the unified

THEORY OF EVERYTHING . BY . DR S. K. K. TOMKA (Extension of The Unified Gravitational and Electromagnetic Theory) (First Edition published on 23rd February 2016)

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Copyright © Surbjit Kaur Kang Tomka. PhD. 2016: . All rights reserved. Without limiting the rights under copyright reserved above, no part of this publication may be reproduced, stored in or introduced into a retrieval system or transmitted in any form or by any means (including photocopying) without the prior written permission of the copyright owner, the author. Please support the work of the author by purchasing your own copy, thank you.

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“To have been blessed and to know that I could be blessed.” * Dedicated to: * My supportive husband, my lovely daughter and my parents. *

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CONTENTS: : Chapter 1: Introduction: A Brief History: : Part I A New Consideration for Light: Chapter 2: Developing Hooke’s Law and Einstein’s formula for Energy Chapter 3: Introduction of a Source Mass, Ms Chapter 4: Developing Newton’s Formula for the attraction between masses and deriving Kepler’s third law Chapter 5 : Establishing a relationship between the Gravitational constant (G), Plank’s constant (h) and the speed of light ( c ). Introduction of the Universal constant, U. Chapter 6: Establishing a relationship between the distance of the source Mass (Ms) and the effective photon mass (mp) i.e. the extension Chapter 7: Introducing two identical charges (Q 1 and Q2 ) Defining Z (impedance) Determining the distance (2 d) between the charges and defining the relationship between the charges and frequency, f Chapter 8: Establishing fundamental relationships for G Chapter 9: Determining the relationship between the electric and magnetic field strengths and G Defining the Unification Constant (or the Fifth Field), η Chapter 10: Determining a theoretical value for h and a calculated value for G Chapter 11: Establishing the relationship between the Source Mass Ms and the effective photon mass mp also the angles between the two photons Chapter 12: Establishing the relationship between the effective photon mass mp, its acceleration, a, and G. Chapter 13: Deriving the expression between the effective photon mass and the restoring force factor. Chapter 14: The de Broglie formula and its relationship to the source Mass (Ms) Chapter 15: The Kinetic Energy (K.E.) and Potential Energy of the effective photon mass (m p). :

Part II: Virtual Capacitors and their properties: Chapter 16: Defining the virtual capacitor Chapter 17: Dimensions (shape, operation) and properties of the virtual capacitor. Chapter 18: Resonance, reactance and inductance of the virtual capacitor Chapter 19: Critical resistance

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Chapter 20: Determining the energy stored by the virtual capacitor and the related power Chapter 21: Using the dimensions of the virtual capacitor to derive G Chapter 22: Exploring the Energy and Power properties of the virtual capacitor and verifying the velocity Chapter 23: The Hidden Dimension Chapter 24: Comparing the photon energy with the energy stored in the virtual capacitor :

Part III: A New Consideration for Time: Chapter 25: Defining frequency and deriving an alternative expression for h Chapter 26: Deriving a fundamental relationship between G and the restoring force factor and an alternative expression for h. :

Part IV: Redefines Electromagnetic relationships: Chapter 27: Defining the permittivity for free space, ε o Chapter 28: Relationship between c2 and c3 Chapter 29: Redefining Z and ε o Chapter 30: Defining the magnetic permeability for free space, μ o Relationship between the Impedance (Z), the Permittivity and the Permeability of Free Space Relationship between Charge (Q1 Q2) and the Dielectric Permittivity (εo) and the Magnetic Permeability (μo) Chapter 31: Defining the Magnetic Field Strength, H Chapter 32: Defining the Electric Field Strength, E, Defining the Separation between Q1 and Q2 (d) in terms of Plank’s constant, Defining H in terms of Plank’s constant Chapter 33: Defining the Magnetic Field, B Chapter 34: The Universe’s Balancing Equation, The Relationship Between Plank’s constant and the Restoring Force Factor ξ Chapter 35: Expressing C for the virtual capacitor in terms of G, H and E Chapter 36: Defining magnetic field strength, H, and the electric field strength E, for the virtual capacitor :

Part V Establishing Expressions and Relationships for the Electric and Magnetic Forces: Chapter 37: Deriving an expression for H / E Chapter 38: Establishing an expression for H Z and H / Z Chapter 39: Deriving an expression for E Z and E / Z Chapter 40: Deriving an expression for H ε o and H / εo

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Chapter 41: Deriving and expression for E ε o Chapter 42: Deriving an expression for E / ε o Chapter 43: Deriving an expression for E Ho Chapter 44: Deriving an expression for E / B Chapter 45: Deriving and expression for E H in terms of B Chapter 46: Deriving and expression for E H / εo2 Chapter 47: Deriving the expression for E / H = Z π Chapter 48: Deriving an expression for E H / Z 2 Chapter 49: Redefining η Chapter 50: Redefining the virtual capacitance C :

Part VI: Commences the expansion of the Unified Approach to other areas of fundamental physics: Chapter 51: Development of Boyle’s law Chapter 52: Redefining Temperature in terms of η The Relationship between Photon Energy and the Temperature Chapter 53: Development of Faraday’s concept static and fluid charge Chapter 54: Density Chapter 55: Introducing Electric Dipole Moment, Introducing Magnetic Dipole Moment and establishing a physical understanding of the constant η, and relationships in terms of temperature Chapter 56: Sound, Hidden variable Chapter 57: Values and definitions:

: Appendices (Additional Chapter) Appendix 1: Simple harmonic system Appendix 2: Introducing the concept of a time evolving system (exploring a possible “wormhole effect”) Appendix 3: Tension and Gravity Appendix 4: Torque (exploring the possibility of the existence of gravitons) Appendix 5: Dimensional analysis – Dimensions [M LT] and new dimensions: [G] and [Q]. Exploring the concept of the continuum. Appendix 6: Exploring Spacetime Appendix 7: Exploring the possibility of Anti Gravity The Relationship between the Magnetic Field Strength and Charge Appendix 8: Distribution of Charge

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Appendix 9: Alternative Equivalent Expressions for G and π applicable to this model Appendix 10: Exploring Phase, Voltage and possible connection to Chaos Theory Appendix 11: The Relationship between the Permittivity of Free Space, ε o, and Volts, The relationship between , εo, the magnetic field strength and the speed of light Appendix 12: The Relationship between the Magnetic Field Strength, H, and Sound, O. Relationship between Sound and Current Appendix 13: The Relationship between Volts and the Source Mass The Relationship between the Charge and the Source Mass The Relationship between the Restoring Force Factor and Frequency The Relationship between the Source Mass and the Universal Gravitational Constant Appendix 14: Exploring Boundary Conditions Appendix 15: Exploring faster than the speed of light Appendix 16: The Relationship between Q o, Q1 and Q2

: References and further reading Amendment notes: List of updates to the version first published on 23 rd February 2016

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CHAPTER 1 “Whilst the great ocean of truth lay all undiscovered before me.” Sir Isaac Newton (1643-1727)

: Introduction: : This book proposes a Single Unified Theory combining the Universal Gravitational Constant, G, and electromagnetism. In doing so, offers a new insight into the order of our Universe.

: : Early scientists were pioneers to understanding the world around us. One of those early scientists was Johannes Kepler (1571-1630) whose studies of planetary motion provided the foundations for the formulation of Kepler’s laws. In the seventeenth century, science rapidly evolved to provide the foundations for much of modern-day physics. Interestingly Robert Hooke published an esoteric anagram in 1660. He published its solution, much later, in 1678,. in Latin. The solution is “ut tension, sic vis”, meaning, as the extension so the force. In other words “extension is proportional to force”. Sir Isaac Newton’s works ‘Philosophiae Naturalis Principia Mathematica’ (Latin for Mathematical Principles of Natural Philosophy), was first published on 5th July 1687. This insightful work included Newtons’ laws of motion, Newton’s law of universal gravitation (a long range attractive interaction that acts on all particles) and a derivation of Kepler’s law of planetary motion, previously derived by Kepler through empirical observation. Sir Isaac Newton was asked how he discovered the law of gravity Newton replied, "By thinking about it all the time.” : In the nineteenth century, in 1831, Michael Faraday made an all important observation, namely that time varying magnetic fields could induce an electric current. This observation established a basis for James Clerk Maxwell, who in 1865 published his work, ‘Dynamical Theory of the Electromagnetic Field”. This was the first unified theory, namely, the unified theory for electromagnetism. Maxell conceived that light was wave-like and traveled at a constant velocity. : Albert Einstein’s mass-energy equivalence equation in its simplified form, E = mc 2, was explored in his seminal work, published on 21st November 1905, ”Does the Inertia of Body Depend upon its Energy”. Later, in 1915 Einstein expanded the theory of special relativity to describe the equivalence of inertial fields and gravity. General Relativity is Einstein’s theory of four dimensional space-time. This incorporates the three dimensions of space and one of time, in which a field is used to describe the curving geometry of the four dimensional space. Einstein believed that electromagnetism and gravity could be combined into a Unified Field Theory.

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: Herman Weyl in 1919 built on Einstein’s General Relativity theory, by introducing a gauge field. This is essentially a fundamental field that cannot be directly measured. However some quantities associated with it can be measured. Theodor Kalmza (1921) extended General Relativity to five dimensions. : In 1926 Oscar Klein proposed that the fourth spatial dimension was curled up into a small unobserved circle. The Kaluza-Klein theory proposed that the gravitational curvature of the extra spatial direction behaves as an additional force similar to electromagnetism. The extension of general relativity to five dimensions, wherein the fifth dimension is real but “very small and curled up”, was essential to the beginnings of string theory. : Modern science continues to work towards unlocking the secrets of our Universe that have previously evaded us. Progress is being made towards establishing a Grand Unification Theory (GUT). GUT is a theory which unifies the electromagnetic force (the interaction, which acts on electrically charged particles with the photon as the mediation particle of this force), with the weak force (a short range interaction on electrons, neutrinos and quarks, responsible for some forms of radioactivity, which is mediated by W and Z bozons) and the strong force (the interaction responsible for holding quarks together to form hadrons and for holding neutrons and also photons together to form an atomic nuclei, with gluons acting as the exchange particles). : There is considerable interest in creating a Theory of Everything (TOE). TOE unifies gravity (its exchange particle is postulated to be the graviton) with the other three interactions, namely, electromagnetism, the strong interaction and the weak interaction.

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In this book, I will propose a Unified Theory in which the Universal Gravitational Constant intrinsically relates to electromagnetic phenomena. In the course of the derivation of this concept a number of fundamental physical relationships are unveiled that provide insights into the origin of the fundamental constants governing the nature of the Universe. This suggests that, not only are these physical constants related in unexpected ways in our Universe, but also that they may have a fixed relationship in any conceivable selfconsistent Universe. This may have significant consequences to the development of string theory and the hypothesized interactions between the multiverses. : I derive a relationship between the Gravitational constant (G) and the speed of light (c), the impedance of free space (Z) and fundamental charge (|Q1| = |Q2|).

G3 

Z c Q1 Q 2 32 π 4

:Equation 29.4:

I develop other fundamental relationships from my theory, establishing a new Universal framework for Physics. : An understanding is developed in the form of a mathematical approach and is best suited for those with an appreciation of physics and mathematics. The book has been broadly subdivided into six parts and the appendices, which are essentially additional chapters. : Part I sets out to establish fundamental relationships between the building blocks of our Universe, using a variety of thought processes. Part II considers a new model for light. Part III evolves the model to enable a new definition for time. Part IV redefines electromagnetism. Part V establishes relationships between the electric and magnetic forces. Part VI commences the extension of the unified approach to other areas of fundamental physics, such as, pressure, density and temperature.

: Appendices. The appendices are essentially additional chapters. In this section I start to consider how concepts such as, ‘worm-hole effects’, the multiverse, the anti-gravity and chaos theory may fit in with this work, these areas may need further development. In this section there is also a dimensional analysis, a deeper examination of charge distribution, an important definition of π is derived and a new interpretation of the permittivity of free space. The consistency of the model is demonstrated through the interrelationship of terms.

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PART I: A New Consideration for Light: :

CHAPTER 2 “It is easier to resist at the beginning than at the end” Leonardo da Vinci (1452-1519) : Developing Hooke’s Law and Einstein’s formula for Energy: : Starting by establishing a restoring force factor, ξ. : I consider two identical coupled photons. To establish the restoring force factor (otherwise known as the spring constant) between the two photons, which are coupled and assumed to be in free space. I use the formulae for simple harmonic motion (like two masses on a spring). The equation for simple harmonic motion {Reference 1, page 1241}:

ω2 

ξ m

:Equation. 2.1:

where m is the mass under consideration and ξ is the spring constant and can be considered as the restoring force factor between the photons. The derivation for Equation 2.1 can be found in Appendix 1. : Where ω is defined by the photon frequency, f, and given by: : ω=2πf

:Equation. 2.2:

{Simple Harmonic Motion ω = 2 π f Reference 1, page 361} As the photons are considered massless, replace m, which denotes the mass by using: : : E = m c2

:Equation 2.3:

E=hf

:Equation 2.4:

E denotes Energy and c is the speed of light. { E = m c2 Reference 1, page1149 } Energy is also expressed by: :

{ E = h f Reference 1, page1024} Where h is Plank's constant. Substituting E = h f, Equation 2.4 into E = m c2, Equation 2.3 and rearranging to give: :

m

hf c2

Now substituting, m = h f / c2, Equation 1.5 and

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:Equation 2.5:

ω = 2πf, Equation 2.2 into ω2 = ξ / m, Equation 2.1 to give: :

ξc2 hf

(2 π f) 2 

:Equation 2.6:

Rearranging Equation 2.6 to solve for ξ to give: :

(2 π) 2 h f 3

ξ

c2

:Equation 2.7:

Equation 2.7 may also be written in terms of the wavelength by substituting for f, where: : λf=c

:Equation 2.8:

{ λ f = c Reference 1, page 1011} The restoring force factor, ξ, between the photons in terms of wavelength becomes: :

(2 π) 2 h c 3

ξ

λ3 c 2

:Equation 2.9:

Equation 2.9 simplifies to: :

ξ

(2 π) 2 h c λ3

: : : : Reference 1: Physics for Scientists and Engineers with Modern Physics, Fourth Edition, Author Serway

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:Equation 2.10:

CHAPTER 3 “If I have seen further than others, it is because, I have stood on the shoulders of giants”. Sir Isaac Newton (16431727) : Introducing a Source Mass (Ms): : Now consider a source mass, Ms, (a dark mass) under the influence of the restoring force factor. Using conventional equations for a mass suspended by two “rigid springs/ strings” with their ends separated by a distance S = 2L. The Force, Fm, is equal to the restoring force factor multiplied by the extension x (Hooke’s Law) i.e.: Fm = ξ x

:Equation 3.1:

{Hooke’s Law Reference 1, page 179} Recall that the restoring force factor, ξ, as given in, Equation 2.7 is: (ξ = (2 π)2 h f 3 / c2), by substituting for ξ into 2.1 gives: :

Fm 

x (2 π) 2 h f 3 c2

:Equation 3.2:

The force, Fm, is also given by Newton’s second law of motion: mass multiplied by its acceleration, a. {Newton’s second law of motion force= (mass) (acceleration) Reference 1, page 112) : Fm = (Ms) (a)

:Equation 3.3:

Substituting for Ms using Equation 2.5 {(Ms = h f / c2)}, Equation 3.3 becomes: :

Fm 

hf

a

:Equation 3.4:

Fm = √ ( (sinФ Tension )2 + (cosФ Tension )2 )

:Equation 3.5:

c2

The force, Fm, is also given by: : Where sin(Ф) is established later. Equating the force Fm, Equations 2.2 and 2.4 gives: :

Fm 

hf c2

a

x (2 π) 2 h f 3 c2

:Equation 3.6:

Solving for the acceleration, a: : a = (x) (2 π)2 f2 : Reference 1: Physics for Scientists and Engineers with Modern Physics, Fourth Edition, Author Serway

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:Equation 3.7:

CHAPTER 4 “Kepler’s laws, although not rigidly true, are sufficiently near the truth to have led to the discovery of the law of attraction of the bodies of the solar system. The deviation from complete accuracy is due to the facts, that the planets are not of inappreciable mass, that, in consequence, they disturb each other’s orbits about the Sun and by their action on the Sun itself, cause the periodic masses of the Sun and Planets; these errors are appreciable although very small, since the mass of the largest of the planets, Jupiter, is less than one thousandth of the Sun’s mass.” Sir Isaac Newton (1643-1727) Isaac Newton realises that the exact measurement of G will be difficult due to the interaction of many masses. :

Developing Newton’s Formula for the attraction between masses and deriving Kepler’s third law: : Consider the attractive force between the source Mass, Ms, and the effective photon mass, mp. The attractive force, FN, between two masses is given by Newton’s formula. {Newton’s formula Reference 1, page 392}

G M1 M2

FN 

r2

:Equation 4.1:

Where G is the universal gravitational constant, M1 and M2 are the masses under consideration and r is the distance between the masses M1 and M2. Using formula 4.1, where the masses under consideration are Ms the source Mass, and the photon effective mass m p. The attractive force, FN, is given by: :

G Ms mp

FN 

r2

:Equation 4.2:

Where r is given by √(L2+x2) , Pythagoras theorem. {Pythagoras Reference 1, page 54} The radial force FR is given by: :

FR 

mp v 2 r

:Equation 4.3:

Where the radial acceleration, a, is given by: :

v2 r

a

:Equation 4.4:

{ a = v2 / r, Reference 1, page 85} The two forces, FN and FR, can be equated so: :

G Ms mp r2



mp v 2 r

Note that mp cancel out: :

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:Equation 4.5:

v2 r

:Equation 4.6:

2πr T

:Equation 4.7:

1 f

:Equation 4.8:

v=2πrf

:Equation 4.9:

G Ms



r2 As the radial velocity of the photon is: :

v T is the period and can be replaced by 1/ frequency. {Period Reference 1, page 361, 466}

T

Substitute Equation 4.9 into Equation 4.6 and simplifying to give: :

G Ms r2

(2 π r f) 2 r

:Equation 4.10:

(2 π f) 2 r 3 G

:Equation 4.11:



Solving the above for Ms: :

Ms  Rearranging to solve for r: :

G Ms

r3 

(2 π f) 2

:Equation 4.12:

Note that Equation 4.12 is identical to Kepler’s third law for planetary motion. {Kepler’s third law can be found in, Reference 1, page 397} : Using Equation 2.5, m = h f / c2, to express Ms: :

hf

Ms 

c2

:Equation 4.13:

Note here that f / Ms is a constant (speed of light squared / Plank’s constant = c2 / h) and the value is given in Chapter 57. By substituting Ms, Equation 4.13, into Equation 4.12, the cube of the distance between the Ms and mp is: :

r3  r3 

Gh f (2 π c f) 2 Gh f (2 π c ) 2

:Equation 4.14:

:Equation 4.15:

The distance r between Ms and mp is: : :r = cube root { (1/ f) [ G h / (2 π c) ]2 }: :

r [

:

Gh f (2 π c )

2

]1/3

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:Equation 4.16:

: Reference 1: Physics for Scientists and Engineers with Modern Physics, Fourth Edition, Author Serway

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CHAPTER 5 “Classical physics has been superseded by quantum theory, quantum theory is verified by experiments. Experiments must be described in terms of classical physics” C. F. von Wiezsacker (1912-2007) :

Establishing a relationship between the Gravitational constant (G), Plank’s constant (h) and the speed of light ( c ): Introduction of the Universal constant, U: : Equating the two expressions for Ms given by, Equation 4.11, Ms = (2 π f)2 r3 / G and Equation 4.13, Ms = h f / c2, provides: :

Ms 

hf c2

(2 π f) 2 r 3 G

:Equation 5.1:

4 π2 c2 f r 3 G

:Equation 5.2:



Solving for Plank’s constant, h: :

h :

h G = 4 π2 c2 f r3

:Equation 5.3:

Collecting constants: :

f r3 

hG 4 π2 c 2

:Equation 5.4:

Let U be the Universal constant (1.24464572522529 x10-62) for a given value of Q (Q1 and Q2) discussed later.

U

hG 4 π2 c2

:Equation 5.5:

Let: : U = f r3

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:Equation 5.6:

CHAPTER 6 Unique concepts generate unique results. :

Establishing a relationship between the distance of the source Mass (Ms) and effective photon mass (mp) the extension: : For a system undergoing simple harmonic motion, the period of oscillation, T, is given by: :

T

r a

2π

:Equation 6.1:

Where g is the acceleration and r is the distance between the two points under consideration. Rearranging 6.1 to solve for the acceleration, a, gives: :

4 π2 r

a

T2

:Equation 6.2:

As 1/T is the frequency, the acceleration can be expressed as: : a = 4 π2 f 2 r

:Equation 6.3:

F = m a = Ms (a)

:Equation 6.4:

F=ξx

:Equation 6.5:

Ms (a) = ξ x

:Equation 6.6:

Using, Force = mass*acceleration: : As earlier Force = restoring force factor * extension: :

In a radial system, x = r, therefore x and r can be equated so: :

: Substituting for ξ, where, ξ = (2 π)2 (h f 3 / c2 ) from Equation 2.7 into the above expression gives: :

Ms a 

4 π2 h f 3 c2

(x)

:Equation 6.7:

Substituting in for a, which is defined in Equation 6.3 (a = 4 π2 f2 r) to give: :

Ms 4 π 2 f 2 r 

4 π2 h f 3 c2

(x)

:Equation 6.8:

Substituting in for Ms, where Ms = (2 π f)2 r3 / G given by Equation 4.11, into the above gives: : :

4 π2 f 2 r3 (4 π 2 f 2 r)  G

4 π2 h f 3 c2

(x)

:Equation 6.9:

Simplify:

r3 (4 π 2 f 2 r)  G

hf c2

(x)

Rearranging the above:

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:Equation 6.10:

4 π2 f 2 r3 (r)  G

hf

(x)

:Equation 6.11:

( Ms) (r) = (Ms) (x)

:Equation 6.12:

x=r

:Equation 6.13:

c2

: As Ms {Ms = (2 π f)2 r3 / G given by Equation 4.11 and (Ms = h f / c2) from Equation 4.13 and:

Hereafter the distance between Ms and mp shall be referred to as r: : : Note that from Equation 6.7 :

Ms a

r (

4 π2 h f 3 c2

:Equation 6.14:

)

Simplify:

Ms a c 2

r

4 π2 h f 3

:Equation 6.15:

As λ f = c from Equation 2.8, the above can be expressed as: :

r

Ms a λ 3 4 π2 c h

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:Equation 6.16:

CHAPTER 7 “Facts which at first seem improbable will, even on scant explanation, drop the cloak, which has hidden them and stand forth in naked and simple beauty.” Galileo Galilei (1564-1642) :

Introducing two identical virtual charges (Q 1 and Q2 ) of equal magnitude (Q) but not necessarily of the same sign: Defining Z (impedance): Determining the distance (2 d) between the charges and defining their relationship with the frequency, f . : Introducing two charges (or energies) Q1 and Q2 (for simplicity assume that they are identical in magnitude, both equal to Q so that Q1 = Q2 = +/-Q, this is considered further in Appendix 8). Determining the distance between the charges (2 d) and their relationship with the frequency f. Let FQ be the force between two charges separated by a distance 2d (Coulomb’s Law 1785). FQ is expressed as: : {FQ Reference 1, page 653}

FQ 

k e Q1 Q 2 (2 d)2

:Equation 7.1:

Where ke is given by: :

1 4 πε o

:Equation 7.2:

1 Zπc

:Equation 7.3:

Zc 4

:Equation 7.4:

Z c (Q 1 Q 2 ) 4 (2 d) 2

:Equation 7.5:

ke  {ke Reference 1 page 654} And εo is given by: :

εo  Where Z π is the impedance of free space. {Impedance of free space Reference 1, page 1003}

ke 

Zπc  4π

Let the product Q1 Q2 = Q2 Substituting ke into FQ i.e. Equation 7.4 into Equation 7.1 gives: :

FQ 

Consider the force required to stretch space by a distance 2d, by two charges of magnitude Q1 and Q2 where, Q = √(Q1 Q2): Force = (restoring force factor)*distance: : FQ = ξ 2 d

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:Equation 7.6:

Equating the two expressions for FQ, i.e. Equation 7.5 and Equation 7.6 gives: :

2dξ 

Z c (Q 1 Q 2 ) 4 (2 d) 2

:Equation 7.7:

d3 

Z c (Q 1 Q 2 ) 32 ξ

:Equation 7.8:

The above simplifies to: :

Substitute for ξ, as given by, Equation 2.7 ξ = (2π)2 h f 3 / c2 into Equation 7.8 above gives: :

Z c (Q1 Q 2 )

d3 

32 (

4 π2 h f 3 c2

:Equation 7.9:

)

The above simplifies to: :

Z c 3 (Q 1 Q 2 )

d3 

128 π 2 h f 3

:Equation 7.10:

Substitute for h in to the above, where h = 4 π² c² f r³ / G given earlier in Equation 5.2: :

d3 

Z c 3 (Q 1 Q 2 ) 4 π2 c2 f r3 128 π f ( ) G

:Equation 7.11:

2 3

Simplifying the above: :

d3 

Z c G (Q 1 Q 2 ) 128 π 2 f 3 (4 π 2 f r 3 )

:Equation 7.12:

Simplifying further: :

d3 

Z c G (Q1 Q 2 )

d3 r 3 f 4 

Z c G (Q1 Q 2 )

512 π 4 f 4 r 3

:Equation 7.13:

Collecting variable terms: :

512 π 4

:Equation 7.14:

Assuming a fixed value of Q1 and Q2. Let U = d3 r3 f 4 which is a constant: (1.23322871544011 x10-42): The Equation 7.14 can also be expressed in terms of the wavelength λ by using Equation 2.8: :

d3 r 3 

Z c G (Q 1 Q 2 ) λ 4 512 π 4 c 4

:Equation 7.15:

Simplifying the above equation: :

d3 r 3 

Z G (Q 1 Q 2 ) λ 4 512 π 4 c 3

It is important to rearrange Equation 7.8 to provide an expression for the impedance Z: :

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:Equation 7.16:

Z

32 d 3 ξ c Q1 Q 2

:Equation 7.17:

Assuming a value for Q1 Q2 as given in Chapter 57 Z is calculated to be equal to 119.9169832 as given in Reference 2. A number of constants arise when Q2 = Q1 Q2 are fixed, these include: : : d3 r3 f 4 : (1.23323193923603 x10-42) [L6 T -4], r3 f : (1.24464572522529 x10-62) [L3 T -1] and: d3 ξ : (2.88384689169287 x10-29) [L5 T -4]: : Although the product of the frequency with the distance (f d) is not absolutely constant it is worth noting that: : f d : 4627352.9408284(6) [L T -1]: : : : Reference 1: Physics for Scientists and Engineers with Modern Physics, Fourth Edition, Author Serway Reference 2: The NIST Reference on Constants Units and Uncertainty (www physics.nist.gov/constants)

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CHAPTER 8 Mathematics is the language in which the universe is written. : Establishing Fundamental Relationships with G: : Establishing the fundamental relationship between Q and G and verifying this relationship. Establishing the fundamental relationship between G and f, also between G and the restoring force factor. : Starting by rearranging Equation 7.14 for Q1 Q2. d3 r3 f 4 = (Q1 Q2) Z G c / [ 512 π4 ], Equation 7.14 and then substituting in Z from Equation 7.17 where Z = 32 d3ξ / (c Q1 Q2): :

512 π 4 d3 r 3 f 4 cGZ

Q1 Q 2 

:Equation 8.1:

Let Q2 = Q1 Q2 is the charge : (Q = √ ( Q1 Q2 ) = 1.60217662089849 x10-19)

c G Z Q1 Q 2

d3 r 3 f 4 

512 π 4

:Equation 8.2:

(let S = d3 r3 f 4 = 1.23323193923603 x10-42 [L6 T -4]) A fundamental relationship is uncovered: :

G4

d3 r 3 f 4 

2

4

G 4 ) 2

:Equation 8.3:

G 1 3 3 1/4 2 (d r )

:Equation 8.4:



(

The above equation is derived in chapter 10. Note that from Equation 8.3 that: :

f

Using the expression given in Equation 8.3 for d3 r3 f4 in terms of G and substituting into Equation 8.2 gives: : :

G4 2

4



π c G Z Q1 Q 2 512 π 5

:Equation 8.5:

Solving the above for G: :

G4 

16 c G Z Q1 Q 2 512 π 4

:Equation 8.6:

Simplifying: : :

G3 

:

16 c Z Q1 Q 2 512 π 4

Page 23 of 178

:Equation 8.7:

Simplifying further: :

G3 

c Z Q1 Q 2 32 π 4

:Equation 8.8:

Substituting the expression for Z from Equation 7.17 [Z = 32 d3 ξ / (c Q1 Q2)] into Equation 8.8 gives: :

G3 

c Q1 Q 2 32 d 3 ξ 32 π 4 c Q1 Q 2

:Equation 8.9:

Simplify: :

G3 

d3 ξ π4

:Equation 8.10:

Rearranging: : π4 G3 = d3 ξ

:Equation 8.11:

The important Equation 8.11 helps to verify the value of Q in the model. It provides a fundamental relationship between the restoring force factor and G. :

:

Page 24 of 178

CHAPTER 9 “Measure what is measurable and make measurable what is not so.” Galileo Galilei (1564-1642) : Determining the relationship between the electric and magnetic field strengths and G. : Consider the relationship between the electric and the magnetic fields strengths. The relationship is provided as: :

E H

:Equation 9.1:

Zo = Z π

:Equation 9.2:

Zo  Where Zo is the impedance of free space: :

Then Equation 9.1 may be written as: :

Zπ

E H

:Equation 9.3:

Using the relationship established earlier for Ms given by Ms = (2 π f)2 r3 / G in Equation 4.11: : And rearranging for r3: :

r3 

G Ms (2 π f) 2

:Equation 9.4:

Substituting for r3, given above, into Equation 7.14, d3 r3 f 4 = (Q1 Q2) c Z G / (512 π4), gives: :

d3 f 4 G Ms (2 π f)

2

c G Z Q1 Q 2



512 π 4

:Equation 9.5:

Cancel down: :

d 3 f 2 Ms 

c Z Q1 Q 2 128 π 2

:Equation 9.6:

Solve for Ms: :

Ms 

c Z Q1 Q 2 128 π 2 d 3 f 2

:Equation 9.7:

As Zo = Z π from Equation 9.2 then Equation 9.7 can be expressed as: :

Ms 

c Z o Q1 Q 2 128 π 3 d3 f 2

:Equation 9.8:

Substituting for d3, where d3 = (Q1 Q2) c Z / (32 ξ) as established earlier in Equation 7.8: : :

Ms 

c Z Q1 Q 2 c Z Q1 Q 2 128 π 2 f 2 ( ) 32 ξ

Simplifying: :

:

Page 25 of 178

:Equation 9.9:

32 ξ c Z Q1 Q 2

c Z Q1 Q 2

Ms 

2

128 π f

2

:Equation 9.10:

Simplifying further: :

ξ

Ms 

4 π2 f 2

:Equation 9.11:

Note that Ms = ξ r / (a), where a is the acceleration and: a = 4 π2 f2 r, given by Equation 6.3: As r3 = G (Ms) / (2 π f)2, established earlier in Equation 9.4 then substituting for Ms, where : Ms = ξ / (4 π2 f 2) given above in Equation 9.11: : :

ξ

G( r  3

4 π2 f 2 (2 π f) 2

) :Equation 9.12:

Simplify: :

r3 

Gξ (4 π 2 f 2 ) 2

:Equation 9.13:

Solve for G: : :

G

r 3 (4 π 2 f 2 ) 2 ξ

:Equation 9.14:

ξ

r 3 (4 π 2 f 2 ) 2 G

:Equation 9.15:

E Hπ

:Equation 9.16:

Rearranging to provide an expression for ξ: :

Rearranging Equation 9.3, Z π = E / H, to give: :

Z :

As Ms = Q1 Q2 c Z / (128 π2 d3 f2) from Equation 9.7 and Z π = E/H, given in Equation 9.16 above, by substituting for Z into the expression for Ms, it can be shown that: : :

Ms 

Q1 Q 2 c

E 128 π d f H 3

3

2

:Equation 9.17:

Using equation Ms= ξ / (4 π2 f 2) from Equation 9.11 in the above gives: :

ξ 2

4π f

2



Q1 Q 2 c

E 128 π d f H 3

3

2

Simplifying gives: :

:

Page 26 of 178

:Equation 9.18:

ξ

Q1 Q 2 c E 32 π d 3 H

:Equation 9.19:

Q1 Q 2 E

:Equation 9.20:

Rearrange to give: :

H

32 π 3 d ξ c

For a defined value of Q, d3 ξ is found to be constant. (d3 ξ = 2.88384689169288 x10-29) So therefore (32 π / c) d3 ξ is a constant: : Let this constant be described as the unification constant (or known as the fifth field), η, where, η = (32 π / c) (d3 ξ): :

η

32 π 3 d ξ c

:Equation 9.21:

32 d 3 ξ c Q1 Q 2

:Equation 9.22:

Deduce that: :

Z Where Z = 119.9169832 [L4 T -3 Q -2] {The value of Z is given in Reference 2} The relationship between η and Z is given by: :

η = Z π Q 1 Q2

:Equation 9.23:

Z π Q1 Q 2

:Equation 9.24:

32 d3 ξ  c

Z Q1 Q 2

:Equation 9.25:

Z

32 d 3 ξ c Q1 Q 2

:Equation 9.26:

Equating Equation 9.21, η = (32 π / c) (d ξ) and 3

Equation 9.23, η = Z π Q1 Q2: :

32 π 3 d ξ c Solving for Z: :

The above agrees with Z = 32 d3 ξ / (c Q1 Q2), Equation 7.17. Let E = 1 then, Equation 9.20 becomes: : H η = Q 1 Q2

:Equation 9.27:

or: :

Hη  Q1 Q 2

1

Substituting an expression for ξ using Equation 9.15, ξ = r3 (4 π2 f 2)2 / G, into : Equation 9.20, H (32π / c) d3ξ = (Q1 Q2) E, to obtain: :

:

Page 27 of 178

:Equation 9.28:

512 π 5 r 3 d3 f 4 H Gc

Q1 Q 2 E

:Equation 9.29:

G c Q1 Q 2 E 512 π 5 H

:Equation 9.30:

Rearranging: :

d3 r 3 f 4 

The above equation provides a relationship between the electromagnetism and the Universal Gravitational constant. From Equation 8.2, d3 r3 f 4 = (Q1 Q2) c Z G / [ (512) π4), note that Equation 9.30 above reduces to the familiar equation: Zπ=E/H:: : : {Note that by substituting Ms, given in Equation 9.9, Ms = Q1 Q2 c Z / ( (128 π2 f 2) (Q1 Q2) c Z / (32 ξ) ), into Equation 9.5 given by, d3 f 4 G Ms / (2 π f)2 = (Q1 Q2) c Z G / (512 π4) also gives: : Ms = ξ / (4 π2 f 2) as per Equation 9.11 and: As r3 = G ξ / (4 π2 f 2)2, Equation 9.13 then: As r3 = G Ms / (4 π2 f2) as per Equation 9.4} : Reference 2: The NIST Reference on Constants Units and Uncertainty (www physics.nist.gov/constants)

:

:

Page 28 of 178

CHAPTER 10 “All truths are easy to understand once they are discovered; the point is to discover them.” Galileo Galilei (1564-1642) : Determining a theoretical value for Plank’s constant, h, and a calculated value for the Universal Gravitational constant, G, from a defined value for Q and the speed of light, c (see Chapter 57). Establishing various expressions for G. : Using d3 r3 f 4 = (Q1 Q2) G c Z / [ 512 π4) ] from Equation 8.2: :

G c Z Q1 Q 2

d3 r 3 f 4 

512 π 4

:Equation 10.1:

d3 r3 f 4 is a constant, S, for a given value of Q and c: : Let S = Q1 Q2 c Z G / (512 π4): : S=

d3 r3 f 4

=

1.23322871544011 x 10-42

:Equation 10.2:

By postulating that the charge Q has a magnitude, given in Chapter 57, (Q = 1.60217662089849 x 10-19) and by rearranging Equation 10.1 for G gives: :

512 π 4 d 3 r 3 f 4 c Z Q1 Q 2

G

:Equation 10.3:

A value for G can be calculated and is given in Chapter 57: : G = 6.66485802996474 x 10-11

:Equation 10.4:

: From Equation 8.8 (where G3 = c Z Q1 Q2 / (32 π4) ):

16 c Z Q1 Q 2

G3 

512 π 4

:Equation 10.5:

And taking the cube root: :

G

(

c Z Q1 Q 2 32 π

4

)1/3

:Equation 10.6:

Alternatively rearranging Equation 10.3 gives: :

d3 r 3 f 4 

c G Z Q1 Q 2 512 π 4

:Equation 10.7:

As Z = 32 d3 ξ / ( c Q1 Q2 ), Equation 7.17: :

d3 r 3 f 4 

c G Q1 Q 2 32 d 3 ξ 512 π 4 c Q1 Q 2

:Equation 10.8:

Simplify: :

r3 f 4 

32 G ξ 512 π 4

Solve for G: :

:

Page 29 of 178

:Equation 10.9:

G

16 π 4 r 3 f 4 ξ

:Equation 10.10:

Note that: : Ms (a) = ξ r where a is the acceleration so this can be rearranged as: r = Ms (a) / ξ, substituting this into expression 10.11 derived later. Ms = ξ / (4 π2 f 2) from Equation 9.11:

G

ξ2 r4 a M3s

:Equation 10.11:

Proof for Equation 10.11 is drawn out in the following chapter. : In Equation 5.2, Plank’s constant, h, is given by: :

h

4 π2 c2 f r 3 G

:Equation 10.12:

Plank’s constant, h, can be calculated and is an absolute constant (i.e. Plank’s constant does not change with the value of Q or G). Plank’s constant does not change with G, as any deviation in G is cancelled out by the corresponding deviation in r. Thus h can be calculate and given in Chapter 57, and is found to be: : h = 6.6260695729 x 10-34 [L4 T -3]

:Equation 10.13:

: Rearranging the expression for G3 where: G3 = 16 Q1 Q2 c Z / (512 π4), Equation 10.5 gives: :

G3  16

c Z Q1 Q 2 512 π 4

:Equation 10.14:

As d3 r3 f 4 = G c Z Q1 Q2 / (512 π4), Equation 10.1 then: :

d3 r 3 f 4 

G G3 16

:Equation 10.15:

The above equation can be stated as: :

d3 r 3 f 4 

G4 24

:Equation 10.16:

As per Equation 8.3: : : Expressing the relationship between Plank’s constant, h, and the constant η: : As η = 32 π d3 ξ / c Equation 9.21 and d3 = G4 / (16 r3 f 4) from rearranging Equation 10.16 by substituting for d3 in the expression for η gives:

η

:

32 π ξ G 4 c 16 r 3 f 4

Page 30 of 178

:Equation 10.17:

Rearrange for r3 and simplifying to obtain:

r3 

2 πξ G 4 cηf4

:Equation 10.18:

As h = 4 π2 c2 f r3 / G Equation 10.12, by substituting for the term r3 in the above equation gives the relationship:

h

4 π 2 c 2 f 2 πξ G 4 G c η f3

:Equation 10.19:

Simplifying gives:

h

8 π 3 c G3 ξ η f3

Note that ξ / f 3 is a constant, the value is given is Chapter 57

:

Page 31 of 178

:Equation 10.20:

CHAPTER 11 “A true friend is one soul in two bodies” Aristotle (384BC-322BC) : Establishing the relationship between the Source Mass, Ms, and the effective photon mass, mp, also the angle, Ф, between the two photons. Simple Harmonic Motion (S.H.M.) for a spring is typically given by ω2 = ξ / Ms, Equation 2.1, {S.H.M. Reference 1, page 381 where ω = 2 π f = 2 π / T}

ξ

f2 

4 π 2 Ms

:Equation 11.1:

Derivation of Equation 11.1 is found in Appendix 1 Note that, f = (1/ [2 π] ) √(ξ / m) Consider Ms the effective source mass as the mass of the photons multiplied by the restoring force factor divided by some function of the angle, 2Ф, between them i.e.: :

m p2 ξ

Ms 

sin2 Φ

:Equation 11.2:

We are familiar with Newton’s expression for the force FN due to the attraction between masses, FN = G M1 M2 / r2 given in Equation 4.1 Consider a system with two source masses Ms: :

ξr

G Ms Ms

ξ r3 

G Ms Ms

r2

:Equation 11.3:

Rearranging: : :Equation 11.4:

Solve for G and multiply both sides by, r / Ms3, to give: :

Gr  Ms

ξ r4 M3s

:Equation 11.5:

Solve for G: :

G

ξ r 4 Ms M3s r

:Equation 11.6:

: We know that from standard physics at resonance (Equation 2.1): :

ω2 

ξ Ms

Where Ms is a mass: :

:

Page 32 of 178

:Equation 11.7:

4 π2 f 2 

ξ Ms

:Equation 11.8:

ξ = 4 π 2 f 2 Ms

:Equation 11.9:

4 π2 f 2 r

:Equation 11.10:

Also the angular acceleration, a, is given by: :

a

v2  r

Where v is the radial velocity of the photon and is given by, v=2πfr

:Equation 11.11:

v = 2 π f r {Reference 1, page 280} Dividing ξ given by Equation 11.9 by the acceleration, a, given by Equation 11.10 above: :

4 π 2 f 2 Ms

ξ  a

4 π2 f 2 r

:Equation 11.12:

Simplifying:

ξ  a

Ms r

:Equation 11.13:

ξ r4 ξ M3s a

:Equation 11.14:

Substitute Equation 11.13 into Equation 11.6, G = (Ms / r ) ξ r4 / Ms3: to give: :

G Simplify the above: :

G

ξ2 r4 a M3s

:Equation 11.15:

Thus the proof for Equation 10.11: : : ξ /a = Ms / r given by Equation 11.13 is rearranged to solve for Ms: :

Ms 

ξ r( ) a

:Equation 11.16:

We now have an expression for Ms, which I hypothesis is valid for the system where the masses under consideration are Ms and mp. As it is the photon, that is experiencing the acceleration then:

mp 

a r( ) ξ

This expression is derived in the following Chapter 12: : By substituting for, a / ξ = r / Ms given be rearranging Equation 11.16 This then gives a relationship between Ms and mp: :

:

Page 33 of 178

:Equation 11.17:

mp 

r2 Ms

:Equation 11.18:

Ms 

r2 mp

:Equation 11.19:

And: :

: Substitute mp given in Equation 11.17, mp= r a / ξ , into Ms (Ms = mp2 ξ / sin2Ф ) as in Equation 11.2: :

Ms 

ξ

r 2 a2

sin2 Φ ξ 2

:Equation 11.20:

Simplify:

r 2 a2

Ms 

ξ sin2 Φ

:Equation 11.21:

It has been shown that G = ξ2 r4 / (a Ms3) in Equation 11.15. Substitute mp into Equation 11.15 replacing Ms with mp, where Ms = mp2 ξ / sin2Ф from Equation 11.2: :

ξ2 r4

G a(

m p2 ξ sin2 Φ

:Equation 11.22:

)

3

Simplify: :

r 4 sin6 Φ

G

a ξ m p6

:Equation 11.23:

: As G = ξ2 r4 / (a Ms3) in Equation 11.15 and substituting for Ms where Ms = r2 / mp as given in Equation 11.19: :

ξ2 r4

G

r2 3 a( ) mp

:Equation 11.24:

Expand the powers: :

G

ξ2 r4 r6 a( 3 ) mp

:Equation 11.25:

Cancel down: :

G

:

ξ2 r2 a( 3 ) mp

Page 34 of 178

:Equation 11.26:

Simplify: :

ξ 2 m p3

G

a r2

:Equation 11.27:

Equation 11.27 G = ξ2 mp3 / [ a (r2) ] and π4 (G)3 = d3 ξ, Equation 8.11 Help to tune the value of Q. : It is noted here that Equation 11.27 only provides an exact solution at discreet values in the model, otherwise the Right hand side of the equation oscillates, wave like, around the value of G. But otherwise gives an excellent approximation. The oscillations may be real or may be due to rounding errors (e.g. taking the cube root of r) in the model. : Sin2Ф can be expressed in a number of ways these include: Using Ms = mp2 ξ / sin2Ф from Equation 11.2 and Ms = r2 / mp as given in Equation 11.19 by equating these two equations: :

sin Φ  2

ξ m p3 r2

:Equation 11.28:

Alternatively using m p = r2 / Ms from Equation 11.18 substituting into the above: :

sin2 Φ 

ξ r4 M3s

:Equation 11.29:

Again using G = ξ2 r4 / (a Ms3), Equation 11.15 and Sin2Ф = ξ r4 / Ms3, Equation 11.29 then: :

sin 2 Φ 

aG ξ

:Equation 11.30:

sin 2 Φ 

Gr Ms

:Equation 11.31:

Using G r / Ms = ξ r4 / Ms3, Equation 10.5 and Sin2Ф = ξ r4 / Ms3, given by Equation 11.29 then: :

Note that by substituting for a = v2 / r = 4 π2 f 2 r, Equation 11.10, into Equation 11.30, sin2Ф = a G / ξ, then: :

sin2 Φ 

4 π2 G f 2 r ξ

:Equation 11.32:

And rearranging for G: :

G

ξ sin 2 Φ 4 π2 f 2 r

:Equation 11.33:

Note that by rearranging Equation 11.19, Ms = r2 / mp, : : r2 = m p Ms and: :

:

Page 35 of 178

:Equation 11.34:

r = √ (mp Ms )

:Equation 11.35:

and: :

mp 

r2 Ms

: Reference 1: Physics for Scientists and Engineers with Modern Physics, Fourth Edition, Author Serway

:

Page 36 of 178

:Equation 11.36:

CHAPTER 12 “All knowledge has its origins in our perceptions.” Leonardo da Vinci (1452-1519) : Establishing the relationship between the photon mass m p, its acceleration, a, and the Universal Gravitational constant G. As mp is given by mp = r a / ξ given in Equation 11.17 And acceleration a is given by a = v2 / r = 4 π2 f2 r, Equation 11.10: : Finding an expression for m p a: :

mp a 

ra (4 π 2 f 2 r) ξ

:Equation 12.1:

Substitute for, a, into the above and on the right hand side of the equation only where, a = v 2 / r = 4 π2 f2 r, Equation 11.10 gives: :

mp a 

r 4 π2 f 2 r (4 π 2 f 2 r) ξ

:Equation 12.2:

r 3 (4 π 2 f 2 ) 2 ξ

:Equation 12.3:

Collecting terms: :

mp a 

Earlier it was established that, G = r3 (4 π2 f 2)2 / ξ, Equation 9.14, therefore: : G = mp a

:Equation 12.4:

G a

:Equation 12.5:

Or alternatively the above can be expressed as: :

mp  As a = v2/r = 4 π2 f2 r, Equation 11.10 then the above becomes: :

G

mp 

4 π2 f 2 r

:Equation 12.6:

The equations, 12.4, 12.5 and 12.6 are not exact at all values but give an exact solution at discrete values only, otherwise it gives an excellent approximation. By rearranging Equation 12.6, m p = G / r (4 π2 f2), gives: : G = mp r (4 π2 f2)

:Equation 12.7:

: Equating the expressions for mp, where mp = r2 / Ms, from Equation 11.18, and Equation 12.5, m p = G / r , gives: :

r2  Ms

G a

:Equation 12.8:

Rearranging the above for a gives: :

a

:

G Ms r2

Page 37 of 178

:Equation 12.9:

Rearranging the above for G gives: :

a r2 Ms

G

:Equation 12.10:

Recall Equation 8.3, d3 r3 f4 = (G / 2)4, by substituting for G using Equation 12.7, G = mp r (4 π2 f 2), gives: :

d3 r 3 f 4 

(

4 π 2 f 2 mp r 2

)4

:Equation 12.11:

Simplifying and collecting fixed and variable terms gives: :

d3

16 π 8 

r f 4 mp

4

:Equation 12.12:

For discreet values of mp. Consider the expression (mp ω)2, where, ω is given by Equation 2.2, ω = 2 π f, so: : (mp ω)2 = 4 π2 f 2 mp2

:Equation 12.13:

: As mp = r2 / Ms, from Equation 11.18, the above can be expressed as

4 π 2 f 2 mp  2

r2 2 ) Ms

:Equation 12.14:

r2 r ) Ms Ms

:Equation 12.15:

4 π2 f 2 (

The above can be expressed as: :

4 π 2 f 2 mp  2

4 π2 f 2 r (

As a = 4 π2 f 2 r, Equation 11.10, then substituting for a in the above gives : :

4 π 2 f 2 mp  2

a(

r2 r ) Ms Ms

:Equation 12.16:

Gr Ms

:Equation 12.17:

As G = a r2 / Ms then: :

4 π 2 f 2 mp  2

Note that Equation 11.31 states sin2Ф = G r / Ms, so sinФ = ω mp

:Equation 12.18:

or as, ω = 2 π f, Equation 2.2, the above is equivalently expressed as: : sinФ = 2 π f mp

:Equation 12.19:

sin φ 2πf

:Equation 12.20:

sin φ 2 π mp

:Equation 12.21:

or rearranged

mp  Similarly the frequency can be expressed as: :

f

:

Page 38 of 178

Further information can be obtained by applying geometrical relationships such as: : sin2Ф + cos2Ф = 1

:Equation 12.22:

using either the definition for sinФ given by Equation 12.19, sinФ = 2 π f mp, or Equation 11.31, sin2Ф = G r / Ms. It may be interesting to explore the relationship: 2 sin2Ф = 1 – cos (2Ф)

:Equation 12.23:

So: :

1-

2Gr  Ms

cos (2 φ)

:Equation 12.24:

Note that sin2Ф can be defined by squaring Equation 12.19, sinФ = 2 π f mp, as: : Sin2Ф = 4 π2 f2 mp2 :

:

Page 39 of 178

:Equation 12.25:

CHAPTER 13 “To understand the universe use your own lamp.” Siddhartha Gautama - Buddha (approximately 563BC-483BC) : Deriving the expression between the effective photon mass and the restoring force factor. (i.e. mp = r (a / ξ), Equation 11.17 Using the Equations, d3 r3 f 4 = (G / 2)4, Equation 8.3 and G = mp a, Equation 12.5 We can equate the expressions to obtain: :

mp a 4 ) 2

:Equation 13.1:

r2 a 4 ) 2 Ms

:Equation 13.2:

r2 4 π2 f 2 r 4 ) 2 Ms

:Equation 13.3:

2 π2 r3 f 2 4 ) Ms

:Equation 13.4:

d3 r 3 f 4 

(

d3 r 3 f 4 

(

As Ms = r2 / mp, Equation 11.19: :

Substitute in for acceleration, a, where a = v 2 / r = 4 π2 f2 r, Equation 11.10: :

d3 r 3 f 4 

(

Becomes: :

d3 r 3 f 4 

(

Expand the power: :

d3 r 3 f 4 

16 π 8 r 12 f 8 Ms4

:Equation 13.5:

Solve for d: :

d3 

16 π 8 r 9 f 4 Ms4

:Equation 13.6:

As ξ = 4 π2 f 2 Ms, Equation 11.9 or rearranged as ξ / Ms = 4 π 2 f 2 Squaring both sides gives (16 π4 f4) = (ξ / Ms)2 And substituting, (ξ / Ms )2, into the above Equation 13.6, to give: :

d3 

π4 r9 ξ2 M6s

As ξ / Ms = 4 π2 f 2, Equation 11.8 squaring both sides of this equation and multiply by r3: :

:

Page 40 of 178

:Equation 13.7:

ξ2 M2s



(4 π 2 f 2 ) 2

:Equation 13.8:

(4 π 2 f 2 ) 2 r 3

:Equation 13.9:

Multiply both sides by r3: :

ξ2 r3 M2s



As r³ = G Ms / (2 π f)², Equation 9.4 or r³ = G ξ / (4 π2 f2)2, Equation 9.13 Substitute for r³ on the right hand side only in the above equation: :

ξ2 r3 M2s



(4 π 2 f 2 ) 2

Gξ (4 π 2 f 2 ) 2

:Equation 13.10:

Simplify: :

ξ2 r3 M2s



Gξ

:Equation 13.11:

Rearrange for G: :

G

ξ r3 M2s

:Equation 13.12:

As Ms = r2 / mp, Equation 11.19 substitute into the above: :

ξ r3

G

r4 ( 2) mp

:Equation 13.13:

Simplify: :

ξ m p2

G

r

:Equation 13.14:

As G = mp a, from Equation 12.4 then: :

ξ mp

a

r

:Equation 13.15:

Rearranging Equation 13.15 provides the expression mp = r (a / ξ) as per Equation 11.17: : : Note that by substituting mp = r (a / ξ), Equation 11.17, into Equation 13.14, G = ξ m p2 / r, gives: :

G

ξ r 2 a2 ξ2 r

:Equation 13.16:

The above simplifies to: :

G

:

a2 r ξ

Page 41 of 178

:Equation 13.17:

The model shows that as r becomes smaller the angular acceleration, a, (a is sometimes referred to as spin) increases as expected (for example, as would be the case in a tornado).

:

Page 42 of 178

CHAPTER 14 In one’s book, are counted all of one’s yesterdays’ and tomorrows’, the threads and strings from one’s own rich tapestry. : The de Broglie formula Defining Tension, Tension, in terms of G. : Consider the De Broglie formula for momentum, p, which is given by: :

p

h λ

:Equation 14.1:

{ de Broglie formula, p = h / λ, Reference 1, page 1218 } As the general formula for momentum is given by mass multiplied by velocity. It can be demonstrated that: :

c Ms 

h λ

:Equation 14.2:

where, c is the speed of light and Ms is the source mass, introduced in chapter 4, which can be expressed as given in Equation 4.13, Ms = h f / c2 Equation 14.4 can be rearranged to solve for Ms, so: :

Ms 

h cλ

: : Reference 1: Physics for Scientists and Engineers with Modern Physics, Fourth Edition, Author Serway

:

Page 43 of 178

:Equation 14.3:

CHAPTER 15 The living force, which may be transmuted from form to form, colour to colour through the cosmic alchemy. : The Kinetic Energy (K.E.) and Potential Energy of the effective photon mass. K.E. is typically given by effective photon mass multiplied by its radial velocity, v, squared: : K.E. = mp v2

:Equation 15.1:

v=2πfr

:Equation 15.2:

v 2 = 4 π 2 f 2 r2

:Equation 15.3:

G 2 v a

:Equation 15.4:

a = v 2 / r = 4 π2 f2 r, Equation 11.10: : Squaring both sides: :

As G = mp a, Equation 12.4 then: mp = G / a: :

K.E.  a = v 2 / r = 4 π2 f2 r, Equation 11.10: :

K.E. 

G v2 v2 ( ) r

:Equation 15.5:

K.E. = G r

:Equation 15.6:

G r = mp v 2

:Equation 15.7:

K.E. = G r

:Equation 15.8:

From Equation 15.1 then: :

:

: Potential Energy (P.E.) is typically given by the product of the mass, acceleration and distance (i.e. P.E. is: m g h): : P.E. = mp a r

:Equation 15.9:

We have already shown that, mp a = G, Equation 12.4 Therefore it is possible to express the potential energy, P.E., as: : P.E. = G r

:Equation 15.10:

P.E. = K.E.

:Equation 15.11:

Thus: : :

:

Page 44 of 178

PART II Virtual Capacitors and their properties: Perception, intuition, then reason, is the path to enlightenment. : Consider that two identical photons each carrying a charge of Q. The charges are separated by a distance (S = 2 d). As before there is a restoring force which is given by the restoring force factor, ξ, multiplied by the separation, 2d, of the charges. Consider that this system has created a virtual capacitor. The two coupled photons will resonate when their electric and magnetic fields are out of phase by 90 degrees. Capacitance is calculated using the impedance for free space and using the frequency of light. For the purpose of this exercise I have considered frequencies in the visible region of the electromagnetic spectrum for my model.

:

Page 45 of 178

CHAPTER 16 “Facts do not cease to exist because they are ignored” Aldous Huxley (1894-1963) : Defining the virtual capacitor. The capacitance, C, of the charged coupled photon system is given by: :

C

εo A S

:Equation 16.1: :

{Reference 1, page 743}: εo is given by, 1 / (Z π c), rearranging the above Equation 16.1 to give the capacitive Area, A. S = 2 d: A=2dCZπc

:Equation 16.2: :

where Zo = Z π is the impedance of free space given by 119.9169832π : To find and compare the area for the virtual capacitor, the separation of the coupled photons is considered to provide the circumference of the virtual capacitor. Therefore the radius, Rad, is calculated by dividing the circumference (photon separation S = 2d) by 2π.

