Jump to content

Quant not Quark


kwrk

Recommended Posts

Hi everybody,

I am working on a model that uses the solution to a 2nd order DE, Ψ(e,ε), e=elementary charge, ε=electric constant, as probability amplitude for the electromagnetic field, EΨ. Applying this to a point charge gives:

- quantized values for particle energy - as function of a partial product of powers over the fine-structure constant α, see table below for spherical + 1st angular symmetry term;

- a numerical approximation of the fine-structure constant 1/α ≈ 4π Γ(1/3)|Γ(-1/3)| ( Γ=Gamma function);

- magnetic moments, calculated from electromagnetic fields; agreement with experiment within factor of 2;

and on a speculative level

- a possibility to quantitatively express gravitational force entirely in electromagnetic terms;

- an indication of a common base for strong force, electromagnetism and mass/gravitation.

The model is quite incomplete, its elaboration primitive and sloppy and I know the standard model already explains everything, but I need only e + ε, the standard model needs an additional 20+ parameters to do so.

More info at  url deleted or Youtube url deleted  “Quant not Quark”

Any idea, comment or critique welcome,

Happy new year,

kwrk

.............W_lit/MeV.......Wn/We lit.......Wn/We calc.......calc/lit.......Πn/1,509

y00...(ν......0,3eV-calc.........-................... 6E-7..................-..............α^3)

.........e.......0,51...............Ref...................-........................-...............-

.........µ.......105,66...........206,8...............206,8.............1,000.........α^(-1)

.........η.......547,86.........1072,1.............1066,0.............0,994.........α^(-1)α^(-1/3)

.........p.......938,27.........1836,2.............1841,5.............1,003.........α^(-1)α^(-1/3)α^(-1/9)

.........n.......939,57.........1838,7.............1841,5.............1,002.........α^(-1)α^(-1/3)α^(-1/9)

.........Λ.......1115,68.......2183,3.............2209,6.............1,012.........α^(-1)α^(-1/3)α^(-1/9)α^(-1/27)

.........Σ.......1192,64.......2333,9.............2348,0.............1,006.........α^(-1)α^(-1/3)α^(-1/9)α^(-1/27)α^(-1/81)

.........Δ.......1232,00.......2411,0.............2420,4.............1,004.........α^(-3/2)

y10. .π.........139,57..........273,1...............298,2.............1,092.........1,44 α^(-1)

........ρ0........775,26........1517,2............1537,6.............1,013.........1,44 α^(-1)α^(-1/3)

........ω0........782,65.......1531,6.............1537,6.............1,004.........1,44 α^(-1)α^(-1/3)

........Σ0.......1383,70.......2707,8.............2656,3.............0,981........1,44 α^(-1)α^(-1/3)α^(-1/9)

........Ω-........1672,45.......3272,9.............3187,2.............0,974........1,44 α^(-1)α^(-1/3)α^(-1/9)α^(-1/27)

........tau......1776,82.......3477,................3491,3.............1,004........1,44 α^(-3/2)

Link to comment
Share on other sites

!

Moderator Note

Rule 2.7 says (in part; emphasis added)

Advertising and spam is prohibited. We don't mind if you put a link to your noncommercial site (e.g. a blog) in your signature and/or profile, but don't go around making threads to advertise it. Links, pictures and videos in posts should be relevant to the discussion, and members should be able to participate in the discussion without clicking any links or watching any videos

IOW, you need to have the discussion here.

 
Link to comment
Share on other sites

It didn't look that strict to me, when I browesed through some topics. Anyway, I'll try.

I have trouble with formulars though. Neither regular Tex nor \( y=x^2 \ as recommended in the math section here works for me.

In the worst case, attaching the file I linked would be ok ?

 

Link to comment
Share on other sites

I'll probably need 2 days to press my 10 page paper into one, but until then: the essential conclusions are already in the table above. The corresponding formula is:

[math]W_n / W_e = 1.509 (y_l^m)^{-1/3} \Pi_{k=0}^n\alpha^{-1/3^k}[/math], y=1 for spherical, y=1/3 for 1st angular symmetry.

I.e. a significant portion of particles in the uds-quark energy range can be given with an accuracy of about 1% as function of two parameters (in this form it is W_electron and [math]\alpha[/math]; 1.509 is a term from W_µ/W_e, too special for the 1st round).

