A System to convert Gravitational Energy directly into Electrical Energy by Professor Dr. Fran De Aquino

A System to convert Gravitational Energy directly into Electrical Energy Fran De Aquino Maranhao State University, Physics Department, S.Luis/MA, Brazil. Copyright © 2012 by Fran De Aquino. All Rights Reserved. We show that it is possible to produce strong gravitational accelerations on the free electrons of a conductor in order to obtain electrical current. This allows the conversion of gravitational energy directly into electrical energy. Here, we propose a system that can produce several tens of kilowatts of electrical energy converted from the gravitational energy. Key words: Modified theories of gravity, Electric fields effects on material flows, Electron tubes, Electrical instruments. PACS: 04.50.Kd, 83.60.Np , 84.47.+w, 07.50.-e.

1. Introduction In a previous paper [1], we have proposed a system to convert gravitational energy into rotational kinetic energy (Gravitational Motor), which can be converted into electrical energy by means of a conventional electrical generator. Now, we propose a novel system to convert gravitational energy directly into electrical energy. It is known that, in some materials, called conductors, the free electrons are so loosely held by the atom and so close to the neighboring atoms that they tend to drift randomly from one atom to its neighboring atoms. This means that the electrons move in all directions by the same amount. However, if some outside force acts upon the free electrons their movement becomes not random, and they move from atom to atom at the same direction of the applied force. This flow of electrons (their electric charge) through the conductor produces the electrical current, which is defined as a flow of electric charge through a medium [2]. This charge is typically carried by moving electrons in a conductor, but it can also be carried by ions in an electrolyte, or by both ions and electrons in a plasma [3]. Thus, the electrical current arises in a conductor when an outside force acts upon the free electrons. This force is called, in a generic way, of electromotive force (EMF). Usually, it is of electrical nature (F = eE ) .

Here, it is shown that the electrical flow can also be achieved by means of gravitational forces (F = me g ) . The Gravitational Shielding Effect (BR Patent Number: PI0805046-5, July 31, 2008 [4]), shows that a battery of Gravitational Shieldings can strongly intensify the gravitational acceleration in any direction and, in this way, it is possible to produce strong gravitational accelerations on the free electrons of a conductor in order to obtain electrical current. 2. Theory From the quantization of gravity it follows that the gravitational mass mg and the inertial mass mi are correlated by means of the following factor [1]: 2 ⎧ ⎡ ⎤⎫ mg ⎪ ⎛ Δp ⎞ ⎪ ⎢ ⎟⎟ − 1⎥⎬ (1) χ= = ⎨1 − 2 1 + ⎜⎜ ⎢ ⎥⎪ mi 0 ⎪ ⎝ mi 0 c ⎠ ⎣ ⎦⎭ ⎩ where mi 0 is the rest inertial mass of the particle and Δp is the variation in the particle’s kinetic momentum; c is the speed of light. When Δp is produced by the absorption of a photon with wavelength λ , it is expressed by Δp = h λ . In this case, Eq. (1) becomes

2 2 ⎧ ⎡ ⎤⎫ ⎛ h mi 0 c ⎞ ⎪ ⎪ ⎢ = ⎨1 − 2 1 + ⎜ ⎟ − 1⎥ ⎬ ⎢ ⎥⎪ mi 0 ⎪ ⎝ λ ⎠ ⎣ ⎦⎭ ⎩ 2 ⎧ ⎡ ⎤⎫ ⎛ λ0 ⎞ ⎪ ⎪ ⎢ (2) = ⎨1 − 2 1 + ⎜ ⎟ − 1⎥ ⎬ ⎢ ⎥⎪ ⎝λ ⎠ ⎪⎩ ⎣ ⎦⎭ where λ0 = h mi 0 c is the De Broglie wavelength for the particle with rest inertial mass mi 0 . It has been shown that there is an additional effect - Gravitational Shielding effect - produced by a substance whose gravitational mass was reduced or made negative [5]. The effect extends beyond substance (gravitational shielding) , up to a certain distance from it (along the central axis of gravitational shielding). This effect shows that in this region the gravity acceleration, g 1 , is reduced at the same proportion, i.e., g1 = χ 1 g where

χ 1 = m g mi 0

and

the

gravity

acceleration before the gravitational shielding). Consequently, after a second gravitational shielding, the gravity will be given by g 2 = χ 2 g 1 = χ 1 χ 2 g , where χ 2 is the value of the ratio m g mi 0 for the second gravitational shielding. In a generalized way, we can write that after the nth gravitational shielding the gravity, g n , will be given by

g n = χ 1 χ 2 χ 3 ...χ n g

(3)

This possibility shows that, by means of a battery of gravitational shieldings, we can make particles acquire enormous accelerations. In practice, this can lead to the conception of powerful particles accelerators, kinetic weapons or weapons of shockwaves. From Electrodynamics we know that when an electromagnetic wave with frequency f and velocity c incides on a material with relative permittivity ε r , relative magnetic permeability μ r and electrical conductivity σ , its velocity is reduced to v = c nr where nr is the index of refraction of the material, given by [6]

ε μ c 2 nr = = r r ⎛⎜ 1 + (σ ωε ) + 1⎞⎟ ⎠ v 2 ⎝ σ >> ωε If , ω = 2πf , Eq. (4) reduces to nr =

μrσ 4πε0 f

(4) (5)

Thus, the wavelength of the incident radiation (See Fig. 1) becomes

λmod =

v c f λ = = = f nr nr

v=c

4π μfσ

(6)

v = c/nr

λ = c/f

λmod = v/f = c/nr f

Fig. 1 – Modified Electromagnetic Wave. The wavelength of the electromagnetic wave can be strongly reduced, but its frequency remains the same.

