Quantum Gravity - Maranhao State University

7/28/2019 Quantum Gravity - Maranhao State University

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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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16

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

=


17/76

17

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


18/76

18

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


19/76

19

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


20/76

20

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.


21/76

21

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


22/76

22

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


23/76

23

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


24/76

24

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


25/76

25

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.


26/76

26

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


27/76

27

( 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 =


28/76

28

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


33/76

33

( )

[ ( )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-


34/76

34

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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35

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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36

( ) ( )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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37

( ) ( )

( )( )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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38

( ) ( )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


39/76

39

( ) ( ) dtdV

M

cVcVF

y

ip

yy

y23

2222 1

2

1

3

=

For , we can write this

expression in the following form

cVy


40/76

40

(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


41/76

41

Real Particles Imaginary Particles(Tardyons) (Tachyons)

Real Inertial Mass Imaginary Inertial Mass

Non-null Null Non-null Null

V


42/76

42

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

==

=

+

=

==


43/76

43

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


44/76

44

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


45/76

45

time interval , can only bedetected if

tEt h . Consequently,

a variation of energy E , during atime interval Et


46/76

46

+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


47/76

47

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


48/76

48

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

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Quantum Gravity - Maranhao State University