THEORY OF ELECTRON WU SHENG-PING

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1 THEORY OF ELECTRON WU SHENG-PING Abstract. This article try to unified the four basic forces by Maxwell equations, the only experimental theory. Self-consistent Maxwell equation with the e-current from matter current is proposed, and is solved to four kinds of electrons and the structures of particles. The static properties and decay are reasoned, all meet experimental data. The equation of general relativity sheerly with electromagnetic field is discussed as the base of this theory. In the end the conformation elementarily between this theory and QED and weak theory is discussed. Contents 1. Unit Dimension of sch 1 2. Self-consistent Electrical-magnetic Fields 2 3. Calculation of Recursive Re-substitution 3 4. Solution 3 5. Electrons and Their Symmetries 5 6. Propagation and Movement 7 7. Balance Formula 8 8. Muon 9 9. Pion 9 1. Pion Neutral Tauon Proton Neutron Mesons Mechanical Quantification Great Unification Conclusion 13 References Unit Dimension of sch A rebuilding of units and physical dimensions is needed. Time s is fundamental. We can define: The unit of time: s (second) The unit of length: cs (c is the velocity of light) Date: Mar 9, 219. Key words and phrases. Maxwell equations; Decay; Electron function; Recursive resubstitution. 1

2 2 WU SHENG-PING The unit of energy: /s (h is Plank constant) The unit dielectric constant ɛ is [ɛ] = The unit of magnetic permeability µ is then [µ] = We can define the unit of Q (charge) as [Q]2 [E][L] = [Q]2 c [E][T ]2 [Q] 2 [L] = c[q] 2 cɛ = cµ = 1 [Q] = [H] = [Q]/[L] 2 = [cd] = [E] Then : C = ( ) 1/2 C is charge SI unit Coulomb. For convenience we can define new base units by unit-free constants c = 1, = 1, [Q] = then all physical units are power of second s n, the units are reduced. Define UnitiveElectricalCharge : σ = σ = C 64e e /σ = e/σ = /64 Define and rebuild the system: s Cs : 1 = m e /e /σ =: β s Cs means the value of the redefined second becomes c seconds. We always use this units system. 2. Self-consistent Electrical-magnetic Fields Try so-called expanded Maxwell equation for the free E-M field in the previously defined units (2.1) A ν = ia µ ν A µ /2 + cc. = J ν with definition ν A ν = (A i ) := (V, A), (J i ) = (ρ, J), (J i ) = ( ρ, J) := ( i ) := ( t, x1, x2, x3 ) := ( i ) := ( t, x1, x2, x3 ) It s deduced by using momentum to express e-current in a electron: the mass and charge have the same movement in electron. The equation 2.1 have symmetry CP T, cc.p T The energy (reference to the section 15) of the field A under Lorentz gauge are (2.2) ε :=< A ν, A ν >=< E, E > /2+ < H, H > /2

3 THEORY OF ELECTRON 3 (2.3) ε t = < A ν, J ν > t =< ρ, V > t 3. Calculation of Recursive Re-substitution We can calculates the solution by recursive re-substitution (RRS) for the two sides of the equation. For the equation ˆP B = ˆP B make the algorithm (It s approximate, the exact solution needs a rate on the start state in the re-substitution for the normalization condition) ˆP ( B k + B n+1 ) = ˆP B k k n k n One can write down a function initially and correct it by re-substitution. Here is the initial state V = V i e ikt, A i = V, µ µ A i = Substituting into equation 2.1. We call the fields correction A n with n degrees of A i the n degrees correction. 4. Solution Firstly 2 A = k 2 A is solved. Exactly, it s solved in spherical coordinate = r 2 ( 2 f k 2 f) = k 2 r 2 f + (r 2 f r ) r + 1 sin θ (sin θf θ) θ + 1 sin 2 θ (f ϕ) ϕ Its solution is f = RΘΦ = R l Y lm Θ = Pl m (cos θ), Φ = cos(α + mϕ) R l = Nj l (kr) j l (r) is spherical Bessel function. j 1 (r) = sin(r) r 2 cos r r j 1 () = dr r 2 j 1 (kr)j 1 (k r) = Ck 2 δ(k k ) Define F (x) := NR 1 (r)y 1,1 (θ, ϕ) F (x) F (x) = δ(r) 4πr 2 with consideration 2 (F (x) F (kx)) 2 F (kx) has a suspending singularity at O. Then F (x) F (x) = 1 sin θ dθ dϕ dr r F (x ) r F (x x ) 4π I lim r r F (x ) =, lim x x r F (x x ) =

