Free Energy

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FREE ENERGY AND ANTIGRAVITATIONAL DEVICES AND SYSTEMS THAT UTILIZE

GRAVITATIONAL MASS FLUCTUATIONS William S. Alek INTALEK, INC., 3506 43rd Place, Highland, IN 46322-3129, USA Phone:(219) 924-2742 http://www.intalek.com/ mailto:[email protected]

ABSTRACT: The purpose of this paper is to reveal a portion of classical physics that contains an intrinsic relativistic phenomenon called Gravitational Mass Fluctuations. A correlation has been established between mass, inductors, and capacitors, thereby linking gravity to electromagnetism. A simplified gravitational mass relativity model called Natural Relativity (NR) Theory is presented and shown to be a primary gravitational effect. This theory is correlated to Einstein's Special Relativity (SR) Theory, and as a consequence, creates a new “Principle of Equivalence Theorem” and a secondary gravitational effect. A temporal rotation operator is introduced using Euler’s Identity, which shows the complex (i.e., time-future) motion of matter. The speed of light c, Planck's constant h, permeability µ0, permittivity ε0, Boltzmann's Constant k, electric charge q, and the Fine Structure Constant α are invariant between equipotential surfaces of gravity because the fluctuation or curvature of the parameters that compose these constants are shown to evaluate to unity gain. In other words, these constants remain constant anywhere with a given gravity well. Gravitomagnetic Theory shows that the magnetic field energy produced by a moving electron is equivalent to its’ special relativistic mass fluctuation, and therefore, is shown to couple to gravity. This motion can either have a typical velocity or a complex (i.e., time-future) velocity. If the velocity is complex, then the special relativistic mass fluctuation of an electron is NEGATIVE, exhibit an antigravitational effect, and produce a complex (i.e., time-future) magnetic field. In addition, the total field energy of a complex magnetic field contained within a volume is NEGATIVE. In the Bohr model of the Hydrogen atom, an Amperian Current is described as an electron circulating around a nucleus at a relativistic speed. This creates a magnetic induction emerging from the center of the nucleus. Canceling this field by applying an external magnetic induction causes the velocity of the electron to become complex. The presence of NEGATIVE RESISTANCE, the production of NEGATIVE ENERGY, and the control of GRAVITY/ANTIGRAVITY occur by fluctuating the mass of an object. The theory presents a conceptual breakthrough in energy and high-speed field propulsion technology, and explores solutions based entirely within the framework of classical physics.

INTRODUCTION Puthoff (1996) coined the phrase, “metric engineering”, and Puthoff, Little and Ibison (2002) consider the vacuum to be a polarizable medium, and that it can be expressed in terms of tensor formulations of curved space-time. The bending of light passing near a massive object is caused by induced spatial variation in the refractive index of the vacuum near the object. This is correlated to changes in permeability µ0 and permittivity ε0 of the vacuum. Changes occurring in the vacuum also affect the mass of objects, the length and bending of rulers, the frequency of clocks, the energy of light, etc. This paper links gravity with electromagnetism by presenting formulations of curved spacetime in terms of classical physics, which are caused by relativistic fluctuations of mass M , inductance L , and capacitance C of an object. For example, when an object with mass M naturally falls downward in a given gravity well, its’ natural relativistic mass M increases due to Newtonian Gravitation, or universal mass attraction. Therefore, the new mass of an object is displaced to a new position within this well, and mass-energy remains conserved. However, by converting this increase in relativistic mass M to energy, a force acts upon the object, and

Gravitational Mass Fluctuations

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the new mass is now displaced to its original position that was higher vertically in the well. The object exhibits an antigravitational effect. The rate of change of this fluctuation could cause the speed of the object to easily exceed the speed of light. This is because the relativistic gravitational mass of the object, which is shown to be convergent, is moving at right angles to a relativistic inertial mass, which is shown to be divergent. Since the speed of the object with relativistic gravitational mass has no known upper limit, the resulting speed through deep space could be enormous and necessitates the use of the warp factor equation.

EVERYTHING IN THIS UNIVERSE IS CURVED! SAME OBJECT WITH MASS OF 1kg AND VOLUME OF 1m3

CURVATURE DUE TO GRAVITY

SAME OBJECT WITH MASS OF 1kg AND VOLUME OF 1m3

OBSERVER ON MOON WOULD SEE SAME OBJECT ON EARTH AS BEING MORE MASSIVE AND SMALLER IN SIZE

r

gMOON

r

g0 EYE

EYE SURFACE OF MOON

SURFACE OF EARTH

OBSERVER ON EARTH WOULD SEE SAME OBJECT ON MOON AS BEING LESS MASSIVE AND LARGER IN SIZE

FIGURE 1. The same sphere changes in mass and size due to changes of gravity.

Shown above are two spheres with equal mass and size. Since the gravity of the Moon g MOON is approximately 1 6 the gravity of the Earth g 0 , an observer on the Moon would measure an identical sphere on the Earth as having more mass and being smaller in volume. Likewise, an observer on the Earth would measure an identical sphere on the Moon as having less mass, and larger in volume. This is due to the curvature of space and time caused by universal mass attraction, or gravity. So, relative to an observer on the Earth, a 1 kg sphere of mass on the Moon has less mass than the same 1 kg sphere of mass on the Earth. And, a 1m3 sphere of volume on the Moon is larger in size than the same 1m3 sphere of volume on the Earth. This relativistic change in mass and volume are referred to as Gravitational Mass Fluctuations, or GMF.

THE MUTUAL EXCLUSION PRINCIPLE Marmet (2001) considers “separately” the influence of a gravitational potential upon matter, and assumes for the moment that kinetic energy is zero. He’s implicitly invoking what I call the mutual exclusion principle, and therefore, considers kinetic energy and gravitational energy independently. The mutual exclusion principle is a tool used to compute the curvature or fluctuation of various parameters related to Gravitational Energy systems. For Kinetic Energy systems, the following flux-based parameters mass M , inductor L , and capacitor C , are invariant between equipotential surfaces of gravity gY . However, for Gravitational Energy systems, and given an equipotential surface of gravity reference, the following temporal-based parameters relativistic mass ±∆M , relativistic inductor ±∆L , and relativistic capacitor ±∆C , fluctuate or curve between equipotential surfaces of gravity. The kinetic energy of Gravitational Energy systems is assumed to be zero. By applying the product rule, the mutual exclusion principle is mathematically expressed as, z (t ) =

d dx dy ( x y) = y + x = yx+ x y dt dt dt

Where, the Flux Coupling Term is, z (t ) = y x

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And the Gravitational Coupling Term or temporal-based fluctuating system is, z (t ) = x y

The first term is regarded as kinetic or fluxes, and therefore, couples to inertia and is Newtonian-based. The second term is regarded as temporal, and therefore, couples to gravity and is non-Newtonian-based. This principle, which has been implicitly used for several centuries, excludes one term from the other. The fixed distance + y and the changing distance + y or +∆y are directed towards the center of gravity. The fixed distance + x and the changing distance + x or +∆x are directed across an equipotential surface of gravity.

GRAVITATIONAL MASS FLUCTUATION FLUCTUATING MASS M ±M

−y

r∓r

g0

+y

FIGURE 2. The fluctuating mass of an object.

The complete ideal momentum model is composed of two terms, fM (t ) =

dpY d ( M vY ) dv dM = = M Y + vY = M vY + vY M dt dt dt dt

(1)

Where, the Flux Coupling Term is M vY , and mass M is invariant within any equipotential surface of gravity gY . The Gravitational Coupling Term is vY M , and changing mass M fluctuates between equipotential surfaces of gravity. For a “mass fluctuating system”, the Gravitational Coupling Term is NOT zero Newtons. So, given an object 2 having mass M moving at constant velocity vY , or vY = 0 m s , the Flux Coupling Term is, fM (t ) = M vY = 0 N

(2)

This removes the Flux Coupling Term, leaving only the Gravitational Coupling Term, fM (t ) = vY M ≠ 0 N

(3)

Since M has units of resistance in mNs m 2 , its direction of change could either be POSITIVE or NEGATIVE. If M is negative, it has units of negative resistance or, M < 0 mNs m 2

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Now, the instantaneous gravitationally induced power PM of a fluctuating mass M is, PM (t ) = vY fM (t ) = vY 2 M

(5)

So, for certain values of M , the total instantaneous power PM can be NEGATIVE or, PM (t ) < 0 Watts

(6)

Then, integrating PM with respect to time when the total power is less than zero watts results in NEGATIVE energy of mass M or, EM (t ) = ∫ PM (t ) dt = vY 2 ∫ M dt = vY 2 M (t ) < 0 Joules

(7)

If M is positive, it has units of positive resistance or, M > 0 mNs m 2

(8)

Now, the instantaneous gravitationally induced power PM of a fluctuating mass M is, PM (t ) = vY fM (t ) = vY 2 M

(9)

So, for certain values of M , the total instantaneous power PM can be POSITIVE or, PM (t ) > 0Watts

(10)

Then, integrating PM with respect to time results in excess POSITIVE energy of mass M or, EM (t ) = ∫ PM (t ) dt = vY 2 ∫ M dt = vY 2 M (t ) > 0 Joules

(11)

So, the energy equivalent of mass (gravitational energy) EM is, EM (t ) = M (t ) vY 2

(12)

By rearranging terms, the mass equivalent of energy M is, M (t ) =

EM (t ) vY 2

(13)

EM vY 2

(14)

EM E v = M 2 Y vY vY

(15)

The fluctuating mass equivalent of energy M is, M =

So, the gravitational momentum model is, fM (t ) = M vY =

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Letting y = dy dt = vY , the Gravitational Force Coupling Term is, fM (t ) =

EM y

(16)

By rearranging terms, the fluctuating energy equivalent of mass EM is, EM = fM (t ) y

(17)

EM (t ) = M (t ) c 2

(18)

EM = M c 2

(19)

The total energy EM contained within matter is,

The change in this total energy EM is,

By rearranging terms, the change in total mass M is, M =

EM c2

(20)

So, the time derivative form of the Gravitational Mass Coupling Term is, fM (t ) y c2

(21)

dM =

fM dy c2

(22)

∆M =

fM ∆y c2

(23)

M =

The derivative form is,

The difference form is,

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GRAVITATIONAL INDUCTIVE MASS FLUCTUATION FLUCTUATING INDUCTANCE L±L

−y

g0

+y IL

-

+ vL

FIGURE 3. The fluctuating inductance of an object.

The complete ideal inductor model is composed of two terms,

ν L (t ) =

dI d dL ( L I L ) = L L + I L = L I L + I LL dt dt dt

(24)

Where, the Flux Coupling Term is L I L , and inductance L is invariant within any equipotential surface of gravity gY . The Gravitational Coupling Term is I LL , changing inductance L fluctuates between equipotential surfaces of gravity.

For an “inductive fluctuating system”, the Gravitational Coupling Term is NOT zero volts. By applying a constant current I L through inductor L , or I L = 0 Amps s , the Flux Coupling Term is,

ν L (t ) = L I L = 0Volts

(25)

This removes the Flux Coupling Term, leaving only the Gravitational Coupling Term,

ν L (t ) = I L L ≠ 0 Volts

(26)

Since L has units of resistance in ohms, Ω , its direction of change could either be POSITIVE or NEGATIVE. If L is negative, it has units of negative resistance or, L < 0Ω

(27)

Now, the instantaneous gravitationally induced power PL of a fluctuating inductor L is, PL (t ) = I L ν L (t ) = I L 2 L

(28)

So, for certain values of L , the total instantaneous power PL can be NEGATIVE or, PL (t ) < 0Watts

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Then, integrating PL with respect to time when the total power is less than zero watts results in NEGATIVE energy of inductor L or, EL (t ) = ∫ PL dt = I L 2 ∫ L dt = I L 2 L(t ) < 0 Joules

(30)

If L is positive, it has units of positive resistance or, L > 0Ω

(31)

Now, the instantaneous gravitationally induced power PL of a fluctuating inductor L is, PL (t ) = I L ν L (t ) = I L 2 L

(32)

So, for certain values of L , the total instantaneous power PL can be POSITIVE or, PL (t ) > 0 Watts

(33)

Then, integrating PL with respect to time results in excess POSITIVE energy of inductor L or, EL (t ) = ∫ PL dt = I L 2 ∫ L dt = I L 2 L(t ) > 0 Joules

(34)

Equate this to the energy equivalent of mass EM , EL (t ) = EM (t )

(35)

Then, the energy equivalent of mass (gravitational energy) EL is, EL (t ) = vY 2 M L (t ) = I L 2 L(t )

(36)

By rearranging terms, the mass equivalent of energy M L is, M L (t ) =

EL (t ) I L 2 L(t ) = vY 2 vY 2

(37)

I L 2 L(t ) M L (t )

(38)

EL I L 2 L = vY 2 vY 2

(39)

vY 2 =

The fluctuating mass equivalent of energy M L is, ML =

So, the gravitational inductor model is, fL (t ) = M L vY =

William Alek

EL EL I L 2 L v = = Y vY 2 vY vY

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Letting y = dy dt = vY , the Gravitational Inductive Force Coupling Term is, fL (t ) =

EL (t ) y

(41)

By rearranging terms, the fluctuating energy equivalent of mass EL is, EL = fL (t ) y

(42)

EL (t ) = M L (t ) c 2

(43)

EL = M L c 2

(44)

The total energy EL contained within matter is,

The change in this total energy EL is,

By rearranging terms, the change in total mass M is, ML =

EL fL (t ) y I L 2 L I 2L M (t ) L = = 2 = 2L = L 2 2 L(t ) c c y I L L(t ) M L (t )

(45)

So, the time derivative form of the Gravitational Inductive Coupling Term is, fL (t ) L(t ) y M L (t ) c 2

(46)

dL =

fL L dy M L c2

(47)

∆L =

f L L ∆y M L c2

(48)

L=

The derivative form is,

The difference form is,

William Alek

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GRAVITATIONAL CAPACITIVE MASS FLUCTUATION FLUCTUATING CAPACITANCE C ±C −y

g0

-

+

+y

VC

iC

+

V

-

FIGURE 4. The fluctuating capacitance of an object.

The complete ideal capacitor model is composed of two terms, iC (t ) =

dV d dC = C VC + VC C (C VC ) = C C + VC dt dt dt

(49)

Where, the Flux Coupling Term is CVC , and capacitor C is invariant within any equipotential surface of gravity gY . The Gravitational Coupling Term is VC C , and changing capacitance C fluctuates between equipotential surfaces of gravity.

For a “capacitive fluctuating system”, the Gravitational Coupling Term is NOT zero amps. By applying a constant voltage across capacitor C , or VC = 0Volts s , the Flux Coupling Term is, iC (t ) = C VC = 0 Amps

(50)

This removes Flux Coupling Term, leaving only the Gravitational Coupling Term, iC (t ) = VC C ≠ 0 Amps

(51)

Since C has units of conductance in mhos, , its direction of change could either be POSITIVE or NEGATIVE. If C is negative, it has units of negative conductance or, C<0

(52)

Now, the instantaneous gravitationally induced power PC of a fluctuating capacitor C is PC (t ) = iC (t ) VC = VC 2 C

(53)

So, for certain values of C , the total instantaneous power PC can be NEGATIVE or, PC (t ) < 0Watts

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Then, integrating PC with respect to time when the total power is less than zero watts results in NEGATIVE energy of capacitor C or, EC (t ) = ∫ PC dt = VC 2 ∫ C dt = VC 2 C (t ) < 0 Joules

(55)

If C is positive, it has units of positive conductance or, C>0

(56)

Now, the instantaneous gravitationally induced power PC of a fluctuating capacitor C is, PC (t ) = iC (t ) VC = VC 2 C

(57)

So, for certain values of C , the total instantaneous power PC can be POSITIVE or PC (t ) > 0 Watts

(58)

Then, integrating PC with respect to time results in excess POSITIVE energy of capacitor C or, EC (t ) = ∫ PC dt = VC 2 ∫ C dt = VC 2 C (t ) > 0 Joules

(59)

Equate this to the energy equivalent of mass EM , EC (t ) = EM (t )

(60)

Then, the energy equivalent of mass (gravitational energy) EC is, EC (t ) = vY 2 M C (t ) = VC 2 C (t )

(61)

By rearranging terms, the mass equivalent of energy M C is, M C (t ) =

EC (t ) C (t ) VC 2 = vY 2 vY 2

(62)

VC 2 C (t ) M C (t )

(63)

EC VC 2 C = vY 2 vY 2

(64)

EC E V 2C v = C = C 2 Y vY vY vY

(65)

vY 2 =

The fluctuating mass equivalent of energy M C is, MC =

So, the gravitational capacitor model is, fC (t ) = M C vY =

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Letting y = dy dt = vY , the Gravitational Capacitive Force Coupling Term is, fC (t ) =

EC y

(66)

By rearranging terms, the fluctuating energy equivalent of mass EC is, EC = fL (t ) y

(67)

EC (t ) = M C (t ) c 2

(68)

EC = M C c 2

(69)

The total energy EC contained within matter is,

The change in this total energy EC is,

By rearranging terms, the change in total mass M is, MC =

EC fC (t ) y VC 2 C V 2C M (t ) C = = = 2C = C 2 2 2 C (t ) c c y VC C (t ) M C (t )

(70)

So, the time derivative form of the Gravitational Capacitive Coupling Term is, fC (t ) C (t ) y M C (t ) c 2

(71)

dC =

fC C dy M C c2

(72)

∆C =

fC C ∆y M C c2

(73)

C=

The derivative form is,

The difference form is,

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THE GRAVITATIONAL COUPLING OF A FLUCTUATING MASS, INDUCTOR OR CAPACITOR y

1kg OBJECT (BEFORE)

22,500.0

GRAVITATIONAL REFERENCE

y0

RADIUS

19,920

g yn = G

ME yn 2

NATURAL MASS FLUCTUATION DUE TO FREE FALL

∆y = y0 − y1

1kg OBJECT (AFTER)

INCREASED GRAVITY

y1

7,529 6,378.1 0.0

2.5

5.0

g y0 = 1.0

7.5

g y1 = 7.0

10.0

g yn

g 0 = 9.8

ACCELERATION DUE TO GRAVITY m sec2

FIGURE 5. The natural mass fluctuation of an object due to gravitational free fall.

Natural universal mass attraction or classic Newtonian gravity is a force f y that acts through a center of mass of the Earth with mass M E and a test mass M separated by a distance y , fy =

G ME M = gy M y2

(74)

So, given, Gravitational constant G = 6.67259 × 10−11 N m 2 kg 2 Mass of the Earth M E = 5.9787 × 1024 kg Radius of Earth yE = 6.3781× 106 m The surface gravity g 0 of the Earth is, −11 2 2 24 G M E ( 6.67259 × 10 N m kg )( 5.9787 × 10 kg ) g0 = = = 9.80665 m sec2 2 6 yE 2 6.3781 10 × m ( )

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The gravity g y0 at GRAVITATIONAL REFERENCE position y0 is, g y0 =

−11 2 2 24 G M E ( 6.67259 × 10 N m kg )( 5.9787 × 10 kg ) = = 1.00537 m sec 2 2 6 y0 2 (19.920 ×10 m )

(76)

The gravity g y1 at position y1 is, −11 2 2 24 G M E ( 6.67259 × 10 N m kg )( 5.9787 × 10 kg ) = = 7.03767 m sec 2 g y1 = 2 6 y12 7.529 10 × m ( )

(77)

Marmet (2001) equates force fM produced by a fluctuating mass dM to the gravitational force f y , f y (t ) = fM (t )

(78)

The derivative form of a fluctuating mass dM is, dM =

fM dy g y M dy = c2 c2

(79)

The derivative form of the gravitational RED SHIFT (or BLUE SHIFT) of mass dM M and the energy equivalent of mass dE E displaced dy within a given gravity well g y is, dM dEM g y dy = = 2 M EM c

(80)

Marmet states the equation above shows a “calculated change of energy levels as a function of gravitational potential is in perfect agreement with the Pound and Rebka and also the Pound and Snider experiments”. Therefore, energy increases as a function of downward or positive displacement within a given gravity well. Let dYg y = g y dy , the exponential solutions of the derivative form of mass and energy equivalent of mass are,

∫

M y1

M y0

1 1 dM = 2 M c

ln ( M ) M 1 = My

y0

∫

1 Yg c2 y

⎛ My ln M y1 − ln M y0 = ln ⎜ 1 ⎜ My ⎝ 0

(

)

(

)

M y1 M y0

=e

g y1 y1

g y0 y0

g y 1 y1

⎞ 1 ⎟ = 2 g y1 y1 − g y0 y0 ⎟ c ⎠

(

⎞ ⎟⎟ ⎠

⎛ g y1 y1 − g y0 y0 ⎜⎜ c2

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g y 0 y0

⎛ g y1 y1 − g y0 y0 ⎜⎜ c2 ⎝

M y1 = M y0 e⎝

(81)

dYg y

)

(83)

(84)

⎞ ⎟⎟ ⎠

(85)

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EM y1 = EM y0 e

Rev 3.2

⎛ g y1 y1 − g y0 y0 ⎞ ⎟⎟ ⎜⎜ c2 ⎝ ⎠

(86)

So, the Pound, Rebka and Snider experiment used Mossbaurer spectroscopy to measure the electromagnetic gravitational RED SHIFT (or BLUE SHIFT) of 14.4 keV gamma rays emitted from Fe57 through a vertical distance of 22.6 m . With the gamma rays emitted upward, they showed the RED SHIFT was within one percent (1%) of this result, SHIFT =

SHIFT =

∆M ∆EM g y1 y1 − g y0 y0 = = M EM c2

(87)

(9.806581 m sec2 )(6.3781226 × 106 m) − ( 9.806650 m sec 2 )( 6.3781000 × 106 m ) (2.99792458 × 108 m sec) 2 SHIFT = −2.465961× 10−15

(88)

(89)

With the gamma rays emitted downward, they showed the BLUE SHIFT was,

( 9.806650 m sec )( 6.3781000 ×10 m ) − (9.806581 m sec )(6.3781226 ×10 m) 2

SHIFT =

6

2

(2.99792458 × 108 m sec) 2 SHIFT = 2.465961× 10−15

6

(90)

(91)

Now, the force fL produced by a fluctuating inductor dL is equated to the gravitational force f y , f y (t ) = fL (t )

(92)

fL L dy g y L dy = M L c2 c2

(93)

The derivative form of a fluctuating inductor dL is, dL =

The derivative form of the gravitational RED SHIFT (or BLUE SHIFT) of inductor dL L displaced dy within a given gravity well g y is, dL g y dy = 2 L c

(94)

Let dYg y = g y dy , the exponential solution of the derivative form of an inductor is,

∫

Ly1

Ly0

1 1 dL = 2 L c

ln ( L ) L 1 = Ly

y0

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∫

g y1 y1 g y0 y0

1 Yg c2 y

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dYg y g y 1 y1

(95)

(96)

g y 0 y0

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⎛ Ly ln Ly1 − ln Ly0 = ln ⎜ 1 ⎜ Ly ⎝ 0

( )

( )

Ly1

=e

Ly0

Rev 3.2

⎞ 1 ⎟ = 2 g y1 y1 − g y0 y0 ⎟ c ⎠

(

⎛ g y1 y1 − g y0 y0 ⎜⎜ c2 ⎝

Ly1 = Ly0 e

)

⎞ ⎟⎟ ⎠

(97)

(98)

⎛ g y1 y1 − g y0 y0 ⎞ ⎜⎜ ⎟⎟ c2 ⎝ ⎠

(99)

So, the gravitational RED SHIFT (or BLUE SHIFT) of an inductor L is, SHIFT =

∆L g y1 y1 − g y0 y0 = L c2

(100)

Now, the force fC produced by a fluctuating capacitor dC is equated to the gravitational force f y , f y (t ) = fC (t )

(101)

fC C dy g y C dy = M C c2 c2

(102)

The derivative form of a fluctuating inductor dC is, dC =

The derivative form of the gravitational RED SHIFT (or BLUE SHIFT) of capacitor dC C displaced dy within a given gravity well g y is, dC dEC g y dy = = 2 C EC c

(103)

Let dYg y = g y dy , the exponential solution of the derivative form of a capacitor is,

∫

C y1

C y0

1 1 dC = 2 C c

ln ( C ) C 1 = Cy

y0

( )

C y1 C y0

=e

g y1 y1 g y0 y0

1 Yg c2 y

⎛ Cy ln C y1 − ln C y0 = ln ⎜ 1 ⎜ Cy ⎝ 0

( )

∫

g y 1 y1

⎛ g y1 y1 − g y0 y0 ⎜⎜ c2 ⎝

(

⎞ ⎟⎟ ⎠

⎛ g y1 y1 − g y0 y0 ⎜⎜ c2

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g y 0 y0

⎞ 1 ⎟ = 2 g y1 y1 − g y0 y0 ⎟ c ⎠

C y1 = C y0 e⎝

(104)

dYg y

)

(106)

(107)

⎞ ⎟⎟ ⎠

(108)

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So, the gravitational RED SHIFT (or BLUE SHIFT) of a capacitor C is, SHIFT =

∆C g y1 y1 − g y0 y0 = C c2

(109)

NATURAL RELATIVITY THEORY M − ∆M DECREASED GRAVITY CURVATURE DUE TO −∆y GRAVITY

DECREASING MASS GRAVITATIONAL REFERENCE

gy

MASS M AT REST

INCREASING MASS

+∆y

INCREASED GRAVITY M + ∆M

TOWARDS CENTER OF GRAVITY

FIGURE 6. A change of relativistic mass due to gravity.

