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Quantum Controller of Gravity
Fran De Aquino
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Quantum Controller of Gravity
Fran De Aquino
Professor Emeritus of Physics, Maranhao State University, UEMA.
Titular Researcher (R) of National Institute for Space Research, INPE
Copyright © 2016 by Fran De Aquino. All Rights Reserved.
A new type of device for controlling gravity is here proposed. This is a quantum device because results
from the behaviour of the matter and energy at subatomic length scale (10-20m). From the technical point of
view this device is easy to build, and can be used to develop several devices for controlling gravity.
Key words: Gravitation, Gravitational Mass, Inertial Mass, Gravity, Quantum Device.
Introduction
Some years ago I wrote a paper [1]
where a correlation between gravitational
mass and inertial mass was obtained. In the
paper I pointed out that the relationship
between gravitational mass, m g , and rest
the weight is equal in both sides of the
lamina. The lamina works as a Gravity
Controller. Since P′ = χP = (χmg )g = mg (χg ) ,
we can consider that
m ′g = χm g or that g ′ = χg
inertial mass, mi 0 , is given by
In the last years, based on these concepts, I
have proposed some types of devices for
controlling gravity. Here, I describe a device,
which acts controlling the electric field in the
Matter at subatomic level Δx ≅ 10 −20 m . This
Quantum Controller of Gravity is easy to build
and can be used in order to test the correlation
between gravitational mass and inertial mass
previously mentioned.
⎧
⎤⎫
⎡
⎛ Δp ⎞
⎪
⎪
⎢
⎟
⎜
− 1⎥ ⎬ =
χ=
= ⎨1 − 2 1 + ⎜
⎟
⎥⎪
⎢
mi 0 ⎪
⎝ mi 0 c ⎠
⎦⎭
⎣
⎩
2
mg
⎧
⎡
⎛ Unr
⎪
= ⎨1 − 2⎢ 1 + ⎜⎜
⎢
mi 0 c 2
⎝
⎪
⎣
⎩
⎧
⎡
⎛ Wn
⎪
= ⎨1 − 2⎢ 1 + ⎜⎜ 2r
⎢
⎝ ρc
⎪⎩
⎣
2
⎤⎫
⎞
⎪
⎟ − 1⎥ ⎬ =
⎟
⎥
⎠
⎦ ⎪⎭
2
⎤⎫
⎞
⎪
⎟⎟ − 1⎥ ⎬
⎥⎪
⎠
⎦⎭
(1)
where Δp is the variation in the particle’s kinetic
momentum; U is the electromagnetic energy
absorbed or emitted by the particle; nr is the
index of refraction of the particle; W is the
density of energy on the particle ( J / kg ) ; ρ is the
(
(
)
matter density kg m 3 and c is the speed of
light.
Also it was shown that, if the weight of a
r
r
r
particle in a side of a lamina is P = mg g ( g
perpendicular to the lamina) then the weight of the
same particle, in the other side of the lamina is
r
r
P ′ = χm g g , where χ = m gl mil0 ( m gl and mil0
are respectively, the gravitational mass and the
inertial mass of the lamina). Only when χ = 1 ,
)
2. The Device
Consider a spherical capacitor, as shown in
Fig.1. The external radius of the inner spherical
shell is ra , and the internal radius of the outer
spherical shell is rb . Between the inner shell and
the outer shell there is a dielectric with electric
permittivity ε = ε r ε 0 . The inner shell works as an
inductor, in such way that, when it is charged with
an electric charge + q , and the outer shell is
connected to the ground, then the outer shell
acquires a electric charge − q , which is uniformly
distributed at the external surface of the outer
shell, while the electric charge + q is uniformly
distributed at the external surface of the inner
shell (See Halliday, D. and Resnick, R., Physics,
Vol. II, Chapter 28 (Gauss law), Paragraph 28.4).
2
-
-q - - - +q +
+
+
+
+
+
+
-
capacitive
V
R10
+
rb
V1=V
+
r
ra
+ + +
- - V2=0
-
f is
the
frequency;
C = 4πε (ra rb rb − ra ) is the capacitance of the
spherical capacitor; R is the total electrical
resistance of the external shell, given
by R = (Δz σS ) + R10 , where Δz σS is the
-
+
+
+
+
reactance;
-
electrical resistance of the shell ( Δz = 5mm is
its thickness; σ is its conductivity and S is its
surface area), and R10 is a 10gigaohms resistor.
