THE OLD CLASSICAL ATOM
THE NEW MODEL ELECTRON
The collected works from; a
BRITGRAV4 Conference talk,
a BRITGRAV5 Conference talk, the
"James Clerk Maxwell 150 Years On Conference" poster presentation
and an Institute of Physics poster presentation
©Dunstan Dunstan 2010, 2011, 2013, 2014, 2015, 2016, 2017 The Hague
MSc Applied Energy in Heat Transfer and Fluid Flow Cranfield,
MSc Instrumentation & Analytical Science UMIST,
Master Of Physics with Astrophysics Kent,
BSc Technology The Open University,
BSc Genetics and Biochemistry The Open University
Dedicated to The Merry Maidens of
Cornwall,
For The Republic of Atlantis and for republicans with The Hague around the World;
An anisotropicdipole gravitonelectron of 1/19 the protoncentre diameter with
a thin shell in a doublehemispherical formation
A proton of 1836 sphericalstationary (gravitonlocked) electrons
formed from a massiveprimordial electron cloud
A lengthvarying (absorptiontime dependent)
particlephoton mass
A gravitonelectron membranecollision (and magneton collision)
causing photon emission
Historical studies of “The
old classical atom” usually begin their personal study with the writer
depicting a relationship between various atomic constants, i.e. by dimensional
analysis, (without giving an illustration of the physical picture that is being
measured) and then continue with a description of the entire atom with only a
few sketches (which the writer hastily reminds the reader to not take
seriously).
The reader is initially
interested in the equations, but the writer fails to complete the picture,
(i.e. he claims that we can never see the Atom) leaving the reader uncertain
that the story is worth reading and is nothing more than a story. This makes you wonder what you are looking at
anyway, since it is well known in Biology that the eye can detect a single
photon.
This leaves us to complete
the task, e.g. one can throw in as many historical anecdotes as one likes in
order to keep the readers’ attention, or draw realistic illustrations of
experiments which can actually be photographed (see The Experiments.pdf).
Two types of stories emerge, one with endless equations that bore
the readers as they do not describe the atom and the second type where the
physics historian depicts how attempts were made to describe the atom, but the
physicists became bored with Classical Physics and gave up (deciding to invent
Quantum Mechanics and General Relativity, which are equally boring and fail to
give us a picture of the atom).
Since we are at a point where
artists are becoming interested in science and scientists are trying to become
artistic, it may be the right time to point out to physics historians what has
been overlooked by scientists in different epochs, i.e. because the moment was
not right for elucidation (due to various conflicting societal, political,
economical, militaristic and no doubt religious issues). This is all probably
due to the Devil.
Leaving religious mythologies
aside, by taking a cue from the Devil, as modernday Satanists advise us, one
can start with the era around the year 1913, i.e. just before World War One
began, when Bohr (having discovered the length of the Hydrogen atom 1^{st}shell
radius), promptly gave up on classical physics and wrote a famouslyoverlooked
memorandum to Rutherford describing his predicament. Bohr wrote that it is admittedly an impossible
task to describe the atom with Classical mechanics. He went on to invent (early) Quantum
Mechanics, which has not answered the question as to what the atom looks like,
nor answered the other questions on the same level concerning the electron,
proton, neutron, graviton, magneton, photon and any other particle which the
reader has in mind.
KEY TO FIGURE 1: A quick overview. In Fig.1 the electron graviton (not shown), is hypothetically
interacting with the electron surface at the electron's approximate centre, i.e. the graviton is
travelling orthogonally into the electron's centre at the velocity of light.
The electron is orbiting at 137th the velocity of light,
(i.e. the velocity of light times the Fine Structure Constant = 2.187 million metres per second).
The graviton hypothetically contributes one half its mass to the electron as it collides with the electron
surface, i.e. 1/2 mass times velocity squared = the energy of the collision.
This is gravitational deceleration (negative acceleration of the graviton), which should equal the inertial
acceleration of the electron's hypothetical magneton rim. It will be shown in the text below how both the
electron's radius and the electron's hypothetical thickness, (i.e. the proton's centre radius/19 divided by
the electroncharge ionisingvolume per 1st shellradius squared orbitalarea times unit radius, "e/radius^{2} x unit length"),
are found to equal approximately 0.212 Newtons for both the electron graviton's force of deceleration and the
electronmagneton rim's force of inertial acceleration. 0.212 Newtons = the Coulomb Force (see text for details)
divided by the cube of the Fine Structure Constant. (See the Poincaré Energy Equation, Equation 1 and Equation 2
for the mathematics about how the electron's gravitondecelerational forces equal the electron's
rimspin inertial forces. i.e. 0.212 Newtons)
0.212 Newtons is from the force equation:
"the electron mass squared x the gravitational constant and divided by the socalled Planck Length squared."
If we look at the classical mechanics of the reaction of photonemission absorptionphenomena, we
can find a relation between the stressstrain equation for circumferential (rim spin) and axial (electron gravitational)
deceleration.
