A Possibility of Gravity Control in Luminescent Materials                               Fran De Aquino                    ...
2where      U is the electromagneticenergy absorbed (or emitted) by the                                              2   ...
3                           2                    aD µσ                   1.Theorymg = mi − 2 1 +         3      ...
4                                                               According to the Quantum                                 ...
5                                                                                                               1       ...
6 R0 , ξ are respectively, the radius and         W = P.∆t = (5300 w)(2.59 × 10 s ) =                                     ...
Eletroluminescent disk (ON)                                 Eletroluminescent disk (OFF)  (organic luminescent disk)      ...
8APPENDIX A    It is important to note that the                               QE0 E02ε 0 S (∆V d ) ε 0 S                  ...
Upcoming SlideShare
Loading in …5
×

Fran de aquino a possibility of gravity control in luminescent materials, 2001

879 views
778 views

Published on

Published in: Technology
0 Comments
0 Likes
Statistics
Notes
  • Be the first to comment

  • Be the first to like this

No Downloads
Views
Total views
879
On SlideShare
0
From Embeds
0
Number of Embeds
3
Actions
Shares
0
Downloads
8
Comments
0
Likes
0
Embeds 0
No embeds

No notes for slide

Fran de aquino a possibility of gravity control in luminescent materials, 2001

  1. 1. A Possibility of Gravity Control in Luminescent Materials Fran De Aquino physics/0109060 Maranhao State University, Physics Department, 65058-970 S.Luis/MA, Brazil. E-mail: deaquino@elo.com.brAbstract It was demonstrated (gr-qc/9910036) that the gravitational and inertialmasses are correlated by an adimensional factor, which depends on the incident ( oremitted)radiation upon the particle. There is a direct correlation between the radiationabsorbed( or radiated) by the particle and its gravitational mass, independently of theinertial mass. Only in the absence of electromagnetic radiation the mentioned factorbecomes equal to one. On the other hand, in specific electromagnetic conditions, itcan be reduced, nullified or made negative. This means that there is the possibilityof the gravitational masses can be reduced, nullified or made negative by means ofelectromagnetic radiation. This unexpected theoretical result was confirmed by anexperiment using Extra-Low Frequency(ELF) radiation on ferromagnetic material(gr-qc/0005107). Recently another experiment using UV light on phosphorescentplastic have confirmed the phenomenon. We present a complete explanation for thealterations of the gravitational field in luminescent materials. This work indicates thatthe alterations of the gravitational field can be sufficiently strong to invert the gravityon luminescent materials.Introduction It is known that the physical conditions, it can be reduced, nullifiedproperty of mass has two distinct or made negative.aspects, gravitational mass mg and The general expression ofinertial mass mi . Gravitational mass correlation between gravitational massproduces and responds to gravitational mg and inertial mass mi , is given byfields. It supplies the mass factors inNewtons famous inverse-square law  2 of gravity(F12=Gmg1mg2 /r122). Inertial   q    m g = mi − 2  1 +   − 1 m i (1)mass is the mass factor in Newtons  mi c   2nd Law of Motion (F=mia).  In a previous paper1 we haveshown that the gravitational mass and where the momentum q , according tothe inertial mass are correlated by an the Quantum Mechanics, is given byadimensional factor, which dependson the incident radiation upon the q = N k = N ω /(ω / k ) = U /(dz / dt) =    particle. It was shown that only in theabsence of electromagnetic radiation =U /v (2)this factor becomes equal to 1 andthat, in specific electromagnetic
