quantum gravity - maranhao state university
TRANSCRIPT
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Mathematical Foundations of the
Relativistic Theory of Quantum Gravity
Fran De AquinoMaranhao State University, Physics Department,S.Luis/MA, Brazil.Copyright 2008-2011 by Fran De Aquino. All Rights Reserved
Abstract: Starting from the action function, we have derived a theoretical background that leads tothe quantization of gravity and the deduction of a correlation between the gravitational and the inertialmasses, which depends on the kineticmomentum of the particle. We show that the strong equivalenceprinciple is reaffirmed and, consequently, Einstein's equations are preserved. In fact, such equationsare deduced here directly from this new approach to Gravitation. Moreover, we have obtained ageneralized equation for the inertial forces, which incorporates the Mach's principle into Gravitation.Also, we have deduced the equation of Entropy; the Hamiltonian for a particle in an electromagneticfield and the reciprocal fine structure constant directly from this new approach. It was also possible todeduce the expression of the Casimir force and to explain the Inflation Period and the Missing Matter,
without assuming existence ofvacuum fluctuations. Thisnew approach to Gravitation will allow us tounderstand some crucial matters in Cosmology.
Key words: Quantum Gravity, Quantum Cosmology, Unified Field.PACs: 04.60.-m; 98.80.Qc; 04.50. +h
Contents
1. Introduction 3
2. Theory 3
Generalization of Relativistic Time 4
Quantization of Space, Mass and Gravity 6
Quantization of Velocity 7
Quantization of Time 7
Correlation Between Gravitational and Inertial Masses 8
Generalization of Lorentz's Force 12
Gravity Control by means of theAngular Velocity 13
Gravitoelectromagnetic fields and gravitational shielding effect 14
Gravitational Effects produced by ELF radiation upon electric current 26
Magnetic Fields affect gravitational mass and the momentum 27
Gravitational Motor 28
Gravitational mass and Earthquakes 28
The Strong Equivalence Principle 30
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Incorporation of the Mach's Principle into Gravitation Theory 30
Deduction of the Equations of General Relativity 30
Gravitons: Gravitational Forces are also Gauge forces 31
Deduction of Entropy Equation starting from the Gravity Theory 31
Unification of the Electromagnetic and Gravitational Fields 32
Elementary Quantum of Matter and Continuous Universal Fluid 34
The Casimir Force is a gravitational effect related to the Uncertainty Principle 35
The Shape of the Universe and Maximum speed of Tachyons 36
The expanding Universe is accelerating and not slowing down 38
Gravitational and Inertial Masses of the Photon 39
What causes the fundamental particles to have masses? 40
Electrons Imaginary Masses 41
Transitions to the Imaginary space-time 44
Explanation for red-shift anomalies 50
Superparticles (hypermassive Higgs bosons) and Big-Bang 51
Deduction of Reciprocal Fine Structure Constant and the Uncertainty Principle 53
Dark Matter, Dark Energy and Inflation Period 53
The Origin of the Universe 59
Solution for the Black Hole Information Paradox 61
A Creators need 63
The Origin of Gravity and Genesis of the Gravitational Energy 64
Explanation for the anomalous acceleration of Pioneer 10 66
New type of interaction 68
Appendix A 71
Allais effect explained 71
Appendix B 74
References 75
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1. INTRODUCTION
Quantum Gravity was originallystudied, by Dirac and others, as the
problem of quantizing GeneralRelativity. This approach presentsmany difficulties, detailed by Isham[1]. In the 1970's, physicists tried aneven more conventional approach:simplifying Einstein's equations byassuming that they are almost linear,and then applying the standardmethods of quantum field theory tothe thus oversimplified equations. Butthis method, too, failed. In the 1980's
a very different approach, known asstring theory, became popular. Thusfar, there are many enthusiasts ofstring theory. But the mathematicaldifficulties in string theory areformidable, and it is far from clear thatthey will be resolved any time soon.At the end of 1997, Isham [2] pointedout several "Structural ProblemsFacing Quantum Gravity Theory". Atthe beginning of this new century,the problem of quantizing thegravitational field was still open.
In this work, we propose a newapproach to Quantum Gravity.Starting from the generalization of theaction function we have derived atheoretical background that leads tothe quantization of gravity. Einstein'sGeneral Relativity equations arededuced directly from this theory of
Quantum Gravity. Also, this theoryleads to a complete description of theElectromagnetic Field, providing aconsistent unification of gravity withelectromagnetism.
2. THEORY
We start with the action for afree-particle that, as we know, isgiven by
=b
adsS
where is a quantity which
characterizes the particle.In Relativistic Mechanics, theaction can be written in the followingform [3]:
dtcVcLdtSt
t
t
t ==2
1
2
1
221
where221 cVcL =
is the Lagrange's function.In Classical Mechanics, the
Lagrange's function for a free-particleis, as we know, given by:where V is the speed of the particleand is a quantity hypothetically [
2aVL =
a 4]given by:
2ma =
where is the mass of the particle.However, there is no distinction aboutthe kind of mass (if gravitationalmass, , or inertial mass ) neither
about its sign
m
gm im
( )
.The correlation between anda can be established based on thefact that, on the limit , therelativistic expression for
cL must be
reduced to the classic expression.The result [2aVL = 5] is: cVL 22= .
Therefore, if mcac =2= , we obtain. Now, we must decide if2aVL =
gmm = or imm = . We will see in this
work that the definition of includes. Thus, the right option is , i.e.,
gm
im gm
.ma g 2=
Consequently, cmg= and the
generalized expression for the actionof a free-particle will have thefollowing form:
( )1=b
ag dscmS
or
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( )21 2222
1
dtcVcmSt
tg =
where the Lagrange's function is
( )31 222 .cVcmL g =
The integral dtcVcmSt
t g
222
12
1 = ,preceded by the plus sign, cannothave a minimum. Thus, the integrandof Eq.(2) must be always positive.Therefore, if , then necessarily
; if , then . The
possibility of is based on the
well-known equation
0>gm
0>t 0
gm 0gm ( ) 0gm ( )
sign if 0
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energy, where is the kineticinertial energy. From Eqs. (7) and (9)we thus obtain
KiE
( )10
1
2
22
20 .cM
cV
cmE i
i
i =
=
For small velocities , weobtain
( cV
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( )162 000 iigig EEEEE =+=+
However Kii EEE += 0 .Thus, (16) becomes
( )170 .EEE Kiig =
Note the symmetry in the equations ofand .Substitution ofiE gE Kiii EEE =0
into (17) yields( )182 Kigi EEE =
Squaring the Eqs.(4) and (7) andcomparing the result, we find thefollowing correlation betweengravitational energy and momentum :
( )192222
2
.cmpc
Eg
g+=
The energy expressed as a function ofthe momentum is, as we know, calledHamiltonian or Hamilton's function:
( )20222 .cmpcH gg +=Let us now consider the problem
of quantization of gravity. Clearly there issomething unsatisfactory about thewhole notion of quantization. It isimportant to bear in mind that thequantization process is a series of rules-of-thumb rather than a well-defined
algorithm, and contains manyambiguities. In fact, for electromagnetismwe find that there are (at least) twodifferent approaches to quantization andthat while they appear to give the sametheory they may lead us to very differentquantum theories of gravity. Here we willfollow a new theoretical strategy: It isknown that starting from the Schrdingerequation we may obtain the well-knownexpression for the energy of a particle inperiodic motion inside a cubical box of
edge length L [ 7 ]. The result now is
( )21,...3,2,18 2
22
== nLm
hnE
g
n
Note that the term 22 8 Lmh g (energy)
will be minimum for where
is the maximum edge length of a cubicalbox whose maximum diameter
maxLL = maxL
( )223maxmax Ld =is equal to the maximum length scale ofthe Universe.
The minimum energy of a particleis obviously its inertial energy at rest
. Therefore we can write22 cmcm ig =
2
2
22
8
cm
Lm
hng
maxg
=
Then from the equation above it followsthat
( )238max
gcL
nhm =
whence we see that there is a minimumvalue for given bygm
( ) ( )248max
mingcL
hm =
The relativistic gravitational mass
( ) 21
221
= cVmM gg , defined in the
Eqs.(4), shows that
( ) ( ) ( )25minmin gg mM =The box normalization leads to the
conclusion that the propagation number
2== kkr
is restricted to the
values Lnk 2= . This is deducedassuming an arbitrarily large but finite
cubical box of volume [3L 8]. Thus, wehave
nL =From this equation, we conclude that
min
max
max
Ln
=
and
minminminmin nL ==
Since 1=minn . Therefore, we can write
that ( )26minmaxmax LnL =From this equation, we thus concludethat
( )27minnLL =or
( )28n
LL max=
Multiplying (27) and (28) by 3 and
reminding that 3Ld= , we obtain
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( )29n
ddorndd maxmin ==
Equations above show that the length(and therefore the space)is quantized.
By analogy to (23) we can also
conclude that
( ) ( )308min
maxmax
cL
hnMg =
since the relativistic gravitational mass,
( ) 21
221
= cVmM gg , is just a
multiple of .gm
Equation (26) tells us that
maxmaxmin nLL = . Thus, Eq.(30) can be
rewritten as follows
( ) ( )318
2
max
maxmax
cL
hnMg =
Comparison of (31) with (24) shows that
( ) ( ) ( )322
minmaxmax gg mnM =
which leads to following conclusion that
( ) ( )332
mingg mnM =
This equation shows that thegravitational mass is quantized.
