Unified Absolute Relativity Theory. António Saraiva

Unified Absolute Relativity Theory António Saraiva – 2009-04-03

[email protected]

Introduction – Everything is relative, including light speed.

From a particular and evident property of the Lorentz’s equations we have derived a theory that agrees with all known experimental data and works for atomic and sub atomic scales, but it also works for gravity at macroscopic scales.

Basis of the theory From the Lorentz’s equations:

       − + = − + = 2 2 2 0 0 2 2 0 0 / 1 / / 1 c v c vx t t c v vt x x _c2_t2 _x2 _c2_t02 _x02 − = −

For n relative frames with v_n relative speeds:

       − + = − + = 2 2 2 2 1 2 1 2 2 2 2 1 2 1 2 / 1 / / 1 c v c x v t t c v t v x x c2t₂2 −x₂2 =c2t₁2 −x₁2 2 v v_n 1 1,t x x₂,t₂ x_n,t_n

       − + = − + = 2 2 2 1 1 2 2 1 1 / 1 / / 1 c v c x v t t c v t v x x n n n n n n 2 2 2 2₁2 ₁2 x t c x t c _n − _n = − 2 2 2 2 v v c v v c v n n x − − =        − + = − + = 2 2 2 2 2 2 2 2 2 / 1 / / 1 c v c x v t t c v t v x x x x n x x n 2 2 2 2 ₂2 ₂2 x t c x t c _n − _n = − So: c2t₁2−x₁2 =c2t₂2−x₂2 =⋅⋅⋅=c2t_n2 −x_n2 c2t_n2 −x_n2 =k (Constant)

The orthodox relativity use 2 S

k = as a variable, but as we have demonstrate k is a constant with only one value for all wave particles. As we can’t give independent values to x and t the space-time doesn’t exist. x and t are not space and time but wavelength and period of a electromagnetic wave.

One direct consequence is that vacuum light speed is variable with the frequency.

Derivation and Generalization of the Planck’s Formula

The Planck’s formula E =hf is not correct for all electromagnetic spectrum.

Magnetic wave equation:

(

)

_

     − = 2 2 2 2 2 0 4 sin c t x x B B π

Energy of the magnetic field: 0 3 2 2µ x B E=

(

)

_      − = 2 2 2 2 2 2 2 0 0 3 4 sin 2 B x c t x x E π µ and c t −x =k 2 2 2 ₄ 2 4 2 0 0 3 16 2 x k B x E π µ = x k B E 0 2 4 2 0 2 16 µ π = And f w x= hf w c E =

General Planck’s formula: hf kf c c E 2 2₋ = 17 1 0 2.9353 10 − × = ms B ; 25 2 2 0 8.8 10 − × = m s E Exact value of k

Equalling the forces of the electron:

₂ 0 2 3 4 4 _e e R q c khf F πε = = and π e e x R 2 137 = ₄ 0 2 2 137 4 hc x q k e ε π = k =1.91555918×10−34m2

Exact mass of the proton:

Energy: E =1.50327736×10−10J kE h c kg hc E m 2 2 2 27 3 1.67271338 10 − × = + =

The Unified Absolute Relativity Theory predicts the mass of the electron: k hc hq m_{e 3} 0 2 137 2 ε π × =

Another formula for k:

(

)

3 0 3 4 3 3 0 137 8 137 ε π µ + = e e x c x h k Electron wavelength: _{6 4} 2 0 11 3 0 3 0 137 2 ) 137 ( q c h x x_{e e} π ε µ ε = + Hydrogen-Deuterium Abundance (1−x)m_H + xm_D =1.6728×10−27

x – Abundance of the deuterium; m_D - Deuterium mass; m_H - Hydrogen mass m_D ≈2m_H 27 27 10 6727 . 1 1 10 6728 . 1 − − × = + × = x m_H x=6×10−5

The wrong abundance of the deuterium is 4 10 1 .

1 × − to keep the wrong Einstein’s

mass of 27 10 6726 . 1 × − . Wrong formula: _{E =mc}2 Right formula: E =mcw

The right abundance is not from the variable abundance in a particular sea water, but from the universe abundance.

Speed of the electromagnetic waves

2 2 kf c w= − For w=0 k c f_M = Matter frequency -- f_{M =2.16607214×10}25Hz

As we can see for low frequency waves, like light, the speed appears as a constant but the speed changes a lot exactly for the frequencies related with subatomic particles, the scale where classic relativity fails.

The small variation of speed for low frequencies is allowed by all known experimental data.

For frequencies greater then f_M the waves have imaginary speeds and wavelengths that mean they are longitudinal waves.

Wavelength of a wave-particle f kf c x 2 2₋ = c w iw f M f

k k m 17 10 38403728 . 1 × − = =λ

Energy of a wave particle

and mw2 =hf _{2 2} kf c hf m − = x ix k i f M f 2 2 _kf c hcf E − = E =mcw

Mass of a wave particle

To explain all the existing particles, frequencies and masses must be positive and negative.

The sum of all that exists is equal to zero.

The masses of the longitudinal photons are negative. The mass has the signal of the charge.

Energy of a wave particle

The energies of the longitudinal photons are imaginary.

m f M f electron neutrino proton neutron boson w.. boson z.. photons l transversa.. photons al longitudin .. E iE f M f GeV i89.58

General wave particle symmetry

Unified force

According to our theory the light speed is variable, so around the particles exists a field of speed variation or an acceleration field. The variation of speed with time is equivalent of the variation of the squared speed with space.

The forces must be explained by only one mechanism and only one formula. For example, the protons have only one force not the electric and the strong.

w= c2 −kf 2 t k t c w= − 2 2 Acceleration: dt dw g = w kf g 3 = Force: F =mg and ₂ w hf m= m f electron negative.. proton negative.. electron positive.. proton positive..