A

2 πc Z dC 

2dC εo

:Equation 16.3: :

: If 2d is the separation of the charges, hypothesis that the separation of the charges is the circumference, so space is wrapped into a circle (imagine a loop of string, which has been cut), then the radius, Rad, of the wrapped space is given by: : 2 d = circumference

:Equation 16.4:

Radius, Rad, is given by: :

2d 2π

:Equation 16.5:

d π

:Equation 16.6:

d π ( )2 π

:Equation 16.7:

R ad  R ad  Area of a circle is typically given by π r2 where r is the radius: : 2 π R ad 

Area of a circle 

d2 π

The Capacitance, C, at resonance, ωresonance is typically equal to : ωresonance = 2 π f: :

:

Page 46 of 178

:Equation 16.8:

1

C

Z o ω resonance

:Equation 16.9:

Where Zo = Z π is the impedance of free space given by 119.9169832 π. The capacitance can be re expressed as: :

C

1 2 πf Z π

C

1 2 π2 Z f

:Equation 16.10:

:Equation 16.11:

As A = 2 d C (Z π c), Equation 16.3. By substituting for C, where, C = 1 / (Z π 2 π f), Equation 16.10, the capacitive area can be expressed as: :

A

2πZc d Zπ2πf

:Equation 16.12:

A

cd πf

:Equation 16.13:

A

λd π

:Equation 16.14:

The above reduces to: :

c is the speed of light and c / f = λ: :

Note that: :

π 

λd  A

λ A ( ) d

:Equation 16.15:

Equating the expressions for the Area A, where A is given by, A = C 2 d Z π c, Equation 16.3 and A = d λ / π, Equation 16.14 gives: :

λd π

2 πc Z C d 

:Equation 16.16:

Solve for the capacitance, C: :

C

λ 2

2π Zc

: : : Reference 1: Physics for Scientists and Engineers with Modern Physics, Fourth Edition, Author Serway :

:

Page 47 of 178

:Equation 16.17:

CHAPTER 17 “A vacuum can only exist, I imagine, by the things which enclose it” Zelda Fitzgerald (1900-1948) : Dimensions and properties of the virtual capacitor: : As the charge moves radially, and the wave propagates through wrapped space from the base of the triangle to its tip, it draws out a cone shape. Therefore as the wave propagates in wrapped space, from the base of the cone to the tip, the capacitive area shrinks. : Assuming that the capacitive cross-sectional area is triangular shaped. The area of a triangle is given by, half the base times height, Ŧ. As the base is 2d, half the base is d. The capacitive area, A, may be written as: : (d) (Ŧ) = A

:Equation 17.1:

dŦ=λd/π

:Equation 17.2:

Ŧ=λ/π

:Equation 17.3:

As A = λ d / π, Equation 16.14: :

Note that λ / Ŧ = π Using distance/ time = speed Let the linear distance traveled be equal to Ŧ π over one cycle. : Then the speed of the wave is: : f (π Ŧ) = c

:Equation 17.4:

: The hypotenuse, Δ, is the length that the charge processes through. Using Pythagoras to calculate the hypotenuse: : : Δ = √ (d2 + Ŧ2)

:Equation 17.5:

Substitute for Ŧ as given by Ŧ = λ / π, Equation 17.3: :

Δ

( d2 

λ2 π2

)

:Equation 17.6:

: The velocity along the hypotenuse, Vhyp, of the individual component which comprise the wave that is moving radially is of interest. Below I consider two possible solutions, firstly where Vhyp = Vzip and secondly where Vhyp = Vspiral. : Hypothesis that the velocity (Vzip) along the hypotenuse can be described as: : Vzip = 2 π (f) (Δ)

:Equation 17.7:

: Then, Vzip, the component that is moving along the length of the hypotenuse in a linear fashion is given by: :

:

Page 48 of 178

Vzip 

2 π f ( d2 

λ2 π2

)

:Equation 17.8:

The square root sign can have a positive or negative number, implying that Vzip can be negative or positive. In this case Vhyp is a constant, but one which is faster than the speed of light; giving rise to a Phantom particle (where c / Vzip = 0.499413189 and by Pythagoras, perhaps, the velocity along the third component). This scenario may be analogous to a commonly occurring physical phenomena. A wave, which has been reflected from the shore and is traveling back out to sea when, it, the reflected wave, encounters a shore bound wave, at the point that they meet, a ‘zip’ wave is created. This zip wave travels faster than either of the two interacting waves. The properties of the phantom particle / wave is explored later. : Note that from rearranging Equation 17.7, Vzip = 2 π (f) (Δ), that the frequency is is given by: :

Vzip

f

2 πΔ

:Equation 17.9:

: However consider an alternative solution. If the velocity along the hypotenuse, Vspiral, is covering an area, so instead is given by: : Vspiral = 2 π (f) (Δ)2

:Equation 17.10:

Then the component moving along the length of the hypotenuse in a radial fashion (similar in fashion to the spiral paths drawn out in a bubble chamber) is given as: :

Vspiral 

2 πf ( d2 

λ2 π2

)

:Equation 17.11:

Then in this case, Vspiral is found to be a variable. Note from rearranging Equation 17.10, Vspiral = 2 π (f) (Δ)2, that the frequency is given by: :

f

Vspiral 2 πΔ2

:Equation 17.12:

More work is required here to explore the possible solutions for, Vhyp. : Hypothesis that the phase (here I mean the angle of polarisation) is given by: :

Phase 

Vspiral π

:Equation 17.13:

As there is an angle of polarization, this implies that there is an elliptical factor. : Note that: :

cΔ  Vspiral

c Vzip

:Equation 17.14:

As c = f λ (Equation 2.8) and Vspiral = 2 π (f) (Δ)2 Equation 17.10, then the above Equation 17.14 can be written as: :

:

Page 49 of 178

cΔ  Vspiral

c  Vzip

fλΔ 2πf Δ

2

λ 2πΔ



:Equation 17.15:

As Δ = √ (d2 + λ2 / π 2) given by Equation 17.6 then Equation 17.15 becomes: :

cΔ  Vspiral

c  Vzip

λ 2 π d2 

:Equation 17.16:

λ2 π2

Which simplifies to: :

cΔ  Vspiral

c  Vzip

λ 4 π 2 d2  4 λ 2

:Equation 17.17:

: An alternative expressions for Vspiral Recall that Vspiral is given by Equation 17.11, Vspiral = 2 π f (d2 + λ2 / π 2) and that Equation 4.15 provides an expression for r3, i.e. r3 = G h / (f (2 π c) 2), so: :

Vspiral r3



λ2 4 π2 c 2 f ) Gh π2

:Equation 17.18:

8 πc 2 f 2 π2 f λ2 (π f d 2  ) Gh π2

:Equation 17.19:

2 π f (d 2 

Rearranging and simplifying gives: :

Vspiral r3



Recognising that f r3 is the universal constant given by Equation 5.6, the above can be rearranged so the expression outside the brackets on the right hand side is a constant: :

8 πc2 f r3 (π 2 f d 2  f λ 2 ) Gh

Vspiral 

:Equation 17.20:

Recall from Equation 2.8 that λ = c / f, substituting this into the above expression and simplifying gives: :

Vspiral 

8 πc2 f r3 c2 (  π 2 f d2 ) Gh f

:Equation 17.21:

Note that, the first term in the brackets, c2 / f, is significantly larger than the second term in the brackets, π 2 f d2. : Verifying that the equations for Vspiral are consistent with the model: Equation 17.11, Vspiral = 2 π f (d2 + λ2 / π 2), can be equated with Equation 17.21 Vspiral = [8 π c2 f r3/ (G h)] [(c2 / f + π 2 f d2)] to give: :

Vspiral 

2 π f (d 2 

λ2 π

) 2



8 πc2 f r3 c2 (  π 2 f d2 ) Gh f

:Equation 17.22:

4c2 r3 c2 (  π 2 f d2 ) Gh f

:Equation 17.23:

Simplifying (d 2 

:

λ2 π

) 2

Page 50 of 178

Solving for λ2 gives: :

4 π2 c2 r 3 c2 (  π 2 f d2 ) - π 2 d2 Gh f

:Equation 17.24:

4 π 2 c 4 r 3 4 π 4 c 2 f r 3 d2  - π 2 d2 Gh f Gh

:Equation 17.25:

λ2  Expanding the right hand side: :

λ2 

Recall that from Equation 4.15, f r3 = G h / (2 π c) 2, by substituting for f r3 in the above equation gives: :

λ2 

4 π 2 c 4 r 3 4 π 4 c 2 G h d2  - π 2 d2 Gh f 4 π2 G h c2

:Equation 17.26:

4 π2 c 4 r 3  π 2 d2 - π 2 d2 Gh f

:Equation 17.27:

4 π2 c 4 r3 Gh f

:Equation 17.28:

Which simplifies to: :

λ2  Reducing to: :

λ2 

So providing a numerically correct expression for λ2, in a unique relationship; by taking the square root gives the expected wavelength λ. Equation 17.28 can easily be rearranged for Plank’s constant h: :

h

4 π2 c 4 r3 G f λ2

:Equation 17.29:

Recall Equation 2.8, c = λ f, so by substituting λ2 = c2 / f2 into the above provides: :

h

4 π2 c4 f 2 r3 G f c2

:Equation 17.30:

Which reduces to: :

h

4 π2 c2 f r3 G

As per Equation 10.12. :

:

Page 51 of 178

:Equation 17.31:

CHAPTER 18 “In experimental philosophy we are to look upon propositions inferred by general induction from phenomena as accurately or very nearly true, notwithstanding any contrary hypotheses that may be imagined, till such time as other phenomena occur, by which they may either be made more accurate, or liable to exceptions. This rule we must follow, that the argument of induction may not be evaded by hypotheses.” Sir Isaac Newton (1643-1727) : Resonance, reactance and inductance of the virtual capacitor: : The natural frequency, f, is given by, f = [ 1 / (2 π) ] [ √ (ξ / m) ] Derived in Appendix 1.

ξ Ms

1 2π

f

:Equation 18.1:

: As Ms = r2 / mp, Equation 11.18 substitute for Ms gives an equivalent expression: : :

ξ

f

1 2π

f

1 2π

ξ mp

ω resonance 

1 LC

:Equation 18.4:

1 C (2 π f)

:Equation 18.5:

1 ωC

:Equation 18.6:

r2 ( ) mp

:Equation 18.2:

Simplify: :

r2

:Equation 18.3:

Further discussion can be found in Appendix 2. : At the resonance frequency ωresonance, {resonance frequency ωresonance Reference 1, page 950}:

Where C is the capacitance and Lind is the inductance. : The inductance at resonance can be calculated using: :

L ind 

1

2

Capacitive reactance ΨC is defined as: :

ψC  {ΨC Reference1, page 970} Inductive reactance ΨL is defined as: :

:

Page 52 of 178

ΨL = ω Lind

:Equation 18.7:

Resistance = √ (Z2 + [ΨL – ΨC]2)

:Equation 18.8:

Resistance = Z

:Equation 18.9:

{ΨL Reference 1, page 969} ΨC and ΨL are found to equal Z π at resonance: : {Resistance Reference 1, page 973}

: : : : Reference 1: Physics for Scientists and Engineers with Modern Physics, Fourth Edition, Author Serway :

:

Page 53 of 178

CHAPTER 19 “I do six impossible things before breakfast every morning” , Lewis Caroll (1832-1898) : The critical resistance Rcritical is generally given by: : {Critical resistance Reference 1, page 953}

R critical 

4 L ind C

:Equation 19.1:

For this system the Rcritical is found to be constant at: 753.460626923541 Ω [L4 T -3 Q -2: Note that: : Rcritical = 2 π Z :

: : : Reference 1: Physics for Scientists and Engineers with Modern Physics, Fourth Edition, Author Serway

:

:

Page 54 of 178

:Equation 19.2:

CHAPTER 20 The capacity to store energy, is a prerequisite to the capacity for life. : Determining the energy stored by the virtual capacitor and the related power: Energy, Ecap, stored by a capacitor is typically given by: {Reference 1, page 750}

E cap 

Q1 Q 2 2C

:Equation 20.1:

The energy stored by the virtual capacitor is presumably the energy that can be released, perhaps in the form of heat or sound. : Power Pwr Power = Energy*frequency: : {Reference 1, page 186}

Pwr 

Q1 Q 2 f 2C

:Equation 20.2:

Power is also given by I2 R, where R is the resistance and I is the current. {Reference 1, page 984} Consider that for this system the solution, Imax2 is given by: : Imax2 = π2 f2 (Q1 Q2)

:Equation 20.3:

the resonant frequency is given by 2 π f, therefore Imax = (2 π f) √( Q1 Q2 / 4): : Substituting Imax2, Equation 20.3 and Resistance = Z, Equation 18.9, into the equation for power given by, Pwr = Q1 Q2 f / (2 C), Equation 20.2, results in the expression: :

Q1 Q 2 f  2C

π 2 f 2 Q1 Q 2 Z

:Equation 20.4:

1  2C

:Equation 20.5:

Simplifying: :

π2 f Z

Solve for C: :

C

1 2 π2 f Z

Power/applied force is a velocity Velocity, which is a constant. The applied force is the distance d multiplied by the restoring force factor, ξ, That is the applied force is ξ d: :

:

Page 55 of 178

:Equation 20.6:

( Velocity 

Q1 Q 2 f ) 2C ξd

:Equation 20.7:

Simplify: :

Velocity 

Q1 Q 2 f 2Cξ d

Velocity = 22557671.711795(9) [L T -1] It should be noted here that, c / Velocity = 13.290044372940(9) : : : : Reference 1: Physics for Scientists and Engineers with Modern Physics, Fourth Edition, Author Serway

:

:

Page 56 of 178

:Equation 20.8:

CHAPTER 21 “That one body should act upon another through a vacuum without the mediation of anything else is so great an absurdity that no one suited to do science.. can ever fall into it, .. Gravity must be caused by an agent.. but whether that agent be material or immaterial I leave to my readers.” Sir Isaac Newton (1643-1727) : Using the dimensions of the virtual capacitor to derive G: : : From Equation 6.15 given by, d3 r3 = (Q1 Q2) Z G λ4 / [512 c3 π4)] substituting in Ŧ = λ / π Equation 17.3 and d3 = (Q1 Q2) c Z / (32 ξ), Equation 7.8 gives: : r3 = G ξ (Ŧ4) / (16 c4)

:Equation 21.1:

r3 / (ξ Ŧ4) = G / (16 c4)

:Equation 21.2:

Rearranging: : provides the constant (5.15689285863993 x10-46) for a given value for Q, Q1 , Q2 : : Substituting Ŧ = λ / π, Equation 17.3 and rearranging gives: :

r3 ξ(

λ



4

)

π4

G 16 c 4

:Equation 21.3:

Simplifying: :

π4 r3 ξλ

4



π4 c4 r3 ξλ

4

G 16 c 4 

G 16

:Equation 21.4:

:Equation 21.5:

Using f = c / λ (Equation 2.8): :

f 4 r3  ξ

G 16 π 4

:Equation 21.6:

f 4 r3 / ξ is a constant for a defined value of Q. Rearrange f 4 r3 / ξ = G / (16 π4), Equation 21.6 for G: :

G

:

16 π 4 f 4 r 3 ξ

Page 57 of 178

:Equation 21.7:

CHAPTER 22 “You can have no dominion greater or less than that over yourself.” Leonardo da Vinci (1452-1519) : Exploring the Energy and Power properties of the virtual capacitor and verifying the velocity. : Using the standard equation for energy stored in a capacitor, Ecap = Q1 Q2 / (2 C)), Equation 20.1 and substituting for C where C = 1/ (2 π2 f Z), Equation 16.11: :

E cap 

Q1 Q 2 1 2 ( ) 2 π2 f Z

:Equation 22.1:

Simplify: Ecap = Q1 Q2 π2 f Z

:Equation 22.2:

: Substitute Z = 32 d3 ξ / (c Q1 Q2 ), Equation 9.22, derived earlier into the above gives: :

E cap 

Q1 Q 2 π 2 f

32 d 3 ξ c Q1 Q 2

:Equation 22.3:

Simplify: :

E cap 

32 π 2 f d3 ξ c

:Equation 22.4:

Ecap / applied force = distance where the applied force is given by Hooke’s law restoring force factor multiplied by the distance. The distance of interest is d and the restoring force factor is ξ. So the applied force is d ξ: :

E cap dξ



32 π 2 f d 2  c

is a distance

:Equation 22.5:

The above can also be expressed as (f / c = λ, Equation 2.8): :

E cap ξd



32 π 2 d 2 λ



is a distance

:Equation 22.6:

Ecap / (d ξ) = 32 π d2 / Ŧ

:Equation 22.7:

Ŧ = λ / π, Equation 17.3 and Ecap / applied force = 32 π2 d2 / λ, Equation 22.6: :

: Ecap / (applied force), provides a distance over which the force is applied. This distance when divided by the wavelength λ gives: :

:

Page 58 of 178

32 π 2 d 2 λ

2

1 13.290044372940(9)



:Equation 22.8:

Power = Ecap * frequency, where Ecap is given by, Ecap = 32 π2 f d3 / c, Equation 22.4: :

32 π 2 f 2 d3 ξ c

Pwr 

:Equation 22.9:

Power / applied force, is a velocity and equal to Velocity: :

32 π 2 f 2 d 2  c

Pwr  dξ

Velocity

:Equation 22.10:

The velocity (22557671.711796 ms-1 ) is a constant (hypothesis that the movement is from base to tip). The velocity, when divided by the speed of light c: : 22557671.7117958 / c = 13.290044372940(9)

:Equation 22.11:

: As Velocity = Q1 Q2 f / (2 C d ξ), Equation 20.8 And Velocity = 32 π2 f 2 d2 / c, Equation 22.10: These equations are numerically equal and so can be equated: :

32 π 2 f 2 d 2 c

:Equation 22.12:

Q1 Q 2  64 C d ξ

π 2 f d2 c

:Equation 22.13:

c Q1 Q 2

C f d3 ξ

:Equation 22.14:

Q1 Q 2 f  2Cdξ Simplify: :

Collect the constants and the variables: :

64 π

2



The relationship given by, C f d3 ξ, is constant for a defined magnitude of Q = √ ( Q1 Q2 ), and has the value: : C f d3 ξ = 1.21832109477884 x10-32 [Q2 L -1 T -1] : Note that as Z = 32 d3 ξ / (c Q1 Q2 ), Equation 9.22, that c Q1 Q2 / ( 64 π2 ) = C f d3 ξ, Equation 22.14, reduces to the familiar form C = 1/ (2 π2 f Z), Equation 20.6 : Verifying the expressions obtained for distance and velocity using, distance / velocity = Time and using, 32 π2 d2 / λ = distance, Equation 22.6, also 32 π2 f 2 d2 / c = velocity, Equation 22.10 and λ f = c from, Equation 2.8,

:

Page 59 of 178

:Equation 22.15:

:

Time 

32 π 2 d 2 ) λ 32 π 2 f 2 d 2 ( ) c (



c λf

2



f f

2



1 f

:Equation 22.16:

Time = 1 / frequency: : When the impedance of the material, between the coupled photons changes, this results in changes to other parameters, such as the velocity.

:

Page 60 of 178

CHAPTER 23 There are some things that cannot be counted on one’s fingers or measured in one’s heart. : The Hidden Dimension: : 2d is the separation between the charges Q1 and Q2. The circumference is considered to be 2d in wrapped space the radius Rad is : : 2 d = 2 π Rad

R ad 

d π

:Equation 23.1: :Equation 23.2:

Area ,Ac ,of a circle with radius Rad is given by π Rad2:Substitute Rad in to the equation for a circle to give: :

Ac 

d π ( )2 π

d2 π

Ac 

:Equation 23.3:

:Equation 23.4:

The ratio of the area of a virtual capacitor, A = λ d / π, Equation 16.14, to the area of a circle, Ac = d2 / π, Equation 23.4 The ratio, X, is the cross sections of the areas: :

λd ) π d2 ( ) π

:Equation 23.5:

λ d

:Equation 23.6:

64.787030908069(7)

:Equation 23.7:

X

(

Simplify: :

X :

λ  d :

The cross section X of the areas, λ / d, is fairly constant at: = 64.787030908069(7) : The frequency*radius of the virtual capacitor is fairly constant velocity. The frequency (f) multiplied by the radius (Rad ) where, Rad = d / π, Equation 23.2 is: :

d f ( )  1472932.1879273(4) π Multiply the above by the ratio of cross section X of the areas, λ / d, to give: :

:

Page 61 of 178

:Equation 23.8:

λ d f  d π

c π

:Equation 23.9:

Note that frequency*d is fairly constant (at : 4627352.9408284(5)) [L T -1] Reduces to: :

λ f d d

c

:Equation 23.10:

(λ) f = c

:Equation 23.11:

Reduces to the familiar form of Equation 2.8:

:

:

Page 62 of 178

CHAPTER 24 There are three types of believers, those who say it is a coincidence, those who think it is serendipity and those who know it is a truism. : Comparing the photon energy with the energy stored in the virtual capacitor: Energy, Ecap stored in the virtual capacitor this is typically given by, Ecap = Q1 Q2 / (2 C), Equation 20.1. Given that the capacitance, C, is given by, C = 1/ (2 π2 f Z), Equation 16.11, then substituting for C into Ecap gives: :

E cap  (

Q1 Q 2 2 2 π2 f Z

:Equation 24.1:

)

: Ecap = π2 Z Q1 Q2 f

Equation 24.2:

Note that η = Z π Q1 Q2 from Equation 9.23 then the energy stored by the virtual capacitor can also be expressed as: Ecap = η π f

Equation 24.3:

As the energy of the photon Ephoton is given by (h f) and the expression for Planks constant, h, where : h = 4 π2 c2 f r3 / G, Equation 10.12: :

4 π2 c2 f 2 r 3 G

E photon 

:Equation 24.4:

The ratio of the energies Ephoton / Ecap is: :

E photon E cap

( 

4 π2 c2 f 2 r3 ) G π 2 Z Q1 Q 2 f

:Equation 24.5:

Simplify:

E photon E cap



4 c2 f r3 Z G Q1 Q 2

:Equation 24.6:

The ratio of the energies Ephoton / Ecap, is found to be a constant (that is equal to: 21.8099551047939). This ratio of energies gives the same numerical value as h / (η π ), which is explored below in Equation 24.12. : Only a fraction of the photon energy is stored in the virtual capacitor. The ratio of the Ephoton given by, Equation 24.4, and the capacitance C, where C is given by, C = 1/ (2 π2 f Z), Equation 16.11 ) is: :

E photon C

:

( 

4 π2 c2 f 2 r3 ) G 1 ( ) 2 π2 f Z

Page 63 of 178

:Equation 24.7:

:

8 π4 c2 Z f r3 f 2 G

:Equation 24.8:

1.56843464533602 x 10 -30

:Equation 24.9:

E photon C



Note that f r3, G, Z, c and π are constants. Where U = f r3, Equation 5.6 So collecting the constants: :

8 π4 c2 Z f r3  G thus the ratio of Ephoton / C is: :

E photon C



(1.56843464533602 x 10 -30 ) (f 2 )

:Equation 24.10:

Consider the reciprocal:

C E photon



6.37578367051284 x 10 29

:Equation 24.11:

f2

: It is interesting to note that when: : : C / Ephoton, = 1,

then,

:f1 = 798485044976600s-1

and

λ = 0.000000375451562788863m: :

then,

: f 2 = 564614189979000s-1

and

λ = 0.000000530968692110183m: :

then,

: f 3 = 461005555661131s-1

and

λ = 0.000000650301182531446m: :

then,

: f 4 = 399242522488300s-1

and

λ = 0.000000750903125577725m: :

: C / Ephoton, = 2, : C / Ephoton, = 3, : C / Ephoton, = 4, : The visible region (violet to red) of the spectrum is apparent. Also note that : f 2 = f 1 / √ 2, f3 = f1 / √ 3 f4 = f1 / √ 4 Consider that the ratio of the energies given in Equation 24.6 is equivalent to h / (η π ) then stating this: :

h  ηπ

4 c2 f r3 Z G Q1 Q 2

:Equation 24.12:

As η = Z π Q1 Q2 from Equation 9.23, then substituting this into the above equation gives: :

h Z π 2 Q1 Q 2



4 c2 f r3 Z G Q1 Q 2

Simplifies to: :

:

Page 64 of 178

:Equation 24.13:



4 c2 f r3 G

:Equation 24.14:

f r3 

4 c2 π2 Gh

:Equation 24.15:

h π2 Rearrange

Where f r3 is the Universal constant, as given in Equation 5.6. : Note that power is given by energy multiplied by the frequency. Then as, Ecap = η π f Equation 24.3, then the power generated by the virtual capacitor is: : Powercap = η π f2 :Equation 24.16: :

:

Page 65 of 178

Part III: A New Consideration for Time: “The diversity of the phenomena of nature is so great and treasures hidden in the heavens so rich precisely in order that the human mind shall never be lacking in fresh nourishment” Johannes Kepler (1571-1630) :

CHAPTER 25 “Time stays long enough for anyone who will use it.” Leonardo da Vinci (1452-1519) : Defining frequency. Also deriving an alternative expression for Plank’s constant, h. : As: d3 r3 f 4 = (Q1 Q2) c Z G / [512 π4) ], Equation 8.2 and d3 r3 = (Q1 Q2) Z G λ4 / [ 512 c3 π4 ], Equation 7.15. First using Equation 7.15 and rearranging for c3: :

Z G Q1 Q 2 λ 4

c3 

514 π 4 d 3 r 3

:Equation 25.1:

Substitute for Z in the above equation, where Z is, Z = 32 d3 ξ / (c Q1 Q2), Equation 7.17 results in: :

c3 

G Q1 Q 2 λ 4

32 d 3 ξ 514 π 4 d 3 r 3 c Q1 Q 2

:Equation 25.2:

Simplify: :

32 G ξ λ 4

c4 

514 π 4 r 3

c4 

G ξ λ4 16 π 4 r 3

:Equation 25.3:

:Equation 25.4:

Rearrange: :

c4 λ

4



Gξ 16 π 4 r 3

:Equation 25.5:

As λ f = c, Equation 2.8 then, c4 / λ4 = f 4, giving: :

f4  f

Gξ 16 π 4 r 3

:Equation 25.6:

1 G ξ 1/4 ( ) 2 π r3

:Equation 25.7:

16 π 4 r 3 Gξ

:Equation 25.8:

As f is typically the inverse of the Period T i.e. 1/T: :

T4 

:

Page 66 of 178

If time is an illusion it is a very good one. : Deriving an alternative expression for Plank’s constant by using Equation 25.6 given as, f4 = G ξ / (16 π4 r3), rearranging this for r3 gives:

r3 

Gξ 16 π 4 f 4

:Equation 25.9:

Substitute the expression for r3 given above in equation 25,9 into the expression for h where h = 4 π2 c2 f r3 / G Equation 5.2 to give:

h

Gξ 4 π2 c2 f G 16 π 4 f 4

:Equation 25.10

Simplifies to:

h

c2 ξ 4 π2 f 3

Note that ξ / f 3 is a constant 2.91054502786985 x10-49 [L2 T -1] :

:

Page 67 of 178

:Equation 25.11

CHAPTER 26 “Do not dwell in the past. Do not dream of the future. Concentrate the mind on the present moment” Siddhartha Gautama - Buddha (approximately 563BC-483BC) : Deriving a fundamental relationship between G and the restoring force factor. Subsequently deriving alternative expressions for Plank’s constant, h. As d3 r3 f 4 = (G / 2)4 from Equation 10.16, rearranging for f: :

G4 1 16 d 3 r 3

f4 

f4 

G4



G4

1 16 d3 r 3

:Equation 26.1:

:Equation 26.2:

As f 4 = G ξ / (16 π4 r3), Equation 25.6 and f 4 = (G)4 [1 / (16 d3 r3)], Equation 26.2, these equations can be equated, giving: :

f4 

Gξ 4

16 π r

3

1 16 d3 r 3

:Equation 26.3:

Simplifying: :

ξ π4



G3 d3

:Equation 26.4:

Simplify: : : ξ d3 = G3 π4

:Equation 26.5:

: This is an important relationship and is very useful in redefining fundamental properties in terms of the Universal Gravitational constant. : Alternative expressions for Plank’s constant may be derived. Commencing by rearranging Equation 26.5 for ξ to give:

ξ

G3 π 4 d3

:Equation 26.6:

Then substituting the expression for ξ into the expression for Plank’s constant given by, h = c2 ξ / 4 π2 f3 Equation 25.11 which gives:

h

c2

G3 π 4

4 π2 f 3

d3

As d3 = Z c Q1 Q2 / (32 ξ ) :Equation 7.8, substituting this expression into Equation 26.7 above gives:

:

Page 68 of 178

:Equation 26.7:

h

c 2 G3 π 4 2

4π f

3

32 ξ Z c Q1 Q 2

:Equation 26.8:

Simplify: :

8 π 2 c G3 ξ

h

:Equation 26.9:

Z Q1 Q 2 f 3

Recall that, ξ / f 3, is a constant. : By substituting the equation 26.9 into, h / (Z π2 Q1 Q2 ) = 4 c2 f r3 / (Z G Q1 Q2 ), Equation 24.13 gives: :

(

8 π 2 c G3 ξ Z Q1 Q 2 f 3

)

Z π 2 Q1 Q 2



4 c2 f r3 Z G Q1 Q 2

:Equation 26.10:

4 c2 f r3 G

:Equation 26.11:

Simplifies to: :

8 c G3 ξ Z Q1 Q 2 f 3



Simplifies: :

2 G4  c Z Q1 Q 2

f 4 r3 ξ

:Equation 26.12:

Recall f 4 r3 / ξ = G / (16 π4 ) Equation 21.6 then equating the above gives: :

f 4 r3  ξ

2 G4  c Z Q1 Q 2

G 16 π 4

:Equation 26.13:

which reduces to the equation: :

G3 

:

c Z Q1 Q 2 32 π 4

Page 69 of 178

:Equation 26.14:

PART IV Redefines Electromagnetic relationships: :

CHAPTER 27 In the matter of science, truth is more important than imagination. : Defining εo is the permittivity for free space (otherwise known as the electric constant): : Using C = C = εo A / S = εo A / (2 d), Equation 16.1, for a parallel plate capacitor. A = λ d / π, Equation 16.14, and C = 1 / (2 π2 f Z), Equation 16.11, substituting the expression for the Area gives: :

C

εo A  2d

εo λ d  2d π

1 2

2π Zf

:Equation 27.1:

Simplifies to: :

εo λ  π

1 2

π Zf

:Equation 27.2:

Rearrange: :

εo 

1 πZf λ

:Equation 27.3:

1 πZc

:Equation 27.4:

(or 1 / εo = π Z f λ ) Utilise c = λ f: from Equation 2.8: :

εo  (or 1 / εo = π Z c )

Substitute for Z, where Z = 32 d3 ξ / (c Q1 Q2), Equation 7.17 to give: :

εo 

1 c Q1 Q 2 π c 32 d 3 ξ

:Equation 27.5:

Simplifies to: :

εo 

Q1 Q 2 32 π d 3 ξ

:Equation 27.6:

or more appropriately expressed below as (1 / εo = 32 π d3 ξ / Q1 Q2 ): :

1  εo

32 π d 3 ξ Q1 Q 2

:Equation 27.7:

εo is the permittivity for free space, the value derived from this model is given in Chapter 57 and agrees with the accepted value.

:

Page 70 of 178

The permittivity for free space can be expressed in terms of the unification constant, η, firstly rearranging Equation 27.4, εo = 1 / π Z c: :

1  εo

πZc

:Equation 27.8:

Substituting in for π Z using, η = Z π Q1 Q2, Equation 9.23, gives: :

1  εo

:

ηc Q1 Q 2

Page 71 of 178

:Equation 27.9:

CHAPTER 28 “Imagination is more important than knowledge.” Albert Einstein (1879-1955) : The relationship between c2 and c3. (Or the relationship between a two dimensional ‘world’ and a three dimensional ‘world’.) : Starting with: :

c2  εo

π Z c3

:Equation 28.1:

c

1 π Z εo

:Equation 28.2:

Reduces to: :

This is the same as Equation 27.4. Substitute Z = 32 d3 ξ / (c Q1 Q2), Equation 7.17 and εo = Q1 Q2 / (32 π ξ d3), Equation 27.6 into Equation 28.1 to give: :

c2

32 π ξ d 3  Q1 Q 2

π c3

32 ξ d 3 c Q1 Q 2

:Equation 28.3:

The above reduces to: : c=c

:Equation 28.4:

2500000

:Equation 28.5:

So verifying that the system is not violated. : It is interesting to note that:

c  Z c / Z = 2500000 [Q2 T2 L -3] exactly.

