Preprint
Hypothesis

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Quantum Relativity (Impact of Energy with Space-Time 4)

Submitted:

03 March 2025

Posted:

04 March 2025

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Abstract
This research answers the knowledge gap regarding the explanation of the quantum jump of the electron. This scientific paper aims to complete Einstein’s research regarding general relativity and attempt to link general relativity to quantum laws.
Keywords: 
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1. Introduction

This research was created for the purpose of answering questions about physics phenomena that have not been answered. Such as explaining the phenomenon of the quantum jump of the electron and the phenomenon of cumulative entanglement. What happens in the phenomenon of the quantum jump of the electron is that when we give the electron energy, this energy causes the electron to move from the energy level that it occupies to the higher energy level without crossing the distance between the two orbits, which leads to the occurrence of the phenomenon of the quantum jump of the electron.[1] (Svidzinsky et al., 2014)
The role of this scientific paper is to provide a scientific explanation of how the quantum leap occurs without crossing the distance between the orbits. The theory of quantum entanglement is a connection between two quantum entangled particles. If one particle is observed, the other particle is affected by it at the same moment. This is what Einstein objected to; because when the electron traveled this distance in the same period of time, this would lead to the existence of a speed faster than the speed of light. Einstein proved it in special relativity. The maximum speed in the universe is the speed of light. Therefore, the phenomenon of quantum entanglement does not agree with Einstein’s laws. After the validity of quantum laws was proven. There has become a conflict between the laws of relativity that apply to the universe and the quantum laws that apply to atoms. This scientific paper aims to resolve this conflict between the laws of relativity and quantum laws. By establishing a law derived from the laws of relativity to apply to quantum laws. (Equation number 1)
This law in equation 1 is known as quantum relativity because it links the laws of relativity and quantum theory. This law is derived from general relativity. The law works to explain the phenomenon of the quantum leap and the phenomenon of quantum entanglement, as it explains that when energy is given to the atom, the atom does not gain energy, but rather space-time gains that energy. We will discuss the interpretation of this theory in detail later.
  • The goal of this scientific research is to answer the explanation of the phenomenon of quantum leap and quantum entanglement and to add some modifications in the Bohr model.

2. Equations

These laws want to explain the results of the final derivation process of this research and what this research wants to prove.
G μ ν + Λ g μ ν = 8 π × h p 4 λ   n m 4 × e 4 G Δ E n 4 T μ ν        
G μ ν + Λ g μ ν = 2 π 6 × 4 × 4 λ   n m 5 × μ 0 × ε 0 3 × e 5     t p 2 Δ E n 5 T μ ν            
G 3 = 2 π 4 × 4 × C 15 λ   n m × e 3 × 8 π     l p × t p 3 Δ E n 3
G 2 = 1   μ 0 × ε 0 5   2 π × l p × t p Δ E n × λ   n m × e   2
Where Gµν represents the Einstein tensor, α is the fine-structure constant, G is the universal gravitational constant, Tµν is the energy-momentum tensor,  λ   n m is the wavelength is (nm), Δ E n is the photon energy is in electron volt, n is the energy level.
i γ μ μ + Γ μ ψ 4 ( G μ ν + Λ g μ ν ) = 8 π h p 4 p × α 4 λ   n m × e 4 G × ψ 4 Δ E n 4 T μ ν                    
i γ μ μ + Γ μ ψ 4 ( G μ ν + Λ g μ ν ) = 2 π 2 × h p 4 × 4 × p × α 4 μ 0 × ε 0 4 × λ   n m × e 5     l p 2 ψ 4 Δ E n 5 T μ ν
p = m × C
Where p represents the momentum.
  G μ ν + Λ g μ ν = 2 π 2 × 4 λ   n m × e     l p 2 Δ E n T μ ν
G μ ν + Λ g μ ν = 2 π 4 × 4 × m p 2 λ   n m × e 3 × μ 0 × ε 0 3   l p × t p 2 Δ E n 3 T μ ν
Where     represents the reduced Planck constant, rn is the Bohr radius, En is the photon energy is in joules,  t p is the Planck time, l p is the Planck length. This law explains the final result of the derivation. This law proves the creation of a relationship that links the photon energy and curvature of space-time.
G 2 =   2   μ 0 × ε 0 5 ×   E p 4
Δ E n = h p μ 0 × ε 0 × λ   n m × e
Where E p represents the Planck energy.
E p 2 = h p × n 2 2 π × Z 2 C 5 G          
Where Ep represents the Planck energy,
Z is the atomic number. This law links Planck energy with other cosmological constants.
E = n 2 × h a 2 × 2 K E Z 2 + n × P × ω k × α × Z
E = m × C 2 + P × C
Where E represents the energy, h( a) is the atomic constant, KE is the kinetic energy, P is the momentum, ω is the angular velocity, C is the speed of light,k is the wave vector, and   α is the fine-structure constant. This law explains the final result of the derivation. This law proves the creation of a relationship that links energy and kinetic energy. That the lost kinetic energy comes out in the form of radiant energy.
G μ ν + Λ g μ ν = 2 π 2 × 4 μ 0 × ε 0 × λ   n m   l p × t p Δ E n T μ ν
G μ ν + Λ g μ ν = 8 π μ 0 × ε 0   1 a p × E p   T μ ν
Where a p represents the Planck acceleration, E p is the Planck energy. This law explains the final result of the derivation. This law proves the creation of a relationship that links the Planck energy and curvature of space-time.
G μ ν + Λ g μ ν = 2 π 5 × 4 × 4 λ   n m × e 4 G Δ E n 4 T μ ν
G μ ν + Λ g μ ν = 2 π 6 × 4 × 4 λ   n m × e 5 × μ 0 × ε 0 2   l p 2 Δ E n 5 T μ ν    
Where ε0 represents the vacuum permittivity, μ 0   is the Vacuum permeability, e is the electron charge. This law works to link the constants ( gravity, electron charge and Planck constant ) into one law.
    l p 2 e 2   =   G C 4 × 4 π × ε 0         1 α    
This is a law that links the constants ( gravity, electron charge and speed ) into one law.
G μ ν + Λ g μ ν = 8 π × ¯ × e 2 × Z 2 n 2 × E 2 T μ ν
¯ = k e × X = e 2 × m e 2 4 π × ε 0 2 × × C × m p 2 = H × × C e 2 = H × k e α
¯ = e 2 × k e m 1 2 × G × C 2
Where ¯ represents the multiverse constant. This law wants to prove is the creation of a relationship that links the curvature of space-time and the energy of the total photon.
n 2 × 2 = k e m 1 × r × e 2
k e m 1 = m e × k e = m e 4 π × ε 0
This law affects the Planck constant and the charge of the electron.
          G μ ν + Λ g μ ν = 4 × 8 × 2 π 9 μ 0 × ε 0 2 × λ   n m 8 × e 8 G   Δ E n 8 T μ ν
          G μ ν + Λ g μ ν = 4 × 8 × 2 π 10 μ 0 × ε 0 4 ×   λ   n m   9 × e 9   l p 2   Δ E n 9 T μ ν              
Where μ 0 Vacuum permeability, p is the momentum,  l p is the Planck length.
Δ E n = h p × 2 k E λ   n m × e × p × α 2
Δ E n = 2 π × G × m p 2 λ   n m × e
Where h p represents the Planck constant, ν is the frequency.
E = 2 π × G × m p 2 × Z λ × α
E = m × C 2
G μ ν + Λ g μ ν = 2 π 4 × 4 × 2 λ   n m × e 3 × μ 0 × ε 0 3 × a p 2   1 Δ E n 3 T μ ν
Where  m p  represents the Planck mass.
Δ E n 4 = h p 4 λ × e 4 × μ 0 × ε 0 2
Δ E n = 2 π × × C λ × e
Δ E n 4 = h p 4 × F p × G λ × e 4
Δ E n 4 = h p 4 × F p × G λ × e 4
2 π × r n = 2 × d × s i n Θ

3. These Laws Have Been Modified from the Mix Planck Laws

  • How quantum entanglement occurs?
What happens is that the electron connects to the other electron through space-time, as space-time acts like a quantum tunnel that connects the two electrons. In this way, the electron does not penetrate the speed of light, But in relation to large objects, you see that it has crossed the speed of light.
  • This hypothesis was based on scientific foundations, the most important of which is:
1) the connection between relativity and quantum mechanics occurs via quantum entanglement and loop gravitational entanglement.
2) quantum entanglement occurs by the contraction of space-time.
3) space-time contraction occurs by space-time absorbing energy.
4) the quantum jump of the electron occurs as a result of the contraction of space-time.

