Quantum Controller of Gravity

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Quantum Controller of Gravity

Fran de Aquino

To cite this version:

(2)

Fran De Aquino

Professor Emeritus of Physics, Maranhao State University, UEMA. Titular Researcher (R) of National Institute for Space Research, INPE

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 m). -20 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,

, and rest

inertial mass,

, is given by

g m 0 i m

( )

1

2

1

2

1

2

1

2 2 2 2 0 2 0 0

⎪⎭

⎪

⎬

⎫

⎪⎩

⎪

⎨

⎧

⎥

⎦

⎤

⎢

⎣

⎡

−

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

+

−

=

⎪⎭

⎪

⎬

⎫

⎪⎩

⎪

⎨

⎧

⎥

⎦

⎤

⎢

⎣

⎡

−

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

+

−

=

⎪⎭

⎪

⎬

⎫

⎪⎩

⎪

⎨

⎧

⎥

⎦

⎤

⎢

⎣

⎡

−

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛ Δ

+

−

=

c

Wn

c

m

Un

c

m

p

m

r i r i i g

ρ

χ

where is the variation in the particle’s kinetic

momentum; is the electromagnetic energy

absorbed or emitted by the particle; is the

index of refraction of the particle;

W

is the density of energy on the particle ;

p Δ U r

n

(

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 particle in a side of a lamina is

P

m

g

r

=

(gr

perpendicular to the lamina) then the weight of the same particle, in the other side of the lamina is

, where

g

m

P

r

′

=

χ

_g

r

χ

=

m

l_g

m

_il₀ ( and are respectively, the

gravitational mass and the

inertial mass of the lamina). Only when

l g

m

l i m0 1 =

χ

,

the weight is equal in both sides of the

lamina. The lamina works as a Gravity

Controller. Since

P′=

χ

P=

( )

χ

mg g=mg

( )

χ

g

,

we can consider that

m_g′ =

χ

m_g

or that

g′=

χ

g

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

(

Δ

x

≅

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.

2. The Device

Consider a spherical capacitor, as shown in Fig.1. The external radius of the inner spherical shell is , and the internal radius of the outer spherical shell is . Between the inner shell and the outer shell there is a dielectric with electric permittivity a

r

b

r

0

ε

=

_r . 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 , which is uniformly distributed at the external surface of the outer

shell, while the electric charge 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).

q

−

q

(3)

2

Fig.1 – Spherical Capacitor - A Device for Controlling Gravity developed starting from a Spherical Capacitor.

-

--

-

--

+

+ +

V

+

q

-q

V

2 =0

V

1 =V

r

b

r

a R10

Under these conditions, the electric field between the shells is given by the vectorial sum of the electric fields

E

r

_aand

E

r

_b, 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

r

_a

<

r

<

r

_b is

2 0

4 r

q

E

_R

=

_a

+

_b

=

πε

_r

ε

. In the nucleus of the capacitor and out of it, the resultant electric field is null because

E

r

_aand

E

r

_bhave opposite directions (See Fig. 2(a)).

Note that the electrostatic force, , between and will move the negative electric

charges in the direction of the positive electric

charges. This causes a displacement, , of the electric field, Fr q −

+

q

x

Δ

b

E

r

, into the outer shell (See Fig. 2 (b)). Thus, in the region with thickness

Δ

x

the intensity of the electric field is not null but equal to

E

_b.

The negative electric charges are accelerated with an acceleration,

a

, in the direction of the positive charges, in such way that they acquire a velocity, given by

r

x

a

v

= 2

Δ

(drift velocity).

The drift velocity is given by [2]

( )

2 2 2 nSe X R V nSe Z V nSe i v= = = + C

where is the positive potential applied on the inner shell (See Fig. 1);

V

fC

X

_C

=

1

2 π

is the

capacitive reactance; f is the frequency;

(

r

a

r

b

r

b

r

a

)

C

=

4 πε

−

is the capacitance of the spherical capacitor; R is the total electrical resistance of the external shell, given by

R

=

(

Δ

z

σ

S

)

+

R

10, where

Δ

z

σ

S

is the

electrical resistance of the shell ( is its thickness;

mm

z

=

5 Δ

σ

is its conductivity and is its surface area), and is a 10gigaohms resistor. Since

S

10

R

S

z

R

₁₀

>>

Δ

σ

, we can write that

. Ω × = ≅ 10 10 1 10 R R (a) (b)

Fig.2 - The displacement,Δx, of the electric field,Er_b, into the outer shell. Thus, in the region with thickness Δx the intensity of the electric field

is not null but equal to E_b.