Substitution of (33) into (13) leads
to quantization of gravity, i.e.,( )
( )
( )34min4
2max
min22
gn
nr
mGn
r
GMg
gg
=
=
==
From the Hubble's law, it follows that
( )2maxmaxmax dH~
lH~
V ==
( )2minminmin dH~
lH~
V ==
whence
min
max
min
max
dd
VV =
Equations (29) tell us that
maxminmax ndd = . Thus the equation
above gives
( )35max
max
minn
VV =
which leads to following conclusion
( )36n
VV max=
this equation shows that velocity is alsoquantized.
From this equation one concludesthat we can have or
maxVV =
2maxVV = , but there is nothing in
between. This shows clearly that
cannot be equal to c (speed of light invacuum). Thus, it follows that
maxV
( )
( )( )
.
..........................................
..........................
max
max
max
max
max
max
max
numberbigaisnwhere
nVVnn
TardyonsnVVnn
cnVVnn
nVVnn
TachyonsVVn
VVn
VVn
x
xx
xx
xx
xx
2211
11
33
22
1
+=+=+=+=
===
==
==
==
==
Then is the speed upper limit ofthe Tardyons and also the speed lowerlimit of the Tachyons. Obviously, this limitis always the same in all inertial frames.
Therefore can be used as a referencespeed, to which we may compare any
speed , as occurs for the relativisticfactor
c
c
V221 cV . Thus, in this factor,
does not refer to maximum propagationspeed of the interactions such as someauthors suggest; is just a speed limitwhich remains the same in any inertialframe.
c
c
The temporal coordinate ofspace-time is now (
is then obtained when ).
Substitution of
0x
tVxmax=
0 ctx =0
cVmax
( lH)~nnVVmax == into thisequation yields ( )( )lxH~nVxt max 00 1== .On the other hand, since lH
~V = and
nVV max= we can write that
nH~
Vl max1= .Thus( ) ( ) maxtH
~ntH
~lx ==0 .
Therefore, we can finally write
( )( ) ( )371 0 ntlxH~nt max==
which shows the quantization oftime.
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From Eqs. (27) and (37) we caneasily conclude that the spacetimeis notcontinuous it is quantized.
Now, let us go back to Eq. (20)which will be called the gravitationalHamiltonian to distinguish it from theinertial Hamiltonian :iH
( )382202 .cmpcH ii +=
Consequently, Eq. (18) can be rewrittenin the following form:
( )392 igi HHH =
whereiH is the variation on the
inertial Hamiltonian or inertial kineticenergy. A momentum variation p yields
a variation iH given by:
( ) ( )404202242
022
cmcpcmcppH iii +++=
By considering that the particle isinitially at rest ( . Then, Eqs. (20),
(38) and (39) give respectively: ,
and
)0=p2cmH gg =
20cmH ii =
20
2
0
11 cmcm
pH i
i
i
+=
By substituting , and into
Eq.(39), we getgH iH iH
( )41112 0
2
0
0 .ii
ig mcm
pmm
+=
This is the general expression of thecorrelation between the gravitational andinertial mass. Note that
for 250cmp i> , the value of
becomes negative.gm
Equation (41) shows that
decreases of for an increase of
. Thus, starting from (4) weobtain
gm
gm
p
( )
( )21 cV
Vmmpp
gg
=+
By considering that the particle is initiallyat rest , the equation above gives( 0=p
( )
( )21 cV
Vmmp
gg
=
From the Eq.(16) we obtain:( ) iiiiiiig EEEEEEEE =+== 0000 22
However, Eq.(14) tells us that gi EE = ;
what leads togig EEE += 0 or gig mmm += 0 .
Thus, in the expression of p we
can replacegg mm for , i.e,0im
( )20
1 cV
Vmp i
=
We can therefore write
( )
( )421
20 cV
cV
cm
p
i
=
By substitution of the expression aboveinto Eq. (41), we thus obtain:
( ) ( )43112 0220 21
iig mcVmm
=
For 0=V we obtain .Then,0ig mm =
( ) ( )minmin 0ig mm =
Substitution of into the quantized
expression of (Eq. (33)) gives(mingm )
)
gM
( )min02
ig mnM =
where is the elementary
quantum of inertial mass to bedetermined.
(min0im
For 0=V , the relativistic
expression 221 cVmM gg = becomes
00 ggg mMM == . However, Eq. (43) shows
that 00 ig mm = . Thus, the quantized
expression of reduces togM
( )min02
0 ii mnm =
In order to define the inertial quantumnumber, we will change n in theexpression above for . Thus we havein
( ) ( )44min02
0 iii mnm=
)
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which shows the quantization of inertialmass; is the inertial quantum number.in
We will change in the quantizedexpression of for in order to
define the gravitational quantum number.Thus, we have
n
gM gn
( ) ( )amnM igg 4402
min=
Finally, by substituting given
by Eq. (43) into the relativistic expressionof , we readily obtain
gm
gM
( ) ( )45112
1
21
22
22
ii
g
g
McVM
cV
mM
=
=
=
By expanding in power series andneglecting infinitesimals, we arrive at:
ig Mc
VM
2
2
1 =
Since 01 22 > cV , the equation abovecan be rewritten as follows:
( )461 2
2
ig Mc
VM
=
Thus, the well-known expression for thesimple pendulum period, ( )( )glMMT gi2= ,can be rewritten in the following form:
cVforc
V
g
lT
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2
3
222
2 22
c
uGMuGMu
d
du gg++=+
hh
E
This leads to the following expression
+=+
2
22
22
2 31
c
uGMu
d
ud g h
h
In the absence of term 2223 cuh , theintegration of the equation should beimmediate, leading to 2 period. In orderto obtain the value of the perturbation wecan use any of the well-known methods,which lead to an angle, for twosuccessive perihelions, given by
22
2262
hc
MG g+
Calculating per century, in the case ofMercury, we arrive at an angle of 43 forthe perihelion advance.
This result is the best theoreticalproof of the accuracy of Eq. (45).
Now consider a relativistic particleinside a gravitational field. The conditionfor it to escape from the gravitational fieldis that its inertialkinetic energy becomesequal to the absolute value of thegravitational energy of the field, which isgiven by
( )221 cVr
mGmrU
gg
=
By substituting and given by Eq.
(43) into this expression, and assumingthat the velocity Vof the particle thatcreates the field is small , we get
gm gm
( cV
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22
0
1 cVr
mGmE
gi
Kg
=
Substitution of the expression of KgE into
this expression gives
22
02
22 11
1
1
cVr
Gmc
cV
i
=
which simplifies to
22022 111
crc
GmcV i
+==
For this expression givescV
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( )00
pi
g
i
g
m
mBVqEq
m
m
dt
d
dt
pd rrrrr
+=
=
That is now the general expression forLorentz's force. Note that it dependson .
g
m
When the force is perpendicular tothe speed, Eq. (5) gives
( ) 221 cVdtVdmdtpd g =rr
.By comparing
with Eq.(46), we thus obtain
( ) BVqEqdtVdcVmirrrr
+= 220 1
Note that this equation is the expressionof an inertial force.
Starting from this equation, well-known experiments have been carried
out in order to verify the relativisticexpression: 221 cVmi . In general,
the momentum variation p isexpressed by tFp = where is the
applied force during a time interval
F
t .Note that there is no restrictionconcerning the nature of the force ,i.e., it can be mechanical,electromagnetic, etc.
F
For example, we can look on themomentum variation p as due toabsorption or emission ofelectromagnetic energy by the particle(by means ofradiation and/or by meansofLorentz's force upon the charge of theparticle).
In the case of radiation (anytype), p can be obtained as follows. Itis known that the radiation pressure ,
, upon an area of avolumedP dxdydA =
dxdydzd =V of a particle( the
incident radiation normal to thesurface )is equal to the energyabsorbed per unit volume
dA dU
( )VddU .i.e.,
( )47dAdz
dU
dxdydz
dU
d
dUdP ===
V
Substitution of ( is the speedof radiation) into the equation abovegives
vdtdz = v
( )( )48
v
dD
v
dAdtdU
d
dUdP ===
V
Since we can write:dFdPdA =
( )49v
dUdFdt=
However we know that dtdpdF= , then
( )50
v
dUdp =
From Eq. (48), it follows that
( )51v
dDddPddU
VV==
Substitution into (50) yields
( )522
v
dDddp
V=
or
=Dp
dDdv
dp0 020
1 VV
whence
( )532
v
Dp
V=
This expression is general for all types ofwaves including non-electromagneticwaves such as sound waves. In thiscase, in Eq.(53), will be the speed ofsound in the medium and the intensityof the sound radiation.
v
D
In the case of electromagneticwaves, the Electrodynamics tells us that
will be given byv
( )
++
===
112
2
rrr
c
dt
dzv
where is the real part of the
propagation vector
rk
kr
; ir ikkkk +==r
;
, and , are the electromagneticcharacteristics of the medium in whichthe incident (or emitted) radiation ispropagating ( 0 r= where r is the
relative dielectric permittivity and;mF/10854.8 120
= 0r= where
r is the relative magnetic permeability
and ;m/H70 104= is the
electrical conductivity). For an atominside a body, the incident (or emitted)radiation on this atom will be propagatinginside the body, and consequently,=body, =body, =body.