Force between two equal particles: ₃ 4 w khf F = or ₃ 0 0 2 2 4 0 2 2 2 ) )( ( ) ( v w vw c c f v c kh F + + − =

Unified force between two electrons

2 3 3 4 e e ee x k x khc w hkf F + = = and x_{e 2.426 10}−12m × = F_ee =1.142×10−12N Electric force: ₃ 4 2 0 2 4 w hkf R q F ₌ e ₌ πε ε R=1.42×10−8m Rydberg constant: R_H =1.0968×107m−1 Rydberg wavelenght: m R_H H 8 10 1174 . 9 1 − × = = λ λ =_H 2.π.R

Force in hydrogen atom

Rydberg constant: R_H =1.096776×107m−1 Rydberg wavelength: m R_H H 8 10 11763 . 9 1 − × = = λ Rydberg frequency: f c Hz H H 15 10 28805 . 3 × = = λ Orbital frequency: f_OR =2f_H

Orbital speed: v=137x_ef_OR and x_e =2.426×10−12m

Bohr radius: m f v R OR B 11 10 3 . 5 . 2 − × = = π Centript acceleration: 22 2 10 035 . 9 × = = R v g x_e =2.426×10−12 m_e =9.11×10−31 g_e =1.1478×1018 _{1.32 10}−15 × = p x _{1.6728 10}−27 × = p m _{7.112 10}27 × = p g g= g_eg_p =9.035×1022 F =m_eg=8.231×10−8N

Unified force = Strong force Two protons: F_PP = m_Pg_P =+12.973N

Two neutrons: F_NN =m_Ng_N =+13.041N

A proton and a neutron: F_PN =m_Pg_N =−13.00N

What about the electric force?

2 0 2 4 13 R q_e πε = R=4.2×10−15m

This is precisely the distance between the proton and the neutron in a deuteron. The strong force is equal to the electric force, that means that the strong force doesn’t exist. The neutrons behaves as a negatively charged particles, it is only neutral for

macroscopic distances. 2πR=nx_P and 15 10 32 . 1 × − = P

x (Compton wavelength of the proton) n = 20 Acceleration: 27 2 10 755 . 7 × = = R v g _v=5.7158×106

The binding energy is not kinetic or potential.

Electric field of the neutron density e Charg .. +

_

Radius m 15 10 1× −

The Particle Values w chf mcw= _{2 2} kf c hf m − = mk c km h h f 2 4 2 2 2 + + − = Neutron: m_=1.6749×10−27kg_{; f =2.27156×10}23_Hz w= c2−kf2 =2.9977532×108ms−1 m f w x= =1.31969×10−15 ; 27 2 3 10 78635 . 7 × − = = ms w kf g Proton: _m=1.6727×10−27 _{; f =2.26858×10}23 8 10 99775365 . 2 × = w ; 15 10 321423 . 1 × − = x g=7.75574×1027 m f M f

ν

e− P+ 0 N − w 0 Z

Neutrino: _m=4×10−36 14 2 10 425571 . 5 × = = h mc f ; w=c 14 2 10 7768 . 9 2 − × = = ∆ c kf w ; x=5.5255467×10−7 2 10 061 . 1 × = g Boson Z E =i91.2GeV k w w c hc E 2 2 − = _{2.08 10}8 −1 × =i ms w f =1.53×1025Hz x=i1.36×10−17m m=−2.34×10−25kg Boson W E=80.4GeV w=2.21×108ms−1 f =1.43×1025Hz x=1.542×10−17m m=1.95×10−25kg Top quark E =174.2GeV _w=1.35×108 f =1.9×1025 x=7.11766×10−18 m=6.9×10−25

Electron E =0.510998918MeV _{f =1.23559×10}20 _{x=2.42631017×10}−12 _{w c} ≈ c−w=∆w ₂ 2 kf w c∆ = _{∆w=5.07×10}−3_ms−1 _{9.10938229 10}−31 × = m 18 2 3 10 253 . 1 × − = = ms w kf g F =mg=1.14×10−12N Monopole 3 4 2 0 2 4 w khf x q F ₌ m ₌ πµ e m q h q 2 =

q_e-- Electric charge;

µ

₀-- Vacuum permeability _{1.2427 10}−17 × = x 25 10 2752 . 1 × = f 8 10 584552 . 1 × = w 25 10 365 . 3 × − = m E=99.8GeV

Mass and Energy Relation

w hcf mcw

E= =

For high energy particles

2 2 2 2 2 2 3 2kE c h kE c h hc m E + + = ∆ ∆

For charged particles with low energy:

∆E=c2∆m

For high energy particles:

E k hc m E 2 3 = ∆ ∆

Absolute Relativity Kinematics

    − = − = 2 2 0 2 2 0 / 1 / / 1 c v t t c v x x xt =x₀t₀ = A (Constant) w=x/t and f =1/t w= Af2 And mw2 =hf f3 =h/mA2 and 2 0 3 0 h/m A f = As f = f₀ 1−v2/c2 m=m₀/(1−v2/c2)3/2 w= x/t 0 2 2 0(1 / ) t c v x w= − w=w₀(1−v2/c2) ) / 1 ( / 1 2 2 0 2 2 0 c v w c v f hc w hcf E − − = =

2 2 0/ 1 v /c E E = − Kinetic energy E_k =E−E₀ and E= E₀/ 1−v2/c2 E₀ =m₀cw₀

(

2 2

)

2 2 2 2 0 0 / 1 1 / 1 v c v c v c cw m E_k − + − = For w₀ =c and v<<c 2 0 2 1 v m E_k = Momentum p=mv ₂ 0 _{2 3/2} ) / 1 ( v c v m p − = and v=v₀(1−v₀2/c2) 2 2 0 0 0 / 1 v c v m p − =

Binding energy of two quarks x _{8 10}−19m × ≈ TeV x q Fx E m _26.4 0 2 = = =

µ

For the moment there’s no particle accelerator with enough energy to separate two quarks or monopoles.