D. J. Dunn in his treatment of classical mechanics, gives the elasticity modulus (measurement)
as being equal to the circumferential (electron rim spin) stress minus the axial (electrongraviton) deceleration stress. The Poisson ratio
is considered as being equal to 1, i.e. so the equation reads: 0.212 Newtons x the Planck Length/ [(4Pi x electron radius squared) x (the electron's
effective radius "thickness")] = the circumferential (electronrim spin) stress minus the axial (electron graviton) deceleration stress
The circumferential (electronrim spin) stress is a maximum of 1.022 MegaelectronVolts and the
axial (electron graviton) deceleration stress is a maximum of half 1.022 MegaelectronVolts, i.e. the 511,002 electronVolts which constitutes the normal massenergy of the
barepure electron with no photon inside of it. This means that the stress caused to the
electron's rim when it is decelerated by protonmagneton capture, combined with the stress caused by the electron's axial graviton
penetrating the deceleratedelectron's (innernegative tension) hemisphericalsurface, causes photon emission.
The numerical value we get, e.g. of 3.822091799 x 10^{13}kgm^{1}s^{2}
equates to the Coulomb Force via the electroncharge volume, the electroncharge volume to 1st shellradius squared ratio, the electron chargetomass ratio, the magnetic constant. the 62,584 magnetons
(which ionise the proton), Wien's photonlength displacement Law and the Fine Structure Constant. This indicates that the change in electron
deceleration due to protonmagneton (electroncapture). causes a change in the ratio of stress between the electron rim and the electron centre, i.e.
a change which causes a different photon length and photon temperature.
The electroncharge volume to electronmass ratio, times the magnetic constant per 4Pi steradian, [e/m_{e} x 4Pi x 10^{7}/4Pi] often shows up as a scaling factor,
i.e. involving gravitational equations which seem too large compared to wellknown standardphysics equations. This scaling factor, which results in units of 17,588 metres squared, is equivalent
to 511,002 electronVolts multiplied by the Fine Structure Constant and divided lastly by the 0.212 Newtons of gravitational force. 511.002 electronVolts is the energy per metre cubed of the standard
electron, e.g. the standard electron orbiting the proton's first shell in the ground state of the
proton at Absolute Zero (0 degrees in degrees Kelvin). The "Electron StressStrain Relationships.pdf", will attempt to simplify the terms used in the preceding
paragraphs.
Electron StressStrain Relationships.pdf
A good idea to start the
physical discussion of the atom involves the electronground stateorbit.
Figure 1 depicts the spinning electron (the small black dot to the right of the
arrows), the spinning magneton (spinning in the same direction as the electron)
and the 1836 stationaryelectron proton (to the left of the arrows). The
spinning magneton rotates as well (Oliver Heaviside Electrical Papers) and is
reabsorbed and reemitted continually by its attachedstationary electron
within the proton. This rotation is necessary, (as classically), any particle
which had no rotation would shear off and break from its connecting path (i.e.
the next junction point of its master particle), if it could not spiral into
the masterparticle junction to avoid a 90degree bend. The stationary
electrons are held in place by the gravitons (Dunstan BRITGRAV4 2004 RAL) and
do not move around as J. J. Thomson thought (at Cambridge university in the
"Cambridge Cavendish years" in the early 1900s), but are stationary (as Walter Kaufmann
thought from his Gottingen university studies during the
"Gottingen years" in the early 1900s).
The
electron in the picture, is travelling into the plane of the diagram and then
orbiting to the left behind the proton. It orbits within all of the 1stshell
magnetons and spincouples with them (only 1 magneton is drawn into the
picture), i.e. winding in the magneton (which is trying to expand all the way
out to the molar radius, at some 7.34 x 10^{10} m).
The electron in the picture,
has just been decelerated by the magnetons which it has contracted with and has
released a photon (the photon is not shown). The photon is spinning the same
way as the electron and the magneton in the diagram but it is travelling out of
the electron towards the viewer, i.e. at 180 degrees from and in the opposite
(linear) direction to the electron. The magneton in the diagram has an orbital
diameter of 5.2917 x 10^{11} m and forms the radius of the 1st shell
of the Hydrogen atom with this magnetonorbital diameter. The following
portabledocument filechart depicts how the
electron graviton would affect Hydrogenatomic orbitalparameters.The magneton and
electron 1stshell parameters are given in the 1stShell portabledocument file
.
Magneton StressStrain Relationships.pdf
A good idea
to start the mathematical discussion of the atom involves the term 8Pi^{2}
m_{e}/h^{2}. This term involves the graviton, electron,
proton, magneton and photon reactions in the atom’s groundstate orbit, (i.e.
where the free electron is decelerated at the molar radius from the velocity of
light to the 1stShell orbit at a lower velocity of c x F.S.C.). Newton’s 1st
and 2nd laws apply to this phenomenon.
M_{e}
= electron mass in (protonbound) groundstate at velocity of c x F.S.C.