4. Derivation of Equations

Completing the derivation of the laws resulting from quantum relativity (quantum world)
G μ ν + Λ g μ ν = 2 π 3 × r n 2 × m e × Z 2 n × λ 4 × ε 0 2 e 4 × G E n 4 T μ ν
n × λ = 2 π × r n
G μ ν + Λ g μ ν = 2 π × n × λ 2 × m e × Z 2 n × λ 4 × ε 0 2 e 4 × G E n 4 T μ ν
G μ ν + Λ g μ ν = 2 π × m e × Z 2 n × λ 2 × ε 0 2 e 4 × G E n 4 T μ ν C 2 C 2
p = m × C
G μ ν + Λ g μ ν = 2 π × p × Z 2 n × λ 2 × ε 0 2 × C 2 e 4 × G E n 4 T μ ν
C = λ × ν
G μ ν + Λ g μ ν = 2 π × p × Z 2 n × λ 2 × ε 0 2 × λ × ν 2 e 4 × G E n 4 T μ ν h p 2 h p 2
p = h p λ
E n = h p × ν
G μ ν + Λ g μ ν = 2 π × Z 2 × p 4 n × λ 2 × ε 0 2 e 4 × G E n 6 T μ ν
p = h p λ
G μ ν + Λ g μ ν = 2 π × Z 2 × h p 4 n 2 × ε 0 2 e 4 × G λ × E n 6 T μ ν
E t = E n × n
G μ ν + Λ g μ ν = 2 π × h p 4 ε 0 2 × λ 6 × E t 2 e 4 × G E n 4 T μ ν
E t = E × α
G μ ν + Λ g μ ν = 2 π × h p 4 ε 0 2 × λ 6 × E × α 2 e 4 × G E n 4 T μ ν
E = m × C 2
G μ ν + Λ g μ ν = 2 π × h p 4 ε 0 2 × λ 6 × m × C 2 × α 2 e 4 × G E n 4 T μ ν
C = λ × ν
p = m × C
G μ ν + Λ g μ ν = 2 π × h p 4 ε 0 2 × λ 6 × p × λ × ν × α 2 e 4 × G E n 4 T μ ν
p = h p λ
G μ ν + Λ g μ ν = 2 π × h p 4 ε 0 2 × λ 6 × h p × ν × α 2 e 4 × G E n 4 T μ ν
G μ ν + Λ g μ ν = 2 π × h p 2 ε 0 2 × λ 6 × ν × α 2 e 4 × G E n 4 T μ ν 2 × h p × C 2 2 × h p × C 2
μ 0 = 2 × α e 2 h p C
G μ ν + Λ g μ ν = 2 π × h p 4 × 4 μ 0 × ε 0 2 × λ 6 × ν 2 × C 2 G E n 4 T μ ν
C = λ × ν
G μ ν + Λ g μ ν = 2 π × h p 4 × 4 × ν 4 μ 0 × ε 0 2 × C 8 G E n 4 T μ ν
μ 0 × ε 0 = 1 C 2
G μ ν + Λ g μ ν = 2 π × h p 4 × 4 × ν 4 C 4 G E n 4 T μ ν
C = λ × ν
G μ ν + Λ g μ ν = 2 π × h p 4 × 4 λ 4 G E n 4 T μ ν e 4 e 4
G μ ν + Λ g μ ν = 2 π × h p 4 × 4 λ   n m × e 4 G Δ E n 4 T μ ν
G μ ν + Λ g μ ν = 2 π 5 × 4 × 4 λ   n m × e 4 G Δ E n 4 T μ ν
kE = m e × v 2 2           m e m e
p = m e × v
kE = p 2 2 m e          
E = n × h a 2 × 2 k E Z 2
h a = 1 α
E t = E × α
E t = n 2 × h a × 2 k E Z 2
E t = E n × n
E n = n × h a × 2 k E Z
E n = n × h p × C 2 π × r n
n × h p × C 2 π × r n = n × h a × 2 k E Z                 p × α p × α    
h p = m e × λ × ν × α × 2 π × r n
h p = p × α × 2 π × r n
h a = 1 α
h p × C = n × h p × 2 k E Z × p × α 2
Δ E n = h p × C λ
Δ E n = h p × 2 k E λ   n m × e × p × α 2
λ   n m = h p × 2 k E Δ E n × e × p × α 2
G μ ν + Λ g μ ν = 2 π × h p 4 × 4 λ [ 4 G E n 4 T μ ν
En is the photon energy is in joules
      l p 2 =   × G C 3          
G =   C 3 ×   l p 2          
G μ ν + Λ g μ ν = 2 π 2 × h p 3 × 4 λ 4 C 3 ×   l p 2 E n 4 T μ ν                         C C        
    μ 0 × ε 0 = 1 C 2  
G μ ν + Λ g μ ν = 2 π 2 × h p 3 × 4 λ 4 × μ 0 × ε 0 2 × C   l p 2 E n 4 T μ ν                                
C = λ × ν
  E n = h p × ν          
G μ ν + Λ g μ ν = 2 π 2 × h p 4 × 4 λ 5 × μ 0 × ε 0 2   l p 2 E n 5 T μ ν    
This is derivation number 1
    G μ ν + Λ g μ ν = 2 π 3 × r n 2 × 4 × m e 4 × Z 2 n 6 × μ 0 × ε 0 2 × λ 2 α 4 × G E n 4 T μ ν
μ 0 × ε 0 = 1 C 2
G μ ν + Λ g μ ν = 2 π 3 × r n 2 × 4 × C 4 × m e 4 × Z 2 n 6 × λ 2 α 4 × G E n 4 T μ ν
E = m × C 2
  G μ ν + Λ g μ ν = 2 π 3 × r n 2 × 4 × E 2 × m e 2 × Z 2 n 6 × λ 2 α 4 × G E n 4 T μ ν
    E t = E × α
    G μ ν + Λ g μ ν = 2 π 3 × r n 2 × 4 × E t 2 × m e 2 × Z 2 n 6 × λ 2 α 2 × G E n 4 T μ ν
  E t = E n × n
G μ ν + Λ g μ ν = 2 π 3 × r n 2 × 4 × m e 2 × Z 2 n 4 × λ 2 α 2 × G E n 2 T μ ν   v 2 v 2
n × = m e × v × r n
G μ ν + Λ g μ ν = 2 π 3 × 4 ×   n × 2 × Z 2 n 4 × λ 2 × v 2 α 2 × G E n 2 T μ ν  
  C 2 = n 2 × v 2 α 2
G μ ν + Λ g μ ν = 2 π 3 × 4 ×   n × 2 × Z 2 n 2 × λ 2 × C 2 G E n 2 T μ ν
C = λ × ν
G μ ν + Λ g μ ν = 2 π 3 × 4 ×   n × 2 × Z 2 n 2 × λ 2 × λ × ν 2 G E n 2 T μ ν                       2 π h p 2 2 π h p 2
E n = h p × ν
G μ ν + Λ g μ ν = 2 π 3 × 4 ×   4 × Z 2 λ 4 G E n 4 T μ ν             e 4 e 4
G μ ν + Λ g μ ν = 2 π 5 × 4 ×   4 λ   n m   4 × e 4 G Δ E n 4 T μ ν            
  l p 2 =   × G C 3          
G =   C 3 ×   l p 2                
G μ ν + Λ g μ ν = 2 π 5 × 4 ×   3 λ   n m 4 × e 4   C 3 ×   l p 2 Δ E n 4 T μ ν                     C C      
    μ 0 × ε 0 = 1 C 2
G μ ν + Λ g μ ν = 2 π 5 × 4 ×   3 λ   n m 4 × e 4 × μ 0 × ε 0 2 × C     l p 2 Δ E n 4 T μ ν                  
C = λ × ν
  E n = h p × ν
G μ ν + Λ g μ ν = 2 π 6 × 4 ×   4 λ   n m 5 × e 5 × μ 0 × ε 0 2     l p 2 Δ E n 5 T μ ν              
This is derivation number 2
G μ ν + Λ g μ ν = 2 π 3 × r n 2 × 4 × m e 4 × Z 2 n 6 × μ 0 × ε 0 2 × λ 2 α 4 × G E n 4 T μ ν
n × λ = 2 π × r n
  G μ ν + Λ g μ ν = 2 π × n × λ 2 × 4 × m e 4 × Z 2 n 6 × μ 0 × ε 0 2 × λ 2 α 4 × G E n 4 T μ ν
  G μ ν + Λ g μ ν = 2 π × 4 × m e 4 × Z 2 n 4 × μ 0 × ε 0 2 α 4 × G E n 4 T μ ν
μ 0 × ε 0 = 1 C 2
    G μ ν + Λ g μ ν = 2 π × 4 × C 4 × m e 4 × Z 2 n 4 α 4 × G E n 4 T μ ν
  E = m × C 2
    G μ ν + Λ g μ ν = 2 π × 4 × E 2 × m e 2 × Z 2 n 4 α 4 × G E n 4 T μ ν
E t = E × α
  G μ ν + Λ g μ ν = 2 π × 4 × E t 2 × m e 2 × Z 2 n 4 α 2 × G E n 4 T μ ν
    E t = E n × n
G μ ν + Λ g μ ν = 2 π × 4 × m e 2 × Z 2 n 2 α 2 × G E n 2 T μ ν                 C 2 C 2
p = m × C
p = h p λ
G μ ν + Λ g μ ν = 2 π × 4 × p 2 × Z 2 n 2 × C 2 α 2 × G E n 2 T μ ν                
C = λ × ν
G μ ν + Λ g μ ν = 2 π × 4 × p 2 × Z 2 n 2 × λ × ν 2 α 2 × G E n 2 T μ ν         h p 2 h p 2
  E n = h p × ν
G μ ν + Λ g μ ν = 2 π × 4 × p 2 × Z 2 × h p 2 n 2 × λ 2 α 2 × G E n 4 T μ ν
p = h p λ
G μ ν + Λ g μ ν = 8 π × h p 4 λ 4 G E n 4 T μ ν        
G μ ν + Λ g μ ν = 8 π × p 4 × Z 4 n 4 α 4 × G E n 4 T μ ν        