  2. 2. 2where U is the electromagneticenergy absorbed (or emitted) by the  2 particle; v is the velocity of the   nhf    mg = mi − 2 1 +  nr  − 1m i (6)incident( or emitted) radiation. It can be  mi c 2   shown that  c In that case, according to the Statisticalv= (3) Mechanics, the calculation of n can ε r µr   1 + (σ ωε ) + 1  2 be made based on the well-known 2   method of Distribution Probability . If allwhere ; ε ,µ and σ, are the the particles inside the body have theelectromagnetic characteristics of the same mass mi , the result isoutside medium around the particle inwhich the incident radiation is ( ε = ε r ε 0 where ε r Npropagating n= a (7)is the relative electric permittivity and Aε 0 = 8.854 × 10−12 F / m ; µ = µ r µ0 where where N / A is the average density of µ r is the relative magnetic permeability incident (or emitted) photons on theand µ 0 = 4π × 10−7 H / m ). For an atom body; a is the area of the surface of ainside a body , the incident(or emitted) particle of mass mi from the body.radiation on this atom will be Obviously the power P of thepropagating inside the body , and incident radiation must beconsequently , σ = σbody , ε = εbody, P = Nhf / ∆t = Nhf , thus we can write 2µ =µbody. N = P / hf 2 . Substitution of N into By substitution of Eqs.(2) and(3) into Eq.(1), we obtain Eq.(7) gives a  P a   n=  = D (8) ( ) 2    U εrµr    hf 2  A  hf 2mg = m −2 1+ 2 i 1+(σ / ωε +1 −1mi )2  mi c 2     where D is the power density of the   2  incident( or emitted) radiation. Thus  U   = m −2 1+ 2 nr  −1m i i (4) Eq.(6) can be rewritten in the following  mc   form:  i In the equation above nr is the  2    aD   refractive index , which is given by: mg = mi − 2 1 +   − 1m i (9)  mi cvf    εµ = r r  1 + (σ ωε ) + 1  cnr =   2 (5) v 2   For σ >> ωε Eq.(3) reduces toc is the speed in vacuum and v is thespeed in medium. 4πf v= (10) If the incident(or emitted) µσradiation is monochromatic and hasfrequency f , we can put U = nhf inEquation(4), where n is the number ofincident (or radiated) photons on the By substitution of Eq.(10) into Eq.(9)particle of mass mi . Thus we obtain we obtain
  3. 3. 3  2    aD µσ    1.Theorymg = mi − 2 1 + 3 − 1mi (11)  mi c 4πf     When the material is luminescent, the radiated photonsThis equation shows clearly that, number n , radiated from theatoms (or molecules) can have their electrons, cannot be calculated by thegravitational masses strongly Eq.(7), because, according to thereduced by means of Extra-Low quantum statistical mechanics, theyFrequency (ELF) radiation. are undistinguishable photons of We have built a system (called varying frequencies and consequentlySystem G) to verify the effects of the follow the Einstein-Bose statistics. InELF radiation on the gravitational mass that case, as we know, the number nof a body . In the system G, a 60Hz ( n is the number of photons withfrequency radiation was produced by frequency between f and f+∆f ) willan ELF antenna. A thin layer of be given byannealed pure iron around the antenna  8πVf 2  1 f + ∆f ∫f df ≅( toroid form ) have absorbed all the n=  v 3  E / mi c 2 ELF radiation. −1  e In this annealed iron toroid µ r = 25000 ( µ = 25000 µ0 ) and  8πV  1 ≅ 3  f 2 ∆fσ = 1.03 × 107 S / m .The power density  v e E / mi c 2 −1 D of the incident ELF radiation reach Thus, assuming ∆f ≅ 1Hz (quasi-monoapproximately 10Kw/m2 . By chromatic ) we obtainsubstitution of these values into Eq.11)  8πVf 2  1  8πVf 2  1it is easy to conclude the obtained n ≅  3  E / m c2 ≅  3  q / mic ≅  v  i  results.  e −1  v  e −1 The experimental setup and the  8πV  1obtained results were presented in a ≅  2  λc / λ (12)paper2 .  vλ  e −1 The experiment above where λc = h / mi c is the Comptonmentioned have confirmed that the wavelength for the particle of mass migeneral expression of correlationbetween gravitational mass and inertial and λ is the average wavelength ofmass (Eq.4) is correct. In practice, this the light emitted from the particle ; V ismeans that the gravitational forces can the volume of the body which containsbe reduced, nullified and inverted by the particle .means of electromagnetic radiation. By substitution of Eq.(12) into Recently another experiment Eq.(6) we obtainusing UV light on phosphorescent  2 materials have confirmed the   8πV  λc λ   mg = mi − 2 1 + 2  λc / λ  − 1mi (13)phenomenon3.   vλ  e − 1  In this paper we present a complete explanation for the For σ << ωε the Eq.(3) reduces toalterations of the gravitational field inluminescent ( photo, electro, thermo c cand tribo )materials. It was shown that v= and λ= ε r µr f ε r µrthe alterations of the gravitational fieldcan be sufficiently strong to invert thegravity on luminescent materials . Consequently Eq.(13) can be rewritten In the following form
