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Electron and proton charges

  1. Oct 17, 2010 #1

    mathman

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    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?
     
  2. jcsd
  3. Oct 17, 2010 #2

    Vanadium 50

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    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.
     
  4. Oct 17, 2010 #3
    So the answer is anthropic principle?
     
  5. Oct 17, 2010 #4

    bcrowell

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    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.

    No, there is nothing in King's experiment that requires the invocation of the anthropic principle.
     
  6. Oct 18, 2010 #5

    tom.stoer

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    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.
     
  7. Oct 18, 2010 #6
    This can't be right. Assume an inequality. Matter would arrange itself so that total charge was neutral.
     
  8. Oct 18, 2010 #7
    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.:biggrin:
     
  9. Oct 18, 2010 #8
    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...
     
  10. Oct 18, 2010 #9

    Vanadium 50

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    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)
     
  11. Oct 18, 2010 #10

    mathman

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    My own wild guess - it is related to the question of what happened to the anti-matter after the big bang.
     
  12. Oct 18, 2010 #11
    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.
     
  13. Oct 18, 2010 #12

    bcrowell

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    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.
     
    Last edited: Oct 18, 2010
  14. Oct 19, 2010 #13

    tom.stoer

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    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:

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

    [tex]G(x)|\text{phys}\rangle = 0[/tex]

    Now one can integrate these equations

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

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

    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.
     
    Last edited: Oct 19, 2010
  15. Oct 19, 2010 #14
    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?
     
    Last edited: Oct 19, 2010
  16. Oct 19, 2010 #15

    tom.stoer

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    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 [tex]\psi[/tex] (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

    [tex]j^\mu = \bar{\psi} \gamma^\mu \psi[/tex]

    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.
     
  17. Oct 19, 2010 #16

    Vanadium 50

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    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.
     
  18. Oct 19, 2010 #17
    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
     
  19. Oct 19, 2010 #18

    tom.stoer

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    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:

    NProton qProton + NElektron qElektron = 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)

    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.

    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 accross all fermion families in combination with the condition of vanishing gauge anomalies.
     
    Last edited: Oct 20, 2010
  20. Oct 20, 2010 #19

    DrDu

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    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.
     
  21. Oct 20, 2010 #20
    Thanks, tom. I get the gist of it. Certainly different than I'm used to running into.
     
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