G μ ν + Λ g μ ν = 8 π × h p 4 λ   n m 4 × e 4 G Δ E n 4 T μ ν        
This is derivation number 3
a = G × m × α × Z 2 R × r n 2
a = v 2 r  
v 2 r = G × m × α × Z 2 R × r n 2
v 2 = G × m × α × Z 2 R × r n n 2 × α n 2 × α
C 2 = n 2 × v 2 α 2  
C 2 = n 2 × G × m × Z 2 R × r n × α m m
C 2 = n 2 × G × m × Z 2 R × r n × α m m
E = m × C 2
E = n 2 × G × m e 2 × Z 2 R × r n × α
E t = E × α
E t = n 2 × G × m e 2 × Z R × r n
E t = E n × n
E n = n × G × m e 2 × Z R × r n
R = m e 2 m p 2
Where m p represents the Planck mass.
E n = n × G × m p 2 × Z r n     2 π 2 π  
n × λ = 2 π × r n
E n = 2 π × G × m p 2 × Z λ
E n = h p × ν
E n = n × h p × C 2 π × r n
n × h p × C 2 π × r n = 2 π × G × m p 2 × Z λ
n × λ = 2 π × r n
h p × C = 2 π × G × m p 2 e
E n = h p × C λ
Δ E n = 2 π × G × m p 2 λ   n m × e
E = 2 π × G × m p 2 × Z λ × α
E = m × C 2
This is derivation number 4
G μ ν + Λ g μ ν = 2 π 3 × r n 2 × m e × Z 2 n 4 × ε 0 2 e 4 × G h p × C 4 T μ ν                                                                      
= h p 2 π
G μ ν + Λ g μ ν = 2 π × r n 2 × m e × Z 2 n 4 × × ε 0 2 × h p 2 e 4 × G C 4 T μ ν                             v 2 v 2    
n × = m e × v × r n
G μ ν + Λ g μ ν = 2 π × Z 2 n 2 × ε 0 2 × v 2 × h p 2 e 4 × G C 4 T μ ν                
v = 2 π × r n T = 2 π × r n × ν × α × Z n 2
G μ ν + Λ g μ ν = n 2 × 2 π ε 0 2 × 2 π × r n × ν × α 2 × h p 2 e 4 × G C 4 T μ ν                
      E n = h p × ν
G μ ν + Λ g μ ν = n 2 × 2 π ε 0 2 × 2 π × r n × α 2 × E n 2 e 4 × G C 4 T μ ν                 4 π 2 × 2 4 π 2 × 2
α 2 = 1 4 π × ε 0 2 e 4 × C 2  
G μ ν + Λ g μ ν = 4 π 2 × n 4 × 2 × 2 π 2 π × r n × α 2 × E n 2 α 2 × G C 2 T μ ν                
C = λ × ν
G μ ν + Λ g μ ν = 4 π 2 × n 4 × 2 × 2 π 2 π × r n × α 2 × E n 2 α 2 × G λ × ν 2 T μ ν                 h p 2 h p 2
      E n = h p × ν
G μ ν + Λ g μ ν = 2 π 3 × 4 × n × 4 r n 2 × λ 2 G E n 4 T μ ν                      
G μ ν + Λ g μ ν = 2 π 3 × 4 × n × 4 r n 2 × λ 2 G E n 4 T μ ν                       2 π 2 2 π 2
n × λ = 2 π × r n
G μ ν + Λ g μ ν = 2 π 5 × 4 × 4 λ 4 G E n 4 T μ ν          
  l p 2 =   × G C 3          
G =   C 3 ×   l p 2          
G μ ν + Λ g μ ν = 2 π 6 × 4 × 3 λ 4 C 3 ×   l p 2 E n 4 T μ ν                   C C      
μ 0 × ε 0 = 1 C 2    
G μ ν + Λ g μ ν = 2 π 6 × 4 × 3 λ 4 × μ 0 × ε 0 2 × C   l p 2 E n 4 T μ ν                            
C = λ × ν
    E n = h p × ν        
G μ ν + Λ g μ ν = 2 π 6 × 4 × 4 λ 5 × μ 0 × ε 0 2 l p 2 E n 5 T μ ν      
G μ ν + Λ g μ ν = 2 π 6 × 4 × 4 λ 5 × μ 0 × ε 0 2 l p 2 E n 5 T μ ν                         e 5 e 5  
G μ ν + Λ g μ ν = 2 π 6 × 4 × 4 λ n m 5 × μ 0 × ε 0 2 × e 5 l p 2 Δ E n 5 T μ ν      
G μ ν + Λ g μ ν = 2 π 5 × 4 × 4 λ n m 4 × e 4 G Δ E n 4 T μ ν                      
p = h p λ
G μ ν + Λ g μ ν = 8 π × p 4 × Z 4 n 4 α 4 × G E n 4 T μ ν        
G μ ν + Λ g μ ν = 2 π 3 × 4 × n × 4 r n 2 × λ 2 G E n 4 T μ ν                       n 2 n 2
n × λ = 2 π × r n
G μ ν + Λ g μ ν = 2 π × 4 × n × 4 r n 4 G E n 4 T μ ν                  
D e r i v a t i o n   o f   P l a n c k   e n e r g y   w i t h   P l a n c k   l e n g t h
E n = h p × C λ
n × λ = 2 π × r n
E n = n × h p × C 2 π × r n
= h p 2 π
E n = n × × C r n
  l p 2 =   × G C 3          
=   C 3 ×   l p 2 G        
E n = n × C 4 × l p 2 r n × G
G = n × C 4 × l p 2 r n × E n
G μ ν + Λ g μ ν = 2 π × 4 × n × 4 r n 5 n × C 4 ×   l p 2 E n 5 T μ ν                  
    μ 0 × ε 0 = 1 C 2
G μ ν + Λ g μ ν = 2 π × 4 × n × 4 μ 0 × ε 0 2 × r n 5 n ×   l p 2 E n 5 T μ ν         e 5 e 5            
G μ ν + Λ g μ ν = 2 π × 4 × n × 4 μ 0 × ε 0 2 × r n 5 × e 5 n ×   l p 2 Δ E n 5 T μ ν                  
G μ ν + Λ g μ ν = 2 π × 4 × n × 4 r n 4 G E n 4 T μ ν                   e 4 e 4  
G μ ν + Λ g μ ν = 2 π × 4 × n × 4 r n 4 × e 4 G Δ E n 4 T μ ν                
This is derivation number 5
          G μ ν + Λ g μ ν = 2 π 3 × r n 2 × 4 × m e 4 × Z 2 n 6 × μ 0 × ε 0 2 × λ 2 α 4 × G E n 4 T μ ν             2 π × r n 2 2 π × r n 2
E n = h p × ν
          G μ ν + Λ g μ ν = 2 π 4 × r n 4 × 4 × m e 4 × Z 2 2 π × n 6 × μ 0 × ε 0 × r n 2 × λ 2 α 4 × G h p × ν 4 T μ ν            
= h p 2 π
          G μ ν + Λ g μ ν = r n 4 × 4 × m e 4 × Z 2 2 π × n 6 × μ 0 × ε 0 × r n 2 × λ 2 α 4 × G × ν 4 T μ ν             v 4 v 4  
n × = m e × v × r n
          G μ ν + Λ g μ ν = 4 × Z 4 2 π × n 2 × μ 0 × ε 0 × r n 2 × λ 2 α 4 × G v × ν 4 T μ ν    
v = 2 π × r n T = 2 π × r n × ν × α × Z n 2
          G μ ν + Λ g μ ν = 4 × Z 4 2 π × n 2 × μ 0 × ε 0 × r n 2 × λ 2 α 4 × G 2 π × r n × ν × α × Z n 2 × ν 4 T μ ν    
          G μ ν + Λ g μ ν = n 2 3 × 4 μ 0 × ε 0 2 × λ 2 G 2 π 5 × r n 6 ×   ν 8 T μ ν                     h p 8 h p 8
E n = h p × ν
          G μ ν + Λ g μ ν = n 2 3 × 4 × h p 8 μ 0 × ε 0 2 × λ 2 G 2 π 5 × r n 6 ×   E n 8 T μ ν                    
n × λ = 2 π × r n
          G μ ν + Λ g μ ν = n × 4 × h p 8 μ 0 × ε 0 2 × λ 7 G r n ×   E n 8 T μ ν                                   2 π 9 2 π 9
n × λ = 2 π × r n
          G μ ν + Λ g μ ν = 4 × 8 × 2 π 9 μ 0 × ε 0 2 × λ 8 G   E n 8 T μ ν  
G =   C 3 ×   l p 2          
= h p 2 π
          G μ ν + Λ g μ ν = 4 × 7 × 2 π 9 μ 0 × ε 0 2 × λ 8 C 3 ×   l p 2   E n 8 T μ ν                   C C        
  μ 0 × ε 0 = 1 C 2
          G μ ν + Λ g μ ν = 4 × 7 × 2 π 9 μ 0 × ε 0 4 × λ 8 × C   l p 2   E n 8 T μ ν                    
C = λ × ν
G μ ν + Λ g μ ν = 4 × 8 × 2 π 10 μ 0 × ε 0 4 × λ 9 l p 2 E n 9 T μ ν
