Academic Open Internet Journal ISSN 1311-4360

www.acadjournal.com

Volume 17, 2006

Electromagnetic interaction of gravity. Proposal for unified field theory.

Author: Nikos Alexandris, Secondary Education Science Teacher

e-mail : nalxhal@yahoo.gr , http://www.geocities.com/nalxhal/time

ABSTRACT

From this theory arises an equation connects Planck’s constant (h), the speed of light (c), the constant of gravity (G), the constant of electricity (K_e=1/4π.ε_ο ) and Boltzmann’s constant (k_bl). The important aspect in this equation is that it arrives at a function with a geometrical content in 10 dimensions of space: forces are added logarithmically. From the following proposals of the theory, we obtain applications with temperatures comparable to the GUT and masses like the Planck mass and the inclusion of the Avogadro number and mass of proton .

Electromagnetic interaction of gravity. Proposal for unified field theory

INTRODUCTION

From this theory arises an equation connects Planck’s constant (h), the speed of light (c), the constant of gravity (G), the constant of electricity (K_e=1/4.π.ε_ο ) and Boltzmann’s constant (k_bl). The important aspect in this equation is that it arrives at a function with a geometrical content in 10 dimensions of space: forces are added logarithmically. From the following proposals of the theory, we obtain applications with temperatures comparable to the GUT and masses like the Planck mass and the inclusion of the Avogadro number and mass of proton .

The resulting mathematical solution is not the solution that can determine the relationship of the forces contained in the equation in its full depth. However, it is valuable as a manner for a solution.

MAIN ARTICLE

HYPOTHESES

1. In some point of the space, electromagnetic oscillations LC are taking place. This space is traversed by a current (i), and is equivalent to condenser (C).This ypothesis is not oblicated but a way for transformations of ypothesis 4 .

2. There is a mass (m) that performs an oscillation and is proportional to some electric

charge (Q): Q=k.m.

In order for the mass to be positive (m>0), the constant k must have the same sign (+ or -) as the electric charge

3. We may consider that there is a mass (m) that performs a harmonic oscillation with period Τ=θ_O.2π.(λ/g)^1/2 , and frequency f =1/T, g: acceleration of gravity.

4. The harmonic oscillation has an oscillation constant (τ), which is proportional with the density of the electric charge and mass: τ = k.t, t = K_e.m.ρ_c . These teansformations are oblicated for this theory .

The energy of the oscillation is proportional to the square of the amplitude (Ε = (1/2).τ.χ²), χ = l_c, λ=2π.l_c

The acceleration of the electric charge due to the Coulomb force is equal to the acceleration of the oscillation:

g_c=g_χ and the density of the electric charge (ρ) is : ρ_c=Q/l_c³.

5. In that area the gravity and the electric power are equalized at the length

l_g = (2π)^1/2.l_c or l_g = l_c and m_c= Q/k =m_g, π = 3,14…, m_g: gravitational mass, θ_c.K_e. k².m_c²/ l_c²= θ_g.G. m_g²/ l_g² , θ_c = θ_g =1, θ_c, θ_g : constants of both interactions .

6. The quantum mechanics laws are in force : E_λc= n₁.N.h.c/ n₂.λ = E.k.m. l_c , with wavelength n₂.λ and E: the intensity of the electric field.

with n_1:energy level, n₂: number of waves. It should be valid that the electric energy is equal to the electromagnetic energy E_c=E_λc, and that oscillation consists of an integer number of fundamental waves λ. The density of the electric charge (ρ) is proportional to the acceleration (g), as an expansion of the Poisson’s equation: ρ_c=a.g_c, g_c=k. E, E: the intensity of the electric field.

7. Thermal power can be equivalent to electromagnetic with wavelength lx :

Ε_lx= Ε_T, Ε_lx= n₁.N.k_bl.Τ=n₂.h.c/lx, , n₁: degrees of freedom, n₂: energy level,

N: number of particles, T: temperature .

The degrees of freedom of the thermal movement are equal to the quantum number of the E_λc oscillation (of that, that is equal to the electric oscillation) and the number of the waves of E_λc is equal to the quantum number of the E_lx oscillation that is equal to the thermal movement.

The above relations between lengths were taken in order to be in accordance with the Planck mass in the applications of mass with the Planck temperature.

The hypotheses

a) θ_O =1/2π, θ_c =θ_g=1 or

b) θ_O =1, θ_c =θ_g=1

These are wo solutions simply to agree with M_Planck, T_Planck (Planck mass and Planck temperature) for n₁=10, n₂=12 and mass of proton .

symbols

frequency (f)

condenser capacity (C )

electric potential (U_e)

potential (U_m)

electric charge (Q)

length (l)

acceleration (g)

speed of light (c)

self-induction coefficient (L)

energy (E)

force (F)

intensity of current (i)

HYPOTHESIS 1

E = (1/2).L.i² Þ

E = (1/2).L.(Q/δt )² Þ

U_e. Q = (1/2).L.(Q/δt )² Þ

L = 2.U_e. δt² /Q, δt = time (1)

HYPOTHESIS 2

If Q = k.m, let be k a constant and m the mass

From Einstein’s theory U_m= c² or in low speed v²/2 so U_m = E/m = E/(Q/k) = (E/Q).k = U_e.k Þ U_e= U_m/k (2)

