kin ect i ccu an tum gravity theory

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Experimental Test for Kinetic Quantum Gravity Theory

Fran De AquinoMaranhao State University, Physics Department, S.Luis/MA, Brazil.

Copyright © 2003 by Fran De Aquino

All Rights Reserved

We propose the followingadditional experiment to check thepossibility of Gravity Control. According to the Eq.(24) ofKinetic Quantum Gravity ( physics/ 0212033) the gravitational mass of aparticle is changed when it absorbs or

emits a photon. The Eq.(30), tell usthat the gravitational mass of theelectron can become negative if it

emits a high-energy photon (γ -ray) with

frequency Hz.hcm f e

202 1021 ×=> .

There are several processes to make

an electron emits a γ -ray. Theproposed experiment is based on theinverse Compton effect. The Comptoneffect is well-known: a high-energy

photon collides with an electron initiallyat rest, producing a photon with energyless than the energy of the incidentphoton and a recoil electron. In theinverse Compton effect a high-energyelectron collides with a low-energy photonproducing a high-energy photon . Thusconsider the arrangement presentedin Fig.1 where electrons are emittedfrom a 100 MeV Betatron with velocity

=0.999986c and collide with infrared photons (wavelength = 10.6µm ) insidea evacuated tube. As shown in Fig.1,

φ+θ=30° where θ≅0°. Thus, theCompton effect theory predicts thatafter the collision each electron emits a

photon with frequency f given by:

( ) ( ) ( )

Hz.

hccvvmhc p f eeee

22

2

1032

1

×≅

≅

′−′≅′≅

At this moment, the gravitationalmass of the electron, gem , according

to the Eq.(30), becomes

ege m.c

hf m 7371

22

−≅−≅

Therefore, according to the Eq.(48),the gravitational force upon theelectron (electron-Earth) will be given

by

( )

( ) ( ) e

gege

ge

gc

hf g

c

hf g

c

hf

gmr

M Gm

r

m M GF

=

=−

−=

=−=

−=−= ⊕⊕

222

22

222µ µ

µ µ µ

This means that the acceleration due

to the gravity upon the electronbecomes

( )repulsiongg e µ =

Before the collision: µ gge −= .

Thus, the electrons go up andstrike the metal plate P with velocity

= s / m. yge 442 ≅ . Then a current of

negative charge, I G , will be observedby the galvanometer G. In order to the trajectory of theelectrons be vertical , the diagram ofthe inverse Compton effect presentedin Fig.1 must be symmetric of thediagram of Compton effect where a

photon with frequency f ≅ 2.3×1022 Hz

strikes an electron at rest producing a

photon with wavelength = 10.6µm

(φ = 30°) and giving to the electron a

recoil velocity v 'e ≅ 0.999 986c withθ≅0° ( recoil angle).



2

Fig.1 - Experimental arrangement for studying the gravitational behavior of

electrons at the Inverse Compton Effect ( photon - electron collision ).

ve' =0.999 986 c

Betatron

100MeV

G

I G

P

f = 2.3×1022 Hz-e

-e-e

g y=1m

ge

Pulsed CO2 laserwavelength = 10.6µm

f =2.83×1013 Hz

hf

hf '

hf

electron

ge

After the collision, the electron emits a

photon hf ( γ - ray ) which changes

the gravity upon the electron (Eq. 24). Under

these circumstances the new acceleration

due to gravity upon the electron is µ gg e = .

Before the collision: µ gge −= .

φ+θ

φ+θ=30°

θ≅0°

slits

Absorption

chamber

0.60m

Lampblack

(infrared absorption)

Pb

W

Metal

Lampblack

E

quartzsteel

steel



3

Fig.2 - If photons hf ' strike again with the electron of acceleration ge it will be deviated of its

vertical trajectory. So that this doesn't happen, the laser beam must be pulsed and in a

such way that ∆t >√2∆y/ge = √2 (δ / cos 30° )/ge ≅ 0.8 milliseconds ( δ = 1mm is

the diameter of the laser beam. See figure above). Therefore, the interval among the pulses must be greater than 0.8 milliseconds . The CO2 laser has a relatively long time

nearly 1 milliseconds. This is sufficient for the present experiment.

δ=1 mm

hf '

hf '

-e

-e

hf ' ge

ve'

30°

∆y=½ge(∆t)2

∆t

30°

ve'

60°

δ∆y

hf '

Laserbeam



4 The idea is that, in the total absence of gravity , the electron wouldremain at rest after the emission of the

γ -ray, because this is its initial positionof the electron at the Compton effect

where the photon strikes the electron at rest (see Fig.3(a)).

Thus, at the present experiment,the only one motion of the electronafter the collision would be causedexclusively by the gravity. Therefore, ifthe acceleration due to the gravityupon the electron has been invertedthen the electrons go up right to theplate P. At this experiment, in agreement

with the Quantum Electrodynamics, wemust consider the effects of thevacuum polarization because the

energies of the electrons are verygreater than their inertial energiesat rest (at the direction of the betatron:

( ) 2291881 cm.cvcvmc' p E eeeeee ≅′−′≅=

at the direction of the plate P:) 2222 73712 cm.cmcmhf cm E eeegee ≅≅= ).

That is to say, we must consider theelectrons inside their polarization clouds (electrons-positrons).

Based on the Eq.(11), the well-known Compton's equation becomes

( )φ λ λ coscm

h

e

−=−′ 1

Note that now we have em and not

only em as at the original expression.