G μ ν + Λ g μ ν = 4 × 8 × 2 π 10 μ 0 × ε 0 4 × λ 9 l p 2 E n 9 T μ ν e 9 e 9
          G μ ν + Λ g μ ν = 4 × 8 × 2 π 10 μ 0 × ε 0 4 ×   λ   n m 9 × e 9   l p 2   Δ E n 9 T μ ν
          G μ ν + Λ g μ ν = 4 × 8 × 2 π 9 μ 0 × ε 0 2 × λ 8 G   E n 8 T μ ν  
p = h p λ
= h p 2 π
          G μ ν + Λ g μ ν = 4 × p 8 × 2 π μ 0 × ε 0 2 α 8 × G   E n 8 T μ ν              
          G μ ν + Λ g μ ν = 4 × p 8 × 2 π μ 0 × ε 0 2 α 8 × G   E n 8 T μ ν
This is derivation number 6
                                                  G μ ν + Λ g μ ν = 8 π × m e 4 × Z 2 n 4 × μ 0 × ε 0 2 α 4 × G E n 4 T μ ν           C 4 C 4
p = m × C
                                                  G μ ν + Λ g μ ν = 8 π × p 4 × Z 2 n 4 × μ 0 × ε 0 2 × C 4 α 4 × G E n 4 T μ ν           h p 4 h p 4
C = λ × ν
E n = h p × ν
                                                  G μ ν + Λ g μ ν = 8 π × p 4 × Z 2 × h p 4 n 4 × μ 0 × ε 0 2 × λ 4 α 4 × G E n 8 T μ ν
p = h p λ
                                                  G μ ν + Λ g μ ν = 8 π × p 8 × Z 2 n 4 × μ 0 × ε 0 2 α 8 × G E n 8 T μ ν
                                                  G μ ν + Λ g μ ν = 8 π × h p 8 × Z 2 n 4 × μ 0 × ε 0 2 × λ 8 α 8 × G E n 8 T μ ν                                 C C        
G =   C 3 ×   l p 2          
= h p 2 π
                                                  G μ ν + Λ g μ ν = 2 π 2 × 4 × h p 7 × Z 2 n 4 × μ 0 × ε 0 2 × λ 8 × C α 8 × C 4 ×   l p 2 E n 8 T μ ν                               h p h p      
                                                                                                μ 0 × ε 0 = 1 C 2
C = λ × ν
                                                  G μ ν + Λ g μ ν = 2 π 2 × 4 × h p 8 × Z 2 n 4 × μ 0 × ε 0 4 × λ 9 α 8 ×   l p 2   E n 9 T μ ν
Another way to derive the law.
G μ ν + Λ g μ ν = 8 π 1 G C 4 T μ ν           m e 4 m e 4    
G μ ν + Λ g μ ν = 8 π m e 4 m e 4 G C 4 T μ ν           ψ 4 ψ 4    
m C ψ = i γ μ μ ψ
Dirac equation
G μ ν + Λ g μ ν = 8 π m e 4 i γ μ μ ψ 4 G ψ 4 1 T μ ν          
i γ μ μ ψ 4 ( G μ ν + Λ g μ ν ) = 8 π m e 4 G ψ 4 T μ ν                     C 8 C 8
E = m × C 2
i γ μ μ ψ 4 ( G μ ν + Λ g μ ν ) = 8 π E 4 C 8 G ψ 4 1 T μ ν           α 4 α 4    
E t = E × α
    E t = E n × n
i γ μ μ ψ 4 ( G μ ν + Λ g μ ν ) = 8 π E n × n 4 C 8 G ψ 4 α 4 T μ ν
E n = h p × ν
i γ μ μ ψ 4 ( G μ ν + Λ g μ ν ) = 8 π h p × ν × n 4 C 8 G ψ 4 α 4 T μ ν         m e 4 m e 4    
E = m × C 2
E t = E × α
  E t = E n × n
i γ μ μ ψ 4 ( G μ ν + Λ g μ ν ) = 8 π h p × ν × m e 4 1 G ψ 4 E n 4 T μ ν         λ 4 λ 4
C = λ × ν
i γ μ μ ψ 4 ( G μ ν + Λ g μ ν ) = 8 π h p × C × m e 4 λ 4 G ψ 4 E n 4 T μ ν
p = m × C
p = h p λ
i γ μ μ ψ 4 ( G μ ν + Λ g μ ν ) = 8 π p × α 8 1 G ψ 4 E n 4 T μ ν
E n = n × h p × C 2 π × r n
i γ μ μ ψ 4 ( G μ ν + Λ g μ ν ) = 8 π 2 π × r n × α 4 × p 8 1 G × ψ 4 n × h p × C 4 T μ ν
h p = m e × λ × ν × α × 2 π × r n
h p = p × α × 2 π × r n
E n = h p × C λ
i γ μ μ ψ 4 ( G μ ν + Λ g μ ν ) = 8 π h p 4 × p 4 λ   n m × e 4 G × ψ 4 Δ E n 4 T μ ν               C 4 C 4      
i γ μ μ ψ 4 ( G μ ν + Λ g μ ν ) = 8 π h p 4 × E n 4 λ   n m × e 4 × C 4 G × ψ 4 Δ E n 4 T μ ν                  
E n = n × h p × C 2 π × r n
i γ μ μ ψ 4 ( G μ ν + Λ g μ ν ) = 8 π h p 4 × n × h p × C 2 π × r n 4 λ   n m × e 4 × C 4 G × ψ 4 Δ E n 4 T μ ν       m e × α 4 m e × α 4          
h p = m e × λ × ν × α × 2 π × r n
h p = p × α × 2 π × r n
i γ μ μ ψ 4 ( G μ ν + Λ g μ ν ) = 8 π h p 4 × h p × C 4 m e × α 4 λ   n m × e 4 × h p 4 G × ψ 4 Δ E n 4 T μ ν  
i γ μ μ ψ 4 ( G μ ν + Λ g μ ν ) = 8 π h p 4 p × α 4 λ   n m × e 4 G × ψ 4 Δ E n 4 T μ ν                     C C      
G =   C 3 ×   l p 2          
= h p 2 π
i γ μ μ ψ 4 ( G μ ν + Λ g μ ν ) = 2 π 2 × h p 3 × 4 × p × α 4   λ   n m × e 4 × C     C 4 ×   l p 2 ψ 4 Δ E n 4 T μ ν             h p h p      
    μ 0 × ε 0 = 1 C 2
C = λ × ν
i γ μ μ ψ 4 ( G μ ν + Λ g μ ν ) = 2 π 2 × h p 4 × 4 × p × α 4 μ 0 × ε 0 4 × λ   n m × e 5     l p 2 ψ 4 Δ E n 5 T μ ν
m C ψ = i γ μ μ ψ
m C ψ = i γ μ μ + Γ μ ψ
μ = μ Γ μ
m C ψ = i γ μ μ + Γ μ ψ
Where μ represents the affine derivative, Γ μ is the connection coefficients, μ is the partial derivative
i γ μ μ + Γ μ ψ 4 ( G μ ν + Λ g μ ν ) = 2 π 2 × h p 4 × 4 × p × α 4 μ 0 × ε 0 4 × λ   n m × e 5     l p 2 ψ 4 Δ E n 5 T μ ν
This is derivation number 7
          G μ ν + Λ g μ ν = 4 × 8 × 2 π 9 μ 0 × ε 0 2 × λ 8 G   E n 8 T μ ν  
          G μ ν + Λ g μ ν = 8 π × h p 8 μ 0 × ε 0 2 × λ 8 G   E n 8 T μ ν  
G =   C 3 ×   l p 2          
= h p 2 π
          G μ ν + Λ g μ ν = 2 π 2 × h p 7 × 4 μ 0 × ε 0 2 × λ 8 C 3 ×   l p 2   E n 8 T μ ν                                 C C        
  μ 0 × ε 0 = 1 C 2
C = λ × ν
          G μ ν + Λ g μ ν = 2 π 2 × h p 7 × 4 μ 0 × ε 0 4 × λ 9     l p 2   E n 8 T μ ν                             h p h p      
          G μ ν + Λ g μ ν = 2 π 2 × h p 8 × 4 μ 0 × ε 0 4 × λ 9     l p 2   E n 9 T μ ν      
          G μ ν + Λ g μ ν = 8 π × h p 8 μ 0 × ε 0 2 × λ 8 G   E n 8 T μ ν  
p = h p λ
          G μ ν + Λ g μ ν = 8 π p 8 μ 0 × ε 0 2 α 8 × G   E n 8 T μ ν  
This is derivation number 8
μ 0 × ε 0 = 1 C 2 m m
E = m × C 2
μ 0 × ε 0 = m E
E = m   μ 0 × ε 0   α α
E t = E × α
  E t = E n × n
E n = m × α × Z   μ 0 × ε 0 × n               C 2 C 2    
p = m × C
E n = p × α × Z × C n                
E n = n × h p × C 2 π × r n
n × h p × C 2 π × r n = p × α × Z × C n  
h p = m e × λ × ν × α × 2 π × r n
h p = p × α × 2 π × r n
h p × C = h p × Z × C n  