It is then evident that the index of
refraction vcnr = will be given by
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( ) ( )54112
2
++==
rrr
v
cn
On the other hand, from Eq. (50) followsthat
rnc
U
c
c
v
Up =
=
Substitution into Eq. (41) yields
( )551121 0
2
20
ir
i
g mncm
Um
+=
If the body is also rotating, with anangular speed around its central axis,then it acquires an additional energyequal to its rotational energy
( )2
21
IEk = . Since this is an increasein the internal energy of the body, andthis energy is basically electromagnetic,we can assume that , such as U,corresponds to an amount ofelectromagnetic energy absorbed by thebody. Thus, we can consider as anincrease in the electromagneticenergy absorbed by the body.Consequently, in this case, we mustreplace in Eq. (55) for( )
kE
kE
kEU=
U
U UU + . If
UU . In this case, if thebody is a Mumetal disk
( )17 .101.2;100000,105 == mSgaussatr with radiusR , ( )2021 RmI i= , the equationabove shows that the gravitational massof the disk is
( ) ( )diskidiskg mf
Rm 0
4413 11012.1121
+
Note that the effect of theelectromagnetic radiation applied uponthe disk is highly relevant, because in theabsence of this radiation the index ofrefraction, present in equations above,
becomes equal to 1. Under thesecircumstances, the possibility of strongly
reducing the gravitational mass of thedisk practically disappears. In addition,the equation above shows that, inpractice, the frequency of theradiation cannot be high, and that
extremely-low frequencies (ELF) aremost appropriated. Thus, if the frequencyof the electromagnetic radiation appliedupon the disk is
f
Hzf 1.0= (See Fig. I (a))and the radius of the disk is ,and its angular speed
mR 15.0=
( )rpmsrad 000,100~/1005.1 4= , theresult is
( ) ( )diskidiskg mm 06.2
This shows that the gravitational mass ofa body can also be controlled by means
of its angular velocity.In order to satisfy the conditionUU
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For >> , the equation above reducesto
fz
1=
In the case of the Mumetal subjected to
an ELF radiation with frequency, the value is .Obviously, the thickness of the Mumetaldisk must be less than this value.
Hzf 1.0= mmz 07.1=
Equation (55) is general for alltypes of electromagnetic fields includinggravitoelectromagnetic fields (See Fig. I(b)).
Acceleration
Gravitoelectric
Field
Gravitomagnetic
Field
Fig. I (a) Experimental set-up in order to measure the
gravitational mass decreasing in the rotating Mumetal
disk. A sample connected to a dynamometer can measure
the decreasing of gravity above the disk. (b)
Gravitoelectromagnetic Field.
(b)
(a)
TransmitterELF electromagnetic radiation
Mumetal disk
Motor
Balance
The Maxwell-like equations forweak gravitational fields are [9]
tDjH
B
t
BE
D
GGG
G
GG
G
+=
=
=
=
0.
.
where GGrGG ED 04 = is the
gravitodisplacement field ( rG is the
gravitoelectric relative permittivity of themedium; G0 is the gravitoelectric
permittivity for free space and gEG =
is the gravitoelectric field intensity); isthe density of local rest mass in the localrest frame of the matter;
GGrGG HB 0= is the gravitomagnetic
field ( rG is the gravitomagnetic relative
permeability, G0 is the gravitomagnetic
permeability for free space and is
the gravitomagnetic field intensity;
GH
GGG Ej = is the local rest-mass
current density in this frame ( G is the
gravitoelectric conductivity of themedium).
Then, for free space we can writethat
===
2000444
r
GMgED GGGGG
But from the electrodynamics we knowthat
24 r
qED
==
By analogy we can write that
2
4 r
MD
g
G
=
By comparing this expression with theprevious expression of , we getGD
21280 ..1098.2
16
1 == mNkgG
G
which is the expression of thegravitoelectric permittivity for freespace.
The gravitomagnetic permeabilityfor free space [10,11] is
kgm
c
GG
26
20 10733
16 == .
We then convert Maxwell-like equations
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for weak gravity into a wave equation forfree space in the standard way. Weconclude that the speed ofGravitationalWaves in free space is
cv
GG
==
00
1
This means that both electromagneticand gravitational plane waves propagateat the free space with the same speed.
Thus, the impedance for free space is
c
Gc
H
EZ GGG
G
GG
16000 ====
and the Poynting-like vector is
GG HESrrr
=
For a plane wave propagating in the
vacuum, we have GGG HZE=
. Then, itfollows that
20
22222
3222
1i
GG
G
hG
ch
ZE
ZS
===
rrr
which is the power per unit area of aharmonic plane wave of angularfrequency.
In classical electrodynamics thedensity of energy in an electromagneticfield, , has the following expressioneW
2
02
12
02
1
HEW rre +=
In analogy with this expression wedefine the energy density in agravitoelectromagnetic field, , as
followsGW
202
1202
1GGrGGGrGG HEW +=
For free space we obtain1== rGrG
200 1 cGG =
cHE GGG 0=
andGGG HB 0=
Thus, we can rewrite the equation of
as followsGW
G
G
G
GGG
G
G
BBBc
cW
0
22
002
122
20
21 1
=
+
=
Since VGG WU = , (Vis the volume of the
particle) and for free space wecan write (55) in the following form
1=rn
( )amcB
mc
Wm
i
G
G
i
G
g
551121
1121
0
2
20
2
0
2
2
+=
+=
where V0im= .
This equation shows how thegravitational mass of a particle is alteredby a gravitomagnetic field.
A gravitomagnetic field, accordingto Einstein's theory of general relativity,arises from moving matter (mattercurrent) just as an ordinary magneticfield arises from moving charges. TheEarth rotation is the source of a veryweak gravitomagnetic field given by
1140 1016
= srad
r
MB
Earth
GEarthG .,
Perhaps ultra-fast rotating stars cangenerate very strong gravitomagneticfields, which can make the gravitationalmass of particles inside and near the starnegative. According to (55a) this will
occur if GG cB 0061.> . Usually,
however, gravitomagnetic fieldsproduced by normal matter are veryweak.
Recently Tajmar, M. et al., [12]have proposed that in addition tothe London moment, ,
(
LB
1110112 = .** emBL ; and
are the Cooper-pair mass and chargerespectively), a rotating superconductorshould exhibit also a large
gravitomagnetic field, , to explain anapparent mass increase of NiobiumCooper-pairs discovered by Tate etal[
*m
*e
GB
13,14]. According to Tajmar and Matos[15], in the case ofcoherent matter,
is given by: where
GB
202 grGcGB = c
is the mass density of coherent matterand gr is the graviphoton wavelength.
By choosing gr proportional to the local
density ofcoherent matter,c
. i.e.,
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cG
gr
gr
cm
02
1=
=
h
we obtain
2
122
00
2
0
=
=
==
cGGcgrGcG
B
and the graviphoton mass, , isgrm
cm cGgr h0=
Note that if we take the case ofnolocal sources ofcoherent matter( )0=c ,the graviphoton mass will be zero.However, graviphoton will have non-zeromass inside coherent matter( )0c .
This can be interpreted as aconsequence of the graviphoton gainingmass inside the superconductor via theHiggs mechanism due to the breaking ofgauge symmetry.
It is important to note that theminus sign in the expression for
can be understood as due to thechange from the normal to thecoherent state of matter, i.e., a switchbetween real and imaginary values
for the particles inside the materialwhen going from the normal to thecoherent state of matter. Consequently,in this case the variable U in (55)must be replaced by and not by
only. Thus we obtain
GB
GiU GU
( )bmncm
Um ir
i
G
g 551121 0
2
20
=
Since VGG WU = , we can write (55b)
for , in the following form1=rn
( )cmc
B
mc
Wm
i
cG
G
i
c
G
g
551121
1121
0
2
20
2
0
2
2
=
=
where V0ic m= is the local densityofcoherent matter.
Note the different sign (insidethe square root) with respect to (55a).
By means of (55c) it is possible tocheck the changes in the gravitationalmass of the coherent part of a givenmaterial (e.g. the Cooper-pair fluid). Thusfor the electrons of the Cooper-pairs wehave
ieeie
ie
eG
ie
ie
eG
Giege
mm
mc
m
mc
Bmm
+=
=
+=
=
+=
2
20
2
2
20
2
4112
112
where e is the mass density of the
electrons.In order to check the changes in
the gravitational mass of neutrons andprotons (non-coherent part) inside thesuperconductor, we must use Eq. (55a)
and [Tajmar and
Matos, op.cit.]. Due to ,
that expression of can be rewritten in
the following form
202 grGGB =
120 =grcG
GB
( )cgrGGB 222
0 ==
Thus we have
( )
innin
in
nG
cnin
in
nG
Gingn
mm
mc
m
mc
Bmm
=
=
+=
=
+=
14
12
112
2
20
22
2
20
2
( )
ippip
ip
pG
cp
ip
ip
pG
Gipgp
mm
mc
m
mc
Bmm
=
+=
=
+=
14
12
112
2
20
22
2
20
2
=
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where n and p are the mass density
ofneutrons and protons respectively.In Tajmars experiment, induced
accelerations fields outside thesuperconductor in the order of g100 , at
angular velocities of aboutwere observed.