U_e = c²/k So (1) Þ

L = 2.(c² /k).(l/c) ² /Q = 2.c² . l² /k.c ² .Q Þ L = 2.l² /k.Q (3)

l = length of current (l_i)

HYPOTHESIS 3

Τ = θ_O. 2π.(λ/g)^1/2, if λ= l_i Þ

f = 1/(2π.(L.C)^1/2) = 1/( 2π.θ_O(l/g)^1/2) Þ

2π.(L.C)^1/2 = 2π.θ_O(λ/g)^1/2 Þ

4π².L.C = 4π².θ_O².λ/g, (3) Þ

(2.λ² /k.Q). 4.π².C = 4π².θ_O².λ/g Þ

C = 4π².θ_O².k.Q/8π².g.λ (4)

λ= wavelength = length of current, This must be set with the introduction of the equation (3)

HYPOTHESIS 4

Energy is E = (1/2).C. U_e²,(2) Þ

E = (1/2).C.( U_m/k)² = (1/2).C.U_m²/k² =

= (1/2).C.(g.χ)²/ k² = (1/2). (C.g²/ k²).χ²,

Þ E = (1/2).τ.χ² , τ = C.g_χ²/ k²,

χ = amplitude length, We accept this with the energy forms: (τ.χ², g_χ) .

a) Law of Coulomb and Newton :

mg_c = K_e. Q²/ l_c², K_e = 1/4π.ε_ο Þ

mg_c = K_e. k.m.Q/ l_c² Þ g_c = k.K_e.Q/ l_c² (5)

if g_c = g_χ Þ τ = C.g_χ²/ k² =

C.( k.K_e.Q/ l_c²)²/ k² = C.K_e²Q.Q/ l_c⁴

= C.K_e²Q.Q/ l_c. l_c³,

If ρ_c = Q/ l_c³ then τ = C.K_e²Q. ρ_c / l_c Þ

K_el =1/4π.ε_ο ,

τ = C.K_e².k.m. ρ_c / l_c (6)

b) The above relation can come up by the use of interactions of the energy E and the density of the electric charge.

(E/l)=θ.ρ_m.l_c³.g_c=θ.Κ_e.Q.ρ_c.l_c

with ρ_c=k.ρ_m,

θ: coefficient of shape that θ/V=l_c^-3, V: volume and g_c=k.E, E : the intensity of the electric field.

MAIN TRANSFORMATIONS

let be t_c = K_e.m. ρ_c then τ = k.(C.K_e/ l_c).t_c

but C.K_e/ l_c=1 because K_eQ²/ l_c²=

U.Q/ l_c Þ K_eQ/ l_c = U, Q = C.U Þ

C.K_e/ l_c =1 So τ = k. t_c (7)

Law of Coulomb F= K_eQ²/ l_c² = K_e(Q.Q/ l_c³). l_c =

(K_e.k.m.ρ_c). l_c=k. t_c . l_c

so τ = k. t_c , this agree by (7) equation .

(6) Þ τ = C.K_e².k.m. ρ_c / l_c, t = K_el.m. ρ_c, (4), ρ_cλ=Q/ λ.l_c² Þ

τ = (4π².θ_O².k.Q/8π².g.λ).(K_e.k/ l_c). t_c =

(4π².θ_O².k².Q.K_e/8π².g.λ.l_c). t_c

= (4π².θ_O².k².Q.K_e/8π².g.λ.l_c²). l_c. t_c

for λ=2π. l_c : ypothesis 4 Þ

τ = (4π².θ_O².k²/16π³.g).K_e.ρ_c. l_c. t_c =

(4π².θ_O².k²/16π³.m.g).(K_e.m. ρ_c ). l_c. t_c =

(θ_O².k²/4π.F). l_c.t_c.t_c ,

Let be β = (θ_O².k²/4π.F). l_c so τ =β. t_c ² (8)

(7),(8) Þ k = β.t_c (9)

In order to be m>0 it should be Q/k>0 so t_c/Q>0 and k/t_c>0 so (7): τ>0, (9): β>0.

1. τ = k. t_c
2. t_c = K_e.m. ρ_c
3. β = (θ_O².k²/4π.F). l_c
4. τ =β. t_c ²
5. k =β.t_c

All the above mathematical analysis is to provide a way to extract the main transformations and is not oblicated .

HYPOTHESIS 5

Since the speed at which the interaction occurs is the speed of the light

(hypothesis 2) there is no relative situation (10)

1) Hypothesis 2, law of Coulomb Þ

F = K_e. k².m_c²/ l_c², for F gravity = F electricity

F gravity = G. m_g²/ l_g² Þ

K_e. k².m_c²/ l_c²= G. m_g²/ l_g² ,

Ypothesis 5.1.a) if m_c = m_g, l_g = (2π)^1/2.l_c Þ

K_e. k². m_c²/l_c²= G.m _c²/ ((2π)^1/2.l_c) ² Þ

2π.K_e. k².m_c² = G. m _c²

G = 2π.K_e.k² (10a)

Ypothesis 5.1.b) if m_c = m_g, l_g = l_c Þ G = K_e.k² (10b)

2) The above relation can arise by assuming the existence of the interactions of the energy E system with mass density ρ_g=m/l_g³, and charge density ρ_c=Q/l_c³, so

(E/l_g)=θ.G.m_g.ρ_g.l_g=(E/l_c)=θ.K_e.Q.ρ_c.l_c,θ: coefficient of shape of the system, θ/V=l^-3, l= l_c or l_g and ρ_c=k.ρ_g.

