Test

[Schindelbeck]

testing

[math]\psi\Psi[/math]

[math]W_n / W_{ref} = (y_l^m)^{-1/3} \Pi_{k=0}^n\alpha_0^{ (-1/3^k)}[/math]

[Schindelbeck]

[math]\psi\Psi[/math]

[math]c_0^2 = (\epsilon_c \mu_c)^-1[/math]

[math]c_0^2 = (\epsilon_c \mu_c)^-1[/math] with  [math]\epsilon_c = (2.998E+8 [m^2/Jm])^-1 = (2.998E+8)^-1 [J/m][/math]

text

[math] g_{µµ} = ({\rho_0} / {r} )^2 \phi(r), - ({\rho_0} / r )^2 \phi(r), − r^2, − r^2 sin^2 \theta [/math]

[math] W_n /W_e ≈ \frac {3}  {2} \Pi_{k=1}^{n} \alpha^{(-3/3^k)} [/math]

[math] \alpha^{1.5} [/math]  [math] \sigma_{min}: \sigma_{min} = (2\Gamma(-1/3) /3)^3 [/math]. Inserting a term [math] (\sigma_{0} / {\sigma_{min}}^{(1/3)} = 4\pi{\Gamma(-1/3)}^2)[/math] in \W_n /W_e

 

 

Edited December 11, 2023 by Schindelbeck

[Schindelbeck]

TK noticed that you can express the equations of EM within the formalism of differential geometry of GR. To get both EM + GR in 1 metric he:
1) added a 5th dimension (scalar [math] \Phi [/math])
2) used the constant of gravitation, G, in his equations.
The resulting disagreement with particle properties is formally due to the constant used, as noted in TK’s original article. The right step to solve this problem is a step back: drop G. This “unifies” EM and particle physics, gravitation can be recovered by series expansion.
To do so requires a set of electromagnetic units adapted to GR, e.g. keeping basic SI: [math]  c_0^2  = (\epsilon_c \mu_c)^{-1} [/math] with  [math]  \epsilon_c = (2.998E+8 [m^2/Jm])^{-1} = (2.998E+8)^{-1} [J/m], \mu_c = (2.998E+8 [Jm/s^2])^{-1} = (2.998E+8)^{-1} [s^2/Jm][/math],  
[math] e_c = 3.110E-18 [J][/math]  and [math] e_c/(4π\epsilon_c) = 7.419E-11 [m]  [/math]
I’ll mainly discuss dimensionless & relative values in the following.

I will work with the following basic conditions:
- electrostatic approximation: among the EM potentials entering Kaluza’s metric only the electric potential, [math] e_c /(4π\epsilon_cr) = \rho_0/r  [/math], will be considered
- flat 5D spacetime [math] G_{AB} = 0[/math],  which is identical to curved 4D and a [math] T_{AB} [/math]  consisting of x5-terms of the metric (Wesson)
- Spin 1/2 [math] \hbar [/math]  as boundary condition for particles.

The last point enters the model via providing an integration limit. This is just a remedy in place of a proper metric including Spin. Einstein-Cartan may be an ansatz, there are others. Anyway the (very) approximate ansatz for the metric I use will be sufficient for the following.

Kaluza’s ansatz gives a) EFE, b) Maxwell equations c) a wave-like equation connecting the scalar  [math] \Phi [/math] with the electromagnetic tensor.
I solve for  [math] \Phi [/math] and put it back in a 4D metric. The solution I use will be [math]  \Phi ~ (\frac{\rho_0}{r})^2 exp(-((\rho_0/r)^3)) = (\frac{\rho_0}{r})^2 \phi(r) [/math]. (Euler-) Integrals over this will feature the incomplete [math] \Gamma  [/math]-functions. Of particular importance will be the limits of [math] \Gamma(+1/3) ≈ 2.679  [/math] appearing in the integral over a point charge, [math] \frac{ \phi(r)}{r^2}  [/math] and [math]  \Gamma(-1/3) ≈ 4.062 [/math] appearing in the integral for a (wave-) length,  [math] \phi(r)dr[/math].
A semi-classical approach for angular momentum will be used to recalculate an integration limit, [math]\sigma [/math]:
[math] J_z = r[/math] x [math]p =  rW/c_0 = \int{}\phi(r) \frac {e_c^2} {(4π\epsilon_c r)}  ≡ 1/2 [\hbar]   [/math]    (*1)
this gives as limit (of the Euler integral) [math]\sigma_0 [/math] for spherical symmetry:
[math] \sigma_0  ≈ 8 (\frac{4 \pi \Gamma(-1/3)^3} {3})^3  [/math] = 1.77E+8[-]    (*2)
(*1 This relation will be the reason for the fine-structure constant,  [math] \alpha [/math], entering the equations.)
For a particle the equation for [math]\Phi [/math] given above will include 2 more coefficients.
- It can be shown that σ0 has to be included due to the existence of a limit implying the exponential to be an approximation for a solution of a quadratic equation and the properties of the Euler integral.
- A constant referring to a ground state, which turns out to be the lightest charged particle, the electron. This constant my be recalculated from the electron energy but it can be given as a function of [math] \alpha [/math] as well
[non-sph.sym.part] *[math]\alpha^9 =  [9/8\alpha]*\alpha^9 = \alpha_{Pl}[/math]
As indicated its value is identical with the ratio of the electron and the Planck energy, denoted [math]\alpha_{Pl}[/math]. ([math]\frac {e_c^2}{(4\pi\epsilon_c)}  = \frac {G W_{Pl}^2}{c_0^4} [/math] as definition for Planck)
This gives:
[math] \Phi = (\rho_0/r)^2 exp(-(\sigma_0 \alpha_{Pl}(\rho_0/r)^3)) = (\rho_0/r)^2 \phi(r)  [/math] in
[math]  g_{µµ} = (\frac {\rho_0} {r})^2 \phi(r), - (\frac {\rho_0} {r} )^2 \phi(r),  −  r^2,   −  r^2 sin^2 \theta [/math].
You may see where you will get gravitation back from, the expansion of [math]\phi(r)[/math]. Due to its origin from a quadratic equation, r in the exponential is limited to particle radius [math]\lambda_c ≈ \rho_0[/math]. For [math] r > \rho_0[/math] the terms related to spin should vanish (i.e. [math]\sigma [/math] and the difference between [math]\lambda_c[/math]  and [math]\rho_0  [/math]) giving  [math]\Phi ≈ [/math]  electromagnetic term  * [math] (1- \alpha_{Pl}) [/math].
(The same reasoning will give correct values for ≈ critical, vacuum density originating from minor terms of the metric.)