New particle

We found a new particle that is the quantum of vacuum and its energy is equal to the potential energy of the particle in the universe gravitational field. Several experiments show a missing mass precisely equal to 310 MeV.

E 310MeV 2 0 0 0  =      = µ ε

ε₀ = Vacuum permittivity; µ₀ = Vacuum permeability

Gravitational potential energy: U U R m GM E 0 0 =

G – Gravitational constant; MU = mass of our universe

RU = Radius of our universe (we leave at the surface of our universe)

E₀ =hf₀ ; f₀ =7.5×1022Hz kg c hf m 28 2 0 0 5.5 10 − × = =

Those properties prove the relation between electromagnetism and gravity.

Derivation of the Lorentz’s equations and their true meaning Transverse effect propagation speed:

Approximation:     = − = − wt ct t v c t v w A A 2 2 0 0 ) ( c wt v c t v w₀ )₀ 2 2 ( − = − With w₀t₀ = x₀ and wt= x 2 2 0 0 / 1 v c vt x x − − =     + = − = t v w t v c ct t w B B ) ( 2 2 0 0 w v t c t w v c2 2 0 0 ( ) + = − −v2 c2x₀ = x+vt / 1

Substituting the value of x:

vt c v vt x x c v + − − = − 2 2 0 0 0 2 2 / 1 / 1 front wave.. emitter 2 2 v c − c c v A t v c t v w₀ ) ₀ 2 2 ( − = − ctA =wt v v B ct t w₀₀ = c v tB (w v)t 2 2 + = −

2 2 2 0 0 / 1 / c v c vx t t − − = Separation: 2 2 0 0 / 1 v c vt x x − + = 2 2 2 0 0 / 1 / c v c vx t t − + =

All the solutions of the systems without v: v C t v c t v w₀ ) ₀ 2 2 ( + = − ctC =wt v D ct t w_{0 0} = c v tD (w v)t 2 2 − = −

One possible general form of the Lorentz’s equations: v vcosα        − + = − − = 2 2 2 0 0 2 2 0 / 1 / cos / 1 cos c v c vx t t c v vt x x α α        − + = − + − = 2 2 2 0 0 0 2 2 2 0 / 1 / cos 2 / 2 1 cos ) / 1 ( c v c vx t t c v vt c sen v x x α α α

For the four equations:

α =0        − + = − + = 2 2 2 0 0 2 2 0 0 / 1 / / 1 c v c vx t t c v vt x x 2 2 2 2 ₀2 ₀2 x t c x t c − = − α =90      − = − = 2 2 0 2 2 0 / 1 / 1 c v t t c v x x xt =x₀t₀ xt t x_{0 0} =− 0 0 0 0 .. .. .. .. t t or ct x ct x or x x = = = = xt t x_{0 0}= 0 0 0 0 .. .. .. .. t t or ct x ct x or x x = = = = 2 0 2 0 2 2 2 2 x t c x t c − = − 2 2 2 2 ₀2 ₀2 x t c x t c − = − 2 2 0 0 / 1 v c vt x x − + = 2 2 0 / 1 v c vt x x − − = 2 2 2 0 0 / 1 / c v c vx t t − + = 2 2 2 0 / 1 / c v c vx t t − − =

New general relativity calculations

Light deflection by the sun

As the space contraction and time dilatation formulas are note equal we conclude that light speed is variable, so we demonstrate that the values of one test of general relativity can be calculated if we consider that the light speed in gravitational fields behaves as in the optical mediums.

This test conceived by Einstein try to calculate the deviation of a light ray from a distant star that passes near the Sun surface and is observed on the Earth.

2 2 0 0 2 2 1 / 1 / x x v c t t v c  _{= −}   =  −  and w=x t/ and w₀ =x₀/t₀ 2 2 0 2 c v w w c − = and w₀ ≈c 2 2 c v w c − = v c v w=− ∆ ∆ 2 (1) Sun center Sun surfice S R 2 2 S R= l +R Earth l i i 1 w w₂ r d

α

On the place defined by the distance l , from the Sun surface, the light ray that passes near the Sun has an incident angle i , a refraction angle r and an angle shift

The refraction plan divide two zones of the space with propagation speeds w1 and w2. According to the laws of refraction:

senr w w seni 2 1 = ; 2 2 S S R l R seni + = ; 1 2 2 2 w w R l R senr S S + = 2 2 cos S R l l i + = ; r=i+dα ; 1 2 2 2 ) ( w w R l R d i sen S S + = + α 1 2 2 2 . cos cos . w w R l R send i d seni S S + = + α α 1 2 . w w R d l R_S +

α

= _S 1 1 2 w w w l R d_{α =} S − w c l R d ₌ S _∆ . α and v c v w=− ∆ ∆ 2 dv c l v R d S 2 . 2 − = α (2)

If we want to put gravity in the relativity equations we must change the linear velocity v by the escape speed as the gravitational potential:

R GM v2 2 S = ; M_S -- Sun mass 4 2 2 2 S S R l GM v + =

(

l R

)

dl l GM dv S S 4 / 5 2 2 2 2 + − =

Substituting v and dv in (2) we get:

(

_{2 2}

)

6/4 2 2 S S S R l dl c R GM d + =

α

(

)

∫

+ = ES D S S S R l dl c R GM 0 4 / 6 2 2 2 2

α

; D_ES= Earth Sun distance

We have only consider the angle deviation of the light ray that comes from the Sun. Considering also the light ray that goes to the Sun, the deviation angle will be double: δ =2α

4 ₂ 1₂ S S S R c R GM =

δ

S S R c GM 2 4 = δ =8.4838561×10−6 =1.75" rad δ

Thus, we have calculated the correct deviation.

Shapiro time delay

This test of general relativity conceived by Irwin Shapiro intends to measure the delay of a radar signal from the Earth to Mars, when the superior conjunction, reflected on Mars and detected on the Earth.