If a lamina with thickness equal to ξ contains n atoms/m3, then the number of atoms per area unit is nξ . Thus, if the electromagnetic radiation with frequency f incides on an area S of the lamina it reaches nSξ atoms. If it incides on the total area of the lamina, S f , then the total number of atoms reached by the radiation is N = nS f ξ . The number of atoms per unit of volume, n , is given by

N0 ρ A

(7)

where N 0 = 6.02 × 10 26 atoms / kmole is the Avogadro’s number; ρ is the matter density of the lamina (in kg/m3) and A is the molar mass(kg/kmole). When an electromagnetic wave incides on the lamina, it strikes N f front atoms,

( )

where Nf ≅ n Sf φm , φ m is the “diameter” of the atom. Thus, the electromagnetic wave incides effectively on an area S = Nf Sm , where S m = 14 πφm2 is the cross section area of one atom.

3 After these collisions, it carries out ncollisions with the other atoms (See Fig.2).

ntotal

photons N collisions

⎛ P ⎞ = ⎜⎜ 2 ⎟⎟(nl Sξ ) ⎝ hf ⎠

(12)

Substitution of Eq. (12) into Eq. (11) yields 2 ⎧ ⎡ ⎤⎫ ⎡⎛ P ⎞ λ0 ⎤ ⎪ ⎪ ⎢ = ⎨1 − 2 1 + ⎢⎜⎜ 2 ⎟⎟(nl Sξ ) ⎥ −1⎥⎬ (13) ⎢ ⎥ mi 0(l ) ⎪ λ ⎥⎦ ⎢⎣⎝ hf ⎠ ⎥⎦⎪⎭ ⎣⎢ ⎩

mg (l )

atom Sm Wave

Substitution of P given by Eq. (9) into Eq. (13) gives

Fig. 2 – Collisions inside the lamina.

Thus, the total number of collisions in the volume Sξ is

Ncollisions= Nf +ncollisions= nlSφm +(nlSξ −nmSφm) =

(8)

= nlSξ

The power density, D , of the radiation on the lamina can be expressed by P P D= = S N f Sm

(9 )

We can express the total mean number of collisions in each atom, n1 , by means of the following equation n1 =

ntotal

photons N collisions

2 ⎧ ⎡ ⎤⎫ ⎡⎛ N f SmD⎞⎛ n Sξ ⎞ 1 ⎤ ⎥⎪ ⎪ ⎢ ⎟⎜ l ⎟ ⎥ −1 ⎬ (14) = ⎨1− 2 1+ ⎢⎜⎜ 2 ⎟⎜ ⎥ ⎟ mi0(l ) ⎪ ⎢ ⎢⎣⎝ f ⎠⎝ mi0(l )c ⎠ λ ⎥⎦ ⎥⎪ ⎢ ⎦⎭ ⎩ ⎣

mg(l )

Substitution of N f ≅ (nl S f )φm and into Eq. (14) results

2 ⎧ ⎡ ⎤⎫ ⎡⎛ nl3S 2f Sm2φm2ξD ⎞ 1 ⎤ ⎪ ⎢ ⎟ ⎥ −1⎥⎪⎬ (15) = ⎨1− 2⎢ 1+ ⎢⎜ 2 ⎥ mi0(l ) ⎪ ⎢⎣⎜⎝ mi0(l )cf ⎟⎠ λ ⎥⎦ ⎢⎣ ⎥⎦⎪ ⎩ ⎭ where mi 0(l ) = ρ (l )V(l ) . Now, considering that the lamina is inside an ELF electromagnetic field with E and B , then we can write that [7]

mg(l )

(10 )

Since in each collision a momentum h λ is transferred to the atom, then the total momentum transferred to the lamina will be Δp = (n1 N ) h λ . Therefore, in accordance with Eq. (1), we can write that 2 ⎧ ⎡ ⎤⎫ λ0 ⎤ ⎡ ⎪ ⎪ ⎢ = ⎨1− 2 1+ ⎢(n1 N) ⎥ −1⎥⎬ = ⎥⎪ mi0(l ) ⎪ ⎢ λ⎦ ⎣ ⎦⎭ ⎩ ⎣ 2 ⎧ ⎡ ⎤⎫ λ0 ⎤ ⎡ ⎪ ⎪ ⎢ = ⎨1− 2 1+ ⎢ntotal photonsNcollisions ⎥ −1⎥⎬ (11) ⎥⎪ λ⎦ ⎣ ⎪⎩ ⎢⎣ ⎦⎭

mg(l )

Since Eq. (8) gives N collisions = nl Sξ , we get

S = N f Sm

n r (l ) E 2 2μ 0 c

(16 )

Substitution of Eq. (16) into Eq. (15) gives 2 ⎧ ⎡ ⎤⎫ ⎡⎛ nr(l )nl3S2f Sm2φm2ξE2 ⎞ 1 ⎤ ⎥⎪ ⎪ ⎢ ⎟ ⎥ −1 (17) = ⎨1− 2 1+ ⎢⎜ ⎜ 2μ0mi0(l )c2 f 2 ⎟ λ ⎥ ⎥⎬ mi0(l ) ⎪ ⎢ ⎢ ⎠ ⎦ ⎥⎪ ⎣⎝ ⎢ ⎦⎭ ⎩ ⎣

mg(l )