Δ E n = h p × C λ
Δ E n = h p × C λ × e
This is derivation number 9
μ 0 × ε 0 = 1 C 2
C = λ × ν
μ 0 × ε 0 = 1 λ × ν 2 2 2
= h p 2 π
μ 0 × ε 0 = 2 π × 2 λ × ν × h p 2
E n = h p × ν
μ 0 × ε 0 = 2 π × 2 λ × E n 2
E n 2 = 2 π × 2 μ 0 × ε 0 × λ 2
E n 4 = 2 π × 2 μ 0 × ε 0 × λ 2 2
E n 4 = 2 π × 4 μ 0 × ε 0 2 × λ 4
p = h p λ
= h p 2 π
E n 4 = p × α 4 μ 0 × ε 0 2
E n = n × h p × C 2 π × r n
n × h p × C 2 π × r n 4 = p × α 4 μ 0 × ε 0 2
h p = m e × λ × ν × α × 2 π × r n
h p = p × α × 2 π × r n
h p × C 4 = h p 4 μ 0 × ε 0 2
Δ E n = h p × C λ
Δ E n 4 = h p 4 λ × e 4 × μ 0 × ε 0 2
This is derivation number 10
  μ 0 × ε 0 = m × α × Z E n × n
μ 0 × ε 0 = 2 π × 2 λ × E n 2
m × α × Z E n × n = 2 π × 2 λ × E n 2
m × α × Z n = 2 π × 2 E n × λ 2
m × α × Z n = 2 π × 2 E n × λ 2               C 2 C 2
E = m × C 2
E × α × Z n = 2 π × × C 2 E n × λ 2
E t = E × α
  E t = E n × n
E n 2 = 2 π × × C 2 λ 2
E n = 2 π × × C λ
E n = n × h p × C 2 π × r n
n × h p × C 2 π × r n = 2 π × × C λ               2 π × m × α 2 π × m × α
h p = m e × λ × ν × α × 2 π × r n
h p = p × α × 2 π × r n
h p × C = 2 π × 2 λ × m × α
p = h p λ
h p × C = 2 π × × p λ × m × α
p = m × C
Δ E n = 2 π × × C λ × e
Δ E n = 13.605693099 e V 1 × Z 2 n 2 2 Z 2 n 1 2
This is derivation number 11
m C ψ = i γ μ μ ψ
Where F p represents the Planck force
μ 0 × ε 0 = 1 C 2
μ 0 × ε 0 2 = 1 F p × G
E n 4 = 2 π × 2 μ 0 × ε 0 × λ 2 2
E n 4 = 2 π × 4 × F p × G λ 4
p = h p λ
E n 4 = p × α 4 × F p × G
E n = n × h p × C 2 π × r n
n × h p × C 2 π × r n 4 = p × α 4 × F p × G
h p = m e × λ × ν × α × 2 π × r n
h p = p × α × 2 π × r n
h p × C 4 = h p 4 × F p × G
Δ E n = h p × C λ
Δ E n 4 = h p 4 × F p × G λ × e 4
This is derivation number 12
C 4 = 1 μ 0 × ε 0 2
F p × G = C 4
F p × G = 1 μ 0 × ε 0 2
E n 4 = 2 π × 4 μ 0 × ε 0 2 × λ 4
E n 4 = 2 π × 4 × F p × G λ 4
This is derivation number 13
E t = E × α
E t = E n × n
  E × α = E n × n
E = m × C 2
E n = h p × ν
m × C 2 × α = h p × ν × n
m × C 2 × α 2 = h p × ν × n 2
F p × G = C 4
F p × G × m × α 2 = h p × ν × n 2
μ 0 × ε 0 2 = 1 F p × G
m × α 2 μ 0 × ε 0 2 = h p × ν × n 2       C 2 C 2
p = m × C
p × α 2 μ 0 × ε 0 2 = h p × ν × n × C 2      
μ 0 × ε 0 = 1 C 2
p × α 2 μ 0 × ε 0 = E n 2      
E n = n × h p × C 2 π × r n
p × α 2 μ 0 × ε 0 = n × h p × C 2 π × r n 2      
n × h p × C 2 π × r n 2 = p × α 2 μ 0 × ε 0      
h p = m e × λ × ν × α × 2 π × r n
h p = p × α × 2 π × r n
h p × C 2 = h p 2 μ 0 × ε 0      
Δ E n = h p × C λ
Δ E n 2 = h p 2 λ × e 2 × μ 0 × ε 0
This is derivation number 14
E p 2 = 2 π × n 2 × 2 × ε 0 2 r n 2 × m e × Z 2 × C 5 e 4 × G                             2 π 2 2 π 2          
C = λ × ν
n × λ = 2 π × r n
E p 2 = 2 π 4 × 2 × ε 0 2 × n 2 m e × Z 2 × ν 5 × λ 3 e 4 × G                               4 π × C 2 4 π × C 2      
α 2 = 1 4 π × ε 0 2 e 4 × C 2  
E p 2 = 2 π 4 × 2 2 × n 2 4 π × C 2 × m e × Z 2 ν 5 × × λ 3 α 2 × G                              
p = m × C
E p 2 = 2 π 4 × 2 2 × n 2 4 π 2 × p × Z 2 ν 5 × × λ 3 α 2 × G                              
= h p 2 π
E p 2 = n 2 2 π × p × Z 2 ν 5 × h p × λ 3 α 2 × G           λ 2 λ 2            
p = h p λ
E p 2 = h p × n 2 2 π × Z 2 ν 5 × λ 5 G          
C = λ × ν
E p 2 = h p × n 2 2 π × Z 2 C 5 G          
This is derivation number 15
  l p 2 =   × G C 3          
α = 1 4 π × ε 0 e 2 × C  
  m p 2 =   × C G          
α = 1 4 π × ε 0 e 2   m p 2 × G  
G = 1 4 π × ε 0 e 2   m p 2 × α  
  l p 2 =   C 3 × 4 π × ε 0         e 2   m p 2 × α    
  m p 2 =   × C G          
  l p 2 =   G C 4 × 4 π × ε 0         e 2 α    
  l p 2 e 2   =   G C 4 × 4 π × ε 0         1 α    
This is derivation number 16
G μ ν + Λ g μ ν = 8 π 1 G C 4 T μ ν
G =   C 3 ×   l p 2          
G μ ν + Λ g μ ν = 8 π C 3 ×   l p 2 C 4 T μ ν                 C C          
      μ 0 × ε 0 = 1 C 2
G μ ν + Λ g μ ν = 8 π μ 0 × ε 0 2 ×   l p 2 C 5 T μ ν       4 4
C = λ × ν
E n = h p × ν
= h p 2 π
G μ ν + Λ g μ ν = 2 π 6 × 4 × 4 λ   n m 5 × μ 0 × ε 0 2 × e 5     l p 2 Δ E n 5 T μ ν            
This is derivation number 17
G μ ν + Λ g μ ν = 8 π 1 G C 4 T μ ν
G =   C 3 ×   l p 2          
G μ ν + Λ g μ ν = 8 π C 3 ×   l p 2 C 4 T μ ν                  
G μ ν + Λ g μ ν = 8 π   l p 2 C T μ ν    
C = λ × ν
    E n = h p × ν
= h p 2 π
G μ ν + Λ g μ ν = 2 π 2 × 4 λ   n m × e     l p 2 Δ E n T μ ν            
This is derivation number 18
λ   n m = 1 R × Z 2 n 2 2 Z 2 n 1 2        
R = 1.0973731731 × 10 7   m 1
Δ E n = h p × C λ
Δ E n = h p × C × α 2 e × λ × Z 2 n 2 2 Z 2 n 1 2
n × λ = 2 π × r n
λ = 2 π × r n n
Δ E n = h p × C × α 2 e × 2 π × r n n × Z 2 n 2 2 Z 2 n 1 2
This is derivation number 19
Δ E n = h p × C × α 2 e × λ × Z 2 n 2 2 Z 2 n 1 2
λ = h p × C × α 2 e × Δ E n × Z 2 n 2 2 Z 2 n 1 2
Δ E n = 13.605693099 e V 1 × Z 2 n 2 2 Z 2 n 1 2
λ = h p × C × α 2 e × 13.605693099 e V 1 × Z 2 n 2 2 Z 2 n 1 2 × Z 2 n 2 2 Z 2 n 1 2
λ = h p × C × α 2 e × 13.605693099 e V 1 = 2 π × r n n
This is derivation number 20
G μ ν + Λ g μ ν = 8 π 1 G C 4 T μ ν
F p × G = C 4
G μ ν + Λ g μ ν = 8 π 1 1 F p T μ ν
F p =   l p × t p      
G μ ν + Λ g μ ν = 8 π 1 l p × t p T μ ν       ν ν  
  E n = h p × ν
= h p 2 π
G μ ν + Λ g μ ν = 2 π 2 × 4 × ν 1   l p × t p E n T μ ν       λ λ  
C = λ × ν
G μ ν + Λ g μ ν = 2 π 2 × 4 × C λ   l p × t p E n T μ ν             C C      
μ 0 × ε 0 = 1 C 2
G μ ν + Λ g μ ν = 2 π 2 × 4 λ × μ 0 × ε 0 × C   l p × t p E n T μ ν  
C = λ × ν
G μ ν + Λ g μ ν = 2 π 2 × 4 λ 2 × μ 0 × ε 0 × ν   l p × t p E n T μ ν                