2M_{e}
= electron mass in its free state at velocity of light
h = Planck’s
constant, Joules per frequency of particle cycle
13.60578693
Volts = magnetic and/or photonic forces per metre squared
Multiplying this term by the 13.605786 Volts that it takes to free the
groundstate electron with magnetism or accelerate it using photons to the
freeelectron state at the velocity of light, yields 2.22 x 10^{39} m^{5}.
This is equal to 1/e x 1stshell_{radius}^{2}
e = electron charge and proton charge
1stshell_{radius} = groundstate
orbitdistance from atom’s centre
Planck’s
constant refers to the amount of energy there is in one cycle of a particle,
e.g. in one photon cycle. Hypothetically speaking, though it has not been
done before, dividing by c^{2} gives the mass per cycle of a photon
(particle). Planck's constant can also be thought of as the change in areal mass per time change.
the electron mass multiplied by the square of the (electron chargevolume divided by
the square of the 1stshell radius) and divided by the atomic time of 752.88 seconds, yields Planck's constant,
i.e. divided by the ratio of Pi multiplied by the square of the Fine Structure constant.
The atomic time is found in several physics equations, e.g when manipulating Joules divided
by Watts or volume divided by Amperes. For example, the cube of the (electron chargevolume
divided by the square of the 1stshell radius) divided by the "atomic time"
equals the 3.311 mAmperes volumetric flow of the 1stshell magneton, i.e.
with a ratio of the Fine Structure constant squared (divided by 4) being applied.
The electron charge refers to the minimum volume (for the 62,584 magnetons to
occupy) in order for the magnetons to ionise the Hydrogen atom when the
electron is orbiting at the 1stshell radius in the ground state, i.e. at
absolute zero. This volume is determined by the molar volume (per
molecule) and the Faraday number. The molar
volume times the Faraday number = e.
The preceding equations give us the
Classical Grand Unified Field Theory, i.e. using Maxwell's Laws from Heaviside,
which can be applied to the Hydrogen Atom, based upon the classical Laws of
Newton, Galileo, Laplace, D'Alembert and Kepler.
The Grand Unified Field Theory.pdf
Grand Unified Field Theory Discussion.pdf
Ask your physicist friends
what the electron charge is and very few of them will tell you that it is the
volume of magnetonspace which causes the charge separation, i.e. this
volume of moving magnetons causes electrons to be attracted down the magnetons’
path toward a proton which has lost its own electron. This is of
course what we call electricity, but very few of your physicistfriends will
tell you how far this magneton path extends from atom to atom in a copper wire,
i.e. how does it overlap the next magneton (cathode to anode) pathway to cause
electricity in a copper wire?
Temperature is another
phenomenon which is much taken for granted, i.e. it is known as degrees Kelvin
but not by dimensional analysis (Amps squared per metre squared). Wilhelm
Wien reported that a photon’s length (in metres) divided into the charge
volume times c^{2}/5, yields the temperature of the atom which emitted
that photon. By dimensional analysis ec^{2}/(5 x the photon
length) yields m^{4}/s^{2}. Yet no one (including your physics friend)
will give units other than the Kelvin to temperature!
Sidgewick reported from
Oxford in 1950 that two protons will only combine to form the Hydrogen two
molecule if they both collide at the same place against the wall of the
container. This means that their collision area divided into their
positive charge current (which pulls a free electron away from its pathway
between atoms in the wall) is an example of what we call temperature.
If we multiply the amperage
of the Hydrogen atom by the amperage of the electroncharge volume in the wall
(which has pulled out the Hydrogen atom’s electron with its 62,584 magnetons,
i.e. the induction volume where the electron orbits circuitously within a
magnetic field continually) and we divide by the 1stshell radius of the
Hydrogen atom multiplied by the radius of our ionised electron in the induction
orbit, e.g. what we will call the simple–harmonic oscillatororbit, then
we get the maximum temperature of Hydrogen, 31,603 degrees Kelvin. The
units are Amps^{2}/m^{2} and we find our atomic scaling
factors, which we shall find explained in the first attached article (see
Maxwell's 150 Years.pdf below).
By the Uniqueness Theorem, we
shall find that all of our atomic equations which are in dimensional units of
Amps^{2}/m^{2}, can be related specifically to 31,603 degrees
Kelvin by atomic (dimensionless) scaling factors. This we can say classically
because temperature is due to an accelerated (or decelerated) charge, i.e. m s^{2}
x m^{3} = Amperes per metre squared. A decelerated free electron will
emit a photon and a photon's length determines the temperature.
Let us take another example
in our tour (or as another might say, our guided walk around the classical
atom). Our first example was 13.605 Volts x 8Pi^{2}m_{e}/h^{2}
= 1/e x 1stshell radius^{2}. This, (by the Uniqueness Theorem) =
1/Hydrogenminimum photonlength^{5} and some scaling factors, i.e.
[4Pi]^{5}/[62,584 x 10^{3} x 17,275 x F.S.C.^{5}] .