1500 srad.
Starting from ( ) rGmg initialg= we
can write that ( ) rmmGgg ginitialg +=+ .
Then we get rmGg g= . For
( ) rGmgg initialg == it follows that
( ) iinitialgg mmm == . Therefore a
variation of gg = corresponds to a
gravitational mass variation 0ig mm =
.Thus correspondsto
ggg4101100 =
04101 ig mm
On the other hand, the totalgravitational mass of a particle can beexpressed by
( ) ( )( )
( )( )
( ) 22
2
2
2
cENmNmNmNm
cENmNmNmN
cENmNmNmN
cENmmN
mmNmmN
cENmNmNmNm
pieeeipppinnni
pieeeipppinnn
pieeippinn
pieeiee
ippippinninn
pgeegppgnng
+++=
=+++
+++=
=++
++
=+++=
where E is the interactionenergy; , , are the number of
neutrons, protons and electronsrespectively. Since and
nN pN eN
ipin mm
pn it follows that pn and
consequently the expression ofreduces to
gm
( ) ( )dcENmNmNmm pieeeipppig 552 20 ++
Assuming that ipppieee mNmN 2
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316103 mkgmp
p /=pV
Starting from the London momentit is easy to see that by preciselymeasuring the magnetic field and theangular velocity of thesuperconductor, one can calculate themass of the Cooper-pairs. This hasbeen done for both classical and high-Tcsuperconductors [17-20]. In theexperiment with the highest precision todate, Tate et al, op.cit., reported adisagreement between the theoreticallypredicted Cooper-pair mass in Niobium
of 99999202 .mm e* = and its
experimental value of ( )210000841. ,where is the electron mass. This
anomaly was actively discussed in theliterature without any apparent solution[
em
21-24].If we consider that the apparent
mass increase from Tatesmeasurements results from an increase
in the gravitational mass of the
Cooper-pairs due to , then we can
write
*gm
GB
( )
***
**
*****
*
**
.
.
.
ii
ii
iginitialggg
i
g
e
g
mm
mm
mmmmm
m
m
m
m
=+=
==
===
==
410840
0000841
00008412
where .410840 = .*From (55c) we can write that
***
*
*
**
ii
i
G
ig
mm
mc
mm
+=
=
+=
2
20
24112
where is the Cooper-pair massdensity.
*
Consequently we can write
4
2
20
2
108404
112 =
= .
*
*
cG
From this equation we then obtain
316103 mkg /*
Note that .* pNow we can calculate the
graviphoton mass, , inside the
Cooper-pairs fluid (coherent part of thesuperconductor) as
grm
kgcm Ggr52
0 104= h*
Outside the coherent matter ( )0=c thegraviphoton mass will be zero
00 == cm cGgr h .
Substitution of p,*=c and
into the expression of
1500 srad.
p gives
4101 p
Compare this value with that one
obtained from the Tajmar experiment.Therefore, the decrease in the
gravitational mass of the superconductor,expressed by (55e), is
SCiSCi
SCipSCiSCg
mm
mmm
,,
,,,
410
This corresponds to a decrease of the
order of in respect to the initialgravitational mass of the superconductor.However, we must also consider the
gravitational shielding effect, producedby this decrease of in thegravitational mass of the particles insidethe superconductor (see Fig. II).
Therefore, the total weight decrease inthe superconductor will be much greater
than . According to Podkletnovexperiment [
%210
%210
%210
25] it can reach up to 1% ofthe total weight of the superconductor
at 16523 srad.. ( )rpm5000 . In thisexperiment a slight decrease (up to
%1 ) in the weight of samples hungabove the disk (rotating at 5000rpm) was
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observed. A smaller effect on the orderof has been observed when thedisk is not rotating. The percentage ofweight decrease is the same for samplesof different masses and chemicalcompounds. The effect does not seem todiminish with increases in elevationabove the disk. There appears to be ashielding cylinder over the disk thatextends upwards for at least 3 meters.No weight reduction has been observedunder the disk.
%.10
It is easy to see that the decreasein the weight of samples hung above thedisk (inside the shielding cylinder overthe disk) in the Podkletnov experiment,is also a consequence of the
Gravitational Shielding Effect showed inFig. II.In order to explain the
Gravitational Shielding Effect, we startwith the gravitational field,
2
R
GMg
g=
r, produced by a particle
with gravitational mass, . The
gravitational flux,gM
g , through a spherical
surface, with area and radiusS R ,
concentric with the mass , is givenby
gM
( )g
g
SSg
GMRR
GM
SgdSgSdg
44 22
==
==== rr
Note that the flux g does not depend on
the radius R of the surface , i.e., it isthe same through any surface concentricwith the mass .
S
gM
Now consider a particle withgravitational mass, , placed into the
gravitational field produced by .
According to Eq. (41), we canhave
gm
gM
10 = ig mm , 00 ig mm, 10 = ig mm ,
etc. In the first case, the gravity
The quantization of the gravitational mass(Eq.(33)) shows that for n = 1 the gravitationalmass is not zero but equal to mg(min).Although the
gravitational mass of a particle is never null,Eq.(41) shows that it can be turned very close tozero.
acceleration, g , upon the particle gm ,
is 2
R
GMgg
g+==
r. This means that
in this case, the gravitational flux, g ,
through the particle will be given bygmgg gSSg === , i.e., it will be
symmetric in respect to the flux when
0ig mm = (third case). In the second case
)0gm , the intensity of thegravitational force between and
will be very close to zero. This isequivalent to say that the gravityacceleration upon the particle with mass
gm gM
gm will be 0g . Consequently we canwrite that 0= Sgg . It is easy to see
that there is a correlation between
0ig mm and gg , i.e.,
_ If 10 = ig mm 1= gg
_ If 10 = ig mm 1= gg
_ If 00
ig
mm 0gg
J ust a simple algebraic form contains therequisites mentioned above, thecorrelation
0i
g
g
g
m
m
=
By making = 0ig mm we get
gg =
This is the expression of the gravitationalflux through
gm . It explains the
Gravitational Shielding Effect presentedin Fig. II.
As gSg = and Sgg = , we obtain
gg =
This is the gravity acceleration inside gm .
Figure II (b) shows the gravitational
shielding effect produced by two particlesat the same direction. In this case, the
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gravity acceleration inside and above thesecond particle will be ifg2 12 ig mm = .
These particles are representativeof any material particles or materialsubstance (solid, liquid, gas, plasma,
electrons flux, etc.), whose gravitationalmass have been reduced by thefactor . Thus, above the substance, thegravity acceleration is reduced at the
same proportion
g
0ig mm= , and,
consequently, gg = , where is the
gravity acceleration below the substance.
g
Figure III shows an experimentalset-up in order to check the factor above a high-speed electrons flux. As we
have shown (Eq. 43), the gravitationalmass of a particle decreases with theincrease of the velocity Vof the particle.
Since the theory says that thefactor is given by the correlation
0ig mm then, in the case of an electrons
flux, we will have thatiege mm= where
as function of the velocity V is
given by Eq. (43). Thus, we can writethat
gem
== 1
1
121
22 cVm
m
ie
ge
Therefore, if we know the velocity V ofthe electrons we can calculate . ( is
the electron mass at rest).iem
When an electron penetrates theelectric field (see Fig. III) an electric
force,
yE
yE EeFrr
= , will act upon the
electron. The direction of will be
contrary to the direction of
EFr
yEr
. The
magnetic force which acts upon the
electron, due to the magnetic fieldBF
r
Br
, is
and will be opposite toeVBFB =r
EFr
because the electron charge is negative.
By adjusting conveniently B we
can make EB FFrr
= . Under these
circumstances in which the total force iszero, the spot produced by the electrons
flux on the surface returns from O toand is detected by the
galvanometerG . That is, there is nodeflection for the cathodic rays. Then it
follows that
O
yeEeVB = since EB FFrr
= .
Then, we get
B
EV
y=
This gives a measure of the velocity ofthe electrons.
Thus, by means of theexperimental set-up, shown in Fig. III, wecan easily obtain the velocity V of theelectrons below the body , in order tocalculate the theoretical value of . The
experimental value of can be obtainedby dividing the weight, gmg P = of
the body for a voltage drop V~
acrossthe anode and cathode, by itsweight, gP mg= , when the voltage
V~
is zero, i.e.,
g
g
P
P =
=
According to Eq. (4), the gravitationalmass, , is defined bygM
221 cV
mM
g
g
=
While Eq. (43) defines by means of
the following expressiongm
022
11
121
ig mcV
m
=
In order to check the gravitational massof the electrons it is necessary to knowthe pressure produced by theelectrons flux. Thus, we have put apiezoelectric sensor in the bottom of theglass tube as shown in Fig. III. Theelectrons flux radiated from the cathodeis accelerated by the anode1 and strikeson the piezoelectric sensor yielding apressure
P
P which is measured bymeans of the sensor.