The signal passes near the Sun’s surface and due to the space-time bending it suffers a delay.

Our calculations consider the space absolute and the light speed variable.

11 10 279 . 2 × = MS

D -- Mars Sun distance; 11

10 5 . 1 × = TS

D --Earth Sun distance;

_{3.779 10}11

× =

MT

D --Mars Earth distance; _{1.989 10}30

× = S M -- Sun’s mass _{6.95 10}8 × = S R -- Sun’s radius.

∆t = time delay ; w = slower light speed w≈c 2 ₂ c w c D t= _MT − ∆      − = − = 2 2 0 2 2 0 / 1 / 1 c v t t c v x x and w=x/t; w₀ =x₀/t₀; w0 ≈c c v c w 2 2 − = c v w c 2 = − ₃ 2 2 c v D t = _MT ∆ Escape speed: R GM v S i 2 2 = and ct D_MT = 2 wt _w∆t cw w c D t= _MT − ∆ 2

R=

(

l−D_MS

)

2+R_S2

(

)

2 2 2 2 S MS S i R D l GM v + − = Average v:

(

)

MT D S MS S D R D l dl GM v MT

∫

+ − = 0 2 2 2 2 _       = ₂ 2 4 log 2 S TS MS MT S R D D D GM v and ₃ 2 2 c v D t= _MT ∆ _       = ∆ 2 ₃ log 4 ₂ S TS MS S R D D c GM t ∆t=247.2µs

The experimental value of ∆t is a little lower than 250µs.

Correction of Mercury’s perihelion precession

We do the derivation and the calculation of the general relativity correction of the Mercury’s perihelion precession, considering that the space is absolute and the light speed is variable in gravitational fields.

Mars Earth Sun 2 2 S MS R D + D_TS2+R_S2 S R R l

Correction of the gravitational force From the formulas of the space contraction and time dilatation:

     − = − = 2 2 0 2 2 0 / 1 / 1 c v t t c v x x xt =x₀t₀ = A ( A = constant ) Doing w= x/t e f =1/t w= Af2

The wave energy is given:

E =mcw and w hcf E= 3 ₂ mA h f = and 2 0 3 0 A m h f = As f = f₀ 1−v2/c2

(

_{2 2}

)

3/2 0 / 1 v c m m − =

This equation is different from the Einstein’s formula. But this one is coherent with the two equations of time dilatation and space contraction.

No one can explain why the Einstein’s formula only can be derived from the time equation, denying the space formula.

We think there is an interpretation problem of the experimental data. All the experiments give not the relation between the masses but the relation between the ratio of the mass by the electric charge.

2 2 0 0 / 1 1 c v q m q m − =

If we consider that the charge is also variable: 2 2 0 / 1 v c q q − = Thus,

(

_{2 2}

)

3/2 0 0 0 / 1 m c v m m m m − − = − = ∆ ₂ 2 0 2 3 c v m m≈ ∆

with

r GM

v2 = 2 ( free fall speed from infinity )

r c GM m m= 3 ₂ ∆ ₂ r GMm F = m r GM F = ∆ ∆ _{2 2 3} 2 2 3 r c m M G F = ∆ _{2 3} 2 2 2 3 r c m M G r GMm F =− −

Orbital movement equation Doing

r

u =1 we have the classical equation of an elliptic orbit

2 2 2 2 ) 1 ( u GMma F u d u d ε θ + =− −

Substituting the value of F

) 1 ( 1 ) 1 ( 3 1 _{2 2 2} 2 2 ε ε θ  = −     − − + a u a c GM d u d (1) As 1 ) 1 ( 3 1 _{2 2} ≈ − − ε a c GM

we can use the classical solution

) 1 ( 1 2 2 2 ε θ +u = a − d u d ) 1 ( cos 1 2 ε θ ε − + = a u

Substituting the value of u in (1)

2 _{2 2 2 2} 1 ₁ ) 1 ( 3 ) 1 ( 3 ) 1 ( sen sen C a c GM a c GM d du a − + + − + − − = − ε θ ε θ ε θ θ ε θ ε (2)

a = orbit major semi axis ; ε = orbit eccentricity

As we can see the terms of this equation are angles and the term responsible for the correction is:

θ ε δ ) 1 ( 3 2 2 − = a c GM

To obtain the value ofδ for a complete orbit we doθ =2π , thus ) 1 ( . 6 2 2 _ε π δ − = a c GM G=6.67×10−11 ; M =1.989×1030 ; c=3×108 ; a =5.787×1010 ; ε =0.2056 7 10 01317 . 5 × − = δ Radians/revolution

The value of the shift in seconds per one hundred years is:

100 2408 . 0 1 3600 180 × × × × = ∆ π δ

0.2408 = revolution period in years ∆=42.94"

Thus, we have the value of the Mercury’s precession.

Saying, after all, that the light speed is constant in the vacuum is false because the vacuum without a gravitational field doesn’t exist. All the space is a huge gravitational field.

Pound-Rebka experiment

The real general relativity calculations are very simple.

In this experiment a gamma ray is emitted from the ground to the top of a tower and the gravitational redshift of the ray is cancelled in the detector with a Doppler shift due to the speed V. So, the speed V is a measure of the gravitational redshift when the frequency is the same.

x₀ =8.61×10−11m ; ∆R=h=22.6m _{7.36 10}−7 −1 × = ∆ = V ms V Gravitational redshift: γ V

2 2 0 1 v /c x x= − v c v x x= ∆ ∆ _{0 2} And R GM v= 2 v GM R ∆R      − = ∆ −3/2 2 1 2 R R GM c x x=− ∆ ∆ ₂0 _{2 0 2} c gh x x=− ∆

(G – gravitational constant; M – earth mass; R – earth radius; 2 8 . 9 − = ms g ) 25 10 12 . 2 × − − = ∆x Doppler effect: c V c x x= ₀ + V c x x= ∆ ∆ 0 ∆x=+2.11×10−25

Relativistic Flyby Anomaly

We found that the flyby anomaly of the earth orbits of satellites can be explained by a relativistic correction from The Unified Absolute Relativity Theory.