(10a),(9) Þ G = 2π.K_e.β².t_c²

(10b),(9) Þ G = K_e.β².t_c²

HYPOTHESIS 6

5.1.a) G = 2π.K_e.β². t_c², t = K_el.m. ρ_c Þ

G = 2π.K_e.β².( K_e.m. ρ_c )² Þ

G = 2π.K_e³.m². ρ_c².β² Þ

if ρ_c = α.g_c, g_c = k.E, E : is the intensity of the electric field (this will modified at calculation of parameter α ) .

G = 2π.K_e³.m².g_c².α².β² Þ

G = 2π.K_e³.(E_c²/ l_c²).α².β² ,

E_c : electricity energy since the origin of m is the parameter t, and arises of

electricity : function (6)

E_λc = n₁.N.h.c/ n₂.λ_, , if E_c=E_λc Þ

G = 2π.K_e³.(( n₁.N.h.c/ n₂.λ)²/ l_c²).α².β² Þ

G = 2π.K_e³.((n₁/n₂)⁴.N².h².c²/λ².l_c²).α².β² , if lx⁴=λ².l_c²Þ

G = 2π.K_e³.((n₁/n₂)².N².h².c²/lx⁴).α².β² Þ

G = 2π. (n₁/n₂)².N².K_e³.h².c².α².β²/lx⁴,

lx² = l_c.λ

5.1.b) (10b) Þ G = (n₁/n₂)².N².K_e³.h².c².α².β²/lx⁴,

lx² = l_c.λ

The length of the electromagnetic oscillation that is equal to the thermal

movement is equal to the length of the gravitational interaction.

HYPOTHESIS 7

5.1.a) G = 2π. (n₁/n₂)².N².K_e³.h².c².α².β²/lx⁴,

let be k_bl Boltzmann’s constant, Ε_lx= Ε_T,

Ε_lx= n₁.N.k_bl.Τ=n₂.h.c/lx Þ

lx= (n₂/n₁).h.c/N.k_bl.T

The n₂ expresses the number of the fundamental waves and n₁expresses the degrees of freedom of the thermal movement that are possessed by n₁ quantum of E_λc.

So G = 2π. (n₁/n₂)².N².K_e³.h².c².α².β²/((n₂/n₁).h.c/N.k_bl.T)⁴ and

G = 2π. (n₁/n₂)².N⁶.K_e³.k_bl ⁴.T⁴.α².β²/(n₂/n₁)⁴.h².c² Þ

2π.α².β² = ( h².c².G / K_e³. k_bl ⁴).( (n₁/n₂)^-6.N^-6/Τ⁴) ,

K_e =1/4π.ε_ο = 8,9875.10⁹.N.m²/C²

h².c².G / K_e³. k_bl ⁴ = 99,6895, (99,68955)^1/4 = 3,1598199 without dimensions

let be π^*= 3,1598199 without dimensions for the time being .

So h².c².G / K_e³. k_bl ⁴ = π^*4 and 2π.α².β² = (n₁/n₂)^-6.N^-6.π^*4/ Τ⁴

5.1.b) α².β² = ( h².c².G / K_e³. k_bl ⁴).( (n₁/n₂)^-6.N^-6/Τ⁴)

CALCULATION OF THE PARAMETERS α, β, OF THE CONSTANT k

AND THE VARIABLE Τ

a) dE /δx = ρ_c/ ε₀ ( Poisson function ), dE is the differential of intensity of the electric field : dE = dF/dQ, for δx = l_c Þ

(dF/dQ)/ l_c = ρ_c/ε_0, hypothesis 2 : dQ = k.dm Þ ρ_c = (dF/k.dm. l_c).ε₀ and the differential of acceleration dg_c = dF/dm Þ

ρ_c = ε₀.g_c/k.l_c Þ ρ_c = α_p.g_c : hypothesis 6 and dg_c = k.dE so

α_p = ε₀/k.l_c (11)

b) From Coulomb’s law : F = θ.K_e.Q.ρ_c.l , θ/V=l^-3 Þ ρ_c = α_c.g_c,

α_c = 4π.ε₀/θ.k.l, is accepted if l.θ = 4π.l_c so α_c = ε₀/k.l_c

ex. l=3.l_c, θ = (4/3).π Þ α_c = ε₀/k.l_c

The equation α_p = ε₀/k.l_c is assumed to agree to the next one, so hypothesis 6 will be:

ρ_c = α.g_c, dg_c = k.dE , E_c=E_λc and concerns the differential of acceleration, alteration of intensity of electric field and the differential of electrical energy .

ypothesis5.1.a)

2π.α².β² = (n₁/n₂)^-6.N^-6.π^*4/ Τ⁴Þ

(2π) ^1/2.α.β = (n₁/n₂)^-3.N^-3.π^*2/ Τ² Þ

(2π) ^1/2.β = (n₁/n₂)^-3.N^-3.π^*2/ α.Τ² , (11) Þ

(2π) ^1/2.β = (n₁/n₂)^-3.N^-3.π^*2/( ε₀/k.l_c).Τ² ,

a = a_p, in order to agree at temperature applications Þ

(2π) ^1/2.β = (n₁/n₂)^-3.N^-3.π^*2.k. l_c /ε₀.Τ² (12)