The establishment of a GRAVITATIONAL REFERENCE is defined as a fixed point of reference within a given gravity well g y . This point may be located in a plane of equipotential surface of gravity, and is used throughout this paper. Natural relativistic changes of mass, volume, frequency, energy, etc. fluctuate or curve as a function of displacement ±∆y from this point of reference. This displacement defines a new point within a new plane of equipotential surface of gravity g ±∆y , and the gravity at that point may be increased or decreased based upon the sign of the displacement. Marmet (2001) invokes the principle of mass-energy conservation regarding the displacement of matter between planes. For example, an object of mass M displaced a distance ∆y changes back to its original mass when returned to its original position within a given gravity well. Therefore, a new and simplified relativity model is introduced. For gravitational energy systems, and given an equipotential surface of gravity reference, the following parameters including relativistic mass ±∆M , relativistic inductance ±∆L , and relativistic capacitance ±∆C , fluctuate or curve between equipotential surfaces of gravity by displacement ±∆y . Again, the kinetic energy of gravitational energy systems is assumed to be zero. So, given a common equipotential surface of gravity reference g y , an increase in gravity causes a natural relativistic increase in mass, energy equivalent of mass (gravitational energy), inductance, and capacitance. Likewise, a decrease in gravity causes a natural relativistic decrease in the same metrics. Given an object with a rest mass M y0 , an equivalent energy of the rest mass EM y0 , an inductance Ly0 , and a capacitance C y0 , the new rest mass M y1 , the new equivalent energy of the rest mass EMy0 , the new inductance Ly1 , and the new capacitance C y1 are,

William Alek

M y1 = γ NR M y0 = M y0 ± ∆M y0

(110)

EMy1 = γ NR EMy0 = EMy0 ± ∆EMy 0

(111)

Ly1 = γ NR Ly0 = Ly0 ± ∆Ly0

(112)

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Rev 3.2

C y1 = γ NR C y0 = C y0 ± ∆C y0

(113)

The natural relativistic gamma γ NR is,

γ NR = e

⎛ g y1 y1 − g y0 y0 ⎜⎜ c2 ⎝

⎞ ⎟⎟ ⎠

(114)

The difference forms and the exponential forms of the natural relativistic mass M y1 model, energy equivalent of mass EMy1 model, inductor Ly1 model, and capacitor C y1 model at position ±∆y within a given gravity well g y are,

M y1 = M y0 ± ∆M y0

⎛ g y1 y1 − g y0 y0 ⎞ ⎟⎟ c2 ⎠

⎛ g y y1 − g y0 y0 ⎞ ⎜⎜ ⎝ = M y0 ⎜ 1 ± 1 ⎟ = M y0 e 2 c ⎝ ⎠

EMy1 = EMy0 + ∆EMy0 = EMy0

(115)

⎛ g y1 y1 − g y0 y0 ⎞ ⎟⎟ c2 ⎠

⎜⎜ ⎛ g y1 y1 − g y0 y0 ⎞ ⎝ ⎜1 ± ⎟ = EMy0 e 2 c ⎝ ⎠

(116)

⎛ g y1 y1 − g y0 y0 ⎞ ⎟⎟ c2 ⎠

⎜⎜ ⎛ g y y1 − g y0 y0 ⎞ ⎝ Ly1 = Ly0 + ∆Ly0 = Ly0 ⎜ 1 ± 1 L e = ⎟ y0 2 c ⎝ ⎠

(117)

⎛ g y1 y1 − g y0 y0 ⎞ ⎟⎟ c2 ⎠

⎛ g y y1 − g y0 y0 ⎞ ⎜⎜ ⎝ C y1 = C y0 ± ∆C y0 = C y0 ⎜ 1 ± 1 ⎟ = C y0 e 2 c ⎝ ⎠

(118)

In summary, Natural Relativity (NR) Theory is the primary gravitational effect.

SPECIAL RELATIVITY THEORY TIME FUTURE DECREASED GRAVITY

t

TIME FORWARD

+ jvx

M − ∆M

−∆y

GRAVITATIONAL REFERENCE

gy CURVATURE DUE TO VELOCITY

INITIAL MASS M, AT REST AND t = 0 sec

+∆y

vx

INCREASED GRAVITY

M + ∆M t

FIGURE 7. A change of relativistic mass due to velocity.

William Alek

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INTALEK, INC.

Rev 3.2

Given the rest mass of an object M , the special relativistic mass M v model presented by Einstein (1905) shows an object moving at velocity vx is, M

Mv =

v2 1 − x2 c

= γ M = M + dM

(119)

The gamma γ is,

γ=

1

(120)

v2 1 − x2 c

Implementing the binomial expansion of the above equation,

γ SR = 1 +

1 vx 2 3 vx 4 5 vx 6 35 vx 8 + + + + ... 2 c 2 8 c 4 16 c 6 128 c8

Using the 1st order term of the expansion shown above where vx

γ SR = 1 +

(121)

c , the new special relativistic gamma γ SR is,

vx 2 2 c2

(122)

The derivative form of the special relativistic mass M v model moving at velocity vx is, ⎛ v2 ⎞ M v = γ SR M = M ± dM = M ⎜ 1 ± x 2 ⎟ ⎝ 2c ⎠

(123)

The derivative form of a fluctuating mass dM is, dM =

M vx 2 2 c2

(124)

An object can move at a real (i.e., time-forward) velocity vx , at an imaginary (i.e., time-future) velocity j vx , or at a velocity that is a combination of the two. The real and imaginary components are rotated about the temporal axis and therefore, can be described as complex motion. The rotation is given as 0° ≤ θ ≤ 90° , where the real axis is θ = 0° and the imaginary time-future axis is θ = 90° . The complex number uses the Euler’s identity e jθ , which functions as a temporal rotation operator. vx = v e jθ = v cos θ + j v sin θ

(125)

So, the derivative form of the inertial RED SHIFT (or BLUE SHIFT) of mass dM M , energy equivalent of mass dEM EM , inductor dL L , or capacitor dC C of an object moving at a real velocity vx or a complex velocity j vx is, SHIFT =

William Alek

dM dEM dL dC vx 2 = = = = 2 M EM L C 2c

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The exponential solution of the derivative form of mass is,

∫

Mv

M0

v2 1 dM = x 2 M 2c

(127)

vx 2 2 c2

(128)

ln ( M ) M = Mv

0

⎛M ln ( M v ) − ln ( M 0 ) = ln ⎜ v ⎝ M0

⎞ vx 2 ⎟= 2 ⎠ 2c

(129)

⎛ vx 2 ⎞

⎜ 2⎟ ⎜ 2c ⎟ Mv = e⎝ ⎠ M0

Mv = M0 e

⎛ vx 2 ⎜ 2 ⎜ 2c ⎝

EMv = EM 0 e Lv = L0 e

(130) ⎞ ⎟ ⎟ ⎠

⎛ vx 2 ⎜ 2 ⎜ 2c ⎝

(131) ⎞ ⎟ ⎟ ⎠

(132)

⎛ vx 2 ⎞ ⎜ ⎟ ⎜ 2 c2 ⎟ ⎝ ⎠ ⎛ vx ⎜ 2 ⎜ 2c ⎝ 2

Cv = C0 e

(133)

⎞ ⎟ ⎟ ⎠

(134)

So, the inertial RED SHIFT (or BLUE SHIFT) of a mass M , an energy equivalent of mass EM , an inductor L , and a capacitor C is, SHIFT =

∆M ∆E M ∆ L ∆ C v x 2 = = = = 2 M EM L C 2c

(135)

The special relativistic gamma γ SR is,

γ SR = e

⎛ vx 2 ⎜ 2 ⎜ 2c ⎝

⎞ ⎟ ⎟ ⎠

(136)

The difference forms and the exponential forms of the special relativistic mass M v model, energy equivalent of mass EMv model, inductor Lv model, and capacitor Cv model of an object moving at a real velocity vx or a complex velocity j vx are,

⎛ v2 M v = M 0 ± ∆M 0 = M 0 ⎜ 1 ± x 2 ⎝ 2c

⎛ vx 2 ⎞

⎜ 2⎟ ⎞ ⎜ 2c ⎟ ⎝ ⎠ = M e ⎟ 0 ⎠

(137)

⎛ vx 2 ⎞

E M v = E M 0 ± ∆E M 0

William Alek

⎜ 2⎟ ⎛ ⎜ 2c ⎟ v2 ⎞ = EM 0 ⎜ 1 ± x 2 ⎟ = EM 0 e ⎝ ⎠ ⎝ 2c ⎠

Page 19

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⎛ v2 Lv = L0 ± ∆L0 = L0 ⎜ 1 ± x 2 ⎝ 2c

Rev 3.2

⎛ vx 2 ⎞

⎜ 2⎟ ⎜ 2c ⎟ ⎞ ⎝ ⎠ ⎟ = L0 e ⎠

(139)

⎛ vx 2 ⎞

⎜ 2⎟ ⎛ ⎜ 2c ⎟ v2 ⎞ Cv = C0 + ∆C0 = C0 ⎜1 ± x 2 ⎟ = C0 e⎝ ⎠ ⎝ 2c ⎠

(140)

A NEW PRINCIPLE OF EQUIVALENCE THEOREM Since the gravitational RED SHIFT (or BLUE SHIFT) of an object due to gravity g y is, SHIFT =

∆M g y1 y1 − g y0 y0 = M c2

(141)

And the inertial RED SHIFT (or BLUE SHIFT) of the same object due to velocity vx is, v2 ∆M = x2 M 2c

(142)

g y y1 − g y0 y0 vx 2 = 1 2 2c c2

(143)

SHIFT =

Then, it follows,

A new Principle of Equivalence Theorem is therefore determined as, ⎛1 1 ⎞ vx 2 = g y1 y1 − g y0 y0 = G M E ⎜ − ⎟ 2 ⎝ y1 y0 ⎠

(144)

So, an object displaced ±∆y within the Earth’s gravity well g y is equivalent to the same object moving at complex velocity vx ,

⎛1 1 ⎞ vx = 2 g y1 y1 − g y0 y0 = 2 G M E ⎜ − ⎟ ⎝ y1 y0 ⎠

(

)

(145)

If the displacement ∆y of an object is POSITIVE, then the object is moving at a real velocity vx . However, if the displacement ∆y is NEGATIVE, then the same object is moving at a complex (i.e., time-future) velocity

−1 vx , or

velocity j vx , where j = −1 . The real and imaginary components are rotated about the temporal axis as a complex velocity. The rotation is given as 0° ≤ θ ≤ 90° , where the real axis is θ = 0° and the imaginary time-future axis is θ = 90° . The complex number uses the Euler’s identity e jθ , which functions as a temporal rotation operator. vx = v e jθ = v cos θ + j v sin θ

(146)

So, given an object moving at a complex velocity vx , the equivalent displacement to position y1 within the Earth’s gravity well g y , where 0 < y1 ≤ ∞ or −1 ≤

William Alek

y0 vx 2 is, 2G ME

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y1 =

1 g y1

⎛ vx 2 g y + ⎜ y0 0 2 ⎝

Rev 3.2

⎞ y0 ⎟= 2 y ⎠ 1 + 0 vx 2G ME

(147)

The equivalent maximum complex velocity vx max at y1 = ∞ is −1 =

y0 vx 2 2G ME

vx max = −

(148)

2G ME y0

(149)

Given the equivalent maximum complex velocity vx max , the minimum mass M min at y1 = ∞ is, ⎛ G ME ⎞

M min

⎜− ⎟ ⎛ G ME ⎞ ⎜ y c2 ⎟ 0 ⎝ ⎠ M e = M y0 ⎜ 1 − = ⎟ y 0 y0 c 2 ⎠ ⎝

(150)

AT REST

ACCELERATING

dM = 0

d 2M > 0

g0

f0

ON THE EARTH

IN SPACE

FIGURE 8. An elevator at rest on the Earth is NOT equivalent to an elevator accelerating in space.

As shown above, the natural relativistic mass fluctuation of an elevator at rest on the surface of the Earth is zero, or dM = 0 . However, the second order special relativistic mass fluctuation of the same elevator accelerating in space is non-zero or d 2 M > 0 , and as a consequence, radiates electromagnetic waves. According to Woodward (1998), radiation reaction is observed in bodies being accelerated based upon Newton’s second law of motion, f0 = m a . Therefore, the gravitational mass can’t be equivalent to its’ inertial mass due to their differences in mass fluctuations. An elevator traveling a distance + dy in free fall and the same elevator moving at a constant velocity vx at right angles to free fall produce virtually no radiation reaction as shown below. So, given this second scenario, these mass fluctuations are considered equivalent, hence, validating the new Principle of Equivalence Theorem.

William Alek

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FREE FALL A GIVEN DISTANCE

CONSTANT VELOCITY

dM > 0

dM > 0

NEAR EARTH

IN SPACE

FIGURE 9. An elevator in free fall above the Earth is equivalent to an elevator moving at constant velocity in space.

In summary, this new Principle of Equivalence Theorem describes an object moving at one-half the square of a real velocity vx is equivalent to the same object having fallen down a displacement + dy within a given gravity well g y . This object naturally acquires more relativistic mass, inductance and capacitance as it moves at a real velocity vx , and augments its’ own gravitation with other objects. On the other hand, the same object moving at one-half the square of a complex velocity j vx is equivalent to the same object having fallen up a displacement −dy in the same gravity well. This object naturally loses more relativistic mass, inductance and capacitance as it moves at a complex velocity j vx , and diminishes its’ own gravitation with other objects. Therefore, special relativity is considered to be a secondary gravitational effect.

M

1.10 INCREASING INERTIAL MASS (RED SHIFT)

INERTIAL MASS (kg)

1.05

M0 e

⎛ 2 ⎞ ⎜ vx ⎟ ⎜⎜ 2 ⎟⎟ ⎝ 2c ⎠

INERTIAL REFERENCE 1.00

vx DECREASING INERTIAL MASS (BLUE SHIFT)

0.95 TIMEFUTURE

TIME FORWARD

0.90 1.0 j × 108

7.5 j × 107

5.0 j × 107

2.5 j × 107

0.0 VELOCITY (m/sec)

2.5 × 107

5.0 × 107

7.5 × 107

1.0 × 108

FIGURE 10. Velocity profile of a 1kg inertial mass.

William Alek

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Rev 3.2

Example 1. Given the velocity profile above of an object having a mass M 0 moving at a special relativistic timeforward velocity vx , compute the new special relativistic inertial mass M v . So, given, Direction of time θ = 0° Mass of object M 0 = 1.0 kg Velocity of object v = 1.0 × 108 m sec The time-forward velocity vx of an object is, vx = v e jθ = (1.0 ×108 m sec ) e j 0° = 1.0 ×108 m sec

(151)

The new special relativistic inertial mass M v is,

Mv = M0 e

⎛ vx 2 ⎜ 2 ⎜ 2c ⎝

⎞ ⎟ ⎟ ⎠

(

= (1.0 kg ) e

)

2 ⎛ 1.0×108 m sec ⎜ ⎜ ⎜ 2 2.99793×108 m sec ⎝

(

)

2

⎞ ⎟ ⎟ ⎟ ⎠

(152)

M v = 1.05721 kg

(153)

M v − M 0 = (1.05721 kg ) − (1.0 kg ) = 0.05721 kg

(154)

The inertial mass of the object was increased by,

Example 2. Given the velocity profile above of an object having a mass M 0 moving at a special relativistic velocity time-future j vx , compute the new special relativistic inertial mass M v . So, given, Direction of time θ = 90° Mass of object M 0 = 1.0 kg Velocity of object v = 1.0 × 108 m sec The time-future velocity vx of an object is, vx = v e jθ = (1.0 × 108 m sec ) e j 90° = 1.0 j × 108 m sec

(155)

The new special relativistic inertial mass M v is,

Mv = M0 e

⎛ vx 2 ⎜ 2 ⎜ 2c ⎝

⎞ ⎟ ⎟ ⎠

(

= (1.0 kg ) e

)

2 ⎛ 1.0 j ×108 m sec ⎜ ⎜ ⎜ 2 2.99793×108 m sec ⎝

(

)

2

⎞ ⎟ ⎟ ⎟ ⎠

(156)

M v = 0.94589 kg

(157)

M v − M 0 = ( 0.94589 kg ) − (1.0 kg ) = −0.05411 kg

(158)

The inertial mass of the object was reduce by,

William Alek

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1.0000000000 GRAVITATIONAL REFERENCE g y0 0.9999999998 GRAVITATIONAL MASS (kg)

DECREASING GRAVITATIONAL MASS (BLUE SHIFT) 0.9999999996

M y0 e

⎛ gy y −gy y ⎞ 0 0⎟ ⎜ 1 1 ⎜⎜ ⎟⎟ c2 ⎝ ⎠

0.9999999994

0.9999999993485 DECREASING GRAVITY

0.9999999992

INCREASING GRAVITATIONAL MASS (RED SHIFT)

y0

y

0.9999999990 M 4.0 × 107

2.0 × 107 0.0 6.3781× 106

6.0 × 107

8.0 × 107

10.0 × 107

DISPLACEMENT FROM CENTER OF EARTH (m)

FIGURE 11. Displacement profile of a 1kg gravitational mass due to Earth’s gravity well.

Example 3. Given the gravitational profile above of an object having a mass M y0 displaced to a position −∆y within Earth’s gravity well g y , compute the new natural relativistic gravitational mass M y1 . So, given, Mass of object M y0 = 1.0 kg Object on Earth’s surface y0 = 6.3781× 106 m Object displaced to y1 = 1.0 × 108 m Speed of light c = 2.99792458 × 108 m sec Gravitational constant G = 6.67260 × 10−11 N m 2 kg 2 Mass of the Earth M E = 5.9787 × 1024 kg The acceleration due to gravity at surface of Earth is, g y0 =

−11 2 2 24 G M E ( 6.67260 × 10 N m kg )( 5.9787 × 10 kg ) = = 9.80665 m sec 2 2 6 y0 2 ( 6.3781×10 m )

(159)

The acceleration due to gravity at altitude y1 = 1.0 × 108 m above the Earth is, g y1 =

William Alek

G M E ( 6.67260 × 10 = y12

−11

N m 2 kg 2 )( 5.9787 × 1024 kg )

(1.0 ×10 m ) 8

2

Page 24

= 0.039894 m sec 2

(160)

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Gravitational Mass Fluctuations

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Rev 3.2

Given the exponential solution of the natural relativistic mass model, the new gravitational mass is,

M y1 = M y0 e

⎛ g y1 y1 − g y0 y0 ⎞ ⎟⎟ ⎜⎜ c2 ⎝ ⎠

(

= (1.0 kg ) e

)(

)(

)(

⎛ 2 8 2 6 ⎜ 0.039894 m sec 1.0×10 m − 9.80665 m sec 6.3781×10 m ⎜ 2 ⎜ 2.99792458×108 m sec ⎝

(

)

) ⎞⎟ ⎟ ⎟ ⎠

M y1 = 0.9999999993485 kg

(161) (162)

The gravitational mass of the object was reduced by, M y1 − M y0 = ( 0.9999999993485 kg ) − (1.0 kg ) = −6.516 × 10−10 kg

(163)

Example 4. Assuming there are no other gravitational influences besides the Earth, compute the new minimum natural relativistic gravitational mass M min of an object at y1 = ∞ . So, given, Mass of object M y0 = 1.0 kg Object on Earth’s surface y0 = 6.3781× 106 m Speed of light c = 2.99792458 × 108 m sec Gravitational constant G = 6.67260 × 10−11 N m 2 kg 2 Mass of the Earth M E = 5.9787 × 1024 kg The minimum gravitational mass M min at y1 = ∞ is,

M min = M y0 e

⎛ G ME ⎞ ⎜− ⎟ ⎜ y c2 ⎟ 0 ⎝ ⎠

(

= (1.0 kg ) e

)(

⎛ −11 2 2 24 ⎜ 6.67260×10 N m kg 5.9787×10 kg ⎜− 2 ⎜ 6.3781×106 m 2.99792458×108 m sec ⎝

(

)(

)

) ⎞⎟ ⎟ ⎟ ⎠

M min = 0.9999999993041 kg

(164) (165)

The gravitational mass of the object was reduced by, M y1 − M y0 = ( 0.9999999993041 kg ) − (1.0 kg ) = −6.959 × 10−10 kg

(166)

HOW GRAVITY AFFECTS THE VOLUME OF OBJECTS An object of volume Vy (length L y , width Wy , and height H y ) contracts as a function of position +∆y within gravity well g y . Likewise, the same the volume Vy dilates as a function of position −∆y in the same gravity well. So, the difference form of the gravitational RED SHIFT (or BLUE SHIFT) of an object of volume ∆V V displaced ∆y = y0 − y1 within a given gravity well g y is, g y y1 − g y0 y0 ∆L ∆W ∆H = = =− 1 L W H c2

William Alek

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The difference forms and exponential forms of the natural relativistic object of volume Vy at position ±∆y is, ⎛ g y1 y1 − g y0 y0 ⎞ ⎟⎟ c2 ⎠

− ⎛ g y y1 − g y0 y0 ⎞ ⎜⎜ ⎝ L y1 = L y0 ∓ ∆L y0 = L y0 ⎜1 ∓ 1 = L e ⎟ y0 c2 ⎝ ⎠

⎛ g y1 y1 − g y0 y0 ⎞ ⎟⎟ c2 ⎠

⎜⎜ − ⎛ g y y1 − g y0 y0 ⎞ ⎝ Wy1 = Wy0 ∓ ∆Wy0 = Wy0 ⎜1 ∓ 1 ⎟ = Wy0 e 2 c ⎝ ⎠

⎛ g y1 y1 − g y0 y0 ⎞ ⎟⎟ c2 ⎠

⎜⎜ − ⎛ g y y1 − g y0 y0 ⎞ ⎝ H y1 = H y0 ∓ ∆H y0 = H y0 ⎜ 1 ∓ 1 = H e ⎟ y 2 0 c ⎝ ⎠

(168)

(169)

(170)

HOW VELOCITY AFFECTS THE VOLUME OF OBJECTS An object of volume Vv (length Lv , width Wv , and height Hv ) contracts moving at a real velocity vx . Likewise, the same object of volume Vv dilates moving at a complex velocity j vx . So, the difference form of the inertial RED SHIFT (or BLUE SHIFT) of an object of volume ∆V V moving at a real velocity vx or a complex velocity j vx is, v2 ∆L ∆W ∆H = = =− x2 2c L W H

(171)

The difference forms and exponential forms of the special relativistic object of volume Vv moving at a real velocity vx or a complex velocity j vx is,

⎛ v2 Lv = L0 ∓ ∆L0 = L0 ⎜ 1 ∓ x 2 ⎝ 2c

⎛ vx 2 ⎞

⎜− 2 ⎟ ⎞ ⎜ 2c ⎟ ⎝ ⎠ ⎟ = L0 e ⎠

(172)

⎛ vx 2 ⎞

⎜− 2 ⎟ ⎛ ⎜ 2c ⎟ v2 ⎞ ⎠ Wv = W0 ∓ ∆W0 = W0 ⎜ 1 ∓ x 2 ⎟ = W0 e⎝ c 2 ⎝ ⎠

⎛ v2 H v = H 0 ∓ ∆H 0 = H 0 ⎜ 1 ∓ x 2 ⎝ 2c

(173)

⎛ vx 2 ⎞

⎜− 2 ⎟ ⎞ ⎜ 2c ⎟ ⎝ ⎠ ⎟ = H0 e ⎠

(174)

HOW GRAVITY AFFECTS THE FREQUENCY OF TIME A mechanical oscillator vibrating at a frequency f y contracts (i.e., slows down) as a function of position +∆y within gravity well g y . Likewise, the same mechanical oscillator vibrating at a frequency f y dilates (i.e., speeds up) as a function of position −∆y in the same gravity well. So, the difference form of the gravitational RED SHIFT (or BLUE SHIFT) of an oscillator vibrating at a frequency ∆f f displaced ∆y = y0 − y1 within a given gravity well g y is, g y y1 − g y0 y0 ∆f =− 1 f c2

William Alek

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Rev 3.2

The difference form and exponential form of the natural relativistic frequency f y of an oscillator at position ±∆y is,

f y1 = f y0 ∓ ∆f y0

⎛ g y1 y1 − g y0 y0 ⎞ ⎟⎟ c2 ⎠

⎜⎜ − ⎛ g y y1 − g y0 y0 ⎞ ⎝ = f y0 ⎜ 1 ∓ 1 = f e ⎟ y0 2 c ⎝ ⎠

(176)

HOW VELOCITY AFFECTS THE FREQUENCY OF TIME A mechanical oscillator vibrating at a frequency f v contracts (i.e., slows down) while moving at a real velocity vx . Likewise, the same mechanical oscillator vibrating at a frequency f v dilates (i.e., speeds up) while moving at a complex velocity j vx . So, the difference form of the inertial RED SHIFT (or BLUE SHIFT) of an oscillator vibrating at a frequency ∆f f while moving at a real velocity vx or a complex velocity j vx is, v2 ∆f =− x2 2c f

(177)

The difference form and exponential form of the special relativistic frequency f v of an oscillator moving at a real velocity vx or a complex velocity j vx is,

⎛ v2 f v = f 0 ∓ ∆f 0 = f 0 ⎜ 1 ∓ x 2 ⎝ 2c

⎛

vx 2 ⎞

⎜− 2 ⎟ ⎞ ⎜ ⎟ ⎝ 2c ⎠ ⎟ = f0 e ⎠

(178)

HOW GRAVITY AFFECTS AN INTERVAL OF TIME A mechanical oscillator vibrating for an interval of time t y contracts (i.e., slows down) as a function of position +∆y within gravity well g y . Likewise, the same mechanical oscillator vibrating for an interval of time t y dilates

(i.e., speeds up) as a function of position −∆y in the same gravity well. So, the difference form of the gravitational RED SHIFT (or BLUE SHIFT) of an oscillator vibrating for an interval of time ∆t t displaced ∆y = y0 − y1 within a given gravity well g y is, g y y1 − g y0 y0 ∆t =− 1 t c2

(179)

The difference form and exponential form of the natural relativistic time interval t y of an oscillator at position ±∆y is, ⎛ g y1 y1 − g y0 y0 ⎞ ⎟⎟ c2 ⎠

− ⎛ g y y1 − g y0 y0 ⎞ ⎜⎜ ⎝ t y1 = t y0 ∓ ∆t y0 = t y0 ⎜1 ∓ 1 ⎟ = t y0 e 2 c ⎝ ⎠

William Alek

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(180)

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INTALEK, INC.