4 4 WU SHENG-PING Figure 1. The function of j 1 lim r r F (x ) r F (x x ) = C(θ, ϕ, θ, ϕ )( sin(r ) r cos(r ))((1 + C ) sin(r ) r cos(r )) + O(r 2 ) k 3/2 F (kx) k 3/2 F (kx) = 1 4π I sin θ dθ dϕ d(kr ) kr F (kx ) kr F (kx kx ) = kδ(kr) 4πk 2 r 2 sin(r )/r has fourier transform on variable r. Because the function F has not Fourier transform, and < F, F > is infinite, we can consider the limit of its forms truncated. Define then We have calculation Ω k (x) := NR 1 (kx)y, < Ω k (x), Ω k (x) >= 1 < Ω n 1 (x), Ω n 1 (x) >= 1 < Ω k Ω 1 Ω k Ω 1 >=< Ω k Ω k Ω 1 Ω 1 > =< Ω k Ω k < Ω 1 Ω 1 >>= < ( 2 Ω k ) Ω k < Ω 1 Ω 1 >> The solution of l = 1, m = 1, Q = e /σ is calculated or tested for electron, V = NR 1 ( k e r)y 1, 1 e iket

5 THEORY OF ELECTRON 5 5. Electrons and Their Symmetries Some states of electrical field A are defined as the core of the electron, which s the start function A i = V that is electrical, for the RRS to get the whole electron function e: e + r : NR 1 (k e r)y 1,1 e iket, Define the symbols: e + l e r e l : NR 1 (k e r)y 1, 1 e iket : NR 1 ( k e r)y 1, 1 e iket : NR 1 ( k e r)y 1,1 e iket ( e l ) := e+ r ( t), ( e + l ) := e r ( t) ( e r ) := e + l ( t), ( e+ r ) := e l ( t) r, l is the direction of the spin. Normalize the electron function with charge and mass, using the equation 2.1 < e µ c i t e µc >= Q c Then < e µ e µ >= m e k ec = m e /Q c < e cµ e µ c >= Q c The magnetic dipole moment of electron is calculated as the first rank of proximation r A/4 + cc. µ z =< A i i φ A i > /4 + cc. = Q e 2m e By the discussion in the section 2 the spin is S z = µ z k e /e = 1/2 Calculate the correction in RRS of the equation 2.1 (5.1) A n = A n 1 i (A i A i )/2 u = (A i (i t A i )) u (i t (A i A i )/2 u ) n 3 i (A i A i )/2 u = (A i (i t A i ))(i t (A i A i )/2) n 3 (i (A i A i )/2) u = δ(t r)/(4πr) The convolution is made in 4-d space. Generally x x 2 t = x x (x, t ) := (x, t r), r 2 = x x and f(x)e ict u = f(x)e ic(t+r) 1 4πr = f(x )e ict 1 4πr = f(x)eict 1 4πr δ(x 2a) f a (x a) g a ( (x a)) = δ(x 2a) g a (x a) f a ( (x a)) with definition f(x) g(x) := f(x + t)g(t)dt