In the case in which the area S f is just the area of the cross-section of the lamina (Sα ) , we obtain from Eq. (17), considering that mi0(l) = ρ(l)Sαξ , the following expression

2 ⎧ ⎡ ⎤⎫ ⎡⎛ nr(l )nl3Sα Sm2φm2 E2 ⎞ 1 ⎤ ⎥⎪ ⎪ ⎢ ⎟ ⎥ −1 ⎬ (18) = ⎨1− 2 1+ ⎢⎜ 2 2 ⎟ ⎥ ⎜ mi0(l ) ⎪ ⎢ 2 μ ρ c f ⎢⎣⎝ 0 (l ) ⎠ λ ⎥⎦ ⎥⎪ ⎢ ⎦⎭ ⎩ ⎣ According to Eq. (6) we have v c (19 ) λ = λ mod = = f nr (l ) f

N 0 ρ air (23) = 5.16×1025 atoms/ m3 AN The parallel metallic plates (p), shown in Fig.3 are subjected to different drop voltages. The two sets of plates (D), placed on the extremes of the tube, are subjected to V ( D )rms = 134.72 V at f = 60 Hz , while the

Substitution of Eq. (19) into Eq. (18) gives

central set of plates (A) is subjected to V( A )rms = 273.98 V at f = 60 Hz . Since

mg(l )

⎧ ⎡ 4φ 4 E4 ⎤⎫ nr4(l )nl6Sα2 Sm m ⎪ ⎢ ⎥⎪ 1 = ⎨1 − 2⎢ 1 + − ⎥⎬ (20) 2ρ 2 c6 f 2 mi0(l ) ⎪ μ 4 ⎢ ⎥⎪ 0 (l ) ⎣ ⎦⎭ ⎩ mg(l )

Note that E = E m sin ωt .The average value for E 2 is equal to

sinusoidaly ( E m

is the maximum value

E m2 because E varies

for E ). On the other hand, E rms = E m

4 Consequently we can change E 4 by E rms , and the equation above can be rewritten as follows m g (l ) = χ= mi0(l )

⎧ ⎡ ⎤⎫ 4φ 4 E4 nr4(l ) nl6 Sα2 S m m rms ⎪ ⎢ ⎥⎪ = ⎨1 − 2⎢ 1 + − 1⎥ ⎬ (21) 2 2 6 2 4μ 0 ρ (l )c f ⎪ ⎢ ⎥⎪ ⎣ ⎦⎭ ⎩

Now consider the system shown in Fig.3. It was designed to convert Gravitational Energy directly into Electrical Energy. Thus, we can say that it is a Gravitational EMF Source. Inside the system there is a dielectric tube ( ε r ≅ 1) with the following characteristics:

α = 8mm (diameter), Sα =πα2 4 = 5.03×10−5 m2 . Inside the tube there is a Lead sphere ( ρ s = 11340 Kg / m 3 ) with 4mm radius and mass M gs = 3.04 × 10 −3 kg . The tube is filled with air at ambient temperature and 1atm. Thus, inside the tube, the air density is (22 ) ρ air = 1 . 2 kg .m − 3 The number of atoms of air (Nitrogen) per unit of volume, n air , according to Eq.(7), is given by

nair =

d = 14mm , then the intensity of the electric field, which passes through the 36 cylindrical air laminas (each one with 5mm thickness) of the two sets (D), is E( D)rms = V( D)rms d = 9.623×103V / m

and the intensity of the electric field, which passes through the 19 cylindrical air laminas of the central set (A), is given by E( A)rms = V( A)rms d = 1.957 × 10 4 V / m

Note that the metallic rings (5mm thickness) are positioned in such way to block the electric field out of the cylindrical air laminas. The objective is to turn each one of these laminas into a Gravity Control Cells (GCC) [5]. Thus, the system shown in Fig. 3 has 3 sets of GCC. Two with 18 GCC each, and one with 19 GCC. The two sets with 18 GCC each are positioned at the extremes of the tube (D). They work as gravitational decelerator while the other set with 19 GCC (A) works as a gravitational accelerator, intensifying the gravity acceleration produced by the mass M gs of the Lead sphere. According to Eq. (3), this gravity, after the 19 th GCC becomes g19 = χ19GMgs r02 , where χ = mg (l ) mi (l ) given by Eq. (21) and r0 = 9mm is the distance between the center of the Lead sphere and the surface of the first GCC of the set (A). The objective of the sets (D), with 18 GCC each, is to reduce strongly the value of the external gravity along the axis of the tube. In this case, the value of the external gravity, g ext , is reduced by the factor χ d18 gext , where χ d = 10−2 . For example, if the base BS of the system is positioned on the Earth

d = 14 mm

Dielectric tube εr ≅ 1

α=8mm Air

ξ = 5 mm

Air 1 atm 300K

Air

ED(rms)

2 Air

Metallic ring (5 mm thickness)

Parallel plate capacitor (p)

Lead sphere (8 mm diameter)

18 Mg

Electric field of capacitor

134.72V 60Hz

Air

r0=9mm 1

Air

Iron rod (8 mm diameter, 100 mm height)

Air

EA(rms)

Aluminum

273.98V 60Hz

Fe +Fp+Fn = 0

Inductor 1 turn (inner radius: r =5mm)

ΣF=0

is = 58.6A, # 1 AWG

xB = 100 mm

Air

i=4A

# 14 AWG

220V 60Hz 128.9 kW

Nucleus μ r = 60 (iron cast)