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Fig. I I The Gravitational Shielding effect.
(a)
mg = mimg < mi
mg < 0
(b)
g < g due to the gravitationalshielding effectproduced by mg 1
Particle 1
mg1
Particle 2
mg 2
mg1 =x mi1 ; x < 1
g
g
P2 = mg 2g= mg 2 (xg )
P1 = mg1g =x mi 1g
g g g
g < gg g < 0
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Let us now deduce the correlation
between P and .geM
When the electrons flux strikesthe sensor, the electrons transfer to ita momentum .
Since
VMnqnQ geeee ==
VFdtFQ 2== , we conclude that
=
e
gen
F
V
dM
2
2
The amount of electrons, , is given
byen
Sdne = where is the amount ofelectrons per unit of volume(electrons/m
3); is the cross-section
of the electrons flux and thedistance between cathode andanode.
S
d
In order to calculate we willstart from the Langmuir-Child law andthe Ohm vectorial law, respectivelygiven by
en
d
VJ
23~
= and VJ c= , ( )ec =
where is the thermoionic current
density;
J
2
3
1610332 = VmA ... isthe called Childs constant; V
~is the
voltage drop across the anode andcathode electrodes, and V is thevelocity of the electrons.
By comparing the Langmuir-Child law with the Ohm vectorial lawwe obtain
Ved
V2
23~
=
Thus, we can write that
edV
SVne
23~
=
and
PVV
edMge
=
23
22~
Where SFP = , is the pressure to bemeasured by the piezoelectricsensor.
In the experimental set-up thetotal force acting on the
piezoelectric sensor is the resultant ofall the forces produced by each
electrons flux that passes througheach hole of area in the grid of the
anode 1, and is given by
F
F
S
( ) 23
22VVM
ed
nSPSnnFF ge
~
===
where is the number of holes in thegrid. By means of the piezoelectricsensor we can measure andconsequently obtain .
n
F
geM
We can use the equationabove to evaluate the magnitude of
the force to be measured by thepiezoelectric sensor. First, we will findthe expression ofV as a function of
F
V~
since the electrons speed Vdepends on the voltage V
~.
We will start from Eq. (46)which is the general expression forLorentzs force, i.e.,
( )0i
g
m
mBVqEq
dt
pd rrrr
+=
When the force and the speedhave the same direction Eq. (6) gives
( ) dtVd
cV
m
dt
pd grr
23
221 =
By comparing these expressions weobtain
( )BVqEq
dt
Vd
cV
mi rrrr
+= 2
322
0
1
In the case of electrons acceleratedby a sole electric field , theequation above gives
( 0=B )
( )ieie m
VecV
m
Ee
dt
Vda
~2
1 22==
rrr
Therefore, the velocity V of theelectrons in the experimental set-upis
( )iem
VecVadV
~2
12 43
22==
From Eq. (43) we conclude that
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Dynamometer (D)
Fig. III Experimental set-up in order to check the factor above a high-speed electrons flux.
The set-up may also check the velocities and the gravitational masses of the electrons.
G
- +
V~
B
+ yV~
Eyy
O
+
iG
R
Fe V eO
+
Anode 1 Filaments Cathode Anode 2
Piezoelectric
sensor
d
Collimators g= g gGrid d Collimators
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0gem when . Substitution
of this value ofV into equation abovegivesV
cV 7450.
KV1479.~
. This is the
voltage drop necessary to be appliedacross the anode and cathodeelectrodes in order to obtain 0gem .
Since the equation above canbe used to evaluate the velocity V ofthe electrons flux for a givenV
~, then
we can use the obtained value ofVtoevaluate the intensity ofB
rin order to
produce yeEeVB = in the
experimental set-up. Then by
adjusting B we can check when theelectrons flux is detected by thegalvanometer . In this case, as wehave already seen, , and
the velocity of the electrons flux iscalculated by means of theexpression
G
yeEeVB =
BEV y= . Substitution of
into the expressions of and
, respectively given by
V gem
geM
iege mcV
m
= 11
12122
and
221 cV
mM
ge
ge
=
yields the corresponding values ofand which can be compared
with the values obtained in theexperimental set-up:
gem geM
=
==
nS
ed
VV
FM
mPPmm
ge
ieiege
22
~ 23
where and are measured by
the dynamometer and ismeasured by the piezoelectricsensor.
P P
D F
If we have and2160 mnS .md 080.= in the experimental set-up
then it follows that23
14
10821 VVMF ge~
. =By varying V
~from 10KV up to 500KV
we note that the maximum value foroccurs whenF KVV 7344.
~ . Under
these circumstances, andcV 70.
iege mM 280. . Thus the maximum
value for isFgfNF 19091 .max
Consequently, for KVV 500=max~
, the
piezoelectric sensor must satisfy thefollowing characteristics:
Capacity 200gf Readability 0.001gf
Let us now return to theexplanation for the findings ofPodkletnovs experiment. Next, wewill explain the decrease of 0.1% in
the weight of the superconductorwhen the disk is only levitating but notrotating.
Equation (55) shows how thegravitational mass is altered byelectromagnetic fields.
The expression of forrn >> can be obtained from (54),
in the form
( )56
4
2
f
c
v
cnr
==
Substitution of (56) into (55) leads to
0
2
14
121 ii
g mcm
U
fm
+=
This equation shows that atoms offerromagnetic materials with very-high can have gravitationalmasses strongly reduced by meansof Extremely Low Frequency (ELF)
electromagnetic radiation. It alsoshows that atoms ofsuperconducting
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materials (due to very-high ) canalso have its gravitational massesstrongly reduced by means of ELFelectromagnetic radiation.
Alternatively, we may put
Eq.(55) as a function of the powerdensity ( or intensity ), , of theradiation. The integration of (51)gives
D
vDU V= . Thus, we can write(55) in the following form:
( )571121 0
2
3
2
i
r
g mc
Dnm
+=
where V0im= .
For >>
, will be given by(56) and consequently (57) becomesrn
( )5814
121 0
2
ig mcf
Dm
+=
In the case of Thermalradiation, it is common to relatethe energy of photons totemperature, T, through therelation,
Thf where is theBoltzmanns constant. On the otherhand it is known that
KJ = /. 2310381
4TD B=
where 42810675 KmwattsB = /.
is the Stefan-Boltzmanns constant.Thus we can rewrite (58) in thefollowing form
( )amc
hTm i
B
g 5814
121 0
23
+=
Starting from this equation, we canevaluate the effect of the thermalradiation upon the gravitational massof the Copper-pair fluid, .
Below the transition temperature, ,CPfluidgm ,
cT
( 50.
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be much greater than . Thiscan explain the smaller effect on theorder of observed in thePodkletnov measurements when thedisk is not rotating.
%310
%.10
Let us now consider an electriccurrent I through a conductorsubjected to electromagneticradiation with power density andfrequency .
D
f
Under these circumstances thegravitational mass of the
electrons of the conductor, accordingto Eq. (58), is given by
gem
ege mcf
Dm
+= 14
121
2
where .kg.me3110119 =
Note that if the radiation uponthe conductor has extremely-lowfrequency (ELF radiation) then
can be strongly reduced. Forexample, if ,and the conductor is made ofcopper
(
gem
Hzf610 2510 m/WD
0 ; and)then
m/S. 71085 =38900 m/kg=
14
cf
D
and consequently .ege m.m 10
According to Eq. (6) the forceupon each free electron is given by
( ) EedtVd
cV
m
Fge
e
rr
r
==2
3221
where E is the applied electric field.Therefore, the decrease of
produces an increase in the velocityof the free electrons and
consequently the drift velocity isalso increased. It is known that thedensity of electric current through
a conductor [
gem
V
dV
J
28] is given by
deVJrr
=
wheree is the density of the
free electric charges ( For cooperconductors ).
Therefore increasing produces anincrease in the electric current
3101031 m/C.e =
dV
I.Thus if is reduced 10 timesgem
)ege m.m 10 the drift velocity is
increased 10 times as well as theelectric current. Thus we concludethat strong fluxes of ELF radiationupon electric/electronic circuits cansuddenly increase the electriccurrents and consequently damagethese circuits.
dV
Since the orbital electronsmoment of inertia is givenby ( ) 2jjii rmI = , where refers to
inertial mass and not to gravitationalmass, then the momentum
im
iIL = ofthe conductor orbital electrons are notaffected by the ELF radiation.Consequently, this radiation justaffects the conductors free electronsvelocities. Similarly, in the case ofsuperconducting materials, themomentum, iIL = , of the orbitalelectrons are not affected by thegravitomagnetic fields.
The vector ( )vUDrr
V= , which wemay define from (48), has the samedirection of the propagation vector k
r
and evidently corresponds to the
Poynting vector. Then can bereplaced byD
r
HErr
.Thus we can write( ) ( )[ ] ( ) 2
21
21
21
21 1 EvvEEBEEHD ==== .
For >> Eq. (54) tells us thatfv 4= . Consequently, we obtain
fED
42
21=
This expression refers to theinstantaneous values of andD E.