  4. 4. 4 According to the Quantum  2  Statistical Mechanics the gas of   8πV  2 3 λc λ   photons inside a luminescent materialmg = mi − 2 1+ 3  f nr λ / λ  −1mi (14)   c  e c −1  has a average number of photons N ,  where 1 1 mi = me = 9.11×10−31kg N= = λc / λ (16)For electrons e E / mi c 2 e −1and λ c ( electrons ) = 2.42 × 10 −12 m .For atomsλ c ( atoms ) << λc ( electrons ) . On the other This means that the UV power P should have the following valuehand, if hc2 λc λ P = Nhf 2 = 2 λ / λ (17)λ >> λc ⇒ ≈1 λ (e c − 1) e λc / λ − 1 For λ = 365 nm (UV light ).The equation above givesThen Eq.(14) reduces to P ≈ 68 W  2    8πV  2 3  mg = mi − 2 1 + 3  f nr  − 1mi (15)   c    From the Electrodynamics we  know that a radiation with frequency f In the Hardeman experiment 3 propagating within a material with electromagnetic characteristics ε, µ 8πV  2 3 and σ has the amplitudes of its waves 3  f nr ≈ 0.658 c  decreased of e−1=0.37 (37%) when it penetrates a distance z, given byConsequently, from Eq.(15) we obtainfor the electrons of the luminescent 1 z= (18)material:  1 + (σ ωε )2 − 1 ω 2 εµ 1   m g ( electrons ) ≈ 0.605mi ( electons ) The radiation is totally absorbed if it penetrates a distance δ≅5z. This means a 39.5% reduction Thus, if we put under UVin gravitational masses of the radiation (λ=365nm , > 68w) a sheet ofelectrons of the atoms of the phosphorescent plastic withphosphorescent material. Thus, the σ<<1S/m and ε>ε0 ; µ>µ0, the Eq.total reduction in gravitational mass of above tell us that z >> 5mm.the phosphorescent material will be Consequently, we can assume thatgiven by the UV radiation at 365nm has a good penetration within the plastic sheetm e − 0 .605 m e above ( thickness = 2mm). × 100 % ≈ − 0 . 011 %me + m p + mn On the other hand, if we assume that the sheet has an indexExactly the value obtained in the of refraction nr~1, thus , according toHardeman experiment. Eq.(15), for f~6×1014Hz (green light Now we will calculate the radiated from the sheet), thepower of UV radiation, necessary to gravitational mass of the electrons ofproduce the reduction of weight, the sheet will be NEGATIVE and givendetected by Hardeman in the byphosphorescent material.
  5. 5. 5   1 2 g 2   8πV  2 3   P = Tω = ( Fr )ω = (mg g )r   =mg = me − 2 1 + 3  f nr  − 1me ≅ r   c     = mg g 3r (21) ≅ −335.1me (19)Thus, the total reduction in where r = R − (R0 + ∆r ) , (see Fig.1-a)gravitational mass of the sheet of and mg is the gravitational mass of thephosphorescent material will be electroluminescent material inside thegiven by left-half of the rotor ( when NEGATIVE,me − 335.1me × 100% ≈ −9.1% obviously ) ( see rotor in Fig.1). It isme + m p + mn easy to show that mg may be written in If V=2m×1.36m×0.002m=0.00544m3 the formwe will have a reduction ofapproximately 100%.  N (Kme )  mg ≅ − mi  N (me + mp + mn ) (22)2.The Gravitational Motor   for Kme > me + mp + mn ; K > 3666.3 For σ >>ωε , as we have seen ,Eq.(3) reduces to Eq.(10),i.e., where m p = mn = 1.67 × 10−27 kg are the 4π fv = masses of the proton and neutron µσ respectively and K , in agreementConsequently Eq.(13), for λ >> λc , with Eq.(20), is given bycan be rewritten in the following form,  8πVf 2 4  K ≈ 2  3 nr  (23)  2   c    8πV  2 4   mg = mi − 2 1 + 3  f nr  −1mi (20)   c    By substitution of Eq.(23) into Eq.(22)  we obtainThe difference between Eq.(20) andEq.(15) is in exponent of the index of  8πVf 2 4   me  refraction nr . mg ≅ −2 3 nr  mi (24) Both Eq.(15) and Eq.(20) tell us  c  me + mp + mn  that luminescent materials with highrefractive indices can be very efficients But the electroluminescent (EL)in gravity control technology . material of the rotor is divided in disks In the particular case of the to reduce the gravitational pressure onGravitational Motor ( presented in a them(see Fig.1-b). These disks(previous paper4 ) these materials can organic luminescent material) aresimplify its construction. between electrodes and submitted to Let us consider figure 1 where suitable alternating voltage ∆V to emitwe present a new design for the blue light( frequency f = 6.5 × 1014 hz ).Gravitational Motor based onelectroluminescent materials. Thus, according to Eq.(24), the The average mechanical power gravitational mass m1 of one EL disk, g of the motor is (with volume V = π R 0 ξ P 2 where