5Ω 80W

17 18

Cable 500AWG

Np =100 turns Ns =10 turns

Air

2.2kV 60Hz

2.5m T

100 mm

20V 1μHz

Relay Initialization source ~

Converter 60Hz / 1μHz

Transformer

Fig. 3 – A Gravitational EMF Source (Developed from a process patented in July, 31 2008, PI0805046-5)

surface, then gext = 9.81m / s

is reduced

to χ d18 gext and, after the set A, it is increased by χ 19 . Since the system is designed for χ = −6.4138 , then the gravity acceleration on the sphere becomes −20 19 18 2 χ χd gext = 2.1×10 m/ s , this value is much

protons (Fp ) and neutrons (Fp ) of the Iron rod, are respectively expressed by the following relations M gs ⎞ ⎛ (28) Fe = mge ae = χ Be me ⎜⎜ χ 7 G 2 ⎟⎟ r0 ⎠ ⎝

(Fe ) ,

M gs ⎞ ⎛ Fp = mgp a p = χ Bp m p ⎜⎜ χ 7 G 2 ⎟⎟ r0 ⎠ ⎝

smaller than gsphere = GMgs rs2 =1.27×10−8 m/ s 2 . The values of χ and χ d , according to Eq. (21) are given by ⎧ ⎡ ⎤⎫ 6 S 2 S 4φ 4 E4 nr4(air) nair α m m ( A)rms ⎥⎪ ⎪ ⎢ −1⎥⎬ = χ = ⎨1 − 2⎢ 1 + 2 c6 f 2 4μ02 ρ air ⎪ ⎢ ⎥⎪ ⎣ ⎦⎭ ⎩ ⎧ ⎫ = ⎨1 − 2⎡⎢ 1 +1.44×10−16 E(4A)rms −1⎤⎥⎬ ⎣ ⎦⎭ ⎩

(24)

⎧ ⎡ ⎤⎫ 6 S 2 S 4φ 4 E4 nr4(air) nair α m m ( D)rms ⎥⎪ ⎪ ⎢ −1⎥⎬ = χ d = ⎨1− 2⎢ 1 + 2 c6 f 2 4μ02 ρ air ⎪ ⎢ ⎥⎪ ⎣ ⎦⎭ ⎩ ⎧ ⎫ = ⎨1 − 2⎡⎢ 1 +1.44×10−16 E(4D)rms −1⎤⎥⎬ ⎣ ⎦⎭ ⎩ where n r ( air ) =

(25)

ε r μ r ≅ 1 , since (σ << ωε ) * ;

nair = 5.16 ×1025 atoms/ m3 , φm = 1.55×10−10 m , Sm = πφm2 4 = 1.88×10−20m2 Since

and

f = 60 Hz .

E ( A )rms = 1 . 957 × 10 4 V / m

E ( D )rms = 9 .623V × 10 3 / m , we get

χ = −6.4138

and

(26)

and

χ d ≅ 10−2

(27)

Note that there is a uniform magnetic field, B , through the Iron rod. Then, the gravitational forces due to the gravitational mass of the sphere (M gs ) acting on electrons *

The electrical conductivity of air, inside the dielectric tube, is equal to the electrical conductivity of Earth’s atmosphere near the land, whose average value is

σ air ≅ 1 × 10 −14 S / m [8].

(29)

M gs ⎞ ⎛ (30) Fn = mgn an = χ Bn mn ⎜⎜ χ 7 G 2 ⎟⎟ r0 ⎠ ⎝ The factors χ B are due to the electrons, protons and neutrons are inside the magnetic field B. In order to make null the resultant of these forces in the Iron (and also in the sphere) we must have Fe = F p + Fn , i.e., m e χ Be = m p χ Bp + m n χ Bn

(31)

It is important to note that the set with 19 GCC (A) cannot be turned on before the magnetic field B is on. Because the gravitational accelerations on the Iron rod and Lead sphere will be enormous χ 19 GM gs r02 ≅ 5.4 × 10 6 m / s 2 , and will

(

)

explode the device. The force Fe is the electromotive force (EMF), which produces the electrical current. Here, this force has gravitational nature. The corresponding force of electrical nature is Fe = eE . Thus, we can write that (32 ) m ge a e = eE The electrons inside the Iron rod (See Fig. 3) are subjected to the gravity acceleration produced by the sphere, and increased by the 19 GCC in the region (A). The result is M gs (33 ) a e = χ 19 g s = χ 19 G 2 r0 Comparing Eq. (32) with Eq.(33), we obtain ⎛ m ge ⎞ 19 M gs ⎟⎟ χ G 2 (34) E = ⎜⎜ r0 ⎝ e ⎠ The electron mobility, μ e , considering various scattering mechanisms can be obtained by solving the Boltzmann equation

in the relaxation time approximation. The result is [9] eτ (35) μe = m ge where τ is the average relaxation time over the electron energies and

m ge

is the

gravitational mass of electron, which is the effective mass of electron. Since τ can be expressed by

τ = m geσ ne 2 [10], then Eq. (35) can be written as follows μe =

(36)

Thus, the drift velocity will be expressed by σ ⎛ m ge ⎞ 19 M gs ⎟χ G 2 (37) v d = μ e E = ⎜⎜ ne ⎝ e ⎟⎠ r0 and the electrical current density expressed by ⎛ m ge ⎞ 19 M gs ⎟⎟ χ G 2 (38 ) j e = ρ qe v d = σ iron ⎜⎜ e r0 ⎝ ⎠ where ρ qe = ne , and m ge = χ Be me . Therefore, Eq. (38) reduces to M gs ⎛m ⎞ (39) je = σ iron ⎜ e ⎟χ Be χ 19G 2 r0 ⎝ e ⎠ In order to calculate the expressions of χ Be , χ Bp and χ Bn we start from Eq. (17), for the particular case of single electron in the region subjected to the magnetic field B (Iron rod). In this case, we must substitute nr (l ) by

n riron = (μ r (iron )σ iron 4πε 0 f )2 ; 1

1 Ve = 1 43 πre3 ( re is the electrons radius), S f

by (SSAe )ρ eVe ( SSAe is the specific surface area for electrons in this case: SSAe = 12 Ae me = 12 Ae ρ e Ve = 2πre2 ρ eVe ),