E n = h p × ν
G μ ν + Λ g μ ν = 2 π 3 × 4 × λ 2 × μ 0 × ε 0   l p × t p E n 2 T μ ν        
F p =   l p × t p
G μ ν + Λ g μ ν = 2 π 3 × 4 × F p λ 2 × μ 0 × ε 0   l p × t p 2 E n 2 T μ ν        
F p × G = C 4
G μ ν + Λ g μ ν = 2 π 3 × 4 × C 4 λ 2 × μ 0 × ε 0 × G   l p × t p 2 E n 2 T μ ν                     C × C ×    
m p 2 = × C G        
μ 0 × ε 0 = 1 C 2
G μ ν + Λ g μ ν = 2 π 3 × 4 × m p 2 λ 2 × μ 0 × ε 0 3 × C ×   l p × t p 2 E n 2 T μ ν                    
C = λ × ν
  E n = h p × ν
G μ ν + Λ g μ ν = 2 π 4 × 4 × m p 2 λ 3 × μ 0 × ε 0 3   l p × t p 2 E n 3 T μ ν                   e 3 e 3      
G μ ν + Λ g μ ν = 2 π 4 × 4 × m p 2 λ × e 3 × μ 0 × ε 0 3   l p × t p 2 Δ E n 3 T μ ν
This is derivation number 21
G μ ν + Λ g μ ν = 8 π 1 G C 4 T μ ν
G =   C 5 ×   t p 2          
G μ ν + Λ g μ ν = 8 π C 5 ×   t p 2 C 4 T μ ν                 C C          
        μ 0 × ε 0 = 1 C 2
G μ ν + Λ g μ ν = 8 π μ 0 × ε 0 3 ×   t p 2 C 5 T μ ν       4 4
C = λ × ν
  E n = h p × ν
= h p 2 π
G μ ν + Λ g μ ν = 2 π 6 × 4 × 4 λ   n m 5 × μ 0 × ε 0 3 × e 5     t p 2 Δ E n 5 T μ ν            
This is derivation number 22
G μ ν + Λ g μ ν = 2 π 4 × 4 × m p 2 λ × e 3 × μ 0 × ε 0 3   l p × t p 2 Δ E n 3 T μ ν        
m p 2 = × C G        
G μ ν + Λ g μ ν = 2 π 4 × 4 × × C λ × e 3 × μ 0 × ε 0 3 × G   l p × t p 2 Δ E n 3 T μ ν        
G μ ν + Λ g μ ν = 8 π 1 G C 4 T μ ν
G μ ν + Λ g μ ν T μ ν = 8 π × G C 4
G μ ν + Λ g μ ν T μ ν = 2 π 4 × 4 × × C λ × e 3 × μ 0 × ε 0 3 × G   l p × t p 2 Δ E n 3    
G 2 = 2 π 4 × 4 × × C 5 λ × e 3 × μ 0 × ε 0 3 × 8 π   l p × t p 2 Δ E n 3                   l p × t p l p × t p    
F p =   l p × t p
G 2 = 2 π 4 × 4 × F p × C 5 λ × e 3 × μ 0 × ε 0 3 × 8 π   l p × t p 3 Δ E n 3                  
F p × G = C 4
G 3 = 2 π 4 × 4 × C 9 λ × e 3 × μ 0 × ε 0 3 × 8 π   l p × t p 3 Δ E n 3                  
    μ 0 × ε 0 = 1 C 2
G 3 = 2 π 4 × 4 × C λ × e 3 × μ 0 × ε 0 7 × 8 π   l p × t p 3 Δ E n 3                  
G 3 = 2 π 4 × 4 × C 15 λ × e 3 × 8 π   l p × t p 3 Δ E n 3
This is derivation number 23
Δ E n = h p × ν × α 2 e × Z 2 n 2 2 Z 2 n 1 2
E n = h p × C λ
Δ h p × ν = h p × ν × α 2 e × Z 2 n 2 2 Z 2 n 1 2
Δ ν = ν × α 2 × Z 2 n 2 2 Z 2 n 1 2
ν = v × n 2 2 π × r n × α × Z
Δ ν = v × n 2 2 π × r n × α × α 2 × Z 2 n 2 2 Z 2 n 1 2
Δ ν = v × n 2 4 π × r n × Z 2 n 2 2 Z 2 n 1 2
Δ ν = 2187691.261 4 π × 5.29177202590 × 10 11 × Z 2 n 2 2 Z 2 n 1 2
ν = Δ ν × 2 α   × 1 Z 2 n 1 2 Z 2 n 2 2
Δ ν = 3.289842008 × 10 15 e V 1 × Z 2 n 2 2 Z 2 n 1 2
This is derivation number 24
ν = v × n 2 2 π × r n × α × Z
v = 2 π × ν × r n × α × Z n 2
C = λ × ν
v = 2 π × C × r n × α × Z n 2 × λ
v = 7.2738949827 × 10 4 λ
This is derivation number 25
ν = Δ ν × 2 α   × 1 Z 2 n 1 2 Z 2 n 2 2               h p h p        
ν = Δ E n × 2 α × h p × 1 Z 2 n 1 2 Z 2 n 2 2
Δ E n = 13.605693099 e V 1 × Z 2 n 2 2 Z 2 n 1 2
ν = 13.605693099 e V 1 × Z 2 n 2 2 Z 2 n 1 2 × 2 α × h p × 1 Z 2 n 1 2 Z 2 n 2 2
ν = 13.605693099 e V 1 × 2 α × h p
ν = 13.605693099 × 1.60217662 × 10 19 × 2 α × h p
This is derivation number 26
G μ ν + Λ g μ ν = 2 π 4 × 4 × m p 2 λ × e 3 × μ 0 × ε 0 3   l p × t p 2 Δ E n 3 T μ ν
F p =   l p × t p = C 4 G = a p × m p  
a p × m p =   l p × t p
l p × t p × m p =   a p
G μ ν + Λ g μ ν = 2 π 4 × 4 × 2 λ   n m × e 3 × μ 0 × ε 0 3 × a p 2   1 Δ E n 3 T μ ν
This is derivation number 27
G μ ν + Λ g μ ν = 8 π 1 G C 4 T μ ν
F p = C 4 G
G μ ν + Λ g μ ν = 8 π 1 1 F p T μ ν
F p =   l p × t p  
G μ ν + Λ g μ ν = 8 π   l p × t p 1 T μ ν                 C C          
Δ E n = h p × C λ
C = 1 μ 0 × ε 0
G μ ν + Λ g μ ν = 2 π 2 × 4 μ 0 × ε 0 × λ   n m   l p × t p Δ E n T μ ν              
This is derivation number 28
G μ ν + Λ g μ ν = 8 π 1 G C 4 T μ ν
F p = C 4 G
G μ ν + Λ g μ ν = 8 π 1 1 F p T μ ν
F p = a p × m p  
G μ ν + Λ g μ ν = 8 π 1 1 a p × m p   T μ ν
m p = E p C 2
G μ ν + Λ g μ ν = 8 π 1 C 2 a p × E p   T μ ν
μ 0 × ε 0 = 1 C 2
G μ ν + Λ g μ ν = 8 π μ 0 × ε 0   1 a p × E p   T μ ν
This is derivation number 29
n × λ = 2 π × r n
n × λ = 2 × d × s i n Θ
Bragg’s law
2 π × r n = 2 × d × s i n Θ
This is derivation number 30
G =   × C   m p 2      
G 2 =   × C   m p 2 2    
C = 1 μ 0 × ε 0
G 2 =   2 μ 0 × ε 0 ×   m p 4          
m p = E p C 2
G 2 =   2 × C 8 μ 0 × ε 0 ×   E p 4          
C = 1 μ 0 × ε 0
G 2 =   2   μ 0 × ε 0 5 ×   E p 4          
This is derivation number 31
Δ E n = h p × C λ   n m × e
C = 1 μ 0 × ε 0
Δ E n = h p μ 0 × ε 0 × λ   n m × e
This is derivation number 32
  l p 2 =   × G C 3                       C C        
× C =     l p 2 × C 4 G                      
Δ E n = h p × C λ   n m × e
Δ E n = 2 π × l p 2 × C 4 λ n m × e × G
C = 1 μ 0 × ε 0
Δ E n = 2 π × l p 2 μ 0 × ε 0 2 × λ n m × e × G
This is derivation number 33
  t p 2 =   × G C 5                 C C          
× C =     t p 2 × C 6 G                      
Δ E n = h p × C λ   n m × e
Δ E n = 2 π × t p 2 × C 6 λ n m × e × G
C = 1 μ 0 × ε 0
Δ E n = 2 π × t p 2 μ 0 × ε 0 3 × λ n m × e × G
This is derivation number 34
F p =   l p × t p = C 4 G = a p × m p  
C 4 G =   l p × t p                       C C          
C 5 G = × C l p × t p
C 5 G 2 =   × C l p × t p   2
Δ E n = h p × C λ   n m × e
C 5 G 2 = Δ E n × λ   n m × e 2 π × l p × t p   2
G 2 = 2 π × l p × t p × C 5 Δ E n × λ   n m × e   2
C = 1 μ 0 × ε 0
G 2 = 1   μ 0 × ε 0 5   2 π × l p × t p Δ E n × λ   n m × e   2
This is derivation number 35