If we multiply this term,
(i.e. 2.22 x 10^{39} m^{5}
which is equal to 1/e x 1stshell_{radius}^{2}), by our
31,603 Amps^{2}/m^{2} and the square of the Hydrogenminimum
photonlength, we end up with 5.8494 x 10^{29} m/s^{2}, i.e.
acceleration. It is well known from the Classical Atom (Newton) that
inertial acceleration equals gravitational acceleration, i.e. where velocity
squared divided by radius = mass times Newton's gravitational constant divided
by radius squared. The maximum inertial
acceleration of the electron and its spin in the ground state is equal to
c^{2}/1stshell radius. 5.8494 x 10^{29} m/s^{2}
= c^{2}/1stshell radius and some scaling factors, e.g. the Fine
Structure constant, 8Pi and 10. The mass of the electron
times Newton's gravitational constant divided by the Planck Length squared,
will give the same answer, (i.e. with factors of 8Pi and 10 being involved).
A stepbystep rigorous proof for
this shall be presented later.
Newton said that matter and
light were obviously interconvertible. We may hypothesise that the
deceleratedfree electron emits light where (due to) the electron dipole
graviton interacting more quickly with the energyabsorption capability of the
electron membrane, as 2 Belgian women researchers (Betty and Yves see BRITGRAV4
Figure 1) and myself have implied. The gravitational constant and the
radius of the graviton, i.e. the Planck Length, must account for this. If we
multiply the electron mass by the gravitational constant, by the electron
charge, the square of the Fine Structure Constant and divided by the square of
the Planck length times 20 Pi, then we get the maximum temperature of Hydrogen,
( i.e. 31,603 degrees ).
If we multiply the electron
mass by the gravitational constant and divided by the square of the Planck
length, we can now say that the electron interacts with itself in the
groundstate orbit as the Coulombforce magnetons of the Hydrogen proton force
the electrondipole graviton to orbit the proton in the groundstate orbit and
couple with the back of the electron. We can hypothesise this because it explains
(classically) why the electron does not spiral into the proton centre as Bohr
said it must, i.e. as it lost energy by charging the proton. Henri Poincaré
(Dernier Pensées 1910) warned us to pay attention to Walter Ritz when he
hypothesised that the electron must undergo spincoupling, i.e. between a
"vorticalspinning electron" and a "vorticalspinning
magneton". Henri Poincaré was a French Mathematical Physics Professor of
the highest order. It was Mr. Henri Poincaré who actually wrote the nowfamous
equation e = mc^{2} (1898 Henri Poincaré). Some of the international,
(i.e. nationalist), press have attributed this equation erroneously (on
purpose) to someone else. This person was forced to admit some 40 years after
the great Frenchman's death that it was Henri Poincaré who wrote e = mc^{2}
and not him.
The equation actually means
(as well) that all energy changes (i.e. according to the 1st law of
Thermodynamics), occur with the second dimension being involved. This means
that photon emission from electrons, magnetons, isotope radiation and breaking
radiation all involve surfacearea equation terms, (i.e. phenomena involving
surfaces of photons or electrons). This seconddimensional term occurs
(Jointly) because the electron has forward (Linear) velocity (Newton's First Law) as well as
sideways (Spin) velocity (D'Alembert's Principle) at the same time. This is noted in the equation J = L
+ S, i.e. where the electron's "Joint" momentum is due to its combined
"Linear" momentum and its "Spin" momentum.
The electron's internal energy
is Mc^{2}. It has forward velocity components of up to "c" metres per second
and sideways velocity components of up to "c" metres per second. Since it has these two
components of velocity it must be a twodimensional object and have a surface
that constitutes its shape, i.e. it must have a very small thickness and
volume. The electron surface (membrane) must have the two velocity components
moving in the surface membrane all the time except when the electron is
travelling at the velocity of light. This is because the forwardlinear velocitycomponent
would be travelling faster than the velocity of light if it travelled at the
velocity of light within the electron when the electron was travelling at the
velocity of light, i.e. as a free electron. The electron would always have its
sidewaysspin velocity component except when it was captured as a Betaparticle
(i.e. by the proton) to form the neutron. If the electron had only two
dimensions to it, it would collapse when it collided with other electrons and
with atoms. It must have a threedimensional component and this we hypothesise
to be the graviton.
The term Mc^{2} stems
from Lazare Carnot's mathematical work during the French Revolution in the
1790s. Lazare Carnot was the father of Sadi Carnot, who is credited by Kelvin
and Clausius with founding the 2nd Law of Thermodynamics, the Law of entropy or
photon (heat) emission. It shows that you can get a lot from Science if you lop
off a few inbreedingmonarchs' heads, i.e. as Diderot noted for Cicero in
classical literature.