(10a) Þ k = (G/2π.K_el )^1/2 _, (9) Þ

β = k/ t_cλ = ((G/2π.K_el)^1/2)/ t_cλ , t_cλ = K_e.m.ρ_cλ Þ

β² = G/2π.K_el (K_el.m. ρ_cλ) ², ρ_cλ = Q/λ. l_c² Þ

β²=G/ 2π.K_e(K_e.m.Q/λ.l_c²) ² Þ

2π.β² = (G / K_e³.m ²).(λ².l_c⁴/ Q ²) (13)

(12),(13)Þ 2π.β² = ((n₁/n₂)^-3.N^-3.π^*2.k. l_c /ε₀.Τ²)²=

(G / K_e³.m ²).(λ².l_c⁴/Q ²) Þ

N^-6.π^*4.k². l_c²/ε₀².Τ⁴= (G / K_e³.m ²).( λ².l_c⁴/ Q ²) Þ

Τ⁴ = (n₁/n₂)^-6.N^-6.π^*4.K _e³.(k².m ²).Q ²/ ε₀².G . λ².l_c² =

(n₁/n₂)^-6.N^-6.π^*4.K_e³.Q ⁴/ ε₀².G . λ².l_c² Þ

Τ⁴ = ((n₁/n₂)^-6.N^-6.π^*4.K_e³/ ε₀².G ).( Q ⁴/ λ².l_c² ) Þ

Τ⁴ = (n₁/n₂)^-6.N^-6.1,38585×10⁶⁴.( Q /λ.l_c )² and

T =(n₁/n₂)^-3/2. N^-3/2.1,085×10¹⁶.Q/ (λ.l_c)^1/2 , λ = l_c , a = a_p (14)

(10a) Þ k =(G/2π.K_e)^1/2 = 3,43745×10^-11 C/Kg

(12) Þ β = (n₁/n₂)^-3.N^-3.π^*2.k.l_c/(2π) ^1/2.ε₀.Τ², a = a_p,

ypothesis 5.1.b) T =(n₁/n₂)^-3/2. N^-3/2.1,085×10¹⁶.Q/ (λ.l_c)^1/2 , λ = l_c , a = a_p

temperature is the same :(14)

(10b) Þ k =(G/K_e)^1/2 = 8,6164×10^-11 C/Kg

(12) Þ β = (n₁/n₂)^-3.N^-3.π^*2.k.l_c/ε₀.Τ², a = a_p,

this equation will not be affected. Equation (14) will be modified at temperature applications, applications 4, 5.

Because β>0 is should be and a>0 and k>0 so Q>0, of course as long as m>0.

Because the positive charges have the same sign, they are expelled and these forces are cancelled out by the attractive gravitational forces.

If it were m<0, k would have the opposite sign (+ or -) of Q and t_c=K_e. m.ρ_c=K_e. m².k/l³ will have the same sign (+ or -) with k and opposite with Q. (9):β>0, (7):τ>0, (14): k>0 so Q<0. The negative charges are creating repulsive forces that would be added to antigravity, so the above relations would not be valid.

APPLICATION TO THE MASSES

Because the above equations were derived using basic mathematics, it is possible to eliminate those coefficients which influence the accurate measurement of the masses, so that they may be compared with the real masses. That’s why three different places are going to be chosen from the above analysis for the extraction of three different masses, and their ratios are going to be calculated. These ratios will be pure numbers. The method is going to be a simple dimensional analysis.

1. Hypothesis 2: Q = k.m, Q = e =1,602×10^-19 C, (10a) :

k=(G/2π.K_el)^1/2 =3,43745×10^-11 C/Kg arises :

ypothesis5.1a) m=e/k > 0 Þ m_eg =4,66094×10^-9 kg > 0 (15)

ypothesis5.1b) m_eg =1,8594×10^-9 kg > 0

In order for the mass to be positive, m>0 the constant k should has the sign (+ or -) of the electric charge, otherwise it will result in negative mass and the situation of antigravity. But already it is accepted that k>0 and e>0 is the charge of proton.

2. Hypothesis 3 : oscillation period T=1/f is in effect T²= θ_O².4π².λ/g Þ

g_λ = 4π².θ_O².λ/ T², T =λ/c Þ g_λ = 4π².θ_O².λ/(λ/c)² ,

hypothesis 4: g_λ = g_c = k.E, (5) Þ

g_c = k. K_e.Q/ l_c² = 4π².θ_O².λ./ (λ/c)² ,ypothesis 2 Þ

k. K_e.k.m_c/ l_c² = 4π².θ_O².c²/λ Þ

m_c = (l_c²/λ).4π².θ_O².c²/k². K_e Þ

a) for θ_O =1/2π, θ_c =θ_g=1 ,

for λ =2π. l_c it will be

m_c = c².l_c/ 2.π.k². K_el , for l_c = λ_Planck

m_c =c².λ_Planck / 2π.k². K_el =2,17671.10^-8 kg = M_Planck > 0 (16)

b) θ_O =1, θ_c =θ_g=1

for 2π.l_c = (λ.λ_Planck/2π)^1/2 Þ m_c = M_Planck > 0

The equivalent mass of the charge is equal to the Planck mass

The relation (16) makes completely compatible the theory GUT with the hypotheses 2, 3, 4.