This anomaly consists of an increase of the speed of the satellites during earth flybys, that can’t be explained with Newton’s physics.

As the Einstein’s general relativity theory doesn’t predict this correction, it is a proof that supports our theory.

Orbital movement equation from UART 2 _{2 2 2 2 sin 1 1} ) 1 ( sin 3 ) 1 ( 3 ) 1 ( C a c GM a c GM d du a − + + − + − − = − ε θ ε θ ε θ ε θ θ ε As ) 1 ( sin 2 ε θ ε θ − − = a d du and doing: C₁ =−1 ) 1 ( sin 3 ) 1 ( 3 2 2 2 2 _ε θ ε ε θ θ − + − = ∆ a c GM a c GM

a = major semi axis;

ε

= eccentricity; M = mass; G = gravitational constant; c = light speed; θ = angle

The first correction is the geodetic effect or the perihelion precession correction. The two effects are the same.

The gravitomagnetism doesn’t exist.

The second correction is the flyby anomaly. As we see the effect cancels for a complete orbit.

Mercury’s Perihelion Precession

) 1 ( 3 2 2 _ε θ θ − = ∆ a c GM

; For one orbit θ =2π

) 1 ( . 6 2 2 _ε π θ − = ∆ a c GM M =2×1030kg; a=5.8×1010m; ε =0.2056 _{5 10}−7 × = ∆θ radians/revolution

The value of the shift in seconds per one hundred years is: 100 2408 . 0 1 3600 180 π θ

α =∆ ; 0.2408= Revolution period in years

α =42.94arc seconds (The same value as Einstein)

Geodetic Effect ) 1 ( . 6 2 2 _ε π θ − = ∆ a c GM ε =0.0014; a=7×106; M =6×1024; T =1.84×10−4years T . 3600 180 π θ α =∆ × α =13.4 arc seconds/year

The Einstein value is α =6.6 , 2 6 . 6 4 . 13 = Flyby Anomaly ) 1 ( sin 3 2 2 _ε θ ε θ − = ∆ a c GM For θ =2π ∆θ =0 The effect cancels for a complete orbit.

Earth Flyby Data Mission Date a (m) ε ∆v (m/s) Galileo Dec90 5×106 2.47 3.92×10−3 Near Jan98 8.57×106 1.81 13.46×10−3 Cassini Aug99 1.58×106 5.8 1.1×10−4 Rosetta Mar05 _2.55×107 _{1.33 1.82 10}−3 ×

Angle for the maximum ∆v

Orbital speed:       + = a R GM v 2 1 and v=vsinθ θ ε ε cos 1 ) 1 ( 2 + − = a R ε ε θ θ ε ) 3 2 cos sin 1 ( 2 2 − + − = a GM v =0 θ d dv ε ε ε ε θ 6 12 ) 3 ( 3 cos 2 2 2 2 + − ± − = Galileo θ =0.603537rad θ θ ε θ ε ε ∆ + − = ∆ 2 ₂ ) cos 1 ( sin ) 1 ( a R and ) 1 ( sin 5 2 2 _ε θ ε θ − = ∆ a c GM 3 10 85 . 2 × − = ∆R ; 3 10 92 . 3 × − = ∆v

The variation of R and ∆vis along the same direction, and there’s a direct relation between them: t R v= ∆ ∆

We don’t know how to derive the value of the time or period t.

For Galileo: t=0.7277 Near θ =0.923 ; ∆R=6.36×10−3 t=0.472 Cassini θ =1.7441 t=26.6 Rosetta θ =1.11 t=4.1

For Galileo and Near we found the relation:

28000 . / 4 3 ε GM a t=

For Cassini and Rosetta:

4300 . / 4 3 ε GM a t= ; 2π 4300 28000 =

Units unification in S. I. system

Everything is made of speed and distance. Time doesn’t exist in nature. There are only 3 distance dimensions.

Definition of mass Wavelength of the electron: x_e =2.426×10−12m

Light speed = c

Particular electron relations:

Electron charge -- q_e ≈ x_e3c2

Planck’s constant -- 5 3

c x

h≈ _e

Magnetic flux quantum -- Φ₀ ≈x_e2c

Inverse permeability -- 2 0 1 c x_e ≈ µ Permittivity -- ε₀ ≈ x_e Electron energy -- E≈ x_e4c4 Electron mass -- m≈x_e4c2 Boltzmann constant -- k_B ≈x_e2

Using: distance = L and speed = V

So, the mass is equal: 4 2

V L

List of units

Mass -- M =L4V2

Time -- T =LV−1

Electric charge -- q=L3V2

Electric dipole moment -- d =qL= M =L4V2

The electric dipole moment is a mass.

Magnetic charge -- LV q h q_{m 0} 2 2 = = Φ = Planck’s constant -- 5 3 V L h=

Magnetic flux quantum -- Φ₀ =q_m = M

Inverse permeability = Density = Electric potential = 1 =LV2 µ

Magnetic current -- _{I LV}2

m =

Magnetic field -- B=V (Magnetic flux density) Electric field -- E =V2

Electric current = Magnetic voltage -- I =L2V3

Permittivity -- ε =L

Force -- _{F = L}3_V4

Magnetic potential = Inverse resistance -- A = LV Ω = 1 = Circulation Gravitational constant -- −3 = L G Pressure -- _LV4 Farad -- L2

Henry -- −2 V Energy -- 4 4 V L E = Moment -- L4V3 Watt -- 3 5 V L

Magnetic field strength -- H =LV3

Electric flux -- L2V2 = Energy

Acceleration = Magnetic current density -- 1 2

V L J a M − = = Energy -- 2       = µ ε E

Electric current density -- _{J V}3

E =

Electric displacement field -- D= 1 =LV2 µ Magnetic current -- 2 LV Im = 3 1 G x_G = Boltzmann constant -- 2 L kB =

The temperature is an energy surface density: ₂ L2V4

L E

T_k = =

For the Sun:

T_k =5800K; Surface -- 2 18 2 10 6 4 R m S = π = × Energy -- E=ST_k = 3.5×1022J

Power surface density at earth -- _1.3 −2

= Wm

P

A=4πD_ES2 =2.8×1023m2 P_{T =3.6×10}23W _{--- E =3.6×10}23_J Units table L-1 L0 L L2 L3 L4 L5 V-1 Thermal Resistance; Electric Resistance Time; Inverse Frequency V0 Distance; Permittivity Surface; Capacitance; Boltzmann Constant Volume; Inverse Gravitational Constant V Frequency; Vorticity Speed; Magnetic Field Magnetic Potential; Conductance; Circulation Magnetic Charge; Magnetic Flux True Magnetic Dipole Moment V2 Acceleration; Current Density Electric Field; Inverse Inductance Magnetic Current; Electric Voltage; Inverse Permeability Electric Flux; Q.M. Probability Electric Charge Mass; Electric Dipole Moment V3 Sound Resistance Electric Current Density; Potential Vorticity Magnetic Field Strength Magnetic Voltage; Electric Current Momentum; False Magnetic Moment Planck Constant; Angular Momentum V4 Pressure; Energy Density Temperature; Surface Tension Force Energy; Torque V5 Luminance Spectral Irradiance Intensity; Irradiance Power The existence There’s the nothing.

But, if so, the nothing can’t exists.

On the logical limit, if nothing exists the nothing can’t exist.

That means the nothing has an intrinsic instability, it can’t allows the existence of it self, so the nothing perpetually oscillate between his symmetric components like

) 1 ( ) 1 (

0= + + − . This oscillation with no initial energy is the existence. So the sum that all that exists is equal to zero.

The nothing and everything

The existence exists forever.

The nothing and everything is made of two things: speed and distance. Speed is equal to magnetic field and distance is equal to permittivity.

The nothing:

The nothing decays to four types of wave-particles:

The total speed and distance remains equal to zero as the energy. Two particles has positive mass and the other two negative mass.

0 Bρ B0 ρ 0 Bρ B0 ρ 0 Bρ B0 ρ 0 Bρ 0 Bρ 0 ερ ερ₀ ερ₀ ερ₀ 0 ερ ε0 ρ 0 ερ ε0 ρ 0 Bρ B0 ρ 0 Bρ 0 Bρ 0 Bρ 0 Bρ 0 Bρ 0 Bρ 0 ερ 0 ερ 0 ερ 0 ερ 0 ερ 0 ερ 0 ερ 0 ερ

Two charged particles and two neutral.

At the surface of our universe

The title means that this is not the only one. The multiplicity of things is a law of the nature. Our universe can be just a simple subatomic particle of another mega universe. The orthodox physics states that the universe has no center, but this corresponds to another hidden centralistic view of our position, because when the physicists make the calculations, they always put us in the center of the “observable universe”.

This universe has a center and we are not living at the center of it, we are living at the surface of our universe, just like we are living at the surface of the earth. It exists already a prove of that. The Hubble constant is not constant with the direction of observation.

We are living at the surface of a black hole that rotates at light speed.

Variable Hubble Constant in an Expanding Universe

The Hubble constant is variable with distance and celeste coordinates, because we are not living at the center of our universe. With a right measurement procedure we can find the location of the universe center.

We must fix a reference distance d:

2.57 1025

2 = ×

= RU

d ; R_U =Local universe radius = 5.14×1025m

R= d2 +R_U2+2dR_U cosα R d Arc α θ = sin sin

Relative expansion speed between A and B:

∆V = 2Rg_U cos(α−θ)−ccosα

g_U =Universe acceleration = 8.74×10−10 ; c = light speed

Hubble frequency:

U R

V

H = 2∆

All sky observation:

observer Local.. c V θ U R

α

A R d B

α

−

θ

18 0 2.61 10 − × = H ; Universe frequency -- H_U =2.91×10−18Hz 18 45 3.66 10 − × = H H₉₀ =5.5×10−18 18 135 5.4 10 − × = H H₁₈₀ =3.42×10−18

In an accelerated expanding universe the Hubble constant is variable.

Rotating Universe

Our universe is eternal as all the existence, it has no beginning. Hubble constant: H 18Hz

0 2.3 10

−

× =

Local gravitational acceleration:

g_U =cH₀ =6.9×10−10 Our universe is rotating with a constant angular speed

ω

: ω=2πH_U ; c=ωR_U =2πH_UR_U 0 _{3.66 10} 19 2 − × = = π H

H_U -- Frequency of the universe

0 45 90

135

The local orbital speed is equal to light speed. R_U = Radius of the local universe Universe period: s Gy H T U U 2.73 10 86.51 1 18 = × = = Radius: m H c R_U 26 0 10 3 . 1 × = =

The local orbital speed is light speed: U U R GM c2 = M_{U 1.76 10}53kg × = Some formulas: U R

ω

Total..universe universe

U U U GM R T 3 2π = ₂ U U U R GM g = 2πR_U =cT_U

Apparent linear expansion

The red shift is due to the relativistic dilatation of the wavelengths by transverse relative speed.

Relative longitudinal speed

v= H₀Rsinα₁−csinα₂

v=H₀Rsinα₁−csinθ and sinα₁ sinθ R R_U =

v= H₀R_U sinθ−csinθ and H₀R_U =c

v=0 U C U R R D 1 α α₁ R H₀ 1 0Rcosα H c 2

α

ccosα2

θ

Relative transverse speed R= R_U2 +D2 −2R_UDcosθ v=H₀Rcosα₁−ccosα₂ R_U2 =R2 +D2 −2DR_U cosα₁ RD R D R _U 2 cos 2 2 2 1 − + = α cosθ 2 2 2 2 0 c D R D R H v₌ + − U ₊ v=H0(D−RU cosθ)+ccosθ v=DH₀

We found the Hubble law but this speed is transverse.

Transverse red shift

2 2 0 v c cx x − = 0 _{2 2 3/2} ) (c v v cv x x − ∆ = ∆

v is the relative speed; ∆v=c−0=c

c v x x ≈ ∆ ₀

For a local speed equal to c the transverse red shift behaves as a longitudinal red shift:

c v c x x₀ = + c v x x ∆ = ∆ ₀

So, a rotating universe with a constant angular speed appears to be expanding. In it the Hubble constant is precisely a constant.

Light dispersion in the interstellar medium and the Pulsar distance

Light dispersion in the interstellar medium of our galaxy, allow us to know the exact distance of Pulsars and other variable emission objects by measuring the time delay between light of different frequencies.

From Lorentz’s equations:

       − + = − + = 2 2 2 0 0 2 2 0 0 / 1 / / 1 c v c vx t t c v vt x x c2t2 −x2 =c2t₀2 −x₀2

For n relative frames:

2 2 2 2 2 2 2 2 2 1 2 1 2 n n x t c x t c x t c − = − =•••= − c2t_n2−x_n2 =k _(Constant)

t and x are the period and the wavelength of the light wave, and k must be different from zero since we know that light speed in the vacuum of our galaxy is variable with the frequency due to the existence of free electrons. c is the light speed in perfect vacuum.

Doing the propagation speed equal to:

t x

w= and the frequency:

t f =1

w=± c2 −kf2

The dispersion time delay between two different frequencies along the distance D is:

_      − = ∆ B A w w D t 1 1

(

₃

)

2 2 2c f f Dk t_=± B − A ∆

The distance of the Pulsar is:

2 2 3 2 A B f f k t c D − ∆ =

If f is the average frequency and B the bandwidth in Hertz, the distance in meter is ( all S.I. units ):

kBf t c D= ∆ 3

According with the data analysis we have done for a great number of Pulsars we found the relation:

kf3.7 =3.7×1020 , so: B tf D e ∆ × = 4 10 3 . 7 ; (e=2.71828)

This is the exact formula of the distance of a Pulsar or a variable star in our galaxy. The interstellar medium seems to be almost uniform. Outside of our galaxy this formula is not correct.

General formulas of light propagation in space _{w= c}2 _−kf n f c knf w n ∆ − = ∆ − 2 1 t D w= ; c D t= t D c w= ∆ ∆ 2 f knf t c D _n ∆ ∆ = ₋₁ 3 2 In our galaxy: n=−1.7 ; k=2.7×1020 2 _{2.7 10}20_/ 1.7 f c w= − × In intergalactic space: n=−0.17 ; 9 10 8 . 1 × = k w= c2−1.8×109/ f0.17

The meaning of the fine structure constant

Normally the fine structure constant is seen as a relation between electric and gravitic forces. As we are going to see that is not the true meaning as the forces must be equal.

The inverse of the fine structure constant is the number of times that the wavelength of the electron makes the perimeter of it orbit.

Angular momentum: π 2 h n vR m L= _e = and n=1 e Rm h v π 2 =

Electric and centript forces between the proton and the electron:

2 0 2 2 4 R q F R v m F e e c πε ε = = = 2 0 2 2 4 R q R v m_{e e} πε = and e Rm h v π 2 = 0₂ 2 e eq m h R π ε

= and the electric energy is:

R q E e 0 2 8πε = so, 2 2 0 4 8 h q m E e e ε = and λ 1 = hc E

We get the Rydberg constant, which is an experimental confirmed value:

7 1 3 2 0 4 10 0974 . 1 8 − × = = m c h q m R e e M ε

So, the radius of the orbit of the electron is: m q m h R e e 11 2 0 2 10 3 . 5 × − = = π ε

And the speed:

6 1 10 2 . 2 2 − × = = ms R m h v e π

The wavelength of the electron is: x_{e 2.426 10}−12m × =

We see that the perimeter:

x_e πR

α 2

1

= 137x_e =2πR

The value is not a perfect integer because the orbit is not perfectly circular.

Meaning of the wave function from quantum mechanics

The exact meaning of the wave function Ψ is not a probability amplitude but a magnetic potential A.

In the Schrodinger equation:

Ψ =−i fΨ dt d π 2 and Ψ = Ψ λ π 2 i dx d

dt dA Eρ= and dx dA Bρ= If Ψ= A:      Ψ = Ψ − = λ π π 2 2 i B f i E ρ ρ f B E λ − = ρ ρ c B E − = ρ ρ (Light speed)

And we know that for an electromagnetic wave: c B E M M ₌ ρ ρ

So, THE WAVE FUNCTION IS A MAGNETIC POTENTIAL.

Speed of the Force from Absolute Relativity

c2t2−x2 =k x=± c2t2 −k w c dt dx V 2 = = and t x w= x t c dt dx V 2 ± = =

Vortex Particle Model

A true fundamental particle is a vortex of nothing, made of speed and distance.

The magnetic field Bρ is a speed. The electric field Eρ is a squared speed. Electric charge = q_e = Eρ×Volume

Magnetic charge = q_m = Bρ×Area=Outflow

Mass = m=q_m2

Magnetic potential = A = Circulation = Γ

For the electron:

The reference length = x_{e 2.4 10}−12m × = Eρ_e =2.4×1018m2s−2 Bρ_e =7.9×109ms−1 w_e ≈c _{7.2 10}−4 2 −1 × = =cx m s A _e

From fluid mechanics: R v_T π 2 Γ = e nx A Eρ= Bρ Eρ _w w

Astronomical Aberration

The Einstein’s aberration formula is wrong and according to relativity theory aberration can’t exist.

Einstein’s aberration formula:

c v c v / . cos 1 / cos ' cos θ θ θ − − =

Einstein’s speed composition formula:

vw c v w c w − − = 2 ₂ ' So: c w' ' cosθ = and c w = θ cos And vw c v w c − − = ₂ ' cosθ According to Einstein w=c cosθ'=1 θ'=0 There’s no aberration.

Let’s see the case that the star is on the zenith, as v and w make 90 degrees one vector must be imaginary: vw c v w c − − = ₂ ' cosθ and v=iV iVw c iV w c − − = ₂ ' cosθ

_{4 2 2} 2 2 2 2 ) ( ) ( ' cos w V c Vc Vw i w V wc c + − + + = θ 2 2 4 2 2 2 2 2 2 ) ( ) ( ' cos w V c Vc Vw w V wc c + − + + = θ According to Einstein w=c cosθ'=1 θ'=0 No aberration.

The classical value:

c v artg = ' θ and _{3 10}4 −1 × = ms v

θ

'=20.6"

If light speed, according to Einstein, is not additive is obvious that, in Einstein’s relativity theory, the aberration must be always zero. And we know that this is not true.

The Compton scattering is not a relativistic phenomena

The relativity theory believers brake, if it’s necessary, all mathematical rules to prove that the theory is valid as for the Compton scattering. But the truth is only one: the Compton scattering is not a relativistic phenomena.

When a photon strikes a rest electron, the photon changes of direction and its wavelength increases (its energy decreases) and the electron gets some speed:

Empirical Compton formula:

λ =λ₀ + (1−cosα)

c m

h e

Official and wrong relativistic derivation (Wikipédia) Citation:

We use that:

(Conservation of energy, where Eγ is the energy of a photon before the collision and Ee

is the energy of an electron before collision — its rest mass). The variables with a prime are used for those after the collision.

And:

(Conservation of momentum, with the pe = 0 because we assume that the electron is at rest.) We then use E = hf = pc: 0 0 hc/λ E =

λ

/ hc E=

α

2 / 2 v m E_e = _e

The cos(θ) term appears because the momenta are spatial vectors, all of which lie in a single 2D plane, thus their inner product is the product of their norms multiplied by the cosine of the angle between them.

substituting pγ with and pγ' with , we derive

Now we fill in for the energy part:

We solve this for pe':

Then we have two equations for , which we equate:

End of citation.

The introduction of cos it’s just a mathematical error. θ

Right derivation

Conservation of the energy: ₀ 2 2 1 v m E E = + _e and m_e2v2 = p_e2 e e m p E E 2 2 0 = + 2 ( 0 ) 2 E E m pe = e −

Conservation of the momentum:

p_e2 = p2 + p₀2−2pp₀cosα p2 + p₀2 −2pp₀cosα =2m_e(E₀ −E) and E₀ = p₀c and E= pc p2 + p₀2 −2pp₀cosα =2cm_e(p₀− p) and 0 0 λ h p = and λ h p= 0 0 2 0 0 0 2 ) 2 ( ) cos ( cos λ λ λ α λ α λ λ e e e e cm h cm h h cm h cm h − − − − − − =

This is the exact equation for Compton scattering. Finding, with a derivation, a empirical equation can be just a question of luck. As we prove, this equation is almost equivalent to the Compton one ( the absolute relativity introduce some, not tested, little corrections ):

α 0

p

p e

For me kg 31 10 11 . 9 × − = and 0 3 10 10m − × = λ

λ₁ =Compton..value ; λ₂ =our..value

α 10 1 10 − × λ λ₂×10−10 0 3.00000000 3.00000000 30 3.00325042 3.00325044 60 3.01213074 3.01213094 90 3.02426148 3.02426227 120 3.03639222 3.03639398 150 3.04527254 3.04527526 180 3.04852296 3.04852608

Evident errors of the theoretical basis of the relativity theory

This paper is based on the original Einstein’s book – Relativity: the special and general theory. To the book citation we have added some commentaries about the existence of several evident errors in the derivations of the basis of the relativity theory. Curiously the relativistic physicians continue to state that the Lorentz’s transformations verify the Einstein’s postulates when it’s possible to prove clearly the contrary.

We think, however, that the relativity theory is partially correct, as proves the experiments, but is necessary to reformulate its theoretical basis.

Relativity: the special and general theory

(The download of this book can be done at – Project Gutenberg) APPENDIX I

SIMPLE DERIVATION OF THE LORENTZ TRANSFORMATION (SUPPLEMENTARY TO SECTION 11)

'

For the relative orientation of the co-ordinate systems indicated in Fig. 2, the x-axes of both systems permanently coincide. In the present case we can divide the problem into parts by considering first only events which are localised on the x-axis. Any such event is represented with respect to the co-ordinate system K by the abscissa x and the time t, and with respect to the system K1 by the abscissa x' and the time t'. We require to find x' and t' when x and t are given. A light-signal, which is proceeding along the positive axis of x, is transmitted according to the equation

x = ct or

x - ct = 0 . . . (1).

Since the same light-signal has to be transmitted relative to K1 with the velocity c, the propagation relative to the system K1 will be represented by the analogous formula

x' - ct' = O . . . (2)

Those space-time points (events) which satisfy (x) must also satisfy (2). Obviously this will be the case when the relation

(x' - ct') = l (x - ct) . . . (3).

is fulfilled in general, where l indicates a constant; for, according to (3), the disappearance of (x - ct) involves the disappearance of (x' - ct').

If we apply quite similar considerations to light rays which are being transmitted along the negative x-axis, we obtain the condition (x' + ct') = µ(x + ct) . . . (4).

By adding (or subtracting) equations (3) and (4), and introducing for convenience the constants a and b in place of the constants l and µ ,

k ' k x ' x v y ' y z