We can say all this due to
the result from the Classical Atom equation where m_{e}G_{o}/P_{L}^{2}
= c^{2}/1stshell radius and a scaling factor, i.e. The Fine Structure
Constant. The inability of Bohr (to use a graviton) in order to explain why the
electron remained in a continual groundstate orbit (i.e. according to
classical physics, as he wrote in the now infamous Rutherford Memorandum), led
him to abandon classical physics. Bohr went on instead to invent (the socalled
"early") Quantum Mechanics, which I quote as being “totally
unnecessary” in this report. It is interesting to note that Bohr could
not make use of the photon as a particle, i.e. since classical cause and effect
relations force one to use a gravitonphoton reaction together. The following
article shows how we could have been saved if we had read Oliver Heaviside’s
original work on gravitation (1893). It is from a poster
presentation I gave at the 150th anniversary of James Clark Maxwell at Aberdeen
University.
The earliermentioned
portable document, "Maxwell's 150 Years", portrays how all atomic
constants and fundamental equations can be depicted by the HeavisideMaxwell
equations. These equations directly describe atomic phenomena or use the atomic scaling factors;
17,275 the ratio of the
inductionorbit radius to the 1stshell radius in the Hydrogen atom (as well as
4Pi times the mass of the electron divided by Planck's constant, all in
dimensionless units), 62,584 (the number of magnetons required to ionise the
Hydrogen atom within the area defined by [Pi x the inductionorbit radius^{2}])
and the Fine Structure Constant. The Fine Structure Constant is not well
defined numerically, (see Wikipedia). It is best defined numerically perchance,
as an element of a ratio. For example, the 1st Shell radius of Hydrogen divided
by the Fine Structure Constant and Pi^{2} yields the radius of one
ionised proton, i.e. ~7.3474 x 10^{10} metre. The 1st Shell radius^{3}
x 17,275 x 62,584 x 10^{3} = the electron charge. The electron charge is
of course the electronproton ionisingvolume made up by the 62,584 parallel
magnetons, (which are part of the 13,605 x 10^{8} magnetons), which
compose 13.605 Volts. 13.605 Volts can ionise the electron and the proton. The
electronproton chargingvolume divided by the Faraday Number equals the molar
volume of one ionised proton. If we divide the molar volume by 4Pi/3 and take
the cube root, we arrive at the radius of one ionised proton. If we then take
the cube root of 17,275 x 62,584 x 10^{3}/{(4Pi/3) x the Faraday
Number}, we get ~13.88 or the reciprocal of The Fine structure Constant x Pi^{2}.
Pi^{2} is important
here for it is the ratio of the molarradius magnetonfrequency due to its
velocity at c to the 1stShell magneton frequency due to its velocity at c x
The Fine Structure Constant.
The Fine Structure Constant
is called the finestructure constant because it is used in equations to
explain why there are slightly different frequencies of red, for example, in
the second shell of Hydrogen, i.e. when an electron is caught by the 2nd shell
of a Hydrogen proton while it is orbiting the Hydrogen 3rd shell of another
proton. As a result, the Hydrogen proton yields up a band of slightly differing
lengths of the red photon, e.g. instead of the exact mathematicallypredicted
photonlength for red. It is assumed then that a change in the radial position
of the electron, i.e. when it releases the red photon, causes the change in the
colour of red. A slight change in the vertical position of the electron, i.e.
as if one hit the proton on the North or South magnetic pole when the electron
is releasing the red photon, causes a very fine difference the spin velocity
and hence the hue of the colour released. This is described as the Hyperfine
Structure Constant.
Photons are emitted when a
captured electron is decelerated to a lowervelocity orbitinglevel which is closer to the
Hydrogen (mathematical) atomic centre. If the proton is spinning backwards or
forwards, or if it is moving away from the captured electron or towards it,
then the photon length will be slightly shorter or longer. This is due to the
time of photon release being slightly shorter or longer.
In Classical Physics, the
decelerated electron always has a slower velocity while its graviton is still constantly travelling at the velocity of light. This phenomenon forces the electron's incoming graviton to convert itself at a faster rate into the electron's surface membrane, e.g. the electricconvection
potential (G. F. C. Searle 1897 Cavendish). The electron membrane cannot
contain this extra gravitonelectricconvection potentialmass within its control volume (Dunstan BRITGRAV 4 2004
Rutherford Appleton Laboratory). The extra mass is emitted as a photon. Matter
and light mass are thus mutually interconvertible by the Law of reversibility
of Light (Isaac Newton Cambridge).
The graviton would be
modelled as a helical coil which attracts matter gravitationally as it pierces it mechanically, (due to
its corkscrewlike construction). The graviton would operate mechanically by travelling through the gaps between subnuclear staticelectrons (e.g. as when we walk upon the surface of a planet) or by actually piercing the surface membranes of subnuclear staticelectrons. As the graviton pierces a surface membrane it would displace matter. This extra matter is the mass anomaly known in nuclear binding (and it is equal in massenergy terms to the radiation emitted when fusion occurs). The equation by which to model the graviton's volume ( for one helical cycle ) is connected
to the gravitational constant. As the previouslymentioned pdf shows, the
Planck length squared times Pi/4, multiplied by the gravitoncycle length gives
the graviton volume for one helical cycle. This volume multiplied by the graviton frequency squared and
divided by the mass of the electron, yields the gravitational constant (with a
coupling factor of 8).