My opinion is that the ypothesis 5.1.b) is better , because agree

to empirical type of proton ,but 5.1.a agree to applications of the

forces in the square of total energy , also ypothesis b) agree that the

forces added logarithmicaly, as presents at the end of this article

APPLICATION TO TEMPERATURES OF THE GUT THEORY

(14) Þ T =(n₁/n₂)^-3/2. N^-3/2.1,085×10¹⁶.Q/ (λ.l_c)^1/2 , λ= l_c Þ

T =(n₁/n₂)^-3/2. N^-3/2.1,085×10¹⁶.Q/ (l_c²)^1/2 Þ

T =(n₁/n₂)^-3/2. N^-3/2.1,085×10¹⁶.Q/ (l_c²)^1/2 Þ

T =(n₁/n₂)^-3/2. N^-3/2.1,085×10¹⁶.Q/l_c

1. PROTON

T= (n₁/n₂)^-3/2.N^-3/2.1,085×10¹⁶.Q/ l_c , λ = l_c = 10^-15.m ,

Q = |e| =1,602×10^-19C

T= (n₁/n₂)^-3/2.N^-3/2.1,085×10¹⁶.Q/ l_c , T = (n₁/n₂)^-3/2.N^-3/2.1,738×10¹² K ≤ 10¹² K

For (n₁/n₂)^-3/2.N^-3/2 ≤ 0,575 so (n₁/n₂).N ≥ (0,575)^-2/3=1,446 then T ≤ 10¹² K

This is in agreement with the theory.

2. QUARK

For λ = l_c = 10^-18.m and Q = |e| then

T= (n₁/n₂)^-3/2.N^-3/2.1,738×10¹⁵ K ≤ 10¹⁶ K, for (n₁/n₂)^-3/2.N^-3/2 ≤ 5,754 or

(n₁/n₂).N ≥ 0,311.

This result is in agreement with the GUT theory, where the quark are kept in hadronians (10¹² K, 10¹⁶ K )

3. particle X : λ = l_c = 10^-30m, Q = |e| as a constant,

T = (n₁/n₂)^-3/2.N^-3/2.1,085×10¹⁶.Q/ l_c Þ T = N^-3/2.1,738×10²⁷ K ≤ 10²⁸ K

For (n₁/n₂)^-3/2.N^-3/2 ≤ 5,754 or (n₁/n₂).N ≥ 0,311 .

Electroweak range (10¹⁶ K, 10²⁸ K), Higgs bosons appearaccording to the theory.

4. (16) Þ l_c = λ_Planck Introducing the length of Planck : λ_Planck, arises a temperature of the big unification (10²⁸ K, 10³² K).

(14) Þ T =(n₁/n₂)^-3/2. N^-3/2.1,085×10¹⁶.Q/ (λ.l_c)^1/2 , Q = |e|, l_c = λ = λ_Planck, are valid T = (n₁/n₂)^-3/2. N^-3/2.1,085×10¹⁶.e/ λ_Planck Þ T =(n₁/n₂)^-3/2. N^-3/2.1,075×10³².K

for T ≤ T_Planck = 1,416×10³² K Þ (n₁/n₂)^-3/2.N^-3/2 ≤ 1,416/1,075 = 1,31663 or

(n₁/n₂).N ≥ 1,31663^-2/3 = 0,832 ≈ 5/6 or 10/12,

But N must be a natural number and at the unification point takes the rate 1.

For N =1 (unification point),T = (10/12)^-3/2.N^-3/2.1,085×10¹⁶. e / λ_Planck Þ

T = (10/12)^-3/2.1,075×10³² K = 1,4137×10³² K ≈T_Planck , n₁=10, n₂=12 . (19)

At the highest Planck temperature, the quantum number of the oscillation that is equivalent to the electric energy is the n₁=10 quantum number, equal to the degrees of freedom of the thermal movement, so n₁ expresses the size at the space. The quantum number of the oscillation with the thermal movement is n₂=12 and is as much as the number of waves of λ oscillation that is equivalent to the electric.

The above relation (19) brings into agreement the GUT theory with the analysis of the hypotheses, since for the derivation of equation (14) of the temperature all the basic equations of the analysis of the hypotheses were used.

Examination of hypotheses 6, 7 for l_c = λ = lx

They could be chosen 1. E = N.hc/λ, N.k_bl.T = h.c/λ 2. E = hc/λ, N.k_bl.T = h.c/λ 3. E = h.c/λ, N.k_bl.T = Ν.h.c/λ . 4. E = Ν.h.c/λ, k_bl.T = h.c/λ.5. E = h.c/λ, k_bl.T = N.h.c/λ. 6. E = N.h.c/λ, k_bl.T = N.h.c/λ.

Case l,

At the case l, l = lx = λ, are valid the same with T = 1,4139×10³² K ≈ T_Planck .

The equation that is contained at the analysis of the hypothesis 7 will help easily the below calculations in any case with the proper change of N places .

G = 2π.(n₁/n₂)².N².K_el³.h².c².α².β²/(h.c/N.k_bl.T)⁴ : N^-6

Also T= N^-6/4.1,085×10¹⁶.Q/l : N^-3/2

Case 2. The basic equation of temperatures turns (there is no N²):

2.π.α².β² = N^-4.π^*4/ Τ⁴, temperature becomes: T = N^-1.1,075×10³².K

and is valid : (1,31648) ^-1 = 0,7596 > 3/4 = 0,75 so (3/4)^-1 =1,333 > 1,31663

consequently the temperature should be multiplied to 1,333, and the equation of

T = 1,333.1,075×10³².K = 1,433×10³².K > T_Planck, is not valid .