5. Method

My name is Ahmed. I have made a theoretical derivation of the equation of general relativity as explained in this research for the purpose of obtaining an equation that can be applied within the quantum world so that it describes the movement of the electron during the quantum jump in the Bohr model. After that, the researcher Samira reviewed the research and verified it, and then she worked on applying this theory to the movement of the electron during the occurrence of the quantum leap, using previous research and matching it with the results of this equation to determine its validity.
  • This part of the research will explain the spectrum of the hydrogen atom in a new way, as the results presented in these tables from previous research match the results extracted from the equation, and this is consistent with the validity of this equation. Because the new equation is consistent with the photon energy equation. We will discuss that part of the research in the results and discussion.
  • Table 5 shows the measurement results tested.[3] (Nanni, 2015)
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  • Table 5 it represents the theoretical and experimental value of the hydrogen atom. Using the photon energy law mentioned above, this table.
Figure 1. Bohr hydrogen atomic model incorporating de Broglie’s .[4] (Jordan, 2024).
Figure 1. Bohr hydrogen atomic model incorporating de Broglie’s .[4] (Jordan, 2024).
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This drawing, taken from previous research, shows how the quantum leap occurs through interference, as my equation showed. When interference occurs between the orbit occupied by the electron and the energy level higher than the electron’s orbit, it occurs in the form of wave interference of this type as a result of a contraction in the fabric of space-time. The black circle represents the orbit occupied by the electron, while the red color represents how interference occurs from the orbit higher to the orbit occupied by the electron in the form of wave interference. In other words, the upper level works to contract, forming a wave equal to the same wave as the level occupied by the electron through the de Broglie equation. n × λ = 2π × r
  • If we make the electron quantum entangled in particle accelerators, then if we make one of these electrons be in a short line and the other be in a long line, when one approaches the speed of light, the other must exceed the speed of light. In other words, the two entangled bodies are in two dimensions, that is, different dimensions, and this happens as a result, a distortion of space-time, which makes during the measurement that the speed is breached, but in reality it does not exceed the speed. This is the same idea as the distortion of the orbits that I explained. Because it is assumed that the electron does not move from its position, however, a distortion occurs in the orbit with the highest energy, and it forms a wave similar to the orbit occupied by the electron, according to De Broglie’s laws. This occurs through the distortion of space-time as a result of the increase in energy.
        i γ μ μ + Γ μ ψ 4 ( G μ ν + Λ g μ ν ) = 2 π 2 × h p 8 × 4 μ 0 × ε 0 4 × λ 9   l p 2 ψ 4 E n 5 T μ ν
    Because the equation connects more than one equation into a single equation. As
    m C ψ = i γ μ μ ψ
    Dirac equation
    G μ ν + Λ g μ ν = 8 π × G C 4 T μ ν
    Δ E n = h p × C λ
    n × λ = 2 π × r n
Figure 2. The observed emission line spectrum of atomic hydrogen in chapter 2 atoms.[5] (Manini, 2020).
Figure 2. The observed emission line spectrum of atomic hydrogen in chapter 2 atoms.[5] (Manini, 2020).
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  • Table 6 shows the measurement results of one of the previous researches related to the spectrum of the hydrogen atom in chapter 2 atoms.[5] (Manini, 2020)
  • This shape is a result of the fact that the electron, after a quantum leap occurred as a result of an interference between the orbital that it occupies and the energy level above it, was in an unstable state. Therefore, when the highest level of energy returns to its position, it releases energy in the form of spectral lines. These lines are determined according to the amount of energy, as shown in the picture.