BRITGRAV4 Figure 1: A typical classical/Quantum
membrane (of specific curvature) intersected by a graviton:
• would be under tension due to local momentum of
graviton. Yves Brihaye and Betti Hartmann have written on a negative tension
existing on the membrane, i.e. which would localise gravitons (Yves Brihaye and
Betti Hartmann 2004).
• would emit a photon as the graviton absorption rate
would change as the electron decelerated (Dick, R. and McArthur, D. M. E.
2002). Conversely, the electron membrane would accelerate as it reabsorbs
converted mass from a photon (Walter Kaufmann 1902, 1906).
• would spin at a velocity equal to its transverse
velocity, e.g. in J = L + S, the linear velocity equals the spin velocity.
• would spincouple with proton magnetons in a
vortical fashion (Poincaré 1910, Ritz 1911), i.e. exhibiting kspace interactions causing
magneton contractions as described by the Fermi vector.
• would follow the continuity equation of the First
law of thermodynamics, i.e. the matter flowing into the electron from the
graviton must flow back out of the electron into the graviton, if no entropy
exists such as light emission.
The 1stshell portable
document file and BRITGRAV4 Figure 1, should give an experimental idea on how
the dipolegraviton complex might interact within the Hydrogen Atom and what
the electron graviton might look like. A deceleratedfree electron will have
its axialspin velocity decelerated if the electron's axialspin velocity and the electron's forwardorbital
velocity are related. The Fine Structure Constant seems to be the parameter
which relates electronaxial spin, forwardorbital velocity and protondistance parameters with one another.
If the earliermentioned graviton does have a helicalcycle length which is directly proportional
to the gravitational constant, then there are 137 graviton cycles
within a single circumferential orbit of the electron in the 1st Shell. If the deceleratedfree
electron is travelling at the velocity of light in the 1st Shell before it is
decelerated, then since it is travelling 137 times faster than the
normalorbiting 1st shell electron, it will cover the distance in 1/137 the
time, i.e. the time which it takes the normalorbiting electron to cover 1
cycle of the alleged graviton length.
This because there are 137 gravitonhelical cycleunits (end to end)
in a single orbit of the electron flight around the proton (within the 1st shell).
This is the equivalent of 1 orbital cycle
for the free electron. If the free electron is decelerated within 1 cycle time
of the graviton unit length, then the decelerated electron would have to absorb
the extra energy of gravitonelectron membraneconversion, i.e. a photon would
have to be emitted. The graviton spin would have to be proportional to the
electron and photon spin, i.e. the spin change in the deceleratedfree electron
(at the velocity of light) due to the decreasing spin change in the proton's
magnetons would cause graviton conversion rates in the electron membrane to
build up photonreleasing pressure until the free electron was decelerated from
the velocity of light to 1/137 the velocity of light in the 1st Shell.
It is important to point out that it is the freeelectron body which is decelerated and not its graviton.
The graviton must always travel at the velocity of light (or it would become tangled up with itself).
Let us look at an example. If
a free electron is decelerated to 3/4 of its speed from the velocity of light due
to its capture by the 2nd Shell of the Hydrogen atom, then it will have lost
1/4 of its forward velocity and release a photon which is 4 times as long as the photon
released when a free electron is decelerated to the 1st Shell. For a free
electron decelerated to the 3rd Shell, 4th Shell, 5th Shell and 6th Shell, the
electron will have lost 1/9 its (forward) velocity, 1/16 its (forward) velocity, 1/25
its (forward) velocity and 1/36 its (forward) velocity. The photons released will be
9 times, 16 times, 25 times and 36 times longer, i.e. 9 times, 16 times,
25 times and 36 times longer than a photon released by a 1stshell electron.
This is shown by experimental data from Aangstrom, Rydberg and others.
The electron's (spin) velocity, on the contrary, should increase,
in order to compensate, i.e. in order to follow conservation of momenta laws,
(as well as conservation of angular momenta laws).
Let us look at another
example. From the 1st Column in our 1stShell pdffile mentioned earlier, we
can see that the number of electron orbits in the singlehydrogen atom’s
groundstate orbit is exactly twice the Hydrogenmaximum photon
emissionfrequency, i.e. twice 3.289 x 10^{15} Hz. From the same chart
the maximum 1stshell magneton orbit frequency is twice the electron frequency,
i.e. 4 times the maximum Hydrogen photonfrequency. It follows that the last
magneton shell in ionised Hydrogen has an orbital frequency of Pi^{2}
times the 1stshell magneton orbitalfrequency.
