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Equation that predicts quark masses

  1. Mar 31, 2004 #1
    The standard model can't predict the rest masses of all known and unknown particles - something is missing from it.I have found an equation by trial and error that predicts all quark masses:


    The following equation generates the masses, in Gev, associated with the six quarks:
    Down, up, strange, charm, bottom, top and predicts the masses of two new quarks labelled X1 and X2.

    M = 12.50 x 10 3pi (n – 5) / 2 0 x ( n – 4 ) 2 x 10 39 ( n – 3 ) / 2 x 10 57 x q n (5)

    M = f (n) q n

    Where n is an odd numbered integer and q is the magnitude of the electric charge associated with the mass. The equation was based on the idea that
    mass = constant x q n and that the constant depends on n and is different for each quark pair - the pairs are next to each other in the table.


    DOWN - 1/3 -1 0.0088

    UP + 2/3 -1 0.0044

    X1 -1/3 +1 0.084

    X2 +2/3 +1 0.16

    STRANGE -1/3 +3 0.21

    CHARM +2/3 +3 1.72

    BOTTOM -1/3 +5 5.20

    TOP +2/3 +5 167.25

    I now understand why this equation works and I am going to submit it to a journal later in the year.The quark at n = +1 doesn't actually exist because of the colour force.This equation enables the rest mass of the muon to be predicted accurately too! The constant k changes for each quark family.
    Last edited: Mar 31, 2004
  2. jcsd
  3. Mar 31, 2004 #2


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    Ok you have a phenomenological mass formula with some (a lot!) unjustified parameters. No big deal.
  4. Mar 31, 2004 #3
    When I take two up quarks from my equation and add their rest masses to a down quark ( constituents of a normal proton), and compare this to the total mass my equation predicts for a proton made from two charm quarks and a strange quark, I get the ratio of the mass of the muon to that of the electron.I only noticed this after I had settled on what the quark masses should be.
  5. Apr 1, 2004 #4


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    Mmm are you telling that
    [tex]{2 m_c + m_s \over 2 m_u + m_d} = {m_\mu \over m_e}[/tex]
    I see, yep, 207 both sides.
    Last edited: Apr 1, 2004
  6. Apr 1, 2004 #5


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    Well, the simple answer is that you are most probably a lier.

    I will refer the unaware public to the table http://pdg.lbl.gov/2002/qxxx.pdf
    where the uncertainty in masses is clearly stated. You have choosen the quotient muon/electron as a hidden input in your empirical table and then waited one day to claim the "discovery".

    Otherwise, please explain the method to find the coefficients in your table and I will apologise.
  7. Apr 1, 2004 #6
    Yes, but what does that matter? It is not like the the leptons are made of quarks...
  8. Apr 3, 2004 #7
    tex formulas...

    Can someone please re-write this equasion in TEX format?

    The currently stated equasion is still extremely vague, and do not want to spend time guessing as to its proper mathematical orientation.

  9. Apr 3, 2004 #8


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    Orion, I believe the key is in the text:
    How is this magnitude, generation dependent, introduced? I guess that kurious used the quotient between leptonic masses as a way to "define" these magnitudes; in this way it is not strange it gets it as an output with 0.1% precision.

    Of course, if the method does not involve this quotient, kurious had got a remarkable, better, REMARKABLE, result. I am afraid it is not so, as he has backed from this thread.
    Last edited: Apr 3, 2004
  10. Apr 4, 2004 #9
    Lepton Lesson...

    I developed an equasion that predicts the lepton masses in Mev along the x-axis intercept: (x = Lepton mass (Mev) when [tex]\Psi(x) = 0[/tex])

    [tex]\Psi(x) = x(x^2 - xE_1 + E_2^2) - E_3^3[/tex]
    [tex]\Psi(x)[/tex] - function amplitude (Mev^3)
    x - energy spectrum (Mev)
    E1 = 1677.789 Mev
    E2 = 435.355 Mev
    E3 = 45.845 Mev

    I will demonstrate how this was accomplished later in thread, however, calculus students should be able to recognize how this formula was compiled as well as demonstrate my point.

    Last edited: Apr 5, 2004
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