Rev 3.2

HOW VELOCITY AFFECTS AN INTERVAL OF TIME A mechanical oscillator vibrating for an interval of time tv contracts (i.e., slows down) while moving at a real velocity vx . Likewise, the same mechanical oscillator vibrating for an interval of time tv dilates (i.e., speeds up) while moving at a complex velocity j vx . So, the difference form of the inertial RED SHIFT (or BLUE SHIFT) of an oscillator vibrating for an interval of time ∆t t while moving at a real velocity vx or a complex velocity j vx is, v2 ∆t =− x2 2c t

(181)

The difference form and exponential form of the special relativistic time interval tv of an oscillator moving at a real velocity vx or a complex velocity j vx is, ⎛

vx 2 ⎞

⎜− 2 ⎟ ⎛ ⎜ 2c ⎟ v2 ⎞ ⎠ tv = t0 ∓ ∆t0 = t0 ⎜1 ∓ x 2 ⎟ = t0 e⎝ ⎝ 2c ⎠

(182)

HOW GRAVITY AFFECTS LINEAR MOMENTUM The momentum p y of an object of mass M y moving at velocity vx increases as a function of position +∆y within gravity well g y . Likewise, the momentum p y of the same object decreases as a function of position −∆y in the same gravity well. So, the difference form of the gravitational RED SHIFT (or BLUE SHIFT) of momentum ∆p p of an object displaced ∆y = y0 − y1 within a given gravity well g y is, ∆p g y1 y1 − g y0 y0 = p c2

(183)

The difference form and exponential form of the natural relativistic momentum p y at position ±∆y is, ⎛ g y1 y1 − g y0 y0 ⎞ ⎟⎟ c2 ⎠

⎛ g y y1 − g y0 y0 ⎞ ⎜⎜ ⎝ p y1 = p y0 ± ∆p y0 = p y0 ⎜ 1 ± 1 ⎟ = p y0 e 2 c ⎝ ⎠

(184)

HOW VELOCITY AFFECTS LINEAR MOMENTUM The momentum pv of an object of mass M 0 increases moving at a real velocity vx . Likewise, the momentum pv of the same object decreases moving at a complex velocity j vx , So, the difference form of the inertial RED SHIFT (or BLUE SHIFT) of momentum ∆p p of an object moving at a real velocity vx or a complex velocity j vx is, ∆p vx 2 = p 2 c2

William Alek

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INTALEK, INC.

Rev 3.2

The difference form and exponential form of the special relativistic momentum pv of an object moving at a real velocity vx or a complex velocity j vx is, ⎛ vx 2 ⎞

⎜ 2⎟ ⎜ 2c ⎟ ⎛ v2 ⎞ pv = p0 ± ∆p0 = p0 ⎜ 1 ± x 2 ⎟ = p0 e⎝ ⎠ ⎝ 2c ⎠

(186)

HOW GRAVITY AFFECTS ANGULAR MOMENTUM The angular momentum S y of an object of mass M y moving at velocity vx with a radius r is invariant as a function of position +∆y within gravity well g y . Likewise, the angular momentum S y of the same object is invariant as a function of position −∆y in the same gravity well. So, the difference form of the gravitational RED SHIFT (or BLUE SHIFT) of angular momentum ∆S S of an object displaced ∆y = y0 − y1 within a given gravity well g y is, ∆S =0 S

(187)

The difference form and exponential form of the natural relativistic angular momentum S y at position ±∆y is,

S y1 = S y0

(188)

HOW VELOCITY AFFECTS ANGULAR MOMENTUM The angular momentum Sv of an object of mass M 0 is invariant moving at a real velocity vx with a radius r . Likewise, the angular momentum Sv of the same object is invariant moving at a complex velocity j vx , So, the difference form of the inertial RED SHIFT (or BLUE SHIFT) of angular momentum ∆S S of an object moving at a real velocity vx or a complex velocity j vx is, ∆S =0 S

(189)

The difference form and exponential form of the special relativistic angular momentum Sv of an object moving at a real velocity vx or a complex velocity j vx is,

William Alek

Sv = S0

(190)

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Gravitational Mass Fluctuations

INTALEK, INC.

Rev 3.2

THE SPACE-TIME MEDIA OR THE AETHER E Zε M TI

E

Zµ

ZG

B

t

FIGURE 12. The space-time media or aether.

According to Puthoff (1996) and Puthoff, Little and Ibison (2002), the vacuum is described as having magnetic permeability µ0 and dielectric permittivity ε0, and acts to impede the propagation of light and the motion of matter. Direct modification of these components changes the nature of light and matter. gy

+∆y gy'

+∆y ' CURVATURE DUE TO GRAVITY

r1

r1 − ∆r1

RAREFYING OF SPACE-TIME MEDIA

M 1 + ∆M 1

r2 − ∆r2

r2

M 2 + ∆M 2

M1

M2

GRAVITATIONAL REFERENCE

FIGURE 13. Two similar objects undergoing natural universal mass attraction.

The “active vacuum” of space, or space-time media (i.e., the aether) is composed of uncondensed relativistic mass. An object M 1 made of matter (i.e., atoms) and given a GRAVITATIONAL REFERENCE point undergoes universal mass attraction (i.e., gravitational free fall) with another object M 2 . Both objects acquire relativistic mass by a natural means from the surrounding space-time media as a function of displacement +∆y between the two objects. This media condenses onto both objects as more relativistic mass, thereby increasing their total mass M n + ∆M n , inductance Ln + ∆Ln , and capacitance Cn + ∆Cn . This action changes the relativistic momentum of both objects resulting with increasing force of attraction. The space-time media between these objects rarefy or relativistic mass condenses out of the media thereby affecting both the magnetic permeability µ0 and the dielectric permittivity ε0 of free space. This rarefaction of media is referred to as a gravity well, and as a consequence, causes the volume of space occupied by both objects and the space between them to be reduced. The space-time media in a rarefying state means gravity between these objects is increasing, which causes light passing near these objects to amplify in energy Eλ y and increase in frequency f λ y as proven by the Pound and Rebka experiment (1964). This behavior of space-time media acts as an impedance upon the natural motion of matter and the propagation of light.

William Alek

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INTALEK, INC.

Rev 3.2

THE GRAVITATIONAL COUPLING OF AN ELECTROMAGNETIC WAVE DECREASED GRAVITY

g −∆y PHOTON A

RED SHIFTING: DECREASING FREQUENCY AND ENERGY

−∆y

GRAVITATIONAL REFERENCE

MONOCHROMATIC LIGHT SOURCE

gy

BLUE SHIFTING: INCREASING FREQUENCY AND ENERGY

+∆y

PHOTON B INCREASED GRAVITY

g +∆y TOWARDS CENTER OF GRAVITY

FIGURE 14. Electromagnetic waves propagating within a given gravity well.

Shown above is the BLUE SHIFTING of an electromagnetic wave due to gravity. Relative to a GRAVITATIONAL REFERENCE point or equipotential surface of gravity within a given gravity well g y , PHOTON A decreases in energy Eλ y and frequency f λ y as it propagates through decreasing gravity g −∆y . Likewise, PHOTON B increases in energy Eλ y and frequency f λ y as it propagates through increasing gravity g +∆y . This effect was demonstrated in the Pound, Rebka and Snider experiment, which used Mossbaurer spectroscopy to measure the electromagnetic gravitational RED SHIFT (or BLUE SHIFT) of 14.4 keV gamma rays emitted from Fe57 through a vertical distance of 22.6 m . Using the difference form of fluctuating energy ∆Eλ of an electromagnetic wave propagating through gravity well g y and the gamma rays emitted upward, the RED SHIFT was within one percent (1%) of this result, SHIFT =

SHIFT =

∆Eλ g y1 y1 − g y0 y0 = Eλ c2

(9.806581 m sec 2 )(6.3781226 × 106 m) − ( 9.806650 m sec 2 )( 6.3781000 × 106 m ) (2.99792458 × 108 m sec) 2

SHIFT = −2.465961× 10−15

William Alek

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(192)

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INTALEK, INC.

Rev 3.2

And with the gamma rays emitted downward, the BLUE SHIFT was,

( 9.806650 m sec )( 6.3781000 ×10 m ) − (9.806581 m sec )(6.3781226 ×10 m) 2

SHIFT =

6

2

6

(194)

(2.99792458 × 108 m sec) 2

SHIFT = 2.465961× 10−15

(195)

The difference form and exponential form of the natural relativistic electromagnetic energy Eλ y at position ±∆y is,

Eλ y1 = Eλ y0 ± ∆Eλ y0 = Eλ y0

⎛ g y1 y1 − g y0 y0 ⎞ ⎟⎟ c2 ⎠

⎜⎜ ⎛ g y1 y1 − g y0 y0 ⎞ ⎝ ⎜1 ± ⎟ = Eλ y0 e 2 c ⎝ ⎠

(196)

Since the energy Eλ of a single photon is, Eλ = h f λ

(197)

Then, the Planck's constant h is, h=

Eλ fλ

(198)

The difference form of the gravitational RED SHIFT (or BLUE SHIFT) of frequency ∆f λ f λ of an electromagnetic wave propagating ∆y = y0 − y1 within a given gravity well g y is, ∆f λ g y1 y1 − g y0 y0 = fλ c2

(199)

The difference form and exponential form of the natural relativistic electromagnetic frequency f λ y at position ±∆y is,

f λ y1 = f λ y0 ± ∆f λ y0 = f λ y0

⎛ g y1 y1 − g y0 y0 ⎞ ⎟⎟ c2 ⎠

⎜⎜ ⎛ g y1 y1 − g y0 y0 ⎞ ⎝ = f e ⎜1 ± ⎟ y λ 0 c2 ⎝ ⎠

(200)

So, the natural relativistic Planck's constant hy ranging from position −∆y to +∆y evaluates to unity gain or,

hy1 =

Eλ y1 f λ y1

=

Eλ y0 e f λ y0 e

⎛ g y1 y1 − g y0 y0 ⎞ ⎜⎜ ⎟⎟ c2 ⎝ ⎠

⎛ g y1 y1 − g y0 y0 ⎞ ⎜⎜ ⎟⎟ c2 ⎝ ⎠

=

Eλ =h fλ

(201)

Therefore, the natural relativistic Planck's constant hy is invariant between equipotential surfaces of gravity or, hy = 6.6260755 × 10 −34 Joule ⋅ sec

William Alek

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Gravitational Mass Fluctuations

INTALEK, INC.

Rev 3.2

The speed of light c is, c = λ fλ

(203)

The difference form of the gravitational RED SHIFT (or BLUE SHIFT) of the wavelength ∆λ λ of an electromagnetic wave displaced ∆y = y0 − y1 within a given gravity well g y is, ∆λ

λ

=−

g y1 y1 − g y0 y0

(204)

c2

The natural relativistic wavelength λ y at position ±∆y is,

⎛

λ y = λ y ∓ ∆λ y = λ y ⎜1 ∓ 1

0

0

0

⎝

⎛ g y1 y1 − g y0 y0 ⎞ ⎟⎟ c2 ⎠

⎜⎜ − g y1 y1 − g y0 y0 ⎞ ⎝ ⎟ = λ y0 e 2 c ⎠

(205)

So, natural relativistic the speed of light c y ranging from position −∆y to +∆y evaluates to unity gain or, ⎛ g y1 y1 − g y0 y0 ⎞ ⎜⎜ − ⎟⎟ c2 ⎠

c y1 = λ y1 f λ y1 = λ y0 e⎝

⎛ g y1 y1 − g y0 y0 ⎞ ⎜⎜ ⎟⎟ c2 ⎠

f λ y0 e⎝

= λ fλ = c

(206)

Therefore, the natural relativistic speed of light c y is invariant between equipotential surfaces of gravity or, c y = 2.99792458 × 108 m sec

(207)

HOW GRAVITY AFFECTS THE PERMEABLILITY OF SPACE-TIME MEDIA The permeability µ0 of space-time media is given as,

µ0 = 4π × 10−7 H m

(208)

The difference form of the gravitational RED SHIFT (or BLUE SHIFT) of inductance ∆L L displaced ∆y = y0 − y1 within a given gravity well g y is, g y y1 − g y0 y0 ∆L =− 1 L c2

(209)

The inductance Ly of space-time media at position ±∆y is, ⎛ g y1 y1 − g y0 y0 ⎞ ⎟⎟ c2 ⎠

⎜⎜ − ⎛ g y y1 − g y0 y0 ⎞ ⎝ Ly1 = Ly0 ∓ ∆Ly0 = Ly0 ⎜1 ∓ 1 = L e ⎟ y0 2 c ⎝ ⎠

William Alek

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Gravitational Mass Fluctuations

INTALEK, INC.

Rev 3.2

The difference form of the gravitational RED SHIFT (or BLUE SHIFT) of length ∆L L displaced ∆y = y0 − y1 within a given gravity well g y is, g y y1 − g y0 y0 ∆L =− 1 L c2

(211)

The length L y of space-time media at position ±∆y is, ⎛ g y1 y1 − g y0 y0 ⎞ ⎟⎟ c2 ⎠

⎜⎜ − ⎛ g y y1 − g y0 y0 ⎞ ⎝ L y1 = L y0 ∓ ∆L y0 = L y0 ⎜1 ∓ 1 L = e ⎟ y0 2 c ⎝ ⎠

(212)

So, the natural relativistic permeability µ y of space-time media ranging from position −∆y to +∆y evaluates to unity gain or,

µy = 1

Ly1 L y1

=

Ly0 e

⎛ g y1 y1 − g y0 y0 ⎜⎜ − c2 ⎝

⎞ ⎟⎟ ⎠

⎛ g y1 y1 − g y0 y0 ⎜⎜ − c2 ⎝

⎞ ⎟⎟ ⎠

L y0 e

=

L = µ0 L

(213)

Therefore, the natural relativistic permeability µ y of space-time media is invariant between equipotential surfaces of gravity or,

µ y = 4π × 10−7 H m

(214)

HOW GRAVITY AFFECTS THE PERMITTIVITY OF SPACE-TIME MEDIA The permittivity ε 0 of space-time media is given as,

ε 0 = 8.85419 × 10−12 F m

(215)

The difference form of the gravitational RED SHIFT (or BLUE SHIFT) of capacitance ∆C C displaced ∆y = y0 − y1 within a given gravity well g y is, g y y1 − g y0 y0 ∆C =− 1 C c2

(216)

The capacitance C y of space-time media at position ±∆y is, ⎛ g y1 y1 − g y0 y0 ⎞ ⎟⎟ c2 ⎠

⎜⎜ − ⎛ g y y1 − g y0 y0 ⎞ ⎝ C y1 = C y0 ∓ ∆C y0 = C y0 ⎜1 ∓ 1 ⎟ = C y0 e 2 c ⎝ ⎠

William Alek

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Gravitational Mass Fluctuations

INTALEK, INC.

Rev 3.2

The difference form of the gravitational RED SHIFT (or BLUE SHIFT) of length ∆L L displaced ∆y = y0 − y1 within a given gravity well g y is, g y y1 − g y0 y0 ∆L =− 1 L c2

(218)

The length L y of space-time media at position ±∆y is, ⎛ g y1 y1 − g y0 y0 ⎞ ⎟⎟ c2 ⎠

⎜⎜ − ⎛ g y y1 − g y0 y0 ⎞ ⎝ L y1 = L y0 ∓ ∆L y0 = L y0 ⎜1 ∓ 1 L = e ⎟ y0 2 c ⎝ ⎠

(219)

So, the natural relativistic permittivity ε y of space-time media ranging from position −∆y to +∆y evaluates to unity gain or,

εy = 1

C y1

L y1

=

C y0 e

⎛ g y1 y1 − g y0 y0 ⎜⎜ − c2 ⎝

L y0 e

⎞ ⎟⎟ ⎠

⎛ g y1 y1 − g y0 y0 ⎞ − ⎜⎜ ⎟⎟ c2 ⎝ ⎠

=

C = ε0 L

(220)

Therefore, the natural relativistic permittivity ε y of space-time media is invariant between equipotential surfaces of gravity or,

ε y = 8.85419 × 10−12 F m

(221)

HOW GRAVITY AFFECTS THE VIRTUAL RESISTANCE OF SPACE-TIME MEDIA The space-time media has virtual resistance or impedance Z 0 , and therefore, isn't capable of absorbing or dissipating electromagnetic energy. Its REAL resistance is infinite or, R0 = ∞ . This media serves to impede the propagation of light and the motion of matter and is calculated as,

µ0 ε0

Z0 =

(222)

Since it has been shown the permeability µ y and permittivity ε y of space-time media are invariant between equipotential surfaces of gravity, it follows the natural relativistic impedance Z y of space-time media ranging from position −∆y to +∆y evaluates to unity gain or, Z y1 =

µy

1

εy

= Z0

(223)

1

Therefore, the natural relativistic impedance Z y of space-time media is invariant between equipotential surfaces of gravity or, Z y = 376.730 Ω

William Alek

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Rev 3.2

HOW GRAVITY AFFECTS THE SPEED OF LIGHT The speed of light c between space-time media is, 1

c=

(225)

µ 0ε 0

Since it has been shown the permeability µ y and permittivity ε y of space-time media are invariant between equipotential surfaces of gravity, it follows the natural relativistic speed of light c y through space-time media ranging from position −∆y to +∆y evaluates to unity gain or, 1

c y1 =

µy ε y 1

=c

(226)

1

Therefore, the natural relativistic speed of light c y through space-time media is invariant between equipotential surfaces of gravity or, c y = 2.997924 × 108 m sec

(227)

HOW GRAVITY AFFECTS BOLTZMANN'S CONSTANT The Boltzmann's Constant k is given as, k=

R N0

(228)

Since the Ideal Gas Constant Ry and Avogadro's Number N y are invariant between equipotential surfaces of gravity, it follows the natural relativistic Boltzmann's Constant k y ranging from position −∆y to +∆y evaluates to unity gain or, k y1 =

Ry1 N y1

=k

(229)

Therefore, the natural relativistic Boltzmann's Constant k y is invariant between equipotential surfaces of gravity or, k y = 1.380658 × 10−23 Joules ° K

(230)

HOW GRAVITY AFFECTS AN ELECTRIC CHARGE A fundamental electric charge q is given as, q=

f E

(231)

The electric force f y increases with the square of a decreasing distance, and the electric field E y also increases with the square of a decreasing distance at position + dy . Likewise, the electric force f y decreases with the square of a

William Alek

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Gravitational Mass Fluctuations

INTALEK, INC.

Rev 3.2

increasing distance, and the electric field E y also decreases with the square of a increasing distance at position −dy , it follows the natural relativistic electric charge q y ranging from position −∆y to +∆y evaluates to unity gain or, q y1 =

f y1 E y1

=q

(232)

Therefore, the natural relativistic electric charge q y is invariant between equipotential surfaces of gravity or, q y = 1.60217733 × 10−19 Coul

(233)

HOW GRAVITY AFFECTS THE FINE STRUCTURE CONSTANT The Fine Structure Constant α is,

α=

q2 2ε0 h c

(234)

Since an electric charge q y , the speed of light c y , the permittivity ε y , and Planck's constant hy are invariant between equipotential surfaces of gravity, it follows the natural relativistic Fine Structure Constant α y ranging from position −∆y to +∆y evaluates to unity gain or,

αy = 1

q y1 2 2 ε y1 hy1 c y1

=α

(235)

Therefore, the natural relativistic Fine Structure Constant α y is invariant between equipotential surfaces of gravity or,

α y = 7.29738 × 10−3 =

William Alek

Page 37

1 137.0356

(236)

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INTALEK, INC.

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A TYPICAL ELECTROMAGNETIC WAVE E

B

FLAT SPACE-TIME MEDIA

t

FIGURE 15. Propagation of electromagnetic wave in flat space-time.

E

B

t

t

FIGURE 16. Typical B and E Fields.

William Alek

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Gravitational Mass Fluctuations

INTALEK, INC.

Rev 3.2

GRAVITATIONAL BLUE SHIFTING OF AN ELECTROMAGNETIC WAVE E

NOTE: SPACE-TIME MEDIA IS MODELED AS UNCONDENSED RELATIVISTIC MASS, INDUCTANCE AND CAPACITANCE THAT CAN BE COMPRESSED OR RAREFIED.

B

RAREFYING SPACE-TIME MEDIA

t

FIGURE 17. Propagation of electromagnetic wave in rarefied space-time.

E

B

t

t

FIGURE 18. Increasing magnitude and frequency of B and E Fields by gravitational function.

William Alek

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GRAVITATIONAL RED SHIFTING OF AN ELECTROMAGNETIC WAVE E

NOTE: SPACE-TIME MEDIA IS MODELED AS UNCONDENSED RELATIVISTIC MASS, INDUCTANCE AND CAPACITANCE THAT CAN BE COMPRESSED OR RAREFIED.

B

COMPRESSING SPACE-TIME MEDIA

t

FIGURE 19. Propagation of electromagnetic wave in compressing space-time.

E

B

t

t

FIGURE 20. Decreasing magnitude and frequency of B and E Fields by gravitational function.

William Alek

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INTALEK, INC.

Rev 3.2

FIGURE 21. A Global Positioning Satellite.