:

Page 72 of 178

CHAPTER 29 “Man will occasionally stumble over the truth, but most of the time he will pick himself up and continue on” Winston Churchill (1874-1965) : Redefining Z and εo: : As Z = 32 d3 ξ / ( c Q1 Q2), Equation 7.17 and as ξ d3 = π4 G3, Equation 26.5, by substituting for ξ d3 in the expression Z gives: :

32 π 4 G 3 c Q1 Q 2

Z

:Equation 29.1:

Similarly as εo = Q1 Q2 / (32 π ξ d3), Equation 27.6 and as ξ d3 = π4 G3, Equation 26.5: by substituting for ξ d3 in the expression for εo gives: :

Q1 Q 2

εo 

32 π π 4 G 3

:Equation 29.2:

(Or, 1 / εo = 32 π π4 G3 / Q1 Q2 ) The Equation 29.2 can be expressed as: :

εo 

Q1 Q 2 32 π 5 G3

:Equation 29.3:

(Or, 1 / εo = 32 π5 G3 / Q1 Q2 ) The Equations 29.1 and 29.3 clearly and simply provide an expression for the unification of electromagnetic phenomena with the forces governing the attraction of masses. : Note Equation 29.1 can be rearranged, to make G the subject of the equation,:

G3 

Z c Q1 Q 2 32 π 4

:Equation 29.4:

As per Equation 26.14. Taking the cube root of both sides:

G

(

Z c Q1 Q 2 32 π 4

)1/3

:Equation 29.5:

G, where G3 = Z c Q1 Q2 / 32 π4 expressed in terms of the unification constant, η, where η = Z π Q 1 Q2, Equation 9.23:

G3 

:

cη 32 π 5

Page 73 of 178

:Equation 29.6:

CHAPTER 30 “An investment in knowledge pays the best interest.” Benjamin Franklin (1706-1790) : Establishing μo the magnetic permeability of free space (otherwise known as the magnetic constant): :

1

c

ε o μo

:Equation 30.1:

{c = 1 / √ (εo μo) Reference 1, page 998}

1

μo 

2

c εo

:Equation 30.2:

Recall εo = Q1 Q2 / (32 π ξ d3), Equation 27.6 and εo = Q1 Q2 / (32 π5 G3), Equation 29.3, these expressions can be substituted into Equation 30.2:

μo  2

c (

1 Q1 Q 2

:Equation 30.3:

32 π ξ d 3

)

32 π ξ d3

μo 

c 2 Q1 Q 2

:Equation 30.4:

as ξ d3 = π4 G3, Equation 26.5: :

μo 

32 π π 4 G 3 c 2 Q1 Q 2

:Equation 30.5:

Simplifies to: :

32 π 5 G3

μo 

c 2 Q1 Q 2

:Equation 30.6:

Substituting Z = 32 π4 G3/ (c Q1 Q2), Equation 29.1 into Equation 30.5 gives: :

μo 

Zπ c

:Equation 30.7:

μo is the magnetic permeability for free space, the value derived from this model (any of the above equations) is given in Chapter 57 and agrees with the generally accepted value. Rearranging Equation 30.7, μo = Z π / c, as

μo  Z

π c

:Equation 30.8:

The magnetic permeability, μo, can be expressed in terms of the unification constant, η, as η = Z π Q 1 Q2, Equation 9.23, then using the above expression: :

μo 

η Q1 Q 2 c

:

:

Page 74 of 178

:Equation 30.9:

Deriving a Relationship between the Impedance (Z), the Permittivity and the Permeability of Free Space : Recall the expression for the permittivity (εo) is given by, εo = Q1 Q2 / (32 π5 G3), Equation 29.3, rearranging this expression and dividing both sides by c to give: :

32 π 5 G 3 c Q1 Q 2

1  c εo

:Equation 30.10:

As Z = 32 π4 G3/ (c Q1 Q2), Equation 29.1, then the above expression can be expressed in terms of Z: :

1  c εo

Zπ

:Equation 30.11:

From Equation 30.1, 1 / c = √ (εo μo ), substituting for 1 / c in the above gives: :

μo ε o



Zπ

:Equation 30.12:

μo  εo

Zπ

:Equation 30.13:

εo :This simplifies to: :

: The Relationship between Charge and the Dielectric Permittivity and the Magnetic Permeability : From and by multiplying By Multiplying Equation 29.1, Z = 32 π4 G3 / (c Q1 Q2), by π, it can then be equated with the Equation 30.13, √ (μo / εo) = Z π, to give: :

μo  εo

32 π 5 G 3 c Q1 Q 2

:Equation 30.14:

Recall that, μo εo = 1 / (c2 ), Equation 30.1, which can be rearranged to give the expression: :

μo 

1 c εo

:Equation 30.15:

Substituting in the above expression for √μo, in to Equation 30.14 , √ (μo / εo ) = 32 π5 G3 / (c Q1 Q2), gives: :

1 c εo

εo

32 π 5 G3 c Q1 Q 2



:Equation 30.16:

Solving for (Q1 Q2) and simplifying gives: : Q1 Q2 = 32 π5 G3 εo

:Equation 30.17:

The product of (Q1 Q2) is a constant. The above equation can be rearranged to provide the ratio: :

Q1 Q 2 G

3



32 π 5 ε o

Also a constant.

:

Page 75 of 178

:Equation 30.18:

: A similar method is applied to obtain an expression for the charge in terms of the magnetic permeability, Recall that, μo εo = 1 / (c2 ), Equation 30.1, which can be rearranged to give the expression: :

εo 

1 c μo

:Equation 30.19:

Substituting in for √ εo, in to Equation 30.14, √ (μo / εo ) = 32 π5 G3 / (c Q1 Q2), gives: :

c μo μo 

32 π 5 G 3 c Q1 Q 2

:Equation 30.20:

Solving for (Q1 Q2) and simplifying: :

Q1 Q 2 

32 π 5 G3 c 2 μo

:Equation 30.21:

: The values for the Universal Gravitational Constant, G, and for the product of (Q1 Q2) can be found in Chapter 57. These values have been calculated using the defined value of c, εo and pi (to a limited number of decimal places) and then honing G, Q1 Q2.to provide the exact match using the model. The product of (Q1 Q2) is a constant. For the simplest case assume that Q1 has the same value as Q2 and by taking the square root to provide a value for the elementary charge where Q1 = Q2. Note that the value of the charges Q1 , Q2 do not necessarily have to be equal, only their product needs to be a constant. The distribution of charge is discussed further in Appendix 8. Note that Equation 30.21 may be rearranged to provide the ratio: :

Q1 Q 2 G3



32 π 5 c 2 μo

Also a constant, with the same value as that of Equation 30.18. : Reference 1: Physics for Scientists and Engineers with Modern Physics, Fourth Edition, Author Serway:

:

Page 76 of 178

:Equation 30.22:

CHAPTER 31 Werner Heisenberg’s (1901-1976) uncertainty principle state that, A particle’s exact position and speed cannot be simultaneously measured to an accuracy of better than: h / (4 π). : Establishing H, the magnetic field strength: : For free space typically: H / εo = 4 π / c2 : As εo = Q1 Q2 / (32 π ξ d3), Equation 27.6 then: :

4 π Q1 Q 2

H

32 π c 2 ξ d 3

:Equation 31.1:

Simplify: :

H

Q1 Q 2 8 c 2 ξ d3

:Equation 31.2:

As ξ d3 = π4 G3 given by Equation 26.5, the above equation can be rewritten as: :

H

Q1 Q 2 8 c 2 π 4 G3

:Equation 31.3:

H = 1.23799014723612 x 10-27 [Q2 T6 L -7] It is noted here that this value for H differs from the value of the magnetic field strength of free space (Ho) by a factor of 4 π i.e. H = 4 π Ho

:

Page 77 of 178

:Equation 31.4

CHAPTER 32 “Electrical force is defined as something which causes motion of electrical charge, an electrical charge is something which exerts electrical force” Arthur Eddington (1882-1944) : Establishing the electric field strength, E, for free space: Starting with the well known expression: :

E  H

Zπ

:Equation 32.1:

E=HZπ

:Equation 32.2:

Rearranged gives: :

Substituting in the expression for H given in Equation 31.2, where H = Q1 Q2 / (8 c2 ξ d³) leads to: :

Z π Q1 Q 2

E

8 c 2 ξ d3

:Equation 32.3:

as ξ d3 = π4 G3, Equation 26.5: :

Z π Q1 Q 2

E

8 c 2 π 4 G3

:Equation 32.4:

Substitute for Z in the above, where Z is given by, Z = 32 π4 G3 / (c Q1 Q2 ), Equation 29.1 provides: :

E

π Q1 Q 2

32 π 4 G 3 8 c 2 π 4 G 3 c Q1 Q 2

:Equation 32.5:

π 32 8 c2 c

:Equation 32.6:

Reduces to: :

E Simplifies to: :

E

4π c3

:Equation 32.7:

This agrees with the generally accepted expression for E for free space. : Establishing the relationship between the Electric field strength (E) and Planks constant (h): : As Z = η / π Q1 Q2, obtained by rearranging Equation 9.23, η = 8 π3 c G3 ξ / (h f 3) obtained by rearranging Equation 10.20 By substitution of Equation 10.20 into Equation 9.23 gives:

Z

8 π 3 c G3 ξ π h f 3 Q1 Q 2



8 π 2 c G3 ξ h Q1 Q 2 f 3

:Equation 32.8:

Substituting the above expression for Z into Equation 32.3 for E where, E = Z π Q1 Q2 / (8 c2 ξ d3), gives:

:

Page 78 of 178

E

8 π 2 c G 3 ξ π Q1 Q 2 h Q1 Q 2 f 3 8 c 2 ξ d 3

:Equation 32.9:

The above simplifies to:

E

π 3 G3 h c f 3 d3

:Equation 32.10:

Equating Equation 32.7, E = 4 π / c3, with the above equation for E gives:

4π

E



c3

π 3 G3 h c f 3 d3

:Equation 32.11:

So giving the constant f 3 d3 as:

f 3 d3 

π 2 c 2 G3 4h

:Equation 32.12:

Or rearranged to express in terms of d:

d3 

π 2 c 2 G3 4h f3

:Equation 32.13:

: The above express can be used to define the magnetic field strength (H) in terms of Plank’s constant (h): : Starting with the expression, H = Q1 Q2 / (8 c2 ξ d3),

Equation 31.2 and substituting for d3 where,

d3 = π2 c2 G3 / (4 h f 3) Equation 32.13, so gives:

H

Q1 Q 2

4h f3

8 c 2 ξ π 2 c 2 G3



h Q1 Q 2 f 3 2 π 2 c 4 G3 ξ

:Equation 32.14:

Where f 3 / ξ is a constant. : Establishing an alternative expression for Plank’s constant, h, starting by rearranging d3 = π2 c2 G3 / (4 h f 3) Equation 32.13, to make h the subject, gives: :

h

π 2 c 2 G3 4 f 3 d3

:Equation 32.15:

From, λ / d = 64.787030908069(7), Equation 23.7 then rearranging this for d and substituting into the above gives

h

π 2 c 2 G3 λ 4 f3 ( )3 64.7870309080697

:Equation 32.16:

Simplifying

h

67983.6128 720116 π 2 c 2 G 3 f 3 λ3

As λ f = c, from Equation 2.8, then c2 / (f3 λ3) = 1 / c, substituting this into the above gives: :

:

Page 79 of 178

:Equation 32.17:

h

67983.6128720116 π 2 G 3 c

:Equation 32.18:

Or: :

ch π 2 G3



67983.6128720116

:Equation 32.19:

6888.1801244747

:Equation 32.20:

Divide both sides by π2 to give: :

ch π 4 G3



As ξ d3 = π4 G3, Equation 26.5, then the above can also be expressed as: :

ch ξ d3



6888.18012 44747

The constant given by Equation 32.20 and 32.21 is dimensionless. Let the constant be denoted by J: : J = 6888.1801244747 So: :

ch ξ d3



J

:Equation 32.21:

:Equation 32.22: :Equation 32.23:

: Alternative Expression for E: : By rearranging Equation 28.1 to give, c3 = c2 / π Z εo and substituting the expression for c3 into Equation 32.7 given by E = 4 π / c3,

E

:

4 π2 Z εo c2

Page 80 of 178

:Equation 32.24:

CHAPTER 33 The magic and mystery of the cosmos is not diminished, but confirmed. : Establishing B the magnetic field: : B = μo H + Mag

:Equation 33.1:

{B Reference 1, page 883} For free space the magnetization vector Mag = 0. Assuming that the magnetization vector is zero for the system then,: B = μo H

:Equation 33.2:

Utilising H = Q1 Q2 / (8 c π G ), Equation 31.3 and 2

4

3

μo = (32 π5 G3) / (c2 Q1 Q2 ), Equation 30.6 to substitute into the expression above for the magnetic field gives: :

B

32 π 5 G 3

Q1 Q 2

2

c Q1 Q 2 8 c 2 π 4 G 3

:Equation 33.3:

Reduces to: :

32 π

B

c28 c2

:Equation 33.4:

Collect terms and simplify: :

4π

B

c4

:Equation 33.5:

Alternatively using the expression μo = (Z π) / (c) from Equation 30.7: :

B

μo H 

Z π Q1 Q 2 c 8 c 2 π 4 G3

:Equation 33.6:

Reduces to: :

B

Z π Q1 Q 2

μo H 

8 c 3 π 4 G3

:Equation 33.7:

Typically E / B = c {E / B = c, Reference 1, page 999} From B = 4 π / (c4), Equation 33.5 and E = ( 4 π / c3), Equation 32.7, then: :

E B

(  (

4π c3 4π c4

) 

c

:Equation 33.8:

c

:Equation 33.9:

)

Using, B = Z π Q1 Q2 / (8 c3 π4 G3), Equation 33.7 and E / B = c, Equation 33.8: :

E

Z π Q1 Q 2 8 c 3 π 4 G3

Simplify: :,

:

Page 81 of 178

E

Z π Q1 Q 2 8 c 2 π 4 G3

: : : : Reference 1: Physics for Scientists and Engineers with Modern Physics, Fourth Edition, Author Serway

:

:

Page 82 of 178

:Equation 33.10:

CHAPTER 34 In reference to a sufficiently powerful intellect: “Nothing is uncertain and the future like the past would be present before its eyes.” Pierre-Simon Laplace (1749-1827) : : Our Universe’s Balancing Equation: : As Z = (32) π4 G3 / (c Q1 Q2), Equation 29.1: :

1  Z π Q1 Q 2

c 32 π G 3 π 4

:Equation 34.1:

: The two sides of this equation are numerically equal and equal to: : : 1.03406693785069 x 1035 [T3 L -4]

:Equation 34.2:

The above value is equal to the reciprocal of the unification constant, i.e. 1 / η, defined in Equation 9.21. :

The Relationship Between Plank’s constant and the Restoring Force Factor ξ: : The restoring force factor ξ can also be expressed in terms of Plank’s constant (h). Recall that, ξ d3 = G3 π4, Equation 26.5, by substituting in the expression for d3, where d3 = π2 c2 G3 / (4 h f 3), Equation 32.13, gives:

ξ

π 4 G3

4h f3 π 2 c 2 G3

:Equation 34.3:

Simplifies to:

ξ

4 π2 h f 3 c2

Note that ξ / f 3, is a constant (see Chapter 57) :

:

Page 83 of 178

:Equation 34.4:

CHAPTER 35 “It will be found …that one universal law prevails in all these phenomena.” Thomas Young (1773-1829) : Expressing C for the virtual capacitor in terms of G, H and E: Using C = εo A / 2d, Equation 16.3 and Area A = d λ / π, Equation 16.14:

C

εo (

λ ) 2π

:Equation 35.1:

Substitute εo = Q1 Q2 / (32 π ξ d3), Equation 27.6 into Equation 29.1 above gives: :

λ Q1 Q 2

C

:Equation 35.2:

2 π 32 π ξ d 3

Substitute λ = c / f, Equation 2.8: :

c Q1 Q 2

C

:Equation 35.3:

2 π f 32 π ξ d 3

As ξ d3 = π4 G3, Equation 26.5: :

c Q1 Q 2

C

:Equation 35.4:

64 π 2 π 4 G3 f

Collect terms: :

C

c Q1 Q 2

:Equation 35.5:

64 π 6 G 3 f

As f = [1/ (2 π) ] [ (G ξ / r3)1/4], Equation 25.7 substitute this into, C = c Q1 Q2 / (2 π f 32 π ξ d3), Equation 35.3 leads to the expression: :

C

c Q1 Q 2 1 G ξ 1/4 2 π 32 π ξ d3 ( )( ) 2 π r3

:Equation 35.6:

Simplifying: :

c Q1 Q 2 Gξ 32 π ξ d3 ( 3 )1/4 r

C

:Equation 35.7:

As ξ d3 = π4 G3, Equation 26.5, the above equation can be written as: :

C



c Q1 Q 2 Gξ 32 π π 4 G 3 ( 3 )1/4 r

:Equation 35.8:

Collecting terms: :

C

c Q1 Q 2 32 π ( 5

:

G13 ξ r3

 1/4

)

c Q1 Q 2 32 π ( 5

G13/4 ξ 1/4 r 3/4

 )

c Q1 Q 2

r 3/4

32 π 5

G13/4 ξ 1/4

Page 84 of 178

Equation 35.9:

: C can be expressed in terms of H: As C = c Q1 Q2 / (64 f π6 G3), Equation 35.5 and, H = Q1 Q2 / (8 c2 π4 G3), Equation 31.3: then: :

C



H c3 8 π2 f

:Equation 35.10:

As E = ( 4π / c3), Equation 32.7: :

C

H 1 E 2πf

Note as E = H Z π, Equation 32.2, then above reduces to the familiar form, C = 1/ (2 π2 f Z), Equation 16.11 :

:

Page 85 of 178

:Equation 35.11:

CHAPTER 36 “It is far better to grasp the universe as it really is, than to persist in delusion, however satisfying and reassuring.” Carl Sagan : Defining magnetic field strength, H, and the electric field strength E, for the virtual capacitor: : HZπ=E

:Equation 36.1:

H / (32 d3 ξ π / (c Q1 Q2) ) = E given by, Equation 29.9, substitute for Z where, Z = 32 d3 ξ / (c Q1 Q2 ) given in Equation 7.17, into equation for H into Equation 36.1 to give: :

E

H

32 π d 3 ξ c Q1 Q 2

:Equation 36.2:

Rearrange the above for H: :

H

E 32 π d 3 ξ ( ) c Q1 Q 2

:Equation 36.3:

Simplify: :

H

E

c Q1 Q 2 32 π d 3 ξ

:Equation 36.4:

Rearrange for H / E gives: :

c Q1 Q 2

H  E

32 π d 3 ξ

E  H

32 π d 3 ξ c Q1 Q 2

:Equation 36.5:

Rearrange for E / H gives: : :

:

:

Page 86 of 178

:Equation 36.6:

Part V Establishing Expressions and Relationships for the Electric and Magnetic Forces. : : “Playing in the sea shore and diverting myself in now and then finding a smoother pebble or a prettier shell than ordinary” Sir Isaac Newton (1643-1727)

: Utilising the expressions listed below to obtain relationships between the Universal Gravitational constant and the electric and magnetic field strengths: : Z = 32 d3 ξ / (c Q1 Q2),

Equation 7.17

: ξ d3 = π4 G3,

Equation 26.5

: εo = Q1 Q2 / (32 π ξ d3),

Equation 27.6

: Z = 32 π4 G3/ (c Q1 Q2 ),

Equation 29.1

: εo = Q1 Q2 / (32 π5 G3),

Equation 29.3

: H = Q1 Q2 / (8 c2 ξ d3),

Equation 31.2

: H = Q1 Q2 / (8 c2 π4 G3),

Equation 31.3

: E = Z π Q1 Q2 / (8 c2 ξ d3),

Equation 32.3

: E = Z π Q1 Q2 / (8 c2 G3 π 4),

Equation 32.4

: E = 4 π / c3,

Equation 32.7

:

:

Page 87 of 178

CHAPTER 37 Appearance precedes reality. : Deriving an expression for H / E: : Using Equation 35.7 to provide an expression for C where, C = c Q1 Q2 / [ (32π ξ d3) (Gξ / r3 )1/4 ] The capacitance C can now be expressed in terms of E and H by substituting in the Equation 36.5, H / E = c Q1 Q2 / (32 d3 ξ π): :

C

(

H ( ) E G ξ 1/4 r3

:Equation 37.1:

)

Solving for H / E: : :

H  E

C(

Gξ r

3

)1/4

:Equation 37.2:

Lifting both sides by to the power of 4: : :

H4 E

4



C4

Gξ r3

:Equation 37.3:

As f = [1/ (2 π) ] [ (G ξ/ r3)1/4], Equation 25.7 then: :

H4 E4



C 4 ( 2 π f )4

:Equation 37.4:

Removing power to the fourth gives: :

H  E

:

2πf C

Page 88 of 178

:Equation 37.5

CHAPTER 38 The portal of knowledge is always open. : Establishing an expression for HZ and H/Z: : Relationships: : : With expressions now for Eo, Ho, Bo, Z, εo and μo the expressions are derived: : An expression for H Z is derived: As ξ d3 = π4 G3, Equation 26.5, H = Q1 Q2 / (8 c2 ξ d³), Equation 31.2, H = Q1 Q2 / (8 c2 π4 G3), Equation 31.3, Z = 32 d3 ξ / c Q1 Q2, Equation 7.17, Z = 32 π4 G3/ (c Q1 Q2 ), Equation 29.1 then: :

HZ

32 π 4 G 3 8 c 2 π 4 G 3 c Q1 Q 2

:Equation 38.1:

32 ξ d 3 8 c 2 ξ d 3 c Q1 Q 2

:Equation 38.2:

Q1 Q 2

Cancel the π4 term in the numerator: :

HZ

Q1 Q 2

Cancel other terms: :

HZ 

4 c3

:Equation 38.3:

An expression for H /Z is derived. As ξ d3 = π4 G3, Equation 26.5, H = Q1 Q2 / (8 c2 ξ d³), Equation 31.2, H = Q1 Q2 / (8 c2 π4 G3), Equation 31.3, Z = 32 d3 ξ / (c Q1 Q2), Equation 7.17, Z = 32 π4 G3 / (c Q1 Q2), Equation 29.1 then: :

H  Z

(Q 1 Q 2 ) 2 256 c (d 3 ξ) 2

:Equation 38.4:

As ξ d3 = π4 G3, Equation 26.5: :

H  Z

(Q 1 Q 2 ) 2 256 c π 8 G 6

:Equation 38.5:

: An alternative Expression for H: By rearranging, Equation 28.1, as c3 = c2 / π Z εo, and substituting this expression for c3 into HZ = 4 / c3, Equation 38.3 gives: :

:

Page 89 of 178

H

:

4 π εo c2

Page 90 of 178

:Equation 38.6:

CHAPTER 39 As I made a path through the intricate labyrinth I may not have taken the most direct route, for there were many meandering paths and cross-ways. : Deriving an expression for E Z and E / Z: : For E Z use, E = 4 π / c3, Equation 32.7 and Z = 32 π4 G3 / (c Q1 Q2), Equation 29.1: :

EZ

4 π 32 π 4 G 3 c 3 c Q1 Q 2

:Equation 39.1:

Collecting terms: :

128 π 5 G3

EZ

c 4 Q1 Q 2

:Equation 39.2:

Using, E = Z π Q1 Q2 / (8c2 G3 π4), Equation 32.4 and Z = 32 π4 G3 / (c Q1 Q2), Equation 29.1: :

EZ

Z π Q1 Q 2 32 π 4 G 3 8 c 2 G 3 π 4 c Q1 Q 2

:Equation 39.3:

32 π 4 G 3 c

:Equation 39.4:

Zπ 4 c2 c

:Equation 39.5:

Simplifying: : :

EZ

Zπ 8 c 2 G3 π 4

Simplifying further: :

EZ Collecting terms: :

EZ

4 πZ c3

:Equation 39.6:

Equating the expressions for E Z given by Equations 39.2 and 39.6 where, E Z = 128 π5 G3 / c4 Q1 Q2, Equation 39.2 and E Z = 4 π Z / (c3), Equation 39.6: :

EZ

128 π 5 G 3 4

c Q1 Q 2



4 πZ c3

Collecting terms: :

:

Page 91 of 178

:Equation 39.7:

32 π 4 G 3  c Q1 Q 2

Z

:Equation 39.8:

Z π Q1 Q 2 1 8 c 2 G3 π 4 Z

:Equation 39.9:

As E = Z π Q1 Q2 / (8 c2 G3 π4), Equation 32.4, then E / Z is: :

E  Z Cancel down: :

E  Z

π Q1 Q 2 8 c 2 G3 π 4

:Equation 39.10:

Alternatively as H = Q1 Q2 / (8 c2 π4 G3), Equation 31.3 and as E = Z π Q1 Q2 / (8 c2 G3 π4) then: :

E  Z

:

πH

Page 92 of 178

:Equation 39.11:

CHAPTER 40 “There are two possible outcomes, if the result confirms the hypothesis, then you have made a measurement. If the result is contrary to the hypothesis, then you have made a discovery.” Enrico Fermi (1901-1954) : Deriving an expression for H εo and H / εo: :

As H = Q1 Q2 / (8 c2 ξ d³), Equation 31.2, H = Q1 Q2 / (8 c2 π4 G3), Equation 31.3, εo = Q1 Q2 / (32 π ξ d3), Equation 27.6 and εo = Q1 Q2 / (32 π5 G3), Equation 29.3 then H εo is: : :

(Q1 Q 2 ) 2

H εo 

256 π c 2 (d 3 ξ) 2

:Equation 40.1:

As ξ d3 = π4 G3, Equation 26.5:

H εo 

(Q1 Q 2 ) 2 256 π c 2 (π 4 G 3 ) 2

:Equation 40.2:

Expanding the powers and collecting terms: :

H εo 

(Q 1 Q 2 ) 2 256 π 9 c 2 G 6

:Equation 40.3:

Using H = Q1 Q2 / (8 c2 ξ d³), Equation 31.2 and εo = Q1 Q2 / (32 π ξ d3), Equation 27.6 or: H = Q1 Q2 / (8 c2 π4 G3), Equation 31.3 and εo = Q1 Q2 / (32 π5 G3), Equation 29.3, to give an expression for H / εo : :

H  εo

:

4π c2

Page 93 of 178

:Equation 40.4:

CHAPTER 41 “Facts mean nothing when they are preempted by appearance. Do not underestimate the power of impression over reality.” Brian Herbert : Deriving and expression for E εo: : Using E = Z π Q1 Q2 / (8 c2 G3 π4), Equation 32.4 and εo = Q1 Q2 / (32 π5 G3), Equation 29.3 then E εo: :

Z π Q1 Q 2

E εo 

2

4

Q1 Q 2

3

8 c π G 32 π 5 G3

:Equation 41.1:

Simplifying: :

Z π (Q 1 Q 2 ) 2

E εo 

256 c 2 π 4 G 3 π 5 G 3

:Equation 41.2:

Cancel π from the numerator: :

Z (Q 1 Q 2 ) 2

E εo 

256 c 2 π 4 G 3 π 4 G 3

:Equation 41.3:

Alternatively using, E = 4 π / c3, Equation 32.7 and εo = Q1 Q2 / (32 π5 G3), Equation 29.3 then: :

E εo 

4π c

3

Q1 Q 2 32 π 5 G3

:Equation 41.4 :

Simplifying: :

Q1 Q 2

E εo 

8 π 4 c 3 G3

:Equation 41.5:

Equating, E εo = Z (Q1 Q2)2 / (256 c2 π4 G3 π4 G3), Equation 41.3 and E εo = Q1 Q2 / (8 π4 c3 G3), Equation 41.5, gives: : :

E εo 

Z (Q 1 Q 2 ) 2 2

4

3

4

256 c π G π G

3

Q1 Q 2



8 π 4 c 3 G3

:Equation 41.6:

Simplifying reduces the above to the familiar equation for Z (Equation 29.1): :

Z Q1 Q 2 4

32 π G

3



1 c

:Equation 41.7:

Rearrange for c: :

c

:

32 π 4 G 3 Z Q1 Q 2

Page 94 of 178

:Equation 41.8:

CHAPTER 42 Sow new concepts, reap the change. : Deriving an expression for E / εo: : Using, E = 4 π / c3, Equation 32.7 and εo = Q1 Q2 / (32 π5 G3), Equation 29.3, then E / εo is : :

(

E  εo

(

4π

) c3 Q1 Q 2 32 π 5 G 3

:Equation 42.1:

)

Simplifying: :

4 π 32 π 5 G 3 c 3 Q1 Q 2

E  εo

:Equation 42.2:

Collecting terms: :

128 π 6 G3

E  εo

c 3 Q1 Q 2

:Equation 42.3:

Alternatively using E = Z π Q1 Q2 / (8 c2 G3 π4), Equation 32.4 and εo = Q1 Q2 / (32 π5 G3), Equation 29.3, then E / εo is : :

E  εo

(

Z π Q1 Q 2 8 c 2 G3 π 4 Q1 Q 2 ( ) 32 π 5 G 3

) :Equation 42.4:

Simplifying: :

E  εo

Z π Q1 Q 2 32 π 5 G 3 8 c 2 G 3 π 4 Q1 Q 2

:Equation 42.5:

Simplifying: :

E  εo

Z π 32 π 8 c2

:Equation 42.6:

Cancel down and collect terms: :

E  εo

4 Z π2 c2

Now equating E / εo = 128 π6 G3 / (c3 Q1 Q2 ), Equation 42.3 and E / εo = 4 Z π2 / c2, Equation 42.7, gives: :

:

Page 95 of 178

:Equation 42.7:

E  εo

128 π 6 G3 c 3 Q1 Q 2



4 Z π2 c2

:Equation 42.8:

Rearranging to give Z: :

32 π 6 c 2 G 3 π 2 c 3 Q1 Q 2



Z

:Equation 42.9:

Simplifies to give the familiar expression for Z: :

Z

:

32 π 4 G 3 c Q1 Q 2

Page 96 of 178

:Equation 42.10:

CHAPTER 43 “Once all truth seekers wandered alone in the darkness, amid the countless, tortuous and looping maze of paths.” : Deriving an expression for E H: : Using, H = Q1 Q2 / (8 c2 π4 G3), Equation 31.3 and E = Z π Q1 Q2 / (8 c2 π4 G3), Equation 32.4 then E H is: :

EH 

Z π Q1 Q 2 2

4

Q1 Q 2

3

8 c π G 8 c 2 π 4 G3

:Equation 43.1:

Simplifying by collecting terms: :

EH

Z π (Q 1 Q 2 ) 2 (8 c 2 π 4 G 3 ) 2

:

:

Page 97 of 178

:Equation 43.2:

CHAPTER 44 “A chasm was created between the worlds, that is bridged with a coat of many colours.” : Deriving an expression for E / B: : Using, E = Z π Q1 Q2 / (8 c2 G3 π4), Equation 32.4 and B = Z π Q1 Q2 / (8 c3 π4 G3), Equation 33.7, E /B is then simply: : :

E  B

( (

Z π Q1 Q 2 8 c 2 π 4 G3 Z π Q1 Q 2 8 c 3 π 4 G3

) :Equation 44.1:

)

Cancel down to give: :

E  B

c

:

:

Page 98 of 178

:Equation 44.2:

CHAPTER 45 One can only say the universe is finite if one can stand outside it. : Deriving an expression to relate E, H and B: : Using, E = Z π Q1 Q2 / (8 c2 G3 π4), Equation 32.4 and H = Q1 Q2 / (8 c2 π4 G3), Equation 31.3: :

EH

Z π Q1 Q 2 2

4

Q1 Q 2

3

8 c π G 8 c 2 π 4 G3

:Equation 45.1:

Simplify: :

EH

Zπ(

Q1 Q 2

)2

:Equation 45.2:

EH=HBc

:Equation 45.3:

2

4

8c π G

3

As B = Z π Q1 Q2 / (8 c3 π4 G3), Equation 33.7 then: :

:

:

Page 99 of 178

CHAPTER 46 In search of inspiration one needs to try new paths. : Deriving an expression for E H / εo2: Using, εo = Q1 Q2 / (32 π ξ d3), Equation 27.6 and ξ d3 = π4 G3, Equation 26.5, εo2 is given by: :

ε o2 

(

Q1 Q 2 32 π ξ d

3

)2 

Q1 Q 2

(

32 π π 4 G 3

)2

:Equation 46.1:

Using εo2 = (Q1 Q2)2 / (32 π π4 G3)2 as above and E H = Z π (Q1 Q2)2 / (8 c2 π4 G3)2, Equation 45.2 then E H / εo2 : :