Case 3. No N at the equation of temperature, the same would be valid for the

fraction of degrees of freedom too. Without any N or any fraction of degrees of freedom the temperature would be 1,31663 times smaller from the Planck

temperature, is not valid.

Case 4. The basic equation of temperatures becomes (there is no N⁴):

2.π.α².β²= N^-2.π^*4/ Τ⁴, and the equation of temperature turns to :

T= N^-1/2.1,075×10³².K So for N^-1/2 is valid :(1,31663)^-2 = 0,576 > 4/7=0,571 the temperature should be multiplied with (4/7) ^-1/2 =1,3228, so Τ=1,3228.1,075×10³².K=1,4228×10³².K which is much bigger that T_Planck and is not valid.

Case 5. The basic equation of temperatures becomes: 2.π.α².β² = N⁴.π^*4/ Τ⁴, and the equation of temperature becomes: T= N¹.1,075×10³².K . So for N¹ is valid :

(1,31663) ¹ < 4/3 = 1,333 333 so the temperature should be multiplied with

1,333, case 2 where T is bigger that T_Planck and is not valid .

Case 6. The basic equation of temperatures becomes: 2.π.α².β² = N².π^*4/ Τ⁴, and the equation of temperature turns :T = N^1/2.1,075×10³².K . So for N^1/2 is valid :

(1,31663) ² =1,733 > 12/7=1,714 or < 7/4 = 1,75 so the temperature should be multiplied with (12/7) ^1/2 = 1,3093 or the (7/4) ^1/2 =1,3228,

T =1,309.1,075×10³².K=1,408x 10³².K, accepted and T =1,3228.1,075×10³².K=

= 1,4228×10³².K, is not valid . The temperature for 12/7 fraction dimensions (with power to 1/2) approaches T_Planck but with smaller approach as concern to fraction of dimensions 10/12 (with power to -3/2).

From the above we see the necessity intoducing the degrees of freedom, since N is a natural number and not decimal and at the unification point it will take the value of 1.

The degrees of freedom will be ordered as following:

a) E = n₁.N.h.c/λ, N.k_bl.T = n₂.h.c/lx, the basic equation of temperature is :

2π.α².β² = (n₂⁴/n₁²).N^-6.π^*4/ Τ⁴ so the temperature is :

T = ( n₂/n₁^1/2).N^-3/2.1,085×10¹⁶.|e|/ λ_Planck Þ T = ( n₂/n₁^1/2).N^-3/2.1,075×10³².K

For N =1 and T = T_Planck it should be : (n₂/n₁^1/2) =1,31663 η n₁^1/2/ n₂ =0,7596 ≈ 3/4 = 9^1/2/4, but as before (case 2) it will arise the temperature bigger than the Planck temperature (is not valid).

b) But since the fraction of dimensions gives the bigger approach to Planck temperature when is in power -3/2, it should at the basic equation of temperatures to be at the power -6.

5. Introducing the Avogadro (N_A) constant, to the quantity of the load, we obtain a temperature near of the big unification T_Planck=1,416×10³² K

If Q = N_A.|e|, λ= l = l_e = 5,291×10^-11m or l = l_e/ N_A, Q = e then

T = N^-3/2.1,085×10¹⁶. N_A.|e|/l_e Þ

T = N^-3/2.1,978×10³¹ ≤ 1,416×10³² KÞ N^-3/2≤1,416/0,1978 =7,158 or

N ≥ 7,158^-2/3 =0,269≈3/11, For N=1 (unification point),

T = (3/11)^-3/2.N^-3/2.1,085×10¹⁶. N_A.|e|/l_e Þ T=(3/11)^-3/2.1,978×10³¹=1,3887×10³² K ≈ T_Planck . For the fraction 3/11 are valid the same as for the fraction 10/12. It follows that the hypotheses 5 and 7 are modified in: E = (n₁/n₂).N.h.c/λ and n₁.N.k_bl.T = n₂.h.c/lx with the introduction of the degrees of freedom and n₁=3, n₂=11 in correspondence, in lower temperature from the highest at length l_e and in particles number N_A.

nature of the basic equation and π*

π* counted 3,1598199 without dimensions .

from the basic identity h².c².G / K_e³. k_bl ⁴= π^*4

h.c = E_c.l, E = energy, l = length, G = E_G.l/m², m = mass, Q = electrical charge

K_e = E_e.l/Q², k_bl = E_T/T, E = f.l, f = force, Q=k.m με k = (G/ K_e)^1/2

h².c².G / K_e³. k_bl ⁴= π^*4 Þ (E_c.l_c)². (E_G.l_g/m²)/( E_e .l_c/Q²)³.(E_T/N.T) ⁴ = π^*4 Þ

(f_c.l_c²)². (f_G.l_g²/m²)/( f_e .l_c²/Q²)³.(f_T.l_T/ N.T) ⁴ = π^*4 Þ l_g= 2π. l_c,

(f_c².f_G. /f_e³ .f_T ⁴). (4π².Q⁶.N⁴.T ⁴/l_T⁴.m²) = π^*4 Þ

(f_c².f_G. /f_e³ .f_T ⁴).( 4π².Q⁶. N⁴.T ⁴/π*⁴.l_T⁴.m²) = 1 Þ

(f_c².f_G. /f_e³ .f_T ⁴). f*⁴=1 and

(4π².Q⁶. N⁴.T ⁴/π*⁴.l_T⁴.m²) = f*⁴, f*= a remnant force Þ

f_c².f_G¹. f*⁴ = f_e³ .f_T ⁴

f_c: electromagnetic force f_G: gravitational force f_e: Coulomb force

f_T: thermal force, f*: remnant force .