Example 5. A Global Positioning Satellite (GPS) transmits an electromagnetic signal at a frequency f λ SAT of ~ 10.23 MHz down to the Earth from an altitude of 20,186.8 km , and has an orbital velocity of 3.874 km sec . The natural relativistic BLUE SHIFT due to gravity and the special relativistic RED SHIFT due to velocity changes the frequency of this transmitted signal. So, given the corrected transmitted frequency f λ SAT , compute the BLUE

SHIFT and RED SHIFT such that a ground-based receiver will read a signal

f λ RX

that is precisely

10230000.000000 Hz . The signal frequency of the satellite is adjustable down to 1 µ Hz .

So, given, Altitude of satellite ∆y = 20.1868 × 106 m Orbital velocity vx = 3.874 × 103 m sec Corrected frequency of satellite f λ SAT = 10229999.995444 Hz Receiver located on surface of Earth y1 = 6.3781× 106 m Speed of light c = 2.99792458 × 108 m sec Gravitational constant G = 6.67260 × 10−11 N m 2 kg 2 Mass of the Earth M E = 5.9787 × 1024 kg The initial radius y0 of the satellite above the Earth is, y0 = y1 + ∆y = ( 6.3781× 106 m ) + ( 20.1868 × 106 m ) = 26.5649 × 106 m

William Alek

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The acceleration due to gravity at altitude y0 = 26.5649 × 106 m above the Earth is, g y0 =

−11 2 2 24 G M E ( 6.67260 × 10 N m kg )( 5.9787 × 10 kg ) = = 0.5653 m sec 2 2 6 y0 2 ( 26.5649 ×10 m )

(238)

The acceleration due to gravity at altitude y1 = 7.529 × 106 m above the Earth is, g y1 =

−11 2 2 24 G M E ( 6.67260 × 10 N m kg )( 5.9787 × 10 kg ) = = 9.80665 m sec 2 2 6 y12 ( 6.3781×10 m )

(239)

Given the exponential solution of the natural relativistic frequency model, the gravitational BLUE SHIFTED frequency is,

f λ y1 = f λ y0 e

⎛ g y1 y1 − g y0 y0 ⎞ ⎜⎜ ⎟⎟ c2 ⎝ ⎠

(

= (10229999.995444 Hz ) e

)(

)(

)(

⎛ 2 6 2 6 ⎜ 9.80665 m sec 6.3781×10 m − 0.5653 m sec 26.5649×10 m ⎜ 2 ⎜ 2.99792458×108 m sec ⎝

(

)

) ⎞⎟ ⎟ ⎟ ⎠

(240)

f λ y1 = 10230000.000854 Hz = f λ

(241)

Given the exponential solution of the special relativistic frequency model, the RED SHIFTED frequency of the BLUE SHIFTED frequency computed above is,

fλv = fλ e

⎛ vx 2 ⎜− 2 ⎜ 2c ⎝

⎞ ⎟ ⎟ ⎠

= (10230000.000854 Hz ) e

( (

)

2 ⎛ 3.874×103 m sec ⎜ ⎜− ⎜ 2 2.99793×108 m sec ⎝

)

2

⎞ ⎟ ⎟ ⎟ ⎠

(242)

f λ v = 10230000.000000 Hz = f λ RX

(243)

So, a ground-based receiver will read a signal that is precisely 10230000.000000 Hz with a satellite frequency f λ SAT given above.

FIGURE 22. A constellation of 24 Global Positioning Satellites (GPS) orbiting the Earth.

William Alek

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Rev 3.2

GRAVITOMAGNETIC THEORY P dB

r

θ

CURRENT ELEMENT

MAGNETIC INDUCTION I

dx

FIGURE 23. The magnetic induction produced by a current element.

A constant positive electric current I must create a stable magnetic field B around a wire. This stable field is due to the flow of electric current shown above. The change of magnetic induction dB at a fixed point P produced by a current element dx is calculated using the Biot-Savart’s Law,

µ0 I dx × r 4π r3

(244)

µ0 I sin (θ ) dx r2 4π

(245)

dB =

Or, dB =

Since charge q is quantized in a single electron e − then, electric current I is defined as quantity N of charges e − passing a fixed point per change of time dt or, I =q=

− dq d ( N e ) = dt dt

(246)

And velocity vx of an electron passing a fixed point is defined as change of distance dx per change of time dt or, vx = x =

William Alek

dx dt

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Then, the electric current I is redefined as, I = vx

d ( N e− )

(248)

dx

So, the change of magnetic induction dB at a fixed point P produced by quantity N of charges e − moving at velocity vx is, dB =

− µ0 vx sin (θ ) d ( N e ) 4π r2

(249)

To find the magnetic induction B produced by a single electron at point P when θ = 90 and N = 1 , then integrate, B = ∫ dB =

µ0 e − vx 4π r 2

(250)

The total energy density uB of magnetic field B contained within volume V is, uB =

UB B2 = V 2 µ0

(251)

Therefore, the total field energy U B of magnetic field B contained within volume V is,

µ0 ( e − ) vx 2 B2 UB = V= V 2 µ0 32 π 2 r 4 2

(252)

The change of magnetic field energy dU B contained within a change of volume d V is,

µ 0 ( e − ) vx 2 2

dU B =

32 π 2 r 4

dV

(253)

The total energy EM contained within matter is, EM = M c 2

(254)

Equate total magnetic field energy U B to the total energy EM contained within matter, U B = EM

(255)

dU B = dM B c 2

(256)

So, the change of magnetic field energy dU B is,

William Alek

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Therefore, the fluctuating magnetic mass dM B contained within a change of volume d V is,

µ 0 ( e ) vx 2 dU dM B = 2 B = dV c 32 π 2 r 4 c 2 − 2

(257)

ω rmax rmin

FIGURE 24. The volume V of an electron is modeled as a hollow spheroid.

Since the energy of an electron is finite, no field component can be present at its’ center. So, the volume of an electron is modeled as a hollow spheroid, π

rmax

0

rmin

V = 2 π ∫ sin (ω ) d ω ∫

r 2 dr = 4 π ∫

rmax

rmin

r 2 dr

(258)

The fluctuating magnetic mass dM B contained within a change of volume d V of a hollow spheroid is,

µ0 ( e − ) vx 2 µ 0 ( e − ) vx 2 ⎛ r 1 ⎞ r 2 dM B = d 4 π ∫ r dr = d⎜ dr ⎟ r c 2 ⎝ ∫r r 2 ⎠ 32 π 2 r 4 c 2 8π 2

(

2

)

max

min

max

(259)

min

Given the radius of a fluctuating magnetic mass dM e ranging from a classic electron radius re to infinity, or re ≤ r ≤ ∞ , the derivative form of a moving electron is,

µ0 ( e − ) vx 2 dM B = c2 8π 2

∫

∞

re

1 dr r2

(260)

The fluctuating magnetic mass dM B is,

µ0 ( e − ) vx 2 dM B = 8 π re c 2 2

William Alek

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Gravitational Mass Fluctuations

INTALEK, INC.

Rev 3.2

So, given the rest mass of an electron M e , the difference form of the special relativistic mass M ev of an electron moving at velocity vx is ⎛ v2 ⎞ M ev = γ SR M e = M e ± dM e = M e ⎜ 1 ± x 2 ⎟ ⎝ 2c ⎠

(262)

Test the fluctuating magnetic mass dM B with the special relativistic mass dM e ,

µ 0 ( e − ) vx 2 M v2 ⇔ e x2 2 8π re c 2 c 2

(263)

The equation reduces to,

µ0 ( e − )

2

8π re

⇔

Me 2

(264)

So, given, Permeability of free space µ0 = 4π × 10−7 H m Fundamental charge of an electron e − = 1.602177 × 10−19 C Classic electron radius re = 2.817941× 10−15 m Rest mass of an electron M e = 9.109390 × 10 −31 kg

( 4π ×10

−7

H m )(1.602177 × 10−19 C )

8 π ( 2.817941× 10−15 m )

2

⇔

9.109390 × 10−31 kg 2

(265)

The fluctuating magnetic mass of a moving electron is identical to the special relativistic mass at any velocity vx , 4.554693 ⇔ 4.554695

(266)

Therefore, the fluctuating magnetic mass dM B is the fluctuating mass dM e of an electron, dM B = dM e

(267)

So, the magnetic mass M B and the mass M e of the electron is,

MB = Me =

µ0 ( e − )

2

4 π re

(268)

And the fluctuating magnetic mass ∆M B and the fluctuating mass ∆M e is,

µ 0 ( e − ) vx 2 ∆M B = ∆M e = 8 π re c 2 2

William Alek

Page 46

(269)

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Gravitational Mass Fluctuations

INTALEK, INC.

Rev 3.2

So, the difference form of the inertial RED SHIFT (or BLUE SHIFT) of the magnetic mass ∆M B M B and the mass ∆M e M e of a particle moving at a velocity vx is, ∆M B ∆M e vx 2 = = 2 2c MB Me

(270)

A particle can move at a real (i.e., time-forward) velocity vx , at an imaginary (i.e., time-future) velocity j vx , or at a velocity that is a combination of the two. The real and imaginary components are rotated about the temporal axis and therefore, can be described as complex motion. The rotation is given as 0° ≤ θ ≤ 90° , where the real axis is θ = 0° and the imaginary time-future axis is θ = 90° . The complex number uses the Euler’s identity e jθ , which functions as a temporal rotation operator. The complex velocity vx is, vx = v e jθ = v cos θ + j v sin θ

(271)

Given the rest mass of an electron M e or the classic electron radius re , the difference forms of the special relativistic magnetic mass M ev model of a particle moving at a complex velocity vx , where 0° ≤ θ ≤ 90° are,

M ev

⎛ v2 = M e ± ∆M e = M e ⎜ 1 ± x 2 ⎝ 2c

⎛ vx 2 ⎞

⎜ 2⎟ ⎞ ⎜ 2c ⎟ ⎝ ⎠ ⎟ = Me e ⎠

− µ0 ( e − ) ⎛ vx 2 ⎞ µ0 ( e ) ⎜⎜⎝ 2 c e = M e ± ∆M e = ⎜1 ± 2 ⎟ = 4 π re ⎝ 2 c ⎠ 4 π re 2

M ev

(272)

2

⎛ vx 2 ⎞ ⎟ 2 ⎟ ⎠

(273)

Now, apply the new Principle of Equivalence Theorem where the fluctuating magnetic mass of a moving electron is equivalent to natural relativistic mass due to the Earth’s gravity well, ⎛1

−

1 ⎞

⎜ ⎟ g y y1 − g y y0 ) µ0 ( e − ) vx 2 ( y1 y0 ⎠ vx 2 ⎝ ∆M e = = Me 2 = Me = G ME Me c2 c2 8 π re c 2 2c

(274)

⎛1 1 ⎞ vx 2 = 2 G M E ⎜ − ⎟ ⎝ y1 y0 ⎠

(275)

⎛1 1 ⎞ vx = 2 G M E ⎜ − ⎟ ⎝ y1 y0 ⎠

(276)

2

1

0

The position y1 of an electron moving at a velocity vx within Earth’s gravity well gY where 0 < y1 ≤ ∞ or −1 ≤

y0 v x 2 is, 2G ME y1 =

William Alek

vx 2 ⎞ 1 ⎛ ⎜ g y0 y0 + ⎟ g y1 ⎝ 2 ⎠

Page 47

(277)

5/22/2005

Gravitational Mass Fluctuations

INTALEK, INC.

y1 =

Rev 3.2

y0 y v2 1+ 0 x 2G M E

(278)

The equivalent maximum complex velocity vx max at y1 = ∞ is, vx max = −

2G ME y0

(279)

Given the equivalent maximum complex velocity vx max , the minimum mass M e min at y1 = ∞ is,

M e min

⎛ G ME = M e ⎜1 − y0 c 2 ⎝

⎛ G ME ⎞

⎜− ⎟ ⎜ y c2 ⎟ ⎞ 0 ⎝ ⎠ M e = ⎟ e ⎠

(280)

So, the difference form of the gravitational RED SHIFT (or BLUE SHIFT) of the magnetic mass ∆M B M B and the mass ∆M e M e of a particle displaced a distance ∆y within a given gravity well gY is,

(

g y1 y1 − g y0 y0 ∆M B ∆M e = = MB Me c2

)

(281)

Given the rest mass of an electron M e or the classic electron radius re , the difference forms of the natural relativistic mass M ey1 model of a particle displaced a distance ∆y within a given gravity well gY are,

M ey1

(

)

⎛ g y1 y1 − g y0 y0 ⎞ ⎛ g y1 y1 − g y0 y0 ⎞ 2 ⎜⎜ ⎟⎟ ⎠ ⎟ = M e e⎝ c = M e ± ∆M e = M e ⎜ 1 ± 2 ⎜ ⎟ c ⎝ ⎠

M ey1 = M e ± ∆M e =

(g ⎜1 ±

2 µ0 ( e − ) ⎛

4 π re ⎜ ⎝

y1

)

⎛ g y −g y ⎞ y1 − g y0 y0 ⎞ µ0 ( e− ) ⎜⎜ y1 1c2 y0 0 ⎟⎟ ⎠ ⎟= e⎝ ⎟ 4 π re c2 ⎠

(282)

2

(283)

In summary, Gravitomagnetic Theory shows that a moving electron produces an increase in relativistic mass that extends from its’ classic radius re to infinity, and couples to gravity. This motion can either have a velocity vx or a complex (i.e., time-future) velocity j vx . If the velocity is complex, then the electron will exhibit an antigravitational effect, and produce a complex (i.e., time-future) magnetic field j B . In addition, the total field energy U B of a complex magnetic field j B contained within a volume V is NEGATIVE. Example 6. An electron e − moving through a wire at a time-forward velocity vx produces a time-forward magnetic induction B at a distance r . Given, Direction of time is forward θ = 0° Velocity of electron e − through a wire v = 1.0 ×10−2 m sec Permeability of free space µ0 = 4π × 10−7 H m Fundamental charge of an electron e − = 1.602177 × 10−19 C Rest mass of an electron M e = 9.109390 × 10−31 kg

William Alek

Page 48

5/22/2005

Gravitational Mass Fluctuations

INTALEK, INC.

Rev 3.2

Radius r = 1.0 m Gravitational constant G = 6.67260 × 10−11 N m 2 kg 2 Radius of surface of Earth y0 = 6.3781× 106 m Mass of the Earth M E = 5.9787 × 1024 kg The time-forward velocity vx , where θ = 0° is, vx = v e jθ = (1.0 × 10−2 m sec ) e j 0° = 1.0 × 10−2 m sec

(284)

The time-forward magnetic induction B at distance r is, B=

−7 −19 −2 µ0 e− vx ( 4 π × 10 H m )(1.602177 ×10 C )(1.0 ×10 m sec ) = 2 4π r 2 4 π (1.0 m )

B = 1.602177 × 10−28 T

(285)

(286)

The POSITIVE fluctuating mass ∆M e of the electron e − is,

(1.0 ×10−2 m sec ) v2 ∆M e = M e x 2 = ( 9.1093897 × 10−31 kg ) 2 2c 2 ( 2.99792458 × 108 m sec )

(287)

∆M e = 5.067782 × 10−52 kg

(288)

2

Applying the new Principle of Equivalence Theorem, y1 =

y0 = y0 v x 2 1+ 2G M E 1+

( 6.3781×10 m ) ( 6.3781×10 m )(1.0 ×10 m sec ) 2 ( 6.67260 × 10 N m kg )( 5.9787 × 10 6

−2

6

−11

2

2

(289)

2

24

kg )

y1 = 6.37809999999490 × 106 m

(290)

So, the equivalent POSITIVE displacement ∆y is gravitational within the Earth’s gravity well is, ∆y = y0 − y1 = 5.0981× 10−6 m

(291)

Example 7. An electron e − moving through a wire at a time-advanced velocity vx produces a time-advanced magnetic induction B at a distance r . Given, Direction of time is advanced θ = 45° Velocity of electron e − through a wire v = 1.0 ×10−2 m sec Permeability of free space µ0 = 4π × 10−7 H m Fundamental charge of an electron e − = 1.602177 × 10−19 C Rest mass of an electron M e = 9.109390 × 10−31 kg Radius r = 1.0 m

William Alek

Page 49

5/22/2005

Gravitational Mass Fluctuations

INTALEK, INC.

Rev 3.2

Gravitational constant G = 6.67260 × 10−11 N m 2 kg 2 Radius of surface of Earth y0 = 6.3781× 106 m Mass of the Earth M E = 5.9787 × 1024 kg The time-advanced velocity vx , where θ = 45° is, vx = v e jθ = (1.0 × 10−2 m sec ) e j 45° = 7.071068 × 10−3 + 7.071068 j × 10−3 m sec

(292)

The maximum time-future velocity vx max within the Earth’s gravity well is,

vx max = −

2 ( 6.67260 × 10−11 N m 2 kg 2 )( 5.9787 × 1024 kg ) 2G ME = − y0 ( 6.3781×106 m ) vx max = 1.11846 j × 104 m sec

(293)

(294)

vx

The time-advanced magnetic induction B at distance r is, B=

−7 −19 −3 −3 µ0 e − vx ( 4 π × 10 H m )(1.602177 × 10 C )( 7.071068 × 10 + 7.071068 j × 10 m sec ) = 2 4π r 2 4 π (1.0 m )

B = 1.132910 × 10−28 + 1.132910 j × 10−28 T

(295)

(296)

The IMAGINARY fluctuating mass ∆M e of the electron e − is, 7.071068 × 10−3 + 7.071068 j × 10−3 m sec ) ( vx 2 −31 ∆M e = M e 2 = ( 9.1093897 × 10 kg ) 2 2c 2 ( 2.99792458 × 108 m sec ) ∆M e = 5.067782 j × 10−52 kg

2

(297)

(298)

Applying the new Principle of Equivalence Theorem, 6.3781× 106 m ) ( y0 y1 = = 2 y v2 6.3781× 106 m )( 7.071068 × 10−3 + 7.071068 j × 10−3 m sec ) ( 1+ 0 x 2G M E 1+ 2 ( 6.67260 × 10−11 N m 2 kg 2 )( 5.9787 × 1024 kg ) y1 = 6.3781× 106 m − 5.0986 j × 10−6 m

(299)

(300)

So, the equivalent IMAGINARY displacement ∆y is shown to be non-gravitational within the Earth’s gravity well is, ∆y = y0 − y1 = 5.0986 j × 10−6 m

(301)

Example 8. An electron e − moving through a wire at a time-future velocity vx produces a time-future magnetic induction B at a distance r .

William Alek

Page 50

5/22/2005

Gravitational Mass Fluctuations

INTALEK, INC.

Rev 3.2

Given, Direction of time is future θ = 90° Velocity of electron e − through a wire v = 1.0 ×10−2 m sec Permeability of free space µ0 = 4π × 10−7 H m Fundamental charge of an electron e − = 1.602177 × 10−19 C Rest mass of an electron M e = 9.109390 × 10−31 kg Radius r = 1.0 m Gravitational constant G = 6.67260 × 10−11 N m 2 kg 2 Radius of surface of Earth y0 = 6.3781× 106 m Mass of the Earth M E = 5.9787 × 1024 kg The time-future velocity vx , where θ = 90° is, vx = v e jθ = (1.0 ×10 −2 m sec ) e j 90° = 1.0 j ×10 −2 m sec

(302)

The maximum time-future velocity vx max within the Earth’s gravity well is,

vx max = −

2 ( 6.67260 × 10−11 N m 2 kg 2 )( 5.9787 × 1024 kg ) 2G ME = − y0 ( 6.3781×106 m ) vx max = 1.11846 j × 104 m sec

(303)

(304)

vx

The time-future magnetic induction B at distance r is, B=

−7 −19 −2 µ0 e − vx ( 4 π × 10 H m )(1.602177 × 10 C )(1.0 j × 10 m sec ) = 2 4π r 2 4 π (1.0 m )

B = 1.602177 j × 10−28 T

(305)

(306)

The NEGATIVE fluctuating mass ∆M e of the electron e − is,

(1.0 j ×10−2 m sec ) v2 ∆M e = M e x 2 = ( 9.1093897 × 10−31 kg ) 2 2c 2 ( 2.99792458 × 108 m sec )

(307)

∆M e = −5.067782 × 10−52 kg

(308)

2

Applying the new Principle of Equivalence Theorem, y0 y1 = = y v2 1+ 0 x 2G M E 1+

( 6.3781×10 m ) ( 6.3781×10 m )(1.0 j ×10 m sec ) 2 ( 6.67260 × 10 N m kg )( 5.9787 × 10 6

−2

6

−11

2

2

y1 = 6.37800000000510 × 106 m

William Alek

Page 51

(309)

2

24

kg )

(310)

5/22/2005

Gravitational Mass Fluctuations

INTALEK, INC.

Rev 3.2

So, the equivalent NEGATIVE displacement ∆y is antigravitational within the Earth’s gravity well is, ∆y = y0 − y1 = −5.0981× 10−6 m

(311)

COMPLEX AMPERIAN CURRENTS y

e− vx e jθ

ELECTRON

NUCLEUS

r

Fv

B e jθ z

TEMPORAL ROTATION 0° ≤ θ ≤ 90°

MAGNETIC INDUCTION

x

CIRCULAR ELECTRON ORBIT

FIGURE 25. The complete complex Bohr model of the Hydrogen atom.

In the Bohr model of the Hydrogen atom, electrons move at relativistic speeds in discrete circular orbits around a nucleus. As a consequence of this motion, a magnetic field B is produced at the center of the orbit. If an opposing external magnetic field is applied to this field, the velocity of the electron will rotate into the imaginary axis or becomes complex. The electron moving at a time-future velocity j vx will create a time-future magnetic field j B that will emerge from the center. The complete complex Bohr model includes the following characteristic equations shown below. These equations contain the real and imaginary components of a moving electron that is rotated about the temporal axis as complex motion. The rotation is given as 0° ≤ θ ≤ 90° , where the real axis is θ = 0° and the imaginary time-future axis is θ = 90° . The complex number uses the Euler’s identity e jθ , which functions as a temporal rotation operator. So, given, Direction of time θ Frequency of orbit f Permeability of free space µ0 Fundamental charge of an electron e − Classic 1st Bohr orbital radius r Rest mass of an electron M e Speed of light c Gravitational constant G Mean radius of surface of Earth y0 Mass of the Earth M E

William Alek

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Gravitational Mass Fluctuations

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Rev 3.2

Mean radius of surface of Sun y0 Mass of the Sun M SUN The complex frequency of orbit f 0 , where 0° ≤ θ ≤ 90° is, f 0 = f e jθ = f cos θ + j f sin θ

(312)

i = e− f0

(313)

The complex Amperian Current i is,

The complex magnetic field B at the center axis of the orbit z = 0 m is, B = µ0

r2 i 2(r2 + z2 )

3

= µ0 2

B = µ0

r 2 e− f0 2(r2 + z2 )

3

(314) 2

e− f0 2r

(315)

The complex angular velocity ω of the electron e − is,

ω = 2 π f0

(316)

vx = r ω = 2 π r f 0

(317)

The complex velocity vx of the electron e − is,

The complex magnetic force Fv of the electron e − directed upon the nucleus is,

Fv = e − vx × B = e− r ω × B = µ0 π ( e− ) f 0 2 2

(318)

The direction of electron motion vx is such that the magnetic force Fv is always an attractive force between the electron and the nucleus. The difference form of the inertial RED SHIFT (or BLUE SHIFT) of the mass ∆M e M e of a particle moving at a velocity vx is, ∆M e vx 2 r 2ω 2 2 π 2 r 2 f 0 2 = 2 = = 2c 2 c2 Me c2

(319)

The special relativistic mass M ev of the circulating electron e − is,

⎛ v2 M ev = M e ± ∆M e = M e ⎜1 ± x 2 ⎝ 2c

William Alek

Page 53

⎛ vx 2 ⎞

⎜ ⎟ ⎞ ⎜ 2 c2 ⎟ ⎝ ⎠ ⎟ = Me e ⎠

(320)

5/22/2005

Gravitational Mass Fluctuations

INTALEK, INC.

Rev 3.2

Applying the new Principle of Equivalence Theorem,

∆M e = M e

2 π r f0 vx rω = Me = Me = Me 2 2 2c 2c c2 2

2

2

2

2

2

(

g y1 y1 − g y0 y0 c

)

2

⎛1 1 ⎞ ⎜ − ⎟ y y = G ME Me ⎝ 1 2 0 ⎠ c

(321)

⎛1 1 ⎞ vx 2 = g y1 y1 − g y0 y0 = G M E ⎜ − ⎟ 2 ⎝ y1 y0 ⎠

(322)

⎛1 1 ⎞ vx = 2 g y1 y1 − g y0 y0 = 2 G M E ⎜ − ⎟ ⎝ y1 y0 ⎠

(323)

(

)

The equivalent displacement to position y1 of an electron moving at a velocity vx within Earth’s gravity well gY where 0 < y1 ≤ ∞ or −1 ≤

y0 vx 2 is, 2G ME y1 =

vx 2 ⎞ 1 ⎛ ⎜ g y0 y0 + ⎟ g y1 ⎝ 2 ⎠

y1 =

(324)

y0 y v2 1+ 0 x 2G M E

(325)

So, the difference form of the gravitational RED SHIFT (or BLUE SHIFT) of the mass ∆M e M e of a particle displaced a distance ∆y within Earth’s gravity well gY is,

(

)

g y1 y1 − g y0 y0 ∆M e G ME ⎛ 1 1 ⎞ = = ⎜ − ⎟ Me c2 c 2 ⎝ y1 y0 ⎠

(326)

Given the rest mass of an electron M e , the difference forms of the natural relativistic mass M ey1 model of a particle displaced a distance ∆y within Earth’s gravity well gY are,

(

)

⎛ g y1 y1 − g y0 y0 ⎞ ⎛ ⎜⎜ ⎟⎟ g y1 y1 − g y0 y0 ⎞ 2 ⎠ ⎟ = M e e⎝ c M ey1 = M e ± ∆M e = M e ⎜1 ± 2 ⎜ ⎟ c ⎝ ⎠ ⎛ G ME ⎛ 1 1 ⎞⎞ ⎜ − ⎟⎟ c 2 ⎜⎝ y1 y0 ⎟⎠ ⎟⎠

⎜⎜ ⎛ G M ⎛ 1 1 ⎞⎞ M ey1 = M e ± ∆M e = M e ⎜1 ± 2 E ⎜ − ⎟ ⎟ = M e e⎝ ⎜ c ⎝ y1 y0 ⎠ ⎠⎟ ⎝

(327)

(328)

In summary, if the real component of the magnetic field is cancelled, or B = 0 T , due to an externally applied magnetic field BEXT , the velocity of the circulating electron is complex, or j vx and a complex magnetic field j B emerges. This complex magnetic field is believed to be present in the Aharonov-Bohm Experiment, which affected the flow of electrons. The complex Amperian Current uses the temporal rotation operator or Euler’s identity e jθ , where 0° ≤ θ ≤ 90° . As the motion of an electron rotates from real to imaginary, or θ → 90° , the electrons special

William Alek

Page 54

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Gravitational Mass Fluctuations

INTALEK, INC.