-
Since
R10 >> Δz σS ,
we
can
write
that
R ≅ R10 = 1 × 1010 Ω .
Fig.1 – Spherical Capacitor - A Device for
Controlling Gravity developed starting from a
Spherical Capacitor.
Under these conditions, the electric field
between the shells is given by the vectorial sum of
r
r
Ea and Eb ,
the electric fields
respectively
produced by the inner shell and the outer shell.
Since they have the same direction in this region,
then one can easily show that the resultant
intensity of the electric field for ra < r < rb is
ε
E R= 0
E= 0 ER=Ea + Eb
Eb
E= 0
E R= 0
Eb
+
Ea
Ea
ra
ER = Ea + Eb = q 4πεr ε 0 r 2 . In the nucleus of the
-
Eb
Ea
rb
V
capacitor and out of it, the resultant electric field
r
r
is null because Ea and Eb have opposite directions
(a)
(See Fig. 2(a)).
r
Note that the electrostatic force, F , between
− q and + q will move the negative electric
charges in the direction of the positive electric
charges. This causes a displacement, Δx , of the
Δx
ε
r
electric field, Eb , into the outer shell (See Fig. 2
(b)). Thus, in the region with thickness Δx the
intensity of the electric field is not null but equal
to Eb .
The negative electric charges are
r
accelerated with an acceleration, a , in the
direction of the positive charges, in such way that
they acquire a velocity, given by v =
(drift velocity).
The drift velocity is given by [2]
2
2
i V Z V R + XC
v=
=
=
nSe nSe
nSe
2aΔx
E R= 0
E= 0 ER=Ea + Eb
Eb
E= 0
Eb
+
Ea
ra
E R= 0
-
F
Ea
Eb
Ea
rb
V
(b)
(2)
where V is the positive potential applied on the
inner shell (See Fig. 1); X C = 1 2πfC is the
Fig.2r - The displacement, Δx , of the electric
field, E b , into the outer shell. Thus, in the region
with thickness Δ x the intensity of the electric field
is not null but equal to E b .
3
If the shells are made with Aluminum, with
the following characteristics: ρ = 2700kg.m −3 ,
A = 27kg / kmol, n = N0ρ A≅ 6×1028m−3 ( N 0 is the
Avogadro’s number N 0 = 6.02 × 1026 kmol−1 ), and
ra =0.1m; rb = 0.105m ; S = 4π (rb + Δz) ≅ 0.152m2 ;
2
(
)
rb −ra =5×10−3m, then R >> XC = 6.8×108 f ohms,
( f > 1Hz ) , and Eq. (2) can be rewritten in the
following form:
i V R10
(3)
≅
= 6.8×10−20V
nSe nSe
The maximum size of an electron has
been estimated by several authors [3, 4, 5].
The conclusion is that the electron must have
a physical radius smaller than 10-22 m * .
Assuming that, under the action of the
r
force F (produced by a pulsed voltage
waveform, V ), the electrons would fluctuate
about their initial positions with the amplitude
of Δx ≅ 1×10−20 m (See Fig.3), then we get
2Δx 2Δx 0.294
(4)
Δt =
=
≅
a
v
V
However, we have that f = 1 ΔT = 1 2Δt .
Thus, we get
(5)
f = 1.7V
Now
consider
Eq.
(1).
The
instantaneous values of the density of
electromagnetic energy in an electromagnetic
field can be deduced from Maxwell’s
equations and has the following expression
v=
W = 12 ε E2 + 12 μH 2
(6)
where E = E m sin ωt and H = H sin ωt are
the instantaneous values of the electric field
and the magnetic field respectively.
