex] as the constraint.

It can be seen that the longitudinal part of the electric field has to be divergent.

OK, lets do it that what. First we restrict to compact one-dim. space = to a circle which is T

[tex]G(x) = \partial_x E(x) + \rho(x)[/tex]

is translated as follows

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

with [tex]p_n \sim n/L[/tex]

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

Of course - as you already said -

[tex]\rho_0 \sim Q[/tex]

The constraint is translated as follows

[tex]G_n|\text{phys}\rangle = 0 \quad \forall n[/tex]

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

[tex]Q|\text{phys}\rangle = 0[/tex]

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 text books do only talk about Lorentz gauge.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 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.

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- #36

tom.stoer

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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.

- #37

tom.stoer

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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

[tex]G^a(x) = \partial_x E^a(x) + \rho^a(x)[/tex]

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.

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

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

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

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

- #38

DrDu

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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?

- #39

tom.stoer

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

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.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?

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

- #40

DrDu

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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?

- #41

tom.stoer

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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.

- #42

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

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.

- #43

tom.stoer

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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.

- #44

DrDu

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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:

[tex] \nabla \cdot \mathbf{E}-\rho+m^2 \phi =0[/tex]

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 [tex] \rho -m^2 \phi [/tex] has to vanish. In the limit m->0, this can either be achieved by [tex] \rho(k=0)=0 [/tex] and finite phi (your proposal) or by an arbitrary total charge and [tex] \phi(k=0)->\infty [/tex].

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.

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:

[tex] \nabla \cdot \mathbf{E}-\rho+m^2 \phi =0[/tex]

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 [tex] \rho -m^2 \phi [/tex] has to vanish. In the limit m->0, this can either be achieved by [tex] \rho(k=0)=0 [/tex] and finite phi (your proposal) or by an arbitrary total charge and [tex] \phi(k=0)->\infty [/tex].

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.

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- #45

tom.stoer

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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.

- #46

DrDu

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But that's the way used e.g. by Zee in "QFT in a Nutshell".I don't think that you can recover QED with massless photons from the Proca theory.

- #47

tom.stoer

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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.

- #48

DrDu

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http://archive.numdam.org/article/AIHPA_1972__16_1_79_0.pdf

where also different forms of the constraint are discussed.

- #49

tom.stoer

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

- #50

DrDu

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