The average value for2
E is equal to22
1mE because E varies sinusoidaly
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( is the maximum value formE E).Substitution of the expression ofinto (58) gives
D
( )amE
fcm ig 591
44121 0
2
23
2
+=
Since 2mrms EE = and2
212
mEE = wecan write the equation above in thefollowing form
( )amE
fcm i
rms
g 59144
121 02
23
2
+=
Note that for extremely-lowfrequencies the value of in this
equation becomes highly expressive.
3f
Since equation (59a)can also be put as a function of
vBE=B ,
i.e.,
( )bmB
cfm ig 591
4121 02
4
2
+=
For conducting materials with;m/S710 1=r ;
the expression (59b) gives
3310 m/kg
04
12
110
121 ig mBf
m
+=
This equation shows that thedecreasing in the gravitational massof these conductors can becomeexperimentally detectable forexample, starting from 100Teslas at10mHz.
One can then conclude that aninteresting situation arises when abody penetrates a magnetic field inthe direction of its center. Thegravitational mass of the bodydecreases progressively. This is dueto the intensity increase of themagnetic field upon the body while itpenetrates the field. In order tounderstand this phenomenon wemight, based on (43), think of theinertial mass as being formed by two
parts: one positive and anothernegative. Thus, when the body
penetrates the magnetic field, itsnegative inertial mass increases, butits total inertial mass decreases, i.e.,although there is an increase ofinertial mass, the total inertial mass
(which is equivalent to gravitationalmass) will be reduced.On the other hand, Eq.(4)
shows that the velocity of the bodymust increase as consequence ofthe gravitational mass decreasingsince the momentum is conserved.Consider for example a spacecraftwith velocity and gravitational
mass . If is reduced to
then the velocity becomes
sV
gM gM gm
sggs VmMV =
In addition, Eqs. 5 and 6 tell us thatthe inertial forces depend on .
Only in the particular case ofgm
0ig mm = the expressions (5) and
(6) reduce to the well-knownNewtonian expression .Consequently, one can conclude thatthe inertial effects on the spacecraftwill also be reduced due to thedecreasing of its gravitational mass.Obviously this leads to a newconcept of aerospace flight.
amF i 0=
Now consider an electriccurrent ftsinii 20= through aconductor. Since the current density,J
r, is expressed by ESddiJ
rrr== ,
then we can write that( ) ftsinSiSiE 20== . Substitution
of this equation into (59a) gives
( )cmftfSc
im ig 591264
121 04
34223
40
+=
sin
If the conductor is a supermalloy rod( )mm40011 then 000100,r = (initial); ;
and . Substitution ofthese values into the equation above
yields the following expression for the
38770 m/kg= m/S. 61061 =26101 mS =
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gravitational mass of the supermalloyrod
( ) ( ) ( )smismg mftfim
+= 12sin1071.5121 4340
12
Some oscillators like the HP3325A(Op.002 High Voltage Output) cangenerate sinusoidal voltages withextremely-low frequencies down to
and amplitude up to20V (into
Hzf6101 =
50 load). The maximumoutput current is .
ppA.080
Thus, for )ppA.A.i 0800400 = and the equationabove shows that the gravitational
mass of the rod becomes negative at
Hz.f610252 >=
( )832 20 cUnmm rig =
Therefore, the action for suchparticle, in agreement with the Eq.(2),is
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( )
[ ( )8412112
1
2
1
2
1
2
1
22222
2222
222
+=
=+=
==
t
t ri
t
t ri
t
t g
.dtcVUncVcm
dtcVccUnm
dtcVcmS
]The integrant function is theLagrangean, i.e.,
( )85121 222220 cVUncVcmL ri +=Starting from the Lagrangean we canfind the Hamiltonian of the particle, bymeans of the well-known generalformula:
( ) .LVLVH =The result is
( ) ( )861
24
1 2222
22
20 .
+
=cV
cVUn
cV
cmH r
i
The second term on the right handside of Eq.(86) results from theparticle's interaction with theelectromagnetic field. Note thesimilarity between the obtainedHamiltonian and the well-knownHamiltonian for the particle in anelectromagnetic field [32]:
( )871 2220 .QcVcmH i +=in which is the electric charge andQ
, the field's scalar potential. Thequantity Q expresses, as weknow, the particle's interaction withthe electromagnetic field in the sameway as the second term on the right
hand side of the Eq. (86).It is therefore evident that it isthe same quantity, expressed bydifferent variables.
Thus, we can conclude that, inultra-high energy conditions( )202 cmcMUn iir > , the gravitationaland electromagnetic fields canbe described by the sameHamiltonian, i.e., in thesecircumstances they are unified !
It is known that starting fromthat Hamiltonian we may obtain a
complete description of theelectromagnetic field. This meansthat from the present theory forgravity we can also derive theequations of the electromagnetic
field. Due to thesecond term on the right hand side ofEq.(86) can be written as follows
2cMpcUn ir =
( )
( )
2200
2
22
22
22
22
144
1
24
1
24
cVr
QQ
R
QQQ
cMcV
cV
cV
cVpc
i
===
=
=
=
whence
( )r
QQcMcV i
0
222
424
=
The factor ( )24 22 cV becomesequal to 2 in the ultra-relativistic case,then it follows that
( )8842 0
2
r
QQ
cMi
=
From (44), we know that there is aminimum value for given byiM
( ) ( )minimini mM = . Eq.(43) shows that
)( ) (minmin 0ig mm = and Eq.(23) gives
( ) maxmaxming cdhcLhm 838 == .
Thus we can write
( ) ( ) ( )89830 maxminmin cdhmM ii ==
According to (88) the valueis correlated to
( )2
2 cM mini
( ) maminmin rQrQQ 02
0 44 = ,i.e.,
( ) ( )90242
0
2
cMr
Qmini
max
min =
where is the minimum electriccharge in the Universe ( thereforeequal to minimum electric charge ofthe quarks, i.e.,
minQ
e31 ); is the
maximum distance between and
maxr
QQ , which should be equal to the so-
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called "diameter", , of the visible
Universe ( where is obtained
from the Hubble's law for , i.e.,
). Thus, from (90) we readily
obtain
cd
cc ld 2= cl
cV =1= H
~clc
( )
( )( )91
96
24
31
120
0
e
dH~
hc
ddhcQ
max
maxcmin
=
==
==
whence we findm.dmax
301043 =
This will be the maximum "diameter" thatthe Universe will reach. Consequently,Eq.(89) tells us that the elementary
quantum of matter is
( ) kgcdhmi73
0 109383== .maxmin
This is, therefore, the smallest indivisibleparticle of matter.
Considering that, the inertial massof the Observable Universe is
kgGHcMU53
03 102 = and that its volume
is ( ) 37930343
34 10 mHcRV UU == ,
where is the Hubbleconstant, we can conclude that thenumber of these particles in theObservable Universe is
1180 1075.1 = sH
( )particles
m
Mn
i
UU
125
min0
10=
By dividing this number by , we getUV
346 /10 mparticlesV
n
U
U
Obviously, the dimensions of the
smallest indivisible particle of matterdepend on its state of compression. Infree space, for example, its volume is
UU nV . Consequently, its radius is
mnR UU153 10 .
If particles with diameterN fill
all space of then . Thus, if
then the number of particles,with this diameter, necessary to fill
all is . Since thenumber ofsmallestindivisible particles of
matter in the Universe is
31m 13 =N
m1510
3
1m particlesN45
10
346 /10 mparticlesVn UU we can concludethat these particles fill all space in theUniverse, by forming a Continuous3Universal Medium or Continuous
Universal Fluid (CUF), the density ofwhich is( ) 327min0 /10 mkg
V
mn
U
iU
CUF=
Note that this density is much smallerthan the density of the Intergalactic
Medium ( )326 /10 mkgIGM .The extremely-low density of the
Continuous Universal Fluid shows that itslocal gravitational mass can be stronglyaffected by electromagnetic fields
(including gravitoelectromagnetic fields),pressure, etc. (See Eqs. 57, 58, 59a,59b, 55a, 55c and 60). The density ofthis fluid is clearly not uniform along theUniverse, since it can be stronglycompressed in several regions (galaxies,stars, blackholes, planets, etc). At thenormal state (free space), the mentionedfluid is invisible. However, at supercompressed state it can become visibleby giving origin to the known mattersince matter, as we have seen, isquantized and consequently, formed byan integer number of elementaryquantum of matter with mass .
Inside the proton, for example, there are( )min0im
( )45
min0 10= ipp mmn elementary quanta
of matterat supercompressed state, withvolume pproton nV and radius
mnR pp303 10 .
Therefore, the solidification of the
matter is just a transitory state of thisUniversal Fluid, which can back to theprimitive state when the cohesionconditions disappear.
Let us now study another aspectof the present theory. By combination ofgravity and the uncertainty principle wewill derive the expression for the Casimirforce.
