Equation that predicts quark masses

[kurious]

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:




EQUATION THAT PREDICTS QUARK REST 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.



QUARK CHARGE n MASS (Gev) )

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.

 

[arivero]

Ok you have a phenomenological mass formula with some (a lot!) unjustified parameters. No big deal.

 

[kurious]

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.

 

[arivero]

Mmm are you telling that
$${2 m_c + m_s \over 2 m_u + m_d} = {m_\mu \over m_e}$$
?
I see, yep, 207 both sides.

 

[arivero]

I only noticed this after I had settled on what the quark masses should be.

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.

 

[Norman]

Mmm are you telling that
$${2 m_c + m_s \over 2 m_u + m_d} = {m_\mu \over m_e}$$
?
I see, yep, 207 both sides.

Yes, but what does that matter? It is not like the the leptons are made of quarks...

 

[Orion1]

tex formulas...

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

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

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

 

[arivero]

Orion, I believe the key is in the text:

q is the magnitude of the electric charge associated with the mass.

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.

 

[Orion1]

Lepton Lesson...



I developed an equation that predicts the lepton masses in Mev along the x-axis intercept: (x = Lepton mass (Mev) when $$\Psi(x) = 0$$)

$$\Psi(x) = x(x^2 - xE_1 + E_2^2) - E_3^3$$
$$\Psi(x)$$ - 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.