For my knowledge, accuracy is pretty good. If any of you has examples from the standard model, giving better results, I'll be truly interested.

I quite often hear - and I've seen that reasoning in this forum too - “standard theory gives the value of the anomalous g factor, g_a, with a precision of 10^-10, you have to beat that”. Not at all.

That's a term for the electron, so let's look at that. I used e as reference in the table above, I could use any other particle, yielding the e energy within the range set in the table.

Magnetic moment of the e is given by: [math]g_a g_D e/(2m_e) \hbar/2[/math]

g_a is from QED, calculating interaction of the electron with the vacuum. Not my business. g_D = Dirac factor = 2 fits perfectly since my model implies an electromagnetic wave with the magnetic part only, i.e. W/2, contributing to the magnetic moment. That leaves W in 2c^2/W to be calculated. Accuracy better than 1%.

But where is the contribution of the standard model, i.e. QCD, to the Bohr magneton ? Or to make it easier, the corresponding values for p, n, etc. ?

QCD is out there for 50 years but when it comes to values for mass/energy and magnetic moments, its performance is poor.

I am admittedly not familiar with QCD and judge it only by its results.

Did I miss something ?

Edited by kwrk
Link to comment
Share on other sites

  • 1 year later...

Update:

Ψ(r) = exp(-(β/(2r3))
is the approximate solution of
(ħc α)2 (r 4πε/ e2) d2Ψ/dr2 - β/2 r-3 dΨ/dr + β/2 r-4 Ψ = 0
Ψ will be used as kind of probability amplitude for the electromagnetic field. The integrals
Ψ2 r-(m+1) dr ≈ Γ(m/3, β/r3) β(-m/3) / 3 ( Γ => Gamma function) may be used for:

I point charge: Wpc = ε ∫ E2 Ψ2 d3r
II photon: Wph = hc / (∫ Ψ2 dr)

Equating I+II (product of the integrals) gives:
Fine-structure constant, α-1 = 4π Γ(+1/3) |Γ(-1/3)|

Their ratio gives quantized particle energies:
Wn/Welectron = 3/2 Π(k=0-n) α^(-1/3k)

n, l.................W_calc/W_lit....α-coefficient (energy)
-1,∞....Planck......0.999.........2/3 α^(-3) (2/3α^(-3))^3 3/2 α^(-1) 2........ [source term]
0, 0.........e...........1.000.........2/3 α^(-3)
1, 0.........µ...........1.000.........α^(-3)α^(-1)
2, 0.........η...........0.993.........α^(-3)α^(-1)α^(-1/3)
3, 0.........p...........1.002.........α^(-3)α^(-1)α^(-1/3)α^(-1/9)
3, 0.........n...........1.000.........α^(-3)α^(-1)α^(-1/3)α^(-1/9)
4, 0.........Λ...........1.011.........α^(-3)α^(-1)α^(-1/3)α^(-1/9)α^(-1/27)
5, 0.........Σ............1.005.........α^(-3)α^(-1)α^(-1/3)α^(-1/9)α^(-1/27)α^(-1/81)
∞,0.........Δ............1.003.........α^(-9/2)
1, 1.........π............1.092.........α^(-3)α^(-1) 1.44
2, 1........ω0...........1.003.........α^(-3)α^(-1)α^(-1/3) 1.44
3, 1........Σ0............0.980........ α^(-3)α^(-1)α^(-1/3)α^(-1/9) 1.44
4, 1........Ω-............0.972........α^(-3)α^(-1)α^(-1/3)α^(-1/9)α^(-1/27) 1.44
5, 1........N1720.....1.005.........α^(-3)α^(-1)α^(-1/3)α^(-1/9)α^(-1/27)α^(-1/81) 1.44
∞,1........tau...........1.003........ α^(-9/2) 1.44
∞,∞.......Higgs.......1.019........ α^(-9/2) 3/2 α^(-1)/2


The relationship with Planck / Gravitation is no numerical coincidence. Expanding the incomplete gamma function of I gives terms for 1.) particle energy as given in table, 2.) Coulomb interaction, 3.) potential energy term of the DE, which as such is responsible for effects associated with strong interaction for r <rparticle, yet for r >> rparticle gives a quantitative term for gravitation.

Precision ~0.001, Parameter: elementary charge, electric constant;

More details in http://doi.org/10.5281/zenodo.801423

Comments + critics welcome,
kwrk

 

Link to comment
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.