An uncertaintyim in
produces an uncertaintyim
p in andp
3 At very small scale.
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therefore an uncertaintygm in ,
which according to Eq.(41) , is givenby
gm
( )92112
2
ii
ig
mcm
pmm
+=
From the uncertainty principle forposition and momentum, we knowthat the product of the uncertainties ofthe simultaneously measurablevalues of the corresponding positionand momentum components is atleast of the magnitude order of ,i.e.,
h
h~rp
Substitution of r~p h into (92) yields
( )931122
i
i
ig mr
cmmm
+=
h
Therefore if
( )94cm
ri
h
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( ) ( )101022AnndA minA ==
It can also be easily shown thatthe minimum volume related tois the volume of a regular tetrahedron
of edge length , i.e.,
mind
mind) ) 331223122 planckminmin lk
~d ==
The maximum volume is the volumeof a sphere of radius , i.e.,mind
( ) ( ) 33343
34
planckminmax lk~
d ==
Thus, the elementary volume333
0 planckVminV lk~
d == must have a
value between min and max , i.e.,
( ) 34
12
2
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( ) ( )
( )( )103
43
4
2
323
=
=
=
=
c
EG
r
hc
c
cmG
r
hc
c
mG
r
hcF ii
However, from the uncertaintyprinciple for energy and time we knowthat
( )104t~E hTherefore, we can write theexpression (103) in the followingform:
( )
( ) ( )1051
1
2
3
33
=
=
=
ctlr
hc
ctc
G
r
hcF
planck
h
From the General Relativity Theorywe know that 00gcdtdr = . If the
field is weak then 200 21 cg = and
( ) ( )222 11 crGmcdtccdtdr =+= .For 122
>>
In this case, ig mm and
ig mm . Thus,
( )( )( )
( )
( )( ) ( )
( )
( ) ( )402
24
2
4
22232
2
4
2
22
2
1920
960
1920
2
1
1
r
hcAl
r
hc
lr
hc
tcr
hc
c
G
r
t
c
G
r
cEcEG
r
mmGF
planck
planck
ii
=
=
=
=
=
=
=
===
hh
whence
( )1071920 4r
hcAF
=
The force will be attractive and itsintensity will be the fourth part of theintensity given by the first expression(102) for the Casimir force.
We can also use this theory toexplain some relevant cosmologicalphenomena. For example, the recentdiscovery that the cosmic expansionof the Universe may be accelerating,and not decelerating as manycosmologists had anticipated [35].
We start from Eq. (6) whichshows that the inertial forces, iF
r,
whose action on a particle, in thecase of force and speed with samedirection, is given by
( )a
cV
mF
gi
rr
23
221
=
Substitution of given by (43) into
the expression above givesgm
( ) ( )am
cVcVF ii
rr022222 1
2
1
32
3
=
whence we conclude that a particlewith rest inertial mass, , subjected
to a force,0im
iFr
, acquires an
acceleration a
r
given by
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( ) ( )022222 1
2
1
32
3 i
i
mcVcV
Fa
=
rr
By substituting the well-known
expression of Hubbles law forvelocity, , ( isthe Hubble constant) into theexpression of , we get theacceleration for any particle in theexpanding Universe, i.e.,
lHV~
= 1181071 = sH .~
ar
( ) ( )02222222 1
2
1
32
3 i
i
mclHclH
Fa
~~
=
rr
Obviously, the distance l increaseswith the expansion of the Universe.Under these circumstances, it is easyto see that the term
( ) ( )2222222 12
1
32
3
clHclH~~
decreases, increasing theacceleration of the expandingUniverse.
Let us now consider thephenomenon of gravitationaldeflection of light.
A distant stars light ray,under the Suns gravitational forcefield describes the usual central forcehyperbolic orbit. The deflection of thelight ray is illustrated in Fig. V, withthe bending greatly exaggerated for abetter view of the angle of deflection.
The distance CS is the
distance of closest approach. Theangle of deflection of the light ray,d ,is shown in the Figure V and is
. 2=
where is the angle of theasymptote to the hyperbole. Then, itfollows that
( ) 22 tantantan ==From the Figure V we obtain
.tan
c
Vy=
Fig. V Gravitational deflection of light about
the Sun.
S C
Photons
Since and are very small we canwrite that
2= andc
Vy=
Then
c
Vy2=
Consider the motion of thephotons at some time after it haspassed the point of closest approach.We impose Cartesian Co-ordinateswith the origin at the point of closestapproach, the x axis pointing along its
path and the y axis towards the Sun.The gravitational pull of the Sun is
t
2r
MMGP
gpgS=
where is the relativistic
gravitational mass of the photon andthe relativistic gravitational mass
of the Sun. Thus, the component in aperpendicular direction is
gpM
gSM
222222
2 sin
tcd
d
tcd
MMG
r
MM
GF
gpgS
gpgS
y
++=
==
According to Eq. (6) the expression ofthe force isyF
( ) dtdV
cV
mF
y
y
gp
y23
221=
By substituting Eq. (43) into thisexpression, we get
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( ) ( ) dtdV
M
cVcVF
y
ip
yy
y23
2222 1
2
1
3
=
For , we can write this
expression in the following form
cVy
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(momentum) to the particle, andconsequently its gravitational mass willbe increased. This means that themotion generates gravitational mass.
On the other hand, if thegravitational mass of a particle is nullthen its inertial mass, according to Eq.(41), will be given by
c
pmi
=
5
2
From Eqs. (4) and (7) we get
Vc
pV
c
Ep
g
=
= 0
2
Thus we have
V
c
pmg
=
202 and V
c
pmi
=
2
0
5
2
Note that, like the gravitational mass, theinertial mass is also directly related to themotion, i.e., it is also generated by themotion.
Thus, we can conclude that is themotion, or rather, the velocity is whatmakes the two types of mass.
In this picture, the fundamentalparticles can be considered asimmaterial vortex of velocity; it is thevelocity of these vortexes that causes thefundamental particles to have masses.
That is, there exists not matter in theusual sense; but just motion. Thus, thedifference between matter and energy
just consists of the diversity of the motiondirection; rotating, closed in itself, in thematter; ondulatory, with open cycle, inthe energy (See Fig. VI).
Under this context, the Higgsmechanism appears as a process, bywhich the velocity of an immaterial vortex
can be increased or decreased by
imaginaryip
)
The Standard Model is the name given tothe current theory of fundamental particles andhow they interact. This theory includes: Stronginteraction and a combined theory of weak andelectromagnetic interaction, known aselectroweak theory. One part of the StandardModel is not yet well established. What causesthe fundamental particles to have masses? Thesimplest idea is called theHiggs mechanism. Thismechanism involves one additional particle,
called the Higgs boson, and one additional forcetype, mediated by exchanges of this boson.
making the vortex (particle) gain or losemass. Ifreal motion is what makes realmass then, by analogy, we can say thatimaginary mass is made by imaginarymotion. This is not only a simplegeneralization of the process based onthe theory of the imaginary functions, butalso a fundamental conclusion related tothe concept ofimaginary mass that, as itwill be shown, provides a coherentexplanation for the materialization of thefundamental particles, in the beginning ofthe Universe.
It is known that the simultaneousdisappearance of a pair(electron/positron) liberates an amount of
energy, , under the form of
two photons with frequency , in such away that
( )2
02 cm realei
f
( ) hfcm realei 222
0 =
Since the photon has imaginary massesassociated to it, the phenomenon of
transformation ofthe energy
into suggests that the imaginary
energy of the photon,m ,
comes from the transformation ofimaginary energy of the electron,
, just as the real energy of
the photon, , results from thetransformation of real energy of theelectron, i.e.,
( )2
02 cm realei
hf2
( )2c
(2
0 cm imaginaryei
hf
( ) ( )
( ) hfcm
cmcm
imaginarypi
realeiimaginaryei
22
22
20
20
20
+=
=+
Then, it follows that
( ) ( )imaginaryipimaginaryei mm =0
The sign (-) in the equation above, is dueto the imaginary mass of the photon tobe positive, on the contrary of theimaginary gravitational mass of thematter, which is negative, as we havealready seen.
http://www2.slac.stanford.edu/vvc/glossary.html#Fundamental%20Particlehttp://www2.slac.stanford.edu/vvc/glossary.html#Fundamental%20Particle -
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Real Particles Imaginary Particles(Tardyons) (Tachyons)
Real Inertial Mass Imaginary Inertial Mass
Non-null Null Non-null Null
V
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Neutron
( )
( ) ( )
( )( )
( )( )
( )
( )( )
( )( )
( )imnin
imni
imni
imn
imgn
realnin
realni
realni
realn
realgn
realniimni
realni
m
mcm
U
m
m
mcm
Um
imm
kgm
0
0
2
20
0
0
2
20
03
2
0
270
1121
1121
106747.1
=
=
+=
=
=
+=
=
=
Proton
( )
( ) ( )
( )( )
( )
( )
( )
( )( )
( )( )
( )impripr
impri
impri
impr
imgpr
realpripr
realpri
realpri
realpr
realgpr
realpriimpri
realpri
m
mcm
Um
m
mcm
Um
imm
kgm
0
0
2
20
0
0
2
2
0
032
0
270
1121
1121
106723.1
=
=
+=
=
=
+=
+=
=
)
where and are
respectively, the real and imaginaryenergies absorbed by the particles.