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If the shells are made with Aluminum, with the following characteristics:

ρ

=

2700

kg

.

m

−3,

kmol kg

A=27 / ,n=N₀

ρ

A≅6×1028m−3( is the

Avogadro’s number ), and

; ; 0

N

1 26 0 6.02 10 − × = kmol N

m

r

a

=

0 .

1 r

b

=

0 .

105 m

(

)

2 2

152 .

0

4 r

z

m

S

=

π

_b

+

Δ

≅

; m r r_b−_a=5×10−3 , then R>>X_C =

(

6.8×108 f

)

ohms, ( f >1Hz), and Eq. (2) can be rewritten in the following form:

( )

3 10 8 . 6 20 10 V nSe R V nSe i v= ≅ = × −

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

force

(produced by a pulsed voltage

waveform,V ), the electrons would fluctuate

about their initial positions with the amplitude

of

Fr

m x≅1×10−20

Δ

(See Fig.3), then we get

( )

4 294 . 0 2 2 V v x a x t= Δ = Δ ≅ Δ

However, we have that

f

=

1 Δ

T

=

1

2 Δ

t

.

Thus, we get

( )

5

7 .

1 V

f

=

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

( )

6 2 2 1 2 2 1 _E _H W= ε + μ

where

E

=

E

_m

sin

ω

t

and

H

=

H

sin

ω

t

are

the instantaneous values of the electric field

and the magnetic field respectively.

It is known that

B=

μ

H

,

E

B

=

ω

k

_r

[

6 ] and

(

)

1

( )

7 1 2 2 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ₊ ₊ = = =

ωε

σ

μ

ε

κ

ω

r r r c dt dz v

where

is the real part of the

propagation vector k

r

k

r

(also called phase

*

Inside of the matter.

constant);

k

=

k

=

k

r

+

ik

i

r

;

ε

,

μ

and

σ

, are

the electromagnetic characteristics of the

medium in which the incident (or emitted)

radiation is propagating (

ε

=

ε

_r

ε

₀

;

m F/ 10 854 . 8 12 0 − × =

ε

μ

=

μ

r

μ

0

where

). It is known that for

free-space

m / H 7 0 4 10 − × =

π

μ

0 =

σ

and

ε

_r

=

μ

_r

=

1 . Then Eq. (7)

gives

c

v

=

From Eq. (7), we see that the index of

refraction

n

_r

=

c

v

is given by

(

)

1 ( )

8

1

2

_⎟

⎠

⎞

⎜

⎝

⎛

₊

=

ε

r

μ

r

σ

ωε

r

v

c

n

V 0 + −

Fig.3 - Controlling the Electric Field in the Matter at subatomic level

(

Δx≅10−20m

)

. t Δ Eb t Δ Eb

m

x

≅

1 ×

10

−20

Δ

Eb Eb Eb − − − + + + F F −

Equation (7) shows that

ω

κ

_r

=

v

. Thus,

v

k

B

(5)

4 H

v

vB

E

=

μ

Then, Eq. (6) can be rewritten in the

following form:

( )

2 2 ₂1 2

( )

9

2

1 _v _H _H

W= ε μμ + μ

For

σ

<<

ωε

, Eq. (7) reduces to

r r

c

v

μ

ε

=

Then, Eq. (9) gives

2 2 2 1 2 2 2 1 c _H _H _H W r r

μ

ε

_⎥ + = ⎦ ⎤ ⎢ ⎣ ⎡ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ =

This equation can be rewritten in the following forms:

( )

10

2

μ

B

W

=

or

( )

11 2 E W =

ε

For

σ

>>

ωε

, Eq. (7) gives

( )

12 2

μσ

ω

= v

Then, from Eq. (9) we get

( )

13 2 2 2 1 2 2 1 2 2 2 1 2 2 1 H H H H H W μ μ μ σ ωε μ μ μ μσ ω ε ≅ ≅ + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = + ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ =

Since

E =vB = v

μ

H

, we can rewrite (13) in

the following forms:

( )

14 2 2

μ

B W ≅ or

( )

15 4 2 E W ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ≅

ω

σ

Substitution of Eq. (15) into Eq. (2), gives

( )