It is known that B = μH , E B = ω k r
[6] and
dz ω
c
(7)
=
v= =
dt κ r
ε r μr ⎛
2
⎞
⎜ 1 + (σ ωε ) + 1⎟
⎠
2 ⎝
where
kr
the
real part of the
r
propagation vector k (also called phase
*
is
Inside of the matter.
r
constant); k = k = k r + iki ; ε , μ and σ,
are
the electromagnetic characteristics of the
medium in which the incident (or emitted)
radiation
is
propagating
( ε = εrε0 ;
ε 0 = 8.854 × 10 −12 F / m ; μ = μ r μ 0 where
μ0 = 4π ×10−7 H / m ). It is known that for freespace σ = 0 and ε r = μ r = 1 . Then Eq. (7)
gives
v=c
From Eq. (7), we see that the index of
refraction nr = c v is given by
ε μ
c
2
nr = = r r ⎛⎜ 1 + (σ ωε ) + 1⎞⎟
⎠
v
2 ⎝
(8)
Δ x ≅ 1 × 10 −20 m
V
+
F
Δt
−
Eb
−
Δt
Eb
−
0
+
Eb
F
−
+
+
−
Eb
Eb
Fig.3 - Controlling the Electric Field in the Matter
−20
at subatomic level Δx ≅ 10 m .
(
)
Equation (7) shows that ω κ r = v . Thus,
E B = ω k r = v , i.e.,
4
E = vB = vμH
Then, Eq. (6) can be rewritten in the
following form:
(
)
(9)
W = 12 ε v2μ μH2 + 12 μH2
For σ << ωε , Eq. (7) reduces to
c
v=
ε r μr
3
⎧ ⎡
⎫
μ ⎛ σ ⎞ E 4 ⎤⎥⎪
⎪ ⎢
⎟⎟ 2 −1 ⎬mi0 =
mg = ⎨1− 2 1+ 2 ⎜⎜
⎥⎪
4c ⎝ 4πf ⎠ ρ
⎪⎩ ⎢⎣
⎦⎭
3
⎧ ⎡
⎛ μ0 ⎞⎛ μrσ ⎞ 4 ⎤⎫⎪
⎪
⎟E −1⎥⎬mi0 =
= ⎨1− 2⎢ 1+ ⎜
⎟⎜
3 2 ⎜ 2 3⎟
π
ρ
256
c
f
⎢
⎥⎦⎪
⎝
⎠
⎝
⎠
⎪⎩ ⎣
⎭
3
⎧ ⎡
⎫
⎤
⎛μ σ ⎞
⎪
⎪
= ⎨1− 2⎢ 1+1.758×10−27 ⎜⎜ r2 3 ⎟⎟E 4 −1⎥⎬mi0
⎥⎦⎪
⎝ρ f ⎠
⎪⎩ ⎢⎣
⎭
(16)
Using this equation we can then calculate the
gravitational mass, m g ( Δx ) , of the region with
Then, Eq. (9) gives
⎡ ⎛ c2 ⎞ ⎤ 2 1
⎟⎟μ⎥μH + 2 μH 2 = μH 2
W = 12 ⎢ε ⎜⎜
ε
μ
⎣ ⎝ r r⎠ ⎦
thickness Δx , in the outer shell. We have already
r
seen that the electric field in this region is Eb ,
whose intensity is given by Eb = q 4πε (rb +Δz) .
Thus, we can write that
2
This equation can be rewritten in the following
forms:
W=
B2
(10)
μ
or
(11)
W = ε E2
For σ >> ωε , Eq. (7) gives
v=
2ω
μσ
(12)
Eb ≅
4πε r
2
b
=
(17)
CV
4πε rb2
where C = 4πε (ra rb rb − ra ) is the capacitance
of the spherical capacitor; V is the potential
applied on the inner shell (See Fig. 1 and 3). Thus,
Eq. (17) can be rewritten as follows
Eb =
Then, from Eq. (9) we get
(18)
ra V
≅ 1.9×102V
rb(rb − ra )
Substitution of ρ = 2700kg.m −3 , σ = 3.5×107 S / m,
⎡ ⎛ 2ω ⎞ ⎤
⎛ ωε ⎞
W = ⎢ε⎜⎜ ⎟⎟μ⎥μH2 + 12 μH2 = ⎜ ⎟μH2 + 12 μH2 ≅
⎝σ ⎠
⎣ ⎝ μσ ⎠ ⎦
1
2
μ r ≅ 1 (Aluminum) and E = E b ≅ 1.9 × 10 2 V
(13)
≅ 12 μH2
into Eq. (16) yields
⎧ ⎡
⎫
1.3×10−2 V4 ⎤⎥⎪
⎪
mg(Δx) = ⎨1− 2⎢ 1+
−
1
⎬mi0(Δx)
3
f
⎢
⎥
⎪⎩ ⎣
⎦⎪⎭
Since E = vB = vμH , we can rewrite (13) in
the following forms:
B2
2μ
(14 )
⎛ σ ⎞ 2
W ≅⎜
⎟E
⎝ 4ω ⎠
(15 )
W ≅
q
or
Substitution of Eq. (15) into Eq. (2), gives
(19)
Equation (5) shows that there is a
correlation between V and f to be obeyed,
i.e., f = 1.7V . By substituting this expression
into Eq. (19), we get
χ=
mg(Δx)
mi0(Δx)
{ [
]}
= 1 − 2 1 + 2.64×10−3V −1
(20)
5
(f
For V = 35.29 Volts
(20) gives
χ=
mg(Δx)
mi0(Δx)
For V = 450 Volts
(20) gives
χ=
χ 2g
(21)
≅ 0.91
= 1.7V = 765Hz ) ,
Eq.