S m by S e = π re2 , ξ by φ m = 2re and mi 0(l ) by me . The result is ⎧⎪

2 ⎤⎫⎪ E4 45.56π 2 re4 nriron − 1 ⎥⎬ (40) μ 02 me2 c 4 f 4 λ2 ⎢⎣ ⎥⎦⎪⎭ ⎪⎩ Electrodynamics tells us that E rms = vBrms = (c n r (iron ) )Brms , and Eq. (19)

⎡

χ Be = ⎨1 − 2⎢ 1 +

gives λ = λmod = (4π μironσ iron f ) . Substitution of these expressions into Eq. (40) yields ⎧⎪ ⎡ ⎤⎫⎪ 45.56π 2r4B4 (41) χBe = ⎨1− 2⎢ 1+ 2 2 2e 2rms −1⎥⎬ μ0 me c f ⎥⎦⎪⎭ ⎪⎩ ⎢⎣ Similarly, in the case of proton and neutron we can write that 4 ⎧ ⎡ ⎤⎫ 45.56π 2rp4 Brms ⎪ ⎪ ⎢ (42) χBp = ⎨1− 2 1+ 2 2 2 2 −1⎥⎬ ⎥⎪ μ0 mpc f ⎪⎩ ⎢⎣ ⎦⎭ 4 ⎧ ⎡ ⎤⎫⎪ 45.56π 2 rn4 Brms ⎪ χBn = ⎨1− 2⎢ 1+ 2 2 2 2 −1⎥⎬ ⎥⎦⎪ μ0 mn c f ⎪⎩ ⎢⎣ ⎭

(43)

The radius of free electron is re = 6.87×10−14 m (See Appendix A) and the radius of protons inside the atoms (nuclei) is rp = 1.2 ×10−15 m , rn ≅ r p , then we obtain from Eqs. (41) (42)

and (43) the following expressions: ⎧ ⎡ ⎤⎫⎪ B4 ⎪ ⎥⎬ (44) − χ Be = ⎨1− 2⎢ 1+ 8.49×104 rms 1 2 ⎢ ⎥ f ⎪⎩ ⎣ ⎦⎪⎭ ⎧ ⎡ ⎤⎫⎪ B4 ⎪ ⎥⎬ (45) − χ Bn ≅ χ Bp = ⎨1− 2⎢ 1+ 2.35×10−9 rms 1 2 f ⎢ ⎥⎦⎪ ⎪⎩ ⎣ ⎭ Then, from Eq. (31) it follows that (46 ) m e χ Be ≅ 2m p χ Bp Substitution of Eqs. (44) and (45) into Eq. (46) gives 4 ⎧ ⎡ ⎤ ⎫⎪ ⎪ 4 Brms ⎢ ⎥⎬ − + × − 1 2 1 8 . 49 10 1 ⎨ 2 f ⎢ ⎥⎦ ⎪ ⎪⎩ ⎣ ⎭ = 3666.3 (47) 4 ⎧ ⎡ ⎤ ⎫⎪ ⎪ −9 Brms ⎢ ⎥⎬ − + × − 1 2 1 2 . 35 10 1 ⎨ f2 ⎢⎣ ⎥⎦ ⎪ ⎪⎩ ⎭ For f = 1μHz , we get

{1 − 2[ 1 + 8.49 × 10 B {1 − 2[ 1 + 2.35 × 10 B 6

−7

4 rms 4 rms

]} = 3666.3 − 1]}

−1

whence we obtain B rms = 2 .5 m T Consequently, Eq. (44) and (45) yields χ Be = −3666.3 and

(48)

(49 ) (50)

(51)

χ Bn ≅ χ Bp ≅ 0.999

In order to the forces Fe and F p have contrary direction (such as occurs in the case, in which the nature of the electromotive force is electrical) we must have χ Be < 0 and χ Bn ≅ χ Bp > 0 (See equations (28) (29) and (30)), i.e., 4 ⎧⎪ ⎡ ⎤⎫⎪ 4 Brms ⎢ ⎥⎬ < 0 1 − 2 1 + 8 . 49 × 10 − 1 ⎨ 2 f ⎢ ⎥⎦⎪⎭ ⎪⎩ ⎣ and 4 ⎧⎪ ⎡ ⎤⎫⎪ −9 Brms ⎢ ⎥⎬ > 0 − + × − 1 2 1 2 . 35 10 1 ⎨ 2 f ⎢ ⎥⎦⎪⎭ ⎪⎩ ⎣ This means that we must have 0.06 f < B rms < 151 .86 f

M gs ⎛m ⎞ je = σ iron ⎜ e ⎟ χ Be χ 19G 2 r0 ⎝ e ⎠ Since σ iron = 1.03 × 10 7 S / m , χ = −6.4138 ,

χ Be = −3666 .3 ,

M gs = 3.04 × 10 −3 kg

and

r0 = 9mm , we obtain

(52)

je = 1.164×106 A / m2 Given that Sα = πα 2 4 = 5.03 × 10−5 m 2 we get

isource = je Sα ≅ 58.6 A

(53)