The frequencies are
determined by velocities and the velocities are determined by the Fine
Structure Constant. If we multiply the velocity, i.e. the velocity of the last
shell magneton in Hydrogen by the time of the last shell's magneton
orbitcycle, then we get the distance, e.g. the molar radius times Pi. Multiplying
by the frequency ratio of the last shell of molar Hydrogen to the Hydrogen
maximum frequency, i.e. multiplying by 4Pi^{2}, gives the Hydrogenminimum
photonlength.
Let us
recap what we have just said in the last two paragraphs. The length of the
photon released by the decelerated electron is due to the exact change in spin
between the free electron and its decelerated spin velocity, i.e. Classical
Mechanics and Classical Physics laws are upheld exactly. There are no
mysterious mathematical coefficients and so the Second Law of Thermodynamics
does not apply, i.e. the entropy itself is in the photon emission. Only the
First Law applies and is needed. The frequency change (its ratio) between the
lastmagneton shell in molar Hydrogen and the Hydrogenmaximum frequency, (i.e.
the photon frequency released in the 1st Shell), determines the length of the
photon released. The outer magneton and the "free" electron, both travel
at or near the velocity of light. The electron spin increases as the "free" electron
is decelerated towards the 1st shell, so the length of the released photon
will be shorter if the "free" electron spins faster for 1 electronrim revolution.
If one had looked at Planck’s
Constant and could have divided by c^{2} Joules per kilogram, one
arrives at the amazing mass of a photon particle, i.e. some 10^{51} kg
per cycle. If we multiply this by the frequency of the proposed graviton, e.g.
some 10^{20} Hz, then we arrive at the mass of the groundstate
electron. From our previous paragraphs on the mass of the electron, we can see
that the mass of a photon was a minimum of ¾ the groundstate electron (for a
1st to 2nd shell transition), 8/9 for a 1st to 3rd shell transition, 15/16 for
a 1st to 4th shell transition and so on up to the mass of the electron.
Assuming that the shell
distances are the same, then there would be about 64 shells from the 1st shell
out to the Hydrogen molar radius at about 7.34 x 10^{10} metre out.
The smallest mass would then be about 1/63^{2} – 1/64^{2}. This
is much larger than 10^{51} kilogram per photon.
One arrives at the amazing
conclusion that the 10^{51} kilogram must be the mass of a photon
particle, i.e. a subparticle.
Classically
speaking, i.e. according to the laws of Classical mechanics, a photoncycle unit
would commence within a gravitoncycle unit, A photoncycle unit would consist
of some 10^{20} individual "helical cylindricallike wires"
of individual single twists. These helicallike twists
would all start near the front of the photon, i.e. within the start of a gravitoncycle unit
which is about to touch the inside of a decelerating electron. The twisted "helicallike"
photon/graviton subunit particles would end near the back of the
same photon. This geometry would allow the graviton/photon subunits to overcome
the surface tension of the electricconvection potential of the electron, i.e. the
surface tension which keeps the electron surface intact.
This would be because the combined pressure of all of the 10^{20}
some photon points in a single graviton cycle unit would pierce the
electricconvection potential of the electron membrane. In turn, the graviton
intersection points with the electricconvection potentialmembrane would be
the propagation points for causing the graviton subunits to turn inwards, i.e.
towards the centre of the electron. The graviton subunits would run into each
other at the centre of the electron hemisphere and reverse direction, (e.g.
their forward direction would gradually become a sideways direction and then
they would turn 90 degrees at the point where they collided at the very centre
of the hemisphericalelectron surface). The photon units would now emanate backwards
out of the electron but with the same spin direction as the electron. The
latter two points are well known in physics laboratories.
If some 1.2355 x 10^{20}
photon subunits compose a photon and the photon is released from within the
graviton, i.e. when the graviton collides into a decelerating electron, then
one can assume that the gravitonunit cyclelength is also composed of 1.2355 x
10^{20} photon subunits. We can assume this as our previous discussion
explained classically how the deceleratedfree electron released a photon when
its velocity was changed from the velocity of light to the Fine Structure
Constant times the velocity of light, (e.g. as in the 1st shell orbit of the Hydrogen atom).
If we have defined a
gravitonunit cyclelength as 2.4263 x 10^{12} m, then this length
divided by 1.2355 x 10^{20} gives 1.9636 x 10^{32} m as the
distance between photon subunits overlapping, e.g. like fibres in a string.
From the “Maxwell’s 150 Years pdf” we defined the volume of a graviton unit
cycle as P_{L}^{2} x Pi x 2.426 x 10^{12} m. If we
divide this volume by 1.2355 x 10^{20} then we get a particle that is
2.426 x 10^{12} m long with a radius of 1.4527 x 10^{45} m.
In the
normal graviton the 1.2355 x 10^{20} photon subunits must overlap one another, i.e.
as fibres do in a string. Classically speaking, this is the only manner by
which there can be coherence to a photon subparticle gravitonunit.