Particle energies
Solving the EFE and integrating for energy, W, gives:
[math] W ≈ 8\pi \Gamma(+1/3)/3 \epsilon_c \rho_0^2 (\sigma_0 \alpha_{Pl} (\rho_0/r)^3)^{(-1/3)}  [/math]
Solutions for different particles will not be orthogonal, I guess they shouldn’t since in general mass is a continuous property. However, there will be a distinguished set of solutions if you relax the condition of orthogonality to “solution not being dependent on parameters of other particles (except ground state)”. Factor [math]  \alpha^9 [/math] for spherical symmetry in [math]  \phi [/math] will give (relative to the electron, We):
[math]  W_n /W_e  ≈  \frac {3} {2} \Pi_{k=1}^{n} \alpha[/math]^[math](-3/3^k) [/math]
a convergent series (giving a range of [math]  \alpha^{1.5} [/math]) from the electron to the Delta-baryon.
The ratio proton/electron will be: [math] 1.5 \alpha^{(-1+1/3+1/9)} = 1.5 \alpha^{-1.444} = 1831 = 0.997 [/math] [exp. value] .
Symmetry is encoded in [math]\sigma [/math], for an approximation of the next spherical harmonic, y1, there will be an additional factor 3^1/3, giving a second series from the pion to the tauon.  At this point the model is to primitive to give more solutions, with one important exception:
The Euler formalism gives a minimum for [math] \sigma_{min}: \sigma_{min} = (2\Gamma(-1/3) /3)^3  [/math]. Inserting the ratio [math] ((\sigma_{0} / \sigma_{min})^{(1/3)} = 4\pi\Gamma(-1/3)^2) ≈ 1.5 \alpha^{-1}[/math] in [math]W_n /W_e [/math] will give as maximum value for energy [math] 1.5^2 \alpha^{-2.5} W_e[/math]) ≈ 254.5 GeV = 1.04 [exp. Value (Higgs VEV)]).
(While *2 for spherical symmetry can be given as a term describing the volume of a sphere, the limit case for Higgs corresponds to a 1D element. This will be important in a more detailed modeling of the fermions in a following post.)

Coupling constants
Equating the energy of a point charge with the energy of a photon will give
[math] \int{}\phi(r) \frac {e_c^2}{(4π \epsilon_c r^2 )} dr = hc_0/(int{}\phi(r) dr) [/math]∫
which can be solved for [math]  \alpha^{-1} = 4π\Gamma(+1/3)\Gamma(-1/3) = 0.998 [/math]  = 0.998 [exp. Value].
Doing the equivalent in 4D-space allows to give both weak coupling constant and [math] \alpha[/math]  in 1 equation
[math] \alpha_N^{-1}≈ \frac {S_N\Gamma(+N-2/N)\Gamma(-N-2/N) } {(N-2)^2} [/math] (SN = N-D-surface area, N={3;4})
This gives a Weinberg angle [math] sin^2\Theta_W = 0.227[/math].
tbc

Edited December 13, 2023 by Schindelbeck

[swansont]

!

Moderator Note

The description of the sandbox:

This forum is provided for members to test BB code, learn how to use the various forum functions, and generally get to grips with the system

IOW it’s not for discussion. Do not expect anyone to respond.

 

[Schindelbeck]

16 minutes ago, swansont said:

!

Moderator Note

The description of the sandbox:

This forum is provided for members to test BB code

 

That's exactly what I did. It took me ~ 20 times to get [math] right. I don't think you want to have that in the regular forum.

[Schindelbeck]