The resistance of the Iron rod is

(54 )

In the case of f = 1μHz = 10 Hz the result is −6

(55) 6.5×10−5 T < Brms < 0.151T Note the cylindrical format (1turn, r = 5mm ) of the inductor (Figs. 3 and 6). By using only 1 turn it is possible to eliminate the capacitive effect between the turns. This is highly relevant in this case because the extremely-low frequency f = 1μHz would strongly increase the capacitive reactance ( X C ) associated to the inductor. When a current i passes through this inductor, the value of B inside the Iron rod is given by B = μ r μ0i xB where x B = 100mm is inductor’s length and μ r = 4000 (very pure Iron). However, the effective permeability is defined as μr(eff) = μr 1+(μr −1) Nm, where N m is the average demagnetizing factor [11]. Since the iron rod has 5mm diameter and 100mm height, then we obtain the factor γ = 100mm / 5mm = 20 which gives N m = 0.02 (See table V[12]). Therefore, we obtain μr (eff ) = 49.4 . Thus, for Brms = 2.5mT (See Eq. (49)) ,i.e., B = μ r (eff ) μ 0 i x B = 2.5mT , the value

of i must be i = 4 A . Then, the resistor in Fig.3. must have 20V 4 A = 5Ω . The dissipated power is 80W. Let us now calculate the current density through the Iron rod (Fig. 3). According to Eq. (39) we have

⎛ xB Rsource = ⎜⎜ ⎝ σ iron Sα

⎞ ⎟⎟ = 1.93 × 10 −4 Ω ⎠

Thus, the dissipated power by the Iron rod is 2 Pd = Rsource isource ≅ 0.66W

(56)

Note that this Gravitational EMF source is a Current Source. As we known, a Current Source is a device that keeps invariable the electric current between its terminals. So, if the source is connected to an external load, and the resistance of the load varies, then the own source will increase or decrease its output voltage in order to maintain invariable the value of the current in the circuit.

Rsource

Vsource

isource

R Load

Fig. 4 – Current Source

Based on Kirchhoff’s laws we can express the electric voltage between the terminals of the Current Source, Vs , by means of the following relation (See Fig.4): V source = R source i source + V where V is the voltage applied on the charge. The transformer T connected to Gravitational EMF Source (See Fig. 3) is

designed † to make the voltage V = 2.2kV @60Hz. Since Rsource isource <<V , then we can write that V source ≅ V . Thus, in the primary circuit, the voltage is V p = V source ≅ V = 2.2kV and the current is i p = isource = 58.6 A ; the winding turns ratio

is N p N s = 10 ; thus, in the secondary circuit the output voltage is Vs = 220V @ 60 Hz and the current is is = 586A . Consequently, the source output power is P = V s i s = 128 .9 kW Note that, in order to initializing the Gravitational EMF Source, is used an external source, which is removed after the initialization of the Gravitational EMF Source. Now it will be shown that this Gravitational EMF source can be miniaturized. We start making x B = 10mm and ξ = 0.5mm ; α = 2mm, d A = 8mm , d D = 16mm and r0 = 1.5mm (See Fig. 5). The sphere with 2mm diameter is now of Tungsten carbide (W+Cobalt) with 15,630kg / m3 density. −6

Then M gs = 6.54×10−5 kg and Sα = 3.14×10 m . 2

Thus, for f = 1μHz Eq.(21) gives ⎧

⎫

(57) χ = ⎨1 − 2⎡⎢ 1 + 2.03×10 −3 E(4A)rms − 1⎤⎥ ⎬ ⎣ ⎦⎭ ⎩ For V A(rms) = 93.5mV and d A = 8mm we get E( A)rms = 11.688V / m , and Eq. (57) yields χ = −6.236 For VD(rms) = 93.5mV and d D = 16mm we get E( D )rms = 5.844V / m and χ d ≅ −0.01 . †

The impedances are respectively,

( = 2πf (μ μ

)

Z p ≅ 2πfL p = 2πf μ r μ 0 N p2 A p l p = 14.36Ω

)

2 −4 Z s ≅ 2πfLs r 0 N s As l s = 6.154 × 10 Ω Z p (total ) = Z p + Z reflected = Z p + (N p N s )Z s = 37.56Ω

where μr = 60 (iron cast), N p = 100, Ns = 10,

l p = l s = 0.18m , φ p = 107.6mm , φs = 136.8mm , A p = 0 .0091 m 2 , A s = 0 .0147 m 2 .

9 Since and σ iron = 1.03×10 S / m = −3666.3 , then the value of je is M gs ⎛m ⎞ je = σ iron ⎜ e ⎟ χ Be χ 19G 2 = r0 ⎝ e ⎠ 7

χ Be

= 5.29 × 105 A / m 2 and

isource = je Sα ≅ 1.66 A The resistance of the iron rod is given by

⎛ xB Rsource = ⎜⎜ ⎝ σ iron Sα

⎞ ⎟⎟ = 3.1× 10 −4 Ω ⎠

Thus, the dissipated power by the Iron rod is 2 Pd = Rsourceisource ≅ 0.9mW

In the case of the miniaturized source, the iron rod has 2mm diameter and 10mm height, then we obtain the factor which gives γ = 10mm / 2mm = 5 N m = 0.06 (See table V[12]). Therefore, we obtain μ r (eff ) = 16.6 . Since