  6. 6. 6 R0 , ξ are respectively, the radius and W = P.∆t = (5300 w)(2.59 × 10 s ) = 6the thickness of the EL disk), is givenby = 1.4 × 1010 j ≅ 3800 Kwh It is important to note that if  π 0 ( ) 16 πR2ξ 2 f 2  me   nr ≅ 2 the power of the motorm ≅ − nr4  ρ (25) 1  me + mp + mn  g 3   c   increases to approximately 112 HP!  8πVf 2  16π (πR02ξ ) f 2 4 3. ConclusionK ≈ 2 3 nr4  = nr  c  c3 We have studied the possibility to control the gravity on luminescent For example, if the rotor has materials and have concluded thatR = 627 mm; L = 1350mm and the EL eletroluminescent materials with highdisks: refractive indices are a new andR0 = 190mm; ρ ≅ 800kg / m3 ;ξ = 45mm; efficient solution for the gravity control technology . Particularly in the case ofχ = 0.2mm and nr ≅ 1 the gravitational motors.then the gravitational mass of each ELdisk (ON) is References 1. De Aquino, F.(2000)“Gravitation andm1 ≅ −4.4kg ; K ≅ 4014 g Electromagnetism:Correlation and Grand Unification”, Journal of New Energy , vol.5, no2 , pp.76-84.If the left-half of the rotor has Los Alamos National Laboratory L−χ preprint no.gr-qc/9910036.Nl = 2 = 60 EL disks(ON), then ξ+χ 2. De Aquino, F. (2000) “Possibility ofthe total gravitational mass mg is Control of the Gravitational Mass by Means of Extra-Low Frequencies Radiation”, Los Alamos National mg = N l m1 = 264kg g Laboratory preprint no.gr-qc/0005107.Thus, according to Eq.(21) the power 3. Hardeman, C. (2001) “ The Aquinoof the motor is /Hardeman Photo-gravity effect”, in http://www.icnet.net/users/chrish/Photo-P = (264) (9.8) (0.627 − (0.190 + 0.002) ≅ 3 gravity.htm 4. De Aquino, F. (2000) “ How to Extract Energy Directly from a ≅ 5.3Kw ≅ 7HP Gravitational Field”, Los Alamos National Laboratory preprint no.gr-qc/0007069.A electric generator coupled at thismotor can produce for one month anamount of electric energy W given by
  7. 7. Eletroluminescent disk (ON) Eletroluminescent disk (OFF) (organic luminescent disk) R0 rRotor R R Motor Axis ∆r (a) Cross-section of the Motorxxxxxxxxxxxxxxxx film electrodes ( thickness = χ = 0.2mm) organic luminescent disk ( thickness =ξ=45mm ) L AC (b) Schematic diagram of the battery of (EL)cellsFig.1 – The Gravitational Motor
  8. 8. 8APPENDIX A It is important to note that the QE0 E02ε 0 S (∆V d ) ε 0 S 2momentum q in Eq.(1) can be also q= = = ( A4) f f fproduced by an Electric and/orMagnetic field if the particle has anelectric charge Q . Assuming that in the dielectric of the capacitor there is N * layers of In that case, combination of dipoles with thickness ξ approximatelyLorentz’s Equation F = QE0 + QV × B ¡ ¡ ¡ ¡ equal to the diameter of the atoms ,i.e., F = mg a (see reference 1, p.78- ¡ N * = d ξ ≅ 1010 d then, according to ¡andEq.(2.05)) gives Eq.(1), for q >> mi c , the gravitational ( Q E0 + V × B ) mass m* of each dipole layer is ¡ ¡ gq = mgV = mg ∆t ( A1 )  q  2q mg m* ≅ −2 g  m c mi ≅ − c ≅ In the particular case of an oscillating  i particle( frequency f , ∆t = 1 / f ) we  ∆V  ε 0 S 2have ≅ −2  ( A5) ( Q E0 + V × B )  d  fc ¢ ¢ q= ( A2) f Thus, the total gravitational mass mg Let us consider a parallel-plate of the dielectric may be written in thecapacitor where d is the distance formbetween the plates; ∆V is the appliedvoltage; E0 = ∆V / d is the external ε S electric field. Inside the dielectric the mg = N *m* ≅ −2 × 1010  0  ∆V 2 g  fcd  ( A6)electric field is E = σ / ε = E0 / ε r where  σ (in C/m2) is the density of electric For example, if we havecharge and ε = ε r ε 0 . ∆V = 50KV ; S = 0.01m ; f ≅ 10 Hz; d = 1mm 2 2 Thus the charge Q on each Eq.(A6) givessurface of the dielectric is given byQ = σS ( S is the area of the surface). mg ≅ −0.15kgThen we have Possibly this is the explanation for theQ = σS = ( Eε ) S = ( Eε r ε 0 ) S = E0ε 0 S ( A3) Biefeld-Brown Effect.Within the field E0 ,the charge Q (or“charge layer”) acquire a momentumq , according to Eq.(A2), given by

×