EH ε o2

Q1 Q 2

Zπ(

2

4

3

8c π G Q1 Q 2 ( )2 4 3 32 π π G



)2 :Equation 46.2:

Simplifying: :

EH ε o2



Zπ(

Q1 Q 2 8 c2 π4 G

)2 ( 3

32 π π 4 G3 2 ) Q1 Q 2

:Equation 46.3:

Cancel down: :

EH ε o2



Z π (32 π π 4 G 3 ) 2 (8 c 2 π 4 G 3 ) 2

:Equation 46.4:

Simplifying: :

EH ε o2

Z π (4 π 5 ) 2



(π 4 c 2 ) 2

:Equation 46.5:

Simplifying: :

EH ε o2



16 Z π 3 c4

:Equation 46.6:

Taking the square root of both sides:

EH ε o2



16 Z π 3 c4

:Equation 46.7:

Simplifying: :

1 εo

EH 

4 c2

Z π3

:

:

Page 100 of 178

:Equation 46.8:

CHAPTER 47 The three phase of someone else’s new concept, it’s wrong, it has its merits, it was obvious. : Deriving the expression for E / H = Z π: : Using, H = Q1 Q2 / (8 c2 π4 G3), Equation 31.3, and E = Z π Q1 Q2 / (8 c2 G3 π4), Equation 32.4: :

E  H

( (

Z π Q1 Q 2 8 c 2 π 4 G3 Q1 Q 2 8 c 2 π 4 G3

) :Equation 47.1:

)

Cancel down to give the familiar expression: :

E  H

:

Zπ

Page 101 of 178

:Equation 47.2:

CHAPTER 48 “Nature is relentless and unchangeable and it is indifferent as to whether its hidden reasons and actions are understandable to one or not.” Galileo Galilei (1564-1642) : Deriving an expression for E H / Z2: E H / Z2: Using, H = Q1 Q2 / (8 c2 π4 G3), Equation 31.3, E = Z π Q1 Q2 / (8 c2 G3 π4), Equation 32.4, E H = (Z π) [Q1 Q2 / (8 c2 G3 π4)]2, Equation 45.2, Z = 32 π4 G3 / (c Q1 Q2 ), Equation 29.1, then E H / Z2: :

EH Z

2



Zπ(

Q1 Q 2 2

4

8c π G

3

)2

1 Z2

:Equation 48.1:

Cancel out Z from the numerator: :

EH Z

2

Q Q π ( 2 1 4 2 3 )2 Z 8c π G



:Equation 48.2:

: As H = Q1 Q2 / (8 c2 π4 G3), Equation 31.3, then substituting H into the above gives: : :

EH Z

2



π Z

:Equation 48.3:

)2 Z 2

:Equation 48.4:

H2

Other relationships can be derived, for example, E H Z2 Using, E H = (Z π ) [Q1 Q2 / (8 c2 G3 π4)]2, Equation 45.2 and Z = 32 π4 G3 / (c Q1 Q2 ), Equation 29.1 then E H Z2: :

E H Z2 

Zπ(

Q1 Q 2 2

4

8c π G

3

As H = Q1 Q2 / (8 c2 π4 G3), Equation 31.3, substituting for H then canceling out Q1 Q2 / (8 c2 π4 G3) from the left hand side and right hand side leaves: :

E

Zπ

Q1 Q 2 8 c 2 π 4 G3

:Equation 48.5:

Therefore: :

E H Z2 

Z2

E2 Zπ

:Equation 48.6 :

E2 π

:Equation 48.7:

Cancel down Z term to give: :

E H Z2 

:

Z

Page 102 of 178

HZ

E π

:Equation 48.8:

E Hπ

:Equation 48.9:

Reduces to the familiar equation: :

Z

:

Page 103 of 178

CHAPTER 49 “To see the world in a grain of sand and heaven in a wild flower Hold infinity in the palm of your hand and eternity in an hour“ William Blake (1757-1827) : Redefining the quintessential element η: As η = (32 π / c) (d3 ξ), Equation 9.21and As ξ d3 = π4 G3, Equation 26.5, η can also be expressed as: :

32 π 4 3 π G c

:Equation 49.1:

E Q1 Q 2 H

:Equation 49.2:

E Q1 Q 2  H

32 π 4 3 π G c

:Equation 49.3:

E H

32 π π 4 G 3 c Q1 Q 2

:Equation 49.4:

32 π ξ d 3 c Q1 Q 2

:Equation 49.5:

η As η = Z π Q1 Q2, Equation 9.23 and E / H = Z π from Equation 47.2 then: :

η Equating Equations 49.1 and 49.2 gives: :

η



Rearrange for E / H: :



Or alternatively as, ξ d3 = π4 G3, Equation 26.5: :

E H



As Z = 32 d3 ξ / (c Q1 Q2 ), Equation 7.17, both Equation 49.3 and Equation 49.4 agree with Equation 48.9, Z = E / (π H) : Recall that the energy stored by the virtual capacitor is given by Ecap = η π f, Equation 24.3 and as η = E / (H Q1 Q2 ) = 32 π π4 G3 / c, Equation 49.3, then Ecap can be expressed as: :

E cap 

E Q1 Q 2 π f  H

32 π 6 G 3 f c

:Equation 49.6:

: η can be expressed in terms of the dielectric constant εo, using η = (32 π5 G3 / c) Equation 49.1 and εo = Q1 Q2 / (32 π5 G3) Equation 29.3 then

η

:

Q1 Q 2 c εo

Page 104 of 178

:Equation 49.7:

CHAPTER 50 “It is not the strongest of the species that survives, not the most intelligent that survives. It is the one that is the most adaptable to change” Charles Darwin : Redefining virtual capacitance C: : Starting with, C = c Q1 Q2 / (2 π f 32 π ξ d3), Equation 35.3: :

c Q1 Q 2

C

2 π f 32 π ξ d 3

:Equation 50.1:

As Z = 32 d3 ξ / (c Q1 Q2 ), Equation 7.17, the capacitance C can be written as: :

1 2πf Z

C

:Equation 50.2:

As Z = E / (π H), Equation 48.9 the above can be expressed as: :

C

Hπ 1 E 2πf

:Equation 50.3:

H 1 E 2f

:Equation 50.4:

Cancel down gives: :

C

As f = [ 1/ (2 π)] [ (G ξ/ r3 )1/4 ], Equation 25.7, substituting for f in the above gives: :

C G ξ 1/4 ( )  2 π r3

H 2E

:Equation 50.5:

2π H 2 E

:Equation 50.6:

πH E

:Equation 50.7:

Simplify: :

C(

Gξ r

3

)1/4 

Simplify: :

C(

Gξ r

3

)1/4 

As Z = E / (π H), Equation 48.9, the above can be written in terms of Z as: :

C(

Gξ r

3

)1/4 

1 Z

:Equation 50.8:

As f = ( 1/ (2 π)) ( (G ξ / r3 )1/4), Equation 25.7, the above can be written in terms of frequency, f, as: :

2π f 

(

Gξ r

3

)1/4

:Equation 50.9:

Using C = c Q1 Q2 / (2 π f 32 π ξ d3), Equation 35.3 and substituting in for 2 π f given above in Equation 50.9 gives: :

:

Page 105 of 178

c Q1 Q 2

C

Gξ

(

r3

1/4

)

:Equation 50.10:

32 π ξ d

3

as ξ d3 = π4 G3, Equation 26.5, the above can be expressed as: : :

c Q1 Q 2

C (

Gξ r3

1/4

:Equation 50.11: 4

32 π π G

)

3

Collect terms: :

c Q1 Q 2

C

13

(

G

ξ

r3

1/4

)

32 π

:Equation 50.12: 5

As εo = Q1 Q2 / (32 π ξ d3), Equation 27.6 substitute into Equation 50.10 to give the above expression in terms of εo: :

C

c εo Gξ ( 3 )1/4 r

:Equation 50.13:

Simplify: :

C

c εo (

r 3 1/4 ) Gξ

:Equation 50.14:

Equating the expressions for the capacitance C, i.e. Equation 50.13 and Equation 50.12: :

c Q1 Q 2

C (

G

13

r3

ξ

1/4

)



32 π

5

c εo G ξ 1/4 ( 3 ) r

:Equation 50.15:

Solve for εo: :

εo 

Q1 Q 2 32 π 5 G3

As per Equation 29.3. :

:

Page 106 of 178

:Equation 50.16:

PART VI Commences the expansion of the Unified Approach to other areas of fundamental physics. Sometimes you have to use your imagination to uncover the truth. :

CHAPTER 51 “In questions of science, the authority of a thousand is not worth the humble reasoning of a single individual.” Galileo Galilei (1564-1642) : Development of Boyle’s law: : Pressure, P, is proportional to the inverse of the volume for an ideal gas: :

1 Vol

:Equation 51.1:

Force Area

:Equation 51.2:

P As Pressure = Applied Force / Area. {Reference 1, page 423} (Note that sound is a function of pressure)

P The applied force is, (ξ )(2 d), and

the Area, A, is as before (which is a triangular cross section) A = d λ / π, Equation 16.14: :

2ξ d A

P

:Equation 51.3:

Volume enclosed, Vol, of the cone is the Area multiplied by π: : Vol = A π

:Equation 51.4:

2ξ d Aπ A

:Equation 51.5:

P Vol = (2 ξ d π)

:Equation 51.6:

Multiplying the expressions for P and enclosed Vol gives: :

P Vol  Cancel down: : As ξ d = π G , Equation 26.5 then: : 3

4

3

ξ dπ

π 5 G3 d2

:Equation 51.7:

Then substituting ξ d π = π5 G3 / d2, into P Vol = (ξ 2 d π): :

P Vol 

2 π 5 G3 d2

As Z = 32 π4 G3 / (c Q1 Q2 ), Equation 29.1 rearranging this equation: :

:

Page 107 of 178

:Equation 51.8:

π 4 G3 

Z c Q1 Q 2 32

:Equation 51.9:

Substitute Equation 51.9, π4 G3 = Z c Q1 Q2 / 32, into P Vol = 2 π5 G3 / d2, Equation 51.8, to give: :

P Vol 

2 π(

Z c Q1 Q 2 32 d2

)

:Equation 51.10:

Simplify: :

P Vol 

π Z c Q1 Q 2 16 d2

: : : : Reference 1: Physics for Scientists and Engineers with Modern Physics, Fourth Edition, Author Serway

:

:

Page 108 of 178

:Equation 51.11:

CHAPTER 52 “In science there is only physics, the rest is just stamp collecting” Lord Kelvin (1824-1907) : INTRODUCING TEMPERATURE Temp: P Vol = β F(Temp)

:Equation 52.1:

Where β is a constant and F(Temp) is some function of temperature Postulate that a function of temperature, F(Temp), is given by: :

Q1 Q 2

F(Temp ) 

4 d2

:Equation 52.2:

Postulate that: : β=πZc

:Equation 52.3:

As εo = 1 / (π Z c), Equation 27.4, then: :

πZc

1 εo

:Equation 52.4:

1  πZc

εo

:Equation 52.5:

Then using Equation 52.3 and Equation 52.4 to give: :

1  β Where εo = 8.8541876204 x 10-12

F(Temp )

P Vol 

εo

:Equation 52.6:

Substituting Equation 52.2, F(Temp) = Q1 Q2 / (16 d2) and β = π Z c, Equation 52.3, into P Vol = β F(Temp) Equation 52.1 gives: :

P Vol 

πZc(

Q1 Q 2

)

:Equation 52.7:

F(Temp) = Temp2

:Equation 52.8:

4 d2

Hypothesis that: :

Then: : 2 Temp 

Q1 Q 2 4 d2

:Equation 52.9:

Taking the square root of both sides: :

Temp

Q1 Q 2



4 d2

:Equation 52.10:

As Q1 Q2 can be positive or negative, if Q1 and Q2 have the same magnitude, Q, then by the Taylor series definition: :

Temp

:



Q1 Q 2 2d





Q 2d

Page 109 of 178

:Equation 52.11:

Temperature can be expressed in terms of Plank’s constant, As d3 = π2 c2 G3 / (4 h f 3), Equation 32.13, then d = ( π2 c2 G3 / (4 h f 3) )1/3. By substituting this expression for d into, Temp = ± Q / (2 d), Equation 52.11 gives:



Temp

(

Q1 Q 2 2

)(

4h f3 π 2 c 2 G3

)1/3

:Equation 52.12:

)1/3

:Equation 52.13:

Simplifies to:

Temp



 Q1 Q 2 f (

h 2

2

2π c G

3

: Expressing a relationship between the temperature, Temp and the impedance Z : As Temp = (1/(2 d))√( Q1 Q2), Equation 52.11 rearranging for d gives:

Q1 Q 2 

2 d Temp

:Equation 52.14:

Squaring both sides gives:

Q1 Q 2 

4 d2 Temp

2

:Equation 52.15:

Now substituting the above expression for (Q1 Q2) into the impedance, Z, where impedance Z, is given by, Z = 32 d3 ξ / (c Q1 Q2),

Equation 7.17 gives: :

Z

32 d3 ξ 4 c d2 Temp

2

:Equation 52.16:

Simplify

8 dξ

Z

c Temp

2

:Equation 52.17:

Collecting constants: :

cZ  8

dξ Temp

2

:Equation 52.18:

The above expression indicates that d ξ / Temp2 is a constant (value given in Chapter 57). (Also note that Temperature / frequency (Temp / f ), is almost a constant. 1.7312020947894(8)x10-26) : Verification that the hypothesis given in Equation 52.9, Temp2 = Q1 Q2 /(4 d2) is consistent with the model: : As Temp = ± Q / (2 d), Equation 52.11 rearranging for d gives: : :

d

:



Q 2 Temp

Page 110 of 178

:Equation 52.19:

Substituting for d into Equation 8.3, given as, d3 r 3 f 4 = (G4 / 24): :

(

G4

Q 3 3 4 ) r f  2 Temp

24

:Equation 52.20:

Simplifies to: :

Temp 

 2(

3

Q3 G

4

)r3 f 4

:Equation 52.21:

: Or the above can be expressed as: :

Temp 

 Q r (2 (

f4 G

4

) )1/3

:Equation 52.22:

: The above equations give the expected answers. It is interesting to note that when the wavelength λ is: λ = 5.8616480922038x10-7 m (and frequency f = 5.11447383541729x1014 s-1) then the temperature numerically equals the permittivity of free space (ε0 = 1 / (c Z π)) indicating a matched system. : Expressing Temperature in Terms of η, the Unification Constant or the Fifth Field: : Recall that the Unification Constant is give by η = Z π Q 1 Q2 given by Equation 9.23 which can be rearranged as: :

Q1 Q 2 

η Zπ

:Equation 52.23:

: The temperature squared, Temp2, is given by, Temp2 = Q1 Q2 /(4 d2), Equation 52.9 substituting for Q1 Q2 using the above gives: : 2 Temp 

η 1 Z π 4 d2

:Equation 52.24:

gives: : 2 Temp 

η 4 π Z d2

:Equation 52.25:

Or

η

:

2 4 π Z d 2 Temp

Page 111 of 178

:Equation 52.26:

The Relationship Between Temperature and Angular Velocity: : As the angular velocity (v) is given by v = 2 π f r, Equation 11.11, and the temperature (Temp) is given by Temp = (1/(2 d))√( Q1 Q2), Equation 52.11, then the ratio of the temperature to the angular velocity is given by: :

Temp v

(Q1 Q 2 )



2d

1 2πfr

:Equation 52.27:

Simplifies to: :

Temp

(Q 1 Q 2 )



v

4 πd r f

:Equation 52.28:

Where ±Q is the square root of the product of Q1 Q2

Temp v





Q 4 πdr f

:Equation 52.29:

: The Relationship between Photon Energy and the Temperature : The photon energy is Plank’s constant, h, given by h = 4 π2 c2 f r3 / G, Equation10.12 or h = 8 π3 c G3 ξ / (η f 3 ) Equation 10.20 multiplied by the frequency f and the temperature is given by Equation 52.11, Temp = (1/(2 d))√( Q1 Q2), the ratio of the photon energy to temperature is given by: :

E photon Temp

8 π 3 c G3 ξ



ηf

2

2d

:Equation 52.30:

Q1 Q 2

Collecting terms and simplifying: :

E photon Temp



16 π 3 c G 3 d ξ η Q1 Q 2

f2

:Equation 52.31:

As the above expression is a constant then ξ d / f 2 must be a constant term, which it is, the value is given in Chapter 57 Alternatively using the expression h = 4 π2 c2 f r3 / G, Equation10.12 to provide Plank’s constant then: :

E photon Temp



4 π2 c2 f 2 r3 G

2d

:Equation 52.32:

Q1 Q 2

Collecting terms and simplifying: :

E photon Temp



8 π2 c2

f 2 r 3d

G Q1 Q 2

Meaning that f 2 r3 d is a constant, the value of which is given in Chapter 57.

:

Page 112 of 178

:Equation 52.33:

CHAPTER 53 To understand the universe is the key to humanity’s destiny. : Development of Faraday’s concept: Introducing Faraday’s concept of positive and negative charges, whereby like charges repel and opposite charges attract and that there are two types of electricity, static and fluid electricity. Consider the case where Q 1 and Q2 are oppositely charged. Then the impedance given by, Z = 32 π4 G3 / (c Q1 Q2 ) from Equation 29.1, becomes a negative number, so allowing the flow of current. : If Q1 = +Q and Q2 = -Q then Z < 0

:Equation 53.1:

: If the charges Q1 and Q2 are similarly charged i.e. both positively or negatively charged. Then the impedance is a positive number, restricting the flow of current. : If Q1 = -Q and Q2 = -Q then Z > 0

:Equation 53.2:

If Q1 = Q and Q2 = Q then Z > 0

:Equation 53.3:

:

:

:

Page 113 of 178

CHAPTER 54 As each divided kingdom unites, the whole story may be told. : Introducing density, ρ: : Density, ρ, is given by the mass (m p ) divided by the volume enclosed, Vol {Reference 1, page 9}

ρ

mp Vol

:Equation 54.1:

Note that, in this system the density essentially resides on the surface, which encloses a volume (like a balloon), thus the density of the enclosed volume is the same as the density at the surface, which encloses the volume. Vol = (π) Area

:Equation 54.2:

Area is the Area, A, of the virtual capacitor and given by, A = C 2 d Z π c = C 2 d / εo, Equation 16.3, then the volume enclosed, Vol is given by substituting in for the Area into the above giving: : Vol = (π) (C 2 d Z π c)

:Equation 54.3:

Vol = 2 π c Z C d

:Equation 54.4:

2

: As εo = 1/ (π Z c), Equation 27.4, then the Vol can be written in terms of εo: :

Vol 

2 πC d εo

:Equation 54.5:

Substituting an expression for the capacitance C into the above, where, C = c εo / (G ξ / r3 )1/4, Equation 50.12, gives: :

Vol 

2 π d c εo ε o G ξ 1/4 ( 3 ) r

:Equation 54.6:

Simplifying and cancel down terms gives: :

Vol 

2 πc d(

r 3 1/4 ) Gξ

:Equation 54.7:

Now consider a term for the mass, in order to calculate the density. Using mp= r a / ξ, Equation 11.17 or mp = G / a, Equation 12.5 or, mp = G / (4 π2 f 2 r), Equation 12.6 Substitute for frequency, f, into mp where f = [ 1/ (2 π) ] (G ξ / r3 )1/4, Equation 25.7 gives: :

mp 

G 4 π2 r

( 2 π(

r 3 1/4 2 ) ) Gξ

Simplify: :

:

Page 114 of 178

:Equation 54.8:

G

mp 

4 π2 (

4 π2 r

r 3 2/4 ) Gξ

:Equation 54.9:

Simplifying and cancel down terms gives: :

G r 3 2/4 ( ) r Gξ

mp 

G

mp 

G

1/2

(

r3 2

r ξ

)1/2

:Equation 54.10:

:Equation 54.11:

Simplify: :

G1/2 r 1/2

mp 

ξ 1/2

:Equation 54.12:

To calculate the density, ρ = mp / Vol, Equation 54.1, use the terms for mp given in Equation 54.12 and for expression for the volume enclosed, Vol , given in Equation 54.7, as below: :

( ρ

G1/2 r 1/2

)

ξ 1/2

r 3 1/4 2 πc d( ) Gξ

:Equation 54.13:

Simplify: :

ρ

G1/2 r 1/2 G1/4 ξ 1/4 ξ 1/2

2 π c d r 3/4

:Equation 54.14:

Simplify: :

G 3/4 r 1/2

ρ

2 π c d r 3/4 ξ 1/4

:Equation 54.15:

Simplify by canceling down: :

G 3/4

ρ

2 π c d r 1/4 ξ 1/4

:Equation 54.16:

The above can be expressed as: :

1 G 3 1/4 ( ) 2 πc d r ξ

ρ

:Equation 54.17:

The above can be equivalently expressed as: :

ρ

1 2 πc r

1/4

1/4

d

(

G3 ξd

3

)1/4

:Equation 54.18:

Recall that ξ d3 = π4 G3, Equation 26.5, then rearranging this gives: :

π4 

ξ d3 G3

So: :

:

Page 115 of 178

:Equation 54.19:

π

(

ξ d3 G

3

)1/4

:Equation 54.20:

Now substituting Equation 54.20, π = (ξ d3 / G3)1/4, into Equation 54.18, ρ = (1 / (2 π c r1/4 d1/4) ) (G3 / (ξ d3) )1/4, gives: :

ρ

1 2

2 π c (d r)1/4

:Equation 54.21:

So, ρ d1/4 r1/4 = 1 / (2 π2 c) is a constant:

Alternative Equivalent Expressions for Density, ρ : Equation 13.17 gives the expression, G = a2 r / ξ, substituting this into Equation 54.17, ρ = (1 / (2 π c d) ) (G3 / (r ξ) )1/4 gives: :

ρ

1 G 2 a 2 r 1/4 ( ) 2 πc d r ξ ξ

:Equation 54.22:

1 G2 a 2 ( 2 )1/4 2 πc d ξ

:Equation 54.23:

1 G a 1/2 ( ) 2 πc d ξ

:Equation 54.24:

The above simplifies to: :

ρ Which becomes: :

ρ :

Alternatively a substitution for G2 can be made into Equation 54.17, where G2 = a4 r2 / ξ2 : :

ρ

1 G a 4 r 2 1/4 ( ) 2 πc d r ξ ξ2

:Equation 54.25:

1 G a4 r ( 3 )1/4 2 πc d ξ

:Equation 54.26:

The above simplifies to: :

ρ :

Alternatively G3 can be substituted into Equation 54.17, where G3 = a6 r3 / ξ3 : :

ρ

1 1 a 6 r 3 1/4 ( ) 2 πc d r ξ ξ3

:Equation 54.27:

1 a6 r 2 ( 4 )1/4 2 πc d ξ

:Equation 54.28:

Which becomes: :

ρ :

Reference 1: Physics for Scientists and Engineers with Modern Physics, Fourth Edition, Author Serway

:

Page 116 of 178

CHAPTER 55 “Let the future tell the truth and evaluate each one according to one’s own work and accomplishments. The present is theirs’; the future, for which I have really worked, is mine.” Nikola Tesla : Introducing Electric Dipole Moment and establishing a physical understanding of the constant η: : The electric dipole moment, DE, is typically given by the potential difference, Volts , multiplied by the separation of the dipole, 2d, so: DE = (Volts) (2 d)

:Equation 55.1:

As potential difference (Volts) is given by the maximum current, Imax, multiplied by the resistance, R, so: (Volts) = Imax R

:Equation 55.2:

As before, use the hypothesis that the relationship for the maximum current Imax2 can be expressed by, Imax2 = π2 f2 (Q1 Q2), Equation 20.3, then: Imax = π f √(Q1 Q2)

:Equation 55.3:

At resonance the resistance, R, is given by R = Z from Equation 18.9, where Z is given by, Z = 32 d3 ξ / (c Q1 Q2),

Equation 7.17, so using,

(Volts) = Imax R, Equation 55.2, gives the expression: :

Volts 

(Q1 Q 2 ) π f

32 d 3 ξ c Q1 Q 2

:Equation 55.4:

Simplifies to: :

Volts 

32 π f d 3 ξ c (Q1 Q 2 )

:Equation 55.5:

: Note that the relationship between the potential difference, Volts, and η, where, η = (32 π d3 ξ) / c, Equation 9.21, is: : Volts = η f / √(Q1 Q2)

:Equation 55.6:

As η = (32 π / c) (π4 G3) , Equation 49.1 then Volts can also be expressed as: :

Volts 

32 π 5 G 3 f c (Q1 Q 2 )

:Equation 55.7:

And as λ = c / f, Equation 2.8, the above expression can be written as: :

Volts 

32 π 5 G 3 λ ( Q1 Q 2 )

:Equation 55.8:

Using DE = (Volts) (2 d), Equation 55.1 gives the expression for the electric dipole moment as: :

DE 

:

32 π f d 3 ξ c (Q1 Q 2 )

2d

64 π f d 3 ξ d c (Q1 Q 2 )



64 π f d 4 ξ c (Q1 Q 2 )

Page 117 of 178

:Equation 55.9:

Simplifies to: :

64 π d 4 ξ

DE 

λ ( Q1 Q 2 )

:Equation 55.10:

As π4 G3 = ξ d3, Equation 8.11, the above equation becomes: :

64 π 5 G 3 d

DE 

λ ( Q1 Q 2 )

:Equation 55.11:

: Other equivalent expressions for the electric dipole moment in terms of the magnetic field strength, H, and the electric field strength, E, can be derived. : By rearranging, H = Q1 Q2 / (8 c2 ξ d³), Equation 31.2, for d3 ξ provides: :

ξ d3 

Q1 Q 2 8 c2 H

:Equation 55.12:

Substitute the above expression for d3 ξ into Equation 55.9, to give the electric dipole moment in terms of H: :

DE 

64 π f Q1 Q 2 d 2

8 c H c ( Q1 Q 2 )



8 π f d ( Q1 Q 2 ) H c3

:Equation 55.13:

: To express the electric dipole moment in terms of E, use, E = 4 π / c3, Equation 32.7, substitute this into the expression for DE where, DE = 64 π f d4 ξ / (c√(Q1 Q2)), Equation 55.9, to give: :

16 c 2 f d 4 ξ

DE 

E

(Q1 Q 2 )

:Equation 55.14:

Or in terms of potential difference, Volts, where Volts = 32 π f d3 ξ / (c √(Q1 Q2)), Equation 55.5, : :

DE 

E Volts c 3 d 2π

:Equation 55.15:

Introducing the Magnetic Dipole Moment and its Relationship with Current: Recall that the electric dipole moment is given by DE = 16 c2 f d4 ξ E / (√(Q1 Q2), Equation 55.14, then the magnetic dipole moment, DH, by analogy will be given by: :

DH 

16 c 2 f d 4 ξ

H

:Equation 55.16:

DE = Z π DH

:Equation 55.17:

(Q 1 Q 2 )

where: :

So as DE = (Volts) (2 d), Equation 55.1 then: :

DH 

:

2 d Volts Zπ

Page 118 of 178

:Equation 55.18:

Recall that from Equation 23.2, Rad = d / π, where Rad is the radius in wrapped space and as Volts / Resistance = Current. In this case the resistance is equal to the impedance of free space so R = Z I = Volts / Z

:Equation 55.19:

so rearranging DH = 2 d Volts / Z π, Equation 55.18, to calculate the current: :

I

DH

π 2d

:Equation 55.20:

2 I d = π DH. Therefore the current (I) is equal to, the magnetic dipole moment (DH) divided by twice the radius (the diameter, 2d) in wrapped space. The shortest distance between two points is a worm-hole, starts to have some physical meaning. : Defining the relationship between the potential difference (Volts) and capacitance (C): : Using the expression for capacitance given by, C = 1 / (2 π2 Z f), Equation 16.11, and Substituting in for Z where, Z = 32 d3 ξ / (c Q1 Q2), Equation 7.17, to give: :

C As Volts = 32 π f d3 ξ / (c √(Q1 Q2)),

1

c Q1 Q 2

2

2 π f 32 d 3 ξ

:Equation 55.21:

Equation 55.5, then Equation 55.21 above, can be expressed as: :

C

1 2π

(Q1 Q 2 ) Volts

:Equation 55.22:

Rearrange for Volts: :

Volts 

(Q 1 Q 2 ) 2 πC

:Equation 55.23:

: It is worth noting that the energy stored in the virtual capacitor, Ecap, can be expressed in term of η: : As Ecap = π2 Z Q1 Q2, Equation 24.2, where Z = η / (π Q1 Q2) Equation 9.23 then: : Ecap = π η f

:Equation 55.24:

Pwr = π η f 2

:Equation 55.25:

Pwr = I Volts

:Equation 55.26:

And the associated Power, Pwr = Ecap f, as: :

Or Power is given by current multiplied by voltage i.e.: :

Or alternatively using, Volts = η f / √(Q1 Q2), Equation 55.6 and I = DH π / (2 d) Equation 55.20, where DH is given by 55.16, DH = 16 c2 f d4 ξ H /√(Q1 Q2)

:

Page 119 of 178

yield the same result i.e.: :

I Volts 

16 c 2 f d 4 ξ

H

(Q1 Q 2 )

π 2d

ηf (Q1 Q 2 )

:Equation 55.27:

Simplifying gives: :

I Volts 

8 c 2 d3 ξ H (Q1 Q 2 )

πηf2

:Equation 55.28:

As H = Q1 Q2 / (8 c2 ξ d3), from Equation 31.2, then Equation 55.28 reduces to: : I Volts = π η f 2

:Equation 55.29:

As per Equation 55.25. Or alternatively the power, Pwr = π η f2, Equation 55.25 and Pwr = I Volts, Equation 55.26, can be equally expressed in terms of electric and magnetic dipole moments as: :

Pwr 

π DH DE 2d 2d

:Equation 55.30:

Which simplifies to: :

Pwr 

π DE DH 4 d2

:Equation 55.31:

The power can be expressed in terms of dielectric constant: Using Pwr = π η f 2, Equation 55.25 and η = Q1 Q2 / (c εo ), Equation 49.7, then: :

Pwr 

π Q1 Q 2 f 2 c εo

:Equation 55.32:

It is worth noting the relationship between volts and energy stored in a capacitor is easily derived as, Volts = η f / √(Q1 Q2), Equation 55.6 and Ecap = π η f, Equation 55.24 so: : Ecap = π √(Q1 Q2) Volts

:Equation 55.33:

Expressing Volts, Current, Power and Magnetic Dipole Moment in terms of Temperature: Recall the expression for temperature is given in Equation 52.14, 2 d Temp = √(Q1 Q2), and that the relationship between the potential difference, Volts, and η, is given by Volts = η f / √(Q1 Q2), Equation 55.6 and that η can be expressed as η = Z π Q1 Q2, Equation 9.23, then Volts can be expressed as: :

Volts 

Z π Q1 Q 2 f

:Equation 55.34:

Q1 Q 2

Which becomes: : Volts = Z π f √(Q1 Q2) :

:

Page 120 of 178

:Equation 55.35:

Recall the expression for temperature is given by, 2 d Temp = √(Q1 Q2), in Equation 52.14 substituting in for the expression √(Q1 Q2) gives: : Volts = 2 π Z f d Temp

:Equation 55.36:

The current, I, in terms of temperature is given by the relationship I = V / R, Equation 55.15, where in this case the resistance is equal to the impedance of free space Z. So: : I = 2 π f d Temp

:Equation 55.37:

The power, Pwr = I Volts, Equation 55.26, in terms of temperature is given by the relationship: : Pwr = 4 π2 Z f2 d2 Temp2

:Equation 55.38:

: Establishing the relationship between the magnetic dipole moment, DH, and temperature, Temp. First recall that the magnetic dipole moment is defined as DH = 2 d Volts / (Z π) Equation 55.18 and Volts is given by, Volts = 2 π Z f d Temp, Equation 55.36, substituting the latter equation into the former equation gives: :

DH 

2 d 2 π Z f d Temp Zπ

:Equation 55.39:

Which simplifies to: : DH = 4 f d2 Temp

:

Page 121 of 178

:Equation 55.40:

Chapter 56 “The knower of the mystery of sound, knows the mystery of the whole universe.” Hazrat Inayat Khan (1882-1927),

: Sound : I introduce the concept of “sound potential”, this is an inherent quality in a system, like a tuning fork that has the potential to make a sound with a particular note. Hypothesis that ‘sound potential’, O, is equal to the pressure, P, multiplied by the charge times the rate of change of charge, i.e.: :

O

P Q1 Q 2 (

δ ) δQ

:Equation 56.1:

: (O, the sound potential can be thought of as a note.). The pressure (P) of the virtual capacitor is given by, P = 2 ξ d / A, Equation 51.3, and where Area (A) is, A = d λ / π, Equation 16.14, so P can be stated as: :

P

2 πξ d dλ

:Equation 56.2:

2 πξ f c

:Equation 56.3:

As λ = c / f, Equation 2.8, the above expression becomes: :

P :

Consider that the rate of change of charge, δ/δQ, is given by: :

δ  δQ

Q1..Q n n

Q

:Equation 56.4:

1

Consider the solution that, Q1 and Q2 have the same magnitude (Q) and sign, then the positive rate of change of charge is: :

δ  δQ

Q1 Q 2  Q1  Q 2

Q 2

:Equation 56.5:

i.e. where Q is 1.60217662089849 x10-19. (Note that if Q1 and Q2 have different signs and the same magnitude then δ/δQ tends to infinity.) Then ‘sound potential’, O, as given by Equation 56.1, O = P √(Q1 Q2 ) (δ/δQ) can be written as: :

O

2 π f ξ Q2 c 2

Where Q = √(Q1 Q2 ), the above simplifies to : :

:

Page 122 of 178

:Equation 56.6:

π f ξ Q2 c

O

:Equation 56.7:

Dimensions of ‘sound potential’, O, are [Q2 L T -4]: : As pressure of the virtual capacitor is given by, P = 2 π f ξ / c, Equation 56.3, Now consider Pd is the pressure, which occurs along the distance, d / π, (d as derived in Chapter 7), then: :

d π

:Equation 56.8:

2f ξd c

:Equation 56.9:

Pd 

P

So:

2πf ξ d  c π

Pd 

The intensity (Intensity) of a wave is given by pressure over area, {Intensity Reference 1, page 483}: ::

Pd A

:Equation 56.10:

2πf ξ cλ

:Equation 56.11:

Intensity  where Area (A) is, A = d λ / π, Equation 16.14,

Intensity 

2f ξ d π  c λd

Substitute λ = c / f, Equation 2.8, into the above and simplify gives: :

Intensity 

2πf2 ξ c2

:Equation 56.12:

Intensity (Intensity) has dimensions [T -4]: Then ‘sound potential’, O, multiplied by intensity, Intensity, is a ‘loudness factor or noise or actual sound’ given by O Intensity, which is: :

O Intensity 

π f ξ Q2 2 π f 2 ξ  c c2

2 π 2 f 3 ξ 2 Q2 c3

:Equation 56.13:

: Dimensions of ‘Loudness factor’, O Intensity, are [Q2 L T -8]: The human ear can detect displacements over the range between approximately, 1 x10-5m to 1 x10-10m. {Reference 1, page 484} : ‘Loudness Factor’ in terms of Plank’s constant, h: : The loudness factor (O Intensity ) can be expressed in terms of h, As O Intensity = 2 π2 f3 ξ2 Q2 / c3, Equation 56.13, and where h = c2 ξ / (4 π2 f3), Equation 25.11:

:

Page 123 of 178

then O Intensity in terms of Plank’s constant (h) is given by: :

2 π 2 f 3 ξ 2 Q2



Q2 ξ 3 2ch

:Equation 56.14:

Loudness Factor  O Intensity 

Q2 ξ 3 2ch

:Equation 56.15:

Q2 ξ 2 ξ 2 ch

:Equation 56.16:

O Intensity 

c3

:

:The above can be expressed as: :

O Intensity 

Q2 ξ 3  2ch

: Recall that c h / ξ d3 = 6888.1801244747, Equation 32.21, rearranging this equation and letting the constant 6888.1801244747 be denoted by the letter J for ease, gives: :

ξ  ch

1 J d3

:Equation 56.17:

Substituting ξ /c h = 1/ (J d3 ) into Equation 56.16 gives: :

O Intensity 

Q2 ξ 2 1 2 J d3

:Equation 56.18:

O Intensity 

Q ξ3 1 2 J d3 ξ

:Equation 56.19:

Q2 ξ 3 1 4 2 J π G3

:Equation 56.20:

Multiply top and bottom by ξ: :

: Recall that ξ d3 = π4 G3, Equation 26.5, then: :

O Intensity  :

Equating Equation 56.20 above with O Intensity = 2 π2 Q2 f 3 ξ2 / c3, Equation 56.13, gives: :

2 π 2 f 3 ξ 2 Q2 c3

Q2 ξ 3 1 2 J π 4 G3



:Equation 56.21:

: Simplifying and substituting in ξ d3 = π4 G3, Equation 26.5: :

2 π2 f 3 c

3



ξ 1 2 J ξ d3

:Equation 56.22:

ξ ξ d3 2 c h ξ d3

:Equation 56.23:

: Substituting in, c h / ξ d3 = J, Equation 32.23, gives: :

2 π2 f 3 c3



Simplifying and collecting terms gives: :

:

Page 124 of 178

f3  ξ

c2 4 π2 h

:Equation 56.24:

Rearranging for the restoring force factor gives: :

ξ

4 π2 h f 3 c2

:Equation 56.25:

f 3 / ξ is indeed a constant, with the value given in Chapter 57. : To have extracted the sword, embedded in stone, to unite the kingdoms. :

Expressing Loudness factor in terms of η: : Recall that, h = 8 π3 c G3 ξ / (η f 3 ), given by Equation 10.20, substituting this into the equation for Loudness factor given in Equation 56.15, as O Intensity = Q2 ξ3 / (2 c h), results in: :

O Intensity 

Q2 ξ 3 η f3 2 c 8 π 3 c G3 ξ

:Equation 56.26:

The above simplifies to: :

O Intensity 

Q2 η f 3 ξ 2 16 π 3 c 2 G 3

:

:

Page 125 of 178

:Equation 56.27:

Hidden variable : c2 / (4 h f 3) : Recall that d3 = π2 c2 G3 / (4 h f3), by rearranging Equation 32.12, and ξ = 4 π2 h f3 / c2, Equation 56.25, by multiplying ξ with d3 gives the familiar: ξ d3 = π4 G3, Equation 26.5. Note that the variable term: c2 / (4 h f 3) cancels out. : : Reference 1: Physics for Scientists and Engineers with Modern Physics, Fourth Edition, Author Serway

:

Page 126 of 178

CHAPTER 57 “A hypothesis, or theory, is clear decisive and positive, but it is believed by no one but the one who created it. Experimental findings, on the other hand, are messy, inexact things, which are believed by everyone except the one who did that work.” Harlow Shapley (1885-1972) : This chapter provides the values and definitions of variables and constants derived in this book. The accuracy of the values in this chapter are limited by the accuracy of π that has been used in the model. The value of c and Z are taken as the predefined accepted values. My model was used to fine tune G, Q and h to determine values that provided exact solutions throughout. ::

SI units: : Amperes: A (m2C /s), Coulombs: C, Joules: J, (= Nm = kgm2s -2), Kilograms: kg, Meters: m, Newton’s: N, (= kg m /s2), Newton metre (Nm) is Joules), Seconds : s, Tesla, T, or [ Ns/(Cm) ], or Nm / A, or (kg / s2A), or [ kg / (s m2 C) ] :

:

Page 127 of 178

Terminology : c: speed of light in free space, π:pi, λ : wavelength (m), f : frequency (s-1) ω: angular frequency given by 2.π.f, εo: permittivity of free space ( C2 / (N m2 ) ), μo : permeability of free space (T m / A):0.00000125663706143592, ke : Coulomb’s constant (N m 2 / C2 ) ke = c Z /4:8987551787.36818, Ms : Effective Source Mass (kg) (=s2 m-3), mp : Effective Photon Mass (m 3s-2kg-1) (=m3 s-2 s -2 m3 = m6 s -4), Q1, Q2 : Charges (C), S, 2d : Separation of Charges Q1, Q2 (m), r : separation of Ms and mp (m), ξ : restoring force factor, C : capacitance (Farads), P : Pressure (N / m2), ρ : Density (kg / m3), H : Magnetic field strength (A / m or m C / s), E : Electric field strength (N / C), B : the magnetic field (Tesla or N s / (C m) ) Volts : volts across the virtual capacitor I : current corresponding to volts across the virtual capacitor :

Fixed known parameters : π : pi:

:3.14159265358979,

: c : The speed of light (m /s) in a vacuum is given by {Reference 2}

:299792458 [LT -2]

: Z π : Impedance of free space (ohms) (Reference 2)

:119.9169832 π,

:

Ho : The magnetic field strength of free space

: ε o / c2

: : Reference 2: The NIST Reference on Constants Units and Uncertainty (www physics.nist.gov/constants)

:

Page 128 of 178

Parameters verified by the model : Z = (32) π4 G3 / (c Q1 Q2), Equation 29.1, is the same as the generally accepted value

:119.9169832

: εo permittivity of free space (the electric constant) [ C2 /( m2 N ) ], εo = Q1 Q2 / (32 π5 G3) Equation 29.3, or equivalently εo =1 / (c.Z.π)

:8.85418781762039x10 -12

:Equation 28.2:

{Reference 2 recommends the value for εo as 8.854187817 x10-12 Fm-1 } : μo = (32 π5 G3) / (c2 Q1 Q2 ), Equation 30.6 or equivalently, μo = (32 π ξ d3) / (c2 Q1 Q2 ),

:0.00000125663706143592 [L3 T -2 Q -2]

:Equation 30.4:

{Reference 2 recommends the value for the magnetic constant as 4 π x10-7} : :

Parameters honed by the model : (Q1 Q2): Charge (C2) (J2)

:  2.5669699245537 x10-38 : [Q2]

when Q1 = Q2 (or Q) : Charge (C) (J)

:  1.60217662089849 x10-19 : [Q]

{Reference 2 recommends the value for Q as: 1.602176208(98) x10-19 } : G : Universal Gravitational constant (N m 2 / kg2)

:6.66485802996474 x10-11

{Reference 2 recommends the value for G as: 6.67408(31) x10-11} : h : Plank’s constant (J.s)

:6.6260695729 x10-34 : [L4 T -3]

{Reference 2 recommends the value for h as: 6.62607040(81) x10-34 } : Reference 2: The NIST Reference on Constants Units and Uncertainty (www physics.nist.gov/constants)

:

:

Page 129 of 178

Some of the constants established in the model : Dimensions for the values are provided. A dimensional analysis is presented in Appendix 5 and Appendix 6. Note that the dimension Q has been introduced to represent Coulombs for ease. : f / Ms = c2 / h :

:1.35639260778764 x1050 [T L -2]

f r3 = G h / (4 π2 c2)

:Reference Equation 4.3:

:1.24464572522555 x10-62 [L3 T -1]

3

:Reference Equation 5.4:

-62

U = f r :Universal Constant

:1.24464572522555 x10

3

[L T -1]

:Reference Equation 5.6:

: d3 ξ

:2.8838468916928(7) x10-29 [L5 T -4]

:Reference Chapter 7:

: fd

:4627352.9408284(6) [L T -1]

:Reference Chapter 7:

: π4 G3 = ξ d3

:2.8838468916928(8) x10-29 [L5 T -4]

:Reference Equation 8.11 and Reference Equation 26.5:

: S = d 3 r3 f 4 = (Q1 Q2) G c Z / [ 512 π4 ]

:1.23322871544011 x10-42 [L6 T -4]

:Reference Equation 10.1:

: d3 r3 f 4 = (G / 2)4

:1.23322871544011 x10-42 [L6 T -4]

:Reference Equation 10.16:

: ξ / f3

:2.91054502786985 x10-49 [L2 T -1]

:Reference Equation 10.20, Equation 25.11 and Equation 34.4:

: :600289429.3387830 [L T -1]

Vzip : Velocity of capacitive area

:Reference Equation 17.8:

: Rcritical :Critical resistance, Rcritical= √ (4L / C)

:753.460626923541 [L4 T -3 Q -2]

:Reference Equation 19.1:

: r3 / (ξ Ŧ4) = G /(16 c4)

:5.15689285863993 x10-46 [L -3 T -4]

:Reference Equation 21.2:

: r3 f 4 / ξ = G / (16 π4)

:4.27633214160055 x10-14 [L]

r f / ξ = 2 G / (c Z Q1 Q2) 3 4

4

-14

:4.27633214160055 x10

:Reference Equation 21.6:

[L]

:Reference Equation 26.12:

: 8 Z π4 c2 f r3 / G

:1.56843464533602 x10-30 [L8 T -6 Q -2]

:Reference Equation 24.9:

: c = 1 / (π Z εo)

:299792458 [L T -1]

:Reference Equation 28.2:

: c/Z

:2500000 [Q2 T2 L -3]

:Reference Equation 28.5:

: Q1 Q2 = 32 π5 G3 εo

:

:2.5669699245537 x10-38 [Q 2]

:Reference Equation 30.17:

Page 130 of 178

: Q1 Q2 / G3 = 32 π5 εo

:8.67057844792915 x10-8 [Q 2 L -3]

:Reference Equation 30.18:

:2.5669699245537 x10-38 [Q 2]

:Reference Equation 30.21:

: Q1 Q2 = 32 π5 G3 / (c2 μo) : Q1 Q2 / G3 = 32 π5 / (c2 μo)

8.67057844792915 x10-8 [Q 2 L -3]

:Reference Equation 30.22:

: H = Q1 Q2 / (8 c2 π4 G3)

:1.23799014723612 x10-27 [Q2 T6 L -7]

:Reference Equation 31.3:

: E/H=Zπ

:376.73031346177 [L4 T -3 Q -2]

:Reference Equation 32.1:

: E = Z π Q1 Q2 / (8 c2 π4 G3) : f 3 d3 = π2 c2 G3 / (4 h) :

:4.66388416230847 x10-25 [T3 L -3] :9.90827100793383 x1019 [L3 T -3]

c h / (π4 G3) = c h / (ξ d3)

:Reference Equation 32.4:

:Reference Equation 32.12:

:6888.1801244747 [Dimensionless]

:Reference Equation 32.20 and 32.21:

: B = 4 π / (c4)

:1.55570430070941 x10-33 [T4 L -4]

:Reference Equation 33.5:

: 1/ (Z π Q1 Q2) = c / (32 π π4 G3)

:1.03406693785069 x1035 [T3 L -4]

:Reference Equation 34.1:

: H Z = 4 / c3

:1.48456043687879 x10-25 [T3 L -3]

:Reference Equation 38.3:

: H / Z = (Q1 Q2)2 / (256 c π8 G6)

:1.03237265831761 x10-29 [Q4 T9 L -11]

:Reference Equation 38.5:

: E Z = 128 π5 G3 / (c4 Q1 Q2 )

:5.59278918738291 x10-23 [L Q -2]

:Reference Equation 39.7:

: E / Z = π Q1 Q2 / (8 c2 G3 π 4)

:3.88926075177353 x10-27 [Q2 T10 L -12]

:Reference Equation 39.10:

: H εo = (Q1 Q2)2 / (256 c2 π9 G6)

:1.09613972799921 x10-38 [Q4 T10 L -12]

:Reference Equation 40.3:

: H / ε o = 4 π / c2

:1.39819729684573 x10-16 [T2 L -2]

:Reference Equation 40.4:

: E εo = Q1 Q2 / (8 π4 c3 G3)

:4.12949063327043 x10-36 [Q2 T7 L -8]

:Reference Equation 41.6:

: E / εo = 128 π6 G3 / ( c3 Q1 Q2 )

:5.2674330592209 x10-14 [L2 T -1 Q -2]

:Reference Equation 42.3:

: E H = Z π (Q1 Q2)2 / (8 c2 G3 π 4)2

:

:5.77384264078846 x10-52 [Q2 T9 L -10]

Page 131 of 178

:Reference Equation 43.2:

: E H / εo2 = 16 Z π3 / (c4)

:7.36491066471848 x10-30 [T Q -2]

:Reference Equation 46.6:

: E H / Z2 = (π / Z) [ Q1 Q2 / (8 c2 G3 π4) ]2

:4.01516646119878 x10-56 [Q6 T15 L -18]

:Reference Equation 48.6:

: E H Z2 = (Z) (E2 / π )

:8.30283355939214 x10-48 [T3 L -2 Q -2]

:Reference Equation 48.8:

: c Z / 8 = d ξ / Temp2 :

: 4493775893.68409 [L5 T -4 Q -2]

ξ d / f 2 :3.82743851387598 x10-8 [L3 T -2] : f 2 r3 d : 5.759415056912 x10-56 [L4 T -2] : f 3 / ξ = c2 / (4 π2 h) :

:Reference Equation 52.18:

:Reference Equation 52.31: :Reference Equation 52.33:

: 3.43578261261903 x1048 [T L -2]

:Reference Equation 56.24:

λ Volts = √(Q1 Q2) / εo :1.80951280219068 x10-8 [L5 T -4 Q -1] :Reference Equation A11.13: : Ms / Volts = 4 π f r3 / Z G √(Q1 Q2) :1.2214441915683 x10-34 [T2 L -2 Q] :Reference Equation A13.2: : η, the unification constant (or the fifth field): η = (32 π / c) (d3 ξ) η = Z π Q 1 Q2

:Reference Equation 9.21: :Reference Equation 9.23:

η = 8 π3 c G 3 ξ / ( h f 3 ) η = c μo Q1 Q2

:Reference Equation 10.20:

:Reference Equation 30.9:

η = (32 π / c) (π G ) 4

3

:Reference Equation 49.1:

η = (E / H) Q1 Q2

:Reference Equation 49.3:

η = Q1 Q2 / (c εo )

:Reference Equation 49.7:

η=4πZd

:Reference Equation 52.26:

2

Temp2

η = Volts √(Q1 Q2) / f

:Reference Equation 55.6:

η = 16 π c G O Intensity / ( Q2 f3 ) 3

2

3

η = 8 π Z c f d Q1 d H 2

2

:Reference Equation 56.27:

:Reference Equation A11.41:

η = 9.67055384324054 x10-36 [L4 T -3] :

:

Page 132 of 178

Appendices: Borrowed words of wisdom are not always true, but sometimes they help. :

Appendix 1 “There is nothing to fear but fear itself” T.S. Roosevelt

: Derivation of ω2 = ξ / m given by Equation 2.1: {Reference 1, page 1241} SHM of a pendulum: :

T

2π

m ξ

:Equation A1.1:

Where T is the period m is a mass ξ is the spring constant (Reference 1, page 365) As the period is equal to the reciprocal of the frequency: : (Reference 1, page 361)

1 f

:Equation A1.2:

ω=2πf

:Equation A1.3:

T And: :

Substituting for T gives: :

ω

2π T

:Equation A1.4:

T

2π ω

:Equation A1.5:

m ξ

:Equation A1.6:

1  ω

m ξ

:Equation A1.7:

ω

ξ m

:Equation A1.8:

Rearranging for T: :

Equating the expressions for the period, T = 2 π √ (m / ξ), Equation A1.1and T = 2 π / ω, Equation A1.5: :

2π  ω

2π

Simplify: :

Rearranging: :

:

Page 133 of 178

Squaring both sides: :

ω2 

ξ m

:Equation A1.9:

Substituting ω for f in the above where, ω = 2 π f, Equation A1.3 gives: :

(2 π f) 2 

ξ m

:Equation A1.10:

1 2π

ξ m

:Equation A1.11:

Rearranging for f: :

f The equation, f = (1/ 2 π ) √ (ξ / m) agrees with {Reference 1, page 365} : : : :

Reference 1: Physics for Scientists and Engineers with Modern Physics, Fourth Edition, Author Serway

:

:

Page 134 of 178

Appendix 2 “We haven’t the money, so we have to think” Earnest Rutherford (1871-1937) : Introducing a time evolving system: Starting with the equation for the natural frequency given in Equation A1.11, f = (1/ 2 π ) √ (ξ / m) If one considers a mass Ms, then the natural frequency, f, is given by: :

1 2π

f

ξ Ms

:Equation A2.1:

Now consider an effect mass mp, which sets up it own resonance frequency, f p, as opposed to having the identical natural resonance, f, of the Source Mass Ms. This may result in a possible ‘wormhole effect’.: :

1 2π

fp 

ξ mp

:Equation A2.2:

Substitute mp into the above equation. Where mp = r2 / (Ms ), Equation 11.19 (rearranged): :

ξ

fp 

1 2π

fp 

1 2π

Ms ξ

1 2π

ξ Ms

r2 ( ) Ms

:Equation A2.3:

Simplify: :

r2

:Equation A2.4:

The ratio of the two frequencies f and f p would be: :

f  fp

1 2π

Ms ξ

:Equation A2.5:

r2

Simplify: :

f  fp

ξ Ms Ms ξ

:Equation A2.6:

r2 Simplify, thus: :

f  fp

r Ms

: The difference between the two frequencies, f and fp, is: :

:

Page 135 of 178

:Equation A2.7:

f - fp 

ξ 1 Ms 2 π

1 2π

Ms ξ r2

:Equation A2.8:

Simplify: :

f - fp 

ξ 2π

(

Ms 1 ) Ms r2

:Equation A2.9:

More neatly expressed as: :

f - fp 

ξ 1/2 r - Ms ( ) 2 π r M1/2 s



(r - Ms ) ξ 2π Ms

:Equation A2.10:

Also the sum of the frequencies gives: :

f  fp 

1 2π

ξ 1  Ms 2 π

Ms ξ r2

:Equation A2.11:

So: :

f  fp 

r  Ms 2π

ξ Ms

:

:

Page 136 of 178

:Equation A2.12:

Appendix 3 “The good Christian should beware of mathematicians and all those who make empty prophecies. The danger already exists, that mathematicians have made a covenant with the devil to darken the spirit and confine man in the bonds of Hell.” Saint Augustine : Using Tension = m v 2 / r {Reference 1, page 146},

Tension 

mp

v2 r

:Equation A3.1:

G

mp

v2 r

:Equation A3.2:

We arrive at a very close approximation for G: : :

As acceleration a = v2 / r = 4 π2 f 2 r, Equation 11.10 G = a mp

:Equation A3.3:

Exactly : Tension Tension is the tension in the “spring or string”, Ф is the angle made between the unextended position (x=0) and the extended position, x. Ms is the mass of the source and a is the acceleration. : Consider that the space between Ms and mp (r ) is like a string or a spring and can experience a tension, Tension then the tension is equal to the restoring force factor multiplied by the distance r. If mp is at an angle to Ms then the tension can be resolved into horizontal and vertical components: : Tension = r ξ

:Equation A3.4

: The horizontal component of the angular tension, Thor, is given by: : Thor = r ξ sin(Ф)

: Equation A3.5:

Where sin Ф can be substituted by Equation 12.19, sin Ф = 2 π f mp : : Thor = 2 π f r ξ mp

: Equation A3.6:

As, ξ mp = r a, from rearranging Equation 11.17, then the above can also be expressed as: : Thor = 2 π f r2 a

: Equation A3.7:

Or alternatively ξ = r a / mp, from rearranging Equation 11.17: :

Thor 

2 π f r 2 a mp mp

As G = mp a, Equation 12.4 then the above becomes: :

:

Page 137 of 178

: Equation A3.8:

2 πf r2 G mp

:Equation A3.9:

1 = sin2Ф + cos2Ф

:Equation A3.10:

cosФ = √(1-sin2Ф)

:Equation A3.11:

Thor  Note the well known expression: :

Rearranging to obtain cosФ gives: : from rearranging Equation 11.2, sin Ф = 2

mp2

ξ / Ms, and substituting into the above, then cosФ is given by: : 2

cosφ 

1-

mp ξ Ms

:Equation A3.12:

Alternatively using the expression, sin Ф = 2 π f mp, Equation 12.19, yields the same result. : Different frequencies exhibit different patterns of vibration, so for example when Tension and Thor are equal then a circular pattern is produced. I am assuming that the tension due to the acceleration of mp is distinct from the component due the tension resulting from the stretching of space between Ms and mp. If this is the case then the wave pattern will change from a vertically oscillating wave (violet, blue and green frequencies), to an elliptical (green, yellow and orange frequencies) and then into a circular pattern (at red frequencies).

:

Page 138 of 178

Appendix 4 “Give me a place to stand and a lever long enough and I shall move the world” Archimedes (287BC-211BC)

: Using Torque (ζ): : ζ = (Force ) (distance)

:Equation A4.1:

{Reference 1, page 146} : As Force = Ms a and (a = 4 π2 f 2 r) defined in Equation 6.3 and distance is given by r: : : ζ = Ms 4 π2 f2 r (r)

:Equation A4.2:

ζ = Ms 4 π2 f2 r2

:Equation A4.3:

ζ 1 = ξ r2

:Equation A4.4:

ζ1 = Ms 4 π2 f2 r2 = ξ r2

:Equation A4.5:

Collect terms:

: And Force = ξ distance (By Hooke’s law): Distance = r:

: Equating equations A4.4 and A4.5 for Torque, ζ, gives: : Ms 4 π 2 f 2 = ξ

:Equation A4.6:

As ( ξ = (2 π) h f / c ) from Equation 2.11 and 2

3

2

r = Ms a c2 / (4 π2 h f 3 ), Equation 6.15 then Equation A4.4 can be written as: :

τ1 

4 π2 h f 3

ξ r2 

c2

(

Ms a c 2 4 π2 h f 3

)2

:Equation A4.7:

Simplifies to: :

τ1 

ξ r2 

M 2s a 2 c 2 4 π2 h f 3

:Equation A4.8:

: Substituting in for Ms, where, Ms = ( 2 π f )2 r3 / G given by Equation 4.11 gives: : :

( τ1 

ξ r2 

4 π2 f 2 r3 2 2 2 ) a c G 4 π2 h f 3

: Substituting the expression for g into the above equation, where, g = 4 π2 f2 r defined in Equation 6.3 gives: :

:

Page 139 of 178

:Equation A4.9:

( τ1 

ξ r2 

4 π2 f 2 r3 2 ) ( 4 π 2 f 2 r )2 c 2 G 4 π2 h f 3

:Equation A4.10:

Simplifies to: :

τ1 

ξ r2 

(4 π 2 ) 3 c 2 f 8 r 8 G2 h f 3

:Equation A4.11:

Simplifying gives: :

τ1 

ξ r2 

64 π 6 c 2 f 5 r 8 G2 h

:Equation A4.12:

Rearranging gives: :

τ1 

ξ r2 

64 π 6 c 2 (f r 3 ) 2 G2 h

f3 r2

:Equation A4.13:

: Collecting the constants ( 64 π6 c2 / ( h G2 ) ) (f r3)2 = 2.91054502786985 x10-49 : ζ1 = ξ r2 = ( 2.91054502786985x10-49 ) (f 3 r2)

:Equation A4.14:

: Consider the interaction due to gravity i.e. : Using Net torque ζnet = F1 d1 - F2 d2: {Reference 1, page 288} 3 3 4

ζnet = ζ1 – ζ2 = F1 d1 – F2 d2

:Equation A4.15:

mp a – G is proportional to ζ2

:Equation A4.16:

4

As 16 d r f = (G) , Equation 8.3: Postulate that ζ2 is the interaction due to gravity (otherwise known as the graviton): : (mp a)4 – 16 d3 r3 f4 = ζ2

:Equation A4.17:

The dimensions of the above equation are discussed in Appendix 5. : ζnet = ζ1 – ζ2

:Equation A4.18:

ζnet = ( 2.91054502786985x10-49 ) (f 3 r2) – [ (mp a)4 – 16 d3 r3 f 4]

:Equation A4.19:

:

: : : Reference 1: Physics for Scientists and Engineers with Modern Physics, Fourth Edition, Author Serway

:

Page 140 of 178

Appendix 5 “The strength of the new theory lies in its consistency and simplicity” Albert Einstein (1879-1955)

: Dimensional Analysis - Dimensions [M L T] and the new dimensions [G] and [Q], Exploring the concept of a continuum: : In order to help define the dimensions for this system I have introduced 2 new dimensions, to represent the Gravitational constant and the charges Q1 and Q2, namely [G] and [Q]. : Length = meters [L] Mass = kilograms [M] Time = seconds [T] Gravitational dimension = [G] Charge = [Q] : Demonstrating that (mp a)4 and 16 d3 r3 f4 are dimensionally equivalent by introducing a new dimension [G]: : G4 = 16 d3 r3 f 4, from equation 10.16 has the dimensions [L3 L3 T -4 = L6 T -4]: G4 has Dimensions [L6 T -4]

:Equation A5.1:

mp = r2 / Ms, Equation 11.18, has the units, L2 M -1: mp has Dimensions [L2 M -1]

:Equation A5.2:

: Ms = ξ r / a is given by rearranging Equation 11.13: a = 4 π2 f2 r, given by Equation 6.3 has dimensions, [L T -2] a has Dimensions [T -2 L]