(realU ( )imU
When neutrons, protons andelectrons were created after the Big-bang, they absorbed quantities ofelectromagnetic energy, respectivelygiven by
( ) ( )
( ) ( )
( ) ( ) ikTUkTU
ikTUkTU
ikTUkTU
eeimaginaryeeereale
prprimaginaryprprprrealpr
nnimaginarynnnrealn
==
==
==
where n , pr and e are the
absorption factors respectively, for theneutrons, protons and electrons;
is the Boltzmannconstant; , and are the
temperatures of the Universe,respectively when neutrons, protonsand electrons were created.
KJk /1038.1 23=
nT prT eT
In the case of the electrons, itwas previously shown that 1.0e .
Thus, by considering that
, we getKTe31102.6
( ) iikTU eeime7105.8 ==
It is known that the protons werecreated at the same epoch. Thus, wewill assume that
( ) iikTU prprimpr7105.8 ==
Then, it follows that21108.1 =e17107.9 =pr
Now, consider the gravitational forces,due to the imaginary masses of twoelectrons, , two protons, , and
one electron and one proton, , all at
rest.
eeF prprF
eprF
( ) ( )( )
( ))(
2
28
2
202
2
203
2
22
2
103.2
3
4repulsion
realei
e
realei
e
imge
ee
rr
mG
r
imG
r
mGF
+=+=
=
==
( ) ( )
( ))(
2
28
2
202
2
2
032
22
2
103.234
repulsionrealpri
pr
realpri
primgpr
prpr
rr
mG
r
im
Gr
mGF
+=+=
=
+
==
( ) ( )
( ) ( )
( ) ( )
)(
2
28
2
00
2
03
203
2
2
103.2
3
4
atraction
realprirealeipre
realprirealei
pre
imgprimgeepr
rr
mmG
r
imim
G
r
mmGF
==
=
+
=
==
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Note that
2
28
20
2 103.2
4 rr
eFelectric
==
Therefore, we can conclude that
)(4 20
2
repulsionr
eFFF electricprpree
+==
and
)(4 20
2
atractionr
eFF electricep
=
These correlations permit to define theelectric charge by means of thefollowing relation:
( ) imGq imaginaryg04=
For example, in the case of theelectron, we have
( )
( )( )
( )( )( )( ) CmG
imG
imG
imGq
realeie
realeie
imaginaryeie
imaginarygee
1903
20
203
20
00
0
106.14
4
4
4
==
==
==
==
In the case of the proton, we get
( )
( )( )
( )( )( )( ) CmG
imG
imG
imGq
realpripr
realpripr
imaginarypripr
imaginarygprpr
1903
20
203
20
00
0
106.14
4
4
4
+==
=+=
==
==
For the neutron,it follows that
( )
( )( )
( )( )( )( )realnin
realnin
imaginarynin
imaginarygnn
mG
imG
imG
imGq
032
0
203
20
00
0
4
4
4
4
=
==
==
==
However, based on the quantization ofthe mass (Eq. 44), we can write that
( ) ( ) 0min02
032 = nmnm irealnin
Since can have only discrete valuesdifferent of zero (See Appendix B), weconclude that
n
n cannot be null.However, it is known that the electric
charge of the neutron is null. Thus, it isnecessary to assume that
( )
( )
( )
( )( )( ) ( )[ ] 04
4
4
4
4
203
2203
20
00
00
0
0
=++=
=+
+
=
=+
+=+=
+
++
imimG
imG
imG
imG
imGqqq
ninnin
imaginarynin
imaginarynin
imaginarygn
imaginarygnnnn
We then conclude that in the neutron,half of the total amount of elementaryquanta of electric charge, , is negative,
while the other halfis positive.
minq
In order to obtain the value ofthe elementary quantum of electriccharge, , we start with theexpression obtained here for theelectric charge, where wechange by its quantized
expression ,
derived from Eq. (44a). Thus, we get
minq
(imaginarygm )
)( ) ( )(min02
imaginaryiimaginaryg mnm =
( )
( )( )
( )( )[ ]
( )min02
032
min0322
0
min02
0
0
4
4
44
i
i
imaginaryi
imaginaryg
mnG
iimnG
imnG
imGq
m=
==
====
This is the quantized expression of theelectric charge.
For 1=n we obtain the value ofthe elementary quantum of electriccharge, , i.e.,minq
( ) CmGq i83
min0032
min 108.34== mm
where is the elementary
quantum of matter, whose valuepreviously calculated, is
.
(min0im )
( ) kgmi73
min0 109.3=
The existenceofimaginary mass
associated to a real particle suggeststhe possible existence of imaginary
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particles with imaginary masses inNature.
In this case, the concept ofwaveassociated to a particle (De Broglieswaves) would also be applied to theimaginary particles. Then, by analogy,the imaginary wave associated to animaginary particle with imaginarymasses and would be
described by the following expressionsim gm
h
rh
r
=
=
E
kp
Henceforth, for the sake of simplicity,we will use the Greek letter to stand
for the word imaginary; pr
is the
momentum carried by the wave andits energy;E 2=k
ris the
propagation number and the
wavelength of the wave; f2=
is the cyclical frequency.According to Eq. (4), the
momentum pr
is
VMp grr
=
where is the velocity of theV particle.By comparing the expressions of
pr
we get
VM
h
g
=
It is known that the variablequantity which characterizes the DeBroglies waves is called wave function,usually indicated by symbol . Thewave function associated with amaterial particle describes the dynamicstate of the particle: its value at aparticular point x, y, z, t is related to theprobability of finding the particle in thatplace and instant. Although doesnot have a physical interpretation, its
square (or
2 * ) calculated for aparticular point x, y, z, t is proportionalto the probability of finding the particlein that place and instant.
Since is proportional to theprobability
2P of finding the particle
described by , the integral of 2 on
the whole space must be finite inasmuch as the particle is somewhere.
On the other hand, if
02 =+
Vd
the interpretation is that the particle willnot exist. However, if
( )1082 =+
Vd
The particle will be everywheresimultaneously.
In Quantum Mechanics, the wavefunction corresponds, as we know,to the displacement y of theundulatory motion of a rope.However, , as opposed to y , is not ameasurable quantity and can, hence,be a complex quantity. For this reason,it is assumed that is described in the
directionx by( )pxEthi
e
=2
0
This is the expression of the wavefunction for a free particle, with totalenergy E and momentum , moving inthe direction
pr
x+ .As to the imaginary particle, the
imaginary particle wave function will be
denoted by and, by analogy theexpression of , will be expressed by:
)xptEhie
=
20
Therefore, the general expression ofthe wave function for a free particle canbe written in the following form
( )( ) ( ) ( )( )
( )( )xptEhi
xptEhi
real
e
e realreal
+
+=
20
20
It is known that the uncertaintyprinciple can also be written as afunction of E (uncertainty in theenergy) and t (uncertainty in thetime), i.e.,
h tE.
This expression shows that avariation of energy E , during a
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time interval , can only bedetected if
tEt h . Consequently,
a variation of energy E , during atime interval Et
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+0.159
V0
Imaginary Body
0 Imaginary Space-time
Virtual Photons ( =V )0.159
Fig. VII Travel in the imaginary space-time. Similarly to the virtual photons,imaginary bodies can have infinite speedin the imaginary space-time.
Ordinary Space-time
cV
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Real particle Real particle
t1 E
t2 c (speed upper limit)
t3
Imaginary Space-time Ordinary Space-time
t3
t2 Vmax(speed upper limit)
Et1
Imaginary particle Imaginary particle
Fig. VIII Virtual Transitions (a) Virtual Transitions of a real particle to the imaginaryspace-time. The speed upper limit forreal particle in the imaginary space-time is c.(b) - Virtual Transitions of an imaginaryparticle to the ordinary space-time. The
speed upper limit forimaginary particle in the ordinary space-time is 11210 smV .max
Note that to occur a virtual transition it is necessary thatt=t1+ t2+ t3
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Thus, if the gravitational mass of theparticle is reduced by means of theabsorption of an amount ofelectromagnetic energy U, for
example, we have
( )
+== 112122
0cmUM
Mi
i
g
This shows that the energyUof theelectromagnetic field remains actingon the imaginary particle. Inpractice, this means thatelectromagnetic fields act on
imaginary particles.The gravity acceleration on aimaginary particle (due to the rest ofthe imaginary Universe) are givenby
.,...,,, njgg jj 321==
where ( ) ( )imaginaryiimaginaryg MM=
and ( )2
jimaginarygjj rGmg = . Thus,
the gravitational forces acting on theparticle are given by
( )
( ) ( )( )( ) .22
2
jgjgjgjg
jimaginarygjimaginaryg
jimaginaryggj
rmGMriGmiM
rGmM
gMF
+==
==
==
Note that these forces are real.Remind that, the Machs principle
says that the inertial effects upon aparticle are consequence of thegravitational interaction of theparticle with the rest of the Universe.Then we can conclude that theinertial forces upon an imaginaryparticle are also real.
Equation (7) shows that , inthe case of imaginary particles, therelativistic mass is
( )( )
11
1
2222
22
=
=
=
=
cVm
cViim
cV
mM
gg
imaginaryg
imaginaryg
This expression shows thatimaginary particles can havevelocities greater than in ourordinary space-time (Tachyons).The quantization of velocity (Eq. 36)shows that there is a speed upperlimit . As we have already
calculated