6 6 WU SHENG-PING hence h(x) f(x) g( x) = h(x) g(x) f( x) (5.2) h(x) f( x) g( x) = h(x) g( x) f( x) Calculate with the properties of RRS start state A ± rl for e± rl A (k e, ϕ) = A( k e, ϕ) A (k e, ϕ) = A + ( k e, ϕ) A l (k e, ϕ) = A r (k e, ϕ) t A(k e, ϕ) = ik e A(k e, ϕ) Correct the first degree term of electron function with magnetic part to find A ν = ( Ω k (x)e ikt, i Ω k (x)e ikt /k) (5.3) < e ν e ν >= ν J ν e = ν e ν = These are Lorentz gauge and current constraint. We have with the discussion of the section 7: The function of e + r < (e + r ) ν, ( e + l ) ν >= < (e + r ) ν, ( e r ) ν >= < (e + r ) ν, ( e l ) ν >= is decoupled with e + l 2 < (e + r ) ν, (e + l ) ν >= The increment of field energy ε on the coupling of e + r, e r mainly between A 2 is ε e = 2 < (e + r ) ν, (e r ) ν > 2e 3 /σ m 1 e = s The condition 2.2 is used. This value of increments on the coupling of electrons are ε e e + r e r e + l e l e + r + e r + e + l + e l + The increment of field energy ε on the coupling of e + r, e l mainly between A 4 is ε x = 2 < (e + r ) ν, (e l ) ν > 1 2 e7 /σ m 1 e = s Other parts are counteracted, including the degree 8. This value of increments on the coupling of electrons are ε x e + r e r e + l e l e + r + e r + e + l + e l +

7 THEORY OF ELECTRON 7 The propagation: 6. Propagation and Movement f 3 The convolution is on space. The particle number normalization: The following are stable propagation: i e i < f, f >= 1 particle electron photon neutino notation e + r γ r ν r structure e + r (e + r + e r ) (e + r + e l ) By the condition of 5.3, their static mass except the couplings is zero. The movement of the propagation is called Movement, ie. the third level wave: f f i e ij Calculate the following coupling system of particle x A x = i e i = c n c e c c differ in charge, spin, and being negative or positive: c = (r, l)(+, )(p, n) The initial and start state for RRS is A i = e x 4 i e i e x := Ω kx f(t)δ(t) f(t) := e ih(t), h R It meets the normalization of particle number. By referencing to the identity 5.2 f is chosen to obtain: or equivalently We have then With the normalization of charge f(t)δ(t) g(t) = f(t)g(t) δ(t) A i = A ν i ( A iν ) = f(t) e ikxt A i Ω kx e ikxt 3 i e i then < A ν i i A iν >= Q x k x n/q x, n := c n 2 c The initial EM energy is ε i =< A ν i A iν > n/e /σ

8 8 WU SHENG-PING The initial MDM is µ z < A ν i ϕ A ν > /4 + cc. 1 2 k x Current is controlled by charge. The corrected field A meets the equation 2.1 with A = A i + Q c n 2 c We notice that the terms in this equation are all delta functions of space, so that the interaction between the decayed blocks are released suddenly all at once, but this decaying process doesn t appear in the solution. Its solution is a solution with electrons and their propagations coupled wholly in grid origin. Consider the second degree correction of the equation 2.1: n=2 A n A ν = ia µ ν A µ /2 + cc. = I s I c For the light decay the self-effect at the initial time I sc := (e x n c e c ) (e x n c e c ) c I s = c I sc the crossing effects at the initial time I c := cc. + (n r+p n r p n r+n n r n )(e x e + r ) µ i (e x e r ) µ +(n l+p n l p n l+n n l n )(e x e + l )µ i (e x e l ) µ +(n r+p n l p n r+n n l n )(e x e + r ) µ i (e x e l ) µ +(n l+p n r p n l+n n r n )(e x e + l )µ i (e x e r ) µ We can easily use the coupling of these currents to calculate the emission of a decay mode. The calculation of strong effect is similar. The reaction 7. Balance Formula A 1i A 2i A 1f A 2f is equivalent of the same energy emission to A 1i + A 2f (x, t) A 1f + A 2i (x, t) by writing them as sums of odd function and even function of t. This means we can shift electron to the other side with the same emission of EM energy for a balance formula.