:Equation A5.3:

ξ has Dimensions [G3 L -3]

:Equation A5.4:

: ξ = π G3 / d3, has the dimensions, [G3 L -3:]

: G = 16 π4 r3 f4 / ξ, given by Equation 10.10, G has the units, [L3 T -4 L3 G -3 = L6 T -4 G -3] : G has Dimensions [L6 T -4 G -3]

:Equation A5.5:

: so Ms has the units, [G3 L -3 L / T -2 L = G3 T2 L -3 = M]: Ms has Dimensions [G3 T2 L -3] : thus mp, which has the units, [L2 M -1 = L2 L3 T -2 G-3 = L5 T -2 G -3]:

:

Page 141 of 178

:Equation A5.6:

mp has Dimensions [L5 T -2 G -3]

:Equation A5.7:

: (mp a), has the units, [L5 T -2 G -3 T -2 L = L6 T -4 G -3], which are the same units of G. : Therefore (mp a )4 will give the same units as G4 QED. : Consider the solution that G3 has the dimensions, [L5 T -4] then,: G3 has Dimensions [L5 T -4] 6

G, which has dimensions, [L T

-4

:Equation A5.8:

-3

G ], has the dimensions of [L], a length : G has Dimensions [L]

:Equation A5.9:

: Then: G3 G = G4 has the dimensions [L5 T -4] [L] = [L6 T -4]

:Equation A5.10:

As per Equation 8.3: It is worth noting then that,: : ξ which has dimensions, [G3 L -3], from Equation A5.4, becomes, [L2 T -4], dimensions of an acceleration squared. ξ has Dimensions [L2 T -4]

:Equation A5.11:

and: Ms, which has dimensions, [G3 T2 L -3], from Equation A5.6, becomes, [L2 T -2], the dimensions of a velocity squared,: Ms has Dimensions [L2 T -2]

:Equation A5.12:

It is should be possible to reduce all units to the dimensions of space-time, for this system: : mp, which has dimensions, [L5 T -2 G -3], from Equation A5.7, becomes, T2, seconds squared: mp has Dimensions [T2]

:Equation A5.13:

: As Plank’s constant, h, where h = c2 ξ / (4 π2 f3), given by Equation 25.11 and c2 has dimensions [L2 T -2] f 3 has dimensions [T -3] ξ has dimensions [L2 T -4] given by Equation A5.11 then: h has Dimensions [L4 T -3]

:Equation A5.14

: Although G is numerically constant, I have introduced the concept of a G, which is dimensionally dynamic according to its power, that is: G has dimensions: [L] :Equation A5.9 G3 has dimensions: [L5 T -4] :Equation A5.8 G4 has dimensions: [L6 T -4] :Equation A5.1

:

Page 142 of 178

This is a reasonable assumption as the Universal Gravitational constant G is numerically constant regardless of the system and how it is wrapped in space and time. In the next chapter the dimensional analysis of G is considered further, in order to draw a logical conclusion. : Using the preceding analysis I now proceed to establish dimensions for impedance (Z), (η), capacitance (C), Induction (Lind), εo the permittivity of free space, μo the permeability of free space, magnetic field strength (H), electric field strength (E), magnetic field (B). For convenience I introduce the dimension [Q], which relates to the charges Q1 and Q2, each of which has dimension [Q]. : As Z, where Z = 32 π4 G3 / (c Q1 Q2 ), Equation 29.1, and G3, which has dimensions (L5 T -4 ), given by, Equation A5.8, then: Z has Dimensions [L4 T -3 Q -2]

:Equation A5.15:

: As η = Z π Q1 Q2, Equation 9.23 and Z has Dimensions [L4 T -3 Q -2] Equation A5.15, then: η has Dimensions [L4 T -3]

:Equation A5.16:

Note that η has the same dimensions as h given by Equation A5.14. : As capacitance, C, where, C = 1 / (2 π2 Z f) Equation 16.11 and Z has dimensions [L4 T -3 Q -2] Equation A5.15 then, C has Dimensions [Q2 T4 L -4]

:Equation A5.17:

: Using the formula for inductance given by Equation 18.5, Lind = (1 / (2 π f)2 )(1/C) ) and as the capacitance C has dimensions [Q2 T4 L -4], Equation A5.17, then, Lind has Dimensions 1/[T -2 Q2 T4 L -4]

:Equation A5.18:

Lind has Dimensions [L4 T -2 Q -2]

:Equation A5.19:

Simplify: : As Resistance, R, where, R = √ ( 4 Lind / C) Equation 19.1 and Lind has dimensions [L4 T -2 Q -2], given by Equation A5.19 and C has dimensions [Q2 T4 L -4] given by Equation A5.17, then, R has Dimensions [L4 T -3 Q -2] Note the same dimensions as impedance. : As εo, where εo = Q1 Q2 / (32 π5 G3), Equation 29.3, and G3 has dimensions [L5 T -4] given by Equation A5.8 then, :

:

Page 143 of 178

:Equation A5.20:

εo has Dimensions [Q2 T4 L -5]

:Equation A5.21:

: As μo, where μo = (32 π5 G3) / (c Q1 Q2 ), Equation 30.6, and G3 has dimensions [L5 T -4] given by Equation A5.8 then, μo has Dimensions [L3 T -2 Q -2]

:Equation A5.22:

: As H, where H = Q1 Q2 / (8 c2 π4 G3), Equation 31.3, and c2 has dimensions [L2 T -2], also G3 has dimensions [L5 T -4] given by Equation A5.8 then, H has Dimensions [Q2 T6 L -7]

:Equation A5.23:

: As E, where E = Z π Q1 Q2 / (8 c2 π4 G3), Equation 32.4, and c2 has dimensions [L2 T -2] and G3 has dimensions [L5 T -4] given by Equation A5.8 and Z has dimensions [L4 T -3 Q -2] given by Equation A5.15 then, E has Dimensions [T3 L -3]

:Equation A5.24:

: As B, where B = 4 π / c4, given by Equation 33.5 and c4 has dimensions [L4 T -4], then, B has Dimensions [T4 L -4]

:Equation A5.25:

-1

(Note that B has dimensions equivalent to Energy ) : Establishing dimensions for the system’s enclosed volume (Vol) pressure (P), temperature (Temp) and density (ρ): : As the enclosed volume of the virtual capacitor given by Vol, where Vol = A π, given by Equation 51.4 then, Vol has Dimensions [L2]

:Equation A5.26:

Note that volume for this system has the dimensions of an area. The volume in this system is created by wrapping surface space, so is an enclosed volume. The dimensional analysis could be repeated with, Vol has Dimensions [L3]

:Equation A5.26.1:

Where the actual volume is considered: As Pressure, P, where P = ξ 2 d / A, given by Equation 51.3 and ξ has dimensions [L2 T -4] given by Equation A5.11 then, P has Dimensions [L T -4]

:Equation A5.27:

: Then P Vol has dimensions, Equation 51.11 has dimensions, P Vol has Dimensions [L3 T -4]

:

Page 144 of 178

:Equation A5.28:

When the pressure is assumed to be distributed evenly throughout the enclosed volume, then Vol has dimensions [L3], and Equation A5.26.1 is expressed as, P Vol has Dimensions [L4 T -4]

:Equation A5.28.1:

: As Temp where, Temp = +/- Q / (2 d), given by Equation 52.11 then, Temp has Dimensions [Q L -1]

:Equation A5.29:

: As density, ρ, where, ρ = mp / Vol, given by Equation 54.1 and mp has dimensions [T2], Equation A5.13 and Enclosed volume, Vol has dimensions [L2] , Equation A5.26 then, (surface) ρ has Dimensions [T2 L -2]

:Equation A5.30:

3

when Vol has dimensions [L ] , Equation A5.26.1 then, ρ has Dimensions [T2 L -3]

:Equation A5.30.1:

so that the density is evenly distributed across the entire volume: Establishing dimensions for the system’s energy and power, i.e. energy of the photon (Ephoton), energy stored in the virtual capacitor (Ecap), the torque (ζ) and the power (Pwr) : As energy stored by the virtual capacitor is, Ecap = π2 Z Q1 Q2, Equation 24.2 and Z has dimensions [L4 T -3 Q -2] given by Equation A5.15 then, Ecap has Dimensions [L4 T -4]

:Equation A5.31:

: As energy photon is, Ephoton = 4 π2 c2 f 2 r3/ G, Equation 24.4 and c2 has dimensions [L2 T -2], f 2 has dimensions [T -2], r3 has dimensions [L3] and G has dimensions [L] given by Equation A5.9 then, Ephoton has Dimensions [L4 T -4]

:Equation A5.32:

: C / Ephoton, Reference chapter 24,:C has dimensions given by, [Q2 T4 L -4] Equation A5.17 then, C / Ephoton has Dimensions [L8 T -8 Q -2]

:Equation A5.33:

: As Torque, ζ1 = ξ r2, given by Equation A4.4, and ξ has dimensions [L2 T -4] given by Equation A5.11 then, ζ has Dimensions [L4 T -4] : As Power, Pwr = Q1 Q2 f / (2 C), given by Equation 20.2, and C has dimensions [Q2 T4 L -4] given by Equation A5.17 then,

:

Page 145 of 178

:Equation A5.34:

Pwr has Dimensions [L4 T -5]

:Equation A5.35:

: As Volts, Volts = √(Q1 Q2) / (εo λ ), given by Equation A11.13, and εo has Dimensions [Q2 T4 L -5] given by Equation A5.21 then, Volts has Dimensions [L4 T -4 Q -1]

:Equation A5.36:

: As Current, I = π f √(Q1 Q2) , given by Equation A11.7 then: I has Dimensions [Q T -1]

:Equation A5.37:

:

Exploring the Concept of a Continuum: : It would be useful to be able to definitively define Q in terms of L and T. The obvious choice would be to define Q as an energy; where for this system the dimensions for energy are given by, [L4 T -4]. However the system seems to be self consistent for any number of dimensional definitions for Q, for example the system is equally valid (i.e. not violated) if Q has the dimensions of acceleration [L T -2]. It appears that Q provides some sort of continuum. : Consider the well known equation for power, which takes the form Power = I2 R. The dimensions of resistance, R, in this system are given by [L4 T -3 Q-2], Equation A5.20. The hypothesis that Imax = π f √(Q1 Q2) Equation 55.3, results in the current, I, possessing the dimensions [Q T -1] and thus Imax2 has dimensions [Q2 T -2]. Then power has dimensions [L4 T -5]. This is consistent with the dimensions of power given in Equation A5.35, where Pwr has dimensions [L4 T -5] : It is reasonable to assume that such a continuum exists in order for energy to transform from one form into another; a concept, which is very familiar.

:

Page 146 of 178

Appendix 6 “The shortest distance between two points is a straight line” Archimedes (287BC-211BC) Or alternatively: The shortest distance between two points is a worm hole. :

Exploring Spacetime: : In the previous chapter a dimensional analysis was undertaken, here I put forward a simple hypothesis to interpret the meaning of spacetime, then demonstrate its validity for this system. : Consider the dimensions [L] and [T] have the following relationship and are interchangeable: : [L] ≡ [T2]

:Equation A6.1:

: Applying the above relationship to, G3 which has dimensions, [L5 T -4], given by Equation A5.8 then: : G3 dimensions that can be expressed as, [L5 L -2] 3

3

G has dimensions that reduce to, [L ]

:Equation A6.2: :Equation A6.3:

Recall G has dimensions [L] : Similarly apply the relationship [L] ≡ [T2], given in A6.1, to G4 which has dimensions: [L6 T -4], given by Equation A5.1 then: : G4 has dimensions that can be expressed as: [L6 L-2] 4

4

G has dimensions that reduce to: [L ]

:Equation A6.4: :Equation A6.5:

: Applying [L] ≡ [T2], given in A6.1, to the restoring force factor, ξ, which has dimensions [L2 T -4], given by Equation A5.11, then: : ξ has dimensions that can be expressed as [L2 L -2]

:Equation A6.6:

ξ becomes dimensionless []

:Equation A6.7:

: To further validate the relationship [L] ≡ [T2], given in A6.1, it can be applied to, ξ d3 = π4 G3, Equation 26.5. As 2d is the separation between the charges Q1 and Q2, d3 has the dimensions of [L3], G3 has dimensions [L3], given by Equation A6.3, and ξ is dimensionless as demonstrated in Equation A6.7 and therefore π4 must also be dimensionless. : The spacetime relationship [L] ≡ [T2], given in A6.1, for this system, can be applied throughout. For example the dimensions for the source Mass (Ms) and the effective photon mass (mp) become [L], the speed of light (c) reduces to dimensions [T], angular acceleration (a) becomes dimensionless and Plank’s constant (h) has dimensions [T5] ≡ [L2 T].

:

Page 147 of 178

Appendix 7 “A mathematical equation stands forever.” Albert Einstein (1879-1955)

: Exploring the concept of Anti-Gravity: : Consider Equation 26.5 : ξ d3 = π4 G3

:Equation 26.5:

Could ξ be negative? Assuming the equations still hold if space was compressed, rather than stretched, then d and so d3 could be negative. : Alternatively consider : :

Z

32 π 4 G 3 c Q1 Q 2

:Equation 29.1:

: Consider if G could be negative, then G3 would also be negative. Q1 and Q2 would both be of the same sign to provide a negative impedance value allowing ‘the flow of electricity’. : The Relationship between the Magnetic Field Strength and the Charge Note that as H = 4 π εo / c2 Equation 38.6 and by rearranging Equation 27.4, εo = 1 / π Z c and substituting into the expression for H gives: :

H

4 Z c3

:Equation A7.1:

and Z is given by Equation 29.1 Z = 32 π4 G3 π / (c Q1 Q2) then: :

H

4 c Q1 Q 2 c 3 32 π 4 G3

:Equation A7.2:

This simplifies to: :

H

Q1 Q 2 8 π 4 c 2 G3

:Equation A7.3:

So: : Q1 Q2 = 8 π4 G3 c2 H

:Equation A7.4:

Q2 = 32 π5 G3 c2 Ho

:Equation A7.5:

Where H = 4 π Ho , Equation 31.4 and Q is defined as Q = √ ( Q1 Q2 ), therefore: :

:

:

Page 148 of 178

Appendix 8 “The laws of nature are but the mathematical thoughts of God” Euclid (435BC-365BC) Thus God must be omnipresent.

: Distribution of Charge: : Consider the possibility that Q1 and Q2 are variables but their product, Q1 Q2, is a constant, so as Q1 increases then Q2 decreases and visa-versa. So when Q1 and Q2 are considered separately, rather than as a product, then how might they vary? : Hypothesis that the charge Q2 varies as the expression: :

Q2 

Q2 f ξ

:Equation A8.1:

Where Q is defined as Q = √ ( Q1 Q2 ), and so Q2 = Q1 Q2, then the above becomes: :

Q1 

ξ f

:Equation A8.2:

Recall that ξ = 4 π2 h f 3 / c2, Equation 56.25. Then: :

Q1 

4 π2 h f 3 c2 f

:Equation A8.3:

Simplifies to: :

Q1 

4 π2 h f 2 c2

:Equation A8.4:

And similarly substituting, ξ = 4 π2 h f3 / c2, Equation 56.25, into the expression for Q2 gives: :

Q2 

Q2 c 2 f 4 π2 h f 3

:Equation A8.5:

Simplifies to: :

Q2 

Q2 c 2 4 π2 h f 2

:Equation A8.6:

So f 2 Q2 is a constant. Also note that by rearranging Equation 56.25, ξ = 4 π2 h f 3 / c2, ξ / f = 4 π 2 h f 2 / c2

:Equation A8.7:

: Also note that the photon energy, h f, is given by rearranging Equation 56.25, ξ = 4 π2 h f 3 / c2, : : h f = ξ c2 / 4 π2 f2

:Equation A8.8:

: Still using the hypothesis that Q1 = ξ / f, Equation A8.2, and substituting this solution for Q1 into the rearranged form of Equation 29.3, Q1 Q2 = 32 π5 G3 εo, then Q2 can be expressed as: :

:

Page 149 of 178

32 π 5 G 3 ε o f ξ

Q2 

:Equation A8.9:

By rearranging Equation 8.11, d3 = π4 G3 / ξ, and substituting this into the above gives:: Q2 = 32 π εo d3 f

:Equation A8.10:

The above can be rearranged to provide an expression for d as follows: :

Q2 32 π ε o f

d3 

:Equation A8.11:

Note that d3 is proportional to Q2 / f. Now rearrange the above to give: :

d2 

Q2 32 π ε o d f

:Equation A8.12:

Note that the term, εo f d, is a constant, thus d2 is proportional to Q2. : Now consider the possibility that the permittivity of free space, εo, which is a constant, is the product of ε1 and ε2, such that ε1 and ε2 are variables, but their product, ε1 ε2, is a constant, so as ε1 increases then ε2 decreases and visa-versa. That is: : εo = ε1 ε2

:Equation A8.13:

So when ε1 and ε2 are considered separately, rather than as a product, then how might they vary?

Q εo

ε1 

Q1

:Equation A8.14:

Or equivalently as: :

εo

Q2

ε1 

Q

:Equation A8.15:

To ensure that the product ε 1 ε 2, is a constant, then ε 2, must therefore be given by: :

Q εo

ε2 

Q2

:Equation A8.16:

Or equivalently as: :

ε2 

Q1 ε o Q

:Equation A8.17:

Now by rearranging Equation A8.11, d2 = Q2 / 32 π εo d f, this can be written as: :

d

Q2 32 π ε o d 2 f

:Equation A8.18:

Substituting in for εo, the permittivity of free space, where εo = ε1 ε2, Equation A8.13 into the above expression gives:

d

:

Q2 32 π (ε 2 d 2 ) ε 1 f

Page 150 of 178

:Equation A8.19:

It is noted here that the term (ε2 d2 ) is a constant, thus d is proportional to Q2 / ε1 f, i.e. d is proportional to 1/f, by substituting in for ε 1. : Consider the special case when, Q1 = Q2 (or ξ / f is equal to Q, then the frequency f pi = 741938715557271 [T -1] and the restoring force factor ξ is ξ = 0.000118871686420531 [L2 T -4]). Note that Q f / ξ = 1, 2, 3, 4…, when f = f pi, f pi / √2, fpi / √3, f pi / √4, respectively. (The same pattern is evident in Chapter 24.) So when the frequency f has the value: f = f pi, the distribution of charge is equal to, Q = Q1 = Q2 when f = f pi / √2, the distribution of charge is equal to, 4Q1 = Q2 and (Q = 2Q1 = Q2/2) when f = f pi / √3, the distribution of charge is equal to, 9Q1 = Q2 and (Q = 3Q1 = Q2/3) when f = f pi / √4, (or = f pi / 2) the distribution of charge is equal to, 16Q1 = Q2 and (Q = 4Q1 = Q2/4) where Q is Q = √ (Q1 Q2 ), given in Chapter 57 : 1.60217662089849 x10-19 [L2 T -3] :

:

Page 151 of 178

The expression Q1 Q2, can be considered as a function of Q and could be replaced throughout the book with f n(q), where the whole function would always take the value of Q2. For example consider if the function f n(q) took the form: : : f n(q) = q1 q2 + q3 q4 + q5 q6 + q7 q8 + q9 q10 + q11 q12 + q13 q14 + q15 q16 = Q2

:Equation A8.20:

: Where each pair could take a fractional value of Q and the sum of all the fractions would make the whole. This would then imply waves of certain frequencies would coexist as a set. The simplest example (if all the values of q were the same fractional value of Q) would conclude that more ‘red’ photons are possible than ‘blue’ photons for the value of the function f n(q). A more complex example would include the possibility of two different values for q (and so two frequencies) in each pair. The physics at this level (of variation) would yield interesting results, it would explore not just two levels of order, but the possibility of multiple levels of interactions.

:

Page 152 of 178

Appendix 9 “Things which are equal to the same thing are also equal to one another” Euclid (435BC-365BC) :

Alternative Equivalent Expressions for G and π applicable to this model: : Recall that d3 r3 f4 = (G / 2)4, Equation 8.3, and d3 ξ = G3 π2, Equation 26.5, then: :

r3 f 4  ξ

G 16 π 4

:Equation A9.1:

Rearranges to give: :

1  ξ

G 16 π 4 r 3 f 4

:Equation A9.2:

Recall that 1 / ξ = c2 / 4 π h f3 , Equation 25.11, so Equating the expressions for 1 / ξ gives: :

G 16 π 4 r 3 f 4



c2 4 π2 h f 3

:Equation A9.3:

which simplifies to: : G h = 4 π2 c2 r3 f

:Equation A9.4:

As Equation 13.17, G = a r / ξ, then 2

G2 

a4 r 2 ξ

:Equation A9.5:

G3 

a6 r 3 ξ6

:Equation A9.6:

And

: Consider Equation 26.5 , ξ d3 = π4 G3, by substituting in for G, given by Equation 13.17 , G = a2 r / ξ, the following equation is derived: :

ξ d3 

π 4 G2

a2 r ξ

:Equation A9.7:

Rearranging, by collecting constant and variable terms, gives: :

π 4 G2 

d3 ξ 2 a2 r

:Equation A9.8:

: Similarly if a substitution for G2 ,given by A9.5 , G2 = a4 r2 / ξ, is made into Equation 26.5, ξ d3 = π4 G3 this gives: :

ξ d3 

π4 G

a4 r 2 ξ

Rearranging by collecting terms gives: :

:

Page 153 of 178

:Equation A9.9:

d3 ξ 2

π4 G 

a4 r 2

:Equation A9.10:

: Now consider substituting for G3 given by Equation A9.6, G3 = a6 r3 / ξ3, into Equation 26.55, ξ d3 = π4 G3,: :

ξ d3 

π4

a6 r 3 ξ3

:Equation A9.11:

Rearranging by collecting variable and constant terms gives: :

d3 ξ 4

π4 

a6 r 3

:Equation A9.12:

Therefore from Equation A9.12 above then taking the quarter root : :

π

ξ

d 3/4 a 3/2 r 3/4

:Equation A9.13:

Also: :

π2 

ξ2

d 3/2 a 3 r 3/2

:Equation A9.14:

Also: :

π3 

ξ3

d 9/4 a 9/2 r 9/4

:Equation A9.15:

And

π5 

ξ5

d15/4 a15/2 r 15/4

:Equation A9.16:

: These alternative expressions, for the quintessential π, allow a myriad of new, but equivalent, equations to express the host of interconnecting relationships and uncover new relationships. For example by substituting in for π2, given by Equation A9.14 into, G h = 4 π2 c2 r3 f, Equation A9.4 gives:

G h  4 c2 f ξ2

(d 3 r 3 ) a3

:Equation A9.17:

And, for example rearranging Equation 11.16, a = ξ r / Ms, and substituting into Equation A9.12, π4 = d3 ξ4 / a6 r3, gives ::

π  4

:

d 3 Ms

6

ξ2 r9

Page 154 of 178

:Equation A9.18:

Appendix 10 If one knows how to look they will discover order everywhere. :

Exploring Phase, Volts and Chaos Theory : Chaos implies that there is some observation of a system that varies unpredictably, with no apparent discernible order or regularity. The Phase (angle of polarization) is given by, Phase = Vspiral / π, Equation 17.13 As Vspiral = 2 π f (d2 + λ2 / π2), Equation 17.11 then : :

Phase 

2f(

λ2

 d2 )

π2

:Equation A10.1:

The plane of polarization (or the phase) rotates in an electric field, so there must exist some function between the two. It is noted here that, (Phase * Volts) * (d / Phase) = electric dipole moment, DE. As Volts varies and is given by, Volts = η f / √(Q1 Q2), Equation 55.6 then: :

Phase Volts 

2f(

λ2 π

2

ηf

 d2 )

Q1 Q 2

:Equation A10.2:

Recall that from Equation 9.23, η = Z π (Q1 Q2), substituting this into the above gives: :

Phase Volts 

2f(

λ2 π

2

 d2 )

Z π Q1 Q 2 f Q1 Q 2

:Equation A10.3:

Simplifying the above expression gives: : Phase Volts 

2 Z π Q1 Q 2 f 2 (

λ2 π2

 d2 )

:Equation A10.4:

Expanding the brackets and recalling that c = λ f, from Equation 2.8: :

Phase Volts 

2 Z c2 π

Q1 Q 2  2 Z π Q1 Q 2 f 2 d 2

:Equation A10.5:

Note that the second term is small compared to the first term. Recall that the impedance (Z) is given by Equation 29.1, Z = 32 π4 G3 / (c Q1 Q2), substituting this into the above equation gives: :

Phase Volts 

2 (32 π 4 G 3 ) c 2 c Q1 Q 2 π

Q1 Q 2  2

(32 π 4 G 3 ) π Q1 Q 2 f 2 d 2 c Q1 Q 2

:Equation A10.6:

Simplifying: :

Phase Volts 

64 π 3 G 3 c Q1 Q 2



64 π 5 G 3

f 2 d2

c Q1 Q 2

Or: :

:

Page 155 of 178

:Equation A10.7:

Phase Volts 

64 π 3 G 3 Q1 Q 2

(c 

π 2 f 2 d2 ) c

:Equation A10.8:

The second term, π2 f 3 d2 / c, varies almost imperceptibly and apparently randomly, until a specific sampling interval (of 1 / c2) is used to reveal hidden orders / patterns. This effect could be related to chaos theory or might simply be due to computation limitations and rounding errors. Further research is needed here. : Note similarly for current, Imax, by dividing Phase Volts = 2 Z c2√(Q1 Q2) / π + 2 π Z √(Q1 Q2) f2 d2, Equation A10.5, by Z gives: :

Phase

Volts  Z

2 c2 π

Q 1 Q 2  2 π Q1 Q 2 f 2 d 2

:Equation A10.9:

recall that (Volts) = Imax R, Equation 55.2, and in this case R is equal to Z, the impedance of free space, so the above equation becomes: :

Phase Imax 

2 Q1 Q 2 (

c2  π f 2 d2 ) π

:Equation A10.10:

The above relationship can also be expressed in terms of magnetic permeability, μo, recall that μo / Z = π / c, Equation 30.8 so

c μo Phase Imax  π

Phase Volts

:Equation A10.11:

Note that Phase Volts is a constant. Multiply through by π and as the phase (angle of polarization) is defined as π Phase = Vspiral, from Equation 17.13, so: : c μo Phase Imax = Vspiral Volts

:Equation A10.12:

Or in terms of the unification constant, η, using η = c μo Q1 Q2, Equation 30.9, then using the above expression: : η Phase Imax = Q1 Q2 Vspiral Volts

:Equation A10.13:

: It is interesting to note the following relationship between the energy stored in a capacitor with the Phase: : Recall, Ecap is the energy stored in the virtual capacitor, from Equation 55.33, Ecap = π √(Q1 Q2) Volts, rearranging, we see that: :

π

E cap Volts Q1 Q 2

:Equation A10.14:

Knowing that, Vspiral = π Phase, Equation 17.13, and using the above, the following relationship may be stated: :

Vspiral 

:

E cap Phase Volts Q1 Q 2

Page 156 of 178

:Equation A10.15:

: A further relationship of interest is that between the wavelength and the phase:

λ Phase



D isp

:Equation A10.16:

where Disp is a constant, which might explain why white light disperses when passing through a prism with a constant refractive index. As λ f = c, from Equation 2.8, then it is true to say that, the frequency multiplied by the phase must also be a constant, such that: :

c D isp

f (Phase ) 

:Equation A10.17:

: Other ways to express PhaseVolts : The Phase multiplied by Volts is a constant and can be expressed in various ways: Recall that the hypotenuse, Δ, is given by Equation 17.6 as, Δ = √(d2 + λ2/ π2), by substituting for λ, where λ = c / f Equation 2.8, gives: :

Δ

(d 2 

c2 π2 f 2

)

:Equation A10.18:

Vspiral = 2 π (f) (Δ)2, Equation 17.10, and Phase = Vspiral / π, then: :

Phase 

2 f (d 2 

c2 π2 f 2

)

:Equation A10.19:

Using the expression Volts = Imax Z, where Imax = π f √(Q1 Q2), Equation 55.3, then Phase Volts can be expressed as: :

Phase Volts 

2 π Z Q f 2 (d 2 

c2 π2 f 2

)

:Equation A10.20:

)

:Equation A10.21:

Where Q = √(Q1 Q2), for ease,

Phase Volts 

2 π Z Q (f 2 d 2 

c2 π2

As c / π = Z / μo, from rearranging Equation 30.8 and 1 / μo = c2 εo then the above can take the form: :

Phase Volts 

2 π Z Q (f 2 d2  c 4 Z 2 ε o ) 2

Note that the term, c4 Z2 εo2 can be written in terms of the electric field E i.e., c4 Z2 εo2 = E c5 / 4 π3.

:

Page 157 of 178

:Equation A10.22:

Appendix 11 ”Nothing is too wonderful to be true, if it is consistent with the laws of nature.” Michael Faraday (1791-1867) :

The Relationship Between Permittivity of Free Space and Voltage : Starting by introducing a new term Qo, which is a constant. Let the “emerald key” Qo be defined as: :

Q

Qo 

16 π 5 G3

:Equation A11.1:

Where Q = √(Q1 Q2) By integrating Qo with respect to Q the expression for permittivity of free space is obtained i.e.:



Q o Q 

Q2 5

32 π G

3

Q1 Q 2



32 π 5 G 3

:Equation A11.2:

: Recall from Equation 29.3 that εo = Q1 Q2 / (32 π5 G3), so therefore: :

εo 

Q

o

Q

:Equation A11.3:

Recall that Volts = 32 π5 G3 / (λ √(Q1 Q2)), Equation 55.8, and Qo is defined by Equation A11.1, where Qo = Q / 16 π5 G3, so Volts can be expressed in terms of Qo: :

2

Volts 

Qo λ

:Equation A11.4

The permittivity for free space is related to Volts by the constant Q o. An expression for the current I, may be obtained from utilizing the well known relationship, Volts = I R, Equation 55.2, where in this case the resistance is given by the impedance of free space, Z, and where Z = 32 π4 G3/ (c Q1 Q2), Equation 29.1, then using, Volts = 32 π5 G3 / (λ √(Q1 Q2)), Equation 55.8, an expression for the current, I, can be obtained:

I

Volts  Z

32 π 5 G 3

c Q1 Q 2

4 3 λ ( Q1 Q 2 ) 32 π G

:Equation A11.5:

Simplifies to : :

I

π c (Q1 Q 2 ) λ

:Equation A11.6:

As λ f = c, from Equation 2.8, the above can be expressed as: : I = π f √(Q1 Q2)

: Note that Qo can be expressed in terms of η, as Volts = 2 / (λ Qo ), Equation A11.14, and η = Volts √(Q1 Q2) / f, Equation 55.6, then: :

:

Page 158 of 178

:Equation A11.7:

η

2 (Q1 Q 2 )

η

2 (Q1 Q 2 )

λ f Qo

:Equation A11.8:

Or, as λ f = c, from Equation 2.8,

c Qo

:Equation A11.9:

: The relationship between the permittivity of free space εo the volts may be derived as follows: Recall that, εo = 1 / (π Z f λ ) from Equation 27.3 and that, Volts = I R, Equation 55.2, where in this case the resistance R is equal to the impedance of free space, Z. Thus: :

I π f λ Volts

εo 

:Equation A11.10:

As the current I is given by, I = π f √(Q1 Q2), Equation A11.7 then the above may be expressed as: :

π f (Q1 Q 2 )

εo 

π f λ Volts

:Equation A11.11:

Which simplifies to: :

(Q1 Q 2 )

εo 

λ Volts

:Equation A11.12:

where (λ Volts) is a constant. By rearranging, the above may be expressed as: : √(Q1 Q2) = εo λ Volts

:Equation A11.13:

Equating the expression given by Equation A11.12, εo = √(Q1 Q2) / (λ Volts) with εo = Q1 Q2 / (32 π5 G3), Equation 29.3, gives: :

Q1 Q 2 5

32 π G

3

(Q1 Q 2 )



λ Volts

:Equation A11.14:

Rearrange for Volts and simplify to give: :

Volts 

32 π 5 G 3 λ (Q 1 Q 2 )

:Equation A11.15:

As per Equation 55.8. The expression for permittivity of free space can be substituting in to previously derived expressions to obtain relationships in terms of Volts. Consider for example, Equation 40.3, H = 4 π εo / c2, by substituting in for εo, given by Equation A11.12, where εo = √(Q1 Q2) / (λ Volts) gives: :

H

4 π (Q1 Q 2 ) c 2 λ Volts

Where the product λ Volts is a constant.