6. Results Obtained

This scientific research aims to prove a theory by comparing the practical results of this theory with the original results and making the comparison in a table. We will discuss that here .
My theory is based on introducing the curvature of spacetime into the equation, but quantum mechanics shows that it is not affected by gravity. How to interact with the curvature of spacetime has not yet been proven. As a result, my equations show a way to conduct an experiment that enables direct interaction with the curvature of spacetime. Therefore, this experiment practically proves that quantum mechanics made a mistake in its concept when it showed gravity does not interact with it. How to conduct an experimental experiment to prove the validity of my equations
Steps to conduct the experiment
1) The place where the experiment will take place must be chosen, and it must be at a high altitude, such as Mount Everest because the higher the altitude, the less gravity.
2) The experiment is about creating a quantum leap for the electron so that we can know the emission lines that represent the fingerprint of the element and compare them at different heights. Let us take the example of the hydrogen atom. After knowing the choice of the element, the device that will measure the spectral lines of the element must be taken to Mount Everest, where the experiment will be conducted.
3) We will excite the element keeping all elements constant as energy and the comparison will be between wavelength and curvature of spacetime. The first measurement is at the bottom of the mountain, that is, before climbing the mountain first. Then we measure in the middle of the mountain, then we test at the top of the mountain and compare the atomic spectra. If my theoretical results are correct, there will be skewing of the spectral lines at different heights due to distortion of the fabric of space-time.
4) If we measure atomic spectra, we also measure the Zeeman, Stark, and magneto-stark effects separately.
  • The reason they were not previously able to measure the curvature of space-time is because my equations show that the effect of energy and wavelength when measured as two variables will cancel each other out, so space-time will not be affected.
  • Gravitational Effect on Atomic Energy Levels
Objective: Measure the effect of gravity on atomic energy levels
Equipment:
• A gas sample (e.g., hydrogen or cesium) in a vacuum chamber.
• A laser to excite electrons at specific energy levels.
• A high-precision spectrometer.
• A variable gravitational field (e.g., using aircraft simulating microgravity).
Procedure:
.1 Measure the atomic spectrum in a normal gravitational environment.
.2 Measure the spectrum in a reduced-gravity environment (e.g., during parabolic flights).
.3 Compare the energy levels and emission lines.
Expected Outcome:
• If the spectrum shifts at different gravitational strengths, it indicates that gravity affects atomic energy levels
  • My equations clearly show that if proven in practical experiments, it indicates that the gravitational constant G is not a cosmic constant in quantum mechanics, but is affected by the wavelength and the energy difference, that is, it is variable. In other words, gravity is not an absolute quantity, but rather the quantum state is influenced by me. For this reason, quantum mechanics is not related to general relativity.
  • My equations explain the effect (magnetic attraction) and Bayfield-Brown effect My equations confirm the effect of electromagnetism on gravity.
  • Well, with these experiments, the Pound-Rebecca experiments, also known as gravitational redshift, will prove what the equation tells you.
G μ ν + Λ g μ ν T μ ν = 8 π × G C 4 = 2.0766474428 × 10 43
E n = E 2 E 1
Δ E n = 13.605693099 e V 1 × Z 2 n 2 2 Z 2 n 1 2
G μ ν + Λ g μ ν = 2 π 6 × 4 × 4 λ   n m × e 5 × μ 0 × ε 0 2   l p 2 Δ E n 5 T μ ν                                                  
G μ ν + Λ g μ ν = 2 π 6 × 1.0545718 × 10 34 4 × 4 λ   n m × e 5 × 4 π × 10 7 × 8.854187813 × 10 12 2   1.616255011 × 10 35 2 Δ E n 5 T μ ν
G μ ν + Λ g μ ν = 6.4231253544 × 10 167 λ   n m × e 5   1 Δ E n 5 T μ ν        
Δ E n 5 = 6.4231253544 × 10 167 λ   n m × e 5   1 G μ ν + Λ g μ ν   T μ ν  
Δ E n 5 = 6.4231253544 × 10 167 λ   n m × e 5   1 2.0766474428 × 10 43        
λ   n m 5 = 3.0930261448 × 10 124 Δ E n × e 5    
λ   n m 5 = 3.0930261448 × 10 124 13.605693099 e V 1 × Z 2 n 2 2 Z 2 n 1 2 × e 5  
λ   n m = 5 3.0930261448 × 10 124 13.605693099 e V 1 × Z 2 n 2 2 Z 2 n 1 2 × 1.60217662 × 10 19 5                                                  
λ = 656.11227252   n m  
This example of a hydrogen atom in the Balmer series.
          G μ ν + Λ g μ ν = 4 × 8 × 2 π 10 μ 0 × ε 0 4 ×   λ   n m   9 × e 9   l p 2   Δ E n 9 T μ ν  
Δ E n = h p × C λ = 1239.8419637 e V n m λ
Photon energy equation.
  Δ E n 9 = 4 × 8 × 2 π 10 μ 0 × ε 0 4 ×   λ   n m   9 × e 9   l p 2           G μ ν + Λ g μ ν T μ ν  
Example of a hydrogen atom.
Δ E n = 9 4 × 8 × 2 π 10 μ 0 × ε 0 4 × λ n m 9 × e 9 l p 2 G μ ν + Λ g μ ν T μ ν
Example of a hydrogen atom in the Balmer series.
We remove the energy level n  
Δ E n = 9 4.8160445107 × 10 223 λ n m 9 × e 9
Δ E n = 9 4.8160445107 × 10 223   656.11227252   9 × 1.60217662 × 10 19 9  
Δ E n = 1.889679597
The unit of measurement for photon energy is electron volt (eV), the wavelength is (nm)
λ   n m = 1 R × Z 2 n 2 2 Z 2 n 1 2        
R = 1.0973731731 × 10 7   m 1
Δ E n = h p × C λ
Δ E n = h p × C × α 2 e × λ × Z 2 n 2 2 Z 2 n 1 2
λ = 2 π × r n n
Δ E n = h p × C × α 2 e × 2 π × r n n × Z 2 n 2 2 Z 2 n 1 2
λ = 2 π × r n n = 2 π × 5.29177202590 × 10 11 × n 2 n
λ = 2 π × r n n = 2 π × 5.29177202590 × 10 11 × n
Δ E n = 6.62607004 × 10 34 × 299792458 × 7.297352563 × 10 3 2 × 1.60217662 × 10 19 × 2 π × 5.29177202590 × 10 11 × Z 2 n 2 2 Z 2 n 1 2
Δ E n = 13.605693099 e V 1 × Z 2 n 2 2 Z 2 n 1 2
Δ E n = h p × C × α 2 e × λ × Z 2 n 2 2 Z 2 n 1 2
Δ E n = 6.62607004 × 10 34 × 299792458 × 7.297352563 × 10 3 2 × 1.60217662 × 10 19 × 3.324918425 × 10 10 × Z 2 n 2 2 Z 2 n 1 2
Δ E n = 13.605693099 e V 1 × Z 2 n 2 2 Z 2 n 1 2
Δ E n = h p × ν × α 2 e × Z 2 n 2 2 Z 2 n 1 2
Δ E n = 6.62607004 × 10 34 × 9.016535737 × 10 17 × 7.297352563 × 10 3 2 × 1.60217662 × 10 19 × Z 2 n 2 2 Z 2 n 1 2
Δ E n = 13.605693099 e V 1 × Z 2 n 2 2 Z 2 n 1 2
The unit of measurement for  Δ E n
photon energy is electron volt (eV)
E n = h p × C λ
λ = 2 π × r n n
E n = n × h p × C 2 π × r n = n × 6.62607004 × 10 34 × 299792458 2 π × 5.29177202590 × 10 11 × n 2
E n = 5.9744197314 × 10 16   J n
Δ E n = h p × ν × α 2 e × Z 2 n 2 2 Z 2 n 1 2
E n = h p × C λ
Δ h p × ν = h p × ν × α 2 e × Z 2 n 2 2 Z 2 n 1 2
Δ ν = ν × α 2 × Z 2 n 2 2 Z 2 n 1 2
ν = v × n 2 2 π × r n × α × Z
Δ ν = v × n 2 2 π × r n × α × α 2 × Z 2 n 2 2 Z 2 n 1 2
Δ ν = v × n 2 4 π × r n × Z 2 n 2 2 Z 2 n 1 2
Δ ν = 2187691.261 4 π × 5.29177202590 × 10 11 × Z 2 n 2 2 Z 2 n 1 2
Δ ν = 3.289842008 × 10 15 e V 1 × Z 2 n 2 2 Z 2 n 1 2
G μ ν + Λ g μ ν = 8 π × p 4 × Z 4 n 4 α 4 × G h p × Δ ν × 2 e V α × 1 Z 2 n 1 2 Z 2 n 2 2 4 T μ ν
G μ ν + Λ g μ ν = 8 π × p 4 × Z 4 n 4 α 8 × G h p × Δ ν × 2 × e 4 × Z 2 n 1 2 Z 2 n 2 2 T μ ν
G μ ν + Λ g μ ν = 8 π × p 4 × Z 4 n 4 α 8 × G Δ E n × 2 × e 4 × Z 2 n 1 2 Z 2 n 2 2 T μ ν
G μ ν + Λ g μ ν = 8 π × p 4 × Z 4 n 4 α 8 × G 13.605693099 e V 1 × Z 2 n 2 2 Z 2 n 1 2 × 2 × e 4 × Z 2 n 1 2 Z 2 n 2 2 T μ ν
G μ ν + Λ g μ ν = 8 π × p 4 × Z 4 n 4 α 8 × G 13.605693099 e V 1 × 2 × e 4 T μ ν
h p = m e × C × α × 2 π × r n
C = λ × ν
h p = m e × λ × ν × α × 2 π × r n
Space-time represents in the equation the force of attraction of the nucleus for the electron. Where we take the hydrogen atom compared to the sodium atom. We find after comparison that the undulations that occur in the sodium atom are higher than those that occur in the hydrogen atom. That is, during the occurrence of the quantum jump of the electron, the higher energy level than the level occupied by the electron undulates. So the number of ripples (ripple amplitude) is higher than that of the hydrogen atom during the occurrence of the quantum jump, and this is consistent with the de Broglie equation. n × λ is represented by a ratio to space-time. It is the number of ripples that occur in the energy level higher than the level occupied by the electron until interference occurs between the two levels, the higher energy level and the level occupied by the electron. In other words, as the number of orbitals occupied by the electron increases, the number of ripples that occur at the higher energy levels increases, causing the curvature (contraction) of the fabric of space-time. The interference between the two levels occurs in a wave form so that the quantum jump of the electron occurs. The photon's energy is represented by a ratio to the fabric of space-time, the force that causes the fabric of space-time to bend (contract). The more energy increases, the more space-time contracts through the occurrence of quantum disturbances at the highest energy level, which makes the highest energy level generate waves similar to the orbital number occupied by the electron. Because of these disturbances that occur at the highest energy level, the two levels interfere with each other, the highest energy level, and the level occupied by the electron. A quantum leap occurs, and this is consistent with the quantum Zeno effect, where the electron will remain fixed in its position. This is what my equation indicates, as I explain that these quantum fluctuations occur through a contraction in the fabric of space-time. This contraction occurs as a result of this tissue absorbing energy. Because of this, contraction affects the energy levels in the atom. This contraction works to contract the energy level higher than the level occupied by the electron. Wave interference occurs between the highest energy level and the level occupied by the electron, and a quantum jump occurs from the observer’s perspective. But from the electron's perspective, it remains fixed in its position.
The Casimir effect is according to a law that states that after all the objects acting on the plates disappear until imaginary particles are detected. My equation proves that there is one thing that was not included in the calculations, which is the effect of space-time. Since the plates have a static mass that works to curve space-time, and the presence of imaginary particles works when they collide with each other, they disappear. But according to the law of conservation of energy, the energy will not disappear and will affect the fabric of space-time, making it turbulent like a water wave, and these disturbances that occur on it form waves. This wave works to impact the panels from moving in and out, and because the external disturbances are higher than the internal ones, they cause the panels to move towards each other.
This relationship shows that although we cannot measure what happens when an electronic quantum jump occurs. This law also shows that there is a relationship between the energy of the photon and the fabric of space-time, even if it is not measured by measuring devices. Because measuring devices are considered primitive devices when making the process of measuring the quantitative world. What is being measured are the spectra of the elements being measured, not what happens to the electron when the quantum jump of the electron to the higher level. Second, Maxwell told Rutherford that the electron changes direction as it orbits the nucleus, so it must lose energy to cause a collision with the nucleus, which it does not. My equation tells me the electron moves in a large circle around the nucleus. A body moving in a large circle whose direction of motion is in a straight line. Thus, the electron moves in a straight line. Newton's law states that an object at rest remains at rest unless acted upon by an external or internal force. Likewise, an object in motion stays in motion unless an external or internal force affects its movement, the electron does not lose energy.
G μ ν + Λ g μ ν = 2 π 5 × 4 × 4 λ   n m × e 4 G Δ E n 4 T μ ν
G μ ν + Λ g μ ν = 2 π 6 × 4 × 4 λ   n m × e 5 × μ 0 × ε 0 2   l p 2 Δ E n 5 T μ ν    
The results of the experimental value were obtained by using the results of previous research on the hydrogen atom. I prove in Table 7 that the results of the equations are identical to their original results in Table 5, which indicates the validity of this law
Δ E n = E 2 E 1
Δ E n = h p × C λ = 1239.8419637 e V n m λ
Photon energy equation.
Δ E n = 13.605693099 e V 1 × Z 2 n 2 2 Z 2 n 1 2
  • These are the results of a relationship between energy and wavelength. The observed results show that whenever the energy increases, the wavelength decreases, as shown by this equation in the hydrogen atom.

7. Conclusions

After the idea of research has been clarified using theoretical and practical scientific evidence to explain the phenomenon of the quantum leap and quantum entanglement from a new perspective, these equations would be used in the following:
1) serving humanity in the advancement of scientific research.
2) using these equations to explore space and quantum world.
3) using these equations in developing communications machines .

Acknowledgements

I would like to thank one of my colleagues at work, who knows Ahmed Mowafak, for his suggestion that I link the equation of quantum relativity to anti-universes, because at this moment I remembered the Dirac equation, so I linked it to the equation of general relativity.

References

  1. Svidzinsky, A. Scully, M. Bohr's molecular model, a century later. Physics T. 2014, 67, 33–39. [Google Scholar] [CrossRef]
  2. Udema, I. I. Renaissance of Bohr's model via derived alternative equation. American J. Mod. Phys 2017, 6, 23–31. [Google Scholar] [CrossRef]
  3. Nanni, L. The hydrogen atom: A review on the birth of modern quantum mechanics. Physics 2015. [Google Scholar] [CrossRef]
  4. Jordan, R. B. Principles of Inorganic Chemistry. Springer N., 2024; pp. 1--18. [CrossRef]
  5. Manini, N. Introduction to the physics of matter: basic atomic, molecular, and solid-state physics. Springer N., 2020; pp. 11--16. [CrossRef]
Table 7. Comparing my theoretical results through my equation with previous results.
Table 7. Comparing my theoretical results through my equation with previous results.
Theoretical value (My work) ---------Experimental value
Spectral Line Energy λ λ
λ(n’=2, n=1) 10.204269824 eV 121.50227268 nm 121.5 nm
λ(n’=3, n=1) 12.093949421 eV 102.51754257 nm 102.5 nm
λ(n’=4, n=1) 12.75533728 eV 97.20181814 nm 97.20 nm
λ(n’=3, n=2) 1.8896795971 eV 656.11227245 nm 656.1 nm
λ(n’=4, n=2) 2.5510674561 eV 486.0090907 nm 486.0 nm
λ(n’=4, n=3) 0.66138785898 eV 1874.6064927 nm 1874.6 nm
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