Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Electron & proton charges precisely equal

  1. Mar 8, 2005 #1
    Sorry if this question has been answered already; I've searched, but couldn't find an answer or discussion.

    Question: If these charges are precisely equal, doesn't that imply that these particles are somehow related? IE, they were created from the same source particle, a particle that had a zero charge, and that somehow split into two particles with precisely equal and opposite charges? (Although, obviously, not equal masses.)

    But if protons are comprised of quarks, and electons are not, how can that be?

    I'm not a physicist, so please keep the answer simple! :smile:
  2. jcsd
  3. Mar 8, 2005 #2


    User Avatar
    Science Advisor
    Homework Helper

    There's not really too much similarity between the 2 particles,one is fundamental,the oher is not.However,what you tried to describe is what we call
    [tex] n\rightarrow p^{+}+e^{-}+\bar{\nu}_{e} [/tex]

  4. Mar 8, 2005 #3

    Andrew Mason

    User Avatar
    Science Advisor
    Homework Helper

    It is one of the great marvels of the universe. The charges exactly cancel each other out so that electrical force is of no consequence over large distances. If it were not the case, the universe would look very different.

    As far as the explanation is concerned, consider that a proton can become a neutron by capturing an electron. But the electron seems to disappear. It causes an up quark to become a down quark which makes the proton a neutron. Electrons can be produced by neutron decay when a down quark becomes and up quark producing a proton with the release of an electron (and a neutrino).

    So, the fact that charges are exactly equal and opposite is explained by quark transformations, all resulting from the mysterious weak interaction.

  5. Mar 8, 2005 #4
    Didn't neutron come first?

    Hi Fastartcee,
    The choice between the first two of the responses to your query is yours to make.

    Kurt’s equation looks a lot like the one offered to me in the 1940s by Martin Deutsch. Let me embellish Kurt’s equation a little and, just for fun, explain how the process is highly exothermic. Very simply: now suppose that the shell of the neutron is neutral and weighs ~ 1836 electron mass units and has non-intrinsic spin-a-half. And that it is hollow except for a tiny singlet positronium orbit. After the neutron has been free (half-life of 614 earth-seconds) a short time, the e+e- orbit that has grown nervous with its idleness, initiates a rather violent “convulsion” that leaves an unaffected nucleon shell whose interior is finally filled with only the charge of the e+e- orbit’s positron. The convulsion occurs as follows: when the positron jumps from orbit radius to dead center the jump converts its mass to 0.511 MeV and that shows up as part of the electron’s escape energy; of course the electron possessed kinetic energy equal to 1/2 its mass equivalent 0.256 MeV. The electron’s total escape energy is 0.78235 MeV, which is about 15.5 KeV, more than accounted for. The intrinsic spin of the electron remains with it. Could the stripping of the positron’s spin account for that part of the electron’s escape energy? Without all my speculations, Kurt’s stated process is highly exothermic and therefore highly spontaneous.

    The second response: “Consider that a proton can become a neutron by capturing an electron. But the electron seems to disappear” fails to reckon with the problem that although the Lyman, Balmer etc series spontaneously make quantum jumps of excited electrons toward the proton the spin-dependent magnetic repulsion dominates at the ground state radius. In short, protons are not very interested in capturing electrons. Such process should be somewhat endothermic. Ipse Dixit, Cheer, Jim
  6. Mar 9, 2005 #5
    Thanks to all, and a couple questions for NEOclassic

    Thanks for the replies, which have helped... a bit. As a group, though, the replies must have been exothermic because I began to perspire in my efforts to comprehend what you were saying. :wink:

    Jim, I don't at all follow "suppose that the shell of the neutron is neutral and weighs ~ 1836 electron mass units and has non-intrinsic spin-a-half. And that it is hollow except for a tiny singlet positronium orbit". If a neutron consists of a shell containing positronium, how does that square with two down quarks and one up quark? Or is the shell simply a manifestation of the quarks?

    And why does the positronium 'get nervous' if a neutron is free, but not if it is a constituent of a nucleus?

    It is just great when a site allows an interested amateur to pose questions and knowledgeable individuals take the time to share their insights.

  7. Mar 10, 2005 #6


    User Avatar

    What does to get nervous mean?
  8. Mar 15, 2005 #7
    Hi Art, your last question is right on, except replacement rather than manifestation. Jim, who was within arms reach of Oppie, Otto Frisch, Bohr, Fermi, Feynman, etc, etc in the 1940s, remained in LANL physics until retirement in 1982 – some 15 years after Murray GellMann invented Quarks which after 40 or so years have yet to be found to exist (according to my 82nd Ed of CRC Handbook). In order for the 6 leptons (three stabilized triplet positroniums pulled together by the Coriolis force) to achieve the desired mass, each must be accelerated to ~ 306 emu equivalent through SR enhancement to trajectory velocities of .99999446 c. With the shell itself being heavy enough – who needs quarks or Higgs?

    For instance, take the very stable deuteron; centered in the shared nucleonic space is the aboriginal “gluon” a hybrid microstructure that is a mini-orbit of two positrons surrounding a massless and spinless negative charge. That alleged structure should very well sooth the positronium’s nerves. Cheers, Jim
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook