Alfredo Dimas Moreira Garcia
description
Transcript of Alfredo Dimas Moreira Garcia
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Undulating Relativity
Author: Alfredo Dimas Moreira Garcia E-mail: [email protected]
ABSTRACT
The Special Theory of Relativity takes us to two results that presently are considered βinexplicableβ to many renowned scientists, to know: -The dilatation of time, and -The contraction of the Lorentz Length. The solution to these have driven the author to the development of the Undulating Relativity (UR) theory, where the Temporal variation is due to the differences on the route of the light propagation and the lengths are constants between two landmarks in uniform relative movement. The Undulating Relativity provides transformations between the two landmarks that differs from the transformations of Lorentz for: Space (x,y,z), Time (t), Speed (u
), Acceleration ( a
), Energy (E), Momentum
( p
), Force ( F
), Electrical Field ( E
), Magnetic Field ( B
), Light Frequency ( y ), Electrical Current ( J
) and
βElectrical Chargeβ ( Ο ).
From the analysis of the development of the Undulating Relativity, the following can be synthesized: - It is a theory with principles completely on physics; - The transformations are linear; - Keeps untouched the Euclidian principles; - Considers the Galileoβs transformation distinct on each referential; - Ties the Speed of Light and Time to a unique phenomenon; - The Lorentz force can be attained by two distinct types of Filed Forces, and - With the absence of the spatial contraction of Lorentz, to reach the same classical results of the special relativity rounding is not necessary as concluded on the Doppler effect. Both, the Undulating Relativity and the Special Relativity of Albert Einstein explain the experience of Michel-Morley, the longitudinal and transversal Doppler effect, and supplies exactly identical formulation to:
Aberration of zenith 2
2
c
v1/
c
vtgΞ± .
Fresnelβs formula )n
1v(1
n
cc'
2 .
Mass (m ) with velocity ( v ) = [resting mass (mo )]/2
2
1c
v .
2.cmE .
Momentum
2
21
.
cvvmop
.
Relation between momentum (p) and Energy (E) 222 .. pcmocE .
Relation between the electric field ( E
) and the magnetic field ( B
) Ec
VB
2.
Biot-Savantβs formula .ΞΌR.Ο2.
I.ΞΌΞΏB
Louis De Broglieβs wave equation v
cu;
u
xtΟΞ³2nis.aΟ(x,t)
2
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Along with the equations of transformations between two references of the UR, we get the invariance of shape to Maxwellβs equations, such as:
0.Ediv;Ρο
ΟEdiv
0.Bdiv
.
t
BERot
.t
E.μο.ΡοBRot;
t
E.μο.Ροj.μοBRot
We also get the invariance of shape to the equation of wave and equation of continuity under differential shape: Other Works: Β§9 Explaining the Sagnac Effect with the Undulating Relativity. Β§10 Explaining the experience of Ives-Stilwell with the Undulating Relativity. Β§11 Transformation of the power of a luminous ray between two referencials in the Special Theory of Relativity. Β§12 Linearity. Β§13 Richard C. Tolman. Β§14 Velocities composition. Β§15 Invariance. Β§16 Time and Frequency. Β§17 Transformation of H. Lorentz. Β§18 The Michelson & Morley experience. Β§19 Regression of the perihelion of Mercury of 7,13β. §§19 Advance of Mercuryβs perihelion of 42.79β. Β§20 Inertia. Β§20 Inertia (clarifications) Β§21 Advance of Mercuryβs perihelion of 42.79β calculated with the Undulating Relativity. Β§22 Spatial Deformation. Β§23 Space and Time Bend. Β§24 Variational Principle. Β§25 Logarithmic Spiral. Β§26 Mercury Perihelion Advance of 42.99". Β§27 Advancement of Perihelion of Mercury of 42.99β "contour Conditions". Β§28 Simplified Periellium Advance. Β§ 29 Yukawa Potential Energy βContinuationβ. Β§ 30 Energy βcontinuationβ.
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Author: Alfredo Dimas Moreira Garcia. e-mail: [email protected]
Translator: Rodolfo Marcos VenΓ’ncio
e-mail: [email protected] Undulating Relativity
Β§ 1 Transformation to space and time
The Undulating Relativity (UR) keep the principle of the relativity and the principle of Constancy of light speed, exactly like Albert Einsteinβs Special Relativity Theory defined: a) The laws, under which the state of physics systems are changed are the same, either when referred to a determined system of coordinates or to any other that has uniform translation movement in relation to the first. b) Any ray of light moves in the resting coordinates system with a determined velocity c, that is the same, whatever this ray is emitted by a resting body or by a body in movement (which explains the experience of Michel-Morley). Letβs imagine first that two observers O and Oβ (in vacuum), moving in uniform translation movement in relation to each other, that is, the observer donβt rotate relatively to each other. In this way, the observer O together with the axis x, y, and z of a system of a rectangle Cartesian coordinates, sees the observer Oβ move with velocity v, on the positive axis x, with the respective parallel axis and sliding along with the x axis while the Oβ, together with the xβ, yβ and zβ axis of a system of a rectangle Cartesian coordinates sees O moving with velocity βvβ, in negative direction towards the xβ axis with the respective parallel axis and sliding along with the xβ axis. The observer O measures the time t and the Oβ observer measures the time tβ (t β tβ). Letβs admit that both observers set their clocks in such a way that, when the coincidence of the origin of the coordinated system happens t = tβ = zero. In the instant that t = tβ = 0, a ray of light is projected from the common origin to both observers. After the time interval t the observer O will notice that his ray of light had simultaneously hit the coordinates point A (x, y, z) with the ray of the Oβ observer with velocity c and that the origin of the system of the Oβ observer has run the distance v t along the positive way of the x axis, concluding that: x2 + y2 + z2 β c2 t2 = 0 1.1 xβ = x β v t. 1.2 The same way after the time interval tβ the Oβ observer will notice that his ray of light simultaneously hit with the observer O the coordinate point A (xβ, yβ, zβ) with velocity c and that the origin of the system for the observer O has run the distance vβtβ on the negative way of the axis xβ, concluding that: xβ2 + yβ2 + zβ2 β c2 tβ2 = 0 1.3 x = xβ + vβ tβ. 1.4 Making 1.1 equal to 1.3 we have x2 + y2 + z2 β c2 t2 = xβ2 + yβ2 + zβ2 β c2 tβ2. 1.5 Because of the symmetry y = yβ end z = zβ, that simplify 1.5 in x2 β c2 t2 = xβ2 β c2 tβ2. 1.6 To the observer O xβ = x β v t (1.2) that applied in 1.6 supplies x2 β c2 t2 = (x β v t)2 β c2 tβ2 from where
tc
vx
c
vtt
22
2 21' . 1.7
To the observer Oβ x = xβ + vβ tβ (1.4) that applied in 1.6 supplies (xβ + vβ tβ)2 β c2 t2 = xβ2 β c2 tβ2 from where
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'tc
'x'v2
c
'v1'tt
22
2
. 1.8
Table I, transformations to the space and time xβ = x β v t 1.2 x = xβ + vβ tβ 1.4 yβ = y 1.2.1 y = yβ 1.4.1 zβ = z 1.2.2 z = zβ 1.4.2
tc
vx
c
vtt
22
2 21'
1.7
'
''2'1'
22
2
tc
xv
c
vtt
1.8
From the equation system formed by 1.2 and 1.4 we find
v t = vβ tβ or '' tvtv (considering t>o e tβ>0) 1.9
what demonstrates the invariance of the space in the Undulatory Relatitivy. From the equation system formed by 1.7 and 1.8 we find
tc
vx
c
v22
2 21 .
'
''2'1
22
2
tc
xv
c
v = 1. 1.10
If in 1.2 xβ = 0 then x = v t, that applied in 1.10 supplies,
2
2
1c
v .
2
2
c'v1 = 1. 1.11
If in 1.10 x = ct and xβ = c tβ then
1'
1.1
c
v
c
v. 1.12
To the observer O the principle of light speed constancy guarantees that the components ux, uy and uz of the light speed are also constant along its axis, thus
uzdt
dz
t
z,uy
dt
dy
t
y,ux
dt
dx
t
x 1.13
and then we can write
tc
vx
c
v22
2 21 =
22
2 21
c
vux
c
v . 1.14
With the use of 1.7 and 1.9 and 1.14 we can write
t
't
'v
v =
tc
vx
c
v22
2 21 =
22
2 21
c
vux
c
v . 1.15
Differentiating 1.9 with constant v and vβ, or else, only the time varying we have
''dtvdtv or dt
dt
v
v '
' , 1.16
but from 1.15 22
2 21
' c
vux
c
v
v
v then
22
2 21'
c
vux
c
vdtdt . 1.17
Being v and vβ constants, the reazons 'v
v and
t
t ' in 1.15 must also be constant because fo this the
differential of tc
vx
c
v22
2 21 must be equal to zero from where we conclude ux
dt
dx
t
x , that is exactly
the same as 1.13.
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To the observer Oβ the principle of Constancy of velocity of light guarantees that the components uβxβ, uβyβ, and uβzβ of velocity of light are also constant alongside its axis, thus
'z'u'dt
'dz
't
'z,'y'u
'dt
'dy
't
'y,'x'u
'dt
'dx
't
'x , 1.18
and with this we can write ,
'
''2'1
22
2
tc
xv
c
v =
22
2 '''2'1
c
xuv
c
v . 1.19
With the use of 1.8, 1.9, and 1.19 we can write
't
t
v
'v =
'tc
'x'v2
c
'v1
22
2
= 22
2 '''2'1
c
xuv
c
v . 1.20
Differentiating 1.9 with vβ and v constant, that is, only the time varying we have
dtvdtv '' or '
'
dt
dt
v
v , 1.21
but from 1.20 22
2 '''2'1
'
c
xuv
c
v
v
v then
22
2 '''2'1'
c
xuv
c
vdtdt . 1.22
Being vβ and v constant the divisions v
v' and
't
t in 1.20 also have to be constant because of this the
differential of '
''2'1
22
2
tc
xv
c
v must be equal to zero from where we conclude ''
'
'
'
'xu
dt
dx
t
x , that is
exactly like to 1.18. Replacing 1.14 and 1.19 in 1.10 we have
22
2 21
c
vux
c
v .
22
2 '''2'1
c
xuv
c
v = 1. 1.23
To the observer O the vector position of the point A of coordinates (x,y,z) is
kzjyixR
, 1.24
and the vector position of the origin of the system of the observer Oβ is
kjivtoR
00' ivtoR
' . 1.25
To the observer Oβ, the vector position of the point A of coordinates (xβ,yβ,zβ) is
kzjyixR
'''' , 1.26
and the vector position of the origin of the system of the observer O is
kjitvoR
00''' itvoR
''' . 1.27
Due to 1.9, 1.25, and 1.27 we have, oRoR ''
. 1.28 As 1.24 is equal to 1.25 plus 1.26 we have
'' RoRR
'' oRRR
. 1.29
Applying 1.28 in 1.29 we have, oRRR ''
. 1.30
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To the observer O the vector velocity of the origin of the system of the observer Oβ is
ivvk0j0ivdt
'oRdv
. 1.31
To the observer Oβ the vector velocity of the origin of the system of the observer O is
ivvkjivdt
oRdv
''00'
'
'' . 1.32
From 1.15, 1.20, 1.31, and 1.32 we find the following relations between v
and 'v
22
2 '''2'1
'
c
xuv
c
v
vv
1.33
22
2 21
'
c
vux
c
v
vv
. 1.34
Observation: in the table I the formulas 1.2, 1.2.1, and 1.2.2 are the components of the vector 1.29 and the formulas 1.4, 1.4.1, and 1.4.2 are the components of the vector 1.30.
Β§2 Law of velocity transformations u
and 'u
Differentiating 1.29 and dividing it by 1.17 we have
K
vu
c
vux
c
v
vuu
c
vux
c
vdt
oRdRd
dt
Rd
22
2
22
2 21
'2
1
'
'
'. 2.1
Differentiating 1.30 and dividing it by 1.22 we have
'
''
'''2'1
''
'''2'1'
''
22
2
22
2 K
vu
c
xuv
c
v
vuu
c
xuv
c
vdt
oRdRd
dt
Rd
. 2.2
Table 2, Law of velocity transformations u
and 'u
K
vuu
'
2.1
'
''
K
vuu
2.2
K
vuxxu
''
2.3
'
'''
K
vxuux
2.4
K
uyyu ''
2.3.1
'
''
K
yuuy
2.4.1
K
uzzu ''
2.3.2
'
''
K
zuuz
2.4.2
K
v'v
1.15
'K
'vv
1.20
22
2 21
c
vux
c
vK
2.5
22
2 '''2'1'
c
xuv
c
vK
2.6
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Multiplying 2.1 by itself we have
22
2
22
2
21
21
'
c
vux
c
v
u
vux
u
vu
u
. 2.7
If in 2.7 we make u = c then uβ = c as it is required by the principle of constancy of velocity of light. Multiplying 2.2 by itself we have
22
2
22
2
'''2'1
'
'''2
'
'1'
c
xuv
c
v
u
xuv
u
vu
u
. 2.8
If in 2.8 we make uβ = c then u = c as it is required by the principle of constancy of velocity of light.
If in 2.3 we make ux = c then c
c
vc
c
v
vcxu
22
2 21
'' as it is required by the principle of constancy of
velocity of light.
If in 2.4 we make uβxβ = c then c
c
cv
c
v
vcux
22
2 '2'1
' as it is required by the principle of constancy of
velocity of light. Remodeling 2.7 and 2.8 we have
22
2 21
c
vux
c
v =
2
2
2
2
'1
1
c
u
c
u
. 2.9
22
2 '''2'1
c
xuv
c
v =
2
2
2
2
1
'1
c
u
c
u
. 2.10
The direct relations between the times and velocities of two points in space can be obtained with the equalities vuxxuu 0''0'
coming from 2.1, that applied in 1.17, 1.22, 1.20, and 1.15 supply
2
222
2
1
'21'
c
v
dtdt
c
vv
c
vdtdt
, 2.11
2
222
2
c
'v1
dt'dt
c
0'v2
c
'v1'dtdt
, 2.12
2
2
22
2
1021c'v
'vv
c'v
c'v
'vv
, 2.13
2
2
22
2
121cv
v'v
cvv
cv
v'v
. 2.14
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Aberration of the zenith To the observer Oβ along with the star uβxβ = 0, uβyβ = c and uβzβ = 0, and to the observer O along with the Earth we have the conjunct 2.3
vux
c
vux
c
v
vux
22
2 21
0 , 2
2
22
21
21
c
vcuy
c
vv
c
v
uyc
, uz = 0,
c0c
v1cvuzuyuxu 2
2
2
22222
exactly as foreseen by the principle of relativity.
To the observer O the light propagates in a direction that makes an angle with the vertical axis y given by
2
2
2
2
c
v1
v/c
c
v1.c
v
uy
uxtangΞ±
2.15
that is the aberration formula of the zenith in the special relativity . If we inverted the observers we would have the conjunct 2.4
v'u'x'
c
2v'u'x'
c
v'1
v'u'x'0
22
2
,
2
2
22
2 c
v'1cu'y'
c
v'2v'
c
v'1
u'y'c
, uβzβ=0,
c0c
v'1cv'u'z'u'y'u'x'u' 2
2
2
22222
2
2
2
2
c
v'1
v'/c
c
v'1c.
v'
u'y'
u'x'tangΞ±
2.16
that is equal to 2.15, with the negative sign indicating the contrary direction of the angles.
Fresnelβs formula Considering in 2.4, ncxu /'' the velocity of light relativily to the water, vv ' the velocity of water in
relation to the apparatus then 'cux will be the velocity of light relatively to the laboratory
nc
v2
c
v
2
11v
n
c
nc
v2
c
v1v
n
c
nc
v2
c
v1
vn/c
c
n/vc2
c
v1
vn/c'c
2
22
1
2
2
2
2
22
2
Ignoring the term 22 c/v we have
nc
v
n
vv
n
c
nc
vv
n
cc
2
21'
and ignoring the term ncv /2 we have the Fresnelβs formula
22
11'
nv
n
c
n
vv
n
cc . 2.17
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Doppler effect
Making 2222 zyxr and 2222 z'y'x'r' in 1.5 we have 222222 t'cr'tcr or
ctr
ct'r'ct'r'ctr
replacing then ctr , '' ctr and 1.7 we find tc
2vx
c
v1ct'r'ctr
22
2
as '
'
k
w
k
wc then
tc
2vx
c
v1t'w'r'k'
k'
1wtkr
k
122
2
where to attend the principle of relativity
we will define tc
2vx
c
v1kk'
22
2
2.18
Resulting in the expression t'w'r'k'wtkr symmetric and invariable between the observers.
To the observer O an expression in the formula of wtkrftr,Ο 2.19
represents a curve that propagates in the direction of R
. To the observer Oβ an expression in the formula of
t'w'r'k'f',t'r'Ο' 2.20
represents a curve that propates in the direction of 'R
.
Applying in 2.18 Ξ»
2Οk ,
Ξ»'
2Οk' , 1.14, 1.19, 1.23, 2.5, and 2.6 we have
K
λλ' e
K'
Ξ»'Ξ» , 2.21
that applied in 'Ξ»y'yΞ»c supply, Ky'y and 'K'yy . 2.22
Considering the relation of Planck-Einstein between energy ( E ) and frequency ( y ), we have to the
observer O hyE and to the observer Oβ 'hy'E that replaced in 2.22 supply
KEE ' and '' KEE . 2.23 If the observer O that sees the observer Oβ moving with velocity v in a positive way to the axis x, emits waves of frequency y and velocity c in a positive way to the axis x then, according to 2.22 and cux the
observer Oβ will measure the waves with velocity c and frequency
cvy'y 1 , 2.24
that is exactly the classic formula of the longitudinal Doppler effect. If the observer Oβ that sees the observer O moving with velocity βvβ in the negative way of the axis xβ, emits waves of frequency 'y and velocity c, then the observer O according to 2.22 and 'v'x'u will measure
waves of frequency y and velocity c in a perpendicular plane to the movement of Oβ given by
2
2
c
v'1Ξ³'Ξ³ , 2.25
that is exactly the formula of the transversal Doppler effect in the Special Relativity.
Β§3 Transformations of the accelerations a
and 'a
Differentiating 2.1 and dividing it by 1.17 we have
222 K
ax
c
vvu
K
a'a
Kdt
KK/dux
c
vvu
Kdt
K/ud
'dt
'ud
. 3.1
Differentiating 2.2 and dividing it by 1.22 we have
222 'K
'x'a
c
'v'v'u
'K
'aa
'K'dt
'K'K/'x'du
c
'v'v'u
'K'dt
'K/'ud
dt
ud
. 3.2
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Table 3, transformations of the accelerations a
and 'a
22 K
ax
c
vvu
K
a'a
3.1
22 'K
'x'a
c
'v'v'u
'K
'aa
3.2
22 K
ax
c
vvux
K
ax'x'a
3.3
22 'K
'x'a
c
'v'v'x'u
'K
'x'aax
3.4
22 K
ax
c
vuy
K
ay'y'a
3.3.1 22 'K
'x'a
c
'v'y'u
'K
'y'aay
3.4.1
22 K
ax
c
vuz
K
az'z'a
3.3.2 22 'K
'x'a
c
'v'z'u
'K
'z'aaz
3.4.2
K
a'a
3.8
'K
'aa
3.9
22
2
c
vux2
c
v1K
3.5
22
2
c
'x'u'v2
c
'v1'K
3.6
From the tables 2 and 3 we can conclude that if to the observer O zeroa.u
and 2222 uzuyuxc ,
then it is also to the observer Oβ zero'a'.u
and 2222 z'u'y'u'x'u'c , thus u
is perpendicular to a
and 'u
is perpendicular to 'a
as the vectors theory requires. Differentiating 1.9 with the velocities and the times changing we have, 'dt'v'dv'tvdttdv , but
considering 1.16 we have, 'dv'ttdv'dt'vvdt 3.7
Where replacing 1.15 and dividing it by 1.17 we have, dtK
dv
'dt
'dv or
K
a'a . 3.8
We can also replace 1.20 in 3.7 and divide it by 1.22 deducing
'K'dt
'dv
dt
dv or
'K
'aa . 3.9
The direct relations between the modules of the accelerations a and aβ of two points in space can be obtained with the vuxvu0'x'a0'x'u0'u
coming from 2.1, that applied in 3.8 and
3.9 supply
2
2
22
2
c
v1
a
c
vv2
c
v1
a'a
and
2
2
22
2
c
'v1
'a
c
0'v2
c
'v1
'aa
. 3.10
That can also be reduced from 3.1 and 3.2 if we use the same equalities
vuxvu0'x'a0'x'u0'u
coming from 2.1.
Β§4 Transformations of the Moments p
and 'p
Defined as uump
and 'uu'm''p
, 4.1
where um and 'u'm symbolizes the function masses of the modules of velocities uu
and 'uu'
.
We will have the relations between um and u'm' and the resting mass mo, analyzing the elastic
collision in a plane between the sphere s that for the observer o moves alongside the axis y with velocity uy = w and the sphere sβ that for the observer Oβ moves alongside the axis yβ with velocity uβyβ = -w. The spheres while observed in relative resting are identical and have the mass mo. The considered collision is symmetric in relation to a parallel line to the axis y and yβ passing by the center of the spheres in the moment of. Collision. Before and after the collision the spheres have velocities observed by O and Oβ according to the following table gotten from table 2
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Sphere Observer O Observer Oβ Before
s
zerouxs , wuys 'vs'x'u ,
2
2
c
'v1ws'y'u
Collision
sβ v'uxs ,
2
2
c
v1w'uys
zero's'x'u , w's'y'u
After
s
zerouxs , wuys 'vs'x'u ,
2
2
c
'v1ws'y'u
Collision
sβ v'uxs ,
2
2
c
v1w'uys
zero's'x'u , w's'y'u
To the observer O, the principle of conservation of moments establishes that the moments uxumpx
and uyumpy , of the spheres s and sβ in relation to the axis x and y, remain constant before and after
the collision thus for the axis x we have
uxs'uys'uxs'muxsuysuxsmuxs'uys'uxs'muxsuysuxsm 22222222 ,
where replacing the values of the table we have
vc
v1wvmv
c
v1wvm
2
2
22
2
2
22
from where we conclude that ww ,
and for the axis y
uys'uys'uxs'muysuysuxsmuys'uys'uxs'muysuysuxsm 22222222 ,
where replacing the values of the table we have
2
22
2
22
2
22
2
22
c
v1w
c
v1wvmwwm
c
v1w
c
v1wvmwwm
,
simplifying we have
2
2
2
222
c
v1
c
v1wvmwm
, where when 0w becomes
2
22
2
2
2
2
222
c
v1
0mvm
c
v1vm0m
c
v1
c
v10vm0m
,
but 0m is equal to the resting mass mo thus
2
2
0
c
v1
mvm
, with a relative velocity 2
2
0
c
u1
mumuv
4.2
that applied in 4.1 supplies 2
2
0
c
u1
umuump
. 4.1
With the same procedures we would have for the Oβ observer
12/172
2
2
0
c
'u1
m'u'm
4.3
and 2
2
0
c
u'1
'um'uu'm''p
. 4.1
Simplifying the simbology we will adopt 2
2
0
c
u1
mumm
4.2
and 2
2
0
c
'u1
m'u'm'm
4.3
that simplify the moments in ump
and 'um''p
. 4.1
Applying 4.2 and 4.3 in 2.9 and 2.10 we have
'K'mmc
'x'u'v2
c
'v1'mm
22
2
and Km'mc
vux2
c
v1m'm
22
2
. 4.4
Defining force as Newton we have dt
umd
dt
and
'dt
'u'md
'dt
'pd'F
, with this we can define then
kinetic energy kk 'E,E as
u
0
2u
0
u
0
u
0
k mududmuu.umdRd.dt
umdRd.FE
,
and
u'
0
2u'
0
u'
0
u'
0
k du'u'm'dm'u''u.'um'd'Rd.dt'
'um'd'Rd'.FE'
.
Remodeling 4.2 and 4.3 and differentiating we have dmcmududmucmumcm 22220
2222 and
dm'cdu'u'm'dm'u'cmu'm'cm' 22220
2222 , that applied in the formulas of kinetic energy
supplies m
m
02
022
k
0
EEcmmcdmcE and 'm
m
02
022
k
0
E'Ecmc'm'dmc'E , 4.5
where 2mcE and 2c'm'E 4.6
are the total energies as in the special relativity and 2oo cmE 4.7
the resting energy. Applying 4.6 in 4.4 we have exactly 2.23. From 4.6, 4.2, 4.3, and 4.1 we find
222o pcmcE and 222
o p'cmcE' 4.8
identical relations to the Special Relativity. Multiplying 2.1 and 2.2 by mo we get
13/172
vc
Ep'pvmum'u'm
c
u1
vm
c
u1
um
c
u'1
'um2
2
2
o
2
2
o
2
2
o
4.9
and 'vc
E''pp'v'm'u'mum
c
u'1
'vm
c
u'1
'um
c
u1
um2
2
2
o
2
2
o
2
2
o
. 4.10
Table 4, transformations of moments p
and 'p
vc
Ep'p
2
4.9 'v
c
E''pp
2
4.10
vc
Epx'x'p
2
4.11 v'
c
E'x'p'px
2
4.12
py'y'p 4.11.1 'y'ppy 4.12.1
pz'z'p 4.11.2 'z'ppz 4.12.2
KE'E 2.23 K'E'E 2.23
2
2
0
c
u1
mumm
4.2
2
2
0
c
u'1
mu'm'm'
4.3
Km'm 4.4 'K'mm 4.4
ok EEE 4.5 ok EE'E' 4.5
2mcE 4.6 2cm'E' 4.6
2oo cmE 4.7 2
oo cmE 4.7
222o pcmcE
4.8 222o p'cmcE'
4.8
Wave equation of Louis de Broglie
The observer Oβ associates to a resting particle in its origin the following properties: -Resting mass mo
-Time ot't
-Resting Energy 2oo cmE
-Frequency hcm
hE
y2
ooo
-Wave function ooo ty2ΟasenΟ with a = constant.
The observer O associates to a particle with velocity v the following:
-Mass 2
2
o
c
v1
mvmm
(from 4.2 where vu )
-Time
2
2
o
22
2
o
c
v1
t
c
vv2
c
v1
tt
(from 1.7 with vux and ot't )
-Energy
2
2
2o
2
2
o
c
v1
cm
c
v1
EE
(from 2.23 with vux and oE'E )
14/172
-Frequency
2
2
2o
2
2
o
c
v1
/hcm
c
v1
yy
(from 2.22 with vux and oy'y )
-Distance x = vt (from 1.2 with xβ = 0)
-Wave function
ux
tyΟ2asenc
v1t
c
v1yΟ2asenty2asenΟ
2
2
2
2
oo with v
cu
2
-Wave length ph
Ξ»phy
pE
vc
Ξ»yu2
(from 4.9 with 0p'p o
)
To go back to the Oβ observer referential where 0x'u'0'u
, we will consider the following variables: -Distance x = vβtβ (from 1.4 with xβ = 0)
-Time 2
2
22
2
c
v'1t'
c
v'02
c
v'1t't (from 1.8 with 0'x'u )
-Frequency 2
2
c
v'1'yy (from 2.22 with 0'x'u )
-Velocity
2
2
c
v'1
v'v
(de 2.13)
that applied to the wave function supplies
t''yasen
c
v'1c
t'v'
c
v'1t'
c
v'1'yΟ2asen
c
vxtyΟ2asenΟ'
2
22
2
2
2
2
2
22
,
but as ot't and oyy' then oΟΟ' .
Β§5 Transformations of the Forces F
and 'F
Differentiating 4.9 and dividing by 1.17 we have
222 c
vu.FF
K
1'F
c
v
dt
dEF
K
1'F
c
v
Kdt
dE
Kdt
pd
dt'
'pd
. 5.1
Differentiating 4.10 and dividing by 1.22 we have
222 c
'v'u'.F'F
K'
1F
c
'v
dt'
dE''F
K'
1F
c
'v
K'dt'
dE'
K'dt'
'pd
dt
pd
. 5.2
From the system formed by 5.1 and 5.2 we have
dt'
dE'
dt
dE or 'u'.Fu.F
, 5.3
that is an invariant between the observers in the Undulating .Relativity.
15/172
Table 5, transformations of the Forces F
and 'F
2c
vu.FF
K
1'F
5.1
2c
'v'u'.F'F
K'
1F
5.2
2c
vu.FFx
K
1x'F'
5.4
2c
v''u'.Fx'F'
'K
1Fx
5.5
KFy/y'F' 5.4.1 K'/y'F'Fy 5.5.1
KFz/z'F' 5.4.2 K'/z'F'Fz 5.5.2
dt
dE
'dt
'dE
5.3
'u'.Fu.F
5.3
Β§6 Transformations of the density of charge Ο , Ο' and density of current J
and 'J
Multiplying 2.1 and 2.2 by the density of the resting electric charge defined as o
o dv
dqΟ we have
vΟJ'JvΟuΟ'uΟ'
c
u1
vΟ
c
u1
uΟ
c
u'1
'uΟ
2
2
o
2
2
o
2
2
o
6.1
and 'vΟ''JJ'vΟ''uΟ'uΟ
c
u'1
'vΟ
c
u'1
'uΟ
c
u1
uΟ
2
2
o
2
2
o
2
2
o
. 6.2
Table 6, transformations of the density of charges Ο , Ο' and density of current J
and 'J
vΟJ'J
6.1 'vΟ''JJ
6.2
ΟvJxx'J' 6.3 'Ο'vJ'x'Jx 6.4
Jy'y'J 6.3.1 'y'JJy 6.4.1
Jz'z'J 6.3.2 'z'JJz 6.4.2
uΟJ
6.5 'u'Ο'J
6.6
2
2
o
c
u1
ΟΟ
6.7
2
2
o
c
u'1
ΟΟ'
6.8
KΟΟ' 6.9 Ξ'Ο'Ο 6.10
From the system formed by 6.1 and 6.2 we had 6.9 and 6.10.
Β§7 Transformation of the electric fields E
, 'E
and magnetic fields B
, 'B
Applying the forces of Lorentz BuEqF
and 'B'u'Eq'F
in 5.1 and 5.2 we have
2c
vu.BuEqBuEq
K
1'B'u'Eq
and
2c
'v'u.'B'u'Eq'B'u'Eq
K'
1BuEq
, that simplified become
2c
vu.EBuE
K
1'B'u'E
and
2c
'v'u'.E'B'u'E
K'
1BuE
from
where we get the invariance of 'u'Eu.E
between the observers as a consequence of 5.3 and the following components of each axis
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222 c
Ezuzv
c
Eyuyv
c
ExuxvuzByuyBzEx
K
1y'B'z'u'z'B'y'u'x'E' 7.1
zBxuxBzuyEK
1'zB''xu''xB''zu''yE' 7.1.1
xByuyBxuzEK
1'xB''yu''yB''xu''zE' 7.1.2
2c
v'z'u'z'E'2c
v'y'u'y'E'2c
v'x'u'x'E'y'B'z'u'z'B'y'u'x'E'
K'
1uzByuyBzEx 7.2
'zB''xu''xB''zu''yE'K'
1zBxuxBzuyE 7.2.1
'xB''yu''yB''xu''zE'K'
1xByuyBxuzE 7.2.2
To the conjunct 7.1 and 7.2 we have two solutions described in the tables 7 and 8.
Table 7, transformations of the electric fields E
, 'E
and magnetic fields B
e 'B
2c
vux1
K
Exx'E'
7.3
2c
x'u'v'1
K'
x'E'Ex
7.4
K
vBz
c
vux
c
v1
K
Eyy'E'
22
2
7.3.1
K'
z'B'v'
c
x'u'v'
c
v'1
K'
y'E'Ey
22
2
7.4.1
K
vBy
c
vux
c
v1
K
Ezz'E'
22
2
7.3.2
K'
y'B'v'
c
x'u'v'
c
v'1
K'
z'E'Ez
22
2
7.4.2
Bxx'B' 7.5 x'B'Bx 7.6
Ezc
vByy'B'
2
7.5.1 z'E'
c
v'y'B'By
2
7.6.1
Eyc
vBzz'B'
2
7.5.2 y'E'
c
v'z'B'Bz
2
7.6.2
KEyy'E' 7.7 K'y'E'Ey 7.8
KEzz'E' 7.7.1 K'z'E'Ez 7.8.1
Ezc
uxBy
2
7.9 z'E'
c
x''uy'B'
2
7.10
Eyc
uxBz
2
7.9.1 y'E'
c
x''uz'B'
2
7.10.1
Table 8, transformations of the electric fields E
, 'E
and magnetic fields B
e 'B
2c
vu.EEx
K
1x'E'
7.11
2c
v''u'.Ex'E'
K'
1Ex
7.12
vBzEyK
1E'y'
7.11.1 z'B'v'y'E'
K'
1Ey
7.12.1
vByEzK
1E'z'
7.11.2 y'B'v'z'E'
K'
1Ez
7.12.2
Bxx'B' 7.13 x'B'Bx 7.14
Byy'B' 7.13.1 y'B'By
7.14.1
Bzz'B' 7.13.2 z'B'Bz 7.14.2
17/172
Relation between the electric field and magnetic field
If an electric-magnetic field has to the observer Oβ the naught magnetic component zero'B
and the
electric component 'E
. To the observer O this field is represented with both components, being the magnetic field described by the conjunct 7.5 and hΓ‘s as components
zeroBx , 2c
vEzBy ,
2c
vEyBz , 7.15
that are equivalent to Evc
1B
2
. 7.16
Formula of Biot-Savart
The observer Oβ associates to a resting electric charge, uniformly distributed alongside its axis xβ the following electric-magnetic properties:
-Linear density of resting electric charge 'dx
dqΟo
-Naught electric current zero'I
-Naught magnetic field zero'uzero'B
-Radial electrical field of module RΟΞ΅2
Ο'z'E'y'E'E
o
o22 at any point of radius 22 z'y'R with
the component zerox'E' . To the observer O it relates to an electric charge uniformly distributed alongside its axis with velocity vux to which it associates the following electric-magnetic properties:
-Linear density of the electric charge
2
2
o
c
v1
ΟΟ
(from 6.7 with u = v)
-Electric current
2
2
o
c
v1
vΟΟvI
-Radial electrical field of module
2
2
c
v1
E'E
(according to the conjuncts 7.3 and 7.5 with
zero'uzero'B
and vux )
-Magnetic field of components zeroBx , 2c
vEzBy ,
2c
vEyBz and module
R2
I
RΟΞ΅2
Ο
c
v1
1
c
v
c
v1
'E
c
v
c
vEB o
o
o
2
22
2
222
where 2
oo c
1
, being in the vectorial form
uR2
IB o
7.17
where u
is a unitary vector perpendicular to the electrical field E
and tangent to the circumference that
passes by the point of radius 22 zyR because from the conjunct 7.4 and 7.6 zeroB.E
.
18/172
Β§8 Transformations of the differential operators Table 9, differential operators
tc
v
xx' 2
8.1
t'c
v'
x'x 2
8.2
yy'
8.1.1
y'y
8.2.1
zz'
8.1.2
z'z
8.2.2
ttc
vx
c
v1
K
1
xK
v
t' 22
2
8.3
t't'c
x'v'
c
v'1
K'
1
x'K'
v'
t 22
2
89.4
From the system formed by 8.1, 8.2, 8.3, and 8.4 and with 1.15 and 1.20 we only find the solutions
otc
x/t
x 2
and ot'c
/t'x'
x' 2
. 8.5
From where we conclude that only the functions Ο (2.19) and Ο' (2.20) that supply the conditions
ot
Ο
c
x/t
x
Ο2
and ot'
Ο'
c
/t'x'
x'
Ο'2
, 8.6
can represent the propagation with velocity c in the Undulating Relativity indicating that the field propagates with definite velocity and without distortion being applied to 1.13 and 1.18. Because of symmetry we can also write to the other axis
ot
Ο
c
/ty
y
Ο2
, ot'
Ο'
c
/t''y
'y
Ο'2
and ot
Ο
c
/tz
z
Ο2
, ot'
Ο'
c
/t''z
'z
Ο'2
. 8.7
From the transformations of space and time of the Undulatory Relativity we get to Jacobβs theorem
K
c
vux1
z,ty,x,
,t'z',y',x'J
2
and
K'
c
x'u'v'1
t',z',y',x'
tz,y,x,J'
2
, 8.8
variables with ux and uβxβ as a consequence of the principle of contancy of the light velocity but are equal ais
J'J and will be equal to one 1J'J when cx'u'ux .
Invariance of the wave equation
The wave equation to the observer Oβ is
zerot'c
1
z'y'x' 2
2
22
2
2
2
2
2
where applying to the formulas of tables 9 and 1.13 we get
zerotc
xuv
c
v1
K
1
xK
v
c
1
zytc
v
x
2
22
2
22
2
2
22
2
from where we find
19/172
zerotc
v
tc
uxv
tc
uxv2
tc
vux2
tc
v2
txc
uxv2
txc
v2
txc
v2
xc
v
tc
uxv2
tc
v
tc
v
txc
uxv4
txc
v2
txc
v2
tc
1
zK
yK
xK
2
2
6
4
2
2
6
22
2
2
6
3
2
2
42
2
4
22
4
22
4
32
22
2
2
2
2
2
6
3
2
2
6
4
2
2
4
22
4
22
4
32
22
2
22
2
2
2
2
2
that simplifying supplies
zerotc
uxv
tc
vux2
tc
v
xc
v
txc
uxv2
tc
1
zK
yK
xK
2
2
6
22
2
2
42
2
4
2
2
2
2
22
4
2
2
2
22
2
2
2
2
2
where reordering the terms we find
zerotc
ux
txc
ux2
xc
v
tc
1
c
vux2
c
v1
zK
yK
xK
2
2
4
22
22
2
2
2
2
2
222
2
2
2
2
2
2
2
8.9
but from 8.5 and 1.13 we have
zerotc
ux
txc
ux2
xtc
xu
xo
tc
x/t
x 2
2
4
22
22
22
22
that applied in 8.9 supplies the wave equation to the observer O zerotc
1
zyx 2
2
22
2
2
2
2
2
. 8.10
To return to the referential of the observer Oβ we will apply 8.10 to the formulas of tables 9 and 1.18, getting
zerot'c
x''uv'
c
v'1
'K
1
x'K'
v'
c
1
'z'yt'c
v'
x'
2
22
2
22
2
2
22
2
from where we find
zero'tc
'v
'tc
'x'u'v
'tc
'x'u'v
'tc
'x'u'v2
'tc
'v2
't'xc
'x'u'v2
't'xc
'v2
't'xc
'v2
'xc
'v
'tc
'x'u'v2
'tc
'v
'tc
'v
't'xc
'x'u'v4
't'xc
'v2
't'xc
'v2
'tc
1
'z'K
'y'K
'x'K
2
2
6
4
2
2
6
22
2
2
6
3
2
2
42
2
4
22
4
22
4
32
22
2
2
2
2
2
6
3
2
2
6
4
2
2
4
22
4
22
4
32
22
2
22
2
2
2
2
2
that simplifying supplies
zero'tc
'x'u'v
'tc
'x'u'v2
'tc
'v
'xc
'v
't'xc
'x'u'v2
'tc
1
'z'K
'y'K
'x'K
2
2
6
22
2
2
42
2
4
2
2
2
2
22
4
2
2
2
22
2
2
2
2
2
where reordering the terms we find
zero'tc
'x'u
't'xc
'x'u2
'xc
'v
'tc
1
c
'x'u'v2
c
'v1
'z'K
'y'K
'x'K
2
2
4
22
22
2
2
2
2
2
222
2
2
2
2
2
2
2
but from 8.5 and 1.18 we have
zero'tc
'x'u
't'xc
'x'u2
'xt'c
x''u
x'o
t'c
/t'x'
x' 2
2
4
22
22
22
22
that replaced in the reordered equation supplies the wave equation to the observer Oβ.
Invariance of the Continuity equation The continuity equation in the differential form to the observer Oβ is
zeroz'
Jz'
y'
Jy'
x'
Jx'
t'
Ο'zero'J.
t'
Ο'
8.11
where replacing the formulas of tables 6, 9, and 1.13 we get
zeroz
Jz
y
JyΟvJx
tc
v
xKΟ
tc
vux
c
v1
K
1
xK
v222
2
20/172
making the operations we find
zeroz
Jz
y
Jy
t
Ο
c
v
x
Οv
t
Jx
c
v
x
Jx
t
Ο
c
vux
t
Ο
c
v
t
Ο
x
Οv2
2
222
2
that simplifying supplies
zeroz
Jz
y
Jy
t
Jx
c
v
x
Jx
t
Ο
c
vux
t
Ο22
where applying ΟuxJx with ux constant we get
zero
z
Jz
y
Jy
x
Jx
t
Οzero
z
Jz
y
Jy
t
ux
c
v
x
Jx
t
Ο
c
vux
t
Ο22
8.12
that is the continuity equation in the differential form to the observer O. To get again the continuity equation in the differential form to the observer Oβ we will replace the formulas of tables 6, 9, and 1.18 in 8.12 getting
zeroz'
z'J'
y'
y'J'v'Ο'x'J'
t'c
v'
x'K'Ο'
t'c
x'u'v'
c
v'1
'K
1
x''K
v'222
2
making the operations we find
zeroz'
z'J'
y'
y'J'
t'
Ο'
c
v'
x'
Ο'v'
t'
x'J'
c
v'
x'
x'J'
t'
Ο'
c
x'u'v'
t'
Ο'
c
v'
t'
Ο'
x'
Ο'v'2
2
222
2
that simplifying supplies
zeroz'
z'J'
y'
y'J'
t'
x'J'
c
v'
x'
x'J'
t'
Ο'
c
x'u'v'
t'
Ο'22
where applying x'u'Ο'x'J' with uβxβ constant we get
zero
z'
z'J'
y'
y'J'
x'
x'J'
t'
Ο'zero
z'
z'J'
y'
y'J'
t'
'x'u'
c
v'
x'
x'J'
t'
Ο'
c
x'u'v'
t'
Ο'22
that is the continuity equation in the differential form to the observer Oβ.
Invariance of Maxwellβs equations
That in the differential form are written this way With electrical charge To the observer O To the observer Oβ
oz
Ez
y
Ey
x
Ex
8.13
o
'
'z
'z'E
'y
'y'E
'x
'x'E
8.14
0z
Bz
y
By
x
Bx
8.15 0
'z
'z'B
'y
'y'B
'x
'x'B
8.16
t
Bz
y
Ex
x
Ey
8.17
't
'z'B
'y
'x'E
'x
'y'E
8.18
t
Bx
z
Ey
y
Ez
8.19
't
'x'B
'z
'y'E
'y
'z'E
8.20
t
By
x
Ez
z
Ex
8.21
't
'y'B
'x
'z'E
'z
'x'E
8.22
t
EzJz
y
Bx
x
Byooo
8.23
't
'z'E'z'J
'y
'x'B
'x
'y'Booo
8.24
21/172
t
ExJx
z
By
y
Bzooo
8.25
't
'x'E'x'J
'z
'y'B
'y
'z'Booo
8.26
t
EyJy
x
Bz
z
Bxooo
8.27
't
'y'E'y'J
'x
'z'B
'z
'x'Booo
8.28
Without electrical charge zero' and zero'JJ
To the observer O To the observer Oβ
0z
Ez
y
Ey
x
Ex
8.29 0
'z
'z'E
'y
'y'E
'x
'x'E
8.30
0z
Bz
y
By
x
Bx
8.31 0
'z
'z'B
'y
'y'B
'x
'x'B
8.32
t
Bz
y
Ex
x
Ey
8.33
't
'z'B
'y
'x'E
'x
'y'E
8.34
t
Bx
z
Ey
y
Ez
8.35
't
'x'B
'z
'y'E
'y
'z'E
8.36
t
By
x
Ez
z
Ex
8.37
't
'y'B
'x
'z'E
'z
'x'E
8.38
t
Ez
y
Bx
x
Byoo
8.39
't
'z'E
'y
'x'B
'x
'y'Boo
8.40
t
Ex
z
By
y
Bzoo
8.41
't
'x'E
'z
'y'B
'y
'z'Boo
8.42
t
Ey
x
Bz
z
Bxoo
8.43
't
'y'E
'x
'z'B
'z
'x'Boo
8.44
2oo c
1
8.45
We demonstrate the invariance of the Law of Gauss in the differential form that for the observer Oβ is
o
'
'z
'z'E
'y
'y'E
'x
'x'E
8.14
where replacing the formulas from the tables 6, 7, 9, and 1.18, and considering uβxβ constant, we get
o22
2
22
2
22
Ξ΅
ΞΟ
K
vBy
c
vux
c
v1
K
Ez
z
K
vBz
c
vux
c
v1
K
Ey
yc
vux1
K
Ex
tc
v
x
making the products, summing and subtracting the term x
Ex
c
v2
2
, we find
o2
2
2
2
22
2
22
2
4
2
22
ΟK
x
Ex
c
v
x
Ex
c
v
z
Byv
z
Ez
c
vux
z
Ez
c
v
z
Ez
y
Bzv
y
Ey
c
vux
y
Ey
c
v
y
Ey
t
Ex
c
uxv
x
Ex
c
vux
t
Ex
c
v
x
Ex
that reordering results
o22
2
222
2 ΟK
c
vux
c
v1
z
Ez
y
Ey
x
Ex
t
Ex
c
1
z
By
y
Bzv
t
Ex
c
ux
x
Ex
c
v
22/172
where the first parentheses is 8.5 and because of this equal to zero , the second blank is equal to
2
ooo c
uxvuxvJxv
gotten from 8.25 and 8.45 resulting in
2o
2o
22
2
o22
2
c
vuxΟ
c
vuxΟ
c
vux
c
v1
Ο
c
vux
c
v1
z
Ez
y
Ey
x
Ex
from where we get oz
Ez
y
Ey
x
Ex
8.13
that is the Law of Gauss in the differential form to the observer O. To make the inverse we will replace in 8.13 the formulas of the tables 6, 7, 9, and 1.13, and considering ux constant, we get
o22
2
22
2
22
Ξ΅
'Ξ'Ο
'K
'y'B'v
c
'x'u'v
c
'v1
'K
'z'E
'z
'K
'z'B'v
c
'x'u'v
c
'v1
'K
'y'E
'yc
'x'u'v1
'K
'x'E
'tc
'v
'x
making the products, adding and subtracting the term 'x
'x'E
c
'v2
2
, we get
o2
2
2
2
22
2
22
2
4
2
22
'K'Ο
'x
'x'E
c
'v
'x
'x'E
c
'v
'z
y'B'v'
'z
'z'E
c
x'u'v'
'z
'z'E
c
v'
'z
'z'E
'y
z'B'v'
'y
'y'E
c
x'u'v'
'y
'y'E
c
v'
'y
'y'E
t'
'x'E
c
x'u'v'
'x
'x'E
c
x'u'v'
t'
'x'E
c
v'
'x
'x'E
that reordering results in
o22
2
222
2
'K'Ο
c
x'u'v'
c
'v1
'z
''z'E
'y
'y'E
x'
'x'E
t'
'x'E
c
1
'z
y'B'
'y
z'B''v
t'
'x'E
c
x'u'
'x
'x'E
c
v'
where the first blank is 8.5 and because of this equals to zero, the second blank is equal to
2
ooo c
'x'u''v'x'u''v'x'J'v
gotten from 8.26 and 8.45 resulting in
2o
2o
22
2
o22
2
c
x'u'v''Ο
c
x'u'v'Ο
c
x'u'v'
c
'v1
'Ο
c
x'u'v'
c
'v1
'z
'z'E
'y
'y'E
x'
'x'E
from where we get o
'
'z
'z'E
'y
'y'E
'x
'x''E
that is the Law of Gauss in the differential form to the Oβ
observer. Proceeding this way we can prove the invariance of form for all the other equations of Maxwell.
Β§9 Explaining the Sagnac Effect with the Undulating Relativity We must transform the straight movement of the two observers O and Oβ used in the deduction of the Undulating Relativity in a plain circular movement with a constant radius. Letβs imagine that the observer O sees the observer Oβ turning around with a tangential speed v in a clockwise way (C) equals to the positive course of the axis x of UR and that the observer Oβ sees the observer O turning around with a tangecial speed vβ in a unclockwise way (U) equals to the negative course of the axis x of the UR. In the moment t = tβ = zero, the observer O emits two rays of light from the common origin to both observers, one in a unclockwise way of arc ctU and another in a clockwise way of arc ctC, therefore ctU = ctC and tU = tC, because c is the speed of the constant light, and tU and tC the time.
23/172
In the moment t = tβ = zero the observer Oβ also emits two rays of light from the common origin to both observers, one in a unclockwise way (useless) of arc ctβU and another one in a clockwise way of arc ctβC, thus ctβU = ctβC and tβU = tβC because c is the speed of the constant light, and tβU and tβC the time. Rewriting the equations 1.15 and 1.20 of the Undulating Relativity (UR):
t
t
v
v '
' =
22
2 21
c
vux
c
v . 1.15
'
'
t
t
v
v =
22
2 '''2'1
c
xuv
c
v . 1.20
Making ux = uβxβ = c ( ray of light projected alongside the positive axis x ) and splitting the equations we have:
c
v1t't 9.1
c
'v1'tt 9.2
c
v1
v'v 9.3
c
'v1
'vv 9.4
When the origin of the observer Oβ detects the unclockwise ray of the observer O, will be at the distance
UC 't'vvt of the observer O and simultaneously will detect its clockwise ray of light at the same point of
the observer O, in a symmetric position to the diameter that goes through the observer O because
CUCU ttctct and CUCU 't't'ct'ct , following the four equations above we have:
vc
R2tR2vtct CCU
9.5
'v2c
R2'tR2't'v2'ct CUC
9.6
When the origin of the observer Oβ detects the clockwise ray of the observer O, simultaneously will detect its
own clockwise ray and will be at the distance U2C2 't'vvt of the observer O, then following the equations
1,2,3 and 4 above we have:
vc
R2tvtR2ct C2C2C2
9.7
c
R2'tR2'ct C2C2
9.8
The time difference to the observer O is:
22CC2vc
Rv4
vc
R2
vc
R2ttt
9.9
The time difference to the observer Oβ is:
c'v2c
'Rv4
'v2c
R2
c
R2't't't CC2
9.10
Replacing the equations 5 to 10 in 1 to 4 we prove that they confirm the transformations of the Undulating Relativity.
24/172
Β§10 Explaining the experience of Ives-Stilwell with the Undulating Relativity We should rewrite the equations (2.21) to the wave length in the Undulating Relativity:
22
2
c
vux2
c
v1
'
and
22
2
c
'x'u'v2
c
'v1
'
, 2.21
Making ux = uβxβ = c ( Ray of light projected alongside the positive axis x ), we have the equations:
c
v1
' and
c
'v1
', 10.1
If the observer O, who sees the observer Oβ going away with the velocity v in the positive way of the axis x,
emits waves, provenient of a resting source in its origin with velocity c and wave length F in the positive
way of the axis x, then according to the equation 10.1 the observer Oβ will measure the waves with velocity c
and the wave length D' according to the formulas:
cv1
FD' and
c'v1
FD' , 10.2
If the observer Oβ, who sees the obsesrver O going away with velocity vβ in the negative way of the axis x,
emits waves, provenient of a resting source in its origin with velocity c and the wave length F' in the
positive way of the axis x, then according to the equation 10.1 the observer O will measure waves with
velocity c and wave lenght A according to the formulas:
cv1
' AF and
c'v1
AF' , 10.3
The resting sources in the origin of the observers O and Oβ are identical thus FF ' .
We calculate the average wave length of the measured waves DA ', using the equations 10.2 and
10.3, the left side in each equation:
cv1'
cv12
12
'F
FAD
2FAD
cv11
cv122
'
We calculate the diffrence between the average wave length and the emited wave length by the sources
F :
F
2F
F cv11
cv12
cv12
cv12
cv11
cv12
F
2F
cv12
cv11
cv12
2F
25/172
cv22
cv
cv211
cv12
2
2F
2
2F
cv
2cv1
1
10.4
Reference http://www.wbabin.net/physics/faraj7.htm
Β§10 Ives-Stilwell (continuation) The Dopplerβs effect transversal to the Undulating Relativity was obtained in the Β§2 as follows: If the observer Oβ, that sees the observer O, moves with the speed βvβ in a negative way to the axis xβ, emits waves with the frequency 'y and the speed c then the observer O according to 2.22 and 'v'x'u will
measure waves of frequency y and speed c in a perpendicular plane to the movement of Oβ given by
2
2
c'v1'yy 2.25
For 'v'x'u we will have zeroux and 1cv1
c'v1
2
2
2
2
with this we can write the relation between
the transversal frequency tyy and the source frequency F'y'y like this
2
2
Ft
cv1
'yy
10.5
With FFtt ''yyc we have the relation between the length of the transversal wave t and the length of
the source wave F'
2
2
F cv1' t 10.6
The variation of the length of the transversal wave in the relation to the length of the source wave is:
2
2F
2
2
F2
2
FF2
2
FFt cv
2
'1
c2v1'1
cv1''
cv1''
t 10.7
that is the same value gotten in the Theory of Special Relativity. Applying 10.7 in 10.4 we have
cv1
t 10.8
With the equations 10.2 and 10.3 we can get the relations 10.9, 10.10, and 10.11 described as follows 2
DA cv1'
10.9
And from this we have the formula of speed D
A
'1
cv
10.10
DAFF '' 10.11
Applying 10.10 and 10.11 in 10.6 we have
2
D
ADAt '
11'
10.12
26/172
From 10.8 and 10.12 we conclude that DtFA ' . 10.13
So that we the values of A and D' obtained from the Ives-Stiwell experience we can evaluate t , F ,
cv and conclude whether there is or not the space deformation predicted in the Theory of Special Relativity.
Β§11 Transformation of the power of a luminous ray between two referencials in the Special Theory of Relativity The relationship within the power developed by the forces between two referencials is written in the Special Theory of the Relativity in the following way:
2c
vux1
vFxu.F'u'.F
11.1
The definition of the component of the force along the axis x is:
dt
duxmux
dt
dm
dt
muxd
dt
dpxFx 11.2
For a luminous ray, the principle of light speed constancy guarantees that the component ux of the light speed is also constant along its axis, thus
uxdt
dx
t
x constant, demonstrating that in two zero
dt
dux and ux
dt
dmFx 11.3
The formula of energy is 2mcE from where we have dt
dE
c
1
dt
dm2
11.4
From the definition of energy we have u.Fdt
dE that applying in 4 and 3 we have
2c
uxu.FFx
11.5
Applying 5 in 1 we heve:
2
2
c
vux1
c
vuxu.Fu.F
'u'.F
From where we find that u.F'u'.F
or dt
dE
'dt
'dE 11.6
A result equal to 5.3 of the Undulating Relativity that can be experimentally proven, considering the βSunβ as the source.
Β§12 Linearity
The Theory of Undulating Relativity has as its fundamental axiom the necessity that inertial referentials be named exclusively as those ones in which a ray of light emitted in any direction from its origin spreads in a straight line, what is mathematically described by the formulae (1.13, 1.18, 8.6 e 8.7) of the Undulating Relativity:
uzdt
dz
t
zuy
dt
dy
t
yux
dt
dx
t
x ,, 1.13
'z'u'dt
'dz
't
'z,'y'u
'dt
'dy
't
'y,'x'u
'dt
'dx
't
'x 1.18
Woldemar Voigt wrote in 1.887 the linear transformation between the referentials os the observers O e Oβ in the following way:
27/172
'Bt'Axx 12.1
'Ft'Ext 12.2 With the respective inverted equations:
tBEAF
Bx
BEAF
F'x
12.3
tBEAF
Ax
BEAF
E't
12.4
Where A, B, E and F are constants and because of the symmetry we donβt consider the terms with y, z and yβ, zβ. We know that x and xβ are projections of the two rays of lights ct and ctβ that spread with Constant speed c (due to the constancy principle of the Ray of light), emited in any direction from the origin of the respective inertials referential at the moment in which the origins are coincident and at the moment where: t = tβ = zero 12.5 because of this in the equation 12.2 at the moment where tβ = zero we must have E = zero so that we also have t = zero, we canβt assume that when tβ = zero, xβ also be equal to zero, because if the spreading happens in the plane yβzβ we will have xβ = zero plus zero't . We should rewrite the corrected equations (E = zero):
'Bt'Axx 12.6
'Ftt 12.7 With the respective corrected inverted equations:
AF
Bt
A
x'x 12.8
F
t't 12.9
If the spreading happens in the plane yβ zβ we have xβ = zero and dividing 12.6 by 12.7 we have:
vF
B
t
x 12.10
where v is the module of the speed in which the observer O sees the referential of the observer Oβ moving alongside the x axis in the positive way because the sign of the equation is positive. If the spreading happens in the plane y z we have x = zero and dividing 12.8 by 12.9 we have:
'vA
B
't
'x or 'v
A
B 12.11
where vβ is the module of the speed in which the observer Oβ sees the referential of the observer O moving alongside the xβ axis in the negative way because the signal of the equation is negative. The equation 1.6 describes the constancy principle of the speed of light that must be assumed by the equations 12.6 to 12.9:
222222 'tc'xtcx 1.6 Applying 12.6 and 12.7 in 1.6 we have:
2222222 'tc'x'tFc'Bt'Ax
From where we have:
28/172
22222
222222 'tc'x
'tc
'ABx2
c
BF'tc'xA
where making A2 = 1 in the brackets in arc and 1'tc
'ABx2
c
BF
22
22
in the straight brackets we have
the equality between both sides of the equal signal of the equation.
Appllying A = 1 in 1'tc
'ABx2
c
BF
22
22
we have
'tc
'Bx2
c
B1F
22
22 12.12
Appllying A = 1 in 12.11 we have 'vB1
B
A
B 12.11
That applied in 12.12 suplies:
't,'xF'tc
'x'v2
c
'v1F
22
2
12.12
as F(xβ, tβ) is equal to the function F depending of the variables xβ and tβ. Applying 12.8 and 12.9 in 1.6 we have:
2
22
2222
F
tc
AF
Bt
A
xtcx
From where we have:
FtcA
Bx2
FcA
B
F
1tc
A
xtcx
22222
2
222
2
2222
where making A2 = 1 in the bracket in arc and 1FtcA
Bx2
FcA
B
F
122222
2
2
in the straight bracket we
have the equality between both sides of the equal signal of the equation.
Applying A = 1 and 12.10 in 1FtcA
Bx2
FcA
B
F
122222
2
2
we have:
t,xF
tc
vx2
c
v1
1F
22
2
12.13
as F(x, t) is equal to the function F depending on the variables x and t. We must make the following naming according to 2.5 and 2.6:
'KF'tc
'x'v2
c
'v1'K
22
2
12.14
K
1F
tc
vx2
c
v1K
22
2
12.15
As the equation to F(xβ, tβ) from 12.12 and F(x, t) from 12.13 must be equal, we have:
29/172
tc
vx2
c
v1
1
'tc
'x'v2
c
'v1F
22
222
2
12.16
Thus:
1'tc
'x'v2
c
'v1
tc
vx2
c
v1
22
2
22
2
or 1'KK 12.17
Exactly equal to 1.10. Rewriting the equations 12.6, 12.7, 12.8 and 12.9 according to the function of v, vβ and F we have:
't'v'xx 12.6
'Ftt 12.7 With the respective inverted corrected equations:
vtx'x 12.8
F
t't 12.9
We have the equations 12.6, 12.7, 12.8 and 12.9 finals replacing F by the corresponding formulae:
't'v'xx 12.6
'tc
'x'v2
c
'v1'tt
22
2
12.7
With the respective inverted final equations:
vtx'x 12.8
tc
vx2
c
v1t't
22
2
12.9
That are exactly the equations of the table I
As F
Bv and B'v then the relations between v and vβ are
F
'vv or F.v'v 12.18
We will transform F (12.12) function of the elements vβ, xβ, and tβ for F (12.13) function of the elements v, x and t, replacing in 12.12 the equations 12.8, 12.9 and 12.18:
F
tc
vtxvF2
c
vF1
'tc
'x'v2
c
'v1F
22
2
22
2
2
22
2
2
2
22
2
2
2
22
c
Fv
tc
vxF21
c
Fv2
tc
vxF2
c
Fv1F
30/172
2
22
2
22
c
Fv
tc
vxF21F
tc
vx2
c
v1
1F1
tc
vxF2
c
FvF
22
22
2
2
222
That is exactly the equation 12.13. We will transform F (12.13) function of the elements v, x, and t for F (12.12) function of the elements vβ, xβ and tβ, replacing in 12.13 the equations 12.6, 12.7 and 12.18:
22
2
2222
2
2
2
222
2
Fc
'v2
F'tc
'x'v2
Fc
'v1
1
'FFtc
't'v'x'v2
F
'v
c
11
1
tc
vx2
c
v1
1F
'tc
'x'v2
c
'v1F1
F'tc
'x'v2
2Fc
'v1F
'F'tc
'x'v2
'Fc
'v1
1F
22
2
222
22
2222
2
That is exactly the equation 12.12. We have to calculate the total diferential of F(xβ, tβ) (12.12):
'dt't
F'dx
'x
FdF
as:
'tc
'v
'K
1
'x
F2
and 't
'x
'tc
'v
'K
1
't
F2
12.19
we have:
'dt't
'x
'tc
'v
'K
1'dx
'tc
'v
'K
1dF
22
where applying 1.18 we find:
o'dt'dt
'dx
'tc
'v
'K
1'dx
'tc
'v
'K
1dF
22 12.20
From where we conclude that F function of xβ and tβ is a constant. We have to calculate the total diferential of F(x, t) (12.13):
dtt
Fdx
x
FdF
as:
tc
v
K
1
x
F2
2
3
and t
x
tc
v
K
1
t
F2
2
3
12.21
we have:
dtt
x
tc
v
K
1dx
tc
v
K
1dF
2
2
32
2
3 12.22
31/172
where applying 1.13 we find:
odtdt
dx
tc
v
K
1dx
tc
v
K
1dF
2
2
32
2
3
From where we conclude that F function of x and t is a constant. The equations 1.13 and 1.18 represent to the observers O and Oβ the principle of constancy of the light speed valid from infinitely small to the infinitely big and mean that in the Undulating Relativity the space and time are measure simultaneously. They shouldnβt be interpreted with a dependency between space and time. The time has its own interpretation that can be understood if we analyze to a determined observer the emission of two rays of light from the instant t=zero. If we add the times we get, for each ray of light, we will get a result without any use for the physics. If in the instant t = tβ = zero, the observer Oβ emits two rays of light, one alongside the axis x and the other alongside the axis y, after the interval of time tβ, the rays hit for the observer Oβ, simultaneously, the points Ax
and Ay to the distance ctβ from the origin, although for the observer O, the points wonβt be hit simultaneously. For both rays of lights be simultaneous to both observers, they must hit the points that have the same radius in relation to the axis x and that provide the same time for both observers (t1
= t2 and tβ1 = tβ2), which means that only one ray of light is necessary to check the time between the referentials. According to Β§ 1, both referentials of the observers O and Oβ are inertial, thus the light spreads in a straight line according to what is demanded by the fundamental axiom of the Undulating Relativity Β§ 12, because of this, the difference in velocities v and vβ is due to only a difference in time between the referentials.
t'xxv 1.2
't'xx'v 1.4
We can also relate na inertial referential for which the light spread in a straight line according to what is demanded by the fundamental axiom of the Undulating Relativity, with an accelerated moving referential for which the light spread in a curve line, considering that in this case the difference v and vβ isnβt due to only the difference of time between the referentials. According to Β§ 1, if the observer O at the instant t = tβ = zero, emits a ray of light from the origin of its referential, after an interval of time t1, the ray of light hits the point A1 with coordinates (x1, y1, z1, t1) to the distance ct1 of the origin of the observer O, then we have:
12
12
2
11 tc
vx2
cv1t't
After hitting the point A1 the ray of light still spread in the same direction and in the same way, after an interval of time t2, the ray of light hits the point A2 with coordinates (x1 + x2, y1 + y2, z1 + z2, t1 + t2) to the distance ct2 to the point A1, then we have:
22
2
22
22
2
12
12
2
2
2
1
1
cvux2
cv1
tc
vx2
cv1
tc
vx2
cv1ux
t
x
t
xux
dtdx
tx
and with this we get:
22
2
22
22
2
2
22 cvux2
cv1t
tc
vx2
cv1t't
21
221
2
2
2122
2
2122
2
21
21
2
2
121 ttc
xxv2
cv1tt
cvux2
cv1tt
cvux2
cv1t
tc
vx2
cv1t't't
The geometry of space and time in the Undulating Relativity is summarized in the figure below that can be expanded to An points and several observers.
32/172
In the figure the angles have a relation ' and are equal to the following segments:
O1 to 'OO is equal to 'OO to Oβ1 1111 't'vvt'OO
O2 to O1 is equal to Oβ1 to Oβ2 211222212122 'O'OOO't'vvt't't'vttv'OO
And are parallel to the following segments: O2 to A2 is parallel to O1 to A1 Oβ2 to A2 is parallel to Oβ1 to A1
'XX is parallel to 11 'XX
The cosine of the angles of inclination and ' to the rays for the observers O and Oβ according to 2.3 and
2.4 are:
cos
cv2
cv1
c/vcos'cos
cvux2
cv1
cv
cux
c'x'u
cvux2
cv1
vux'x'u
2
2
22
2
22
2
Kc/vcos
'cos 12.23
And with this we have: K
sen'sen
12.24
'cosc'v2
c'v1
c/'v'coscos
c'x'u'v2
c'v1
c'v
c'x'u
cux
c'x'u'v2
c'v1
'v'x'uux
2
2
22
2
22
2
'Kc/'v'cos
cos 12.25
33/172
And with this we have 'K'sen
sen 12.26
The cosine of the angle with intersection of rays equal to:
K'
'coscv'1
K
coscv1
K'cx'u'v'1
Kcvux1
cos22
12.27
And with this we have: K''sen
c'v
Ksen
cvsen
12.28
The invariance of the cos shows the harmony of all adopted hypotheses for space and time in the
Undulating Relativity. The cos is equal to the Jacobians of the transformations for the space and time of the picture I, where the
radicals
tcvx2
cv1K
22
2
and 'tc'x'v2
c'v1'K
22
2
are considered variables and are derived.
K
cvux1
Ktc
vx1
tcvx
cv1
K100
Kc/v
01000010v001
t,z,y,x
't','z,'y,'x
x'xJcos
22
22
22j
i
8.8
'K
c'x'u'v1
'K'tc'x'v1
'tc'x'v
c'v1
'K100
'Kc/'v
01000010'v001
't,'z,'y,'x
t,z,y,x
'xx'Jcos
22
22
22l
k
8.8
Β§13 Richard C. Tolman The Β§4 Transformations of the Momenta of Undulating Relativity was developed based on the experience conducted by Lewis and Tolman, according to the reference [3]. Where the collision of two spheres preserving the principle of conservation of energy and the principle of conservation of momenta, shows that the mass is a function of the velocity according to:
2
2
o
cu
1
mm
where om is the mass of the sphere when in resting position and uuuu
the module of its speed.
Analyzing the collision between two identical spheres when in relative resting position, that for the observer Oβ are named Sβ1 and Sβ2 are moving along the axis xβ in the contrary way with the following velocities before the collision: Table 1 Esphere Sβ1 Esphere Sβ2
'v'x'u 1 'v'x'u 2
zero'y'u 1 zero'y'u 2
zero'z'u 1 zero'z'u 2
For the observer O the same spheres are named S1 and S2 and have the velocities
zerouzuy,ux,ux ii21 before the collision calculated according to the table 2 as follows:
34/172
The velocity 1ux of the sphere S1 is equals to:
2
2
22
2
21
2
2
11
c'v31
'v2
c'v'v2
c'v1
'v'v
c
'x'u'v2
c'v1
'v'x'uux
.
The transformation from vβ to v according to 1.20 from Table 2 is:
2
2
22
2
21
2
2
c'v31
'v
c'v'v2
c'v1
'v
c
'x'u'v2
c'v1
'vv
.
That applied in 1ux supplies:
v2
c'v31
'v2ux
2
21
The velocity 2ux of the sphere S2 is equal to:
zero
c'v'v2
c'v1
'v'v
c
'x'u'v2
c'v1
'v'x'uux
22
2
22
2
2
22
Table 2 Sphere S1 Sphere S2
v2
c'v31
'v2ux
2
1
zeroux2
zerouy1 zerouy2
zerouz1 zerouz2
For the observers O and Oβ the two spheres have the same mass when in relative resting position. And for the observer Oβ the two spheres collide with velocities of equal module and opposite direction because of
this the momenta 21 'p'p null themselves during the collision, forming for a brief time 't only one
body of mass
210 'm'mm .
According to the principle of conservation of momenta for the observer O we will have to impose that the momenta before the collision are equal to the momenta after the collision, thus:
wmmuxmuxm 212211
Where for the observer O, w is the arbitrary velocity that supposedly for a brief time t will also see the
masses united 21 mmm moving. As the masses im have different velocities and the masses vary
according to their own velocities, this equation cannot be simplified algebraically, having this variation of masses: To the left side of the equal sign in the equation we have:
v2uxu 1
35/172
2
2
o
2
2
o
2
21
o
2
2
o1
cv41
m
c
v21
m
c
ux1
m
c
u1
mm
zerouxu 2
0
2
2
o
2
22
o
2
2
o2 m
czero
1
m
c
ux1
m
cu
1
mm
To the right side of the equal sign in the equation we have:
wu
2
2
o
2
2
o
2
2
o1
cw1
m
c
w1
m
c
u1
mm
2
2
o
2
2
o
2
2
o2
cw1
m
c
w1
m
c
u1
mm
Applying in the equation of conservation of momenta we have:
wmwmwmmuxmuxm 21212211
w
cw1
mw
cw1
m0.mv2
cv41
m
2
2
0
2
2
00
2
2
0
From where we have:
2
2
2
2
2
2
0
2
2
0
cw1
w
cv41
v
cw1
wm2
cv41
vm2
2
2
cv31
vw
As vw for the observer O the masses united 21 mmm wouldnβt move momentarily alongside to the
observer Oβ which is conceivable if we consider that the instants 'tt are different where supposedly the masses would be in a resting position from the point of view of each observer and that the mass acting with velocity 2v is bigger than the mass in resting position. If we operate with these variables in line we would have:
wmwmwmmuxmuxm 21212211
2
2
0
2
2
0
2
2
00
2
2
2
2
0
cw1
wm2w
cw1
mw
cw1
m0.m
c'v31
'v2
c'v31
'v2c11
m
36/172
2
2
0
2
2
22
0
cw1
wm2
c'v31
'v4c11
c'v31
'vm2
2
2
0
2
2
2
0
cw1
wm2
c'v4
c'v31
'vm2
2
2
0
2
2
0
cw1
wm2
c'v1
'vm2
From where we conclude that 'vw which must be equal to the previous value of w, that is:
2
2
cv31
v'vw
A relation between v and vβ that is obtained from Table 2 when v2ux1 that corresponds for the observer O
to the velocity acting over the sphere in resting position.
Β§14 Velocities composition Reference β Millennium Relativity URL: http://www.mrelativity.net/MBriefs/VComp_Sci_Estab_Way.htm Letβs write the transformations of Hendrik A. Lorentz for space and time in the Special Theory of Relativity:
2
2
cv1
vtx'x
14.1a
2
2
cv1
'vt'xx
14.3a
y'y 14.1b 'yy 14.3b
z'z 14.1c 'zz 14.3c
2
2
2
cv1
cvxt
't
14.2
2
2
2
cv1
c'vx't
t
14.4
From them we obtain the equations of velocity transformation:
2cvux1
vux'x'u
14.5a 2c'x'vu1
v'x'uux
14.6a
2
2
2
cvux1
cv1uy
'y'u
14.5b
2
2
2
c'x'vu1
cv1'y'u
uy
14.6b
2
2
2
cvux1
cv1uz
'z'u
14.5c
2
2
2
c'x'vu1
cv1'z'u
uz
14.6c
Letβs consider that in relation to the observer Oβ an object moves with velocity:
c50,0s/km10.5,1'x'u 5 .
37/172
And that the velocity of the observer Oβ in relation to the observer O is:
c50,0s/km10.5,1v 5 .
The velocity ux of the object in relation to the observer O must be calculated by the formula 14.6a:
c80,0s/km10.4,2
10.0,3
10.5,1.10.5,11
10.5,110.5,1
c'x'vu1
v'x'uux 5
25
55
55
2
.
Where we use c00,1s/km10.0,3c 5 .
Considering that the object has moved during one second in relation to the observer O s00,1t we can
then with 14.2 calculate the time passed to the observer Oβ:
s693,0't75,060,0
10.0,3
10.5,11
10.0,3
10.4,2.10.5,1100,1
cv1
cvux1t
cv1
cvxt
't
25
25
25
55
2
2
2
2
2
2
.
To the observer O the observer Oβ is away the distance d given by the formula:
km10.5,100,1.10.5,1vtd 55 . To the observer Oβ the observer O is away the distance dβ given by the formula:
km10.03923,175,060,010.5,1'vt'd 5.5 .
To the distance of the object OO 'd,d in relation to the observers O and Oβ is given by the formulae:
km10.4,200,1.10.4,2uxtd 55O .
km10.03923,175,060,0.10.5,1't'x'u'd 55
O .
To the observer O the distance between the object and the observer Oβ is given by the formula:
km10.90,010.5,110.4,2ddd 555O .
To the observer O the velocity of the object in relation to the observer Oβ is given by:
s/km10.90,0s00,1km10.90,0
td 5
5
(=0,30c)
Relating the times t and tβ using the formula 2
2
cv1t't is only possible and exclusively when vux and
zero'x'u what isnβt the case above, to make it possible to understand this we write the equations 14.2 and 14.4 in the formula below:
2
2
cv1
coscv1t
't
14.2
2
2
cv1
'coscv1't
t
14.4
38/172
Where ctxcos and
'ct'x'cos .
The equations above can be written as:
,tf't e ','t'ft 14.7
In each referential of the observers O and Oβ the light propagation creates a sphere with radius ct and 'ct
that intercept each other forming a circumference that propagates with velocity c. The radius ct and 'ct
and the positive way of the axis x and 'x form the angles and ' constant between the referentials. If for
the same pair of referentials te angles were variable the time would be alleatory and would become useless
for the Physics. In the equation ,tf't we have tβ identical function of t and , if we have in it
constant and tβ varies according to t we get the common relation between the times t and tβ between two referentials, however if we have t constant and tβ varies according to we will have for each value of
one value of tβ and t between two different referentials, and this analysis is also valid for ','t'ft .
Dividing 14.5a by c we have:
coscv1
cvcos
'cos
cvux1
cv
cux
c'x'u
2
. 14.8
Where cux
ctxcos and
c'x'u
'ct'x'cos .
Isolating the velocity we have:
'coscos1
'coscos
cv
or
21
c
'x'uxu'x'uux
v
14.9
From where we conclude that we must have angles and ' constant so that we have the same velocity
between the referentials. This demand of constant angles between the referentials must solve the controversies of Herbert Dingle.
Β§15 Invariance The transformations to the space and time of table I, group 1.2 plus 1.7, in the matrix form is written like this:
tzyx
K
v
't'z'y'x
00001000010
001
15.1
That written in the form below represents the same coordinate transformations:
ctzyx
K
c/v
'ct'z'y'x
00001000010
001
15.2
We call as:
4
3
2
1
'cx'x'x'x
'ct'z'y'x
'x'x i ,
K
c/v
ij
00001000010
001
,
4
3
2
1
cxxxx
ctzyx
xx j 15.3
39/172
That are the functions ctzyxxcxxxxxxxx iijii ,,,',,,''' 4321 15.4
That in the symbolic form is written:
xx .' or in the indexed form jij
i4
1j
jij
i x'xx'x
15.5
Where we use Einsteinβs sum convention. The transformations to the space and time of table I, group 1.4 plus 1.8, in the matrix form is written:
't'z'y'x
'K
'v
tzyx
00001000010
001
15.6
That written in the form below represents the same coordinate transformations:
'ct'z'y'x
'K
c/'v
ctzyx
00001000010
001
15.7
That we call as:
4
3
2
1
cxxxx
ctzyx
xx k ,
'K
c/'v
'' kl
00001000010
001
,
4
3
2
1
'cx'x'x'x
'ct'z'y'x
'x'x l 15.8
That are the functions ',',','',',','' 4321 ctzyxxcxxxxxxxx kklkk 15.9
That in the symbolic form is written:
''. xx or in the indexed form lkl
k4
1l
lkl
k 'x'x'x'x
15.10
Being 42
1
2
2 21xcvx
cvK (1.7),
42
1
2
2 21'xc'x'v
c'v'K (1.8) and 1'K.K (1.10).
The transformation matrices ij and kl'' have the properties:
il
jjlijklij I
'K
c/'v
K
c/v
'''.
1000
010000100001
00001000010
001
00001000010
0014
1
15.11
jk
iikjilkji
tt I
'Kc/'vKc/v
'''
1000
010000100001
00010000100001
00010000100001
4
1
15.12
Where jit is the transposed matrix of ij and lk
t '' is the transpose matrix of kl'' and
is the Kroneckerβs delta.
kj
lljklijkl I
K
c/v
'K
c/'v
'''.
1000010000100001
00001000010
001
00001000010
0014
1
15.13
40/172
li
kkilkjilk
tt I
Kc/v'Kc/'v
'''
1000010000100001
00010000100001
00010000100001
4
1
15.14
Where lkt '' is the transposed matrix of kl'' and ji
t is the transposed matrix of ij
and is the Kroneckerβs delta.
Observation: the matrices ij and kl' are inverse of one another but are not orthogonal, that is: klji '
and lkij ' .
The partial derivatives j
i
xx ' of the total differential j
j
ii dx
x'x'dx
of the coordinate components that
correlate according to jii xxx '' , where in the transformation matrix ij the radical K is
considered constant and equal to: Table 10, partial derivatives of the coordinate components:
jj
i
xx
xx 1'' 1
x'x1
1
0x'x2
1
0x'x3
1
cv
xx
4
1'
j
2
j
i
x'x
x'x
0x'x1
2
1x'x2
2
0x
'x3
2
0x'x4
2
j
3
j
i
x'x
x'x
0x'x1
3
0x'x2
3
1x'x3
3
0x'x4
3
j
4
j
i
x'x
x'x
0'1
4
xx 0
x'x2
4
0x'x3
4
Kxx
4
4'
The total differential of the coordinates in the matrix form is equal to:
4
3
2
1
4
3
2
1
00001000010
001
cdxdxdxdx
K
c/v
'cdx'dx'dx'dx
15.15
That we call as:
4
3
2
1
'cdx'dx'dx'dx
'dx'dx i ,
K
c/v
x
'xAA
j
iij
00001000010
001
,
4
3
2
1
cdxdxdxdx
dxdx j 15.16
Then we have jj
ii
j
jij
i dxxxdxdxAdxAdxdx
''''4
1
15.17
The partial derivatives l
k
xx'
of the total differential ll
kk dx
xxdx ''
of the coordinate components that
correlate according to lkk xxx ' , where in the transformation matrix kl'' the radical 'K is
considered constant and equal to:
41/172
Table 11 partial derivatives of the coordinate components:
l
1
l
k
'xx
'xx
1'xx
1
1
0'xx
2
1
0'xx
3
1
cv
xx ''41
l
2
l
k
'xx
'xx
0'xx
1
2
1'xx
2
2
0'xx
3
2
0'xx
4
2
l
3
l
k
'xx
'xx
0'xx
1
3
0'xx
2
3
1'xx
3
3
0'xx
4
3
l
4
l
k
'xx
'xx
0'14
xx 0
'xx
2
4
0'xx
3
4
''44
Kxx
The total differential of the coordinates in the matrix form is equal to:
4
3
2
1
4
3
2
1
00001000010
001
'cdx'dx'dx'dx
'K
c/'v
cdxdxdxdx
15.18
That we call as:
4
3
2
1
cdxdxdxdx
dxdx k ,
'K
c/'v
'x
x'A'A
l
kkl
00001000010
001
,
4
3
2
1
'cdx'dx'dx'dx
'dx'dx l 15.19
Then we have: ll
kk
l
lkl
k dxxxdxdxAdxdxAdx ''
''''4
1
15.20
The Jacobians of the transformations 15.15 and 15.18 are:
K
K
c/v
x,x,x,x
'x,'x,'x,'x
x'xJj
i
00001000010
001
4321
4321
15.21
'K
'K
c/'v
'x,'x,'x,'x
x,x,x,x
'xx'J
l
k
00001000010
001
4321
4321
15.22
Where 2
1
2
2 21cvux
cvK (2.5),
2
1
2
2 21c
'x'u'vc'v'K (2.6) and 1'K.K (1.23).
The matrices of the transformation A and 'A also have the properties 15.11, 15.12, 15.13 and 15.14 of the matrices and ' .
From the function lkk xxx ''' where the coordinates correlate in the form lkk xxx ' we
have l
k
kl xx
xx ''
described as:
42/172
1
4
41
3
31
2
21
1
111 '''''' xx
xxx
xxx
xxx
xxx
xxk
k
2
4
42
3
32
2
22
1
122 '''''' xx
xxx
xxx
xxx
xxx
xxk
k
3
4
43
3
33
2
23
1
133 '''''''
xx
xxx
xxx
xxx
xxx
xxk
k
4
4
44
3
34
2
24
1
144 ''
''''' xx
xxx
xxx
xxx
xxx
xxk
k
That in the matrix form and without presenting the function becomes:
2
1
2
2
4
4
3
4
2
4
21
4
4
3
3
3
2
3
1
3
4
2
3
2
2
2
1
2
4
1
3
1
2
1
1
1
43214321l
c
'x'u'v
c
'v1'K
1
'x
x0'x
x0'x
x
'Kc
'v
'x
x
0'x
x1'x
x0'x
x0'x
x
0'x
x0'x
x1'x
x0'x
x
'v'x
x0'x
x0'x
x1'x
x
xxxx'x'x'x'x'x
Where replacing the items below:
221
4
cv
'Kc
'v'xx
K
v'v'xx
4
1
2
1
2
2
4
4
2
1
2
2
4
41111
cvux
cv
Kx'x
c'x'u'v
c'v
'K'xx
Observation: this last relation shows that the time varies in an equal form between the referentials. We get:
2
1
2
2
4
4
3
4
2
4
21
4
4
3
3
3
2
3
1
3
4
2
3
2
2
2
1
2
4
1
3
1
2
1
1
1
43214321l
c
vux
c
v1K
1
'x
x0'x
x0'x
x
c
v
'x
x
0'x
x1'x
x0'x
x0'x
x
0'x
x0'x
x1'x
x0'x
xK
v
'x
x0'x
x0'x
x1'x
x
xxxx'x'x'x'x'x
That is the group 8.1 plus 8.3 of the table 9, differential operators, in the matrix form.
From the function jii xxx '''' where the coordinates correlate in the form jii xxx '' we
have j
i
ij xx
xx
'
'''
described as:
43/172
1
4
41
3
31
2
21
1
111
''''
'''
'''
'''
'''
xx
xxx
xxx
xxx
xxx
xxi
i
2
4
42
3
32
2
22
1
122
''''
'''
'''
'''
'''
xx
xxx
xxx
xxx
xxx
xxi
i
3
4
43
3
33
2
23
1
133
''''
'''
'''
'''
'''
xx
xxx
xxx
xxx
xxx
xxi
i
4
4
44
3
34
2
24
1
144
''''
'''
'''
'''
'''
xx
xxx
xxx
xxx
xxx
xxi
i
That in the matrix form and without presenting the function becomes:
2
1
2
2
4
4
3
4
2
4
21
4
4
3
3
3
2
3
1
3
4
2
3
2
2
2
1
2
4
1
3
1
2
1
1
1
43214321j
c
vux
c
v1K
1
x
'x0x
'x0x
'x
Kc
v
x
'x
0x
'x1x
'x0x
'x0x
'x
0x
'x0x
'x1x
'x0x
'x
vx
'x0x
'x0x
'x1x
'x
'x'x'x'xxxxxx
'
Where replacing the items below:
'K
'vvx'x
4
1
221
4
c'v
Kc
vx'x
2
1
2
2
4
4
2
1
2
2
4
41111
c'x'u'v
c'v
'K'xx
cvux
cv
Kx'x
Observation: this last relation shows that the time varies in an equal form between the referentials. We get:
2
1
2
2
4
4
3
4
2
4
21
4
4
3
3
3
2
3
1
3
4
2
3
2
2
2
1
2
4
1
3
1
2
1
1
1
43214321j
c
'x'u'v
c
'v1'K
1
x
'x0x
'x0x
'x
c
'v
x
'x
0x
'x1x
'x0x
'x0x
'x
0x
'x0x
'x1x
'x0x
'x'K
'v
x
'x0x
'x0x
'x1x
'x
'x'x'x'xxxxxx
'
That is the group 8.2 plus 8.4 from the table 9, differential operators in the matrix form. Applying 8.5 in 8.3 and in 8.4 we simplify these equations in the following way:
44/172
Table 9B, differential operators with the equations 8.3 and 8.4 simplified:
411 xcv
xx' 2
8.1 411 'xcv'
x'x 2
8.2
22 x'x
8.1.1 22 'xx
8.2.1
33 x'x
8.1.2 33 'xx
8.2.2
44 xcK
'xc
8.3B 44 'xc'K
xc
8.4B
zeroxc
uxx 2
4
1
1 8.5 zero
'xc'x'u
x' 2
4
1
1 8.5
The table 9B, in the matrix form becomes:
Kc/vxcxxx'xc'x'x'x
00010000100001
43214321 15.23
'Kc/'v'xc'x'x'xxcxxx
00010000100001
43214321 15.24
The squared matrices of the transformations above are transposed of the matrices A and Aβ.
Invariance of the Total Differential
In the observer O referential the total differential of a function kx is equal to:
4
3
2
1
43214
43
32
21
1
cdxdxdxdx
xcxxxdx
xdx
xdx
xdx
xdx
xxd k
kk 15.25
Where the coordinates correlate with the ones from the observer Oβ according to lkk xxx ' , replacing the
transformations 15.24 and 15.18 and without presenting the function we have:
4
3
2
1
4321
00001000010
001
00010000100001
'cdx'dx'dx'dx
'K
c/'v
'Kc/'v'xc'x'x'x
dxx
d kk
15.26
The multiplication of the middle matrices supplies:
42
12100
01000010
001
00001000010
001
00010000100001
'dxc'dx'vc/'v
c/'v
'K
c/'v
'Kc/'v
15.27
Result that can be divided in two matrices:
42
1
42
1 200
00000000
000
1000010000100001
2100
01000010
001
'dxc'dx'vc/'v
c/'v
'dxc'dx'vc/'v
c/'v
15.28
45/172
That applied to the total differential supplies:
4
3
2
1
42
14321200
00000000
000
1000010000100001
'cdx'dx'dx'dx
'dxc'dx'vc/'v
c/'v
'xc'x'x'xdx
xd k
k
15.29
Executing the operations of the second term we have:
444
1
24
11
42
4
3
2
1
42
14321
2
200
00000000
000
'dx'x'dx
'dxc'v'dx
'x'v'dx
'xc'v
'cdx'dx'dx'dx
'dxc'dx'vc/'v
c/'v
'xc'x'x'x
Where applying 8.5 we have:
zero'dx'x'dx
'dxc'v'dx
'x'dx'dx
c'v'dx
'xc'v
4
44
1
24
44
1
21
42
21
Then we have:
zero
'cdx'dx'dx'dx
'dxc'dx'vc/'v
c/'v
'xc'x'x'x
4
3
2
1
42
14321200
00000000
000
15.30
With this result we have in 15.29 the invariance of the total differential:
'd'dx'x
'
'cdx'dx'dx'dx
'xc'x'x'xdx
xd l
lk
k
4
3
2
1
4321
1000010000100001
15.31
In the observer Oβ referential the total differential of a function ix' is equal to:
4
3
2
1
4321
4
4
3
3
2
2
1
1
'cdx'dx'dx'dx
'xc
'
'x
'
'x
'
'x
''dx
'x
''dx
'x
''dx
'x
''dx
'x
''dx
'x
''x'd i
i
i 15.32
Where the coordinates correlate with the ones from the observer O referential according to jii xxx '' ,
replacing the transformations 15.23 and 15.15 and without presenting the function we have:
4
3
2
1
4321
00001000010
001
00010000100001
cdxdxdxdx
K
c/v
Kc/vxcxxx
'dx'x
''d i
i
15.33
The multiplication of the middle matrices supplies:
42
12100
01000010
001
00001000010
001
00010000100001
dxcvdxc/v
c/v
K
c/v
Kc/v
15.34
Result that can be divided in two matrices:
46/172
42
1
42
1 200
00000000
000
1000010000100001
2100
01000010
001
dxcvdxc/v
c/v
dxcvdxc/v
c/v
15.35
That applied to the total differential supplies:
4
3
2
1
42
14321200
00000000
000
1000010000100001
cdxdxdxdx
dxcvdxc/v
c/v
xcxxx'dx
'x
''d i
i
15.36
Executing the operations of the second term we have:
444
1
24
11
42
4
3
2
1
42
14321
2
200
00000000
000
dxxdx
dxcvdx
xvdx
xcv
cdxdxdxdx
dxcvdxc/v
c/v
xcxxx
Where applying 8.5 we have:
zerodxxdx
dxcvdx
xdxdx
cvdx
xcv
4
44
1
24
44
1
21
42
21
Then we have:
zero
cdxdxdxdx
dxcvdxc/v
c/v
xcxxx
4
3
2
1
42
14321200
00000000
000
15.37
With this result we have in 15.36 the invariance of the total differential:
ddxx
cdxdxdxdx
xcxxx'dx
'x
''d j
ji
i
4
3
2
1
4321
1000010000100001
15.38
Invariance of the Wave Equation
The wave equation to the observer O is equal to:
0
1000010000100001
1
4
3
2
1
1
4321432124
2
22
xc
x
x
x
xcxxxxc1
xxxxc 2
2
22
2
2
2
2
2 15.39
47/172
Where applying 15.24 and the transposed from 15.24 we have:
4
3
2
1
432124
2
22
00001000010
001
1000010000100001
00
010000100001
1
'xc
'x
'x
'x
'K
c'v
'Kc'v'xc'x'x'xxc
15.40
The multiplication of the three middle matrices supplies:
2
12100
01000010
001
00001000010
001
1000010000100001
00
010000100001
c'x'u'v
c'v
c'v
'K
c'v
'Kc'v
15.41
Result that can be divided in two matrices:
2
1
2
1 200
00000000
000
1000010000100001
2100
01000010
001
c'x'u'v
c'v
c'v
c'x'u'v
c'v
c'v
15.42
That applied in the wave equation supplies:
4
3
2
1
2
1432124
2
22
200
00000000
000
1000010000100001
1
'xc
'x
'x
'x
c'x'u'v
c'v
c'v
'xc'x'x'xxc
15.43
Executing the operations of the second term we have:
24
2
2
1
2412412
4
3
2
1
2
14321
2
200
00000000
000
'xc'x'u
c'v
'x'xc'v
'x'xc'v
'xc
'x
'x
'x
c'x'u'v
c'v
c'v
'xc'x'x'x
Executing the operations we have:
24
2
2
1
2412
22'xc
'x'uc'v
'x'xc'v
Where applying 8.5 we have:
zero'xc
'x'uc'v
'x'xc'x'u
c'v
24
2
2
1
2442
1
2
22
48/172
Then we have:
zero
'xc
'x
'x
'x
c'x'u'v
c'v
c'v
'xc'x'x'x
4
3
2
1
2
14321
200
00000000
000
15.44
With this result we have in 15.43 the invariance of the wave equation:
24
2
22
4
3
2
1
432124
2
22 1
1000010000100001
1
'x
'
c'
'xc
'x
'x
'x
'xc'x'x'xxc
15.45
The wave equation to the observer Oβ is equal to:
0
1000010000100001
1
4
3
2
1
4321432124
2
22
'xc
'x
'x
'x
'xc'x'x'x'x
'
c1
'x
'
'x
'
x'
'
'x
'
c'
2
2
22
2
2
2
2
2 15.46
Where applying 15.23 and the transposed from 15.23 we have:
4
3
2
1
432124
2
22
00001000010
001
1000010000100001
00
010000100001
1
xc
x
x
x
K
cv
Kcvxcxxx'x
'
c'
15.47
The multiplication of the three middle matrices supplies:
2
12100
01000010
001
00001000010
001
1000010000100001
00
010000100001
cvux
cv
cv
K
cv
Kcv
15.48
Result that can be divided in two matrices:
2
1
2
1 200
00000000
000
1000010000100001
2100
01000010
001
cvux
cv
cv
cvux
cv
cv
15.49
49/172
That applied in the wave equation supplies:
4
3
2
1
2
1432124
2
22
200
00000000
000
1000010000100001
1
xc
x
x
x
cvux
cv
cv
xcxxx'x
'
c'
15.50
Executing the operations of the second term we have:
24
2
2
1
2412412
4
3
2
1
2
14321
2
200
00000000
000
xcux
cv
xxcv
xxcv
x
x
x
x
cvux
cv
cv
xcxxx
Executing the operations we have:
24
2
2
1
2412
22xc
uxcv
xxcv
Where applying 8.5 we have:
zeroxc
uxcv
xxcux
cv
24
2
2
1
2442
1
2
22
Then we have:
zero
x
x
x
x
cvux
cv
cv
xcxxx
4
3
2
1
2
14321
200
00000000
000
15.51
Then in 15.50 we have the invariance of the wave equation:
24
2
22
4
3
2
1
432124
2
22 1
1000010000100001
1
xc
xc
x
x
x
xcxxx'x
'
c'
15.52
Invariance of the equations 8.5 of linear propagation
Replacing 2.4, 8.2, 8.4B in 8.5 we have:
zero
'x'K
'K
'v'x'uc'xc
'v'xxc
uxx
4
1
242142
1
1
1
50/172
Executing the operations we have:
zero'xc
'v'xc
'x'u'xc
'v'xxc
uxx
4242
1
42142
1
1
That simplified supplies the invariance of the equation 8.5:
zero'xc
'x'u'xxc
uxx
42
1
142
1
1
Replacing 2.3, 8.1, 8.3B in 8.5 we have:
zero
xK
K
vuxcxc
vx'xc
'x'u'x
4
1
242142
1
1
1
Executing the operations we have:
zeroxc
vxc
uxxc
vx'xc
'x'u'x
4242
1
42142
1
1
That simplified supplies the invariance of the equation 8.5:
zeroxc
uxx'xc
'x'u'x
42
1
142
1
1
The table 4 in a matrix from becomes:
c/Epxpxpx
K
c/v
c/'E'px'px'px
3
2
1
3
2
1
00001000010
001
15.53
c/'E'px'px'px
'K
c/'v
c/Epxpxpx
3
2
1
3
2
1
00001000010
001
15.54
The table 6 in a matrix form becomes:
cJxJxJx
K
c/v
'c'x'J'x'J'x'J
3
2
1
3
2
1
00001000010
001
15.55
'c'x'J'x'J'x'J
'K
c/'v
cJxJxJx
3
2
1
3
2
1
00001000010
001
15.56
Invariance of the Continuity Equation
The continuity equation to the observer O is equal to:
zero
cJxJxJx
xcxxxx
Ο
xJx
xJx
xJx
x
ΟJ.
3
2
1
432143
3
2
2
1
1
4
15.57
51/172
Where replacing 15.24 and 15.56 we have:
zero
'c'x'J'x'J'x'J
'K
c/'v
'Kc/'v'xc'x'x'xx
ΟJ.
3
2
1
43214
00001000010
001
00010000100001
15.58
The product of the transformation matrices is given in 15.27 and 15.28 with this:
'c'x'J'x'J'x'J
c'x'u'vc/'v
c/'v
'xc'x'x'xx
ΟJ.
3
2
1
2
143214200
00000000
000
1000010000100001
15.59
Executing the operations of the second term we have:
42
1
14
1
23
2
1
2
14321
2
200
00000000
000
'x
'
c'x'u'v
'x
''v
'x'Jx
c'v
'c'x'J'x'J'x'J
c'x'u'vc/'v
c/'v
'xc'x'x'x
Where replacing 11 'x'u''Jx and 8.5 we have:
zero'x'
c'x'u'v'
'xc'x'u'v
'x'
c'x'u'v
42
1
42
1
42
1 2
Then we have:
zero
'c'x'J'x'J'x'J
c'x'u'vc/'v
c/'v
'xc'x'x'x
3
2
1
2
14321200
00000000
000
15.60
With this result we have in 15.59 the invariance of the continuity equation:
43
2
1
43214
1000010000100001
'x
'Ο'J.
'c'x'J'x'J'x'J
'xc'x'x'xx
ΟJ.
15.61
The continuity equation to the observer Oβ is equal to:
zero
'c'x'J'x'J'x'J
x'cx'x'x'x'
Ο'
x'x''J
x'x''J
x'x''J
x'
Ο''J.
3
2
1
432143
3
2
2
1
1
4
15.62
Where replacing 15.23 and 15.55 we have:
zero
cJxJxJx
K
c/v
Kc/vxcxxxx'
Ο''J.
3
2
1
43214
00001000010
001
00010000100001
15.63
52/172
The product of the transformation matrices is given in 15.34 and 15.35 then we have:
cJxJxJx
cvuxc/v
c/v
xcxxxx'
Ο''J. 3
2
1
2
143214200
00000000
000
1000010000100001
15.64
Executing the operations of the second term we have:
42
1
14
1
23
2
1
2
14321
2
200
00000000
000
xcvux
x
v
xJx
cv
cJxJxJx
cvuxc/v
c/v
xcxxx
Where replacing 11 uxJx and 8.5 we have:
zeroxc
vuxxc
uxvxc
vux
42
1
42
1
42
1 2
Then we have:
zero
cJxJxJx
cvuxc/v
c/v
xcxxx
3
2
1
2
14321200
00000000
000
15.65
With this result we have in 15.64 the invariance of the continuity equation:
43
2
1
43214
1000010000100001
x
ΟJ.
cJxJxJx
xcxxxx'
Ο''J.
15.66
Invariance of the line differential element:
That to the observer O is written this way:
242322212 cdxdxdxdxds 4321 cdxdxdxdx
1000010000100001
4
3
2
1
cdxdxdxdx
15.67
Where replacing 15.18 and the transposed from 15.18 we have:
4
3
2
1
43212
00001000010
001
1000010000100001
00
010000100001
'cdx'dx'dx'dx
'K
c'v
'Kc'v
'cdx'dx'dx'dxds 15.68
The multiplication of the three central matrices supplies:
42
12100
01000010
001
00001000010
001
1000010000100001
00
010000100001
'dxc'dx'v
c'v
c'v
'K
c'v
'Kc'v
15.69
53/172
Result that can be divided in two matrices:
42
1
42
1 200
00000000
000
1000010000100001
2100
01000010
001
'dxc'dx'v
c'v
c'v
'dxc'dx'v
c'v
c'v
15.70
That applied in the line differential element supplies:
4
3
2
1
42
1
43212
200
00000000
000
1000010000100001
'cdx'dx'dx'dx
'dxc'dx'v
c'v
c'v
'cdx'dx'dx'dxds 15.71
Executing the operations of the second term we have:
zero'cdx'dx'dx
c'v'dx
c'v'cdx
c'cdx'dx'v
'cdx'dx'dx'dx
'dx'dx
c'v
c'v
c'v
'cdx'dx'dx'dx
44
1
214
41
4
3
2
1
4
1
2
4321 2
200
00000000
000
Then we have:
zero
'cdx'dx'dx'dx
'dx'dx'
c'v
c'v
c'v
'cdx'dx'dx'dx
4
3
2
1
4
1
2
4321
200
00000000
000
15.72
With this result we have in 15.71 the invariance of the line differential element:
224232221
4
3
2
1
43212
1000010000100001
'ds'cdx'dx'dx'dx
'cdx'dx'dx'dx
'cdx'dx'dx'dxds
15.73
To the observer Oβ the line differential element is written this way:
242322212 'cdx'dx'dx'dx'ds 4321 'cdx'dx'dx'dx
1000010000100001
4
3
2
1
'cdx'dx'dx'dx
15.74
Where replacing 15.15 and the transposed from 15.15 we have:
4
3
2
1
43212
00001000010
001
1000010000100001
00
010000100001
cdxdxdxdx
K
cv
Kcv
cdxdxdxdx'ds 15.75
54/172
The multiplication of the three central matrices supplies:
42
12100
01000010
001
00001000010
001
1000010000100001
00
010000100001
dxcvdx
cv
cv
K
cv
Kcv
15.76
Result that can be divided in two matrices:
42
1
42
1 200
00000000
000
1000010000100001
2100
01000010
001
dxcvdx
cv
cv
dxcvdx
cv
cv
15.77
That applied in the line differential element supplies:
4
3
2
1
42
1
43212
200
00000000
000
1000010000100001
cdxdxdxdx
dxcvdx
cv
cv
cdxdxdxdx'ds 15.78
Executing the operations of the second term we have:
zerocdxdxdx
cvdx
cvcdx
ccdxvdx
cdxdxdxdx
dxdx
cv
cv
cv
cdxdxdxdx
44
1
214
41
4
3
2
1
4
1
2
4321 2
200
00000000
000
Then we have:
zero
cdxdxdxdx
dxdx
cv
cv
cv
cdxdxdxdx
4
3
2
1
4
1
2
4321
200
00000000
000
15.79
With this result we have in 15.78 the invariance of the line differential element:
224232221
4
3
2
1
43212
1000010000100001
dscdxdxdxdx
cdxdxdxdx
cdxdxdxdx'ds
15.80
In Β§7 as a consequence of 5.3 we had the invariance of 'u'.Eu.E
where now applying 7.3.1, 7.3.2, 7.4.1,
7.4.2 and the velocity transformation formulae from table 2 we have new relations between Ex and 'x'E distinct from 7.3 and 7.4 and with them we rewrite the table 7 in the form below:
55/172
Table 7B
uxvKExxE
1''
7.3B
'''1
'''
xuvKxEEx
7.4B
KEyy'E' 7.3.1 K'y'E'Ey
7.4.1
KEzz'E' 7.3.2 K'z'E'Ez
7.4.2
Bxx'B' 7.5 x'B'Bx 7.6
Ezc
vByy'B'
2
7.5.1
z'E'c
v'y'B'By
2
7.6.1
Eyc
vBzz'B'
2
7.5.2
y'E'c
v'z'B'Bz
2
7.6.2
Ezc
uxBy
2
7.9
z'E'c
x''uy'B'
2
7.10
Eyc
uxBz
2
7.9.1
y'E'c
x''uz'B'
2
7.10.1
1''
'11
xuv
uxv
With the tables 7B and 9B we can have the invariance of all Maxwellβs equations. Invariance of the Gaussβ Law for the electrical field:
0'
'z'z'E
'y'y'E
'x'x'E
8.14
Where applying the tables 6, 7B and 9B we have:
02 1
KzKEz
yKEy
ux/vKEx
tcv
x
Where simplifying and replacing 8.5 we have:
011
zEz
yEy
ux/vEx
xuxv
x
That reordered supplies:
011
zEz
yEy
ux/vEx
uxv
x.
That simplified supplies the invariance of the Gaussβ Law for the electrical field. Invariance of the Gaussβ Law for the magnetic field:
zero'z'z'B
'y'y'B
'x'x'B
8.16
Where applying the tables 7B and 9B we have:
0222
Ey
cvBz
zEz
cvBy
yBx
tcv
x
56/172
That reordered supplies:
02
tBx
zEy
yEz
cv
zBz
yBy
xBx
Where the term in parenthesis is the Faraday-Henryβs Law (8.19) that is equal to zero from where we have the invariance of the Gaussβ Law for the magnetic field. Invariance of the Faraday-Henryβs Law:
't'z'B
'y'x'E
'x'y'E
8.18
Where applying the tables 7B and 9B we have:
Ey
cvBz
tK
ux/vKEx
yKEy
tcv
x 22 1
That simplified and multiplied by ux/v1 we have:
uxv
tBz
yEx
uxv
xEy
11
Where executing the products and replacing 7.9.1 we have:
tEy
cux
xEy
uxv
tBz
yEx
xEy
2
As the term in parenthesis is the equation 8.5 that is equal to zero then we have the invariance of the Faraday-Henryβs Law. Invariance of the Faraday-Henryβs Law:
't'x'B
'z'y'E
'y'z'E
8.20
Where applying the tables 7B and 9B we have:
tBxKK
zEy
KyEz
That simplified supplies the invariance of the Faraday-Henryβs Law. Invariance of the Faraday-Henryβs Law:
't'y'B
'x'z'E
'z'x'E
8.22
Where applying the tables 7B and 9B we have:
Ez
cvBy
tKKEz
tcv
xux/vKEx
z 221
That simplified and multiplied by ux/v1 we have:
uxv
tEz
cv
uxv
tBy
uxv
tEz
cv
uxv
xEz
zEx 1111
22
57/172
That simplifying and making the operations we have:
tBy
xEz
uxv
tBy
xEz
zEx
Where applying 7.9 we have:
tEz
cux
xEz
uxv
tBy
xEz
zEx
2.
As the term in parenthesis is the equation 8.5 that is equal to zero then we have the invariance of the Faraday-Henryβs Law. Invariance of the Ampere-Maxwellβs Law:
't'z'E'z'J
'y'x'B
'x'y'B
ooo
8.24
Where applying the tables 6, 7B and 9B we have:
KEzt
KJzyBxEz
cvBy
tcv
x
00022
That simplifying and making the operations we have:
tEz
cv
ctBy
cv
xEz
cv
tEz
cvux
ctEz
cv
ctEzJz
yBx
xBy
2
2
222222
2
20001211
Where simplifying and applying 7.9 we have:
tEz
cux
cv
xEz
cv
tEz
cvux
ctEzJz
yBx
xBy
2222200021
That reorganized supplies
xEz
tEz
cux
cv
tEzJz
yBx
xBy
22000
As the term in parenthesis is the equation 8.5 that is equal to zero then we have the invariance of the Ampere-Maxwellβs Law: Invariance of the Ampere-Maxwellβs Law:
't'x'E'x'J
'z'y'B
'y'z'B
ooo
8.26
Where applying the tables 6, 7B and 9B we have:
ux/vKEx
tKvJxEz
cvBy
zEy
cvBz
y
100022
Making the operations we have:
ux/vtEx
cvux
cvc
zEz
yEy
cvJx
zBy
yBz
1121
22
2
002
020
58/172
Replacing in the first parenthesis the Gaussβ Law and multiplying by
uxv1 we have:
tEx
cvux
ctEx
cv
cxEx
uxcv
xEx
cvJx
zBy
yBz
uxv
tExJx
zBy
yBz
222
2
22
2
200002111
Where replacing uxJx , 7.9.1, 7.9 and 8.5 we have:
tEx
cvux
ctEx
cv
ctEx
ccv
xEx
cvux
zEz
cux
yEy
cux
uxv
tExJx
zBy
yBz
222
2
222
2
20220002111
That simplified supplies:
tEx
cvux
cxEx
cvc
zEz
yEy
cv
tExJx
zBy
yBz
2222
0200021
Replacing in the first parenthesis the Gaussβ Law we have:
tEx
cvux
cxEx
cv
xEx
cv
tExJx
zBy
yBz
222200021
That reorganized makes:
tEx
cux
xEx
cv
tExJx
zBy
yBz
220002
As the term in parenthesis is the equation 8.5 that is equal to zero then we have the invariance of the Ampere-Maxwellβs Law: Invariance of the Ampere-Maxwellβs Law:
't'y'E
'y'J'x'z'B
'z'x'B
ooo
8.28
Where applying the tables 6, 7B and 9B we have:
KEyt
KJyEycvBz
tcv
xzBx
00022
Making the operations we have:
tEy
cv
ctBz
cv
xEy
cv
tEy
cvux
ctEy
cv
ctEy
JyxBz
zBx
2
2
222222
2
20001211
Where simplifying and applying 7.9.1 we have:
tEy
cux
cv
xEy
cv
tEy
cvux
ctEy
JyxBz
zBx
2222200021
That reorganized makes:
xEy
tEy
cux
cv
tEy
JyxBz
zBx
22000
As the term in parenthesis is the equation 8.5 that is equal to zero then we have the invariance of the Ampere-Maxwellβs Law:
59/172
Invariance of the Gaussβ Law for the electrical field without electrical charge:
zero'z'z'E
'y'y'E
'x'x'E
8.30
Where applying the tables 7B and 9B we have:
zerozKEz
yKEy
ux/vKEx
tcv
x
12
Where simplifying and replacing 8.5 we have:
zeroz
EzyEy
ux/vEx
xuxv
x
11
That reorganized makes:
zeroz
EzyEy
ux/vEx
uxv
x
11 .
That simplified supplies the Gaussβ Law for the electrical field without electrical charge. Invariance of the Ampere-Maxwellβs Law without electrical charge:
't'z'E
'y'x'B
'x'y'B
oo
8.40
Where applying the tables 7B and 9B we have:
KEzt
KyBxEz
cvBy
tcv
x
0022
Making the operations we have:
tEz
cv
ctBy
cv
xEz
cv
tEz
cvux
ctEz
cv
ctEz
yBx
xBy
2
2
222222
2
2001211
Where simplifying and applying 7.9 we have:
tEz
cux
cv
xEz
cv
tEz
cvux
ctEz
yBx
xBy
222220021
That reorganized makes:
xEz
tEz
cux
cv
tEz
yBx
xBy
2200
As the term in parenthesis is the equation 8.5 that is equal to zero then we have the invariance of the Ampere-Maxwellβs Law without electrical charge: Invariance of the Ampere-Maxwellβs Law without electrical charge:
't'x'E
'z'y'B
'y'z'B
oo
8.42
Where applying the tables 7B and 9B we have:
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ux/vKEx
tKEz
cvBy
zEy
cvBz
y
10022
Making the operations we have:
ux/vtEx
cvux
cv
zEz
yEy
cv
zBy
yBz
1121
22
2
002
Replacing in the first parenthesis the Gaussβ Law without electrical charge and multiplying by ux/v1 we
have:
tEx
cvux
ctEx
cv
cxEx
uxcv
xEx
cv
zBy
yBz
uxv
tEx
zBy
yBz
222
2
22
2
2002111
Where replacing 7.9, 7.9.1 and 8.5 we have:
tEx
cvux
ctEx
cv
ctEx
ccv
xEx
cv
zEz
cux
yEy
cux
uxv
tEx
zBy
yBz
222
2
222
2
222002111
That simplified supplies:
tEx
cvux
cxEx
cv
zEz
yEy
cv
tEx
zBy
yBz
22220021
Replacing in the first parenthesis the Gaussβ Law without electrical charge we have:
tEx
cvux
cxEx
cv
xEx
cv
tEx
zBy
yBz
22220021
That reorganized makes:
tEx
cux
xEx
cv
tExJx
zBy
yBz
220002
As the term in parenthesis is the equation 8.5 that is equal to zero then we have the invariance of the Ampere-Maxwellβs Law without electrical charge: Invariance of the Ampere-Maxwellβs Law without electrical charge:
't'y'E
'x'z'B
'z'x'B
oo
8.44
Where applying the tables 6, 7B and 9B we have:
KEyt
KEycvBz
tcv
xzBx
0022
Making the operations we have:
tEy
cv
ctBz
cv
xEy
cv
tEy
cvux
ctEy
cv
ctEy
xBz
zBx
2
2
222222
2
2001211
Where simplifying and applying 7.9.1 we have:
tEy
cux
cv
xEy
cv
tEy
cvux
ctEy
xBz
zBx
222220021
61/172
That reorganized makes:
xEy
tEy
cux
cv
tEy
xBz
zBx
2200
As the term in parenthesis is the equation 8.5 that is equal to zero then we have the invariance of the Ampere-Maxwellβs Law without electrical charge:
Β§15 Invariance (continuation)
A function wtf krf 2.19
Where the phase is equal to wtkr 15.81
In order to represent an undulating movement that goes on in one arbitrary direction must comply with the wave equation and because of this we have:
zero
fk
fzyx
r
kf
r
zyxr
r
k
2
22
2
2222
2
2222
23
15.82
That doesnβt meet with the wave equation because the two last elements get nule but the first one doesnβt. In order to overcome this problem we reformulate the phase of the function in the following way. A unitary vector such as
kcosjcosicosn 15.83
where ct
x
r
xcos ,
ct
y
r
ycos ,
ct
z
r
zcos 15.84
has the module equal to 1222 coscoscosn.nnn
. 15.85
Making the product
rr
r
r
zyxzcosycosxcoskzjyix.kcosjcosicosR.n
2222
15.86
we have zcosycosxcosR.nr
that applied to the phase supplies a new phase
wtzcoskycoskxcoskwtR.nkwtkr
15.87
with the same meaning of the previous phase .
Replacing zcosycosxcosR.nr
e c
wk in the phase multiplied by β1 we also get another
phase in the form
c
zcosycosxcostw
c
rtwkrwtwtkr
1 15.88
with the same meaning of the previous phase 1 .
Thus we can write a new function as:
c
zcosycosxcostwf
f 15.89
62/172
That replaced in the wave equation with the director cosine considered constant supplies:
zero
c
wfcos
c
wfcos
c
wfcos
c
wf
2
2
2
22
2
2
2
22
2
2
2
22
2
2
2
2
15.90
that simplified meets the wave equation. The positive result of the phase in the wave equation is an exclusive consequence of the director cosines being constant in the partial derivatives showing that the wave equation demands the propagation to have one steady direction in the space (plane wave). For the observer O a source located in the origin of its referential produces in a random point located at the
distance 222 zyxctr of the origin, an electrical field E
described by:
kEzjEyiExE
15.91
Where the components are described as:
f.EEz
f.EEy
f.EEx
zo
yo
xo
15.92
That applied in E
supplies:
fEfEjEiEkfEjfEifEE ozoyoxozoyoxo
. 15.93
with module equal to f.EEf.EEEE ozoyoxo
222 15.94
Being kEjEiEE zoyoxoo
15.95
The maximum amplitude vector Constant with the components Exo, Eyo, Ezo 15.96
And module 222
zoyoxoo EEEE 15.97
Being f a function with the phase equal to15.87 or 15.88.
Deriving the component Ex in relation to x and t we have:
ct
kxfE
r
kxfE
x
wtkrfE
x
fE
x
Exxoxoxoxo
15.98
wf
Et
wtkrfE
t
fE
t
Exxoxoxo
15.99
that applied in 8.5 supplies
zero
tc
t/x
x
fEzero
t
fE
c
t/x
x
fEzero
t
Ex
c
t/x
x
Exxoxoxo
222
zero
tc
t/x
xzero
tc
t/x
x
fExo
22 15.100
demonstrating that it is the phase that must comply with 8.5.
63/172
zeroc
wk
ct
xzerow
c
t/x
ct
kxzero
t
wtkr
c
t/x
x
wtkrzero
tc
t/x
x
222
as c
wk then Ex complies with 8.5.
As the phase is the same for the components Ey and Ez then they also comply with 8.5.
As the phases for the observers O and Oβ are equal 't'w'r'kwtkr then the components of the
observer Oβ also comply with 8.5.
zero't
't'w'r'k
c
't/'x
'x
't'w'r'k
t
wtkr
c
t/x
x
wtkr
22 15.101
The components relatively to the observer O of the electrical field are transformed for the referential of the observer Oβ according to the tables 7, 7B and 8. Applying in 8.5 a wave function written in the form:
sincossincos iwtkxiwtkxee iwtkxi 15.102
where 1i . Deriving we have:
coskisenkx
end
coswisenwt
15.103
or ikex
and iwet
15.104
That applied in 8.5 supplies:
zerocoswisenwc
t/xcoskisenkzero
tc
t/x
x
22
that is equal to:
zerotc
xwiki
tc
xwk
cossin
22
or zerowec
t/xkezero
tc
t/x
xii
22
where we must have the coefficients equal to zero so that we get na identity, then:
tc
xwkzero
tc
xwiki
tc
xwkzero
tc
xwk
22
22
tc
xwkzerowe
c
t/xke ii
22
Where applying ckw we have:
64/172
ct
x
tc
xck
tc
xwk
22
Then to meet with the equation 8.5 we must have a wave propagation along the axis x with the speed c.
If we apply ukw and t
xv we have:
v
cu
c
vuk
tc
xwk
2
22 .
A result also gotten from the Louis de Broglieβs wave equation.
Β§16 Time and Frequency
Considering the Doppler effect as a law of physics. We can define a clock as any device that produces a frequency of identical events in a series possible to be enlisted and added in such a way that a random event n of a device will be identical to any event in the series of events produced by a replica of this device when the events are compared in a relative resting position. The cyclical movement of a clock in a resting position according to the observer O referential sets the time in this referential and the cyclical movement of the arms of a clock in a resting position according to the observer Oβ sets the time in this referential. The formulas of time transformation 1.7 and 1.8 relate the times between the referentials in relative movement thus, relate movements in relative movement. The relative movement between the inertial referentials produces the Doppler effect that proves that the frequency varies with velocity and as the frequency can be interpreted as being the frequency of the cyclical movement of the arms of a clock then the time varies in the same proportion that varies the frequency with the relative movement that is, it is enough to replace the time t and tβ in the formulas 1.7 and 1.8 by the frequencies y and yβ to get the formulas of frequency transformation, then:
Ky'yKt't 1.7 becomes 2.22
'K'yy'K'tt 1.8 becomes 2.22
The Galileoβs transformation of velocities vu'u
between two inertial referentials presents intrinsically three defects that can be described this way: a) The Galileoβs transformation of velocity to the axis x is vux'x'u . In that one if we have cux then
vc'x'u and if we have c'x'u then vcux . As both results are not simultaneously possible or else
we have cux or c'x'u then the transformation doesnβt allow that a ray of light be simultaneously observed by the observers O and Oβ what shows the privilege of an observer in relation to the other because each observer can only see the ray of light running in its own referential (intrinsic defect to the classic analysis of the Sagnacβs effect). b) It cannot also comply to Newtonβs first law of inertia because a ray of light emitted parallel to the axis x from the origin of the respective inertial referentials at the moment that the origins are coincident and at the moment in which t = tβ = zero will have by the Galileoβs transformation the velocity c of light altered by v to the referentials, on the contrary of the inertial law that wouldnβt allow the existence of a variation in velocity because there is no external action acting on the ray of light and because of this both observers should see the ray of light with velocity c. c) As it considers the time as a constant between the referentials it doesnβt produce the temporal variation between the referentials in movement as it is required by the Doppler effect. The principle of constancy of light velocity is nothing but a requirement of the Newtonβs first law, the inertia law. Newtonβs first law, the inertia law, is introduced in Galileoβs transformation when the principle of constancy of light velocity is applied in Galileoβs transformation providing the equation of tables 1 and 2 of the Undulating Relativity that doesnβt have the three defects described.
65/172
The time and velocity equations of tables 1 and 2 can be written as:
cosc
v
c
vt't
21
2
2
1.7
cosc
v
c
v
v'v
21
2
2
1.15
'cosc
'v
c
'v'tt 2
12
2
1.8
'cosc
'v
c
'v
'vv
21
2
2
1.20
The distance d between the referentials is equal to the product of velocity by time this way:
't'vvtd 1.9 It doesnβt depend on the propagation angle of the ray of light, being exclusively a function of velocity and time, that is, the propagation angle of the ray of light, only alters between the inertial referential the proportion between time and velocity, keeping the distance constant in each moment, to any propagation angle. The equations above in a function form are written as:
't,'v'et,ved 1.9
,t,vf't 1.7
,vg'v 1.15
','t,'v'ft 1.8
','v'gv 1.20
Then we have that the distance is a function of two variables, the time a function of three variables and the velocity a function of two variables. From the definition of moment 4.1 and energy 4.6 we have:
uc
Ep
2
16.1
The elevated to the power of two supplies:
22
2
2
2
pEc
cu 16.2
Elevating to the power of two the energy formula we have:
4202
222
2
2
2
202 cm
cuEE
cu1
cmE
Where applying 16.2 we have:
66/172
220
2420
22
22242
02
222 cmpcEcmp
EcEEcm
cuEE 4.8
From where we conclude that if the mass in resting position of a particle is null zeromo the particle
energy is equal to pcE . 16.3
That applied in 16.2 supplies:
cup
cpc
cup
Ec
cu 2
2
2
2
22
2
2
2
2
16.4
From where we conclude that the movement of a particle with a null mass in resting position zeromo will
always be at the velocity of light cu . Applying in pcE the relations yhE and yc we have:
hppyyh and in the same way
'
h'p
16.5
Equation that relates the moment of a particle with a null mass in resting position with its own way length. Elevating to the power of two the formula of moment transformation (4.9) we have:
vpxcE2v
cEp'pv
cEp'p
22
4
222
2
Where applying pcE and cuxpcosppx we find:
Kp'p
c
vux
c
vp'p
c
uxvp
c
cpv
c
cpp'p
22
2
22
4
222 2
12 16.6
Where applying 16.5 results in:
K'K
h
'
hKp'p
or inverted 'K
' 2.21
Where applying yc and ''yc we have:
Ky'y or inverted 'K'yy 2.22
In Β§ 2 we have the equations 2.21 and 2.22 applying the principle of relativity to the wave phase.
17 Transformation of H. Lorentz For two observers in a relative movement, the equation that represents the principle of constancy of light speed for a random point A is:
2222222222 tczyx'tc'z'y'x 17.01
In this equation canceling the symmetric terms we have: Nesta cancelando os termos simΓ©tricos obtemos:
222222 tcx'tc'x 17.02
67/172
That we can write as: ctxctx'ct'x'ct'x 17.03
If in this equation we define the proportion factors and as:
Bctx'ct'x
Actx'ct'x
17.04
where we must have 1. to comply 17.03.
The equations 17.04 where first gotten by Albert Einstein. When a ray of light moves in the plane yβzβ to the observer Oβ we have xβ = zero and x = vt and such conditions applied to the equation 17.02 supplies:
2
222222 10
cvt'ttcvt'tc 17.05
This result will also be supplied by the equations A and B of the group 17.04 under the same conditions:
Bctvtcvct
Actvtcvct
2
2
2
2
10
10
17.06
From those we have:
cvcv
1
1 and
cvcv
1
1 17.07
Where we have proven that 1. .
From the group 17.04 we have the Transformations of H. Lorentz:
ctx'x
22 17.08
ctx'ct22 17.09
'ct'xx22 17.10
'ct'xct22 17.11
Indexes equations 2
,2
and 2
:
2
2
2
21
12
1
2
11
11
1
1
1
1
cv
cv
cv
cv
cv
cv
cvcv
cvcv
17.12
2
2
2
21
21
2
11
11
1
1
1
1
cv
cv
cv
cv
cv
cv
cv
cv
cvcv
cvcv
17.13
68/172
2
2
2
21
21
2
11
11
1
1
1
1
cv
cv
cv
cv
cv
cv
cv
cv
cvcv
cvcv
17.14
Sagnac effect
When both observersβ origins are equal the time is zeroed (t = tβ = zero) in both referentials and two rays of light are emitted from the common origin, one in the positive direction (clockwise index c) of the axis x and xβ with a wave front Ac and another in the negative direction (counter-clockwise index u) of the axis x and xβ with a wave front Au. The propagation conditions above applied to the Lorentz equations supply the tables A and B below: Table A Equation Clockwise ray (c) Equation Counter-clockwise ray (u) Sum of rays Result Result Condition
cc ctx Condition uu ctx
17.08 cc ct'x 17.08
uu ct'x
cc x'x
uu x'x ucuc xx'x'x
17.09 cc ct'ct 17.09
uu ct'ct ucuc ctct'ct'ct
cc 'ct'x
uu 'ct'x
Table B Equation Clockwise ray (c) Equation Counter-clockwise ray (u) Sum of rays Result Result Condition
cc 'ct'x Condition uu 'ct'x
17.10 cc 'ctx 17.10
uu 'ctx
cc 'xx
uu 'xx ucuc 'x'xxx
17.11 cc 'ctct 17.11
uu 'ctct ucuc 'ct'ctctct
cc ctx
uu ctx
We observe that the tables A and B are inverse one to another. When we form the group of the sum equations of the two rays from tables A and B:
B'ct'ctctctD
Actct'ct'ct'D
ucuc
ucuc
17.15
Where to the observer Oβ cu AA'D is the distance between the front waves Au and Ac and where to the
observer O cu AAD is the distance between the front waves Au and Ac.
In the equations 17.15 above, due to the isotropy of space and time and the front waves cu AA of the
two rays of light being the same for both observers, the sum of rays of light e times must be invariable between the observers, which we can express by:
t'tctct'ct'ctD'D ucuc 17.16
This result that generates an equation of isotropy of space and time can be called as the conservation of space and time principle.
69/172
The three hypothesis of propagation defined as follows will be applied in 17.15 and tested to prove the conservation of space and time principle given by 17.16: Hypothesis A: If the space and time are isotropic and there is no movement with no privilege of one observer considered over the other in an empty space then the propagation geometry of rays of light can be given by:
uc 'ctct and cu 'ctct 17.17
This hypothesis applied to the equation A or B of the group 17.15 complies to the space and time conservation principle given by 17.16. The hypothesis 17.17 applied to the tables A and B results in:
Dctct
CctctBQuadro
B'ct'ct
A'ct'ctAQuadro
cu
uc
cu
uc
17.18
Hypothesis B: If the space and time are isotropic but the observer O is in an absolute resting position in an empty space then the geometry of propagation of the rays of light is given by:
ctctct uc 17.19
That applied to the table A and B results in:
D'ctct
C'ctctBQuadro
Bct'ct
Act'ctAQuadro
u
c
u
c
17.20
B'ct'ct
A'ct'ct
cu
uc
2
2
17.21
Summing A and B in 17.20 we have:
2
2
2
2uc
cv
1
t't
cv
1
D'D
2D'D
2ct2'ct'ct
17.22
This result doesnβt comply with the conservation of space and time principle given by 17.16 and as D'D it results in a situation of four rays of light, two to each observer, and each ray of light with its respective independent front wave from the others. Hypothesis C: If the space and time are isotropic but the observer Oβ is in an absolute resting position in an empty space then the propagation geometry of the rays of light is given:
'ct'ct'ct uc 17.23
That applied to the tables A and B results in:
D'ctct
C'ctctBQuadro
Bct'ct
Act'ctAQuadro
u
c
u
c
17.24
70/172
Bctct
Actct
c2
u
u2
c
17.25
Summing C and D in 17.24 we have:
2
2
2
2uc
cv
1
'tt
cv
1
'DD
2'DD
2'ct2ctct
17.26
This result doesnβt comply with the conservation of space and time principle exactly the same way as hypothesis B given by 17.16 and as D'D D'D it results in a situation of four rays of light, two to each observer and each ray of light with its respective independent front wave from the others.
Conclusion
The hypothesis A, B and C are completely compatible with the demand of isotropy of space and time as we can conclude with the geometry of propagations. The result of hypothesis A is contrary to the result of hypothesis B and C despite of the relative movement of the observers not changing the front wave Au relatively to the front wave Ac because the front waves have independent movement one from the other and from the observers. The hypothesis A applied in the transformations of H. Lorentz complies with the conservation of space and time principle given by 17.16 showing the compatibility with the transformations of H. Lorentz with the hypothesis A. The application of hypothesis B and C in the transformations of H. Lorentz supplies the space and time deformations given by 17.22 and 17.26 because the transformations of H. Lorentz are not compatible with the hypothesis B and C. For us to obtain the Sagnac effect we must consider that the observer Oβ is in an absolute resting position, hypothesis C above and that the path of the rays of light be of R2 :
R'ct'ct'ct uc 2 17.27
For the observer O the Sagnac effect is given by the time difference between the clockwise ray of light and the counter-clock ray of light uc ttt that can be obtained using 17.24 (C-D), 17.27 and 17.14:
22
2
2
4
1
22
vcc
Rv
cv
cv
cR'tttt uc
17.28
Β§9 The Sagnac Effect (continuation)
The moment the origins are the same the time is zeroed (t = tβ = zero) at both sides of the referential and the rays of light are emitted from the common origin, one in the positive way (clockwise index c) of the axis x and xβ with a wave front Ac and the other one in the negative way (counter clockwise index u) of the axis x and xβ with wave front Au. The projected ray of light in the positive way (clockwise index c) of the axis x and xβ is equationed by
cc ctx and cc 'ct'x that applied to the Table I supplies:
cccc
cc Kct'ctcv
ct'ct
1 (1.7) ccc
ccc 'K'ctct
c'v
'ctct
1 (1.8) 9.11
c
cc
c
cc K
v'v
cv
v'v
1
(1.15) c
cc
c
cc 'K
'vv
c'v
'vv
1
(1.20) 9.12
71/172
From those we deduct that the distance between the observers is given by:
ccccc 't'vtvd 9.13
Where we have:
111
cc
cc 'KKc'v
cv
9.14
The ray of light project in the negative way (counter clockwise index u) of the axis x and xβ is equationed by
uu ctx and uu 'ct'x : that applied to the Table I gives:
uuuu
uu Kct'ctcv
ct'ct
1 (1.7) uuu
uuu 'K'ctct
c'v
'ctct
1 (1.8) 9.15
u
uu
u
uu K
v'v
cv
v'v
1
(1.15) u
uu
u
uu 'K
'vv
c'v
'vv
1
(1.20) 9.16
From those we deduct that the distance between the observers is given by:
uuuuu 't'vtvd 9.17
Where we have:
111
uu
uu 'KKc'v
cv
9.18
We must observe that at first there is no relationship between the equations 9.11 to 9.14 with the equations 9.15 to 9.18. With the propagation conditions described we form the following Tables A and B: Table A
Equation Clockwise ray of light (c)
Equation Counter clockwise ray of light (u)
Sum of the rays of light
Result Result
Condition cc ctx Condition uu ctx
1.2 ccc Kct'x 1.2 uuu Kct'x
ccc Kx'x uuu Kx'x uuccuc KxKx'x'x
1.7 ccc Kct'ct 1.7 uuu Kct'ct uuccuc KctKct'ct'ct
cc 'ct'x uu 'ct'x
Table B
Equation Clockwise ray of light (c)
Equation Counter clockwise ray of light (u)
Sum of the rays of light
Result Result
Condition cc 'ct'x Condition uu 'ct'x
1.4 ccc 'K'ctx 1.4 uuu 'K'ctx
ccc 'K'xx uuu 'K'xx uuccuc 'K'x'K'xxx
1.8 ccc 'K'ctct 1.8 uuu 'K'ctct uuccuc 'K'ct'K'ctctct
cc ctx uu ctx
72/172
We observe that for the rays of light with the same direction the Tables A and B are inverse from each other. Forming the equations group of the sum of the rays of light of the Tables A and B:
B'K'ct'K'ctctctD
AKctKct'ct'ct'D
uuccuc
uuccuc 9.19
Where for the observer Oβ cu AA'D is the distance between the wave fronts Au and Ac and where for
the observer O cu AAD is the distance between the wave fronts Au and Ac.
In the equations above 9.19 due to the isotropy of the space and time and the wave fronts cu AA of the
rays of light being the same for both observers, the sumo of the rays of light and of times must be invariable between the observers, which is expressed by:
t'tctct'ct'ctD'D ucuc 9.20
This result that equations the isotropy of space and time can be called as the space and time conservation principle. The three hypothesis of propagations defined next will be applied in 9.19 and tested to prove the compliance of the conservation of space and time principle given by 9.20. With these hypotheses we create a bond between the equations 9.11 to 9.14 with the equations 9.15 to 9.18. Hypothesis A: If the space and time are isotropic and there is movement with any privilege of any observer over each other in the empty space then the propagation geometry of the rays of light is equationed by:
B'KK'vv'tt'ctct
A'KK'vv'tt'ctct
cucucucu
ucucucuc 9.21
With those we deduct that the distance between the observers is given by:
uuuuccccuc 't'vtv't'vtvdd 9.22
Results that applied in the equations A or B of the group 9.19 complies with the conservation of space and time principle given by 9.20, showing that the Doppler effect in the clockwise and counter clockwise rays of light are compensated in the referentials. Hypothesis B: If the space and time are isotropic but the observer O is in an absolute resting position in the empty space then the propagation geometry of the rays of light is equationed by:
Cvttvtv
Bvvv
Actctct
uucc
uc
uc
9.23
With those we deduct that the distance between the observers is given by:
uuccuc 't'v't'vvtdd 9.24
Results that applied in the equations A or B of the group 9.19 complies with the conservation of space and time principle given by 9.20, showing that the Doppler effect in the clockwise and counter clockwise rays of light are compensated in the referentials.
73/172
Hypothesis C: If the space and time are isotropic but the observer O is in an absolute resting position in the empty space then the propagation geometry of the rays of light is equationed by:
C't'v't'v't'v
B'v'v'v
A'ct'ct'ct
uucc
uc
uc
9.25
With those we deduct that the distance between the observers is given by:
uuccuc tvtv't'vdd 9.26
Results that applied in the equations A or B of the group 9.19 complies with the conservation of space and time principle given by 9.20, showing that the Doppler effect in the clockwise and counter clockwise rays of light are compensated in the referentials. In order to obtain the Sagnac effect we consider that the observer Oβ is in an absolute resting position, hypothesis C above and that the rays of light course must be of R2 :
R'ct'ct'ct uc 2 9.27
Applying the hypothesis C in 9.11 and 9.15 we have:
c'v'tt'K'tt cccc 1 9.28
c'v'tt'K'tt uuuu 1 9.29
For the observer O the Sagnac effect is given by the time difference between course of the clockwise ray of light and the counter clock ray of uc ttt that can be obtained making (9.28 β 9.29) and applying 9.27
making:
24211c
'Rvc't'v
c'v't
c'v'tttt uc
9.30
The equation ctv
ctv
c't'vt uucc 222 is exactly the result obtained from the geometry analysis of
the propagation of the clockwise and counter clockwise rays of light in a circumference showing the coherence of the hypothesis adopted by the Undulating Relativity. In 9.30 applying 9.12 and 9.16 we have the final result due to cv and uv :
u
u
c
cuc
cvc
Rv
cvc
Rv
cRv
ctvttt
222
44'4''2 9.31
The classic formula of the Sagnac effect is given as:
224
vcRvttt uc
9.32
From the propagation geometry we have:
cvtt 2 9.33
74/172
The classic times would be given by:
cRt 2 9.34
vcRtc
2 9.35
vcRtu
2 9.36
Applying 9.34, 9.35 and 9.36 in 9.33 we have:
2422cRv
cR
cvt 9.37
cvcRv
vcR
cvtc
2422 9.38
cvcRv
vcR
cvtu
2422 9.39
The results 9.37, 9.38 and 9.39 are completely different from 9.32.
Β§18 The Michelson & Morley experience The traditional analysis that supplies the solution for the null result of this experience considers a device in a resting position at the referential of the observer Oβ that emits two rays of light, one horizontal in the xβ direction (clockwise index c) and another vertical in the direction yβ. The horizontal ray of light (clockwise index c) runs until a mirror placed in xβ = L at this point the ray of light reflects (counter clockwise index u) and returns to the origin of the referential where xβ = zero. The vertical ray of light runs until a mirror placed in yβ = L reflects and returns to the origin of the referential where yβ = zero. In the traditional analysis according to the speed of light constancy principle for the observer Oβ the rays of light track is given by:
L'ct'ct uc 18.01
For the observer Oβ the sum of times of the track of both rays of light along the xβ axis is:
cL
cL
cL't't't uc'x
2 18.02
In the traditional analysis for the observer Oβ the sum of times of the track of both rays of light along the yβ axis is:
cL
cL
cL't't't 'y
2 18.03
As we have cL't't 'y'x2 there is no interference fringe and it is applied the null result of the
Michelson & Morley experience. In this traditional analysis the identical track of the clockwise and counter clockwise rays of light in the equation 18.01 that originates the null result of the Michelson & Morley experience contradicts the Sagnac effect that is exactly the time difference existing between the track of the clockwise and counter clockwise rays of light. Based on the Undulating Relativity we make a deeper analysis of the Michelson & Morley experience obtaining a result that complies completely with the Sagnac effect. Observing that the equation 18.01 corresponds to the hypothesis C of the paragraph Β§9.
75/172
Applying 18.01 in 9.19 we have:
B'K'KL'LK'LKctctD'K'ct'K'ctctctD
AKctKctLL'DKctKct'ct'ct'D
ucucucuuccuc
uuccuuccuc 18.04
From 18.04 A we have:
uuucccu
uc
c tvcttvctL'Dcv
ctcv
ctL'D
2112 18.05
Where applying 9.26 we have:
cLtttctctL'D ucxuc22 18.06
In 18.04 B we have:
c'v
c'v
LctctD ucuc 11 18.07
Where applying 9.25 B we have:
cLtttLctctD ucxuc22 18.08
The equations 18.06 and 18.08 demonstrate that the Doppler effect in the clockwise and counter clockwise rays of light compensate itself in the referential of the observer O resulting in:
cLt't't x'x'y2 18.09
Because of this, according to the Undulating Relativity in the Michelson & Morley experience we can predict that the clockwise ray of light has a different track from the counter clockwise ray of light according to the formula 18.08 obtaining also the null result for the experience and matching then with the Sagnac effect. This supposition cannot be made based on the Einsteinβs Special Relativity because according to 17.26 we have: x'x t't 18.10
76/172
Β§19 Regression of the perihelion of Mercury of 7,13β Let us imagine the Sun located in the focus of an ellipse that coincides with the origin of a system of coordinates (x,y,z) with no movement in relation to denominated fixed stars and that the planet Mercury is in a movement governed by the force of gravitational attraction with the Sun describing an elliptic orbit in the plan (x,y) according to the laws of Kepler and the formula of the Newton's gravitational attraction law:
r
r
kr
r
10.28,310.98,110.67,6r
r
mGMF
22
233011
2oo
19.01
The sub index "o" indicating mass in relative rest to the observer. To describe the movement we will use the known formulas:
rrr
19.02
Λdt
drr
dt
dr
dt
rrd
dt
rdu
19.03
22
2
dt
dr
dt
dru.uu
19.04
Λdt
dr
dt
d
dt
dr2r
dt
dr
dt
rd
dt
rrd
dt
rd
dt
uda
2
22
2
2
2
2
2
2
19.05
The formula of the relativity force is given by:
22
2
2/3
2
2
o2
2
3
2
2
o
2
2o
2
2o
c
u
dt
udua
c
u1
c
u1
mu
dt
du
c
u
c
u1
ma
c
u1
m
c
u1
um
dt
dF
19.06
In this the first term corresponds to the variation of the mass with the speed and the second as we will see later in 19.22 corresponds to the variation of the energy with the time. With this and the previous formulas we obtain:
Λdt
drr
dt
dr
c
1
dt
dr
dt
d
dt
dr2
dt
dr
dt
dr
dt
rd
dt
dr
Λdt
dr
dt
d
dt
dr2r
dt
dr
dt
rd
c
u1
c
u1
mF
22
22
2
2
2
22
2
2
2
2
2/3
2
2
o
19.07
Λdt
d
c
r
dt
dr
dt
d
dt
dr2
dt
dr
dt
dr
dt
rd
dt
dr
dt
dr
dt
d
dt
dr2
c
u1
rdt
dr
c
1
dt
dr
dt
d
dt
dr2
dt
dr
dt
dr
dt
rd
dt
dr
dt
dr
dt
rd
c
u1
c/u1
mF
22
22
2
2
2
2
2
2
22
22
2
22
2
2
2
2
2/322
o
19.08
77/172
In this we have the transverse and radial component given by:
rdt
dr
c
1
dt
dr
dt
d
dt
dr2
dt
dr
dt
dr
dt
rd
dt
dr
dt
dr
dt
rd
c
u1
c/u1
mF 22
22
2
22
2
2
2
2
2/322
or
19.09
Λdt
d
c
r
dt
dr
dt
d
dt
dr2
dt
dr
dt
dr
dt
rd
dt
dr
dt
dr
dt
d
dt
dr2
c
u1
c/u1
mF 22
22
2
2
2
2
2
2
2/322
oΛ
19.10
As the gravitational force is central we should have to null the traverse component zeroFΛ
so we have:
zeroΛdt
d
c
r
dt
dr
dt
d
dt
dr2
dt
dr
dt
dr
dt
rd
dt
dr
dt
dr
dt
d
dt
dr2
c
u1
c/u1
mF 22
22
2
2
2
2
2
2
2/322
oΛ
19.11
From where we have:
2
2
2
2
2
2
2
2
dt
dr
c
11
dt
d
dt
dr
c
r
dt
dr
dt
rd
dt
dr
dt
d
dt
dr2
2
2
2
2
2
2
2
2
22
dt
dr
c
11
dt
dr
dt
rd
dt
dr
c
1
dt
dr
dt
dr
dt
d
dt
drr2
19.12
From the radial component rF
we have:
rdt
dr
c
1
dt
dr
dt
rd
dt
dr
dt
d
dt
dr2
dt
dr
dt
dr
c
u1
dt
dr
dt
rd
c/u1
mF
22
2
2
2
2
2
22
2
2
2/322
or
19.13
That applying 19.12 we have:
rdt
dr
c
1
dt
dr
c
11
dt
d
dt
dr
c
r
dt
dr
dt
dr
c
u1
dt
dr
dt
rd
c/u1
mF
22
2
2
2
22
2
2
2/322
or
19.14
That simplifying results in:
r
dt
dr
c
11
dt
dr
dt
rd
c
u1
mF
2
2
2
2
2
2
2o
r
19.15
This equaled to Newton's gravitational force results in the relativistic gravitational force:
rr
kr
r
mGMr
dt
dr
c
11
dt
dr
dt
rd
c
u1
mF
22oo
2
2
2
2
2
2
2o
r
19.16
78/172
As the gravitational force is central it should assist the theory of conservation of the energy (E) that is written as:
pk EEE = constant. 19.17
Where the kinetic energy (Ek) is given by:
11
1
2
2
222
cu
cmcmmcE ook 19.18
And the potential energy (Ep) gravitational by:
r
k
r
mGME oop
19.19
Resulting in:
rk
cu
cmE o 11
1
2
2
2 Constant . 19.20
As the total energy (E) it is constant we should have:
zerodt
dE
dt
dE
dt
dE pk . 19.21
Then we have:
dt
du
c
u1
um
dt
dE
2
3
2
2
ok
19.22
dt
dr
r
k
dt
dE2
p 19.23
Resulting in:
dt
dr
r
k
dt
du
c
u1
umzero
dt
dr
r
k
dt
du
c
u1
umzero
dt
dE
dt
dE
dt
dE2
2
3
2
2
o2
2
3
2
2
opk
19.24
This applied in the relativistic force 19.06 and equaled to the gravitational force 19.01 results in:
rr
ku
dt
dr
r
k
c
1a
c
u1
mF
222
2
2o
19.25
In this substituting the previous variables we get:
79/172
rr
kΛdt
drr
dt
dr
dt
dr
r
k
c
1Λdt
dr
dt
d
dt
dr2r
dt
dr
dt
rd
c
u1
mF
2222
22
2
2
2
2o
19.26
From this we obtain the radial component rF
equals to:
2
2
22
2
2
2
2
2o
r r
k
dt
dr
r
k
c
1
dt
dr
dt
rd
c
u1
mF
19.27
That easily becomes the relativistic gravitational force 19.16.
From 19.26 we obtain the traverse component F
equals to:
zerodt
d
dt
dr
r
k
c
1
dt
dr
dt
d
dt
dr2
c
u1
mF
22
2
2
2o
Λ
19.28
From this last one we have:
2
2
22o2
2
22
c
u1
dt
dr
rc
k
m
1
dt
dr
dt
dr
dt
d
dt
drr2
19.29
As the gravitational force is central it should also assist the theory of conservation of the angular moment that is written as:
prL
constant. 19.30
kdt
dr
c
u1
mΛrdt
dr
c
u1
mΛdt
drr
dt
dr
c
u1
mrr
c
u1
umrprL 2
2
2o2
2
2o
2
2o
2
2o
19.31
kLkdt
dr
c
u1
mL 2
2
2o
= constant. 19.32
zero
dt
Ldzero
dt
kLd
dt
kLd
dt
kLd
dt
kLd
dt
Ld
19.33
Resulting in L that is constant.
In 19.33 we had zerodt
kd because the movement is in the plane (x,y).
80/172
Deriving L we find:
zerodt
dr
dt
d
dt
drr2
c
u1
m
dt
dr
dt
du
c
u1
um
c
1
dt
dr
c
u1
m
dt
d
dt
dL2
22
2
2o2
2
3
2
2
o2
2
2
2o
19.34
From that we have:
2
2
22
2
22
c
1
dt
du
c
u1
u
dt
dr
dt
dr
dt
d
dt
drr2
19.35
Equaling 19.12 originating from the theory of the central force with 19.29 originating from the theory of conservation of the energy and 19.35 originating from the theory of conservation of the angular moment we have:
2
2
22
2
22o
2
2
2
2
2
2
2
2
22
c
1
dt
du
c
u1
u
c
u1
dt
dr
rcm
k
dt
dr
c
11
dt
dr
dt
rd
dt
dr
c
1
dt
dr
dt
dr
dt
d
dt
drr2
19.36
From the last two equality we obtain 19.24 and from the two of the middle we obtain 19.16. For solution of the differential equations we will use the same method used in the Newton's theory.
Let us assume r
1w 19.37
The differential total of this is drr
1dwdr
r
wdw
2
19.38
From where we have d
dr
r
1
d
dw2
e
dt
dr
r
1
dt
dw2
19.39
From the module of the angular moment we have 2
2
2o c
u1
rm
L
dt
d
19.40
From where we have 2
2
2o c
u1
d
dr
rm
L
dt
dr
19.41
Where applying 19.39 we have 2
2
o c
u1
d
dw
m
L
dt
dr
19.42
That derived supplies
2
2
o2
2
c
u1
d
dw
m
L
dt
d
d
dt
dt
d
dt
rd
19.43
81/172
Where applying 19.40 and deriving we have:
2
2
2
2
2
2
2
2
22o
2
2
2
o2
2
2o
2
2
c
u1
d
d
d
dw
c
u1
d
wd
c
u1
rm
L
c
u1
d
dw
m
L
d
d
c
u1
rm
L
dt
rd
19.44
In this with 19.36 the radical derived is obtained this way:
2
2
2o
2
2
22o
2222
2
c
u1
dt
dw
cm
k
c
u1
dt
dr
rcm
k
dt
du
c
u
c/u1
1
c
u1
dt
d 19.45
2
2
2o
2
2
22o
2222
2
c
u1
d
dw
cm
k
c
u1
d
dr
rcm
k
d
du
c
u
c/u1
1
c
u1
d
d
19.46
That applied in 19.44 supplies:
2
22
2o
2
2
2
2
2
2
22o
2
2
2
c
u1
d
dw
cm
k
c
u1
d
wd
c
u1
rm
L
dt
rd
19.47
Simplified results:
2
2
2
2
22o
222
3
2
2
223o
2
2
2
d
wd
c
u1
rm
L
d
dw
c
u1
rcm
kL
dt
rd
19.48
Let us find the second derived of the angle deriving 19.40:
2
2
2o
2
2
3o
2
2
2o
2
2
c
u1
dt
d
rm
L
c
u1
dt
dr
rm
L2
c
u1
rm
L
dt
d
dt
d 19.49
In this applying 19.42 and 19.45 and simplifying we have:
2
3
2
2
423o
2
2
2
32o
2
2
2
c
u1
d
dw
rcm
kL
c
u1
d
dw
rm
L2
dt
d
19.50
Applying in 19.04 the equations 19.40 and 19.42 and simplifying we have:
2
2
2
2
2o
22
r
1
d
dw
c
u1
m
Lu
19.51
The equation of the relativistic gravitational force 19.16 remodeled is:
2o
2
22
22
2
2
rm
k
dt
dr
c
11
c
u1
dt
dr
dt
rd
19.52
In this applying the formulas above we have:
2o
2
2
2
o22
22
2
2
2o
2
2
2
2
22o
222
3
2
2
223o
2
rm
k
c
u1
d
dw
m
L
c
11
c
u1
c
u1
rm
Lr
d
wd
c
u1
rm
L
d
dw
c
u1
rcm
kL
82/172
2o
2
2
2
o22
2
32o
2
2
2
2
2
22o
22
2
2
223o
2
rm
k
c
u1
d
dw
m
L
c
11
c
u1
rm
L
d
wd
c
u1
rm
L
d
dw
c
u1
rcm
kL
2
2
2
223o
2
2o
2
2
32o
2
2
2
2
2
22o
22
2
2
223o
2
d
dw
c
u1
c
1
rm
kL
rm
k
c
u1
rm
L
d
wd
c
u1
rm
L
d
dw
c
u1
rcm
kL
2o
2
2
32o
2
2
2
2
2
22o
2
rm
k
c
u1
rm
L
d
wd
c
u1
rm
L
2
22
o2
2
c
u1L
km
r
1
d
wd
2
2
2
2
2
2o
o2
2
c
u1
dt
dr
c
u1
m
km
r
1
d
wd
242
o
2
2
o
2
2
dt
drm
c
u1km
r
1
d
wd
2
24
o
2
2
2
2
2
dt
drm
c
u1k
r
1
d
wd
482
o
2
22
22
22
2
2
dt
drm
c
u1k
r
1
d
wd
r
2
d
wd
482
o
22
2
482
o
2
22
22
2
2
dt
drm
uc
k
dt
drm
k
r
1
d
wd
r
2
d
wd
482
o
22
2
2
482
o
2
22
22
2
2
dt
drm
dt
dr
dt
dr
c
k
dt
drm
k
r
1
d
wd
r
2
d
wd
83/172
482
o
2
2
2
482
o
2
2
2
482
o
2
22
22
2
2
dt
drm
dt
dr
c
k
dt
drm
dt
dr
c
k
dt
drm
k
r
1
d
wd
r
2
d
wd
2622
o
2
282
o
2
2
2
482
o
2
22
22
2
2
dt
drcm
k
dt
drm
d
dr
c
k
dt
drm
k
r
1
d
wd
r
2
d
wd
2622
o
2
282
o
2
22
2
482
o
2
22
22
2
2
dt
drcm
k
dt
drm
d
dwr
c
k
dt
drm
k
r
1
d
wd
r
2
d
wd
2622
o
2
242
o
2
2
2
482
o
2
22
22
2
2
dt
drcm
k
dt
drm
d
dw
c
k
dt
drm
k
r
1
d
wd
r
2
d
wd
In this we will consider constant the Newton's angular moment in the form:
dt
drL 2
19.53
That it is really the known theoretical angular moment.
2222o
22
222o
2
42o
2
22
22
2
2
Lrcm
k
d
dw
Lcm
k
Lm
k
r
1
d
wd
r
2
d
wd
2222
o
22
222o
2
42o
22
2
22
2
2
wLcm
k
d
dw
Lcm
k
Lm
kww2
d
wd2
d
wd
2
2
22
22
2
2
Awd
dwABww
d
wd2
d
wd
zeroBw1Ad
dwAw
d
wd2
d
wd 2
2
2
22
2
2
19.54
Where we have:
222o
2
Lcm
kA 19.55
42o
2
Lm
kB 19.56
84/172
The equation 19.54 has as solution:
Qcos1D
1wA1cos1
D
1w o
19.57
Where we consider zeroo .
It is denominated in 19.57 A1Q2 . 19.58
The equation 19.58 is function only of A demonstrating the intrinsic union between the variation of the mass with the variation of the energy in the time, because both as already described, participate in the relativistic force 19.06 in this relies the essential difference between the mass and the electric charge that is invariable and indivisible in the electromagnetic theory. From 19.57 we obtain the ray of a conical:
Qcos1
Dr
A1cos1
D
w
1r
19.59
Where is the eccentricity and D the directory distance of the focus.
Deriving 19.57 we have
D
QsenQ
d
dw 19.60
That derived results in
D
QcosQ
d
wd 2
2
2
19.61
Applying in 19.54 the variables we have:
zeroBw1AddwAw
dwd2
dwd 2
2
2
22
2
2
.
zeroB
D
Qcos11A
D
QsenQA
D
Qcos1
D
QcosQ2
D
QcosQ2
2
222
2
24
19.62
zeroB
D
Qcos11A
D
QcosQA
D
QA
D
QcosQ2
D
QcosQ2
D
QcosQ2
2
22
2
2
2
22
2
2
2
24
zeroB
D
Qcos1A
D
Qcos1A2
D
1A
D
QcosQA
D
QA
D
QcosQ2
D
QcosQ2
D
QcosQ2
2
2222
22
2
2
2
22
2
2
2
24
zeroB
D
1A
D
AQ
D
Qcos
D2
DA2
D
Q2
D
Qcos1AAQQ2Q
222
22
2
2224
19.63
In this applying in the first parenthesis A1Q2 we have:
zero1AAAA22AA211AA1AA12A11AAQQ2Q 222224
85/172
In 19.63 applying in the second parenthesis A1Q2 we have:
zero
D
2
D
A2
D
A12
D
2
D
A2
D
Q2 2
The rest of the equation 19.63 is therefore:
zeroB
D
1A
D
AQ222
2
19.64
The data of the elliptic orbit of the planet Mercury is [1]: Eccentricity of the orbit 206,0 . Larger semi-axis = a = 5,79.1010m.
Smaller semi-axis m80,305.160.658.56206,0110.79,51ab 2102 .
m00,600.955.442.55206,0110.79,51aD 2102 .
m00,165.561.140.269
206,0
206,0110.79,51aD
2102
.
The orbital period of the Earth (PT) and Mercury (PM) around the Sun in seconds are:
.s.,PT 710163
.s.,PM 610607 The number of turns that Mercury (mo) makes around the Sun (Mo) in one century is, therefore:
7941510607
10163100
6
7
,.,.,
N . 19.65
Theoretical angular moment of Mercury:
3021030112o
222 10.73221293742,7206,0110.79,510.98,110.67,61aGMdt
drL
19.66
.10.65,210.32,710.0,3
10.98,110.67,6
Lc
GM
Lcm
mGMA 8
3028
230211
22
20
222o
2o0
19.67
22230
230211
4
20
42o
2o0 10.25,3
10.32,7
10.98,110.67,6
L
GM
Lm
mGMB
19.68
23.013.000.000,110.63,21A1Q 8 19.69
Applying the numeric data with several decimal numbers to the rest of the equation 19.63 we have:
30222
8
2
28
222
2
10.976,810.25,300,600.955.442.55
110.65,2
00,165.561.140.269
23.013.000.000,110.65,2B
D
1A
D
AQ
19.70
Result that we can consider null.
86/172
We will obtain the relativistic angular moment of the rest of the equation 19.63 in this applying the variables we have:
zero
L
GM
Lc
GM1
D
1
Lc
GM1
DLc
GMB
D
1A
D
AQ4
20
22
20
2222
20
222
20
222
2
19.71
zeroGMDcLc
GM1cL
Lc
GM1GML 2
0222
22
2024
22
202
022
zeroGMDcLc
GMcLcL
Lc
GMGMLGML 2
0222
22
202424
22
202
0222
022
zeroGMDcGMLcLc
GMGML 2
02222
0224
2
4022
022
zeroGMDcc
GMLGM1Lc 2
0222
2
40222
0242 19.72
2
20
2222
402222
022
02
2
c2
GMDcc
GMc4GM1GM1
L
2
20
22440
240
2220
22
c2
GMDc4GM4GM1GM1L
2
20
22440
240
4220
22
c2
GMDc4GM4GM21GM1L
2
20
22440
240
440
240
20
22
c2
GMDc4GM4GMGM2GMGM1L
2
20
22440
240
440
20
22
c2
GMDc4GM2GMGMGM1L
30
2
20
22440
2220
22
10.83221292732,7c2
GMDc4GM1GM1L
19.73
This last equation has the exclusive property of relating the speed c to the denominated relativistic angular moment that is smaller than the theoretical angular moment 19.66. The variation of the relativistic angular moment in relation to the theoretical angular moment is very small and given by:
00,509.503.72
110.38,1
10.73221293742,7
10.73221293742,710.83221292732,7L 8
30
3030
. 19.74
That demonstrates the accuracy of the principle of constancy of the speed of the light.
87/172
In reality, the equation 19.06 provides a secular retrocession perihelion of Mercury, which is given by in
.rad10.46,323.013.000.000,079,41521Q
179,4152 5
19.75
Converting for the second we have:
"13,700,600.3.00,180.10.46,3 5
. 19.76
This retrocession, is not expected in Newtonian theory is due to relativistic variation of mass and energy and is shrouded in total observed precession of 5599. "
88/172
§§19 Advance of Mercuryβs perihelion of 42.79β If we write the equation for the gravitational relativity energy ER covering the terms for the kinetic energy, the potential energy Ep and the resting energy:
p
2
2
2o2
op
2
2
2oR E
cu
1
cmcmE1
cu
1
1cmE
. 19.77
Being the conservative the gravitational force its energy is constant. Assuming then that in 19.77 when the radius tends to infinite, the speed and potential energy tends to zero, resulting then:
2op
2
2
2o
R cmE
cu
1
cmE
19.78
Writing the equation to the Newtonβs gravitation energy EN having the correspondent Newtonβs terms to the 19.77:
2o
2o
2o
N cmcmrk
2
umE 19.79
Where 2
um 2o is the kinetic energy,
rk
the potential energy and 2ocm the resting energy or better saying
the inertial energy. From this 19.79 we have:
r
GM2u
rm
mGM2
rmk2
urk
2
umcmcm
rk
2
um o2
o
oo
o
22
o2o
2o
2o 19.80
Deriving 19.79 we have:
zerocmrk
2
um
dtd
dtdE 2
o
2oN
zerodtdr
rk
dtdu
2
u2m2
o
dtdr
r
GM
dtdr
rmk
dtduu o
o22
dtdr
r
GM
dtduu o
2
2o
r
GM
drdu
u
19.81
89/172
Making the relativity energy 19.78 equal to the Newtonβs energy 19.79 we have:
2o
2o
p
2
2
2o
NR cmrk
2
umE
cu
1
cmEE
19.82
o
2o
o
oo
o
2o
o
p
2
2
o
2o
m
cm
rm
mGM
2m
um
m
E
cu1m
cm
19.83
In that denominating the relativity potential ( ) as:
o
p
m
E 19.84
We have:
2o2
2
2
2c
r
GM
2u
cu1
c
2
2
22o2
cu1
ccr
GM
2u
19.85
In this one replacing the approximation:
2
2
2
2 c2u
1
cu
1
1
19.86
We have:
2
222o2
c2u1cc
r
GM
2u
That simplified results in the Newtonβs potential:
r
GM
2u
ccr
GM
2u o
222o
2 19.87
Replacing 19.84 and the relativity potential 19.85 in the relativity energy 19.78:
2
2
22o
2
o
2
2
2o
R
cu
1
cc
r
GM
2u
m
cu
1
cmE 19.88
We have the Newtonβs energy 19.79:
2o
oo2
oN cm
r
mGM
2
umE
90/172
Deriving the relativity potential 19.85 we have the relativity gravitational acceleration modulus exactly as in the Newtonβs theory:
drd
a
2
2
22o
2
cu
1
cc
r
GM
2u
drd
drd
a
2
2
22o
2
cu
1
cdrd
cr
GM
2u
drd
a
Where we have:
zeromE
drd
crGM
2u
drd
o
N2o2
. Because the term to be derived is the Newtonβs energy divided
by mo that is 2o2
o
N cr
GM
2u
mE
that is constant, resulting then in:
2
2
2
cu
1
cdrd
a
drdu
cu
1
ua
2
3
2
2
In this one applying 19.81 we have:
2o
2
3
2
2 r
GM
cu
1
1a
19.89
The vector acceleration is given by 19.05:
Λ2Λ2
22
2
2
dt
dr
dt
d
dtdrr
dt
dr
dtrda
91/172
The relativity gravitational acceleration modulus 19.89 is equal to the component of the vector radius ( r ) thus we have:
2o
2
3
2
2
2
2
2
r
GM
cu
1
1dtd
rdtrd
a
19.90
Being null the transversal acceleration we have:
zeroΛdtd
rdtd
dtdr
2 2
2
19.91
zerodtd
rdtd
dtdr
22
2
That is equal to the derivative of the constant angular momentum dtd
rL2 19.92
zerodtd
rdtd
dtdr
r2dtd
rdtd
dtdL
2
222
19.93
Rewriting some equations already described we have:
r1
w
drr1
dwdrrw
dw2
ddr
r1
ddw
2
or
ddw
rddr 2 and
dtdr
r1
dtdw
2
ddw
Ldtdr
ddw
rrL
ddr
rL
dtdr
ddt
dtd
dtdr 2
22
2
2
2
2
22
2
dwd
rL
ddw
Ldd
rL
ddw
Ldtd
ddt
dtd
dtdr
dtd
dtrd
19.94
From 19.90 we have:
2o
2
2
2
2
2
r
GM
dtd
rdtrd
c2u3
1
In this one we 19.94 the speed of 19.80 and the angular momentum we have:
2o
2
22
2
2
2o
2 r
GM
rL
rdwd
rL
r
GM2
c23
1
2o
2
2
2o
L
GM
r1
dwd
r1
c
GM31
92/172
2o
2o
2
2
2o
L
GM
r1
r1
c
Gm31
dwd
r1
c
GM31
zeroL
GM
r1
c
GM3
r1
r1
dwd
c
GM3
dwd
2o
22o
2
2
2o
2
2
zeroBr1
Ar1
r1
dwd
Adwd
22
2
2
2
zeroBAwwwdwd
Adwd 2
2
2
2
2
zeroBwAwwdwd
Adwd 2
2
2
2
2
19.95
Where we have:
2
3
c
GMA o
2L
GMB o 19.96
The solution to the differential equation 19.95 is:
QD
wQD
w o
cos11cos11 . 19.97
Where we consider zeroo
Then the radius is given by:
Qcos1D
rQcos1
Dw1
r
19.98
Where is the eccentricity and D the focus distance to the directory.
Deriving 19.97 we have
DQsenQ
ddw and
D
QcosQ
d
wd 2
2
2
19.99
Applying the derivatives in 19.95 we have:
zeroBwAwwdwd
Adwd 2
2
2
2
2
zeroBQcos1D1
Qcos1DA
Qcos1D1
DQcosAQ
DQcosQ 2
22
22
zeroBQcos
D
1
D
1QcosQcos21
D
AQcos1
D
QcosAQ
D
QcosQ 22222
22
zeroBQcosD1
D1
QcosDA
Qcos2DA
DA
QcosD
QcosAQD
QcosAQD
QcosQ
22222222
2
2
2
22
93/172
zeroB
D1
DA
DQcosA
DQcosAQ
1DA2
DAQ
QDQcos
222
2
2
2222
zero
AB
DA1
DAA
ADQcosA
ADQcosAQ
1DA2
DAQ
QAD
Qcos222
2
2
2222
zero
AB
DA1
D1
DQcos
DQcosQ
A1
D2
DQ
AQ
DQcos
222
2
2
2222
zero
AB
DA1
D1
A1
D2
DQ
AQ
DQcos
1QD
Qcos22
222
2
2
19.100
The coefficient of the squared co-cosine can be considered null because 1Q and D2 is a very large number:
zero1QD
Qcos 22
2
19.101
Resulting from the equation 19.100:
zero
AB
DA1
D1
A1
D2
DQ
AQ
DQcos
22
22
19.102
Due to the unicity of the equation 19.102 we must have the only solution that makes it null simultaneously the parenthesis and the rest of the equation, that is, we must have a unique solution for both the following equations:
zeroA1
D2
DQ
AQ 22
and zeroAB
DA1
D1
22
19.103
These equations can be written as:
D2
A1
Q1
D1
A1
ba2
19.104
ADB
D1
A1
ca
19.105
In these ones the common term D1
A1
a
must have a single solution then we have:
ADB
D2
A1
Q1
cb2
19.106
With 19.96 and the theoretical momentum we have:
2
3
c
GMA o
2L
GMB o o
2 DGML 1L
DGMDB
2o
19.107
94/172
It is applied in 19.105 and 19.106 resulting in:
A1
D1
A1
ca
19.108
A1
D2
A1
Q1
cb2
19.109
From 19.108 we have the mistake made in 19.105:
zeroD1
A1
D1
A1
19.110
zero10.80,100,600.955.442.55
1D1 11
19.111
From 19.109 we have Q:
2o22
2 c
GM3
D2
1QDA2
1QA1
D2
A1
Q1
19.112
It is applied in 19.104 resulting in 19.110:
zeroD1
A1
D1
A1
D2
A1
DA2
1
1D1
A1
D2
A1
Q1
D1
A1
2
From 19.112 we have:
599.920.999.999,0
10.300,600.955.442.55
10.98,110.67,661
Dc
GM61Q 28
3011
2o
19.113
That corresponds to the advance of Mercuryβs perihelion in one century of:
79,415.00,000.296.1.1
Q179,415. 42,79β 19.114
Calculated in this way: In one trigonometric turn we have "00,000.296.16060360 seconds. The angle in seconds ran by the planet in one trigonometric turn is given by:
QQ 00,000.296.100,000.296.1 .
If 00,1Q we have a regression. 00,000.296.1 .
If 00,1Q we have an advance. 00,000.296.1 .
95/172
The angular variation in seconds in one turn is given by:
00,000.296.11100,000.296.100,000.296.1
QQ .
If zero we have a regression.
If zero we have an advance.
In one century we have 415,79 turns that supply a total angular variation of:
79,415.00,000.296.1.1
Q179,415. 42,79β
If zero we have a regression.
If zero we have an advance.
Β§20 Inertia Imagine in an infinite universe totally empty, a point O' which is the beginning of the coordinates of
the observer O'. In the cases of the observer Oβ being at rest or in uniform motion the law of inertia requires that the spherical electromagnetic waves with speed c issued by a source located at point O' is always observed by O', regardless of time, with spherical speed c and therefore the uniform motion and rest are indistinguishable from each other remain valid in both cases the law of inertia. To the observer Oβ the equations of electromagnetic theory describe the spread just like a spherical wave. The image of an object located in Oβ will always be centered on the object itself and a beam of light emitted from O' will always remain straight and perpendicular to the spherical waves.
Imagine another point O what will be the beginning of the coordinates of the observer which has the same properties as described for the inertial observer O'.
Obviously two imaginary points without any form of interaction between them remain individually and together perfectly meeting the law of inertia even though there is a uniform motion between them only detectable due to the presence of two observers who will be considered individually in rest, setting in motion the other referential.
The intrinsic properties of these two observers are described by the equations of relativistic transformations.
Note: the infinite universe is one in which any point can be considered the central point of this universe.
(Β§ 20 electronic translation)
Β§20 Inertia (clarifications) Imagine in a totally empty infinite universe a single point O. Due to the uniqueness properties of O a
radius of light emitted from O must propagate with velocity c. If this ray propagates in a straight line, then O is defined as the origin of an inertial frame because it is either at rest or in a uniform rectilinear motion. However, in the hypothesis of propagation of the light ray being a curve the movement of O must be interpreted as the origin of an accelerated frame. Therefore the propagation of a ray of light is sufficient to demonstrate whether O is the origin of an inertial frame or accelerated frame.
Now imagine if in the universe described above for the inertial reference frame O there is another inertial frame O' that does not have any kind of physical interaction with O. In the absence of any interaction between O and O' the uniqueness properties are inviolable for both points and rays of light emitted from O and O' have the same velocity c. It is impossible for the velocity of light emitted from O to be different from the velocity of light emitted from O' because each reference exists as if the other did not exist. Being O and O' the origin of inertial frames the propagation of light rays occurs in a straight line with velocity c and the relations between times t and t' of each frame are given by table I.
96/172
Β§21 Advance of Mercuryβs perihelion of 42.79β calculated with the Undulating Relativity Assuming vux
(2.3) zeroxu
cvv
cv
vv
cvux
cv
vuxxu
''2121
''
22
2
22
2
vux zeroxu '' 21.01
(1.17) 2
2
22
2
22
21'2121'
cvdtdt
cvv
cvdt
cvux
cvdtdt
(1.22)
2
2
22
2
22
2 '1'0'2'1''''2'1'cvdtdt
cv
cvdt
cxuv
cvdtdt
2
21'
cvdtdt 2
2'1'cvdtdt 21.02
1'11 2
2
2
2
cv
cv 21.03
2
2'1
'
cv
vv
2
21
'
cv
vv
21.04
'dtdt 'vv ''dtvvdt 21.05
(1.33)
2
2
22
2
22
2 '1
'0'2'1
''''2'1
'
cvvv
cv
cvv
cxuv
cv
vv
(1.34)
2
2
22
2
22
21
'2121
'
cvvv
cvv
cvv
cvux
cvvv
2
2'1
'
cvvv
2
21
'
cv
vv
21.06
'Λ rrrr
rrrr
Λ' rrr '
21.07
'ΛΛ rdrrdrdrrd
rdrrdrdrrd
ΛΛ' 21.08
drrdrrrrdrrdr ΛΛΛΛΛ
drrdrrrrdrrdr ΛΛΛΛ'Λ
21.09
ΛΛΛdtdrr
dtdr
dtrrd
dtrdv
22
2
dtdr
dtdrvvv
21.10
Λ
'Λ''
Λ'''
dtdrr
dtdr
dtrrd
dtrdv
22
2
'''''
dtdr
dtdrvvv
21.11
97/172
Λ2ΛΛ2
22
2
2
2
2
2
2
dtdr
dtd
dtdrr
dtdr
dtrd
dtrrd
dtrd
dtvda
21.12
Λ'''
2Λ'''
Λ''
''' 2
22
2
2
2
2
2
2
dtdr
dtd
dtdrr
dtdr
dtrd
dtrrd
dtrd
dtvda
21.13
2
21
'
cv
vv
21.06
2
22
2
2
2
2
21
'11
'1
''''
cv
vdtd
cv
cv
vdtd
dtdt
cv
vdtd
dtvda
21.14
2
2
2
2
2
22
211
1
1'1'''
cv
dtdv
dtvd
cv
cvc
vdtvda
dtdv
cv
cvv
dtvd
cv
cvc
vdtvda 2
21
22
21
2
2
2
2
2
22
2 21211
1
1'1'''
2
2
22
2
2
22
2
1
111
1'1'''
cv
dtdvv
cvdt
vdcv
cvc
vdtvda
2
2
22
2
2
2
2
2
2
22
2
1
1111
1
1
1'1'''
cv
dtdvv
cvdt
vdcv
cv
cv
cvc
vdtvda
22
2
23
2
22
21
1
1'1'''
cv
dtdvv
dtvd
cv
cvc
vdtvda
22
2
23
2
22
2
2
21
1''
'1'1
'''
cv
dtdvv
dtvd
cv
cv
mdtvd
cv
m
cv
amam ooo
''
'1'1
''''
2
2
2
2 dtvd
cv
m
cv
amamF oo
21.15
98/172
22
2
23
2
21
1cv
dtdvv
dtvd
cv
cv
mF o
06.19 21.16
22
2
23
2
22
2
2
21
1''
'1'1
''''cv
dtdvv
dtvd
cv
cv
mFdtvd
cv
m
cv
amamF ooo
21.17
rdrrkrdFrdFEk
Λ.'' 2 21.18
rdrrkrd
cv
dtdvv
dtvd
cv
cv
mrd
dtvd
cv
mrdFrdFE oo
k
.1
1
'''
'1
.''. 222
2
23
2
22
2 21.19
rdrrk
cv
dtrdvdv
dtrdvd
cv
cv
mdtrdvd
cv
mE ook
.1
1'''
'1222
2
23
2
22
2
rdrrk
cvvvdvvvd
cv
cv
m
cv
vvdmE ook
.1
1'1
''222
2
23
2
22
2
drrk
cvvdvvdv
cv
cv
m
cv
dvvmE ook 22
2
2
2
23
2
22
21
1'1
''
drrk
cv
cv
cv
vdvm
cv
dvvmE ook 22
2
2
2
23
2
22
21
1'1
''
drrk
cv
vdvm
cv
dvvmE ook 2
23
2
22
2
1'1
'' dr
rk
cv
vdvm
cv
dvvmdE oo
k 223
2
22
2
1'1
''
21.20
tconsrk
cv
cmcvcmE o
ok tan1
'1
2
2
2
2
22
21.21
tconsrk
cvcmE oR tan'12
22 tcons
rk
cv
cmE oR tan
12
2
2
21.22
rkvm
cmrk
cv
cmE o
oo
R
2
1
22
2
2
2
2
2
2
2
01
cmk
c
cmE o
oR
21.23
99/172
rcmk
cmE
cv oo
R 1
1
122
2
2
2cm
EH
o
R 222 cGM
cmmGM
cmkA o
o
oo
o
21.24
rAH
cv
1
1
1
2
2
3
23
2
2
1
1
1
rAH
cv
21.25
kdtdrr
dtdr
dtdrr
dtdrrrvrL ΛΛΛΛΛΛ 22
21.26
kdtdrr
dtdr
cvdt
drrdtdr
cv
rr
cvvrvrL ΛΛΛ
''11Λ
'Λ''1
1Λ'1' 22
2
2
2
2
2
2
21.26
kLkdtdrL Λ2
= constant
dtdrL 2 21.27
rdrrkdr
rk
cv
vdvm
cv
dvvmdE oo
k
.
1'1
''22
23
2
22
2
21.20
vrrk
dtrdr
rk
dtvdv
cv
mvF
dtdE ok
..
1
2223
2
2
rrk
cv
amF o Λ
1
223
2
2
21.28
rrk
dtdr
dtd
dtdrr
dtdr
dtrd
cv
mF o ΛΛ2Λ
122
22
2
2
23
2
2
21.29
zerodtdr
dtd
dtdr
cv
mF o
Λ2
1
2
2
23
2
2Λ
21.30
rrkr
dtdr
dtrd
cv
mF or ΛΛ
12
2
2
2
23
2
2Λ
21.31
2rL
dtd
ddwL
dtdr
2
2
2
2
2
2
dwd
rL
dtrd
ddw
rL
dtd
3
2
2
2 2 21.32
100/172
rrkr
rLr
dwd
rL
cv
mF or ΛΛ
12
2
22
2
2
2
23
2
2Λ
21.33
23
2
2
2
2
2
23
2
21
1rGM
rL
dwd
rL
cv
o
22
2
2
2
23
2
2
1
1
1rGM
rL
rdwd
cv
o
22
2
23
2
2
1
1
1LGM
rdwd
cv
o
21.34
22
2311
LGM
rdwd
rAH o
21.35
22
2 113LGM
rdwd
rAH o
222
2
2
2 13131LGM
rA
rdwdA
rH
dwdH o
zeroLGM
AwwdwdAHw
dwdH o 2
22
2
2
233
2cmE
Ho
R 222 cGM
cmmGM
cmkA o
o
oo
o
2LGM
B o 21.36
zeroBAwwdwdAHw
dwdH 2
2
2
2
233
21.37
QDr
w
cos111
DQQsen
ddw
DQQ
dwd
cos2
2
2 21.38
zeroBQD
AQDD
QQAQD
HD
QQH
222
cos113cos11cos3cos11cos
21.39
zeroBQQ
DAQ
DQ
DAQQ
DH
DH
DQHQ
22
22
22 coscos213cos1cos3cos11cos
zeroBQDAQ
DA
DA
QDQ
DAQ
DQ
DAQ
DQH
DH
DQHQ
22222222
222
cos3cos233
coscos3cos3cos1cos
101/172
zeroBD
QADQ
DA
DA
DQAQ
DQ
DAQ
DQH
DH
DQHQ
2
2
22
2
22
22
cos3cos63
cos3cos3cos1cos
zeroBDA
DH
DQA
DQAQ
DQ
DA
DQ
DAQ
DQH
DQHQ
222
2
2
22
22
31cos3cos3
cos6cos3coscos
zeroBDA
DH
DQAAQ
DQ
DA
DAQHHQ
222
22
22 31cos33cos63
zeroAB
DAA
DAH
ADQ
DA
DAQHHQ
ADQAAQ
333
31
3cos63
3cos33 22
22
2
22
zeroAB
DDAH
DQ
DDQ
AH
AHQ
DQQ
31
3cos2
33cos1
22
22
2
22
21.40
12Q zeroD
QQ 2
22 cos1
21.41
zeroAB
DDAH
DQ
DDQ
AH
AHQ
31
3cos2
33 22
22
21.42
zero
AB
DDAHzero
DQ
31
3cos
22
zeroDD
QAH
AHQzero
DQ
2
33cos 22
zeroDD
QAH
AHQ
2
33
22 zero
AB
DDAH
31
3 22 21.43
DAH
QDAHba
2
311
3 2
ADB
DAHca
31
3
21.44
12Q 12
2
2 cmcm
cmE
Ho
o
o
R 12 o
oo
DGMDGM
LDGM
DB
zeroDDA
HDA
Hba
12
3111
3 zero
DADAca
1
311
31
ADB
DAH
Qcb
32
312
21.45
12 o
oo
DGMDGM
LDGM
DB 21.46
102/172
ADA
HQ
cb312
312
DAHQ62 21.47
HQQ The regression is a function of positive energy that governs the movement.
12
2
2
cmcm
cmE
Ho
o
o
R DAQ612 Regression 21.48
zeroDDA
DADA
ba
1231
61
1131 21.49
zeroDD
QAH
AHQDA
2
333
22 zero
AB
DDAHDA
31
33
2222
21.43
2cmE
Ho
R 2c
GMA o 2L
GMB o
zeroAAQDHDHQ 6322 zeroDBDADH 3
zeroAAQDADAQ 6333 22 DADH 3
zeroAAQDDQAQ 333 222
zeroADDQ 32 DAQ312
This regression is not governed by the positive energy
103/172
2
2'1
'
cvvv
21.06
2
22
2
2
2
2
2 '1
''
1'1
''
''1
'
cvv
dtd
cv
cvv
dtd
dtdt
cvv
dtd
dtvda
21.50
2
2
2
2
2
22
2 '1'
''''1
'1
11cv
dtdv
dtvd
cv
cvc
vdtvda
'''2'1
21'
'''1
'1
11 2
21
22
21
2
2
2
2
2
22
2
dtdv
cv
cvv
dtvd
cv
cvc
vdtvda
2
2
22
2
2
22
2 ''''
'1
1'''1
'1
11cv
dtdvv
cvdt
vdcv
cvc
vdtvda
2
2
22
2
2
2
2
2
2
22
2 ''''
'1
1'1'''1
'1
1'1
11cv
dtdvv
cvc
vdtvd
cv
cv
cvc
vdtvda
22
2
23
2
22
2 ''''
'''1
'1
11cv
dtdvv
dtvd
cv
cvc
vdtvda
22
2
23
2
22
2
2
2'
'''
'''1
'111cv
dtdvv
dtvd
cv
cv
mdtvd
cv
m
cv
amam ooo
dtvd
cv
m
cv
amamF oo
2
2
2
211
21.51
22
2
23
2
2
''''
'''1
'1
'cv
dtdvv
dtvd
cv
cv
mF o
21.52
22
2
23
2
22
2
2
2'
'''
'''1
'1
'11
cv
dtdvv
dtvd
cv
cv
mF
dtvd
cv
m
cv
amamF ooo
21.53
104/172
'Λ''.. 2 rdrrkrdFrdFEk
21.54
'Λ'''''
'''1
'11
''.. 222
2
23
2
22
2rdr
rkrd
cv
dtdvv
dtvd
cv
cv
mrd
dtvd
cv
mrdFrdFE oo
k
21.55
'Λ'''''
''''1
'11222
2
23
2
22
2rdr
rk
cv
dtrddvv
dtrdvd
cv
cv
mdtrdvd
cv
mE ook
drrk
cvvdvvvvd
cv
cv
mvvd
cv
mE ook 222
2
23
2
22
2'''''''1
'11
drrk
cvdvvvvd
cv
cv
m
cv
vdvmE ook 22
2
2
2
23
2
22
2''''''1
'11
drrk
cv
cv
cv
dvvm
cv
vdvmE ook 22
2
2
2
23
2
22
2
''1'1
''
1
drrk
cv
dvvm
cv
vdvmE ook 2
23
2
22
2'1
''
1 dr
rk
cv
dvvm
cv
vdvmdE oo
k 223
2
22
2'1
''
1
21.56
tconsrk
cv
cmcvcmE o
ok tan'1
1
2
2
2
2
22
21.57
tconsrk
cvcmE oR tan12
22 tcons
rk
cv
cmE oR tan
'12
2
2
21.58
rkvm
cmrk
cv
cmE o
oo
R
2'
'1
22
2
2
2
2
2
2
2
01
cmk
c
cmE o
oR
21.59
rcmk
cmE
cv oo
R 1'1
122
2
2
21.60
2cmE
Ho
R 222 cGM
cmmGM
cmkA o
o
oo
o
21.61
105/172
rAH
cv
1'1
1
2
2
3
23
2
2
1
'1
1
rAH
cv
21.62
kdtdrr
dtdr
dtdrr
dtdrrrvrL Λ
'ΛΛ
'Λ
'Λ'
Λ''' 22
21.63
k'dt
drΛr
dtd
r
cv
Λdtd
rrdtdr
cv
rr
cvvrr'vx'r'L
22
2
2
2
2
2
21
1
1
1
1
21.63
kLkdtdrL ''
' 2
'
' 2
dtdrL 21.64
'Λ'1
''
122
23
2
22
2rdr
rkdr
rk
cv
dvvm
cv
vdvmdE oo
k
21.56
'Λ''Λ
'''
'1
''' 22
23
2
2vr
rk
dtrdr
rk
dtdvv
cv
mvF
dtdE ok
rrk
cv
amF o Λ
'1
'' 2
23
2
2
21.65
rrk
dtdr
dtd
dtdrr
dtdr
dtrd
cv
mF o ΛΛ
'''2Λ
'''1
' 22
22
2
2
23
2
2
21.66
zerodtdr
dtd
dtdr
cv
mF o
Λ'''
2'1
' 2
2
23
2
2Λ
21.67
rrkr
dtdr
dtrd
cv
mF o
r ΛΛ'''1
' 2
2
2
2
23
2
2Λ
21.68
rrGM
rdtdr
dtrd
cv
o ΛΛ'''1
12
2
2
2
23
2
2
2'
' rL
dtd
d
dwLdtdr ''
2
2
2
2
2
2 '' d
wdrL
dtrd
ddw
rL
dtd
3
2
2
2 '2' 21.69
106/172
2
2
22
2
2
2
23
2
2
''
'1
1rGM
rLr
dwd
rL
cv
o
23
2
2
2
2
2
23
2
2
''
'1
1rGM
rL
dwd
rL
cv
o
22
2
2
2
23
2
2
'1
'1
1rGM
rL
rdwd
cv
o
22
2
23
2
2 '1
'1
1LGM
rdwd
cv
o
21.70
22
23
'11
LGM
rdwd
rAH o
21.71
22
2
'113
LGM
rdwd
rAH o
222
2
2
2
'13131
LGM
rA
rdwdA
rH
dwdH o
zeroLGM
AwwdwdAHw
dwdH o 2
22
2
2
2
'33
2cmE
Ho
R 2cGM
A o 2'LGM
B o 21.72
zeroBAwwdwdAHw
dwdH 2
2
2
2
233
21.73
QDr
w
cos111
DQQsen
ddw
DQQ
dwd
cos2
2
2 21.38
zeroBQD
AQDD
QQAQD
HD
QQH
222
cos113cos11cos3cos11cos
21.74
zeroBQQ
DAQ
DQ
DAQQ
DH
DH
DQHQ
22
22
22 coscos213cos1cos3cos11cos
zeroBQDAQ
DA
DA
QDQ
DAQ
DQ
DAQ
DQH
DH
DQHQ
22222222
222
cos3cos233
coscos3cos3cos1cos
107/172
zeroBD
QADQ
DA
DA
DQAQ
DQ
DAQ
DQH
DH
DQHQ
2
2
22
2
22
22
cos3cos63
cos3cos3cos1cos
zeroBDA
DH
DQA
DQAQ
DQ
DA
DQ
DAQ
DQH
DQHQ
222
2
2
22
22
31cos3cos3
cos6cos3coscos
zeroBDA
DH
DQAAQ
DQ
DA
DAQHHQ
222
22
22 31cos33cos63
zeroAB
DAA
DAH
ADQ
DA
DAQHHQ
ADQAAQ
333
31
3cos63
3cos33 22
22
2
22
zeroAB
DDAH
DQ
DDQ
AH
AHQ
DQQ
31
3cos2
33cos1 22
22
2
22
21.75
12Q zeroD
QQ 2
22 cos1
21.76
zeroAB
DDAH
DQ
DDQ
AH
AHQ
31
3cos2
33 22
22
21.77
zero
AB
DDAHzero
DQ
31
3cos
22
zeroDD
QAH
AHQzero
DQ
2
33cos 22
zeroDD
QAH
AHQ
2
33
22 zero
AB
DDAH
31
3 22 21.78
DAH
QDAHba
2
311
3 2 ADB
DAHca
31
3
21.79
12Q 12
2
2 cmcm
cmE
Ho
o
o
R 1'2
o
oo
DGMDGM
LDGM
DB
zeroDDA
HDA
Hba
12
3111
3 zero
DADAca
1
311
31
ADB
DAH
Qcb
32
312
21.80
1'2
o
oo
DGMDGM
LDGM
DB 21.81
108/172
ADA
HQ
cb312
312
DAHQ62 21.82
HQQ The advance is a function of negative energy that governs the movement
12
2
2
cmcm
cmE
Ho
o
o
R DAQ
DAQ
6161 22 Advance 21.83
zeroDDA
DADA
ba
1231
61
1131 21.84
2cmE
Ho
R 2cGM
A o 2'L
GMB o
zeroDD
QAH
AHQ
2
33
22 zero
AB
DDAH
31
3 22 21.78
zeroDD
QAH
AHQDA
2
333
22 zero
AB
DDAHDA
31
33
2222
zeroAAQDHDHQ 6322 zeroDBDADH 3 21.85
1'2
o
oo
DGMDGM
LDGM
DB DADH 3 21.86
zeroAAQDADAQ 6333 22 21.87
zeroAAQDDQAQ 333 222
zeroADDQ 32 DAQ312 21.88
This advance is not governed by negative energy
zeroAAQDHDHQ 6322 21.85
zeroAAQDHDAQ 633 22 21.89
zeroAAQDHDQAQ 633 222
zeroADHDQ 62 DAHQ62 21.90
zeroAB
DDAH
DQ
DDQ
AH
AHQ
31
3cos2
33 22
22
21.77
zeroAB
DDAH
DQ
DDQ
AH
AHQDA
31
3cos2
333
22
2222
zeroADAB
DDA
DADAH
DQ
DDA
DDAQ
ADAH
ADAHQD
333
33cos3.23
33
33 22
22
222222
109/172
zeroDBDADHDQAAQDHDHQD 3cos6322
zeroDADDQAAQDHDHQD 3cos6322
zeroDA
DQAAQDHDHQ
3cos6322 21.91
DAQ312
zeroDA
DQAA
DADHDH
DA
3cos633131
zeroDA
DQA
DAAADH
DADHDH
3cos63333
zeroDA
DQA
DAADHAHDH
3cos6933
2
zeroDA
DQA
DAAH
3cos393
2
12
2
2 cmcm
cmE
Ho
o
o
R
zeroDA
DQA
DAA
3cos393
2
zero
DA
DQ
DA
3cos9 2
zero
ADQ
31cos
21.92
zeroDA
DQAAQDHDHQ
3cos6322 21.91
DAQ612
zeroDA
DQAA
DADHDH
DA
3cos636161
zeroDA
DQA
DAAADH
DADHDH
3cos66336
zeroDA
DQA
DAADHAHDH
3cos61836
2
1'2
o
oo
DGMDGM
LDGM
DB 12
2
2 cmcm
cmE
Ho
o
o
R
110/172
zeroDA
DQA
DAAH
3cos3186
2
12
2
2 cmcm
cmE
Ho
o
o
R
zeroDA
DQA
DAA
3cos3186
2
zeroDA
DQ
DAA
A
3cos183
31 2
zeroDD
QDA
1cos61
zeroDD
QDA
1cos61
zeroDD
1cos2 21.93
zeroDA
DQAAQDHDHQ
3cos6322 21.91
12Q 12
2
2
cmcm
cmE
Ho
o
o
R
zeroDA
DQAADD
3cos63
zeroDA
DQA
3cos3
zeroDD
Q
1cos 21.94
DAQ612 12Q
DAQ312
AD
QDD
QDD
QQ31cos1cos1cos2
21.95
111/172
Energy Newtonian (EN)
rkum
E oN
2
2
2
22222
rL
dtdr
dtd
rdtdru
rk
rL
dtdrm
E oN
2
22
2
rmk
rL
dtdr
m
E
oo
N 1222
22
zerom
E
rmk
rL
dtdr
o
N
o
212
2
22
2rL
dtd
ddwL
dtdr
2
2
2
2
2
2
dwd
rL
dtrd
ddw
rL
dtd
3
2
2
2 2
zerom
E
rmk
rL
ddwL
o
N
o
2122
22
zeroLm
E
rLmk
rddw
o
N
o
222
22121
zeroLm
E
rLmk
rddw
o
N
o
222
22121
zeroLm
Ew
Lmkw
ddw
o
N
o
22
22
22
22Lmkxo
2
2
Lm
Ey
o
N
zeroyxwwddw
22
QDr
w
cos111
DQQsen
ddw
DQQ
dwd
cos2
2
2
zeroyQD
xQDD
QQsen
cos11cos11
22
zeroyQD
xD
xQQD
QDQ
cos11coscos211cos1 22
222
2
2
112/172
zeroyDQx
DxQ
DQ
DDQ
DQ
DQ
coscos1cos211cos 22
2222222
2
2
2
2
zeroyDQx
Dx
DQ
DQ
DDDQQ
DQ
coscoscos21cos
2
2
222
22
2
2
zeroyDx
DDQ
DQx
DQ
DDQQ
DQ
222
2
2
22
2
2 1coscos2coscos
zeroyDx
DDQ
DQx
DDQQ
222
2
2
22 1cos2cos1
12 Q zero
D
QQ 2
22 cos1
zeroyDx
DDDQx
D
22211cos2
zeroxD
2 zeroy
Dx
DD
22211
22Lmkxo
2
2
Lm
Ey
o
N
oo
oo
o
DGMLLmmGM
DLmk
Dxzerox
D
2
221222
zeroyDDxD
DD
DD 22
22
22
22
2
22
zeroyDDx 222 1
22 DxD
DDx
kDE
DGMmE
DLmE
DyD N
oo
N
o
N
222 222
2222
zerokDEN 2
212 12
2 DkEN
2111
Da
akEN 2
113/172
Β§22 Spatial deformation
2
21
'
cv
tt
'tt
2
221
112
cvc
Lvc
Lvc
Lttt cLt '2'
2
2
2
2
2
21'
1
'2
112
cvLL
cv
cL
cvc
Lt
LL '
This is the spatial deformation. The length L' at rest in the reference frame of the observer O' is greater than the length L that is moving with velocity relative v on reference frame the observer O. Now compute to the observer O' the distance '' vtd between 'OO :
cLvvtd '2''
Thus we obtain the velocity v: '2''2'Lcdv
cLvd .
Now compute to the observer O the distance vtd between 'OO :
2
221
112
cvc
Lvttvvtd
Thus we obtain the velocity v:
2
2
2
21
2112
cv
Lcdv
cvc
Lvd .
The speed v is the same to both observers so we have:
2
21
2'2'
cv
Lcd
Lcdv
Where applying the relation 2
21'
cvLL we obtain:
2
2
2
2
2
21'1
1'2'2'
cvdd
cv
cvL
cdLcd
'dd .
Where the distance d and dβ varies inversely with the distances L and Lβ.
114/172
In general, we obtain (14.2, 14.4):
2
2
2
1
1'
cvcvuxd
d
or
2
2
2
1
''1'
cvcxvud
d
zeroxu ''
2
2
2
1
01'
cvcvd
d
2
21
'
cv
dd
cxu ''
2
2
2
1
1'
cvcvcd
d
cvcv
dd
1
1'
vxu ''
2
2
2
1
1'
cvcvvd
d
2
21'
cvdd
vux
2
2
2
1
1'
cvcvvd
d
2
21'
cvdd
cux
2
2
2
1
1'
cvcvcd
d
cvcv
dd
1
1'
zeroux
2
2
2
1
01'
cvcvd
d
2
21
'
cv
dd
115/172
Β§23 Space and Time Bend Variables with line 'r,'y,'x,'v,'t
etc ...They are used in Β§21.
Geometry of space and time in the plan xyxy .
xfy
'ctx 'rd'.rd'dsy
'ctf'ds
'cdtdx 'rd'.rd'dsdy
j'dsi'ctjyixr
j'yi'x'r
j'dsi'cdtjdyidxrd
j'dyi'dx'rd
dyry
dxrx
rrd.rdr
j'vicj'dt'dsi
'dt'cdtj
'dtdy
i'dt
dx'dtrdv
j'dt'dy
i'dt'dx
'dt'rd'v
c'dt
dx 'v'dt'ds
'dtdy cosvc vsen'v
'dt'ds
cc'dt'ds
'dtdx'dt
dy
dxdy
tg 1 2
2
22
2111
'dt'sd
c'dt'ds
c'dtd
cdxdy
dxd
dxyd
'vcv icc
j'v'v
'dt'vd
'dtcd
'dtvda
zero
'dtcd
'aa'dt'vd
'dtvd
222222 'ds'dtcdydxj'dsi'cdtj'dsi'cdtjdyidxjdyidxrd.rdds
222 'ds'dtcds 222 'dtcds'ds
c'vc'dt'dsc
'dtdsv
22
22
222
2
cvc'dt
ds'dt'ds'v
116/172
dsd
theoretical curve
dxdy
tg dxdy
arctg 2
2
2
2
2
2
2
2
11
1
1
'dt'ds
c
'dt'sd
c
dxdydxyd
dxd
2
2
2111
'dt'ds
cdxdy
dxds
23
2
2
2
2
2
2
2
2
2
2
2
2
11
1
11
11
1
'dt'ds
c
'dt'sd
c
'dt'ds
c
'dt'ds
c
'dt'sd
c
dxdsdxd
dsd
23
2
2
2
23
2
2
2
2
2
1
1
11
1
c'v
'dt'vd'v
c
'dt'ds
c
'dt'sd
'dt'ds
cdsd'v'v
dsd
'dt'ds
'dt'ds
23
2
2
2
1
1
c'v
'dt'vd'v
cdsd'v'v
23
2
2
2
c'v1
'dt'vd
c1
dsd
'rdrrkdr
rk
c'v
'dv'vm
cv
vdvmdE oo
k
22
23
2
22
2
11
21.56
'vrrk
'dt'rdr
rk
c'v
'dt'vd'v
ccm
'v'.F'dt
dE ok
2223
2
2
2
2
1
'vrrk
dsd'vcm'v'.F
'dtdE
ok
22
rrk
dsd
cm'F o 22
r
rcmk
dsd
o221
117/172
Β§24 Variational Principle
ttanconsrk
cv1
cmE
2
2
2o
k
21.21
ttanconsrk
cv1
cm
cv1cm
cv1
vmE
2
2
2o
2
22
o
2
2
2o
k
2o2
22
o
2
2
2o cm
rk
cv1cm
cv1
vm
2
2
o2
22
o
cv1
vmcv1cm
dvdp
rk
cv1cmL
2
22
o Lagrangeana.
2o
2
2
2o cmL
cv1
vm
What is the initial energy of the particle of mass mo.
2ocmLpv
rk
cv1cmcmpvL
2
22
o2
o
Variational Principle
2
1
t
t
dtt,tx,txLSAção uxdtdxx This is the velocity component in x axis.
zerodtt,x,xLS2
1
t
t
Variation of the action along the X axis.
Building the variable x'x in the range 21 ttt we have seen this when x'xzero and
where zero we will have the conditions:
zerodtd t zerot1 zerot2 zero
dd
dtd
x'x x'x d'dx
d
'xd zero
ddx
zerodxd
Then we have a new function 2
1
2
1
t
t
t
t
dtt,'x,'xFdtt,x,xGI and where:
2
1
2
1
t
t
t
t
dtt,x,xLdtt,'x,'xFLFx'xx'xzero
2
1
2
1
t
t
t
t
dtt,x,xLdtt,'x,'xFLFx'xx'xzero
118/172
So we have 2
1
t
t
dtt,'x,'xFI that provides derived:
zerodt
'xFdt
'xFdt
d'xd
'xt,'x,'xF
dtd
'dx'x
t,'x,'xFdI
2
1
2
1
2
1
2
1
t
t
t
t
t
t
t
t
'xF
dtd
'xF
dtd
'xF
dtd'x
F'x
Fdtd
'xF
dtd
zerodt
'xF
dtd
'xF
dtddt
'xFdt
'xFdt
'xF
dI
2
1
2
1
2
1
2
1
t
t
t
t
t
t
t
t
zerodt
'xF
dtd
'xFddt
'xF
dI
2
1
2
1
2
1
t
t
t
t
t
t
zerot'x
Ft'x
F'x
F'x
Fd 12
t
t
t
t
2
1
2
1
zerodt
'xF
dtd
'xFdt
'xF
dtddt
'xF
dI
2
1
2
1
2
1
t
t
t
t
t
t
zero
'xF
dtd
'xFzerozerodt
'xF
dtd
'xF
dI
2
1
t
t
zeroxL
dtd
xLLFx'xx'xzero
xL
dtd
xL
This is the X axis component
rk
cv1cmL
2
22
o
rk
cv1cm
xdtd
rk
cv1cm
x 2
22
o2
22
o
zerocv1cm
x 2
22
o
zero
rk
x
222222
zyxdtdz
dtdy
dtdxv
2
22
o cv1cm
xdtd
rk
x This is the X axis component
32
111
rxk
rx
r1k
xrr1kr
xk
rk
x
2222 zyxr
119/172
222
2
2
o2
121
2
22
o2
22
o zyxxd
d
cv1
vmxd
dvcv2
cv1
21cm
cv1cm
x
2
2
o
222
2
2
o121
222
2
2
o2
22
o
cv1
xm
zyxx
cv1
vmx2zyx
21
cv1
vmcv1cm
x
dtdv
cv2
cv1
21x
cv1
dtxd
cv1
m
cv1
dtdx
cv1
dtxd
cv1
m
cv1
xmdtd
2
121
2
2
2
2
2
2o
2
2
2
2
2
2o
2
2
o
dtdv
cv
cv1
x
cv1
cv1
cv1
dtxd
cv1
mdtdv
cv
cv1
xcv1
dtxd
cv1
m
cv1
xmdtd
2
2
2
2
2
2
2
2
2
2
2o
2
2
22
2
2
2o
2
2
o
22
2
23
2
2
o2
2
2
2
2
2
2
2
2
2
2o
2
2
o
cx
dtdvv
dtxd
cv1
cv1
mdtdv
cv
cv1
x
cv1
cv1
cv1
dtxd
cv1
m
cv1
xmdtd
icx
dtdvvx
cv1
cv1
mi
rxk
22
2
23
2
2
o3
X axis
jcy
dtdvvy
cv1
cv1
mj
ry
k22
2
23
2
2
o3
Y axis
kcz
dtdvvz
cv1
cv1
mk
rzk
22
2
23
2
2
o3
Z axis
rr
krr
kkzjyixr
kkrzkj
r
yki
rxk
233333
rrkk
cz
dtdvvz
cv1
cv1
mj
cy
dtdvvy
cv1
cv1
mi
cx
dtdvvx
cv1
cv1
m222
2
23
2
2
o22
2
23
2
2
o22
2
23
2
2
o
rrkk
cz
dtdvvz
cv1j
cy
dtdvvy
cv1i
cx
dtdvvx
cv1
cv1
m222
2
22
2
22
2
23
2
2
o
120/172
rrkk
cx
dtdvvkz
cv1j
cy
dtdvvjy
cv1i
cx
dtdvvix
cv1
cv1
m222
2
22
2
22
2
23
2
2
o
rrkkzjyix
dtdv
cvkzjyix
cv1
cv1
m222
2
23
2
2
o
dtvdkzjyix
dtdkzjyixa
kzjyixv
rrk
cv
dtdvv
dtvd
cv1
cv1
mF
222
2
23
2
2
o
= 21.16
rrk
cv
dtdvv
dtvd
cv1
cv1
mF
222
2
23
2
2
o
= 21.19
Β§24 Variational Principle Continuation
tetanconsrk
cv1
cm
c'v1cmE
2
2
2o
2
22
ok
21.21
tetanconsrk
cv1
cm
cv1cm
cv1
vm
c'v1cmE
2
2
2o
2
22
o
2
2
2o
2
22
ok
tetanconsrk
rk
rk
cv1
cmrk
cv1cm
cv1
vmrk
c'v1cm
rkE
2
2
2o
2
22
o
2
2
2o
2
22
ok
tetanconscmrk
cv1cm
cv1
vmrk
c'v1cm
rkE 2
o2
22
o
2
2
2o
2
22
ok
2
22
oc'v1cm'T
2
22
ocv1cmT
rkEp
2
2
2o
cv1
vmpv
'p'v
c'v1
'vm'vv
cv1
vmpv
2
2o
2
2o
.
2
2
c'v1'pp
2
2
cv1p'p
pppkR ETpvE'TEEE
'TEk TpvEk Tpv'T 'T'v'pT
121/172
pE'T'L pETL
Lpv'LEEE pkR
Lpv'L 'L'v'pL 'v'ppv'LL
vm
c'v1
'vm
c'v1cm
'dvd
'dv'dT'p o
2
2
o2
22o
'vm
cv1
vm
cv1cm
dvd
dvdTp o
2
2
o2
22o
rdkdzjdyidxk'dzj'dyi'dx'rd
21.08
2
2
2
2
2
2
cv1
vdtrd
cv1
1kdtdzj
dtdy
idtdx
cv1
1k'dt'dzj
'dt'dy
i'dt'dx
'dt'rd'v
2
2
2
2
x
2
2'x
cv1
x
cv1
vdtdx
cv1
1'dt'dx'v'x
xm
c'v1
'xm
c'v1cm
'xdd
'xd'dT'p o
2
2
o2
22ox
'xm
cv1
xm
cv1cm
xdd
xddTp o
2
2
o2
22ox
rkzjyixk'zj'yi'x'r
21.07
x'x y'y z'z
1x'x
1
y'y
1z'z
zeroxL
dtd
xL
zeroxL
'dtd
dt'dt
'xL
x'x
xL
dtd
xL
'L'v'pL 'xmp
xT
xL
ox
zero'xm'dt
d
c'v1
1'L'v'p'xx
L'dt
ddt
'dt'x
Lx'x
o
2
2
zero'dt'xd
c'v1
m'x'L
'x'v'p
'x'p
'v'dt'xd
c'v1
m'L'v'p
'x2
2
o
2
2
o
zero'x'p
zero'x'v
rk
c'v1cm'L2
22
o
zero'dt'xd
c'v1
m'x'L
2
2
o
2222222 zyx'z'y'xr.r'r'.rr
122/172
32
1112
22
or
'xkr'x
r1k
'xrr1kr
'xk
rk
'xrk
c'v1cm
'x'x'L
zero
c'v1
'xmr
'xk'dt'xd
c'v1
m'x'L
2
2
o3
2
2
o
3
2
2
o
r'xk
c'v1
'xm
'r
rk'r
rkk'zj'yi'x
rkk
r'zkj
r'y
kir
'xk233333
'rrk
c'v1
'amk
c'v1
'zmj
c'v1
'ymi
c'v1
'xm2
2
2
o
2
2
o
2
2
o
2
2
o
rrk'r
rk
c'v1
'am22
2
2
o
r'r rrk
c'v1
'am2
2
2
o
= 21.19
Β§25 Logarithmic spiral
zeroBAw3wd
wdA3Hwd
wdH 22
2
2
2
aer 21.37
Qcos1D1
r1w
D
QQsenddw
D
QcosQ
dwd
2
2
2
21.38
a
ae
e1
r1w
aaeddw
a22
2ea
dwd
zeroBeA3eeAa3HeeHa2aaa2aa2
zeroBAe3eAa3HeeHa a2a22aa2
zeroBAe3a1Hea1 a22a2
zeroBHea1Ae3a1 a2a22
zeroBHwa1Aw3a1 222
zeroa1
BHwAw32
2
2
22
2
a
a1AB12H
A61
A6H
A3.2a1
BA3.4HH
r1ew
zeroa1
Ba1
AB12HA61
A6HH
a1AB12H
A61
A6HA3
222
2
22
123/172
zeroa1
Ba1
AB12HA6H
A6H
a1AB12H
A61
a1AB12H
A61
A6H2
A6HA3
222
2
2
22
22
2
zeroa1
Ba1
AB12HA6H
A6H
a1AB12H
A361
a1AB12H
A61
A6H2
A6HA3
222
2
22
222
2
zeroa1
Ba1
AB12HA6H
A6H
a1AB12H
A361
a1AB12H
A18H
A36HA3
222
2
22
222
22
2
zeroa1
Ba1
AB12HA6H
A6H
a1AB12H
A121
a1AB12H
A6H
A12H
222
2
22
22
2
zeroa1
BA6
Ha1
AB12HA12
1A12
H2
2
22
2
zeroa1
BA6
Ha1
BA12
HA12
H2
2
2
22
Β§25 Logarithmic Spiral (Continuation)
2o
2
23
'L
GMr1
dwd
r1AH
21.71
2o
2
23
'L
GMr1
dwd
r1AH
Br1
dwd
r1AH
2
23
2o
R
cm
EH
2o
c
GMA
2o
'L
GMB
zeroBr1
dwd
r1AH
2
23
zeroBr1
dwd
r1A
r1HA3
r1AH3H
2
2
33
2223
r1AH3H
r1A
r1HA3
r1AH3H
23
3
3
2
223 zeror1A
r1HA3
3
3
2
2
124/172
zeroBr1
dwd
r1AH3H
2
223
zeroBwd
wdwAH3H2
223
zeroBwAH3wd
wdAH3wHd
wdH 222
223
2
23
Qcos1D1
r1w
D
QQsen
ddw
D
QcosQ
dwd
2
2
2
21.38
zeroBQcos1D1AH3
Qcos1D1
DQcosQ
AH3Qcos1D1H
DQcosQ
H
22
223
23
zeroBQcosQcos21D1AH3
QcosD1
DQcosQ
AH3D1
DQcosQ
AH3QcosD1H
D1H
DQcos
QH
2222
2
22
223323
zeroBQcosQcos21D
AH3
D
QcosQAH3
D
Qcos
DQAH3
D
QcosH
DH
D
QcosQH
2222
2
2
222
223
323
zeroBQcosD
AH3Qcos2D
AH3D
AH3
D
QcosQAH3
D
Qcos
DQAH3
D
QcosH
DH
D
QcosQH
2222
2
22
2
22
2
2
222
223
323
zeroB
D
QcosAH3
DQcos
DAH6
DAH3
D
QcosQAH3
D
Qcos
DQAH3
D
QcosH
DH
D
QcosQH
2
22
2
22
2
2
222
223
323
zero
AH3B
D
Qcos
AH3AH3
D
Qcos
DAH3AH6
DAH3AH3
D
Qcos
AH3
QAH3D
Qcos
DAH3
QAH3D
Qcos
AH3H
DAH3H
DQcos
AH3
QH
22
2
2
2
2
2
222
2
2
2
2
22
2
22
2
3
2
3
2
23
zero
AH3B
D
QcosD
QcosD2
D1
D
QcosQ
D
Qcos
DQ
D
Qcos
A3H
DA3H
D
Qcos
A3HQ
22
2
22
2
22
22
125/172
zero
AH3B
D1
DA3H
DQcos
D2
DQcos
DQ
DQcos
A3H
DQcos
A3HQ
D
QcosQ
D
Qcos
222
22
2
22
2
2
zero
AH3B
D1
DA3H
D
Qcos
D2
DQ
A3H
A3HQ
D
QcosQ1
222
22
2
22
1cmcm
cmE
H2
o
2o
2o
R DA61Q
2
zero
)1(A3
B
D
1DA3
)1(
D
Qcos
D2
DQ
A3
)1(
A3
Q)1(
D
QcosQ1
222
22
2
22
zero
A3B
D
1DA3
1D
Qcos
D2
DQ
A31
A3Q
D
QcosQ1
22
22
2
22
zero
DA3DB
D
1DA3
1D
Qcos
D2
DQ
A31
A3Q
D
QcosQ1
22
22
2
22
1'L
DGMDB
2o
zero
DA31
D
1DA3
1D
Qcos
D2
DQ
A31
A3Q
D
QcosQ1
22
22
2
22
zero
D
1D
Qcos
D2
DQ
A31
A3Q
D
QcosQ1
22
22
2
22
DA61Q
2
zero
D
1D
Qcos
D2
DA61
D1
A31
DA61
A31
D
Qcos
DA611
222
2
zero
D
1D
Qcos
D2
DA6
D1
D1
A31
DA6
A31
A31
D
Qcos
DA611
222
2
zero
D
1D
Qcos
D1
D
A6D2
D
Qcos
DA6
22222
2
zero
D
1D
Qcos
D
A6D1
D
Qcos
D1A6
22222
2
126/172
zero
D1
D1
D
Qcos
D1
DA61
D
Qcos
D1A6
2
2
zero
D1
D
Qcos
DA61
D
QcosA6
2
2
A6.2
D1.A6.4
DA61
DA61
D
Qcos
2
A12DA24
DA6
DA6121
DA61
D
Qcos
22
A12
DA24
DA6
DA121
DA61
D
Qcos
2
A12
DA36
DA361
DA61
D
Qcos 22
2
A12
DA36
DA361
DA61
D
Qcos 22
2
DA361
D
A36DA361
22
2
zero
D
A3622
2
2
o
c
GMA
14
2
28
3011
2
2
2o
2222
2
10.55,210.99792458,2
10.989,110.67,6
600.955.442.5536
c
GM
D
36
D
A36
A12
DA361
DA61
DQcos
A12
DA181
DA61
A12DA36
211
DA61
A12DA361
DA61
DQcos
D1
A12DA12
A12DA181
DA61
A12DA181
DA61
DQcos
zero
D1
DQcos
127/172
zero
D1
D
Qcos
DA61QzeroM)Q(rzero o
zero
D
1D
Qcos
D2
DQ
A31
A3Q
D
QcosQ1
22
22
2
22
1QzeroMr o
1c
zeroG
D61
c
GM
D61
DA61Q
22o
zero
D
1D
Qcos
D2
D1
A31
A31
D
Qcos11
222
2
zero
D
1D
Qcos
D1
22
zeroD1
D
Qcos
zero
D1
D
QcosQcos1
D1
r1w1QzeroMr o
The presence of Q in the formula Qcos1
D)Q(rr
, allows it to also describe a spiral.
Β§25 Logarithmic Spiral Continuation II
zero
D
1D
Qcos
D2
DQ
A31
A3Q
D
QcosQ1
22
22
2
22
DA61
DA121
QzeroM)Q(rzero o
128/172
zeroD1
DQcos
D2
DA61DA121
D1
A31
DA61DA121
A31
DQcos
DA61DA121
1222
2
zeroDA61
D1
DQcos
DA61
D2
DA121
D1
DA61
A31
DA121
A31
DQcos
DA121
DA61
222
2
zeroDA6
D1
D1
DQcos
DA6
D2
D2
DA12
D1
D1
DA6
A31
A31
DA12
A31
A31
DQcos
DA121
DA61
22222
2
zeroDA6
D1
D1
DQcos
D1
DQcos
DA6
22222
2
DA62
DA6
D1
D1
DA64
D1
D1
DQcos 2222
2
DA12
DA6
D1
D1
DA24
D1
D1
DQcos 222222
DA12
DA6
D1
DA24
D1
DA24
D1
D1
DQcos 222222
DA12
DA6
DA24
DA241
D1
D1
DQcos
DA12
DA144
DA1221
D1
D1
DQcos 22
2
DA12
DA121
D1
D1
DQcos
2
DA12
DA121
D1
D1
DQcos
DA12
DA12
D1
D1
D1
DQcos
129/172
DA12
DA12
D1
D1
D1
DQcos
DA12
DA12
D1
D1
D1
DQcos
DA12DA12
D1
DQcos
D1
DQcos
zeroD1
DQcos
zeroD1
DQcos
DA61DA121
QzeroM)Q(rzero o
DA61
DA61DA121
Q2
DA61Q2
2
o
c
GMA
40,568.469.460.5520563593,0100,000.227.909.571aD 22
42.535.089,477.110.99792458,2
10.9891,1.10.6740831,6c
GMA 28
3011
2o
1.920.999.999,0
DA61DA121
Q
1.920.999.999,0
DA61Q
1,276.789.102.53-14
Q00,000.296.100,000.296.1Q. 1Q Advance 1Q Regression
00,000.296.11Q1
zero Advance zero Regression
130/172
"544.893.549.103,000,000.296.11
DA61DA121
121
"997.876.549.103,000,000.296.11
DA61
121
139.316.210,415969,87
004.363.256,365100PMPT.100N
"7.034.984.994,42139.316.210,415x544.893.549.103,0N.
"2.164.977.994,42139.316.210,415x997.876.549.103,0N.
By definition zero
DA61DA121
QzeroM)Q(rzero o
1QzeroMr o
Qcos1zero
D1
DQcos
Qcos1zero
D1
DQcos
Se If 1Q
cos1
cos1
Energy Newtonian (EN)
zeroy
Dx
D1
D
QD
QcosD2x
D
QcosQ1 222
2
2
22
zero
D1
DQcos
Qcos1D1
r1w1Qr
zeroyDx
D1
D
QD1
D2x
D1
D1Q1 222
22
zeroyDx
D1
D
Q
D2
Dx
D1Q1 222
2
22222
zeroyDx
D1
D
Q
D2
Dx
D
Q
D1
222
2
2222
2
22
131/172
zeroyDx2
D4
D
Q
D
Q222
2
22
2
1Q2
zeroyDx2
D4
D1
D1
22222
zeroyDDDx2
DD4
DD
DD 2222
22
22
2
22
22
22
zeroyDDx241 222
D2x
2
o
N
Lm
E2y DGML2 1D1
a1 2
zeroyDDD2241 222
zeroyD1 222
zeroLm
E2D1 2o
N222 zeroDGMmE2
D1oo
N222
zeromGM
E2D1
oo
N2 kE21
D1 N2
a2kEN
132/172
Β§26 Advancement of the PeriΓ©lio of Mercury of 42,99 "
Supposing vux
(2.3) zeroxu
cvv
cv
vv
cvux
cv
vuxxu
''2121
''
22
2
22
2
vux zeroxu '' 21.01
(1.17) 2
2
22
2
22
21'2121'
cvdtdt
cvv
cvdt
cvux
cvdtdt
(1.22)
2
2
22
2
22
2 '1'0'2'1''''2'1'cvdtdt
cv
cvdt
cxuv
cvdtdt
2
21'
cvdtdt 2
2'1'cvdtdt 21.02
1'11 2
2
2
2
cv
cv 21.03
2
2'1
'
cv
vv
2
21
'
cv
vv
21.04
'dtdt 'vv ''dtvvdt 21.05
(1.33)
2
2
22
2
22
2 '1
'0'2'1
''''2'1
'
cvvv
cv
cvv
cxuv
cv
vv
(1.34)
2
2
22
2
22
21
'2121
'
cvvv
cvv
cvv
cvux
cvvv
2
2'1
'
cvvv
2
21
'
cv
vv
21.06
'Λ rrrr
rrrr
Λ' rrr '
21.07
'ΛΛ rdrrdrdrrd
rdrrdrdrrd
ΛΛ' 21.08
drrdrrrrdrrdr ΛΛΛΛΛ
21.09
21.10
21.11
drrdrrrrdrrdr ΛΛΛΛ'Λ
ΛΛΛdtdrr
dtdr
dtrrd
dtrdv 22
2
dtdr
dtdrvvv
Λ
'Λ''
Λ'''
dtdrr
dtdr
dtrrd
dtrdv 22
2
'''''
dtdr
dtdrvvv
133/172
21.12
21.13
21.06
21.50
21.51
Λ2ΛΛ2
22
2
2
2
2
2
2
dtdr
dtd
dtdrr
dtdr
dtrd
dtrrd
dtrd
dtvda
Λ'''
2Λ'''
Λ''
''' 2
22
2
2
2
2
2
2
dtdr
dtd
dtdrr
dtdr
dtrd
dtrrd
dtrd
dtvda
2
2'1
'
cvvv
2
22
2
2
2
2
2 '1
''
1'1
''
''1
'
cvv
dtd
cv
cvv
dtd
dtdt
cvv
dtd
dtvda
2
2
2
2
2
22
2 '1'
''''1
'1
11cv
dtdv
dtvd
cv
cvc
vdtvda
'''2'1
21'
'''1
'1
11 2
21
22
21
2
2
2
2
2
22
2
dtdv
cv
cvv
dtvd
cv
cvc
vdtvda
2
2
22
2
2
22
2 ''''
'1
1'''1
'1
11cv
dtdvv
cvdt
vdcv
cvc
vdtvda
2
2
22
2
2
2
2
2
2
22
2 ''''
'1
1'1'''1
'1
1'1
11cv
dtdvv
cvc
vdtvd
cv
cv
cvc
vdtvda
22
2
23
2
22
2 ''''
'''1
'1
11cv
dtdvv
dtvd
cv
cvc
vdtvda
22
2
23
2
22
2
2
2
''''
'''1
'111cv
dtdvv
dtvd
cv
cv
mdtvd
cv
m
cv
amam ooo
dtvd
cv
m
cv
amamF oo
2
2
2
211
134/172
21.52
21.53
21.54
21.55
21.56
21.57
21.58
22
2
23
2
2
''''
'''1
'1
'cv
dtdvv
dtvd
cv
cv
mF o
22
2
23
2
22
2
2
2
''''
'''1
'1
'11
cv
dtdvv
dtvd
cv
cv
mF
dtvd
cv
m
cv
amamF ooo
'Λ''.. 2 rdrrkrdFrdFEk
'Λ'''''
'''1
'11
''.. 222
2
23
2
22
2rdr
rkrd
cv
dtdvv
dtvd
cv
cv
mrd
dtvd
cv
mrdFrdFE oo
k
'Λ'''''
''''1
'11222
2
23
2
22
2rdr
rk
cv
dtrddvv
dtrdvd
cv
cv
mdtrdvd
cv
mE ook
drrk
cvvdvvvvd
cv
cv
mvvd
cv
mE ook 222
2
23
2
22
2
'''''''1'11
drrk
cvdvvvvd
cv
cv
m
cv
vdvmE ook 22
2
2
2
23
2
22
2
''''''1'11
drrk
cv
cv
cv
dvvm
cv
vdvmE ook 22
2
2
2
23
2
22
2
''1'1
''
1
drrk
cv
dvvm
cv
vdvmE ook 2
23
2
22
2'1
''
1dr
rk
cv
dvvm
cv
vdvmdE oo
k 223
2
22
2'1
''
1
teconsrk
cv
cmcvcmE o
ok tan'1
1
2
2
2
2
22
teconsrk
cvcmE oR tan1 2
22 tecons
rk
cv
cmE oR tan
'1 2
2
2
135/172
21.59
21.60
21.61
21.62
21.63
kdtdrr
dtdr
cvdt
drrdtdr
cv
rr
cvvrvrL Λ
'ΛΛ
11ΛΛ
11Λ
1''' 22
2
2
2
2
2
2
21.63
21.64
21.56
21.65
21.66
21.67
rkvm
cmrk
cv
cmE o
oo
R
2'
'1
22
2
2
2
2
2
2
2
01
cmk
c
cmE o
oR
rcmk
cmE
cv oo
R 1'1
122
2
2
2cmE
Ho
R 222 cGM
cmmGM
cmkA o
o
oo
o
rAH
cv
1'1
1
2
2
3
23
2
2
1
'1
1
rAH
cv
kdtdrr
dtdr
dtdrr
dtdrrrvrL Λ
'ΛΛ
'Λ
'Λ'
Λ''' 22
kLkdtdrL ''
' 2
'' 2
dtdrL
'Λ'1
''
122
23
2
22
2rdr
rkdr
rk
cv
dvvm
cv
vdvmdE oo
k
'Λ''Λ
'''
'1
''' 22
23
2
2vr
rk
dtrdr
rk
dtdvv
cv
mvF
dtdE ok
rrk
cv
amF o Λ
'1
'' 2
23
2
2
rrk
dtdr
dtd
dtdrr
dtdr
dtrd
cv
mF o ΛΛ
'''2Λ
'''1
' 22
22
2
2
23
2
2
zerodtdr
dtd
dtdr
cv
mF o
Λ'''
2'1
' 2
2
23
2
2Λ
136/172
21.68
21.69
21.70
21.71
Β§25 Logarithmic Spiral continued
2o
2
23
'L
GMr1
dwd
r1AH
21.71
2o
2
23
'L
GMr1
dwd
r1AH
Br1
dwd
r1AH
2
23
2o
R
cm
EH
2o
c
GMA
2o
'L
GMB
zeroBr1
dwd
r1AH
2
23
rrkr
dtdr
dtrd
cv
mF o
r ΛΛ'''1
' 2
2
2
2
23
2
2Λ
rrGM
rdtdr
dtrd
cv
o ΛΛ'''1
12
2
2
2
23
2
2
2'
' rL
dtd
ddwL
dtdr ''
2
2
2
2
2
2 '' d
wdrL
dtrd
ddw
rL
dtd
3
2
2
2 '2'
2
2
22
2
2
2
23
2
2
''
'1
1rGM
rLr
dwd
rL
cv
o
23
2
2
2
2
2
23
2
2
''
'1
1rGM
rL
dwd
rL
cv
o
22
2
2
2
23
2
2
'1
'1
1rGM
rL
rdwd
cv
o
22
2
23
2
2 '1
'1
1LGM
rdwd
cv
o
22
23
'11
LGM
rdwd
rAH o
137/172
zeroBr1
dwd
r1A
r1HA3
r1AH3H
2
2
33
2223
r1AH3H
r1A
r1HA3
r1AH3H
23
3
3
2
223 zeror1A
r1HA3
3
3
2
2
zeroBr1
dwd
r1AH3H
2
223
zeroBwd
wdwAH3H2
223
zeroBwAH3wd
wdAH3wHd
wdH 222
223
2
23
Qcos1D1
r1w
D
QQsen
ddw
D
QcosQ
dwd
2
2
2
21.38
The first hypothesis to obtain a particular solution of the differential equation is to assume the infinite radius
r , thus obtaining:
1QcoszeroQcos1D1
r1w
D
QD
QcosQD
QcosQd
wd222
2
2
zeroBwAH3wd
wdAH3wHd
wdH 222
223
2
23
zerow D
Qd
wd2
2
2
1
cmcm
cmE
H2
o
2o
2o
R
zeroBzero1A3zeroD
Q1A3zero1D
Q1 222
232
3
zeroBD
Q2
zeroDBD
DQ2
zero1Q2 1Q2
This result shows that in infinity the influence of the central mass is zero zeroMo .
The second hypothesis to obtain another particular solution of the differential equation is obtained by
observing that the angle Q of the equation 1Qcos indicates the direction of the infinite radius
r where the influence of the central mass is zero zeroMo and 1Q2 therefore the direction of the
center of mass is given by the angle Q that replaced in the equation 1Qcos results in the
new equation 1Qcos that indicates direction opposite the direction of the infinite radius which is
the direction of the center of mass.
1Qcos QcosQcos 1Qcos 1Qcos
138/172
D211
D1Qcos1
D1
r1w
DQ
DQcosQ
DQcosQ
dwd
222
2
2
D2w
DQ
dwd
2
2
2
1
cmcm
cmE
H2
o
2o
2o
R
zeroBwAH3wd
wdAH3wHd
wdH 222
223
2
23
zeroBD21A3
D2
DQ1A3
D21
DQ1
22
223
23
zeroBD2A3
D2
DQA3
D2
DQ 222
zeroBD4A3
D2
DQA3
D2
DQ
22
22
zeroBDA12
DAQ6
D2
DQ
2222
22
zeroDBD
A12DDAQ6D
DD2
DDQ
2222
22
1DGM
DGM
'L
DGMDB
o
o2
o
zero1DA12
DAQ62Q
22
zeroDA12
DAQ61Q
22
DA121
DAQ6Q
22
DA61DA121
Q2
Applying the results of the second hypothesis in the differential equation:
zeroBwAH3wd
wdAH3wHd
wdH 222
223
2
23
Qcos1D1
r1w
D
QQsen
ddw
D
QcosQ
dwd
2
2
2
21.38
zeroBQcos1D1AH3
Qcos1D1
DQcosQ
AH3Qcos1D1H
DQcosQ
H
22
223
23
139/172
zeroBQcosQcos21D1AH3
QcosD1
DQcosQ
AH3D1
DQcosQ
AH3QcosD1H
D1H
DQcos
QH
2222
2
22
223323
zeroBQcosQcos21D
AH3
D
QcosQAH3
D
Qcos
DQAH3
D
QcosH
DH
D
QcosQH
2222
2
2
222
223
323
zeroBQcosD
AH3Qcos2D
AH3D
AH3
D
QcosQAH3
D
Qcos
DQAH3
D
QcosH
DH
D
QcosQH
2222
2
22
2
22
2
2
222
223
323
zeroB
D
QcosAH3
DQcos
DAH6
DAH3
D
QcosQAH3
D
Qcos
DQAH3
D
QcosH
DH
D
QcosQH
2
22
2
22
2
2
222
223
323
zero
AH3B
D
Qcos
AH3AH3
D
Qcos
DAH3AH6
DAH3AH3
D
Qcos
AH3
QAH3D
Qcos
DAH3
QAH3D
Qcos
AH3H
DAH3H
DQcos
AH3
QH
22
2
2
2
2
2
222
2
2
2
2
22
2
22
2
3
2
3
2
23
zero
AH3B
D
QcosD
QcosD2
D1
D
QcosQ
D
Qcos
DQ
D
Qcos
A3H
DA3H
D
Qcos
A3HQ
22
2
22
2
22
22
zero
AH3B
D1
DA3H
DQcos
D2
DQcos
DQ
DQcos
A3H
DQcos
A3HQ
D
QcosQ
D
Qcos
222
22
2
22
2
2
zero
AH3B
D1
DA3H
D
Qcos
D2
DQ
A3H
A3HQ
D
QcosQ1
222
22
2
22
1cmcm
cmE
H2
o
2o
2o
R
zero
)1(A3
B
D
1DA3
)1(
D
Qcos
D2
DQ
A3
)1(
A3
Q)1(
D
QcosQ1
222
22
2
22
zero
A3B
D
1DA3
1D
Qcos
D2
DQ
A31
A3Q
D
QcosQ1
22
22
2
22
140/172
zero
DA3DB
D
1DA3
1D
Qcos
D2
DQ
A31
A3Q
D
QcosQ1
22
22
2
22
1DGM
DGM
'L
DGMDB
o
o2
o
zero
DA31
D
1DA3
1D
Qcos
D2
DQ
A31
A3Q
D
QcosQ1
22
22
2
22
zero
D
1D
Qcos
D2
DQ
A31
A3Q
D
QcosQ1
22
22
2
22
DA61
DA121
QzeroM)Q(rzero o
zeroD1
DQcos
D2
DA61DA121
D1
A31
DA61DA121
A31
DQcos
DA61DA121
1222
2
zeroDA61
D1
DQcos
DA61
D2
DA121
D1
DA61
A31
DA121
A31
DQcos
DA121
DA61
222
2
zeroDA6
D1
D1
DQcos
DA6
D2
D2
DA12
D1
D1
DA6
A31
A31
DA12
A31
A31
DQcos
DA121
DA61
22222
2
zeroDA6
D1
D1
DQcos
D1
DQcos
DA6
22222
2
DA62
DA6
D1
D1
DA64
D1
D1
DQcos 2222
2
DA12
DA6
D1
D1
DA24
D1
D1
DQcos 222222
DA12
DA6
D1
DA24
D1
DA24
D1
D1
DQcos 222222
DA12
DA6
DA24
DA241
D1
D1
DQcos
141/172
DA12
DA144
DA1221
D1
D1
DQcos 22
2
DA12
DA121
D1
D1
DQcos
2
DA12
DA121
D1
D1
DQcos
DA12
DA12
D1
D1
D1
DQcos
DA12
DA12
D1
D1
D1
DQcos
DA12
DA12
D1
D1
D1
DQcos
DA12DA12
D1
DQcos
D1
DQcos
Where applying the result of the second hypothesis
1Qcos1Qcos :
D1
D11
That it is an identity demonstrating that the result of the second hypothesis is correct.
DA61
DA61DA121
Q2
DA61Q2
2
o
c
GMA
40,568.469.460.5520563593,0100,000.227.909.571aD 22
42.535.089,477.110.99792458,2
10.9891,1.10.6740831,6c
GMA 28
3011
2o
142/172
1.920.999.999,0
DA61DA121
Q
1.920.999.999,0
DA61Q
1,276.789.102.53-14
Q00,000.296.100,000.296.1Q. 1Q Advance 1Q Retrocess
00,000.296.11Q1
zero Advance zero Retrocess
"544.893.549.103,000,000.296.11
DA61DA121
1
21
"997.876.549.103,000,000.296.11
DA61
121
139.316.210,415969,87
004.363.256,365100PMPT.100N
"7.034.984.994,42139.316.210,415x544.893.549.103,0N.
"2.164.977.994,42139.316.210,415x997.876.549.103,0N.
Newtonian Energy EN
rkum
E oN
2
2
2
22222
rL
dtdr
dtd
rdtdru
rk
rL
dtdrm
E oN
2
22
2
rmk
rL
dtdr
m
E
oo
N 1222
22
zerom
E
rmk
rL
dtdr
o
N
o
212
2
22
143/172
zeroyDx
DDQ
DQx
DQ
DDQQ
DQ
222
2
2
22
2
2 1coscos2coscos
zeroyDx
DDQ
DQx
DDQQ
222
2
2
22 1cos2cos1
2rL
dtd
ddwL
dtdr
2
2
2
2
2
2
dwd
rL
dtrd
ddw
rL
dtd
3
2
2
2 2
zerom
E
rmk
rL
ddwL
o
N
o
2122
22
zeroLm
E
rLmk
rddw
o
N
o
222
22121
zeroLm
E
rLmk
rddw
o
N
o
222
22121
zeroLm
Ew
Lmkw
ddw
o
N
o
22
22
22
22Lmkxo
2
2
Lm
Ey
o
N
zeroyxwwddw
22
QDr
w
cos111 D
QQsenddw
DQQ
dwd
cos2
2
2
zeroyQD
xQDD
QQsen
cos11cos11
22
zeroyQD
xD
xQQD
QDQ
cos11coscos211cos1 22
222
2
2
zeroyDQx
DxQ
DQ
DDQ
DQ
DQ
coscos1cos211cos 22
2222222
2
2
2
2
zeroyDQx
Dx
DQ
DQ
DDDQQ
DQ
coscoscos21cos
2
2
222
22
2
2
144/172
Newtonian Energy EN
zeroy
Dx
D1
D
QD
QcosD2x
D
QcosQ1 222
2
2
22
zero
D1
DQcos
Qcos1D1
r1w1Qr
zeroyDx
D1
D
QD1
D2x
D1
D1Q1 222
22
zeroyDx
D1
D
Q
D2
Dx
D1Q1 222
2
22222
zeroyDx
D1
D
Q
D2
Dx
D
Q
D1
222
2
2222
2
22
zeroyDx2
D4
D
Q
D
Q222
2
22
2
1Q2
zeroyDx2
D4
D1
D1
22222
zeroyDDDx2
DD4
DD
DD 2222
22
22
2
22
22
22
zeroyDDx241 222
D2x
2
o
N
Lm
E2y DGML2 1D1
a1 2
zeroyDDD2241 222
zeroyD1 222
zeroLm
E2D1 2o
N222 zeroDGMmE2
D1oo
N222
zeromGM
E2D1
oo
N2 kE21
D1 N2
a2kEN
145/172
Β§27 Advancement of Perihelion of Mercury of 42.99β "contour Conditions" Let us start from the equation expressing the equilibrium of forces:
21.65
On the right side we have the gravitational force
οΏ½οΏ½ defined by Newton, on the left side we have the
physical description of Force οΏ½οΏ½ =
of the Undulating Relativity.
The physical properties of equation 21.65 require its validity when its radius varies from a radius greater than zero to an infinite radius, so the radius varies from π§πππ < π β€ β, and so we have two distinct boundary conditions. The first boundary condition is when the radius is infinite π = β and the gravitational force is zero, which means that the particle is at rest with π£β² = π§πππ and π = π§πππ and the second boundary condition is when the radius is greater which is zero and smaller than infinity π§πππ < π < β which means that the particle is in motion due to the influence of a gravitational force 21.65 with π£β² β π§πππ and π β π§πππ. In Β§26 following the calculations is substituted in 21.65, the equality, 21.62, 21.69 and
2o
R
cm
EH
2o
c
GMA
2o
'L
GMB , more π€ = .
After these substitutions we obtain the differential equation:
zeroBwAH3wd
wdAH3wHd
wdH 222
223
2
23
27.1
This equation has to be valid for the same boundary conditions as equation 21.65, that is, it has to be valid from a radius r greater than zero (π > π§πππ) to an infinite radius (π§πππ < π β€ β). Your solution is given by:
Qcos1D1
r1w
27.2
Which should cover the two contour conditions already described. Applying solution 27.2 in differential equation 27.1 we have:
zeroBwAH3wd
wdAH3wHd
wdH 222
223
2
23
Qcos1D1
r1w
D
QQsen
ddw
D
QcosQ
dwd
2
2
2
21.38
zeroBQcos1D1AH3
Qcos1D1
DQcosQ
AH3Qcos1D1H
DQcosQ
H
22
223
23
zeroBQcosQcos21D1AH3
QcosD1
DQcosQ
AH3D1
DQcosQ
AH3QcosD1H
D1H
DQcos
QH
2222
2
22
223323
rrk
cv
amF o Λ
'1
'' 2
23
2
2
146/172
zeroBQcosQcos21D
AH3
D
QcosQAH3
D
Qcos
DQAH3
D
QcosH
DH
D
QcosQH
2222
2
2
222
223
323
zeroBQcosD
AH3Qcos2D
AH3D
AH3
D
QcosQAH3
D
Qcos
DQAH3
D
QcosH
DH
D
QcosQH
2222
2
22
2
22
2
2
222
223
323
zeroB
D
QcosAH3
DQcos
DAH6
DAH3
D
QcosQAH3
D
Qcos
DQAH3
D
QcosH
DH
D
QcosQH
2
22
2
22
2
2
222
223
323
zero
AH3B
D
Qcos
AH3AH3
D
Qcos
DAH3AH6
DAH3AH3
D
Qcos
AH3
QAH3D
Qcos
DAH3
QAH3D
Qcos
AH3H
DAH3H
DQcos
AH3
QH
22
2
2
2
2
2
222
2
2
2
2
22
2
22
2
3
2
3
2
23
zero
AH3B
D
QcosD
QcosD2
D1
D
QcosQ
D
Qcos
DQ
D
Qcos
A3H
DA3H
D
Qcos
A3HQ
22
2
22
2
22
22
zero
AH3B
D1
DA3H
DQcos
D2
DQcos
DQ
DQcos
A3H
DQcos
A3HQ
D
QcosQ
D
Qcos
222
22
2
22
2
2
zero
AH3B
D1
DA3H
D
Qcos
D2
DQ
A3H
A3HQ
D
QcosQ1
222
22
2
22
1cmcm
cmE
H2
o
2o
2o
R
zero
)1(A3
B
D
1DA3
)1(
D
Qcos
D2
DQ
A3
)1(
A3
Q)1(
D
QcosQ1
222
22
2
22
zero
A3B
D
1DA3
1D
Qcos
D2
DQ
A31
A3Q
D
QcosQ1
22
22
2
22
zero
DA3DB
D
1DA3
1D
Qcos
D2
DQ
A31
A3Q
D
QcosQ1
22
22
2
22
1DGM
DGM
'L
DGMDB
o
o2
o
147/172
zero
DA31
D
1DA3
1D
Qcos
D2
DQ
A31
A3Q
D
QcosQ1
22
22
2
22
zero
D
1D
Qcos
D2
DQ
A31
A3Q
D
QcosQ1
22
22
2
22
27.3
This equation must have solution for the same two contour conditions of 21.65. Solution of 27.3 for the first boundary condition which is when the radius is infinite r = β, and the gravitational force is zero which means that the particle is at rest and we have π£β² = π§πππ and π = π§πππ.
Applying 1Q2 in 27.3 we get:
(1 β 1 )(β )
+ β β +(β )
+ = π§πππ.
(β )
+ = π§πππ π =(β )
27.4
Equation 27.4 is exactly equal to the result of equation 27.2 when the radius is infinite π = β, w = zero and Q = 1, as shown in 27.5:
π€ = = [1 + ππππ (β π)] = [1 + ππππ (β 1)] =(β )
+ = π§πππ 27.5
Therefore in 27.4 we have an exact result that describes how in infinity the eccentricity Ξ΅ is related to the angle β of the infinite radius of the particle, being π β₯ 1 which means that the motion from infinity will be or parabolic with π = 1 or hyperbolic with π > 1. Note that by definition π > π§πππ. Solution of 27.3 for the second boundary condition which is when the radius is greater than zero and less than infinity π§πππ < π < β which means that the particle is in motion due to the influence of a gravitational force with π£β² β π§πππ and π β π§πππ.
Applying π = in 27.3 we have:
zero
D
1D
Qcos
D2
DQ
A31
A3Q
D
QcosQ1
22
22
2
22
27.3
zeroD1
DQcos
D2
DA61DA121
D1
A31
DA61DA121
A31
DQcos
DA61DA121
1222
2
zeroDA61
D1
DQcos
DA61
D2
DA121
D1
DA61
A31
DA121
A31
DQcos
DA121
DA61
222
2
zeroDA6
D1
D1
DQcos
DA6
D2
D2
DA12
D1
D1
DA6
A31
A31
DA12
A31
A31
DQcos
DA121
DA61
22222
2
zeroDA6
D1
D1
DQcos
D1
DQcos
DA6
22222
2
148/172
DA62
DA6
D1
D1
DA64
D1
D1
DQcos 2222
2
DA12
DA6
D1
D1
DA24
D1
D1
DQcos 222222
DA12
DA6
D1
DA24
D1
DA24
D1
D1
DQcos 222222
DA12
DA6
DA24
DA241
D1
D1
DQcos
DA12
DA144
DA1221
D1
D1
DQcos 22
2
DA12
DA121
D1
D1
DQcos
2
DA12
DA121
D1
D1
DQcos
DA12
DA12
D1
D1
D1
DQcos
DA12
DA12
D1
D1
D1
DQcos
DA12
DA12
D1
D1
D1
DQcos
DA12DA12
D1
DQcos
D1
DQcos
149/172
β(β )
+ = π§πππ π =(β )
27.6
In the theory of conic for hyperbole we have π = , equating to 27.6 we have π =
=
(β ) This results
π = π. πππ (β π)) which is the correct formula, of the greater half axis of hyperbola. Therefore in 27.6 we have an exact result that describes how in the course of π§πππ < π < β the eccentricity Ξ΅ is related to the angle β of the particle, being π β₯ 1 which means that the motion will be or parabolic with π = 1 or hyperbolic with π > 1. Note that by definition π > π§πππ
Β§28 Simplified Periellium Advance
Perihelion Retrogression π > π Imagine that the sun and Mercury are two particles, with the Sun being at the origin of a coordinate system and Mercury lying at a point A on the xy plane. The vector radius r = rr connecting the origin to point A will describe Mercury's motion in the xy plane. In the description of the movement of the planet Mercury to the observer O' corresponds to the variables with line for the observer O as without line being used a single radius r = rr and a single coordinate system for both observers. Time t' is a function of time t that is tβ² = tβ²(t) and time t is a function of time t' that is t = t (t ').
dt = dtβ² 1 + dtβ² = dt 1 β 21.02
1 β 1 + = 1 21.03
vβ² = v = 21.04
dt > πtβ² vβ² > v vdt = vβ²πtβ² 21.05 r = rr πr = drr + rdr οΏ½οΏ½. πr = drr. r + rr. dr = ππ 28.01 The radius can be considered a function of time tβ² = tβ²(t) ie r = π(tβ²) = π[tβ²(t)] or it can be considered a function of time t = t(tβ²) ie r = π(t) = π[t(tβ²)]. r = r(tβ²) = r[tβ²(t)] r = r(t) = r[t(tβ²)] 28.02
vβ² =
= r + rβ
β v =
= r + rβ
β
v =
=
=
=
v =
=
=
=
vβ² =
v =
28.03
οΏ½οΏ½ =
=
=( )
= β rβ
r + 2β
+ rβ
β 28.04
οΏ½οΏ½β² =
=
=( )
= β rβ
r + 2β
+ rβ
β 28.05
Both speeds and accelerations are positive.