Rev 3.2

relativistic mass M v decreases and radius rv increases. In addition, the electrons radius of the orbit r (i.e., classic 1st Bohr radius) around the nucleus and the magnetic force Fv acting upon the nucleus remain constant or invariant during this rotation.

y

e− vx ELECTRON

NUCLEUS

r

Fv

B z

MAGNETIC INDUCTION

x

CIRCULAR ELECTRON ORBIT

TEMPORAL ROTATION θ = 0°

FIGURE 26. The time-forward Bohr model of the Hydrogen atom.

Example 9. In the time-forward Bohr model of the Hydrogen atom the electron e − circulates around the nucleus at a relativistic velocity vx as shown above. This creates a magnetic induction B emerging from the center of the nucleus. So, given, Direction of time is forward θ = 0° Frequency of orbit f = 6.8 × 1015 Hz Permeability of free space µ0 = 4π × 10−7 H m Fundamental charge of an electron e − = 1.602177 × 10−19 C Classic 1st Bohr orbital radius r = 5.291773 × 10−11 m Rest mass of an electron M e = 9.1093897 × 10−31 kg Speed of light c = 2.99792458 × 108 m sec Gravitational constant G = 6.67260 × 10−11 N m 2 kg 2 Mean radius of surface of Earth y0 = 6.3781× 106 m Mass of the Earth M E = 5.9787 × 1024 kg The time-forward frequency of orbit f 0 , where θ = 0° is, f 0 = f e jθ = ( 6.8 × 1015 Hz ) e j 0° = 6.8 × 1015 Hz

(329)

The time-forward Amperian Current i is,

William Alek

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Gravitational Mass Fluctuations

INTALEK, INC.

Rev 3.2

i = e − f 0 = (1.602177 × 10−19 C )( 6.8 × 1015 Hz ) = 1.089481× 10-3 Amps

(330)

The time-forward magnetic field B at the center axis of the orbit z = 0 m is, B = µ0

B = µ0

r2 i 2(r2 + z2 )

3

= µ0 2

r 2 e− f0 2(r2 + z2 )

3

(331) 2

(1.089481×10−3 Amps ) = 12.935946 T i = ( 4π × 10−7 H m ) 2r 2 ( 5.291773 × 10−11 m )

(332)

The time-forward angular velocity ω of the electron e − is,

ω = 2 π f 0 = 2 π ( 6.8 × 1015 Hz ) = 4.272566 × 1016 Hz

(333)

The time-forward velocity vx of the electron e − is, vx = r ω = ( 5.291773 × 10 −11 m )( 4.272566 × 1016 Hz ) = 2.260945 × 106 m sec

(334)

The time-forward magnetic force Fv of the electron e − directed upon the nucleus is, Fv = e − vx B = (1.602177 × 10−19 C )( 2.260945 ×106 m sec ) (12.935946 T )

(335)

Fv = 4.685962 × 10−12 N

(336)

The direction of electron motion vx is such that the magnetic force Fv is always an attractive force between the electron and the nucleus. The POSITIVE fluctuating mass ∆M e of the electron e − is, 2.260945 × 106 m sec ) ( vx 2 −31 ∆M e = M e 2 = ( 9.1093897 × 10 kg ) = 2.590585 × 10−35 kg 2 2c 2 ( 2.99792458 × 108 m sec ) 2

(337)

The increased special relativistic mass M v of the electron e − is, M v = M e + ∆M e = ( 9.1093897 × 10−31 kg ) + ( 2.590585 × 10−35 kg ) = 9.109649 × 10 −31 kg

(338)

Applying the new Principle of Equivalence Theorem, y0 = y1 = y v2 1+ 0 x 2G M E 1+

( 6.3781×10 m ) ( 6.3781×10 m )( 2.260945 ×10 m sec ) 2 ( 6.67260 × 10 N m kg )( 5.9787 × 10 kg ) 6

6

−11

y1 = 156.0784 m

William Alek

Page 56

6

2

2

2

(339)

24

(340)

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Gravitational Mass Fluctuations

INTALEK, INC.

Rev 3.2

So, the equivalent POSITIVE displacement ∆y is gravitational within the Earth’s gravity well is, ∆y = y0 − y1 = 6.377944 × 106 m

(341)

y

e− vx e j 45°

ELECTRON

NUCLEUS

r

Fv EXTERNAL MAGNETIC INDUCTION

B e j 45°

BEXT z

MAGNETIC INDUCTION

x

CIRCULAR ELECTRON ORBIT

TEMPORAL ROTATION θ = 45°

FIGURE 27. The time-advanced Bohr model of the Hydrogen atom.

Example 10. In the time-advanced Bohr model of the Hydrogen atom, the real magnetic field created by an electron circulating at relativistic speeds is partially cancelled by an externally applied magnetic field BEXT . The electron reacts by rotating its velocity into the imaginary axis as shown above. As a consequence of this complex velocity v e j 45° , a complex magnetic field B e j 45° emerges. So, given, Direction of time θ = 45° Frequency of orbit f = 6.8 × 1015 Hz Permeability of free space µ0 = 4π × 10−7 H m Fundamental charge of an electron e − = 1.602177 × 10−19 C Classic 1st Bohr orbital radius r = 5.291773 × 10−11 m Rest mass of an electron M e = 9.1093897 × 10−31 kg Speed of light c = 2.99792458 × 108 m sec Gravitational constant G = 6.67260 × 10−11 N m 2 kg 2 Mean radius of surface of Earth y0 = 6.3781× 106 m Mass of the Earth M E = 5.9787 × 1024 kg Mean radius of surface of Sun y0 = 6.96 × 108 m Mass of the Sun M SUN = 1.98892 × 1030 kg The time-advanced frequency of orbit f 0 , where θ = 45° is, f 0 = f e jθ = ( 6.8 × 1015 Hz ) e j 45° = 4.808326 × 1015 + 4.808326 j × 1015 Hz

William Alek

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(342)

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Gravitational Mass Fluctuations

INTALEK, INC.

Rev 3.2

The time-advanced Amperian Current i is, i = e − f 0 = (1.602177 × 10−19 C )( 4.808326 × 1015 + 4.808326 j × 1015 Hz )

(343)

i = 7.703791× 10 −4 + 7.703791 j × 10-4 Amps

(344)

The time-advanced magnetic field B at the center axis of the orbit z = 0 m is, B = µ0

r2 i 2(r2 + z2 )

3

= µ0 2

r 2 e− f 0 2(r2 + z2 )

3

(345) 2

7.703791× 10−4 + 7.703791 j × 10-4 Amps ) ( i −7 = ( 4π × 10 H m ) B = µ0 2r 2 ( 5.291773 × 10−11 m )

(346)

B = 9.147095 + 9.147095 j T

(347)

The time-advanced angular velocity ω of the electron e − is,

ω = 2 π f 0 = 2 π ( 4.808326 × 1015 + 4.808326 j × 1015 Hz )

(348)

ω = 3.02116 × 1016 + 3.02116 j × 1016 Hz

(349)

The time-advanced velocity vx of the electron e − is, vx = r ω = ( 5.291773 × 10−11 m )( 3.02116 × 1016 + 3.02116 j × 1016 Hz )

(350)

vx = 1.598729 × 106 + 1.598729 j × 106 m sec

(351)

The maximum time-future velocity vx max within the Earth’s gravity well is,

vx max

2 ( 6.67260 × 10−11 N m 2 kg 2 )( 5.9787 × 1024 kg ) 2G ME = − = − y0 ( 6.3781×106 m ) vx max = 1.11846 j × 104 m sec

vx

(352)

(353)

The time-advanced velocity vx of the electron e − far exceeds coupling to Earth’s gravity well! The maximum time-future velocity vx max within the Sun’s gravity well is,

vx max

2 ( 6.67260 × 10−11 N m 2 kg 2 )(1.98892 × 1030 kg ) 2 G M SUN = − = − y0 ( 6.96 x 108 m ) vx max = 6.17542 j × 105 m sec

William Alek

Page 58

vx

(354)

(355)

5/22/2005

Gravitational Mass Fluctuations

INTALEK, INC.

Rev 3.2

The time-advanced velocity vx of the electron e − far exceeds coupling to Sun’s gravity well! The time-advanced magnetic force Fv of the electron e − directed upon the nucleus is, Fv = e − vx B

(356)

Fv = (1.602177 × 10−19 C )(1.598729 ×106 + 1.598729 j × 106 m sec ) ( 9.147095 + 9.147095 j T ) Fv = 4.685962 j × 10−12 N

(357)

The direction of electron motion vx is time-advanced or rotated into the future such that the magnetic force Fv is always an attractive force between the electron and the nucleus. The IMAGINARY fluctuating mass ∆M e of the electron e − is, 1.598729 × 106 + 1.598729 j ×106 m sec ) ( vx 2 −31 ∆M e = M e 2 = ( 9.1093897 × 10 kg ) 2 2c 2 ( 2.99792458 × 108 m sec ) ∆M e = 2.590585 j × 10−35 kg

2

(358)

(359)

The special relativistic mass M v of the electron e − is, M v = M e + ∆M e = ( 9.1093897 × 10−31 kg ) + ( 2.590585 j × 10−35 kg )

(360)

M v = 9.1093897 × 10−31 + 2.590585 j × 10 −35 kg

(361)

Applying the new Principle of Equivalence Theorem, y1 =

( 6.3781×106 m ) y0 = 2 y v2 6.3781× 106 m )(1.598729 ×106 + 1.598729 j × 106 m sec ) ( 1+ 0 x 2G M E 1+ 2 ( 6.67260 × 10−11 N m 2 kg 2 )( 5.9787 × 10 24 kg ) y1 = 3.819579 × 10−3 − 156.0822 j m

(362)

(363)

So, the equivalent IMAGINARY displacement ∆y is shown to be non-gravitational within the Earth’s gravity well or, ∆y = y0 − y1 = 6.3781× 10−6 + 156.0822m

William Alek

Page 59

(364)

5/22/2005

Gravitational Mass Fluctuations

INTALEK, INC.

Rev 3.2

y

e− j vx ELECTRON

NUCLEUS

r

Fv EXTERNAL MAGNETIC INDUCTION BEXT z

jB

MAGNETIC INDUCTION

x

CIRCULAR ELECTRON ORBIT

TEMPORAL ROTATION θ = 90°

FIGURE 28. The time-future Bohr model of the Hydrogen atom.

Example 11. In the time-future Bohr model of the Hydrogen atom, the real magnetic field created by an electron circulating at relativistic speeds is being cancelled by an externally applied magnetic field BEXT . The electron reacts by rotating its velocity into the imaginary axis as shown above. As a consequence of this complex velocity j vx , a complex magnetic field j B emerges. So, given, Direction of time θ = 90° Frequency of orbit f = 6.8 × 1015 Hz Permeability of free space µ0 = 4π × 10−7 H m Fundamental charge of an electron e − = 1.602177 × 10−19 C Classic 1st Bohr orbital radius r = 5.291773 × 10−11 m Rest mass of an electron M e = 9.1093897 × 10−31 kg Speed of light c = 2.99792458 × 108 m sec Gravitational constant G = 6.67260 × 10−11 N m 2 kg 2 Mean radius of surface of Earth y0 = 6.3781× 106 m Mass of the Earth M E = 5.9787 × 1024 kg Mean radius of surface of Sun y0 = 6.96 × 108 m Mass of the Sun M SUN = 1.98892 × 1030 kg The time-future frequency of orbit f 0 , where θ = 90° is, f 0 = f e jθ = ( 6.8 × 1015 Hz ) e j 90° = 6.8 j × 1015 Hz

(365)

The time-future Amperian Current i is, i = e − f 0 = (1.602177 × 10−19 C )( 6.8 j × 1015 Hz ) = 1.089481 j × 10-3 Amps

William Alek

Page 60

(366)

5/22/2005

Gravitational Mass Fluctuations

INTALEK, INC.

Rev 3.2

The time-future magnetic field B at the center axis of the orbit z = 0 m is, B = µ0

r2 i 2(r2 + z2 )

3

= µ0 2

r 2 e− f 0 2(r2 + z2 )

3

(367) 2

(1.089481 j ×10 Amps ) = 12.935946 j T i = ( 4π × 10−7 H m ) 2r 2 ( 5.291773 × 10−11 m ) −3

B = µ0

(368)

The time-future angular velocity ω of the electron e − is,

ω = 2 π f 0 = 2 π ( 6.8 j × 1015 Hz ) = 4.272566 j × 1016 Hz

(369)

The time-future velocity vx of the electron e − is, vx = r ω = ( 5.291773 × 10 −11 m )( 4.272566 j × 1016 Hz ) = 2.260945 j × 106 m sec

(370)

The maximum time-future velocity vx max within the Earth’s gravity well is,

vx max = −

2 ( 6.67260 × 10−11 N m 2 kg 2 )( 5.9787 × 1024 kg ) 2G ME = − y0 ( 6.3781×106 m ) vx max = 1.11846 j × 104 m sec

vx

(371)

(372)

The time-future velocity vx of the electron e − far exceeds coupling to Earth’s gravity well. The maximum time-future velocity vx max within the Sun’s gravity well is,

vx max

2 ( 6.67260 × 10−11 N m 2 kg 2 )(1.98892 × 1030 kg ) 2 G M SUN = − = − y0 ( 6.96 x 108 m ) vx max = 6.17542 j × 105 m sec

vx

(373)

(374)

The time-advanced velocity vx of the electron e − far exceeds coupling to Sun’s gravity well! The magnetic force Fv of the electron e − directed upon the nucleus is, Fv = e − vx B = (1.602177 × 10−19 C )( 2.260945 j × 106 m sec ) (12.935946 j T )

(375)

Fv = −4.685962 × 10−12 N

(376)

The direction of electron motion vx is time-future or rotated into the future such that the magnetic force Fv is always an attractive force between the electron and the nucleus.

William Alek

Page 61

5/22/2005

Gravitational Mass Fluctuations

INTALEK, INC.

Rev 3.2

The NEGATIVE fluctuating mass ∆M e of the electron e − is,

( 2.260945 j ×106 m sec ) = −2.590585 ×10−35 kg v2 ∆M e = M e x 2 = ( 9.1093897 × 10−31 kg ) 2 2c 2 ( 2.99792458 × 108 m sec ) 2

(377)

The decreased special relativistic mass M v of the electron e − is, M v = M e + ∆M e = ( 9.1093897 × 10−31 kg ) + ( −2.590585 × 10−35 kg ) = 9.109131× 10 −31 kg

(378)

Applying the new Principle of Equivalence Theorem, y1 =

y0 = y0 vx 2 1+ 2G M E 1+

( 6.3781×10 m ) ( 6.3781×10 m )( 2.260945 j ×10 m sec ) 2 ( 6.67260 × 10 N m kg )( 5.9787 × 10 kg ) 6

6

2

6

−11

2

2

(379)

24

y1 = −156.0860 m

(380)

So, the equivalent NEGATIVE displacement ∆y is antigravitational within the Earth’s gravity well is, ∆y = y0 − y1 = 6.378256 × 106 m

(381)

COMPLEX ELECTRON DRIFT VELOCITY SEGMENT OF COPPER WIRE ELECTRON DRIFT

E e jθ e vx e

r

I e jθ

e−

−

vx e jθ

jθ

A’

e− vx e jθ

A

x TEMPORAL ROTATION 0° ≤ θ ≤ 90°

FIGURE 29. The complete complex electron drift velocity model in a copper wire.

If a copper wire is connected to a battery, an electric field E will be set up at every point within the wire. This field E will act on electrons and will give them a resultant motion. An electric current I is established if a net charge qCu passes through any cross sectional area A of the conductor in time t . The electric field that acts on the electrons doesn’t produce a net acceleration because the electrons keep colliding with the atoms that make up the conductor. The electrons, therefore, move at an average drift velocity vx . If the electron drift velocity j vx is timefuture, the associated electric field j E and magnetic field j B are also time-future. The complete complex electron drift velocity model includes the following characteristic equations shown below. These equations contain the real and imaginary components of a moving electron that is rotated about the temporal axis as a complex particle. The

William Alek

Page 62

5/22/2005

Gravitational Mass Fluctuations

INTALEK, INC.

Rev 3.2

rotation is given as 0° ≤ θ ≤ 90° , where the real axis is θ = 0° and the imaginary time-future axis is θ = 90° . The complex number uses the Euler’s identity e jθ , which functions as a temporal rotation operator. So, given, Direction of time θ Current flowing through a conductor I Fundamental charge of an electron e − Radius of copper wire r Density of conductor material ( 20°C ) Datom Number of conduction electrons per atom of conductor katom Avogadro’s Number N 0 Atomic weight of conductor material Watom Segment length of conductor x Speed of light c Rest mass of an electron M e Radius of surface of Earth y0 Gravitational constant G Mass of the Earth M E Resistivity of conductor material ( 20°C ) ρ atom

The complex current I 0 flowing through a conductor, where 0° ≤ θ ≤ 90° is, I 0 = I e jθ = I cos θ + j I sin θ

(382)

The complex current density J 0 is, J0 =

I0 I = 02 A πr

(383)

The volume V of a segment of a conductor is,

V = A x = π r2x

(384)

The quantity of conduction electrons natom in a volume of conductor is, natom =

Datom N 0 katom Watom

(385)

The net charge qatom in a volume of a conductor is, qatom = natom V e −

(386)

The complex velocity vx is, x t

(387)

Page 63

5/22/2005

vx =

William Alek

Gravitational Mass Fluctuations

INTALEK, INC.

Rev 3.2

The complex current I 0 flowing through a conductor is,

I0 =

qatom natom V e − = = π r 2 natom e − vx x t vx

(388)

So, the complex drift velocity vx of an electron moving through a conductor is, vx =

I0 J0 = − π r natom e natom e− 2

(389)

The fluctuating mass ∆M e of the electron e − is, ∆M e = M e

vx 2 2 c2

(390)

Applying the new Principle of Equivalence Theorem,

∆M e = M e

2

vx = Me 2 c2

(

g y1 y1 − g y0 y0

)

c2

⎛1 1 ⎞ ⎜ − ⎟ y y = G ME Me ⎝ 1 2 0 ⎠ c

(391)

⎛1 1 ⎞ vx 2 = g y1 y1 − g y0 y0 = G M E ⎜ − ⎟ 2 ⎝ y1 y0 ⎠

(392)

⎛1 1 ⎞ vx = 2 g y1 y1 − g y0 y0 = 2 G M E ⎜ − ⎟ ⎝ y1 y0 ⎠

(393)

(

)

The equivalent displacement to position y1 of an electron moving at a velocity vx within Earth’s gravity well gY where 0 < y1 ≤ ∞ or −1 ≤

y0 vx 2 is, 2G ME y1 =

vx 2 ⎞ 1 ⎛ ⎜ g y0 y0 + ⎟ g y1 ⎝ 2 ⎠

y1 =

y0 y v2 1+ 0 x 2G M E

(394)

(395)

The resistivity ρ Atom of a conductor is given as,

ρ Atom =

William Alek

V V E x = x= I 2 J I A πr

Page 64

(396)

5/22/2005

Gravitational Mass Fluctuations

INTALEK, INC.

Rev 3.2

The resistance R of a segment of conductor is, R=

V x x = ρ Atom = ρ Atom π r2 I A

(397)

SEGMENT OF COPPER WIRE ELECTRON DRIFT

E

e−

e−

vx

vx

e−

r

I

A’

vx A

x TEMPORAL ROTATION θ = 0°

FIGURE 30. The time-forward electron drift velocity in a copper wire.

Example 12. A time-forward electric current I is established if a net charge qCu passes through any cross sectional area A of the conductor in time-forward t . The electrons move at an average time-forward drift velocity vx . So, given, Direction of time θ = 0° Current through copper wire I = 10.0 Amps Fundamental charge of an electron e − = 1.602177 × 10−19 C Radius of 10AWG copper wire r = 1.294 × 10−3 m Density of copper conductor ( 20°C ) DCu = 8.92 × 106 gm m3 Number of conduction electrons per atom of copper kCu = 1 electron atom Avogadro’s Number N 0 = 6.0221367 × 1023 atoms mole Atomic weight of copper conductor WCu = 63.546 gm mole Segment length x = 1 m Speed of light c = 2.99792458 × 108 m sec Rest mass of an electron M e = 9.1093897 × 10−31 kg Gravitational constant G = 6.67260 × 10−11 N m 2 kg 2 Mean Radius of surface of Earth y0 = 6.3781× 106 m Mass of the Earth M E = 5.9787 × 1024 kg

Resistivity of copper conductor ( 20°C ) ρCu = 1.68 × 10−8 Ω m The time-forward current I 0 flowing through a conductor, where θ = 0° is, I 0 = I e jθ = (10.0 Amps ) e j 0° = 10.0 Amps

William Alek

Page 65

(398)

5/22/2005

Gravitational Mass Fluctuations

INTALEK, INC.

Rev 3.2

The time-forward current density J 0 is, J0 =

(10.0 Amps ) I0 I = 02 = = 1.901× 106 Amps m 2 2 A πr π (1.294 × 10−3 m )

(399)

The volume V of a segment of copper wire is,

V = A x = π r 2 x = π (1.294 × 10−3 m ) (1m ) = 5.261× 10−6 m3 2

(400)

The quantity of conduction electrons nCu in a volume of copper wire is, nCu =

nCu =

(8.92 ×10

6

DCu N 0 kCu WCu

gm m3 )( 6.0221367 × 1023 atoms mole ) (1 electron atom )

( 63.546 gm

mole )

nCu = 8.4533 × 1028 electrons m3

(401)

(402)

(403)

The net charge qCu in a volume of copper wire is, qCu = nCu V e − = ( 8.4533 × 1028 electrons m3 )( 5.261× 10−6 m3 )(1.60217733 × 10−19 C )

(404)

qCu = 7.126 × 104 C

(405)

The time-forward velocity vx is, vx =

x t

(406)

The time-forward current I 0 flowing through a copper wire is,

I0 =

qCu nCu V e − = = π r 2 nCu e − vx = π r 2 nCu e − vx = 10.0 Amps x t vx

(407)

So, the time-forward drift velocity vx of an electron moving through a copper wire is,

William Alek

1.901× 106 Amps m 2 ) ( I0 J0 = = vx = π r 2 nCu e − nCu e − ( 8.4533 × 1028 electrons m3 )(1.60217733 × 10−19 C )

(408)

vx = 1.403 × 10−4 m sec

(409)

Page 66

5/22/2005

Gravitational Mass Fluctuations

INTALEK, INC.

Rev 3.2

The POSITIVE fluctuating mass ∆M e of the electron e − is,

(1.403 ×10−4 m sec ) v2 ∆M e = M e x 2 = ( 9.1093897 × 10−31 kg ) 2 2c 2 ( 2.99792458 × 108 m sec )

(410)

∆M e = 9.980 × 10−56 kg

(411)

2

The POSITIVE fluctuating mass of an electron is almost 25 orders of magnitude below its’ rest mass M e . Applying the new Principle of Equivalence Theorem, y1 =

y0 = y0 vx 2 1+ 2G M E 1+

( 6.3781×10 m ) ( 6.3781×10 m )(1.403 ×10 m sec ) 2 ( 6.67260 × 10 N m kg )( 5.9787 × 10 6

−4

6

−11

2

2

(412)

2

24

kg )

y1 = 6.3781× 106 m

(413)

So, the equivalent POSITIVE displacement ∆y is gravitational within the Earth’s gravity well is, ∆y = y0 − y1 = 1.8627 × 10−9 m

(414)

Given a time-forward voltage V and a time-forward current I , the resistivity ρCu of copper wire is given as,

ρCu =

V V E x = 1.68 × 10−8 Ω m = x= I 2 J I A πr

(415)

The resistance R of a segment of copper wire is, R=

William Alek

(1 m ) V x = ρCu = (1.68 × 10−8 Ω m ) = 3.193 × 10−3 Ω 2 −3 I A π (1.294 × 10 m )

Page 67

(416)

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Gravitational Mass Fluctuations

INTALEK, INC.