9 THEORY OF ELECTRON 9 The core of muon is 8. Muon µ r : e µ (e r e + r e l ) From the equation 15.1, µ is approximately with mass 3m e /e /σ = 3 64m e [3.2][1] (The data in bracket is experimental by the referenced lab), spin S e (electron spin), MDM µ B k e /k µ. The main channel of decay with balance formula µ r e r ν r, e r e r + ν l e µ e r + δ 1/2 (x + v 2 t) ν r δ 1/2 (x v 1 t) e r + e µ ( t) ν r Reference to the discussion of the section 6 to find the main EM emission is ε x m e 1 = k µ [ s][1] s The core of pion is 9. Pion π + l : e π (ϕ) (e + r ± e + l ) + e π( ϕ) e l It s approximately with mass 3 64m e [4.2][1], spin S e and MDM µ B k e /k π +. Decay Channels: π + l e + l + ν r, e + l e + l + ν l The mean life approximately is ε x /2 = s [( s][1] The precise result is calculated with successive decays. 1. Pion Neutral The core of pion neutral is like a atom π : ((e + r + e + l ), (e r + e l )) It has mass approximately 4 64m e [4.2][1], zero spin, and zero MDM. Its decay modes are π γ r + γ l The loss of energy is 2ε e = s The following particle is similar to π [ s][1] (e x ( ϕ) (ne + r n e l ) + e x (e + r + e + l ), e x ( ϕ) ( ne l + n e + r ) + e x (e r + e l ))

10 1 WU SHENG-PING The core of tauon maybe 11. Tauon τ r : e τ ( ϕ) (ne + r ne l ) + e τ (e r e l e + l ) Its mass approximately 53 64m e [54][1] (n = 5), spin S e and MDM µ B m e /k τ. It has decay mode with a couple of neutrinos counteracted e r e l e + l e r γ r e τ e r + δ 1/2 (x + v 2 t) γ l δ 1/2 (x v 1 t) e r + e τ ( t) γ l The main EM emission is ε e m e k τ 1 = s [ s; BR..17][1] Perhaps, it s a mixture with distinct coefficients n. The following particle is similar to τ e τ ( ϕ) ( ne + r e + l e l + e r ) + e τ (ne l e + r e r e + l e l ) The core of proton may be like 12. Proton p r : e p ( 4e + r 3e r 2e l ) The mass is 29 64m e [29][1] that s very close to the real mass. calculated as 3µ N, spin is S e. The proton thus designed is eternal. The MDM is 13. Neutron Neutron is the atom of a proton and a electron and a neutrino, Neutrino circles around proton with n = (p + r, ν l, e l ) m ν ωr 2 = 1 The interaction is between their (proton and neutrino) magnetic fields of A 2 (gross current) m ν ω 2 r = 1 4π 3 2 e2 σk e /(m p r 2 ) m ν = e 7 /σ m e/2 The EM energy emitted by neutrino is approximately 1 2 ( 6 4π )2 e /σ m 3 e/(m 2 pe 2 /σ ) = 1 673s