2
χg
mg(Δx)
mi0(Δx)
For V = 1200 Volts
(20) gives
χ=
(f
= 1.7V = 60 Hz ) † , Eq.
(f
(22)
≅ 0.04
= 1.7V
= 2040 Hz ) , Eq.
1
Δx
10 GΩ
V f
mg(Δx)
mi0(Δx)
(23)
≅ −1.1
+
-
g
χ
Pulsed
In this last case, the weight of the shell with
r
r
thickness Δx will be PΔx ≅ −1.1mi 0 (Δx ) g ; the
sign (-) shows that it becomes repulsive in respect
to Earth’s gravity. Besides this it is also
intensified 1.1 times in respect to its initial value.
It was shown that, if the weight of a particle
r
r r
in a side of a lamina is P = mg g ( g perpendicular
to the lamina) then the weight of the same
particle, in the other side of the lamina is
r
r
P ′ = χm g g , where χ = m gl mil0 ( m gl and mil0
are respectively, the gravitational mass and the
inertial mass of the lamina) [1]. Only
when χ = 1 , the weight is equal in both sides
of the lamina. The lamina works as a Gravity
Controller. Since P′ = χP = (χmg )g = mg (χg ) ,
we can consider that
m ′g = χm g or that g ′ = χg
Now consider the Spherical Capacitor
previously mentioned. If the gravity below the
capacitor is g , then above the first hemispherical
shell with thickness Δx (See Fig.4) it will
become χg , and above the second hemispherical
shell with thickness Δx , the gravity will be χ 2 g .
†
Note that the frequency
(See text above Eq. (3)).
f must be greater than 1Hz
Fig.4 – The shell with thickness Δ x works as a
Quantum Controller of Gravity.
Since the voltage V is correlated to the
frequency f by means of the expression
f = 1.7V (Eq. (5)), then it is necessary to put
a synchronizer before the pulse generator (See
Fig.5), in order to synchronize V with f .Thus,
when we increase the voltage, the frequency
is simultaneously increased at the same
proportion, according to Eq. (5).
6
QCG
Mechanical dynamometer
-3P
-2P
-P
Pulse Generator
V,
f
Synchronizer
Resistor
V
10 giga ohms
0
P
0–1. 2kV
0–2. 04kHz
g
Fig.4 – Experimental Set-up using a Quantum Controller of Gravity (QCG).
7
References
[1] De Aquino, F. (2010) Mathematical Foundations
of the Relativistic Theory of Quantum Gravity,
Pacific Journal of Science and Technology, 11 (1),
pp. 173-232.
Available at https://hal.archives-ouvertes.fr/hal-01128520
[2] Griffiths, D., (1999). Introduction to Electrodynamics
(3 Ed.). Upper Saddle River, NJ: Prentice-Hall, p. 289.
[3] Dehmelt, H.: (1988). A Single Atomic Particle Forever
Floating at Rest in Free Space: New Value for Electron
Radius. Physica Scripta T22, 102.
[4] Dehmelt, H.: (1990). Science 4942 539-545.
[5] Macken, J. A. Spacetime Based Foundation of Quantum
Mechanics and General Relativity. Available at
http://onlyspacetime.com/QM-Foundation.pdf
[6] Halliday, D. and Resnick, R. (1968) Physics, J. Willey &
Sons, Portuguese Version, Ed. USP, p.1118.