Rev 3.2

SEGMENT OF COPPER WIRE ELECTRON DRIFT

E e j 45° e vx e

vx e j 45°

j 45°

r

I e j 45°

e−

−

A’

e− vx e j 45°

A

x TEMPORAL ROTATION θ = 45°

FIGURE 31. The time-advanced electron drift velocity in a copper wire.

Example 13. A time-advanced electric current I is established if a net charge qCu passes through any cross sectional area A of the conductor in time-advanced t . The electrons move at an average time-advanced drift velocity vx . So, given, Direction of time θ = 45° Current flow through copper wire I = 10.0 Amps Fundamental charge of an electron e − = 1.602177 × 10−19 C Radius of 10AWG copper wire r = 1.294 × 10−3 m Density of copper conductor ( 20°C ) DCu = 8.92 × 106 gm m3 Number of conduction electrons per atom of copper kCu = 1 electron atom Avogadro’s Number N 0 = 6.0221367 × 1023 atoms mole Atomic weight of copper conductor WCu = 63.546 gm mole Segment length x = 1 m Speed of light c = 2.99792458 × 108 m sec Rest mass of an electron M e = 9.1093897 × 10−31 kg Gravitational constant G = 6.67260 × 10−11 N m 2 kg 2 Mean Radius of surface of Earth y0 = 6.3781× 106 m Mass of the Earth M E = 5.9787 × 1024 kg

Resistivity of copper conductor ( 20°C ) ρCu = 1.68 × 10−8 Ω m The time-advanced current I 0 flowing through a conductor, where θ = 45° is, I 0 = I e jθ = (10.0 Amps ) e j 45° = 7.071 + 7.071 j Amps

(417)

The time-advanced current density J 0 is, J0 =

William Alek

I0 I ( 7.071 + 7.071 j Amps ) = 02 = = 1.344 × 106 + 1.344 j × 106 Amps m 2 2 −3 A πr π (1.294 × 10 m )

Page 68

(418)

5/22/2005

Gravitational Mass Fluctuations

INTALEK, INC.

Rev 3.2

The volume V of a segment of copper wire is,

V = A x = π r 2 x = π (1.294 × 10−3 m ) (1m ) = 5.261× 10−6 m3 2

(419)

The quantity of conduction electrons nCu in a volume of copper wire is, nCu =

nCu =

(8.92 ×10

6

DCu N 0 kCu WCu

(420)

gm m3 )( 6.0221367 × 1023 atoms mole ) (1 electron atom )

( 63.546 gm

(421)

mole )

nCu = 8.4533 × 1028 electrons m3

(422)

The net charge qCu in a volume of a copper wire is, qCu = nCu V e − = ( 8.4533 × 1028 electrons m3 )( 5.261× 10−6 m3 )(1.60217733 × 10−19 C )

(423)

qCu = 7.126 × 104 C

(424)

The time-advanced velocity vx is, vx =

x t

(425)

The time-advanced current I 0 flowing through a copper wire is,

I0 =

qCu nCu V e − = = π r 2 nCu e − vx = 7.071 + 7.071 j Amps x t vx

(426)

So, the time-advanced drift velocity vx of an electron moving through a copper wire is, vx =

(1.344 ×106 + 1.344 j ×106 Amps m2 ) I0 J0 = = π r 2 nCu e − nCu e − ( 8.4533 × 1028 electrons m3 )(1.60217733 × 10−19 C ) vx = 9.923 × 10−5 + 9.923 j × 10−5 m sec

(427)

(428)

The fluctuating mass ∆M e of the electron e − is,

( 9.923 ×10−5 + 9.923 j ×10−5 m sec ) v2 ∆M e = M e x 2 = ( 9.1093897 × 10−31 kg ) 2 2c 2 ( 2.99792458 ×108 m sec ) ∆M e = 9.980 j × 10−56 kg

William Alek

Page 69

2

(429)

(430)

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Gravitational Mass Fluctuations

INTALEK, INC.

Rev 3.2

The fluctuating mass of an electron is almost 25 orders of magnitude beyond its’ rest mass M e and is imaginary. Applying the new Principle of Equivalence Theorem, y1 =

( 6.3781×106 m ) y0 = 2 y v2 6.3781× 106 m )( 9.923 × 10 −5 + 9.923 j × 10−5 m sec ) ( 1+ 0 x 2G M E 1+ 2 ( 6.67260 × 10−11 N m 2 kg 2 )( 5.9787 × 1024 kg ) y1 = 6.3781× 106 m

(431)

(432)

So, the equivalent IMAGINARY displacement ∆y is shown to be non-gravitational within the Earth’s gravity well is, ∆y = y0 − y1 = 1.0041 j × 10−9 m

(433)

Given a time-advanced voltage V and a time-advanced current I , the resistivity ρCu of copper wire is given as,

ρCu =

V V E x = 1.68 × 10−8 Ω m = x= I 2 J I A πr

(434)

The resistance R of a segment of copper wire is, R=

(1 m ) V x = ρCu = (1.68 × 10−8 Ω m ) = 3.193 × 10−3 Ω 2 −3 I A π (1.294 × 10 m )

(435)

SEGMENT OF COPPER WIRE ELECTRON DRIFT

jE

e−

e−

j vx

j vx

A’

e−

r

j vx

jI

A

x TEMPORAL ROTATION θ = 90°

FIGURE 32. The time-future electron drift velocity in a copper wire.

Example 14. A time-future electric current + j I is established if a net charge qCu passes through any cross sectional area A of the conductor in time-future t . The electrons move at an average time-future drift velocity + j vx .

William Alek

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5/22/2005

Gravitational Mass Fluctuations

INTALEK, INC.

Rev 3.2

So, given, Direction of time θ = 90° Current flow through copper wire I = 10.0 Amps Fundamental charge of an electron e − = 1.602177 × 10−19 C Radius of 10AWG copper wire r = 1.294 × 10−3 m Density of copper conductor ( 20°C ) DCu = 8.92 × 106 gm m3 Number of conduction electrons per atom of copper kCu = 1 electron atom Avogadro’s Number N 0 = 6.0221367 × 1023 atoms mole Atomic weight of copper conductor WCu = 63.546 gm mole Segment length x = 1 m Speed of light c = 2.99792458 × 108 m sec Rest mass of an electron M e = 9.1093897 × 10−31 kg Gravitational constant G = 6.67260 × 10−11 N m 2 kg 2 Mean Radius of surface of Earth y0 = 6.3781× 106 m Mass of the Earth M E = 5.9787 × 1024 kg

Resistivity of copper conductor ( 20°C ) ρCu = 1.68 × 10−8 Ω m The time-future current I 0 flowing through a conductor, where θ = 90° is, I 0 = I e jθ = (10.0 Amps ) e j 90° = 10.0 j Amps

(436)

The time-future current density J 0 is, J0 =

I0 I (10.0 j Amps ) = 02 = = 1.901 j × 106 Amps m 2 2 −3 A πr π (1.294 × 10 m )

(437)

The volume V of a segment of copper wire is,

V = A x = π r 2 x = π (1.294 × 10−3 m ) (1m ) = 5.261× 10−6 m3 2

(438)

The quantity of conduction electrons nCu in a volume of copper wire is, nCu =

nCu =

(8.92 ×10

6

DCu N 0 kCu WCu

gm m3 )( 6.0221367 × 1023 atoms mole ) (1 electron atom )

( 63.546 gm

mole )

nCu = 8.4533 × 1028 electrons m3

(439)

(440)

(441)

The net charge qCu in a volume of a copper wire is, qCu = nCu V e − = ( 8.4533 × 1028 electrons m3 )( 5.261× 10−6 m3 )(1.60217733 × 10−19 C )

William Alek

Page 71

(442)

5/22/2005

Gravitational Mass Fluctuations

INTALEK, INC.

Rev 3.2

qCu = 7.126 × 104 C

(443)

The time-future velocity vx is, vx =

x t

(444)

The time-future current I 0 flowing through a copper wire is,

I0 =

qCu nCu V e − = = π r 2 nCu e − vx = π r 2 nCu e − vx = 10.0 j Amps x t vx

(445)

So, the time-future drift velocity vx of an electron moving through a copper wire is, vx =

(1.901 j ×106 Amps m2 ) I0 J0 = = π r 2 nCu e − nCu e − ( 8.4533 × 1028 electrons m3 )(1.60217733 × 10−19 C ) vx = 1.403 j × 10 −4 m sec

(446)

(447)

The NEGATIVE fluctuating mass ∆M e of the electron e − is,

(1.403 j ×10−4 m sec ) v2 ∆M e = M e x 2 = ( 9.1093897 × 10−31 kg ) 2 2c 2 ( 2.99792458 × 108 m sec )

(448)

∆M e = −9.980 × 10−56 kg

(449)

2

The NEGATIVE fluctuating mass of an electron is almost 25 orders of magnitude below its’ rest mass M e . Applying the new Principle of Equivalence Theorem, y1 =

y0 = y0 vx 2 1+ 2G M E 1+

( 6.3781×10 m ) ( 6.3781×10 m )(1.403 j ×10 m sec ) 2 ( 6.67260 × 10 N m kg )( 5.9787 × 10 6

−4

6

−11

2

2

(450)

2

24

kg )

y1 = 6.3781× 106 m

(451)

So, the equivalent NEGATIVE displacement ∆y is antigravitational within the Earth’s gravity well is, ∆y = y0 − y1 = −9.3132 × 10−10 m

William Alek

Page 72

(452)

5/22/2005

Gravitational Mass Fluctuations

INTALEK, INC.

Rev 3.2

Given a time-future voltage V and a time-future current I , the resistivity ρCu of copper wire is given as,

ρCu =

V V E x = 1.68 × 10−8 Ω m = x= I 2 J I A πr

(453)

The resistance R of a segment of copper wire is, R=

(1 m ) V x = ρCu = (1.68 × 10−8 Ω m ) = 3.193 × 10−3 Ω 2 −3 I A π (1.294 × 10 m )

(454)

COMPLEX RESISTOR IR

+

+ Ve jθ

VR

-

R

-

FIGURE 33. The complete complex resistor.

Given a complex voltage source VS with a temporal rotation operator e jθ , where 0° ≤ θ ≤ 90° is acting upon the voltage, a complex direct current flows through resistor R . A complex voltage VR appears across the resistor. The resulting complex instantaneous power PR is dissipated or absorbed by the resistor. So, given, Direction of time θ Voltage supply V Resistor R The complex voltage supply VS is, VS = V e jθ = V cos θ + jV sin θ

(455)

The complex current I R flowing through resistor R is, IR =

VS R

(456)

The complex voltage VR across resistor R is, VR = I R R = VS

William Alek

Page 73

(457)

5/22/2005

Gravitational Mass Fluctuations

INTALEK, INC.

Rev 3.2

The real resistance R is, R=

VS IR

(458)

The complex instantaneous power PR dissipated and/or absorbed by the resistor is,

PR = VR I R = I R 2 R =

VR 2 R

(459)

Example 15. Given a time-forward voltage source VS and a known resistor value R , compute the time-forward current and power dissipated by the resistor. So, given, Direction of time θ = 0° Voltage Source V = 10.0Volts Resistor R = 2.5 Ω The time-forward voltage VS is, VS = V e jθ = (10.0Volts ) e j 0° = 10.0Volts

(460)

The time-forward current I R is, IR =

VS (10.0 Volts ) = = 4.0 Amps R ( 2.5 Ω )

(461)

The time-forward instantaneous power PR dissipated by the resistor is, V 2 (10.0Volts ) PR = R = = 40.0Watts R ( 2.5 Ω ) 2

(462)

Example 16. Given a time-advanced voltage source VS and a known resistor value R , compute the time-advanced current and power being dissipated and absorbed by the resistor. So, given, Direction of time θ = 45° Voltage Source V = 10.0Volts Resistor R = 2.5 Ω The time-advanced voltage VS is, VS = V e jθ = (10.0 Volts ) e j 45° = 7.0711 + 7.0711 jVolts

(463)

The time-advanced current I R is, IR =

William Alek

VS ( 7.0711 + 7.0711 jVolts ) = = 2.8284 + 2.8284 j Amps R ( 2.5 Ω )

Page 74

(464)

5/22/2005

Gravitational Mass Fluctuations

INTALEK, INC.

Rev 3.2

The time-advanced instantaneous power PR dissipated and absorbed of the resistor is, VR 2 ( 7.0711 + 7.0711 jVolts ) = = 40.0 jWatts R ( 2.5 Ω ) 2

PR =

(465)

The resistor is dissipating and absorbing an equal amount of heat. The resistor is therefore, temperature neutral or adiabatic. Example 17. Given a time-future voltage source VS and a known resistor value R , compute the time-future current and instantaneous power absorbed by the resistor. So, given, Direction of time θ = 90° Voltage Source V = 10.0Volts Resistor R = 2.5 Ω The time-future voltage VS is, VS = V e jθ = (10.0Volts ) e j 90° = 10.0 jVolts

(466)

The time-future current I R is, IR =

VS (10.0 jVolts ) = = 4.0 j Amps R ( 2.5 Ω )

(467)

The time-future instantaneous power PR absorbed by the resistor is, V 2 (10.0 jVolts ) PR = R = = −40.0Watts R ( 2.5 Ω ) 2

(468)

COMPLEX INDUCTOR iL t ≥ 0sec

S

+ vR

+ Ve jθ

R

-

-

L

+

vL FIGURE 34. The complete complex magnetizing inductor.

Given a complex voltage source VS with a temporal rotation operator e jθ , where 0° ≤ θ ≤ 90° is acting upon the voltage, when switch S closes at t = 0sec , a complex direct current iL flows through resistor R and magnetizes

William Alek

Page 75

5/22/2005

Gravitational Mass Fluctuations

INTALEK, INC.

Rev 3.2

inductor L . A complex voltage vR appears across the resistor and a complex voltage vL appears across inductor L . The resulting complex instantaneous power PR is dissipated and/or absorbed by the resistor, the complex instantaneous power PL stored in the inductor, and the complex energy EL stored in the inductor. So, given, Direction of time θ Time t Voltage supply V Inductor L Resistor R The complex voltage supply VS is, VS = V e jθ = V cos θ + jV sin θ

(469)

The complex voltage across the resistor R is, vR ( t ) = iL ( t ) R

(470)

The complex voltage across the inductor L is, vL ( t ) = L

diL dt

(471)

Letting t0 = 0.0sec , the complex current iL flowing through the resistor R and the inductor L is, VS = vR ( t ) + vL ( t ) = iL ( t ) R + L

iL ( t )

0

(473)

VS L diL − R R dt

(474)

1 iL ( t ) −

VS R

diL = −

R t dt L ∫t0

(475)

R t t L t0

(476)

iL ( t )

V ⎞ ⎛ ln ⎜ iL ( t ) − S ⎟ R ⎠0 ⎝

V ⎛ ln ⎜ iL ( t ) − S R ⎝

William Alek

(472)

VS L diL = iL ( t ) + R R dt iL ( t ) =

∫

diL dt

⎞ ⎛ VS ⎟ − ln ⎜ − R ⎠ ⎝

=−

VS ⎛ ⎜ iL ( t ) − R ⎞ ⎟ = ln ⎜ ⎠ ⎜⎜ − VS ⎝ R

Page 76

⎞ ⎟ R ⎟ = − ( t − t0 ) L ⎟⎟ ⎠

(477)

5/22/2005

Gravitational Mass Fluctuations

INTALEK, INC.

iL ( t ) − −

iL ( t ) =

VS R

Rev 3.2

VS R R R = e − L ( t − t0 ) = e − L t

(478)

R − t ⎞ VS VS − RL t VS ⎛ − e = ⎜1 − e L ⎟ R R R⎝ ⎠

(479)

The complex instantaneous power PR dissipated and/or absorbed by the resistor R is,

PR ( t ) = vR ( t ) iL ( t ) = R iL 2 ( t ) =

R − t ⎞ VS 2 ⎛ L − 1 e ⎜ ⎟ R ⎝ ⎠

2

(480)

The complex instantaneous power PL stored in the inductor L is, PL ( t ) = vL ( t ) iL ( t ) = L iL

PL ( t ) =

R − t ⎞ VS 2 ⎛ VS 2 L 1 − − e ⎜ ⎟ R ⎝ ⎠ R

diL = VS iL ( t ) − R iL 2 ( t ) dt

2R R R R − t ⎞ − t ⎞⎛ − t ⎞⎞ − t ⎞ ⎛ ⎛ VS 2 ⎛ VS 2 ⎛ − RL t L L L L 1 − = 1 − 1 − 1 − = − e e e e e ⎜ ⎟ ⎜ ⎟ ⎜⎜ ⎜ ⎟ ⎟⎟ ⎜ ⎟ R ⎝ ⎝ ⎠ ⎠⎝ ⎝ ⎠⎠ R ⎝ ⎠

(481)

2

(482)

Letting t0 = 0.0sec , the complex energy EL stored in the inductor L is, EL ( t ) = ∫ PL dt = L ∫ iL t

t

t0

t0

i (t ) 2 2 diL 1 dt =L ∫ iL diL = L ⎡ ⎡⎣iL ( t ) ⎤⎦ − ⎡⎣iL ( t0 ) ⎤⎦ ⎤ i t ( ) ⎣ ⎦ 0 dt 2

2 R R ⎡V ⎛ − t ⎞⎤ − t0 1 ⎡ ⎡ VS ⎛ L EL ( t ) = L ⎢ ⎢ ⎜ 1 − e ⎟ ⎥ − ⎢ S ⎜ 1 − e L 2 ⎢ ⎢⎣ R ⎝ ⎠ ⎥⎦ ⎢⎣ R ⎝ ⎣

EL ( t ) =

⎞⎤ ⎟⎥ ⎠ ⎥⎦

2

⎤ LV 2 S ⎥= ⎥ 2 R2 ⎦

R − t ⎞ LVS 2 ⎛ L 1 − e ⎜ ⎟ 2 R2 ⎝ ⎠

2 2 R R ⎡⎛ − t ⎞ − t0 ⎞ ⎤ ⎛ L L ⎢⎜ 1 − e ⎟ − ⎜ 1 − e ⎟ ⎥ ⎢⎣⎝ ⎠ ⎝ ⎠ ⎥⎦

(483)

(484)

2

(485)

Example 18. Given a time-forward voltage source VS , a known resistor value R and inductor value L , compute the time-forward current and power dissipated by the resistor, and the energy stored in the inductor.

So, given, Direction of time θ = 0° Time 0.0sec ≤ t ≤ 1.0sec Voltage supply V = 10.0Volts Inductor L = 470 mH Resistor R = 2.5 Ω The time-forward voltage supply VS is, VS = V e jθ = (10.0Volts ) e j 0° = 10.0Volts

William Alek

Page 77

(486)

5/22/2005

Gravitational Mass Fluctuations

INTALEK, INC.

Rev 3.2

The time-forward current iL flowing through the resistor R and the inductor L at t = 1.0sec is, ( 2.5Ω ) R − t ⎞ VS ⎛ (10.0Volts ) ⎛⎜ − ( 470 mH ) t ⎞⎟ L 1− e iL ( t ) = ⎜ 1 − e ⎟ = ⎟ R⎝ ( 2.5 Ω ) ⎜⎝ ⎠ ⎠

(487)

iL (1.0sec ) = 3.980 Amps

(488)

The time-forward instantaneous power PR dissipated by the resistor R at t = 1.0sec is, 2 ( 2.5Ω ) 2 R − t ⎞ VS 2 ⎛ (10.0Volts ) ⎛⎜ − ( 470 mH ) t ⎟⎞ L PR ( t ) = 1− e ⎜1 − e ⎟ = ⎟ R ⎝ ( 2.5 Ω ) ⎝⎜ ⎠ ⎠

2

(489)

PR (1.0sec ) = 39.609Watts

(490)

The time-forward instantaneous power PL stored in the inductor L at t = 0.130sec and at t = 1.0sec is, 2R − t ⎞ V 2 ⎛ −Rt (10.0Volts ) PL ( t ) = S ⎜ e L − e L ⎟ = R ⎝ ( 2.5 Ω ) ⎠

2

2 ( 2.5 Ω ) ⎛ − ( 2.5 Ω) t t ⎞ − 470 mH ) 470 mH ) ( ( ⎜e ⎟ −e ⎜ ⎟ ⎝ ⎠

(491)

PL ( 0.130sec ) = 10.0 Watts

(492)

PL (1.0sec ) = 0.195Watts

(493)

The time-forward energy EL stored in the inductor L from t0 = 0.0sec to t = 1.0sec is, R − t ⎞ LVS 2 ⎛ ( 470 mH )(10.0Volts ) L − 1 EL ( t ) == e ⎜ ⎟ = 2 2 2R ⎝ 2 ( 2.5 Ω ) ⎠ 2

2

( 2.5 Ω ) ⎞ ⎛ t − 470 mH ) ( ⎜1 − e ⎟ ⎜ ⎟ ⎝ ⎠

EL (1.0sec ) = 3.723 Joules

2

(494)

(495)

Example 19. Given a time-advanced voltage source VS , a known resistor value R and inductor value L , compute the time-advanced current and power dissipated and absorbed by the resistor, and the energy stored in the inductor.

So, given, Direction of time θ = 45° Time 0.0sec ≤ t ≤ 1.0sec Voltage supply V = 10.0Volts Inductor L = 470 mH Resistor R = 2.5 Ω The time-advanced voltage supply VS is, VS = V e jθ = (10.0Volts ) e j 45° = 7.071 + 7.071 jVolts

William Alek

Page 78

(496)

5/22/2005

Gravitational Mass Fluctuations

INTALEK, INC.

Rev 3.2

The time-advanced current iL flowing through the resistor R and the inductor L at t = 1.0sec is, ( 2.5Ω ) R − t ⎞ VS ⎛ ( 7.071 + 7.071 jVolts ) ⎜⎛ − ( 470 mH ) t ⎟⎞ L 1− e iL ( t ) = ⎜ 1 − e ⎟ = ⎜ ⎟ R⎝ ( 2.5 Ω ) ⎠ ⎝ ⎠

(497)

iL (1.0sec ) = 2.815 + 2.815 j Amps

(498)

The time-advanced instantaneous power PR dissipated and absorbed by the resistor R at t = 1.0sec is, 2 ( 2.5Ω ) 2 R − t ⎞ VS 2 ⎛ ( 7.071 + 7.071 jVolts ) ⎜⎛ − ( 470 mH ) t ⎟⎞ L PR ( t ) = 1− e ⎜1 − e ⎟ = ⎜ ⎟ R ⎝ ( 2.5 Ω ) ⎠ ⎝ ⎠

2

(499)

PR (1.0sec ) = 39.609 jWatts

(500)

The time-advanced instantaneous power PL stored in the inductor L at t = 0.130sec and at t = 1.0sec is, 2 ( 2.5 Ω ) ( 2.5 Ω ) 2 2R − t ⎞ VS 2 ⎛ − RL t ( 7.071 + 7.071 jVolts ) ⎛⎜ − ( 470 mH ) t − ( 470 mH ) t ⎞⎟ L −e PL ( t ) = e ⎜e − e ⎟= ⎜ ⎟ R ⎝ ( 2.5 Ω ) ⎠ ⎝ ⎠

(501)

PL ( 0.130sec ) = 10.0 jWatts

(502)

PL (1.0sec ) = 0.195 jWatts

(503)

The time-advanced energy EL stored in the inductor L from t0 = 0.0sec to t = 1.0sec is, 2 ( 2.5 Ω ) 2 R − t ⎞ LVS 2 ⎛ ( 470 mH )( 7.071 + 7.071 jVolts ) ⎛⎜ − ( 470 mH ) t ⎞⎟ L 1− e EL ( t ) == ⎜1 − e ⎟ = 2 ⎜ ⎟ 2 R2 ⎝ 2 ( 2.5 Ω ) ⎠ ⎝ ⎠

EL (1.0sec ) = 3.723 j Joules

2

(504)

(505)

Example 20. Given a time-future voltage source VS , a known resistor value R and inductor value L , compute the time-future current and power absorbed by the resistor, and the negative energy stored in the inductor.

So, given, Direction of time θ = 90° Time 0.0sec ≤ t ≤ 1.0sec Voltage supply V = 10.0Volts Inductor L = 470 mH Resistor R = 2.5 Ω The time-future voltage supply VS is, VS = V e jθ = (10.0 Volts ) e j 90° = 10.0 jVolts

William Alek

Page 79

(506)

5/22/2005

Gravitational Mass Fluctuations

INTALEK, INC.