Instead of forces, the above relation can be written with the form of energy interactions:

(E/l)_c².(E/l)_G¹. (E/l)*⁴ = (E/l)_e³ .(E/l)_T ⁴

The following analysis is valid for both relations in relative conditions. For practical reasons it will be used the first relation.

This can be the basis of the hypothesis that forces are added logarithmically. They are probably applied to an exponential loop.

4π².Q⁶. N⁴.T ⁴/ l⁴.m²= π*⁴.f*⁴ Þ 4π².Q⁶. N⁴.T ⁴/f*⁴.l_T⁴.m²= π*⁴ and

f*= (2π) ^1/2.Q^3/2.N.T /π*.l_T.m^1/2

so π*²= 2π.Q³. T ²/f*².l_T².m and π*²=(3,1598199)²=9,9844618.Cb³.Kelvin ²/Joule².kg or Cb³.Kelvin ².sec⁴ /kg³.m⁴

APPLICATIONS TO THE FORCES

From the above relation of forces, we may obtain applications of masses and energy level

f_c².f_G¹. f*⁴ = f_e³ .f_T ⁴

f_c: electromagnetic force f_G: gravitation force f_e: Coulomb force

f_T: thermal force, f*: remnant force .

We may suppose that one solution of the above function of the forces is:

1. f_c².f_G¹= f_e³ and 2. f*⁴ = f_T ⁴

1. In order to find the entire energy, we should count the mass of the system.

f_c².f_G¹= f_e³ Þ (h.c/l_c)² .(G.m²/l_g²) = (K_e.(k.m)² /l_c²)³, l_g = (2π)^1/2.l_c Þ

(-(2π) ^1/2.h.c/ l_g)².( G.m²/l_g) = ((2.π) ^1/2.K_e.(k.m)² )l_g)³ Þ

m⁴ = (h.c)². G/((2π) ^1/2.(K_e.k² )³ ) Þ

m_cge = 1,7209×10^-7.kg

We will examine if the sum of energies of interactions is valid.

The entire energy : Ε_cge = -E_c - E_g+E_e = -f_c.l_c - f_g.l_g + f_e.l_c Þ

Ε_cge = (h.c/l_c - G.m²/l_g + (K_e.(k.m)² /l_c ),

l_g = (2π)^1/2.l_c (hypothesis 5.1.a) Þ

Ε_cge = (-(2π) ^1/2.h.c - G.m² + (2π) ^1/2.K_e.(k.m)² ).(1/l_g) Þ

Ε_cge = A.(1/l_g),

A = (-(2π)^1/2.h.c – G.m² +(2.π) ^1/2.K_e.(k.m)² ), A = -nc - ng + ne

nc=(2π)^1/2.h.c, ng = G.m², ne=(2π) ^1/2.K_e.(k.m)²

In order for the forces to be equally important, the A-factor must have an order of magnitude equal to 10^-26 (h.c), so the mass should have an order of magnitude of 10^-7 (G.m²), so it should be m_cge. Specifically, the algebraic addition of the factors is to multiply each one with the number of the dimensions that concerns each interaction. But there is a more important reason: the A-factor must have such a value as to give n₁=10 and n₂=12.

Before this, we should calculate the value of N in the thermodynamic equations.

We consider that the system is a black body, so we have:

n₁.N.k_b.T=n₂.h.c/l_g (hypothesis 7) and the force of monochromatic emission

P = h.c/l_g.δt = a.σ.S.Τ⁴, σ = 5,6704×10^-8 watt/m².K⁴, δt = l_g/c, S = l² άρα :

T = (n₂/n₁).h.c/l_g.k_b.N και h.c²/l_g² = = a.σ. l_g².((n₂/n₁).h.c/l_g.k_b.N)⁴ Þ

for a=1, (n₁/n₂)⁴.N⁴ = a.σ.h³.c²/k_b⁴ = 40,8 Þ (n₁/n₂).N = 2,527

for n₁=10, n₂ =12 Þ N = 3,03 so l_g = (n₂.h.c)/ n₁.N.k_b.T Þ

T.l_g = 5,755×10^-3.m.K .

We notice that the number N is the index of the electric force or to sum of the indices of gravity and electromagnetic force. This supports the hypothesis that the indices are dimensions at the space.

n₁.N.k_b.T = n₂.h.c/l_g = Ε_cge = A/ l_g ( hypothesis 7 ),

N=3 so n₁=E_cge/N.k_b.T Þ n₁=A/N.k_b. l_g.T, l_g.T =5,755×10^-3.m.K .

In order for the number n₁ to be n₁»10, the A-factor should be:

Α = 3.(-(2.nc)²-ng²+(3.ne)²) ^1/2 = 2,5041×10^-24.J.m/kg²

2.nc = 9,9585×10^-25.J.m/kg², ng = 1,976×10^-24.J.m/kg², 3.ne =2,365×10^-24.J.m/kg²

The calculations arise from ypothesis 5.1.a

if A_c=(-(2.nc)²-ng²+(3.ne)²) ^½ then Ε_cge = 3.Ac/ l_g

So n₁=A/N.k_b.5,755×10^-3 = 10,50 and n₂=N.A/h.c=12,60, the deviation from the required prices 10 and 12 is 4,7% .