οΏ½οΏ½ =
=
150/172
οΏ½οΏ½ =
= F = 1 β
+ v
28.06
πΈ = β« οΏ½οΏ½β². ππ = β« οΏ½οΏ½. ππ = β« β οΏ½οΏ½. ππ 28.07
πΈ = β«
. ππ = β« 1 β
+ v
. ππ = β« β οΏ½οΏ½. ππ
πΈ = β« = β« = β« β ππ ππΈ = οΏ½οΏ½. ππ = = = β ππ 28.08
πΈ = π c 1 + = = + ππππ π‘πππ‘π πΈ = π c 1 + β = π c 28.09
πΈ = β = π c = 1 + = 1 + π΄ 28.10
In this first variant relativistic kinetic energy is greater than inertial energy > π c .This causes
Mercury's perihelion to recede. The planet seems heavier due to the movement.
= οΏ½οΏ½.
= = = β ππ = β οΏ½οΏ½.
= οΏ½οΏ½. v =
=
= β ππ = β οΏ½οΏ½. v
= οΏ½οΏ½. v = .
=.
= β οΏ½οΏ½. v οΏ½οΏ½ =
= β οΏ½οΏ½ 28.11
οΏ½οΏ½ = β rβ
r + 2β
+ rβ
β = β οΏ½οΏ½
οΏ½οΏ½β = 2β
+ rβ
β = π§πππ = πβ
= 2πβ
+ rβ
= π§πππ
οΏ½οΏ½ = β rβ
οΏ½οΏ½ = β οΏ½οΏ½
β rβ
= β
β
= = βLβ =
β
β β r = β
β + r =
β + = π΄ = π΅ =
151/172
= 1 + π΄ = 1 + 3π΄ + 3π΄ + π΄ β 1 + 3π΄ 3π΄ + π΄ β π§πππ
1 + 3π΄β
+ = π΅ 28.12
1 + 3π΄β
+ 1 + 3π΄ β π΅ = π§πππ
β + 3π΄
β + + 3π΄ β π΅ = π§πππ
β + 3π΄
β π€ + w + 3π΄π€ β π΅ = π§πππ
β + w + 3π΄
β π€ + 3π΄π€ β π΅ = π§πππ 28.13
π€ = = [1 + ππππ (β π)] β
=(β )
β
=(β )
(β )
+ [1 + ππππ (β π)] + 3π΄(β )
[1 + ππππ (β π)] + 3π΄ [1 + ππππ (β π)] β π΅ = π§πππ
βπ(β )
+ +(β )
β 3π΄π(β )
β 3π΄π(β )
+ +(β )
+ 3π΄(β )
β π΅ = π§πππ
(3π΄ β 3π΄π )(β )
+ 1 β π β 3π΄π +(β )
+ + β π΅ = π§πππ
β(β )
+ β β +(β )
+ + β = π§πππ
ππ·π΅ = = = 1
(1 β π )(β )
+ β β +(β )
+ + β = π§πππ
(1 β π )(β )
+ β β +(β )
+ = π§πππ π = 28.14
1 β(β )
+ β β +(β )
+ = π§πππ
1 +6A
ππ·β 1 β
12A
ππ·
πππ (β π)
π·+
1
3π΄1 +
6A
ππ·β
1
3π΄1 +
12A
ππ·β
1
ππ·1 +
12A
ππ·+
2
ππ·1 +
6A
ππ·
πππ (β π)
π·
+1
π π·1 +
6A
ππ·= π§πππ
β(β )
+ + β β β β + +(β )
+ + = π§πππ
β(β )
+ β(β )
+ + = π§πππ
6A(β )
+(β )
β β = π§πππ
(β )=
Β± .
.
(β )=
Β±
.
(β )=
Β±
152/172
(β )=
Β±
(β )
= =
(β )
= π β(β )
= π§πππ 28.15
For hyperbole eccentricity (π) is defined as π =(β )
where (β ) is the angle of the asymptote.
Advance of the Periellium π < π
dt = dtβ² 1 + dtβ² = dt 1 β dt > ππ‘β²
1 β 1 + = 1
v = vβ² = vβ² > v
r = rr πr = drr + rdr οΏ½οΏ½. πr = drr. r + rr. dr = ππ
v =
= r + rβ
β vβ² =
= r + rβ
β
v =
vβ² =
οΏ½οΏ½ =
=
=( )
= β rβ
r + 2β
+ rβ
β
οΏ½οΏ½β² =
=
=( )
= β rβ
r + 2β
+ rβ
β
οΏ½οΏ½ =
=
οΏ½οΏ½ =
= οΏ½οΏ½β² = 1 +
β vβ²
28.16
πΈ = β« οΏ½οΏ½. ππ = β« οΏ½οΏ½β². ππ = β« β οΏ½οΏ½. ππ
πΈ = β«
. ππ = β« 1 +
β vβ²
. ππ = β« β οΏ½οΏ½. ππ
πΈ = β« = β« = β« β ππ ππΈ = οΏ½οΏ½β². ππ = = = β ππ 28.17
πΈ = βπ c 1 β = β = + ππππ π‘πππ‘π πΈ = βπ c 1 β β = βπ c 28.18
πΈ = β β = βπ c = 1 β = 1 β π΄ 28.19
In this second variant relativistic kinetic energy is smaller than inertial energy < π c .This causes the
advance of Mercury's perihelion. The planet really is lighter due to movement.
153/172
= οΏ½οΏ½β².
= = = β = β οΏ½οΏ½.
= οΏ½οΏ½β². vβ² =
=
= β οΏ½οΏ½. vβ²
= οΏ½οΏ½β². vβ² =.
= .
= β οΏ½οΏ½. vβ² οΏ½οΏ½β² =
= β οΏ½οΏ½ 28.20
οΏ½οΏ½β² = β rβ
r + 2β
+ rβ
β = β οΏ½οΏ½
οΏ½οΏ½β²β = 2β
+ rβ
β = π§πππ = πβ
= 2πβ
+ rβ
= π§πππ
οΏ½οΏ½β² = β rβ
οΏ½οΏ½ = β οΏ½οΏ½
β rβ
= β
β
= = βLβ²β =
β
β β r = β
β + r =
β + = π΄ = π΅ =
= 1 β π΄ = 1 β 3π΄ + 3π΄ β π΄ β 1 β 3π΄ 3π΄ β π΄ β π§πππ
1 β 3π΄β
+ = π΅ 28.21
1 β 3π΄β
+ 1 β 3π΄ β π΅ = π§πππ
β β 3π΄
β + β 3π΄ β π΅ = π§πππ
β β 3π΄
β π€ + w β 3π΄π€ β π΅ = π§πππ
β + w β 3π΄
β π€ β 3π΄π€ β π΅ = π§πππ 28.22
π€ = = [1 + ππππ (β π)] β
=(β )
β
=(β )
βπ πππ (β π)
π·+
1
ππ·[1 + ππππ (β π)] β 3π΄
βπ πππ (β π)
π·
1
ππ·[1 + ππππ (β π)] β 3π΄
1
ππ·[1 + ππππ (β π)] β π΅ = π§πππ
154/172
βπ(β )
+ +(β )
+ 3π΄π(β )
+ 3π΄π(β )
β +(β )
+ 3π΄(β )
β π΅ = π§πππ
βπ(β )
+ +(β )
+ 3π΄π(β )
+ 3π΄π(β )
β β(β )
β 3π΄(β )
β π΅ = π§πππ
(3π΄π β 3π΄)(β )
+ 1 β π + 3π΄π β(β )
+ β β π΅ = π§πππ
β(β )
+ β + β(β )
+ β β = π§πππ
(π β 1)(β )
+ β + π β(β )
+ β β = π§πππ
ππ·π΅ = = = 1
(1 β π )(β )
+ β + β π +(β )
β + + = π§πππ
(1 β π )(β )
+ β + β π +(β )
+ = π§πππ π = 28.23
1 β(β )
+ β + β +(β )
+ = π§πππ
1 β β 1 +(β )
+ β 1 β + 1 β β 1 β + 1 β(β )
+ 1 β =
π§πππ
(β )+ β + + β β + + β
(β )+ β = π§πππ
(β )
+ β(β )
+ β = π§πππ
6A(β )
β(β )
+ β = π§πππ
(β )=
Β± .
.
(β )=
Β±
.
(β )=
Β±
(β )
=Β±
(β )
= =
(β )
= π β(β )
= π§πππ 28.24
For hyperbole eccentricity (π) is defined as π =(β )
where (β ) is the angle of the asymptote.
155/172
The movements of the ellipses will focus F '(left) on the origin of the frame.
All ellipses are described by the equation π = π(π‘) =( )
=( )
=.
. ( ) In these the angle
vector radius (tQ), indicates the position of the planet Mercury in all ellipses, the movement of Mercury in the ellipses is counterclockwise, with the value of Q being the cause of perihelion advancement or retraction. The first ellipse in blue represents retrogression of the perihelion, where we have Q = 1.1. The second red ellipse represents the advancement of the perihelion, in this we have Q = 0.9. In this ellipse the perihelion and aphelion advance in the trigonometric sense, that is, counterclockwise which is the same direction as the planet's movement in the ellipse. The fifth ellipse in green represents a stationary ellipse Q = 1.
156/172
Β§ 29 Yukawa Potential Energy
Newton's gravitational potential energy E
F = β r F = F = F. F = β r β r = rr = =
F = βFr E = β F = = zerok
F = β r = β r
Yukawa potential energy E
E = βk = βkr e zerok π β₯ π§πππ 29.01
Potential Core EnergyπΈ
Breaking apart E = βk we get:
E = βk = β = πΈ πΆ πΆ = zerok π β₯ π§πππ
a = zero β C = 1 β E = πΈ r = β β E = zero dE
dr=
d
dr(βkr e ) = βk (β1)r
dr
dre + (r )e βa
dr
dr
dE
dr=
d
dr(βkr e ) = βk(βr e β ar e ) = k
e
r+ ak
e
r
dE
dr=
d
drβk
e
π= π
e
r+ ak
e
r
πΈ = dE = d βke
π= π
e
r+ ak
e
rππ = β k
e
π+ ππππ π‘πππ‘π
F = β r = βk + a r Attractive force
F = ma =
= 21.51
Fβ² = maβ² =
= 21.15
First variant.
F =
= βk + a r 21.51
E = β« F. dr = β«. dr = β« βk + a r . dr
E = β« F. dr = β« dv .
= β« βk + a dr
E = β« F. dr = β« dv . v = β β« π + ak dr
157/172
E = β« F. dr = β« = β β« π + ak dr dE = F. dr = = βk + a dr
E = βm c 1 βv
c= β βπ
e
π+ ππππ π‘πππ‘π
E = βm c 1 β = k + ππππ π‘πππ‘π E = βm c 1 β = k β m c
E = βm c 1 β β k = βm c E = βm c 1 β( )
β k = βm c
1 β = β π΄ =
1 β = 1 β A
dE = F. dr =m vdv
1 βvc
= βke
r+ a
e
rdr
dE
ππ‘= F.
dr
ππ‘=
m vdvππ‘
1 βvc
= βke
r+ a
e
rοΏ½οΏ½
dr
ππ‘
dE
ππ‘= F. v =
m vdvππ‘
1 βvc
= βke
r+ a
e
rοΏ½οΏ½v
F =m a
1 βvc
= βke
r+ a
e
rοΏ½οΏ½
οΏ½οΏ½ = β rβ
r + 2β
+ rβ
β = βk + a οΏ½οΏ½
οΏ½οΏ½β = 2β
+ rβ
β = π§πππ = πβ
= 2πβ
+ rβ
= π§πππ
F = β πβ
οΏ½οΏ½ = βk + a οΏ½οΏ½
β πβ
= β + a 1 β
β πβ
= β + a 1 β
β πβ
= β + a 1 β A
β
= = βLβ
=β
β β π = β + a 1 β A
β + = π + a 1 β A π΅ =
158/172
β + = Bπ + a 1 β A π΅ =
β + = Be 1 β A + aBre 1 β A
β + = Be β ABe + aBre β aABre
β + = Be β AB + aBre β aABe w = π = w
β + w = Be β ABwe + aBw e β aABe
β + w = Be + aBw e β ABwe β aABe
β + w = (1 + aw )Be β (w + a)ABe
β + w = (1 + aw ) β (w + a)Ae Be
r =(β )
w = = πcos(β ) β
= βπsen(β ) β
= βπcos(β )
βπcos(β ) + πcos(β ) = (1 + aw ) β (w + a)Ae Be zero = (1 + aw ) β (w + a)Ae Be
(1 + aw ) β (w + a)Ae = π§πππ w = = πcos(β ) π = w =(β )
1 +(β )
β [πcos(β ) + a]Ae = π§πππ
πcos(β ) 1 +(β )
β πcos(β )[πcos(β ) + a]Ae = π§πππ
πcos(β ) + a β π cos (β )Ae β πcos(β )aAe = π§πππ βπcos(β ) β a + π cos (β )Ae + πcos(β )aAe = π§πππ π cos (β )Ae β πcos(β ) + πcos(β )aAe β π = π§πππ π cos (β )Ae β πcos(β ) 1 β aAe β π = π§πππ π cos (β )Ae β πcos(β ) 1 β aAe β π = π§πππ
πcos(β ) =1 β aAe Β± 1 β aAe β 4Ae (βπ)
2Ae
πcos(β ) =1 β aAe Β± 1 β 2aAe + a A e + 4πAe
2Ae
πcos(β ) =1 β aAe Β± 1 + 2aAe + a A e
2Ae
πcos(β ) =1 β aAe Β± 1 + aAe
2Ae
πcos(β ) =1 β aAe Β± 1 + aAe
2Ae
159/172
πcos(β ) =1 β aAe β 1 β aAe
2Ae
πcos(β ) =βaAe β aAe
2Ae
w = = πcos(β ) = = = βπ π = E = constante
dE
dr=
d
drβk
e
π= π
e
r+ ak
e
r= π§πππ
π + ak = π§πππ = βa π = E = constante
E = βk = βkr e = βk(βπ)e = ake
πcos(β ) =1 β aAe Β± 1 + aAe
2Ae
πcos(β ) =1 β aAe + 1 + aAe
2Ae
w =1
π= πcos(β ) =
2
2Ae=
1
Ae
Second variant.
Fβ² =
= βk + a r 21.15
E = β« Fβ². dr = β«
. dr = β« βk + a r . dr
E = β« Fβ². dr = β« dvβ²
= β« βk + a dr
E = β« Fβ². dr = β« dvβ² . vβ² = β β« π + ak dr
E = β« Fβ². dr = β«1+
vβ²2
c2
= β β« π + ak dr dE = Fβ². dr =1+
vβ²2
c2
= βk + a dr
E = m c 1 +vβ²
c= β βπ
e
π+ ππππ π‘πππ‘π
E = m c 1 + = k + ππππ π‘πππ‘π E = m c 1 + = k + m c
E = m c 1 + β k = m c E = m c 1 +( )
β k = m c
1 + = + π΄ =
1 + = 1 + A
160/172
dE = Fβ². dr =m vβ²dvβ²
1 +vβ²c
= βke
r+ a
e
rdr
dE
ππ‘β²= Fβ².
dr
ππ‘β²=
m vβ²dvβ²ππ‘β²
1 +vβ²c
= βke
r+ a
e
rοΏ½οΏ½
dr
ππ‘β²
dE
ππ‘β²= Fβ². vβ² =
m vβ²dvβ²ππ‘
1 +vc
= βke
r+ a
e
rοΏ½οΏ½vβ²
Fβ² =m aβ²
1 +vβ²c
= βke
r+ a
e
rοΏ½οΏ½
οΏ½οΏ½β² = β rβ
r + 2β
+ rβ
β = βk + a οΏ½οΏ½
οΏ½οΏ½β²β = 2β
+ rβ
β = π§πππ = πβ
= 2πβ
+ rβ
= π§πππ
Fβ² = β πβ
οΏ½οΏ½ = βk + a οΏ½οΏ½
β πβ
= β + a 1 +
β πβ
= β + a 1 +
β πβ
= β + a 1 + A
β
= = βLβ²β
=β
β β π = β + a 1 + A
β + = π + a 1 + A π΅ =
β + = Bπ + a 1 + A π΅ =
β + = Be 1 + A + aBre 1 + A
β + = Be + ABe + aBre + aABre
β + = Be + AB + aBre + aABe w = π = w
β + w = Be + ABwe + aBw e + aABe
β + w = Be + aBw e + ABwe + aABe
β + w = (1 + aw )Be + (w + a)ABe
161/172
β + w = (1 + aw ) + (w + a)Ae Be
r =(β )
w = = πcos(β ) β
= βπsen(β ) β
= βπcos(β )
βπcos(β ) + πcos(β ) = (1 + aw ) + (w + a)Ae Be zero = (1 + aw ) + (w + a)Ae Be
(1 + aw ) + (w + a)Ae = π§πππ w = = πcos(β ) π = w =(β )
1 +(β )
+ [πcos(β ) + a]Ae = π§πππ
πcos(β ) 1 +(β )
+ πcos(β )[πcos(β ) + a]Ae = π§πππ
πcos(β ) + a + π cos (β )Ae + πcos(β )aAe = π§πππ π cos (β )Ae + πcos(β ) + πcos(β )aAe + π = π§πππ π cos (β )Ae + πcos(β ) 1 + aAe + π = π§πππ
πcos(β ) =β 1 + aAe Β± 1 + aAe β 4Ae (π)
2Ae
πcos(β ) =β 1 + aAe Β± 1 + 2aAe + a A e β 4πAe
2Ae
πcos(β ) =β 1 β aAe Β± 1 β 2aAe + a A e
2Ae
πcos(β ) =β 1 β aAe Β± 1 β aAe
2Ae
πcos(β ) =β 1 β aAe Β± 1 β aAe
2Ae
π€ =1
π= πcos(β ) =
β1 + aAe + 1 β aAe
2Ae= π§πππ
πcos(β ) =β 1 β aAe Β± 1 β aAe
2Ae
πcos(β ) = w = = πcos(β ) π = w =(β )
w =1
π= πcos(β ) =
β2 + 2aAe
2Ae=
β1 + aAe
Ae= π β
1
Ae
Β§ 29 Yukawa Potential Energy βContinuationβ
Newton's gravitational potential energy E
F = β r F = F = F. F = β r β r = rr = =
F = βFr E = β F = = zerok
162/172
F = β r = β r
Yukawa potential energy E
E = βk = βkr e zerok zeroa
= (βkr e ) = βk (β1)r e + (r )e βa
dE
dr=
d
dr(βkr e ) = βk(βr e β ar e ) = k(r e + ar e ) = k
e
r+ a
e
r
dE
dr=
d
drβk
e
π= k
e
r+ a
e
r
πΈ = ππΈ = π d βe
π= π
e
r+ a
e
rππ = β k
e
π
F = β r = βk + a r Attractive force
οΏ½οΏ½ =
= οΏ½οΏ½β² = 1 +
β vβ²
28.16
οΏ½οΏ½ =
= οΏ½οΏ½ = 1 +
β v
= βk + a r
πΈ = β« οΏ½οΏ½. ππ = β« οΏ½οΏ½β². ππ = β« βk + a οΏ½οΏ½. ππ
πΈ = β«
. ππ = β« 1 +
β vβ²
. ππ = β« βk + a οΏ½οΏ½. ππ
πΈ = β« = β« = β« βk + a ππ
ππΈ = οΏ½οΏ½β². ππ = = = βk + a ππ 28.17
πΈ = βπ c 1 β = β = β βk + ππππ π‘πππ‘π 28.18
πΈ = βπ c 1 β β k = βπ c 28.18
πΈ = β β k = βπ c = 1 β = 1 β π΄ 28.19
= 1 β π΄ π΄ = 28.19
= οΏ½οΏ½ .
= = = βk + a = βk + a οΏ½οΏ½.
163/172
= οΏ½οΏ½β². vβ² =
=
= βk + a οΏ½οΏ½. vβ²
= οΏ½οΏ½β². vβ² =.
= .
= βk + a οΏ½οΏ½. vβ² 28.20
οΏ½οΏ½β² =
= βk + a οΏ½οΏ½ 28.20
οΏ½οΏ½β² = β rβ
r + 2β
+ rβ
β = βk + a οΏ½οΏ½
οΏ½οΏ½β²β = 2β
+ rβ
β = π§πππ = πβ
= 2πβ
+ rβ
= π§πππ
οΏ½οΏ½β² = β rβ
οΏ½οΏ½ = βk + a οΏ½οΏ½
β rβ
= βk + a
β
= = βLβ²β
=β
β β r = β + a
β + r = + a
β + = + a
β + = π + a π΅ =
β + = π΅π + a π΅ =
= 1 β π΄ = 1 β 3π΄ + 3π΄ β π΄ β 1 β 3π΄
3π΄ β π΄ β π§πππ π΄ =
1 β 3π΄β
+ = π΅π + a 28.21
1 β 3π΄β
+ 1 β 3π΄ = π΅π + a
β β 3π΄
β + β 3π΄ = π΅π + a
β β 3π΄
β + β 3π΄ = π΅e + πaBe
164/172
β β 3π΄
β + β 3π΄ = π΅ + aBe π€ =
β π€ β 3π΄
β e π€ + π€ β 3π΄e π€ = π΅e w + aBe
β π€ + π€ = 3π΄
β e π€ + 3π΄e π€ + π΅e w + aBe
β π€ + π€ = e 3π΄
β π€ + 3π΄π€ + π΅w + aB 28.22
π€ = = π₯π β + π¦π β β
= ππ₯π β β ππ¦π β β
= βπ₯π β β π¦π β i = ββ1
βπ₯π β β π¦π β π₯π β + π¦π β + π₯π β + π¦π β = e 3π΄π π€
πβ π€ + 3π΄π€ + π΅w + aB
βπ₯π β β π¦π β π₯π β + π¦π β = βπ₯π β π₯π β + βπ₯π β π¦π β + βπ¦π β π₯π β + βπ¦π β π¦π β
= (βπ₯2π2πβ β π₯π¦ β π¦π₯ β π¦2πβ2πβ ) = β(π₯2π2πβ + 2π₯π¦ + π¦2πβ2πβ ) = β(π₯ππβ + π¦πβπβ )2
β π₯π β + π¦π β + π₯π β + π¦π β = e 3π΄π π€
πβ π€ + 3π΄π€ + π΅w + aB
π§πππ = e 3π΄π π€
πβ π€ + 3π΄π€ + π΅w + aB
3π΄π π€
πβ π€ + 3π΄π€ + π΅w + aB = zero
3π΄ βπ₯π β β π¦π β π₯π β + π¦π β + 3π΄ π₯π β + π¦π β + π΅ π₯π β + π¦π β + aB = zero
3π΄ βπ₯π β β π¦π β π₯π β + π¦π β = β3π΄ π₯π β + π¦π β π₯π β + π¦π β = β3π΄ π₯π β + π¦π β
β3π΄ π₯π β + π¦π β + 3π΄(π₯ππβ + π¦πβπβ )3 + π΅(π₯ππβ + π¦πβπβ ) + aB = zero π΅ π₯π β + π¦π β + aB = zero π₯π β + π¦π β + a = zero
π€ =1
π= π₯π β + π¦π β = βa
165/172
Β§30 Energy In Β§28 simplified calculation of the perihelion retraction we obtain:
E = β« = β« = β« β dr dE = F. dr = = = β dr 28.08
E = m c 1 + = = + constant E = m c 1 + β = m c 28.09
E = β = m c = 1 + = 1 + A 28.10
In this first variant relativistic kinetic energy is greater than inertial energy > m c . This causes
Mercury's perihelion to recede. The planet seems heavier due to the movement.
E = β« = β« = β« β dr 28.08
E = β« = β« = β« β dr
E = m c 1 + = =
E = m c 1 + β m c 1 +( )
= β( )
= β
E = m c 1 + β m c = β m c = m c 1 + β₯ m c β₯ m c 30.1
Defining potential energy as E = β : 30.2
And applying in 1 we have E = m c 1 + β m c = β m c = βE 30.3
In 3 we have the energy conservation principle written as E + E = zero 30.4
With 3 the kinetic energy equal to E = m c 1 + β m c = β m c 30.5
n 3 the lowest energy of the system is the inertial energy of rest E = m c 30.6
In 3 the highest energy of the system is E = m c 1 + = 30.7
Now defining p = = m vβ² vp = = m vvβ² T = m c 1 + T = βm c 1 β 30.8
And knowing that E = c m c + p = = + m c 1 β = vp β T 30.9
If in 9 m = zero So E = cp. 30.10
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Applying 8 and 9 in 7 we obtain the greatest energy of the written system as:
E = c m c + p = T = vp β T This we have vp = Tβ² + T 30.7b With 8 and 9 we can write 3 in the form:
E = m c 1 + β m c = + m c 1 β β m c = βE 30.3b
E = T β E = vp β T β E = βE 30.3c From 3c we can define the resting inertial energy E = m e E = m in the form: E = Tβ² + E = m c and E = vp β T + E = m c . 30.11
Defining Lagrangean as L = T β E = βm c 1 β + 30.12
This Lagrangian meets
= according to Β§24.
So from 11 temos: E = Tβ² + E = m c and E = vp β L = m c 30.13 In Β§28 simplified calculation of perihelion advance we obtain:
E = β« = β« = β« β dr dE = Fβ². dr = = = β dr 28.17
E = βm c 1 β = β = + constant E = βm c 1 β β = βm c 28.18
E = β β = βm c = 1 β = 1 β A 28.19
In this second variant relativistic kinetic energy is smaller than inertial energy < m c . This causes the
advance of Mercury's perihelion. The planet really is lighter due to movement.
E = β« = β« = β« β dr 28.17
E = β« = β« = β« β dr
E = βm c 1 β = β =
E = βm c 1 β β βm c 1 β( )
= β β β( )
= β
E = βm c 1 β β (βm c ) = β β (βm c ) =
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E = βm c 1 βv
c+ m c = β
m c
1 +vc
+ m c =k
r
E = m c βm c 1 β = m c β = m c β₯ m c 1 β m c β₯ 30.14
Where applying 2 the definition of potential energy E = β :
We get E = m c βm c 1 β = m c β = βE 30.15
In 15 we have the principle of conservation of energy written as E + E = zero. Being 15 the kinetic energy equal to:
E = m c βm c 1 β = m c β 30.16
In 15 the biggest energy of the system is the inertial energy of rest E = m c 30.17
At 15 the lowest energy in the system is Eβ² = m c 1 β = 30.18
Now defining p = = m v v p = = m vβ²v 30.19
And knowing that E = c m c β p = = β + m c 1 + = βv p + Tβ² 30.20
If in 20 m = zero so E = icpβ² 30.21 Proving 20:
E = = c m c β p = c
β·
m c β
β
ββ
β
ββ
= c m c β = m c β
E = = c m c β p = c 1 + β vβ² =
E = m c 1 βv
c= c m c β p = c m c β (m v) = c m c β m v = m c c β v = m c 1 β
v
c
Applying 8, 19 and 20 to 18 results in the lowest energy of the written system as:
E = c m c β p = βT = βv p + Tβ² This we have v p = Tβ² + T 30.18b With 19 and 20 we can write 15 in the form:
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E = m c βm c 1 β = m c + β m c 1 + = βE 30.15b
E = E + T = E + vβ²pβ² β Tβ² = βE 30.15c From 15c we can define the resting inertial energy E = m e E = m in the form: E = βT β E = m c and E = βv p + T β E = m c . 30.22
Defining Lagrangean as L = T β E = m c 1 + + 30.23
This Lagrangian meets
= prova no final.
E = βT β E = m c E = βv p + L = m c 30.24 Rewriting 11 e 22: E = Tβ² + E = m c and E = vp β T + E = m c . 30.11 E = βT β E = m c and E = βv p + T β E = m c . 30.22 Equating E of 11 with E of 22 we have: E = vp β T + E = βT β E = m c
This we get vp = β2E βE = = 30.25
Matching βE = the kinetic energy of 3b we have:
E = m c 1 + β m c = + m c 1 β β m c = βE = = 30.3d
In 3d we should have:
+ m c 1 β β m c =
m v + m c 1 β β m c 1 β = m v
m v + m c β m c β m c 1 β = zero
m v + m c β m v β m c 1 β = zero
m c β m v β m c 1 β = zero
m c 1 β β m c 1 β = zero
The approximation 1 β β 1 β is the cause of Mercury's perihelion setback.
m c 1 β β m c 1 β = zero Result that proves that E = .
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Equating E of 11 with E of 22 we have: E = T + E = βv p + Tβ² β E = m c
This we get v p = β2E βE = = 30.26
MatchingβE = the kinetic energy of 15b we have:
E = m c βm c 1 β = m c + β m c 1 + = βE = = 30.15d
In 15d we should have:
m c + β m c 1 + =
m c + β m c 1 + = zero
m c 1 + + m v β m c 1 + = zero
m c 1 + + m v β m c β m v = zero
m c 1 + β m c β m v = zero
m c 1 + β m c 1 + = zero
The approximation 1 + β 1 + is the cause of the advance of Mercury's perihelion.
m c 1 + β m c 1 + = zero Result that proves that E = .
From 25 and 26 results vp = vβ²pβ² 30.27 Applying 25 to E_0 of 22:
E = βT β E = βT + = m c 1 β + = 1 β + 30.28
E = βT β E = βT + = 1 β β 1 β = m c 30.28
Applying 25 to E of 11:
E = vp β T + E = β T β = βT + Result already obtained on 28.
Applying 26 in E of 11
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E = T + E = T β = m c 1 + β = 1 + β 30.29
E = T + E = T β = 1 + β 1 + = m c 30.29
The approximation that exists in 28 and 29 is the cause of the advance and setback of Mercury's perihelion. Applying 26 to E of 22
E = βv p + T β E = β + T + = T β Result already obtained on 29.
Proof that
= Lβ² = Tβ² β E = m c 1 + +
Fβ² =
= Fβ² =
m c 1 + =
v = = xβ² + y + z ds = |ds| = dx + dy + dz r = x + y + z
Fβ² = = k (r ) = k(β1)r = βk = βk =
Fβ² = Fβ² Δ± + Fβ² Θ· + Fβ² k = βk Δ± β k Θ· β k k = β xΔ± + yΘ· + zk = β r = β r = 19.01
pβ² =
m c 1 + = m c 1 + 2
=
=
xβ² + y + z = (xβ² + y + z ) 2xβ²
=
=
pβ² =
m c 1 + =
=
=
Fβ² = =
=
1 + β xβ² 1 + = xβ² 1 + β xβ² 1 +
1 + = 1 + 2 = =
Fβ² = =
= xβ² 1 + β xβ²
Fβ² = =
= xβ² 1 + β xβ²
Fβ² = Fβ² Δ± + Fβ² Θ· + Fβ² k
Fβ² = xβ² 1 + β xβ² Δ± + yβ² 1 + β yβ² Θ· + zβ² 1 + β zβ² k
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Fβ² = xβ² 1 + Δ± β xβ² Δ± + yβ² 1 + Θ· β yβ² Θ· + zβ² 1 + k β zβ² k
Fβ² = 1 + xβ²Δ± + yβ²Θ· + zβ²k Δ± β xβ² + yβ²Θ· + zβ²k
Fβ² = 1 +
β vβ² = 28.16
Β§30 Energy Continuation
With 3d, 6, and 9 we get:
E = E β E = β m c = + m c 1 β β m c = = vp 30.30
This we get: = 30.31
That applied in 8 results: p = β E = p βm v 30.32
If at 32 m = zero then: E = = vp β v = c 30.33
For a particle with velocity c = λγ and zero resting mass m = zero we have:
E = hΞ³ p = 30.34
Applying c = λγ and 34 in 33 results:
E = = = β E = 30.35
From 34 and 35 we have: E = 2E 30.36 Applying 36 in 30 we have:
E = E β E β E = 2E β E β E = E = 30.37
Applying E = m c in 37 we obtain:
E = E = = m c β m = = 30.38
With 33 and 38 we get: E = = m c β π = 2m π 30.39
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"Although nobody can return behind and perform a new beginning, any one can begin now and create a new end"
(Chico Xavier)
Bibliography [1] Alonso & Finn, Vol. I e II, Ed. Edgard BlΓΌcher LTDA, 1972. [2] Robert Resnik, IntroduΓ§ao Γ Relatividade Especial, Ed. Univ. de S. Paulo e Ed. PolΓgono S.A., 1971. [3] E. Terradas Y R. Ortiz, Relatividad, Cia. Ed. Espasa-Calpe Argentina S. A., 1952. [4] Abraham Pais, βSutil Γ© o Senhor...β A ciΓͺncia e a vida de Albert Einstein, Ed. Nova Fronteira, 1995. [5] Marcel Rouault, FΓsica AtΓ΄mica, Ao Livro TΓ©cnico LTDA, 1959. [6] H. A. Lorentz, A. Einstein e H. Minkowisk, O PrincΓpio da Relatividade, edição da Fundação Calouste Gulbenkian, 1989. [7] L. Landau e E. Lifchitz, Teoria do Campo, Hemus β Livraria Editora LTDA. [8] Robert Martin Eisberg, Fundamentos da FΓsica Moderna, Ed. Guanabara Dois S.A., 1979. [9] Max Born, FΓsica AtΓ΄mica, edição da Fundação Calouste Gulbenkian, 1986. http://www.wbabin.net/physics/faraj7.htm