This means that the Compton effect issimilar for positrons ( anti-matter). The Eq.(30) tells us that thegravitational mass of the positrons and

electrons are similar, and given by:

2

2

c

hf mge −=

Thus, the gravitational forces upon thepositrons will also be repulsive.

If ne is the number of electronsat the direction of the plate P, andaround each one of these electrons

there are N ep virtual positrons andvirtual electrons then the total

momentum Q p which they willtransfer to the plate P will be given by:

⊕ ⊕

⊕

⊕ ⊕ ⊕

Fig.4 - Polarization of the Vacuum. Virtual po sit rons and virtual electrons around the

electron.

-

Incident photon

(a)

Emitted photon

(b)

Fig. 3 - Symmetric Compton Effect.

φ

θ

ve E e

'

-e

electron at rest

pe'

E '

E =hf

p=hf / c

Scattered

photon

Recoil

electron

φ

θ

ve'

E e'

-e

The electron should remain practically at

rest after the collision with the photon. (by analogy with Fig.2 (a))

pe'

E =hf

p=hf / c

incident

electron

Incident

photon

p '

E ' p '

E ' = hf

'

p' =hf

' /c

E e' =√ m e

2c

4+p

' e

2c

2

pe' =meve

' / √ 1 - v

' e

2 / c

2

( high-energy )

( high-energy )



5

( )

( ) ygc

hf N n

ygm N nV M Q

eepe

egeepege p

22

2

2

=

===

and the force pF upon the plate by:

( )

( )

( ) eepe

epe

geepege p

gc

hf N n

gc

hf N n

gm N ng M F

=

=

−=

===

2

2

2

2

Then at the electrometer E will beobserved a voltage proportional to that

force. We know that the leptons should have length scale less than

10-19

m [1]. This means that a electronhas, at the maximum, "radius"

r e~10-19

m. The plausible relation given

by Brodsky and Drell [2] for thesimplest composite theoretical model

of the electrons, cer g

=− 2 or

ce DIRAC r gg

− , where m.c

131093 −×=

and m.g1010112 −×=− [3] gives an

electron radius mr e2210−≈ .On the other

hand, based on the uncertainty principle, we can evaluate the "radius"

r ∆ of the polarization cloud around theelectron, i.e.,

mcm.cm

c

E

c~r

ege

15

210

7371

−≅≅=

∆∆

Mathematically, the particles maximum

(electrons and positrons) inside thecloud is given by

3

er

r ~ ∆

The quantity N ep should have the sameorder of magnitude ( due to the

distribution of polarization), thus

assuming that r e<10-19

m, we can write

12

3

10>

eep r

r

~ N

∆

Therefore if ( ) A~;~ne µ 11013 , the

values of Q p and F p will be the

followings:

N .F and s / kgm.Q p p03300150 >>

This force is sufficiently intense

to be detected by the electrometer E . It is important to note that byincreasing the intensity of the electrons

beam from the betatron the force F pcan be strongly increased. Let us now consider a newsituation for the arrangementpresented in Fig.1. See Fig.5(a). We

have introduced a cathode C and an

anode A to accelerate (electrically) the

electrons to the plate P because nowthe direction of the plate P is to 90°

with respect to acceleration due to

gravity eg

.

Assuming that the accelerationdue to the electric field between A,C is

µ

gga ee −=>> , the electrons strike the

plate P with velocity yae2= . The

force pF upon the plate is now

( )( )

( ) E e N n

m

E em N n

am N na M F

epe

ge

geepe

egeepeege p

−=

=−=

===

It is easily verified that this system canworks as a powerful thrust engine inany direction . Note that the system presented

in Fig.1 can also work as an injector ofelectrons and positrons with

22 chf mge −= into a magnetic toroidal

chamber where magnetic fields give tothe electrons-positrons flux a toroidal form (analogous to the well-knowsystem of confined toroidal plasma,Tokamak ). See Fig.5(b). The electrons-positrons toroidalflux will then have a total gravitational

mass ge M such that



7

(a)

(b)

Fig. 5 - (a) A new situation for the arrangement presented in Fig.1. (b) Toroidal flux

of positrons and electrons with negative gravitational mass.

-e

P A

ae

W hf '

v'e

ge

+ _

g

-e

( )TOROIDg M

ge M

Positrons-electrons flux

C



8

Electric Field E

ON

OFF

Relativistic electrons

Laser

Heavy Electrons ( force beam ) ON

Heavy Electrons ( force beam ) OFF

Fig. 6 - Propulsion in the direction of heavy electrons flow.

+

_

++

__



9

Betatron 1 Betatron 2

Relativistic electrons ON OFF

Laser

Heavy Electrons ( force beam )

Fig. 7 - The System Sun



10

Betaron 1 Betaron 2

Betatron 1 Betaron 2

Fig.8 - Schematic Diagram of the System Sun.





11

REFERENCES

[1] Fritzsch, H. (1984) Quarks-Urstoff

unserer Welt, R. Piper GmbH&Co.

KG,München, Portuguese version (1990), Ed. Presença, Lisboa, p.215.

[2] Brodsky, S . J., and Drell, S . D., (1980)

Anomalous Magnetic Moment and Limits

on Fermion Substructure, Phys. Rev. D,

22, 2236.

[3] Dehmelt, H.G.,(1989) Experiments with

an isolated subatomic particle at rest ,

Nobel Lecture, p.590.

kin ect i ccu an tum gravity theory

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