11 THEORY OF ELECTRON Mesons We can define kinds of energy decreases of decay interaction EM side EM weak side weak strong abbrevation L LS W W S S emission unit ε e ε e m e / k x ε x ε x m e / k x m e We analyze the mesons [1] as the following name mass(m ev ) emission type ratio construction η keV L π class, perhaps K ± s W.88 e x (ϕ) (ne + r e l + e + l ) +e x ( ϕ) ( ne l + e r ) KS 1 1 s unclear KL s unclear D s LS 1.5 (( ne + r + n e r e + l e l ) B s LS 1, (ne l n e + l e + r e r )) D ± S s LS 1.3 τ class D ± s LS.6 τ B ± S s LS 1 τ class Υ keV unclear J/ϕ keV unclear others S 15. Mechanical Quantification The mechanic feature of the electromagnet fields is T is stress-energy tensor, T ij = F k i F kj g ij F µν F µν /4 T ij = mu i u j, u = dx/ds T is quantum expression of the energy, by Lorentz transform it s easy to get the quantum expression of momentum: A ν i A ν + cc. mu The observed mass in mass-center frame is M = dv T (15.1) M = ε It s obtained by moving the frame and then checking the energy. The physicals have operators as the following < t A ν t A ν > mechaincal energy (it s conserved) E < A ν A ν > EM energy or analytic kenitic energy ε < A ν A ν > analytic static mass m < A ν J ν > interaction potential F, Γ The decayed part is independent (uncoupled) of the undecayed part for the equation 2.1, hence the undecayed part A meets the same condition A ν = ia µ ν A µ/2 + cc.

12 12 WU SHENG-PING In principle, we can solve this equation to find the decay process. For the partial field A n of n independent particles decaying to stable state, F =< A ν n A νn >= nγe CΓt =< A µ n ia ν n µ A νn /2 + cc. > For some number n and charge Q (15.2) 1 = dt < A ν 1 A ν1 >= Q σ ( then 1 = 1 = ( 1 = n 3 ( ( dt < A ν n A νn >= dt < A ν n A νn >) 3 = ( dt < A µ 1 iaν 1 µ A ν1 /2 + cc. >), Q = 1 dt < A µ 1 iaν 1 µ A ν1 /2 + cc. >, σ = 1 dt < A ν 1 A ν1 >) 3 = n 2 ( dt < A ν 1 A ν1 >) 3 = ( Q σ = n 1/2 Substitute it into the equation 15.2 to find hence (15.3) 1 = n = 1 dt < A µ 1 iaν 1 µ A ν1 /2 + cc. >) 2 dt < A µ 1 iaν 1 µ A ν1 /2 + cc. >) 2 dt < A µ 1 iaν 1 µ A ν1 /2 + cc. >) 2 dt < A ν A ν > It s the normalization of one decaying particle, and it leads to the result between decay life and EM emission (also an interaction potential): C = 1 The distribution shape of decay can be explain as e Γt/2 e x i e i Ω x i e i e Γt/2 ikxt, < t < is the real wave of the particle x near the initial time. Expand it in that time span Ω x i e i Ce ikt dk k k x iγ/2 With calculation, we find the emission of each branch wave is the same.

13 THEORY OF ELECTRON Great Unification The General Theory of Relativity is (16.1) R ij 1 2 Rg ij = 8πGT ij /c 4 Firstly we redefine the unit second as S to simplify the equation 16.1 Then R ij 1 2 Rg ij = T ij R ij 1 2 Rg ij = F k i F kj g ij F µν F µν /4 We observe that the co-variant curvature is R ij = F ikf k j + g ij F µνf µν /8 17. Conclusion Fortunately, this model explained all the effects in the known world: strong, weak and electromagnetic effects, and even subclassify them further if not being to add new ones. In this model the only field is electromagnetic field, and this stands for the philosophical with the point of that unified world is from an unique source, all depend on a simple hypothesis: the current of matter in a system can be devised to analysis the e-charge current. My description of particles is compatible with QED elementarily, and only contributes to it with theory of consonance state in fact. But my theory isn t compatible to the theory of quarks. In fact, the electron function is a good promotion for the experimental models of proton and electron that went up very early. Underlining my calculations a fact is that the electron has the same phase (electron resonance), which the BIG BANG theory would explain, all electrons are generated in the same time and place, the same source. References [1] K. Nakamura et al. (Particle Data Group), JPG 37, 7521 (21) (URL: address: sunylock@139.com 8 Hanbei Rd, Tianmen, Hubei Province, The People s Republic of China. Postcode: 4317