:

Page 159 of 178

:Equation A11.16:

The above equation can be expressed in terms of frequency using Equation 2.8, λ = c / f, giving: :

4 π f (Q1 Q 2 )

H

c 3 Volts

:Equation A11.17:

It is noted here that this value for H differs from the value of the magnetic field strength of free space (Ho) by a factor of 4π, that is Ho = H / (4π) Equation 31.4. Equation A11.16 can be rewritten in terms of Ho, the magnetic field strength of free space: :

(Q1 Q 2 )

Ho 

c 2 λ Volts

:Equation A11.18:

And similarly Equation A11.17 can be rewritten in terms of Ho as: :

f (Q1 Q 2 )

Ho 

c 3 Volts

:Equation A11.19:

The model can be shown to be consistent by comparing Equation 38.3, H = 4 / Z c3, with Equation A11.17, H = 4 π f √(Q1 Q2) / (c3 Volts ), then: :

1  Z

π f (Q1 Q 2 ) Volts

:Equation A11.20:

Rearranging: : Volts = Z π f √(Q1 Q2)

:Equation A11.21:

As the current, I, is defined by Equation A11.7, I = π f √(Q1 Q2), the model is consistent, in that Volts = I R, where the resistance, R, in this case is Z, the impedance. : The relationship between permittivity of free space, the magnetic field strength and the speed of light: : Recall that Ho = H / (4π) Equation 31.4. and so Ho = √(Q1 Q2) / (c2 λ Volts) Equation A11.18, by rearranging Equation A11.18.

(Q1 Q 2 )

c 2 Ho 

λ Volts

:Equation A11.22:

As Equation A11.12, εo = √(Q1 Q2) / (λ Volts) then Equation A11.22 can be written as: : εo = c2 Ho = c2 H / (4 π)

:Equation A11.23:

: By substituting, εo = 1 / (μo c2 ) from rearranging Equation 30.1, into the above equation then equation A11.23, εo = c2 Ho , gives an expression for the permeability of free space in terms of magnetic field strength of free space.

1 μo c

2



Ho c 2

Rearranging gives: :

:

Page 160 of 178

:Equation A11.24:

1 Ho c 4

μo 

:Equation A11.25:

: The relationship between the Magnetic (H) and Electric (E) Field strengths, impedance (Z) and the speed of light : Recall Equation A11.19, Ho = f √(Q1 Q2) / (c3 Volts ), rearranging gives: :

f (Q1 Q 2 )

c 3 Ho 

Volts

:Equation A11.26:

As Z π = Volts / f √(Q1 Q2) from Equation A11.21 then the above becomes: : c3 Ho = Z π = c3 H / (4 π)

:Equation A11.27:

: The relationship between the magnetic field strength and current: Consider the expression, Volts = Z π f √(Q1 Q2), Equation A11.21, by rearranging this to give: :

Volts f (Q1 Q 2 )



Zπ

:Equation A11.28:

As E / H = Z π, Equation 32.1, this expression can be equated to Equation A11.28 above, Volts / ( f √(Q1 Q2) ) = Z π. So: :

Volts

E  H

:Equation A11.29:

f (Q1 Q 2 )

Rearranging gives: : E ( f √(Q1 Q2) ) = H Volts

:Equation A11.30:

Multiply both sides by π, to obtain an expression in terms of current I, where I = π f √(Q1 Q2), Equation A11.7, then E I = π H Volts

:Equation A11.31:

E I = 4 π2 Ho Volts

:Equation A11.32:

As H = 4 π Ho, Equation 31.4, then: :

: Expressing the Magnetic Field Strength (H) and the Electric Field Strength (E) in terms of Charge Using Equation A11.23, εo = c2 H / (4 π) and substituting for, εo, where εo = Q2 / 32 π d3 f from rearranging Equation A8.10 gives: :

c2 H  4π

Q2 32 π d 3 f

:Equation A11.33:

Simplifying and rearranging gives: :

H

:

Q2 8 c 2 d3 f

Page 161 of 178

:Equation A11.34:

Or H in terms of Q1 using c2 /4 π = π h f2 / Q1, by rearranging Equation A8.4, and substituting this term into Equation A11.23, εo = c2 H / (4 π) gives: :

Hπhf2 Q1

εo 

:Equation A11.35:

Rearranging gives: :

ε o Q1

H

πhf2

:Equation A11.36:

As E = H Z π, Equation 32.2, by substituting for H as given above then E can be expressed as: :

E

Zπ

ε o Q1 πhf2

:Equation A11.37:

Reduces to: :

E

Z ε o Q1 hf2

:Equation A11.38:

Or E in terms of Q2 is obtained by substituting H = Q2 / (8 c2 d3 f) Equation A11.34 into Equation 32.2 E = H Z π giving:

E

Z π Q2 8 c 2 d3 f

:Equation A11.39:

Note that from Equation A11.34, H = Q2 / (8 c2 d3 f) and Equation 9.23 rearranged, Q2 = η / Z π Q1, then: :

H

η 8 π Z c 2 f d3

:Equation A11.40:

Rearranging gives: : η = 8 π Z c2 f d Q1 d2 H As η is a constant and f d is also a constant then Q1 d2 must be a constant, which it is. :

:

Page 162 of 178

:Equation A11.41:

Appendix 12 In the beginning there was a sound. :

The Relationship between Magnetic Field Strength and Sound : Recall that the magnetic field strength, H, is given by H = 4 π f √(Q1 Q2) / (c3 Volts ), Equation A11.17, then by rearranging to give: :

π f (Q1 Q 2 ) c

H c 2 Volts 4



:Equation A12.1:

Let Q = √(Q1 Q2) where Q1 and Q2 have the same magnitude (Q) the above Equation A12.1 can be expressed as: :

H c 2 Volts 4

πfQ  c

:Equation A12.2:

Recall the expression for sound potential, O, is given by O = π f Q2 ξ / c, Equation 56.7, then the sound potential can be expressed in terms of H: :

O

H c 2 Q Volts ξ 4

:Equation A12.3:

As Intensity is defined by Equation 56.12, Intensity = 2 π f2 ξ / c2, Equation 56.12, and Loudness factor is defined as the product of the sound potential O and the Intensity, the Loudness factor can be expressed in terms of the magnetic field strength as: :

O Intensity 

2 π f 2 ξ H c 2 Q Volts ξ 4 c2

:Equation A12.4:

π H Q f 2 ξ 2 Volts 2

:Equation A12.5:

Simplifying gives: :

O Intensity 

As O Intensity = 2 π2 Q2 f3 ξ2 / c3, Equation 56.13, then by equating with A12.5 above, the magnetic field strength, H, can be verified: :

O Intensity 

π H Q f 2 ξ 2 Volts  2

2 π 2 Q2 f 3 ξ 2 c3

:Equation A12.6:

Solving for H to give: :

H

4 πQ f c 3 Volts

:Equation A12.7:

And as, H = 4 π Ho , Equation 31.4, therefore Ho must be equal to: :

Ho 

Qf 3

c Volts

As per Equation A11.2.

:

Page 163 of 178

:Equation A12.8:

Equation A12.8 implies that the frequency divided by the Volts is a constant, which it is. Rearranging Volts = Z π f √(Q1 Q2), Equation A11.21, to give: :

Zπ

Volts Qf

:Equation A12.9:

where Q = √(Q1 Q2). And so by substituting A12.9 into Ho = Q f /c3 Volts, Equation A12.8 gives: :

1

Ho 

Z π c3

:Equation A12.10:

:

Defining a Relationship between Sound and Temperature : Recall that Loudness Factor is O Intensity = π H Q f 2 ξ2 Volts / 2, Equation A12.5, and Temperature (Temp ) for the system is given by, Temp = ± Q / (2 d), Equation 52.11, where Q is defined as Q = √(Q1 Q2). Then the ratio of the Loudness Factor, O Intensity, to the temperature is given by: :

π H Q f 2 ξ 2 Volts 2 d 2 Q

:Equation A12.11:

O Intensity = π H Volts d f2 ξ2 Temp

:Equation A12.12:

O Intesity Temp



Simplifying gives: :

:

Defining a Relationship between Sound and Current : Recall that the current (I) is defined by Equation , I = 2 π f d Temp, by rearranging and substituting into Equation A12.12, O Intensity = π H Volts d f2 ξ2 Temp, for the term Temp gives: :

O Intensity 

π H d f 2 ξ 2 Volts

I 2 πf d

:Equation A12.13:

I Volts 2

:Equation A12.14:

The expression, for loudness factor, above simplifies to: :

O Intensity 

H f ξ2

Where I Volts is Power: :

:

Page 164 of 178

Appendix 13 “From my mind to your mind, from my thoughts to your thoughts” Gene Roddenberry (Star Trek) :

The Relationship between Volts and the Source Mass : Recall that the source mass Ms is described by Equation 4.11, Ms = 4 π2 f r3 / G, the volts is given by Equation 55.6, Volts = η f / √(Q1 Q2) Equation 55.6, where η is given by Equation 9.23, η = Z π (Q1 Q2), consider the ratio of the source mass to volts: :

Ms  Volts

Q1 Q 2 4 π2 f 2 r3 G Z π Q1 Q 2 f

:Equation A13.1:

The above simplifies to: :

Ms  Volts

4 π f r3

:Equation A13.2:

Z G Q1 Q 2

As the term f r3 is a constant then the ratio Ms / Volts is a constant (the value provided in Chapter 57). :

The Relationship between the Charge and the Source Mass : Consider Equation 30.21, Q1 Q2 = 32 π5 G3 / (c2 μo), by rearranging Equation 4.13 such that 1 / c2 = Ms / h f, and substituting the expression into Equation 30.21 gives: :

Q1 Q 2 

32 π 5 G 3 M s μo hf

:Equation A13.3:

It is clear that, f / Ms (or inversely Ms / f ) must be a constant, which it is, the value is given in Chapter 57: :

The Relationship between the Restoring Force Factor and Frequency : As, f 2 = (ξ / 4 π2 Ms) Equation 11.1, rearranging this in the form: :

ξ  f

4 π 2 f Ms

:Equation A13.4:

Q1 = 4 π2 Ms

:Equation A13.5:

Note here that from Equation A8.2, ξ / f = Q1 therefore: : From, Equation 11.16, Ms = ξ r / a, rearranging this for the restoring force factor gives: :

ξ

Ms a r

Then f / ξ using the above equations: :

:

Page 165 of 178

:Equation A13.6:

f  ξ

fr Ms a

:Equation A13.7:

From Equation A13.3 it was found that, f / Ms, is a constant, therefore f / ξ varies as r / a, where a is the acceleration. Substituting for the acceleration ,a, given by Equation 11.10, a = v2 / r = 4 π2 f 2 r, (where v = 2 π f r, Equation 11.11) into Equation A13.7 gives: :

f  ξ

f r Ms 4 π 2 f 2 r

:Equation A13.8:

f 1 Ms 4 π 2 f 2

:Equation A13.9:

Simplifying gives: :

f  ξ :

The Relationship between the Source Mass and the Universal Gravitational Constant : As G = mp a, Equation 12.4, substituting for mp using Equation 11.19, mp = r2 / Ms, then: :

G

a r2 Ms

:Equation A13.10.:

Ms 

a r2 G

:Equation A13.11:

Rearranging for Ms gives: :

The above equation, Ms = a r2 / G, can be written in terms of velocity v, using Equation 11.10, a = v2 / r, to give: :

Ms 

v2 r G

:Equation A13.12:

from Equation 11.11, v = 2 π f r, then Ms = v2 r / G, as above, can be expressed as: :

Ms 

(4 π 2 f 2 r 2 ) r G

:Equation A13.13:

4 π2 f 2 r3 G

:Equation A13.14:

Which becomes: :

Ms 

Recall that f r3 is a constant, so Ms is proportional to f, the frequency. Rearranging Equation A13.14, Ms = 4 π2 f 2 r3 / G, for G gives: :

G

(4 π 2 ) (f r 3 ) (

f ) Ms

:Equation A13.15:

The Universal Gravitational constant, G, is made up of the product of three constant elements, 4 π2, f r3 and f / Ms. Or equivalently as Ms = ξ r / a, Equation 11.16, and the acceleration a = 4 π2 f 2 r, Equation 11.10, then: :

:

Page 166 of 178

ξ

Ms 

4 π2 f 2

:Equation A13.16:

Substituting Equation A13.16, Ms = ξ / 4 π2 f2 into Equation A13.15 gives: :

G

(4 π 2 ) (f r 3 ) (

4 π2 f 3 ) ξ

:Equation A13.17:

f3 ) ξ

:Equation A13.18:

So: :

G

(16 π 4 ) (f r 3 ) (

Where f 3 / ξ is a constant. So again G is the product of several constants. Note that from Equation 29.3, Q1 Q2 = 32 π5 εo G3 : Consider the expressions, Z = 32 d3 ξ / (c Q1 Q2 ) given in Equation 7.17 π4 G3 = ξ d3, Equation 8.11, Z = 32 π4 G3/ (c Q1 Q2), Equation 29.1, d3 = G4 / (16 r3 f 4) from rearranging Equation 10.16 ξ = 4 π2 f 2 Ms, Equation 11.9: Ms = G r / sin2Ф by rearranging Equation 11.31 Sin Ф = 2 π f mp Equation 12.19: Allows the impedance, Z, to be expressed in a plethora of different combinations. : The model is beautifully consistent and interconnected. A myriad of definitions can be developed, that provide consistent answers. :

:

Page 167 of 178

Appendix 14 “As above so below”. Hermes Trismegistus, The Emerald Tablet :

Boundary Conditions: Is there an elastic limit or boundary condition inherent in the system? The model suggests that the elastic limit, or boundary condition, would be evident when, sin2Ф = G r / Ms, Equation 11.31 is equal to one, i.e. sin2Ф =1, giving : :

Gr Ms

1

:Equation A14.1:

Or as, Ms = r2 / mp, Equation 11.36, and

G r mp

1

r2

:Equation A14.2:

Equation 12.1 gives, r2 / mp = ξ / 4 π2 f 2 Then by substituting these terms into Equation A14.1 gives the condition for the elastic limit when: :

1

4 π2 G r f 2 ξ

:Equation A14.3:

Or by canceling the r terms in Equation A14.2, 1 = G r mp / r2 to give: :

G mp

1

r

:Equation A14.4:

So when sin Ф = 1, then sin Ф = 2 π f mp, Equation 12.19, becomes, mp = 1 / (2 π f) , substituting this value of mp into the above gives: :

1

G 2πf r

:Equation A14.5:

G 2

:Equation A14.6:

G ( )4 2

:Equation A14.7:

Rearranging: :

πfr Increasing both sides to the power of four: :

π4 f 4 r 4 

As, r3 d3 f 4 = G4 / 16 from Equation 10.16, the R.H.S. of the above expression can be replaced, then when sin Ф = 1: :

π4 f 4 r 4 

r 3 d3 f 4

:Equation A14.8:

π4 r = d 3

:Equation A14.9:

Simplifying: : By multiplying both sides of the above equation by the restoring force factor ξ and dividing both sides by π4: :

:

Page 168 of 178

ξ d3

ξr

π4

:Equation A14.10:

Rearranging Equation 8.11, G3 = d3 ξ / π4, and substituting to give an alternative boundary condition when sin Ф = 1: : ξ r = G3

:Equation A14.11:

Is an alternative boundary condition which defines the “elastic limit” of the system. : Defining the Trigonometric Relationships : As sin2Ф = ξ r4 / Ms 3 Equation 11.29, then squaring both sides : :

ξ2 r8

sin4 φ 

Ms

6

:Equation A14.12:

As, Ms 6 = π4 ξ2 r9 / d3 by rearranging Equation A9.18, then by substituting for Ms 6 in the above gives:

d3

sin 4 φ 

π4 r

:Equation A14.13:

And so by taking the square root of the above: :

d3

sin2 φ 

π4 r

:Equation A14.14:

Where the square root can be negative or positive, by Taylor’s theorem. As, sin2Ф = G r / Ms, Equation 11.31, then it is also true that by substituting for sin2Ф into the above gives: :

Gr  Ms

1 π2

π2 G 

Ms

d3 r

:Equation A14.15:

The above can be simplified as: :

d3 r3

:Equation A14.16:

: Further as sin2Ф = (1 / π2) √(d3/r), Equation A14.14, then sinФ can be expressed as its square root, that is: :

sinφ 

1 d 3/4 π r 1/4

:Equation A14.17:

Thus as, sinФ = 2 π f mp, Equation 12.9, then it is also true that by substituting for sinФ into the above gives: :

1 d 3/4 π r 1/4

:Equation A14.18:

1 d3/4 2 f mp r 1/4

:Equation A14.19:

2 π f mp  The above can be simplified as: :

π2 

Note that by the above with Equation 9.14, π2 = ξ2 d3/2 / (a2 r3/2) and Substituting, G / a = mp, Equation 12.4, into Equation A14.18, then it can be shown that: :

:

Page 169 of 178

a4

G

r 5/4

2 f ξ 2 d 3/4

:Equation A14.20:

Further, as sin2Ф = (1 / π2) √(d3/r), Equation A14.14 then: :

d3

sin 4 φ 

π4 r

:Equation A14.21:

Rearranging Equation A14.11, 1 = G3 / ξ r, it can be shown that: :

G3 ξr

sin4 φ 

:Equation A14.22:

Equating Equation A14.21, sin4Ф = d3 / (π2 r), with the above gives: :

G3  ξr

d3 π4 r

:Equation A14.23:

Which simplifies to the familiar, π4 G3 = ξ d3, Equation 8.11. : It can also be deduced that as, sinФ = 2 π f mp, Equation 12.9, then sin3Ф = 8 π3 f3 mp3

:Equation A14.24:

Substituting, G / a = mp, Equation 12.4, into the above gives: :

sin 3 φ 

G3

8 π3 f 3

a3

:Equation A14.25:

And the acceleration, a, is defined as, a = 4 π2 f 2 r, Equation 11.10 so the above becomes: :

sin 3 φ 

8 π3 f 3

G3 64 π 6 f 6 r 3

:Equation A14.26:

Simplifying, collecting terms and recalling that f r3 is a constant, the above can be expressed as: :

G3

f 2 sin 3 φ 

8 π3 f r 3

:Equation A14.27:

Then f 2 sin3Ф is a constant, which it is. sinФ, sin2Ф, sin3Ф, sin4Ф… are equal at the ‘elastic limit’, after which point these trig functions disperse. : The Relationship between Density and sinФ And sin Ф = (1 / π) (d3/4 / r1/4) Equation A14.17 and π = ξ1/4 d3/4 / G3/4 from rearranging Equation 8.11. Then: :

sinφ 

G 3/4

d 3/4

d 3/4 ξ 1/4 r 1/4

:Equation A14.28:

Simplifies to: :

sinφ 

:

G 3/4 r 1/4 ξ 1/4

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:Equation A14.29:

Recall that the density is given by ρ = (1 / (2 π c d) ) (G3 / r ξ)1/4, Equation 54.16, therefore: :

sin φ 2 πc d

ρ

:Equation A14.30:

The Relationship between Angular Velocity and Ф As, sin2Ф = G a / ξ, Equation 11.30, and as the acceleration a is defined as, a = v 2 / r, Equation 11.10, then: :

sin2 φ 

G v2 ξr

:Equation A14.31:

G ξr

:Equation A14.32:

And

sinφ 

:

v

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Appendix 15 You have to believe in the magic of the universe to be able to tap into it. : The Momentum and the Kinetic Energy of the Phantom Particle, (faster than the speed of light). : The linear momentum, Mtum zip, of the phantom particle with mass m p and velocity Vzip is given by Mtumzip = mp Vzip

:Equation A15.1:

substituting mp = sinФ / 2 π f Equation 12.20 and Vzip = 2 π f (√(d + λ / π )), then the linear momentum is: : 2

Mtum zip 

2

2

sin φ λ2 2 π f (d 2  2 ) 2πf π

:Equation A15.2:

:The above simplifies to: :

Mtum zip 

(d 2 

λ2

) sin φ

:Equation A15.3:

1 (π 2 f 2 d 2  c 2 ) sin φ πf

:Equation A15.4:

π2

Where λ =c / f from Equation 2.8 so: :

Mtum zip  Note that (f2 d2) is a constant.

The linear kinetic energy (KEzip) of the phantom particle with mass m p and velocity Vzip is given by: :

m p Vzip

KE zip 

2

2

:Equation A15.5:

Substituting mp = sinФ / 2 π f Equation 12.20 and Vzip = 2 π f (√(d2 + λ2 / π2)) then the kinetic energy is given by: :

KE zip 

1 sin φ λ2 4 π 2 f 2 ( d2  2 ) 2 2 πf π

:Equation A15.6:

Simplifying: :

KE zip 

π f ( d2 

λ2 π2

) sin φ

:Equation A15.7:

) sin φ

:Equation A15.8:

Where λ =c / f from Equation 2.8 so: :

KE zip 

π f ( d2 

c2 π2 f 2

:

:

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Appendix 16 Use knowledge to the greater good of humanity. The Relationship between Qo and Q1 and Q2 : By rearranging Equation A11.1, Qo = Q / 16 π5 G3, recall that Q is defined as Q = √ (Q1 Q2), by squaring A11.1

Q1 Q 2

Qo  2

(16 π 5 G 3 ) 2

:Equation A16.1:

Recall Equation A8.9, Q2 / (16 π5 G3) = 2 εo f / ξ, by substituting the expression for Q2 into the above then: :

Qo  2

2 εo f Q1 ξ (16 π 5 G 3 )

:Equation A16.2:

As Q1 = ξ / f, Equation A8.2 then: :

εo

Qo  2

8 π 5 G3

:Equation A16.3:

Rearranging the above gives: :

8 π 5 G3 

εo Qo

2

:Equation A16.4:

Multiply both sides by 2 gives: :

16 π 5 G3 

2 εo Qo

2

:Equation A16.5:

: Equation A8.9, Q2 / (16 π5 G3) = 2 εo f / ξ, can now be expressed as: : 2

Q2 (

Qo ) 2 εo

2 εo f ξ

Qo 

4 εo Q1 Q 2

:Equation A16.7:

2 εo Qo

:Equation A16.8:

:Equation A16.6:

: As Q1 = ξ / f, Equation A8.2 then: : 2

2

Or: :

Q1 Q 2 

It is interesting to note that, Volts = 2 / (Qo λ), from Equation A11.4, so: : Qo2 λ2 Volts2 = 4

:

:

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:Equation A16.9:

References “To avoid being influenced by others, one must close themselves to the external world to come to know thyself.” : Reference 1 conveniently provides all the required fundamental physics required for this work in one source. Reference 1: Physics for Scientists and Engineers with Modern Physics, Fourth Edition, Author Serway Reference 2: The NIST Reference on Constants Units and Uncertainty (www physics.nist.gov/constants) : Other material which may be of interest to read: : 1) The Unified Gravitational and Electromagnetic Theory SKK Tomka, abstract APPC – AIP (2016) 2) Taking the Quantum Leap, Fred Alan Wolf 3) Physical Review Letters, 11 February 2016 Observation of Gravitational Waves from a Binary Black Hole Merger.

4) New Scientist, 02 January 2016 , Squeezing new particles from the LHC.

:

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Amendment Notes On a journey of discovery the un-trodden path leads to hidden treasures. : The changes to each new edition are usually to add new concepts or insights that align with my model. Amendments between this edition and the last published edition on Smashwords of ‘Theory of Everything’ are as follows: Addition of Equations: 11.32, 11.33, 11.34, 11.35,11.36, 12.7, 12.8, 12.9, 12.10, 12.11, 12.12, 12.13, 12.14, 12.15, 12.16, 12.17, 12.18, 12.19, 12.20, 12.21, 12.22, 12.23, 12.24, 12.25, 52.30, 52.31, 52.32, 52.33, A3.3, A3.4, A3.5, A3.6, A3.7, A3.8, A3.9, A3.10, A3.11, A3.12. A9.19, A10.18, A10.19, A10.20, A10.21, A10.22, A11.33, A11.34, A11.35, A11.36, A11.37, A11.38, A11.39, A1140, Insertion of Equation A13.5 Appendix 14 is added – boundary conditions Appendix 15 is added – Faster than the speed of light – the phantom particle Appendix 16 is added – The relationship between Qo, Q1 and Q2 Amendment to Equation 11.2 (sin Φ) becomes (sin2Φ) this follows through in Equations 11.20, 11.28, 11.29, 11.30 Equation 3.5 replaced. Reworked Appendix A8 to provide consistency of the definition of Q (i.e. Q = √(Q1 Q2) and expanded this chapter. Chapter 14 – Section on Tension deleted: : Amendments made in previous editions published on Smashwords: : This book is an update to the First Edition published on 23rd February 2016, which was called the Unified Gravitational and Electromagnetic Theory. The theory has been extended beyond this point and so the new title ‘Theory of Everything’ has been assigned. Appendix 6 is inserted - spacetime Appendix 8 is added - charge distribution Appendix 9 is added - alternative expressions G and π Addition of Appendix 10 – Chaos Addition of Appendix 11 – Permittivity of Free space a new understanding. Addition of Appendix 12: Addition of Appendix 13: An amendment has been made to Equation 52.2 (i.e. 16 replaced by 4) and so the amendment follows through to Equation 52.7, 52.9, 52.10, 52.11, 52.12, 52.13, 52.14, 52.16, 52.17 (now 52.18) The following equations have been added: 13.16, 13.17 17.10, 17.11, 17.12, 17.13, 17.14, 17.15, 17.16, 17.17, 17.18, 17.19, 17.20, 17.21, 17.22, 17.23, 17.24, 17.25, 17.26, 17.27, 17.28, 17.29, 17.30, 17.31

:

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19.2 27.8, 27.9 29.6 30.8, 30.9, 30.10, 30.11, 30.12, 30.13, 30.14, 30.15, 30.16, 30.17, 30.18, 30.19, 30.20, 30.21, 30.22 31.4 32.22, 32.23, 32.24 38.6 49.7 52.17, 52.19, 52.20, 52.21, 52.22 52.23, 52.24, 52.25, 52.26, 52.27, 52.28, 52.29 54.18, 54.19, 54.20, 54.21, 54.22, 54.23, 54.24 55.7, 55.8, 55.10, 55.11, 55 12, 55.13, 55.14, 55.15, 55.16, 55.17, 55.18, 55.19, 55.20, 55.23, 55.24, 55.25, 55.26, 55.27, 55.28, 55.29, 55.30, 55.31, 55.32, 55.33, 55.34, 55.35, 55.36, 55.37, 55.38, 55.39, 55.40 56.26, 56.27 A5.26.1, A5.28.1, A5.30.1 A7.1, A7.2, A7.3, A7.4, A7.5 A11.17, A11.18, A11.19, A11.20, A11.21, A11.22, A11.23, A11.24, A11.25, A11.26, A11.27, A11.28, A11.29, A11.30, A11.31, A11.32, A12.13, A12.14: Equation 11.11 has been inserted although originally included the equation was unnumbered. Insertion of Equation 54.18, 54.19, 54,20, 54.21, A9.16 Correction in Equation 56.1 (the amendment is followed through the chapter.) Correction to typo in Equation 55.36, 55.37 Quotations have been added:

:

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Even Earlier Amendments Notes: “It was curiosity that killed and saved Schroedinger’s cat.”

: Its primary modification is in the formatting of the equations, which the reader should now find much easier to follow. This current edition also has had some minor typing errors removed. Also: A reference to Kepler’s 3rd law is added for Equation 4.12 (March 2016) Extension to the dimensional analysis in Appendix 5 Insertion of Chapter 55 on electric dipole moment Insertion of Chapter 56 on sound Inclusion of dimensions in Chapter 57 Appendix 6 is added, The following equations have been added, Equations: 10.17, 10.18, 10.19, 10.20, 14.3, 14.4, 14.5 24.3, 24.11, 24.12, 24.13, 24.14, 24.15, 24.16 25.9, 25.10, 25.11, 26.6, 26.7, 26.8, 26.9, 26.10, 26.11, 26.12, 26.13, 26.14 28.5, 29.3 and 29.4 32.8, 32.9, 32.10, 32.11, 32.12, 32.13, 32.14, 32.15, 32.16, 32.17, 32.18, 32.19, 32.20, 32.21 34.3, 34.4 49.6 52.12, 52.13, 52.14, 52.15, 52.16, 52.17 :

MESSAGE TO THE READER : Should you find that your downloaded pdf file has any corrupted subscripts or superscripts (they should all be numbers or Arial letter font), then please contact the platform provider. : Finally, there is further work needed to explore these concepts in order to augment this study. Also to continue to work to expand these concepts into other areas of physics and science. :

:

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*** “Hence arises the beauty and the order we see in our Universe.” ** Dr S K K Tomka *

:

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