Rev 3.2

The time-future iL flowing through the resistor R and the inductor L at t = 1.0sec is, ( 2.5Ω ) R − t ⎞ VS ⎛ (10.0 jVolts ) ⎛⎜ − ( 470 mH ) t ⎞⎟ L 1− e iL ( t ) = ⎜ 1 − e ⎟ = ⎟ R⎝ ( 2.5 Ω ) ⎜⎝ ⎠ ⎠

(507)

iL (1.0sec ) = 3.980 j Amps

(508)

The time-future instantaneous power PR absorbed by the resistor R at t = 1.0sec is, 2 ( 2.5Ω ) 2 R − t ⎞ VS 2 ⎛ (10.0 jVolts ) ⎛⎜ − ( 470 mH ) t ⎟⎞ L PR ( t ) = 1− e ⎜1 − e ⎟ = ⎟ R ⎝ ( 2.5 Ω ) ⎝⎜ ⎠ ⎠

2

(509)

PR (1.0sec ) = −39.609 Watts

(510)

The time-future instantaneous power PL stored in the inductor L at t = 0.130sec and at t = 1.0sec is, 2( 2.5 Ω ) ( 2.5 Ω ) 2 2R − t ⎞ VS 2 ⎛ − RL t (10.0 jVolts ) ⎛⎜ − ( 470 mH ) t − ( 470 mH ) t ⎟⎞ L −e PL ( t ) = e ⎜e − e ⎟= ⎟ R ⎝ ( 2.5 Ω ) ⎜⎝ ⎠ ⎠

(511)

PL ( 0.130sec ) = −10.0 Watts

(512)

PL (1.0sec ) = −0.195Watts

(513)

The time-future energy EL stored in the inductor L from t0 = 0.0sec to t = 1.0sec is, 2 ( 2.5 Ω ) 2 R − t ⎞ LVS 2 ⎛ ( 470 mH )(10.0 jVolts ) ⎛⎜ − ( 470 mH ) t ⎞⎟ L 1− e EL ( t ) = ⎜1 − e ⎟ = 2 ⎜ ⎟ 2 R2 ⎝ 2 ( 2.5 Ω ) ⎠ ⎝ ⎠

EL (1.0sec ) = −3.723 Joules

2

(514)

(515)

iL S

t ≥ 0sec

+

+ vL

vR

L

R

-

FIGURE 35. The complete complex demagnetizing inductor.

Given a complex energy EL stored in inductor L with a temporal rotation operator e jθ , where 0° ≤ θ ≤ 90° is acting upon the voltage, when switch S closes at t = 0sec , a complex direct current iL flows through resistor R .

William Alek

Page 80

5/22/2005

Gravitational Mass Fluctuations

INTALEK, INC.

Rev 3.2

The inductor L demagnetizes into the resistor. A complex voltage vR appears across the resistor R . The resulting complex instantaneous power PR and energy ER are dissipated and/or absorbed by the resistor. So, given, Direction of time θ Time t Initial current through inductor I Inductor L Resistor R The complex current I 0 through the inductor L is, I 0 = I e jθ = I cos θ + j I sin θ

(516)

At t = 0sec , the voltage V0 across the resistor R is, V0 = I 0 R

(517)

vR ( t ) = R iL ( t )

(518)

diL dt

(519)

The complex voltage vR across the resistor R is,

The complex voltage vL across the inductor L is, vL ( t ) = − L

Letting t0 = 0.0sec , the complex current iL flowing through the resistor R and the inductor L is, vR ( t ) = vL ( t ) diL dt

(521)

L diL R dt

(522)

iL ( t ) R = − L iL ( t ) = −

(520)

1 R diL = − dt iL ( t ) L

∫

i(t )

I0

(523)

1 R t diL = − ∫ dt iL ( t ) L t0

ln ( iL ( t ) )

iL ( t ) I0

=−

R t L

t t0

⎛ i (t ) ⎞ R ln ( iL ( t ) ) − ln ( I 0 ) = ln ⎜ L ⎟ = − ( t − t0 ) L ⎝ I0 ⎠

William Alek

Page 81

(524)

(525)

(526)

5/22/2005

Gravitational Mass Fluctuations

INTALEK, INC.

iL ( t ) I0

=e

−

Rev 3.2

R ( t − t0 ) L

iL ( t ) = I 0 e

(527)

R − t L

(528)

The complex instantaneous power PR dissipated and/or absorbed by the resistor R is,

PR ( t ) = vR ( t ) iL ( t ) = R iL 2 ( t ) = R I 0 2 e

−

2R t L

(529)

Letting t0 = 0.0sec , the complex energy ER dissipated and/or absorbed by the resistor R is, ER ( t ) = ∫ PR dt = R I 0 2 ∫ e t

t

t0

t0

−

2R t L

dt

2R 2R 2R t0 t⎤ − − L I 0 2 ⎡ − 2LR t0 ⎛ −L ⎞ − L t 2 ⎛ −L ⎞ L L − == − ER ( t ) = R I 0 2 ⎜ e R I e e e ⎢ ⎥ 0 ⎜ ⎟ ⎟ 2 ⎣ ⎝ 2R ⎠ ⎝ 2R ⎠ ⎦

ER ( t ) =

L I02 2

2R − t⎤ ⎡ L 1 − e ⎢ ⎥ ⎣ ⎦

(530)

(531)

(532)

Example 21. Given a time-forward voltage V across inductor L , a known resistor value R and inductor value L , compute the time-forward current flowing through the resistor, and the power and energy dissipated by the resistor.

So, given, Direction of time θ = 0° Time 0.0sec ≤ t ≤ 1.0sec Initial current through inductor I = 4.0 Amps Inductor L = 470 mH Resistor R = 2.5 Ω The time-forward current I 0 through the inductor L is, I 0 = I e jθ = ( 4.0 Amps ) e j 0° = 4.0 Amps

(533)

The time-forward current iL flowing through the resistor R and the inductor L at t = 0.0sec and at t = 1.0sec is, iL ( t ) = I 0 e

William Alek

R − t L

= ( 4.0 Amps ) e

−

( 2.5 Ω ) t ( 470 mH )

(534)

iL ( 0.0sec ) = 4.0 Amps

(535)

iL (1.0sec ) = 0.020 Amps

(536)

Page 82

5/22/2005

Gravitational Mass Fluctuations

INTALEK, INC.

Rev 3.2

The time-forward instantaneous power PR dissipated by the resistor R at t = 0.0sec and at t = 1.0sec is, PR ( t ) = R I 0 2 e

−

2R t L

= ( 2.5 Ω )( 4.0 Amps ) e 2

−

2 ( 2.5 Ω ) t ( 470 mH )

(537)

PR ( 0.0sec ) = 40.0Watts

(538)

PR (1.0sec ) = 9.592 × 10−4 Watts

(539)

The time-forward energy ER dissipated by the resistor R at t = 1.0sec is, ER ( t ) =

L I02 2

2R − t⎤ ⎡ ( 470 mH )( 4.0 Amps ) L − 1 e ⎢ ⎥= 2 ⎣ ⎦

2

2 ( 2.5 Ω ) ⎛ t ⎞ − ⎜1 − e ( 470 mH ) ⎟ ⎜ ⎟ ⎝ ⎠

ER (1.0sec ) = 3.760 Joules

(540)

(541)

Example 22. Given a time-advanced voltage V across inductor L , a known resistor value R and inductor value L , compute the time-advanced current flowing through the resistor, and the power and energy dissipated and absorbed by the resistor.

So, given, Direction of time θ = 45° Time 0.0sec ≤ t ≤ 1.0sec Initial current through inductor I = 4.0 Amps Inductor L = 470 mH Resistor R = 2.5 Ω The time-advanced current I 0 through the inductor L is, I 0 = I e jθ = ( 4.0 Amps ) e j 45° = 2.828 + 2.828 j Amps

(542)

The time-advanced current iL flowing through the resistor R and the inductor L at t = 0.0sec and at t = 1.0sec is, iL ( t ) = I 0 e

R − t L

= ( 2.828 + 2.828 j Amps ) e

−

( 2.5 Ω ) t ( 470 mH )

(543)

iL ( 0.0sec ) = 2.828 + 2.828 j Amps

(544)

iL (1.0sec ) = 0.014 + 0.014 j Amps

(545)

The time-advanced instantaneous power PR dissipated and absorbed by the resistor R at t = 0.0sec and at t = 1.0sec is, PR ( t ) = R I 0 e 2

−

2R t L

= ( 2.5 Ω )( 2.828 + 2.828 j Amps ) e 2

PR ( 0.0sec ) = 40.0 jWatts

William Alek

Page 83

−

2 ( 2.5 Ω ) t ( 470 mH )

(546) (547)

5/22/2005

Gravitational Mass Fluctuations

INTALEK, INC.

Rev 3.2

PR (1.0sec ) = 9.592 j × 10−4 Watts

(548)

The time-advanced energy ER dissipated and absorbed by the resistor R at t = 0.0sec and at t = 1.0sec is, 2R − t⎤ LI 2 ⎡ ( 470 mH )( 2.828 + 2.828 j Amps ) ER ( t ) = 0 ⎢1 − e L ⎥ = 2 ⎣ 2 ⎦

2

2 ( 2.5 Ω ) ⎛ t ⎞ − ⎜1 − e ( 470 mH ) ⎟ ⎜ ⎟ ⎝ ⎠

ER (1.0sec ) = 3.760 j Joules

(549)

(550)

Example 23. Given a time-future voltage V across inductor L , a known resistor value R and inductor value L , compute the time-future current flowing through the resistor, and the power and energy absorbed by the resistor.

So, given, Direction of time θ = 90° Time 0.0sec ≤ t ≤ 1.0sec Initial current through inductor I = 4.0 Amps Inductor L = 470 mH Resistor R = 2.5 Ω The time-future current I 0 through the inductor L is, I 0 = I e jθ = ( 4.0 Amps ) e j 90° = 4.0 j Amps

(551)

The time-future current iL flowing through the resistor R and the inductor L at t = 0.0sec and at t = 1.0sec is, iL ( t ) = I 0 e

R − t L

= ( 4.0 j Amps ) e

−

( 2.5 Ω ) t ( 470 mH )

(552)

iL ( 0.0sec ) = 4.0 j Amps

(553)

iL (1.0sec ) = 0.020 j Amps

(554)

The time-future instantaneous power PR absorbed by the resistor R at t = 0.0sec and at t = 1.0sec is, PR ( t ) = R I 0 2 e

−

2R t L

= ( 2.5 Ω )( 4.0 j Amps ) e 2

−

2 ( 2.5 Ω ) t ( 470 mH )

(555)

PR ( 0.0sec ) = −40.0 Watts

(556)

PR (1.0sec ) = −9.592 × 10−4 Watts

(557)

The time-future energy ER absorbed by the resistor R at t = 0.0sec and at t = 1.0sec is, 2R − t⎤ LI 2 ⎡ ( 470 mH )( 4.0 j Amps ) ER ( t ) = 0 ⎢1 − e L ⎥ = 2 ⎣ 2 ⎦

William Alek

Page 84

2

2( 2.5 Ω ) ⎛ t ⎞ − ⎜ 1 − e ( 470 mH ) ⎟ ⎜ ⎟ ⎝ ⎠

(558)

5/22/2005

Gravitational Mass Fluctuations

INTALEK, INC.

Rev 3.2

ER (1.0sec ) = −3.760 Joules

(559)

COMPLEX CAPACITOR iC t ≥ 0sec

S

+ vR

+ Ve jθ

R

-

-

C

+

vC FIGURE 36. The complete complex charging capacitor.

Given a complex voltage source VS with a temporal rotation operator e jθ , where 0° ≤ θ ≤ 90° is acting upon the voltage, when switch S closes at t = 0sec , a complex direct current flows through resistor R and charges capacitor C . A complex voltage vR appears across the resistor and a complex voltage vC appears across capacitor C . The

resulting complex power PR is dissipated and/or absorbed by the resistor and the complex energy EC is stored in the capacitor. So, given, Direction of time θ Time t Voltage supply V Capacitor C Resistor R The complex voltage supply VS is, VS = V e jθ = V cos θ + jV sin θ

(560)

The complex voltage vR across the resistor R is, vR ( t ) = iC ( t ) R

(561)

The complex current iC through a capacitor C is, iC ( t ) = C

dvC dt

(562)

Letting t0 = 0.0sec , the complex voltage vC across the capacitor C is, VS = vR ( t ) + vC ( t ) = iC ( t ) R + vC ( t )

William Alek

Page 85

(563)

5/22/2005

Gravitational Mass Fluctuations

INTALEK, INC.

dvC dt

(564)

1 dt RC

(565)

vC ( t ) = VS − R C 1

vC ( t ) − VS

∫

vC ( t )

dvC = −

1

vC ( t ) − VS

0

dvC = − vC ( t )

ln ( vC ( t ) − VS )

0

Rev 3.2

=−

1 t dt R C ∫t0

(566)

1 t RC

(567)

⎛ v ( t ) − VS ln ( vC ( t ) − VS ) − ln ( −VS ) = ln ⎜ C ⎝ −VS vC ( t ) − VS −VS

=e

vC ( t ) = VS − VS e

−

−

1 ( t − t0 ) RC

1 t RC

=e

−

t t0

⎞ 1 ( t − t0 ) ⎟=− R C ⎠ 1 t RC

(568)

(569)

1 − t ⎞ ⎛ = VS ⎜1 − e RC ⎟ ⎝ ⎠

(570)

The complex instantaneous power PR dissipated and/or absorbed by the resistor R is, PR ( t ) = vR ( t ) iC ( t ) =

vR 2 ( t ) R

(V =

S

− vC ( t ) )

2

R

=

2 t VS 2 − RC e R

(571)

The complex instantaneous power PC stored in the capacitor C is, PC ( t ) = vC ( t ) iC ( t ) = C vC

2 dvC (VS vC ( t ) − vC ( t ) ) = dt R

1 1 1 1 2 − t ⎞ − t ⎞ − t ⎞⎛ − t ⎞⎞ − t ⎞ ⎛ V2⎛ V2⎛ V2⎛ V2⎛ −1t PC ( t ) = S ⎜ 1 − e RC ⎟ − S ⎜ 1 − e RC ⎟ = S ⎜ 1 − e RC ⎟ ⎜1 − ⎜ 1 − e RC ⎟ ⎟ = S ⎜ e RC − e RC ⎟ ⎜ ⎟ R ⎝ R ⎝ ⎠ R ⎝ ⎠ ⎠⎝ ⎝ ⎠⎠ R ⎝ ⎠

(572)

2

(573)

Letting t0 = 0.0sec , the complex energy EC stored in the capacitor C is, EL ( t ) = ∫ PL dt = C ∫ vC t

t

t0

t0

v (t ) 2 2 dvC 1 dt = C ∫ vC dvC = C ⎡ ⎡⎣ vC ( t ) ⎤⎦ − ⎡⎣ vC ( t0 ) ⎤⎦ ⎤ v ( t0 ) ⎦ dt 2 ⎣

2 2 1 1 ⎤ CV 2 ⎡ ⎛ − t ⎞⎤ − t0 ⎞ ⎤ 1 ⎡⎡ ⎛ S RC RC ⎢ EC ( t ) = C ⎢VS ⎜ 1 − e ⎟ ⎥ − ⎢VS ⎜1 − e ⎟⎥ ⎥ = 2 ⎢ ⎣⎢ ⎝ 2 ⎥ ⎥ ⎢ ⎥ ⎠ ⎝ ⎠ ⎦ ⎣ ⎦ ⎣ ⎦

1 − t ⎞ C VS 2 ⎛ EC ( t ) = ⎜ 1 − e RC ⎟ 2 ⎝ ⎠

William Alek

Page 86

2 2 1 1 ⎡⎛ − t ⎞ − t0 ⎞ ⎤ ⎛ ⎢⎜ 1 − e RC ⎟ − ⎜ 1 − e RC ⎟ ⎥ ⎢⎣⎝ ⎠ ⎝ ⎠ ⎥⎦

(574)

(575)

2

(576)

5/22/2005

Gravitational Mass Fluctuations

INTALEK, INC.

Rev 3.2

Example 24. Given a time-forward voltage source VS , a known resistor value R and capacitor value C , compute the time-forward current and power dissipated by the resistor, and the energy stored in the inductor.

So, given, Direction of time θ = 0° Time 0.0sec ≤ t ≤ 1.0sec Voltage supply V = 10.0Volts Capacitor C = 470 µ F Resistor R = 1.0 k Ω The time-forward voltage supply VS is, VS = V e jθ = (10.0Volts ) e j 0° = 10.0Volts

(577)

The time-forward voltage vC across the capacitor C at t = 1.0sec is, 1 1 − t ⎞ ⎛ − t ⎞ ⎛ vC ( t ) = VS ⎜1 − e RC ⎟ = (10.0 Volts ) ⎜ 1 − e (1.0 k Ω )( 470 µ F ) ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠

(578)

vC (1.0sec ) = 8.809 Volts

(579)

The time-forward instantaneous power PR dissipated by the resistor R at t = 0.0sec and at t = 1.0sec is,

V 2 − 2 t (10.0 Volts ) − (1.0 k Ω)( 470 µ F ) t PR ( t ) = S e RC = e R (1.0 k Ω )

(580)

PR ( 0.0sec ) = 0.10 Watts

(581)

PR (1.0sec ) = 1.419 × 10−3 Watts

(582)

2

2

The time-forward instantaneous power PC stored in the capacitor C at t = 0.33sec and at t = 1.0sec is, PC ( t ) =

VS 2 R

1 2 t − t ⎞ ⎛ − RC (10.0Volts ) RC e e − ⎜ ⎟= (1.0 k Ω ) ⎝ ⎠

2

2 − t ⎞ ⎛ − (1.0 k Ω)(1470 µ F ) t ⎜e − e (1.0 k Ω)( 470 µ F ) ⎟ ⎜ ⎟ ⎝ ⎠

(583)

PC ( 0.33sec ) = 0.0245Watts

(584)

PC (1.0sec ) = 0.0105Watts

(585)

The time-forward energy EC stored in the capacitor C from t0 = 0.0sec to t = 1.0sec is, 1 − t ⎞ C VS 2 ⎛ ( 470 µ F )(10.0Volts ) EC ( t ) = ⎜ 1 − e RC ⎟ = 2 ⎝ 2 ⎠ 2

William Alek

Page 87

2

1 − t ⎞ ⎛ (1.0 k Ω )( 470 µ F ) ⎜1 − e ⎟ ⎜ ⎟ ⎝ ⎠

2

(586)

5/22/2005

Gravitational Mass Fluctuations

INTALEK, INC.

Rev 3.2

EC ( t ) = 0.0182 Joules

(587)

Example 25. Given a time-advanced voltage source VS , a known resistor value R and capacitor value C , compute the time-advanced current and power dissipated and absorbed by the resistor, and the energy stored in the inductor.

So, given, Direction of time θ = 45° Time 0.0sec ≤ t ≤ 1.0sec Voltage supply V = 10.0Volts Capacitor C = 470 µ F Resistor R = 1.0 k Ω The time-advanced voltage supply VS is, VS = V e jθ = (10.0Volts ) e j 45° = 7.071 + 7.071 jVolts

(588)

The time-advanced voltage vC across the capacitor C at t = 1.0sec is, 1 1 − t ⎞ ⎛ − t ⎞ ⎛ vC ( t ) = VS ⎜1 − e RC ⎟ = ( 7.071 + 7.071 jVolts ) ⎜ 1 − e (1.0 k Ω)( 470 µ F ) ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠

(589)

vC (1.0sec ) = 6.229 + 6.229 jVolts

(590)

The time-advanced instantaneous power PR dissipated and absorbed by the resistor R at t = 0.0sec and at t = 1.0sec is, PR ( t ) =

VS 2 − RC2 t ( 7.071 + 7.071 jVolts ) − (1.0 k Ω )( 470 µ F ) t e e = R (1.0 k Ω ) 2

2

(591)

PR ( 0.0sec ) = 0.10 jWatts

(592)

PR (1.0sec ) = 1.419 j × 10−3 Watts

(593)

The time-advanced instantaneous power PC stored in the capacitor C at t = 0.33sec and at t = 1.0sec is, PC ( t ) =

1 2 t − t ⎞ VS 2 ⎛ − RC ( 7.071 + 7.071 jVolts ) − e RC ⎟ = ⎜e R ⎝ (1.0 k Ω ) ⎠

2

2 − t ⎞ ⎛ − (1.0 k Ω )(1470 µ F ) t ⎜e − e (1.0 k Ω )( 470 µ F ) ⎟ ⎜ ⎟ ⎝ ⎠

(594)

PC ( 0.33sec ) = 0.0245 jWatts

(595)

PC (1.0sec ) = 0.0105Watts

(596)

The time-advanced energy EC stored in the capacitor C from t0 = 0.0sec to t = 1.0sec is,

William Alek

Page 88

5/22/2005

Gravitational Mass Fluctuations

INTALEK, INC.

C VS 2 EC ( t ) = 2

Rev 3.2

1 − t ⎞ ⎛ ( 470 µ F )( 7.071 + 7.071 jVolts ) RC 1 e − ⎜ ⎟ = 2 ⎝ ⎠ 2

2

1 − t ⎞ ⎛ ⎜1 − e (1.0 k Ω )( 470 µ F ) ⎟ ⎜ ⎟ ⎝ ⎠

2

EC ( t ) = 0.0182 j Joules

(597)

(598)

Example 26. Given a time-future voltage source VS , a known resistor value R and capacitor value C , compute the time-future current and power absorbed by the resistor, and the energy stored in the inductor.

So, given, Direction of time θ = 90° Time 0.0sec ≤ t ≤ 1.0sec Voltage supply V = 10.0Volts Capacitor C = 470 µ F Resistor R = 1.0 k Ω The time-future voltage supply VS is, VS = V e jθ = (10.0 Volts ) e j 90° = 10.0 jVolts

(599)

The time-future voltage vC across the capacitor C at t = 1.0sec is, 1 1 − t ⎞ ⎛ − t ⎞ ⎛ vC ( t ) = VS ⎜1 − e RC ⎟ = (10.0 jVolts ) ⎜ 1 − e (1.0 k Ω)( 470 µ F ) ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠

(600)

vC (1.0sec ) = 8.809 jVolts

(601)

The time-future instantaneous power PR absorbed by the resistor R at t = 0.0sec and at t = 1.0sec is, V 2 − 2 t (10.0 jVolts ) − (1.0 k Ω)( 470 µ F ) t PR ( t ) = S e RC = e R (1.0 k Ω )

(602)

PR ( 0.0sec ) = −0.10 Watts

(603)

PR (1.0sec ) = −1.419 × 10−3 Watts

(604)

2

2

The time-future instantaneous power PC stored in the capacitor C at t = 0.33sec and at t = 1.0sec is, PC ( t ) =

1 2 t − t ⎞ VS 2 ⎛ − RC (10.0 jVolts ) − e RC ⎟ = ⎜e R ⎝ (1.0 k Ω ) ⎠

2

2 − t ⎞ ⎛ − (1.0 k Ω)(1470 µ F ) t ⎜e − e (1.0 k Ω)( 470 µ F ) ⎟ ⎜ ⎟ ⎝ ⎠

(605)

PC ( 0.33sec ) = −0.0245Watts

(606)

PC (1.0sec ) = −0.0105Watts

(607)

The time-future energy EC stored in the capacitor C from t0 = 0.0sec to t = 1.0sec is,

William Alek

Page 89

5/22/2005

Gravitational Mass Fluctuations

INTALEK, INC.

C VS 2 EC ( t ) = 2

Rev 3.2

1 − t ⎞ ⎛ ( 470 µ F )(10.0 jVolts ) RC 1 e − ⎜ ⎟ = 2 ⎝ ⎠ 2

2

1 − t ⎞ ⎛ ⎜ 1 − e (1.0 k Ω)( 470 µ F ) ⎟ ⎜ ⎟ ⎝ ⎠

EC ( t ) = −0.0182 Joules

2

(608)

(609)

iC S

t ≥ 0sec

+

+ vC

vR

C

R

-

FIGURE 37. The complete complex discharging capacitor.

Given a complex energy EC stored in capacitor C with a temporal rotation operator e jθ , where 0° ≤ θ ≤ 90° is acting upon the voltage, when switch S closes at t = 0sec , a complex direct current iC flows through resistor R . The capacitor C discharges into the resistor. A complex voltage vR appears across the resistor R . The resulting complex instantaneous power PR and energy ER are dissipated and/or absorbed by the resistor. So, given, Direction of time θ Time t Initial voltage across capacitor V Capacitor C Resistor R At t = 0sec , the voltage V0 across the resistor R is, V0 = I 0 R

(610)

vR ( t ) = R iC ( t )

(611)

dvC dt

(612)

The complex voltage vR across the resistor R is,

The complex current iC through a capacitor C is, iC ( t ) = −C

Letting t0 = 0.0sec , the complex voltage vC across the capacitor C is, vC ( t ) = vR ( t ) = R iC ( t ) = − R C

William Alek

Page 90

dvC dt

(613)

5/22/2005

Gravitational Mass Fluctuations

INTALEK, INC.