So the sum of the energies is not valid, but the sum of the squares of the energies multiplied by the square of the number of the spatial dimensions of each corresponding interaction:

Ε_cge² = -4.E_c²- E_g² +9.E_e²

E_c =-3.h.c/ l_{c ,} E_g= -3.G.m² / l_g , E_e=3.(K_e.(k.m)² /l_c

2. f*⁴ = f_T ⁴ and f*= (2.π) ^1/2.Q^3/2.N.T /π*.l_T.m^1/2

so f* = f_T , for Q = |e|, T = T_Planck = 1,416×10³².K Þ

E*= f*.l_T = (2π) ^1/2.Q^3/2.N.T_Planck /π*.m^1/2

Þ m.c² = (2π) ^1/2.Q^3/2.N.T_Planck /π*.m^{1/2 ,} for Q = |e| , m = m_eg Þ

N = π*.m^3/2 .c² /(2π) ^1/2.|e|^3/2.T_Planck Þ N = 3,9713 or N = 4

Thus, the electric energy of the remnant force is equivalent to the mass of the electric charge of positron at the Planck temperature The degrees of freedom of the thermal movement that is equivalent to the remnant force are 4 and corresponded to the dimensions of the space.

It may be noticed that the degrees of freedom in both cases are the index of the basic force or the sum of the indices of the forces in each site of equivalency. The force of the electric charge was defined in 3 dimensions of space and remnant force in 4. This fact shows the validity of the theory and its agreement with the real data, as the degrees of freedom arose from universal constants. The forces at that situation of unification are 5 and the sum of the indices 14. But we have already calculated n₁=10 as a sum of the dimensions of space and the remnant force consist of electric and thermal interactions, so it is not fundamental. The indices of the fundamental forces have a sum of 10.

The final hypothesis for the nature of equation (h².c².G / K_el³. k_bl ⁴ = π^*4 ) is that it constitutes a fundamental relationship according to the GUT theory and it describes a relationship among the electromagnetic, gravitational, Coulomb and thermal forces.

Their dimensions are as follows:

Gravitational force : 1 dimension

Electromagnetic force : 2 dimensions

Coulomb force : 3 dimensions

Thermal force : 4 dimensions

Total: 1+2+3+4 = 10

EMPIRICAL TYPES (inclusion of the Avogadro number)

There is an empirical relation that connects the quantities :

M_Planck, N_A: Avogadro number, , l_g (length of gravity interaction)=λ_Planck.(2π)^1/2.

l_e=5,291×10^-11.m

(m_eg.l_e² .N_A^-2+M_Planck.l_g² )/2=p.(m_eg.M_Planck)^1/2 .l_e.l_g. N_A^-1

p=1,0000086 (17)

The quantities m.l² are moments of inertia and the second member contains a middle rate of moment of inertia. The average is equal to geometric average so:

m_eg.l_e² .N_A^-2 = M_Planck.l_g² (18)

m_eg > 0, M_Planck> 0

From the above relation we obtain the length : l_e/N_A=8,788×10^-35.m

The tendency of inactivity of the equivalent mass of the electric charge of the electron at the length of the electric charge l_e is equal to the tendency of inactivity of the Planck mass. This relation makes compatible the hypothesis 2: Q=k.m with the theory GUT.

The interesting part of the above relation is that though the Planck mass presupposes temperatures that do not occur our empiric world, m_eg as an equivalent mass of the charge of the electron, which is valid everywhere. Even the neutral particles may considered to have mutual-confutation charges and the same is valid and for each point of a filed that can create mutual-confutation particles. The taking from the space of m_eg transfers the unification relations to the compatible space of the observed particles.

Empirical form of angular momentum that it relevant to the mass of proton.

(N_A.m_p. (m_eg/(2π)^1/2)) ^1/2 .c. λ_plank = p.10.h, p = 1,0065

m_eg > 0

This (m_eg/(2π)^1/2 is m_eg of ypothesis 5.1.b

m_p=mass of proton, l_g (length of gravity interaction)=λ_Planck.(2.π)^1/2.

Since m_eg contains the constant k of the hypothesis 2, which means that has a real existence to the above formation angular momentum of the proton

END

First copyright: 27-03-2003

BIBLIOGRAPHY

1. An Introduction to Nuclear Physics, 1992 W. N. COTTINGHAM & D. A. GREENWOOD.

2. General Physics, Electricity, 1974, 5^th edition, K. A. ALEXOPOULOS, Greece

3. Modern Physics, 1989 by Saunders College Publishing

Raymond A. Serway, Clement J. Moses, & Curt A. Moyer

4. Particle and Cosmological Physics, 2003 K. E. VAGIONAKIS, University of Ioannina, Greece.

5. Themes of Physics I, 1983, N. A. OIKONOMOU M.Sc., Ph.D., University of Thessaloniki, Greece.

Nikos Alexandris, Secondary Education Science Teacher, Address: 10, NIKIS 33 str. 570 13 Oreokastro, Thessaloniki, GREECE, e-mail: nalxhal@yahoo.gr

Technical College - Bourgas,

All rights reserved, © March, 2000