16 1 10 758 . 1 1 2 1 1 256 1 2 1 1 4 4 1 2 1 0 4 3 2 3 27 0 4 3 2 3 2 3 0 0 2 4 3 2 i r i r i g m E f m E f c m E f c m ⎪⎭ ⎪ ⎬ ⎫ ⎪⎩ ⎪ ⎨ ⎧ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ − ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ × + − = = ⎪⎭ ⎪ ⎬ ⎫ ⎪⎩ ⎪ ⎨ ⎧ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ − ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + − = = ⎪⎭ ⎪ ⎬ ⎫ ⎪⎩ ⎪ ⎨ ⎧ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ − ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ + − = −

ρ

σ

μ

ρ

σ

μ

π

μ

ρ

π

σ

μ

Using this equation we can then calculate the gravitational mass, , of the region with thickness

( )x g

m _Δ

x

Δ

, in the outer shell. We have already seen that the electric field in this region is

E

r

_b, whose intensity is given by

E

_b

=

q

4 πε

(

r

_b

+

Δ

z

)

2. Thus, we can write that

( )

17

4

_b2 _b2 b

r

CV

r

q

E

πε

=

≅

where

C

=

4 πε

(

r

a

r

b

r

b

−

r

a

)

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

(

r

)

1 .

9

10

2

V

( )

18 r

V

r

E

a b b a b

=

₋

≅

×

Substitution of

ρ

=

2700

kg

.

m

−3,

σ

=3.5×107S /m,

1 ≅

r

μ

(Aluminum) and into Eq. (16) yields

V E E b 2 10 9 . 1 × ≅ = ( )

1

2

1

1 .

3

10

3

1

0( )

( )

19

4 2 x i x g

m

f

V

m

_Δ − Δ

⎪⎭

⎪

⎬

⎫

⎪⎩

⎪

⎨

⎧

⎥

⎦

⎤

⎢

⎣

⎡

−

×

+

−

=

Equation (5) shows that there is a

correlation between V and

to be obeyed,

i.e.,

f V

f =1.7

. By substituting this expression

into Eq. (19), we get

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For

V

=

35 .

29 Volts

(

f

=

1 .

7 V

=

60 Hz

)

†, Eq. (20) gives ( ) ( )

( )

21 91 . 0 0 ≅ = Δ Δ x i x g m m

χ

For

V

=

450 Volts

(

f

=

1 .

7 V

=

765 Hz

)

, Eq. (20) gives ( ) ( )

( )

22 04 . 0 0 ≅ = Δ Δ x i x g m m

χ

For

V

=

1200

Volts

(

f

=

1 .

7 V

=

2040

Hz

)

, Eq. (20) gives

V f

Fig.4 – The shell with thickness Δx works as a

Quantum Controller of Gravity.

Δx g 1 2 ( ) ( )

( )

23 1 . 1 0 − ≅ = Δ Δ x i x g m m

χ

In this last case, the weight of the shell with thickness

Δ

x

will be

P

r

_Δ_x

≅

−

1 .

1 m

_i₀_{( )}_Δ_x

g

r

; 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 in a side of a lamina is ( perpendicular to the lamina) then the weight of the same particle, in the other side of the lamina is

, where

g

m

P

g

r

=

gr

g

m

P

r

′

=

χ

_g

r

χ

=

m

l_g

m

_il₀ ( and are respectively, the

gravitational mass and the

inertial mass of the lamina)

[

l g

m

m_il₀

1]

. Only

when

χ

=1

, the weight is equal in both sides

of the lamina.

The lamina works as a Gravity

Controller. Since P′=

χ

P=

( )

χ

mg g=mg

( )

χ

g ,

we can consider that

mg′ =

χ

mg or that g′=

χ

g

Now consider the Spherical Capacitor previously mentioned. If the gravity below the capacitor isg, then above the first hemispherical shell with thickness (See Fig.4) it will become

x

Δ

g

χ

, and above the second hemispherical shell with thickness

Δ

x

, the gravity will be

χ

2

g

.

†

Note that the frequency must be greater than 1Hz (See text above Eq. (3)).

f

χg χ2 g χ + - Pulsed 10 G Ω

Since the voltage V is correlated to the

frequency

f

by means of the expression

V

(7)

6

Pulse Generator V, f Synchronizer V 0–1. 2kV 0–2. 04kHz g

Fig.4 – Experimental Set-up using a Quantum Controller of Gravity (QCG).

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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.