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.