1

vC ( t )

∫

vC ( t )

V0

1

dvC = −

vC ( t )

1 dt RC

dvC = − vC ( t )

ln ( vC ( t ) )

V0

Rev 3.2

(614)

1 t dt R C ∫t0 1 t RC

=−

(615)

t

(616)

t0

⎛ v (t ) ⎞ 1 ln ( vC ( t ) ) − ln (V0 ) = ln ⎜ C ( t − t0 ) ⎟=− V R C ⎝ 0 ⎠

vC ( t ) V0

=e

−

1 ( t − t0 ) RC

vC ( t ) = V0 e

−

=e

−

1 t RC

(617)

(618)

1 t RC

(619)

The complex instantaneous power PR dissipated and/or absorbed by the resistor R is, PR ( t ) = vR ( t ) iC ( t ) =

vR 2 ( t ) R

2

=

V0 2 − R C t e R

(620)

Letting t0 = 0.0sec , the complex energy ER dissipated and/or absorbed by the resistor R is,

V2 ER ( t ) = ∫ PR dt = 0 t0 R t

ER ( t ) =

V0 2 R

∫

t

t0

−

e

2 t RC

dt

2 2 2 2 − t ⎞ ⎛ − R2C t0 ⎛ − R C ⎞ − R C t V0 ⎛ − R C ⎞ − R C t0 1 2 RC − = − e e C V e e ⎜ ⎟ 0 ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ R ⎝ 2 ⎠ 2 ⎝ 2 ⎠ ⎝ ⎠ 2 − t ⎞ ⎛ 1 ER ( t ) = C V0 2 ⎜ 1 − e R C ⎟ ⎜ ⎟ 2 ⎝ ⎠

(621)

(622)

(623)

Example 27. Given a time-forward voltage V across capacitor C , a known resistor value R and capacitor value C , compute the time-forward current flowing through the resistor, and the power and energy dissipated by the resistor.

So, given, Direction of time θ = 0° Time 0.0sec ≤ t ≤ 1.0sec Initial voltage across capacitor V = 10.0Volts Capacitor C = 470 µ F Resistor R = 1.0 k Ω The time-forward voltage V0 across the capacitor C is,

William Alek

Page 91

5/22/2005

Gravitational Mass Fluctuations

INTALEK, INC.

Rev 3.2

V0 = V e jθ = (10.0 Volts ) e j 0° = 10.0Volts

(624)

The time-forward voltage vC across the capacitor C is,

vC ( t ) = V0 e

−

1 t RC

= (10.0 Volts ) e

−

1

(1.0 k Ω )( 470 µ F )

t

(625)

vC ( 0.0sec ) = 10.0 Volts

(626)

vC (1.0sec ) = 1.191Volts

(627)

The time-forward instantaneous power PR dissipated by the resistor R at t = 0.0sec and at t = 1.0sec is, V 2 − 2 t (10.0 Volts ) − (1.0 k Ω )( 470 µ F ) t PR ( t ) = 0 e RC = e R (1.0 k Ω )

(628)

PR ( 0.0sec ) = 0.10 Watts

(629)

PR (1.0sec ) = 1.419 × 10−3 Watts

(630)

2

2

The time-forward energy ER dissipated by the resistor R at t = 0.0sec and at t = 1.0sec is,

ER ( t ) =

2 − t ⎞ CV0 2 ⎛ ( 470 µ F )(10.0Volts ) RC − 1 e ⎜ ⎟= 2 ⎝ 2 ⎠

2

2 − t ⎞ ⎛ ⎜1 − e (1.0 k Ω )( 470 µ F ) ⎟ ⎜ ⎟ ⎝ ⎠

ER (1.0sec ) = 0.023 Joules

(631)

(632)

Example 28. Given a time-advanced voltage V across capacitor C , a known resistor value R and capacitor value C , compute the time-advanced current flowing through the resistor, and the power and energy dissipated and absorbed by the resistor.

So, given, Direction of time θ = 45° Time 0.0sec ≤ t ≤ 1.0sec Initial voltage across capacitor V = 10.0Volts Capacitor C = 470 µ F Resistor R = 1.0 k Ω The time-advanced voltage V0 across the capacitor C is, V0 = V e jθ = (10.0Volts ) e j 45° = 7.071 + 7.071 jVolts

(633)

The time-advanced voltage vC across the capacitor C is,

vC ( t ) = V0 e

William Alek

−

1 t RC

= ( 7.071 + 7.071 jVolts ) e

Page 92

−

1

(1.0 k Ω )( 470 µ F )

t

(634)

5/22/2005

Gravitational Mass Fluctuations

INTALEK, INC.

Rev 3.2

vC ( 0.0sec ) = 7.071 + 7.071 jVolts

(635)

vC (1.0sec ) = 0.842 + 0.842 jVolts

(636)

The time-advanced instantaneous power PR dissipated and absorbed by the resistor R at t = 0.0sec and at t = 1.0sec is, V 2 − 2 t ( 7.071 + 7.071 jVolts ) − (1.0 k Ω )( 470 µ F ) t PR ( t ) = 0 e RC = e R (1.0 k Ω )

(637)

PR ( 0.0sec ) = 0.10 jWatts

(638)

PR (1.0sec ) = 1.419 j × 10−3 Watts

(639)

2

2

The time-advanced energy ER dissipated and absorbed by the resistor R at t = 0.0sec and at t = 1.0sec is, 2 − t ⎞ CV0 2 ⎛ ( 470 µ F )( 7.071 + 7.071 jVolts ) ER ( t ) = ⎜1 − e RC ⎟ = 2 ⎝ 2 ⎠

2

2 − t ⎞ ⎛ (1.0 k Ω )( 470 µ F ) ⎜1 − e ⎟ ⎜ ⎟ ⎝ ⎠

ER (1.0sec ) = 0.023 j Joules

(640)

(641)

Example 29. Given a time-future voltage V across capacitor C , a known resistor value R and capacitor value C , compute the time-future current flowing through the resistor, and the power and energy absorbed by the resistor.

So, given, Direction of time θ = 90° Time 0.0sec ≤ t ≤ 1.0sec Initial voltage across capacitor V = 10.0Volts Capacitor C = 470 µ F Resistor R = 1.0 k Ω The time-future voltage V0 across the capacitor C is, V0 = V e jθ = (10.0 Volts ) e j 90° = 10.0 jVolts

(642)

The time-future voltage vC across the capacitor C is,

vC ( t ) = V0 e

−

1 t RC

= (10.0 jVolts ) e

−

1

(1.0 k Ω )( 470 µ F )

t

(643)

vC ( 0.0sec ) = 10.0 jVolts

(644)

vC (1.0sec ) = 1.191 jVolts

(645)

The time-future instantaneous power PR absorbed by the resistor R at t = 0.0sec and at t = 1.0sec is,

William Alek

Page 93

5/22/2005

Gravitational Mass Fluctuations

INTALEK, INC.

PR ( t ) =

Rev 3.2

V0 2 − RC2 t (10.0 jVolts ) − (1.0 k Ω )( 470 µ F ) t = e e R (1.0 k Ω ) 2

2

(646)

PR ( 0.0sec ) = −0.10 Watts

(647)

PR (1.0sec ) = −1.419 × 10−3 Watts

(648)

The time-future energy ER absorbed by the resistor R at t = 0.0sec and at t = 1.0sec is,

ER ( t ) =

2 − t ⎞ CV0 2 ⎛ ( 470 µ F )(10.0 jVolts ) RC − 1 e ⎜ ⎟= 2 ⎝ 2 ⎠

ER (1.0sec ) = −0.023 Joules

William Alek

Page 94

2

2 − t ⎞ ⎛ ⎜1 − e (1.0 k Ω )( 470 µ F ) ⎟ ⎜ ⎟ ⎝ ⎠

(649)

(650)

5/22/2005

Gravitational Mass Fluctuations

INTALEK, INC.

Rev 3.2

TESLA’S COMPLEX FIELD GENERATOR EQUIVALENT CIRCUIT PWR

+ VS -

iG

A”

A” G

+

-

N

+

-

S

N

−∆E = −∆M Fe c 2

M

-

S

S

µ0 H M

BG

iM e jθ , θ > 0°

M

+

N

M Fe + ( −∆M Fe )

F t ≥ 0sec + -

iK e jθ , θ > 0°

H

K

L’

-

H

+

L

-

+ jθ

vK e , θ > 0° R -

L’ iR e jθ , θ > 0°

+ jθ

vR e , θ > 0° FIGURE 38. Nikola Tesla’s US patent 568,176.

Nikola Tesla was the first to develop the phenomenon of complex fields back in the 1880's. He devised a series of machines patented in the 1890's that greatly amplify this phenomenon, which he later called RADIANT ENERGY. As shown above, the pivoting magnetic domains created by Amperian Currents of the ferromagnetic material are ordered in the direction of field BG by magnetizing coil G . Magnetizing the high inductance coils M create an opposing field µ0 H M that acts upon the ordered domains of the material, thus canceling or partially canceling the real magnetic field created by the Amperian Currents. An imaginary magnetic field j BM emerges due to this cancellation and couples back into the magnetizing direct current as iM e jθ , where θ > 0° . Therefore, the magnetizing direct current becomes complex because the circulating motion of the electrons is rotating into the imaginary axis.

William Alek

Page 95

5/22/2005

Gravitational Mass Fluctuations

INTALEK, INC.

Rev 3.2

As shown above, before switch F is closed, the capacitors H are charged with a complex direct current iM e jθ produced by an opposing flux from coils M . The complex field energy is stored in capacitors H . At the moment of switch F closure, t = 0sec , the complex direct current flows through coil K , rapidly discharging capacitors H . A very large complex electric potential vL e jθ is observed across the secondary coil L .

William Alek

Page 96

5/22/2005

Gravitational Mass Fluctuations

INTALEK, INC.

Rev 3.2

MASS FLUCTUATION TECHNOLOGIES Since a theoretical link was established between gravity and electromagnetism, two mass fluctuation technologies are presently under investigation. Both technologies are electrical devices with the first being inductive-based, and the second being capacitive-based. Shown below is a simplified schematic diagram that highlights their operation. TIME-FUTURE AMPERIAN CURRENTS

TIME-FUTURE AMPERIAN CURRENTS DECREASED MASS −∆M L

INDUCTOR L

-

+

v'L

i'L

RLOAD

-

v'C

FOR COP > 1.0 vC < v 'C

iC

RLOAD

D1

RLOAD

S1

i'C

RLOAD

S1

+

-

+

vLOAD S1

-

+

vC

FOR COP > 1.0 iL < i 'L D1

-

+

+

vL iL

DECREASED MASS −∆M C

CAPACITOR C

vLOAD

S1

RS

+ VS -

+ VS -

MAGNETIZATION PHASE

DEMAGNETIZATION PHASE

RS

+ VS -

+ VS -

CHARGE PHASE

DISCHARGE PHASE

CAPACITIVE CHARGE / DISCHARGE CYCLE

INDUCTIVE MAGNETIZATION / DEMAGNETIZATION CYCLE

FIGURE 39. Two mass fluctuating systems.

These systems are cyclic, and alter the local gravity well. The mass of these systems is converted to excess field energy during the magnetizing/charging phase. During the demagnetizing/discharging phase, excess electrical energy is collected, and mass is restored after this phase. Then, the cycle begins again. As a consequence, clocks runs faster due to broken symmetry of mass-energy conservation in the proximity of these devices because mass is converted to time-future energy.

A CONCEPT VEHICLE THAT UTILIZES GRAVITATIONAL MASS FLUCTUATIONS

TUNGSTEN

GRAVIMETRIC ENERGY

+ + + + + + + + + + + + + + VP + + + + + + + + + + + + + + + + + + + + + FLUCTUATING + MASS ++ + M ++ + + CLOCKS RUN FASTER + ++ + + + + + + + + + + + ++ + + + + + + + + + + + + + + + GRAVIMETRIC ENERGY

+ + + + + + + + + + + + + + + + + + + + + ++ + + + + FLUCTUATING MASS + ++ M + + + + INSULATING + + + LAYER + + + + + + + + + + + + + + + + + + + +

+ + + + + + + + + + + + + + + R + + L + +

FIGURE 40. A cutaway view of a capacitive-based mass fluctuating concept vehicle.

William Alek

Page 97

5/22/2005

Gravitational Mass Fluctuations

INTALEK, INC.

Rev 3.2

Shown above is a concept vehicle for tunneling through vast distances of space. The first step concerning this antigravitational system is the controlled fissioning or disassociation of tungsten metal. This metal is used as a fuel source of gravitational energy bound as mass within the element. As the matter of the metal disassociates, it radiates away this gravitational energy in the form of a temporal field within the vehicle. This causes clocks to speedup and the mass of the vehicle to decrease. The second step begins by switching off the fissioning process, thereby turning off the temporal field. Mass converted to electrical energy is amplified as field energy in the outer hull, which can be stored, radiated away, or used to power the vehicle. The vehicle seeks a new equipotential surface of gravity that corresponds to its’ new mass, thus producing lift. Then, the cycle begins again.

ENERGIZING THE GRAVITATIONAL PROPULSION UNIT (GPU) y

1kg VEHICLE (AFTER)

22,500.0

DECREASED GRAVITY

y1

RADIUS

19,920

g yn = G

ME yn 2

VEHICLE UNDERGOES A RELATIVISTIC MASS REDUCTION

∆y = y0 − y1

1kg VEHICLE (BEFORE)

GRAVITATIONAL REFERENCE

y0

7,529 6,378.1 0.0

2.5

5.0

g y1 = 1.0

7.5

g y0 = 7.0

10.0

g yn

g0 = 9.8

ACCELERATION DUE TO GRAVITY m sec2

FIGURE 41. Vehicle is undergoing a negative gravitational mass fluctuation above the Earth.

The diagram above shows a 1 kg vehicle undergoing a negative gravitational mass fluctuation in a given an equipotential surface of gravity reference. A typical gravity profile of the Earth shown above is based on Newton’s gravity. This system decreases the relativistic mass of the vehicle such that it displaces or vectors in height to a new equipotential surface of gravity above the Earth. Deactivating this system causes the same vehicle to naturally fall based on universal mass attraction. Example 30. Assuming there are no other gravitational influences besides the Earth and given a vehicle of mass 1 kg positioned at an initial radius y0 as shown in the diagram above, compute the new gravitational mass M y1 of

the vehicle displaced ∆y away from the Earth.

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So, given, Mass of vehicle M y0 = 1.0 kg Vehicle at initial radius y0 = 7.529 × 106 m Displacement of vehicle ∆y = y0 − y1 = −12.391× 106 m Speed of light c = 2.99792458 × 108 m sec Gravitational constant G = 6.67260 × 10−11 N m 2 kg 2 Mass of the Earth M E = 5.9787 × 1024 kg The final radius y1 of the vehicle above the Earth is, y1 = y0 − ∆y = ( 7.529 × 106 m ) − ( −12.391× 106 m ) = 19.920 × 106 m

(651)

The acceleration due to gravity at altitude y0 = 7.529 × 106 m above the Earth is, g y0 =

−11 2 2 24 G M E ( 6.67260 × 10 N m kg )( 5.9787 × 10 kg ) = = 7.038 m sec 2 2 6 y0 2 ( 7.529 ×10 m )

(652)

The acceleration due to gravity at altitude y1 = 19.920 × 106 m above the Earth is, g y1 =

−11 2 2 24 G M E ( 6.67260 × 10 N m kg )( 5.9787 × 10 kg ) = = 1.005 m sec 2 2 6 y12 (19.920 ×10 m )

(653)

Given the exponential solution of the natural relativistic mass model, the new gravitational mass is,

M y1 = M y0 e

⎛ g y1 y1 − g y0 y0 ⎞ ⎜⎜ ⎟⎟ c2 ⎝ ⎠

(

= (1.0 kg ) e

)(

)(

)(

⎛ 2 6 2 6 ⎜ 1.005 m sec 19.920×10 m − 7.038 m sec 7.529×10 m ⎜ 2 8 ⎜ 2.99792458×10 m sec ⎝

(

)

) ⎞⎟

⎟ ⎟ ⎠

(654)

M y1 = 0.9999999996333 kg

(655)

M y1 − M y0 = ( 0.9999999996333 kg ) − (1.0 kg ) = −3.667 × 10−10 kg

(656)

The gravitational mass was reduced by,

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CLOSED-LOOP GRAVITATIONAL FLIGHT CONTROL SYSTEM

DECREASED GRAVITY

y1

VECTOR VEHICLE TO NEW POSITION IN GRAVITY WELL

−∆y '

θ

x0

−∆y

+∆x '

y0

GRAVITATIONAL REFERENCE

+∆x

x1 ALTIMETER ACQUIRE HEIGHT h MICROWAVES

INERTIAL REFERENCE

KNOWN SURFACE GRAVITY g0

SURFACE OF EARTH

FIGURE 42. Acquire position information, then energize the GPU and vectored to new position.

The first step for a Gravitational Flight Control System, or GFCS, is to acquire vehicle height h information above a known surface gravity g 0 . Once this height information is acquired, the second step involves calculating the current mass fluctuation M cur . The third step requires the pilot to determine what the next desired position will be, so, predicted mass fluctuation M pre is calculated and transmitted to a Gravitational Propulsion Unit, or GPU. The GPU is energized causing the mass of the vehicle to fluctuate, and in turn, vectors to the desired position. This system can run “closed-loop” by implementing a software algorithm called a high-speed Proportional-Integral-Derivative (PID) control loop. It acquires new height information and computes the error difference between current and desired position. The PID calculates and transmits in real-time an error value to the GPU. The error value sent to the GPU determines the rate of mass fluctuation. This rate may exceed the speed of light because the vehicle isn’t traversing space by an inertial means, but by a gravitational means. Therefore, warp factor equation shown by Whitfield (1968) could be used. The warp factor W equation is, v =W3 c

W=

3

v c

=

3

(657) ∆y T c

(658)

Where, W is the warp factor, c is the speed of light, and v is the velocity.

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1000

TIMES THE SPEED OF LIGHT

800

600

400 3

v c

200

0

6 4 WARP FACTOR W

2

8

10

FIGURE 43. The warp factor vs. the speed of light.

Example 31. Assuming there are no other gravitational influences besides the Earth and given a GPU with a NEGATIVE mass fluctuation rate ∆M RATE ( t ) operating for a period of time T , compute the displacement ∆y of a

vehicle leaving from the surface of the Earth at warp factor W with a vehicle mass of 1.0 kg . So, given, Initial mass of vehicle M y0 = 1.0 kg Mass fluctuation rate ∆M RATE ( t ) = −1.0 × 10−4 gm sec Operating time T = 1.0 × 10−3 s ec Vehicle located at initial radius Y0 = 6.3781× 106 m Speed of light c = 2.99792458 × 108 m sec Gravitational constant G = 6.67260 × 10−11 N m 2 kg 2 Mass of the Earth M E = 5.9787 × 1024 kg The new gravitational mass of the vehicle M y1 operating a GPU for a period of time T is,

William Alek

M y1 = M y0 + ( ∆M RATE ( t ) × T ) = (1.0 kg ) + ⎡⎣( −1.0 × 10−4 gm sec )(1.0 × 10−3 sec ) ⎤⎦

(659)

M y1 = 0.9999999999000 kg

(660)

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Given the exponential form of the natural relativistic mass M ∆y model, compute the new radius y1 within a given gravity well g y is,

M y1 = M y0 e

⎛ g y1 y1 − g y0 y0 ⎞ ⎜⎜ ⎟⎟ c2 ⎝ ⎠

M y1 M y0 ⎛ M y1 c 2 ln ⎜ ⎜M ⎝ y0

y1 =

=e

= M y0 e

⎛ G ME G ME − ⎜ y0 ⎜ y1 ⎜ c2 ⎜ ⎝

⎛ G ME G ME − ⎜ y0 ⎜ y1 ⎜ c2 ⎜ ⎝

⎞ ⎟ ⎟ ⎟ ⎟ ⎠

(661)

⎞ ⎟ ⎟ ⎟ ⎟ ⎠

(662)

⎞ G ME G ME − ⎟⎟ = y1 y0 ⎠

(663)

y0 ⎛ My y c 1+ 0 ln ⎜ 1 G M E ⎜⎝ M y0 2

(664)

⎞ ⎟⎟ ⎠

( 6.3781×10 m ) ( 6.3781×10 m )( 2.99792458 ×10 m sec ) ln ⎛ 0.9999999999000 kg ⎞ 1+ ⎟ 1.0 kg ( 6.67260 ×10 N m kg )( 5.9787 ×10 kg ) ⎜⎝ ⎠ 6

y1 =

6

2

8

−11

2

2

(665)

24

y1 = 7.4484 × 106 m

(666)

∆y = y0 − y1 = ( 6.3781× 106 m ) − ( 7.4484 × 106 m )

(667)

∆y = −1.0703 × 106 m

(668)

The warp factor W is,

W=

3

v c

=

3

∆y T = c

( −1.0703 ×10 m ) (1.0 ×10 sec ) 6

−3

3

( 2.99792458 ×10

8

m sec )

==

W = 1.5284

−1.0703 × 109 m sec 3

( 2.99792458 ×10

8

m sec )

(669)

(670)

Leaving from the surface of the Earth, the vehicle is displaced over 665 miles above the Earth within 1 millisecond.

CONCLUSION The parameters of space and time identified as mass, volume, frequency, time, temperature, and energy are functions of gravity. Therefore, controlling the mass of a space vehicle, for example, controls gravity, but more precisely, controls its’ current position within a given gravity well. The control of how fast mass fluctuates controls the speed of the vehicle through this well. Whether the vehicle implementing GMF is above the surface of the Earth or

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traveling at warp factor 1.53 through deep space, the physical constants where shown in previous sections to remain invariant within the vehicle. The vehicles’ Gravitational Propulsion Unit, or GPU, using one of the two devices that are under investigation control mass by an electrical means, thus utilizing the theoretical link between gravity and electromagnetism presented in this paper. Either device can change its’ own mass by converting it to excess electrical energy, and as a consequence, the broken symmetry of mass-energy conservation causes the gravitational energy equivalent of mass to be radiated away as a temporal field. A force is created and acts antigravitationally on the device. This changes the frequency of clocks within the vehicle relative to clocks outside of the vehicle. The vehicle will require a Gravitational Flight Control System, or GFCS, capable of controlling the GPU. This control system acts as an interface between man and the GPU. The development of this system can be directly implemented from the mathematical formulations in this paper. An additional requirement is the development of real-time navigation software programs that maps the gravity of the entire Sol System, and includes the nearby stars.

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http://www.intalek.com/Index/Projects/Research/RadiantPressure.pdf Tom Van Flandern, “The Speed of Gravity - What the Experiments Say”, 1998, corrections 2002. On the web at: http://www.intalek.com/Index/Projects/Research/TheSpeedofGravity-WhattheExperimentsSay.htm Tom Bearden, “Scalar Waves”, NEW ENERGY NEWS, July 1998. On the web at: http://www.intalek.com/Index/Projects/Research/SCALAR_WAVES.htm Nikola Tesla, “Apparatus for Producing Electric Currents of High Frequency and Potential”, US Patent 568,176, issued September 22, 1896. On the web at: http://www.intalek.com/Index/Projects/Patents/00568176.pdf Neil Ashby, “Relativistic Effects on Satellite Clocks”, 2003. On the web at: http://www.intalek.com/Index/Projects/Research/RelativisticEffectsOnSatelliteClocks.pdf Robert A. Nelson, “Practical Relativistic Timing Effects in GPS and Galileo”, 2003. On the web at: http://www.intalek.com/Index/Projects/Research/gps.ppt Hayt & Kemmerly, “Engineering Circuit Analysis”, 5th edition, McGraw Hill, 1993, ISBN 0-07-027410-X. Col. William McLyman, “Transformer and Inductor Design Handbook”, 2nd edition, Marcel Dekker, 1988, ISBN 0-8247-7828-6. Weidner and Sells, “Elementary Classical Physics, Volume One”, 2nd edition, 1974. Shadowitz, “The Electromagnetic Field”, 1999, ISBN 0-486-65660-8. Vincent Del Toro, “Electromechanical Devices for Energy Conversion and Control Systems”, Prentice-Hall 1968. Halliday and Resnick, “Physics, Part II”, Wiley 1962, ISBN 0-471-34523-7. Francois Cardarelli, “Scientific Unit Conversion”, 2nd edition, Springer, 1999, ISBN 1-85233-043-0. Stephen Whitfield, “The Making of Star Trek”, 1968, ISBN 0345315545.

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