Vs = V A( rms) = V D( rms) = 93.5mV

and the resistance of the resistor R1 is 21.6mΩ / 31mW (See Fig.5), then the current from the first source is i = 1.2A. Thus, we get B = μr (eff ) μ0i xB = 2.5mT . Since the current through the second source is isource = 1.66 A , and, if the voltage required by the charge, is V = 3.7V (usual lithium batteries’ voltage), then the source voltage is given by Vsource = Rsource i source + V ≅ 3.7V Consequently, the miniaturized source can provide the power:

P = V source i sorce = (3 . 7V )1 . 66 A ≅ 6 . 1W This is the magnitude of the power of lithium batteries used in mobiles. Note that the miniaturized source of Gravitational EMF does not need to be recharged and it occupies a volume (8mm x 70mm x 80mm. See Fig.6) similar to the volume of the mobile batteries. In addition, note that the dimensions of this miniaturized source can be further reduced (possibly down to a few millimeters or less).

dA = 8 mm dD = 16 mm

Metallic ring (0.5 mm thickness)

ξ = 0.5 mm

Air Dielectric tube εr ≅ 1

α=2mm

Air

1 atm 300K

ED(rms) 93.5mV Tungsten carbide sphere (2 mm diameter) Lead sphere (2 mm diameter)

dA = 8 mm dD = 16 mm

1μHz

2 Air

EA(rms)

Printed circuit board

93.5mV 1μHz

xB = 10 mm

Vs=93.5mV i=1.2 A

93.5mV 1μHz

Initialization source

Air

is=1.66A Vs=V=3.7V

80mm

Air

1μHz

V=3.7V

2.5m T

93.5mV

rectifier

Air

2.5m T Cylindrical Inductor 1 turn (inner radius: r = 2mm)

R1 21.6mΩ / 31mW

Full wave

r0=1.5mm Air

Iron rod (2 mm diameter, 10 mm height)

3.7V DC 6.1W

Air

70 mm

Fig. 5 – A Miniaturized Source of Gravitational EMF

11 Iron rod (8 mm diameter, 100mm height)

18mm

134,72V/60Hz

Aluminum

273.98V/60Hz

Inductor 1 turn (inner radius: R=5mm)

18mm 3 8 3

1 12 1

Cross-section

2.2kV/60Hz

220V 60Hz 128.9 kW

μ r μ0i

= 2.5mT 2r i = 4 A; r = 5mm μ r ≅ 49.4

5Ω 80W

20V 1μHz Initialization source

Transformers and Converter (60Hz/1μHz) (inside)

High-power Source 12mm 2 1

Cross-section

μ r μ0i

93.5mV

= 2.5mT

2r i = 1.2 A; r = 2mm μ r ≅ 16.6

93.5mV

1μHz

3.7V DC 6.1W 80mm

Full wave rectifier

Initialization source •

••

6 1

70mm

8mm thickness

Low-power Source

Fig. 6 – Schematic Diagram in 3D of the Gravitational EMF Sources

Appendix A: The “Geometrical Radii” of Electron and Proton It is known that the frequency of oscillation of a simple spring oscillator is

f =

1 2π

K m

( A1)

where m is the inertial mass attached to the spring and K is the spring constant (in N·m−1). In this case, the restoring force exerted by the spring is linear and given by

F = − Kx

( A2)

where x is the displacement from the equilibrium position. Now, consider the gravitational force: For example, above the surface of the Earth, the force follows the familiar Newtonian function, i.e., F = − GM g ⊕ m g r 2 , where M g ⊕ the mass of Earth is, m g

Making x = r = R⊕ , and substituting (A4) into (A1) gives

1 GMg⊕ χ 2π R⊕3

( A5)

In the case of an electron and a positron, we substitute M g ⊕ by m ge , χ by χ e and R⊕ by

Re , where Re is the radius of electron (or positron). Thus, Eq. (A5) becomes

1 Gmge χe 2π Re3

( A6)

The value of χ e varies with the density of energy [1]. When the electron and the positron are distant from each other and the local density of energy is small, the value of χ e becomes very close to 1. However, when the electron and the positron are penetrating one another, the energy densities in each particle become very strong due to the proximity of their electrical charges e and, consequently, the value of χ e strongly increases. In order to calculate the value of χ e under these conditions ( x = r = Re ), we start from the expression of correlation between electric charge q and gravitational mass, obtained in a previous work [1]:

is the

gravitational mass of a particle and r is the distance between the centers. Below Earth’s surface the force is linear and given by GMg⊕mg ( A3) F =− r R⊕3 where R⊕ is the radius of Earth. By comparing (A3) with (A2) we obtain GMg⊕ ⎛ r ⎞ K K ( A4) = = 3 ⎜ ⎟ mg χ m R⊕ ⎝ x ⎠

q = 4πε 0 G m g (imaginary ) i where

m g (imaginary )

the

( A7 ) imaginary

gravitational mass, and i = − 1 . In the case of electron, Eq. (A7) gives qe = 4πε0G mge(imaginary) i =

(

)

= 4πε0G χ e mi0e(imaginary)i =

( 4πε G (

= 4πε0G − χe =

2 3

)

mi0e(real)i 2 =

)

χe mi0e(real) = −1.6 ×10−19 C ( A8)

where we obtain

χ e = −1.8 × 10 21

( A9)

This is therefore, the value of χ e increased by the strong density of energy produced by the electrical charges e of the two particles, under previously mentioned conditions.

Given that m ge = χ e mi 0 e , Eq. (A6) yields

1 Gχ e2 mi0e f= 2π Re3

( A10)

From Quantum Mechanics, we know that

( A11)

hf = mi 0 c 2

where h is the Planck’s constant. Thus, in the case of mi 0 = mi 0e we get mi 0e c 2 f= h

( A12)

By comparing (A10) and (A12) we conclude that mi 0e c 2 1 = h 2π

Gχ e2 mi 0e Re3

( A13)

Isolating the radius Re , we get: 1

⎛ G ⎞ 3 ⎛ χeh ⎞ 3 ⎟⎟ ⎜ Re = ⎜⎜ ⎟ = 6.87×10−14 m 2 m ⎝ i 0e ⎠ ⎝ 2π c ⎠

( A14)

Compare this value with the Compton sized electron, which predicts Re = 3.86 × 10 −13 m and also with standardized result recently obtained of Re = 4 − 7 × 10 −13 m [13]. In the case of proton, we have q p = 4πε0G mgp(imaginary) i =

(

)

= 4πε0G χ p mi0 p(imaginary)i =

( 4πε G(

= 4πε0G − χ p =

2 3

)

mi0 p(real)i 2 =

)

χ p mi0 p(real) = −1.6×10−19 C ( A15)

where we obtain

χ p = −9.7 × 1017

13 Thus, the result is

( A16)

⎛ G ⎞3 ⎛ χ ph ⎞ 3 ⎟ ⎜ ⎟ = 3.72×10−17 m ( A17) Rp = ⎜ ⎜ mi0 p ⎟ ⎜ 2π c 2 ⎟ ⎠ ⎝ ⎠ ⎝ Note that these radii, given by Equations ( A14) and ( A17) , are the radii of free electrons and free protons (when the particle and antiparticle (in isolation) penetrate themselves mutually). Inside the atoms (nuclei) the radius of protons is well-known. For example, protons, as the hydrogen nuclei, have a radius given by R p ≅ 1.2 × 10−15 m [14, 15]. The strong

increase in respect to the value given by Eq. (A17) is due to the interaction with the electron of the atom.

14 Appendix B: An Experimental Setup for Testing a GCC with Air Nucleus

Digital Force Gauge

(± 2 N ; 0.01N )

888888

g′ = χ g Dielectric tube (Acrylic)

Acrylic plates

5 mm GCC 10 mm

5 mm

g φ = 8 mm d = 14 mm

Rectangular plate

Dielectric tube Metallic ring

d = 14 mm 20 mm

Metallic plate

8 mm

Metallic rings

DETAILS OF PARTS 5 mm

30 mm Metal

5 mm

Metal

5 mm 8 mm

Acrylic

20 mm

10 mm

Metal

5 mm

12 mm

16 DETAIS OF THE BOX 26 mm

Acrylic 3 plates 20mm x 26 mm x 2mm 2 plates 20mm x 20mm x 2mm 2 plates 20mm x 20mm x 1mm (inner plates) 4 plates 20mm x 5mm x 2mm Aluminum 2 plates 20mm x 10mm

20 mm

2 mm

26 mm

20 mm

5 mm

20mm

2 mm

20 mm 10 mm

1 mm

20 mm

5 mm 2 mm

26 mm

20 mm

17 References [1] De Aquino, F. (2010) Mathematical Foundations of the Relativistic Theory of Quantum Gravity, Pacific Journal of Science and Technology, 11 (1), pp. 173232. [2] Valkengurg, V., (1992) Basic Electricity, Prompt Publications, 1-38. [3] Fischer-Cripps, A., (2004) The electronics companion. CRC Press, p. 13, ISBN 9780750310123. [4] De Aquino, F. (2008) Process and Device for Controlling the Locally the Gravitational Mass and the Gravity Acceleration, BR Patent Number: PI0805046-5, July 31, 2008. [5] De Aquino, F. (2010) Gravity Control by means of Electromagnetic Field through Gas at Ultra-Low Pressure, Pacific Journal of Science and Technology, 11(2) November 2010, pp.178-247, Physics/0701091. [6] Quevedo, C. P. (1977) Eletromagnetismo, McGrawHill, p. 270. [7] Halliday, D. and Resnick, R. (1968) Physics, J. Willey & Sons, Portuguese Version, Ed. USP, p.1124. [8] Chalmers, J.A., (1967) Atmospheric Electricity, Pergamon press, Oxford, London; Kamsali, N. et al., (2011) Advanced Air Pollution, chapter 20, DOI: 10.5772/17163, edited by Farhad Nejadkoorki, Publisher: InTech, ISBN 978-953-307-511-2, under CC BY-NCSA 3.0 license. [9] Kundu, J. et al., (2007) Semiconductor Physics, Quantum Electronics& Optoelectronics, 10, 1, pp.1-3. [10] Alonso, M. and Finn, E. (1967) Fundamental University Physics, Addison-Wesley Co., Portuguese version Ed Edgard Bl端cher (1972), p. 153. [11] Marshall, S. V. and Skitec, G.G. (1980 ) Electromagnetic Concepts and Applications, Prentice-Hall, NJ, Second Edition, p.287 [12] Chen, D. et al., (1991) Demagnetizing Factors for Cylinders, IEE Transactions on Magnetics, Vol. 27, Number,4, p. 3601. Table V, p. 3610. [13] Mac Gregor. M. H., (1992) The Enigmatic Electron. Boston: Klurer Academic, 1992, pp. 4-5. [14] N.D. Cook (2010). Models of the Atomic Nucleus (2nd ed.). Springer. p. 57 ff.. ISBN 978-3-642-14736-4. [15] K.S. Krane (1987). Introductory Nuclear Physics. Wiley-VCH. ISBN 0-471-80553-X.