The value of the radius of
the photon subunit particle is mathematically significant. If we multiply Pi
times the square of this 1.4257 x 10^{45} m by twice the radius of the
universe (the graviton orbital radius of 1.1784 x 10^{56} m), we get
1.5626 x 10^{33} m^{3}. If we divide this volume into the
proposed overlapping distance of the photon subparticle units, e.g. 2.426 x 10^{12}
m divided by 1.2355 x 10^{20} cycles, we get 4Pi m^{2}, e.g. a
steradian per m^{2}. The metre squared term can be explained by the
derivation we got for the radius of the universe in the “Maxwell’s 150 Years
pdf”, i.e. we got 4Pi x m_{e} x G_{0}/1 m^{2} to be
dimensionally the reciprocal of the number of gravitons per metre square per
steradian to give 1 ms^{2} acceleration. Multiplying the reciprocal of
eq. 17 (i.e. in the “Maxwell’s 150 Years
pdf”), by c^{2} m^{2}s^{2} gives the radius of the
universe per steradian, e.g. in metres squared.
This is the unit of all
matter, i.e. it must flow through the graviton, the electron membrane, the
magneton and all parts of the proton and neutron. In fact, by the 1st Law of
Thermodynamics, it must commute through all nucleons via their gravitons and be
released as photons, whether by synchrotron emission, photon release, gammaray
emission of isotopes or braking radiation.
BRITGRAV4 Figure 2: A depiction (for discerning
mathematicians) of the orbitalground stateelectron. The 9 graviton units
depicted (not to scale) are proposed to emanate out of the front of the
hemisphere, travel out to the edge of the Cosmos and circle back in behind the
back of the electron hemisphere. This depiction of the orbital electron is from the BRITGRAV4
Annual Conference at the RutherfordAppleton Laboratory in 2004. Further research indicates that the
electronrim thickness is more likely to be the electron radius/(the electron charge/the square of the 1stshell radius).
The gravitoncycle units (not to scale) should be approximately 2.42631 x 10 ^{12} metre.
(See text or author for details)
The electron itself is the
most elusive little devil. As the table on inertial acceleration shows, e.g.
column 1 row 5, every time that one tries to put the electron radius, i.e.
7.415649545 x 10^{17} m, into an equation, the coefficient 19 (or 38)
shows up. One thus gets the protoncentre radius (1.4809 x 10^{15} m)
or the fusionapproach radius (2.8179 x 10^{15} m). The
fusionapproach radius of 2.8179 x 10^{15} m, is the measurement from
the centre of the proton to the points where the (nonorbital) magnetons form the
fusion barrier. The fusion barrier is mathematically the value of velocity
which a nucleon must have to allow its own gravitons to pierce the
(nonorbital) magneton layer at the fusionapproach distance and to pierce the
proton’s electrons. When the nucleon’s gravitons pierce the proton’s static
electrons, they displace a certain amount of staticelectron mass, i.e. the
mass anomaly of fusion. This energy quantum is always equal to the massenergy
of the radiation released from within the proton’s (nonorbital) magnetons,
e.g. when the approaching nucleon’s gravitons pierce the (nonorbital)
magnetons’ barrier at the fusionapproach distance. Hence, most classical
equations have always described the fusionapproach radius of 2.8179 x 10^{15}
m as the electron radius when it is the radius of orbit of the proton's (i.e.
the static electrons') magnetons. It is the closest point of approach of a
nucleon to another nucleon if the nucleons combined velocities are less than
the fusionapproach velocity. This is the Thomson crossover section radius,
e.g. where Rutherford showed Helium atoms bouncing off Gold.
BRITGRAV4 Figure 3: A depiction of
the cutoff section of the equatorial plane in a hypotheticalproton centre. If
one packs 1836 staticelectron spheroids in 6 concentric layers, then one
discovers that one has constructed a crystal. This crystal would have a flathexagonal
Northpole top and 6 trapezoidal sides sloping down to the equator
and 6 trapezoidal sides sloping down to the flat hexagonal Southpole bottom.
(see BRITGRAV4 Figure 4)
BRITGRAV4 Figure 4: A depiction of the
hypotheticalproton centre. The magnetons emanating and returning vertically (a
few shown in the diagram) would be locked into their 1stto6thshell radii by
the (not shown) horizontally outgoing and incoming gravitons (in the equatorial
plane) of the protoncentre crystal. (see PROTON.pdf)
BRITGRAV4 Figure 5: A
depiction (for discerning physicists) of the neutron centre. The 9 graviton
units depicted (in figure 2 emanating out of the front of the
doublehemisphere) are now counted as 18 i.e. 9 outgoing gravitons and 9
incoming gravitons make 18 in total. This allows the staticspheroidal
electrons in the neutron centre to obey the closepacking laws for spheres. The
arrows at the graviton unit ends indicate the direction of the 18 gravitons
which connect the central betaparticle to the surrounding static electrons.
(See text for details)
BRITGRAV4 Figure 5 depicts
the absolute centre of the neutron together with the first shell of the Neutron
(and proton), i.e. the 18 surrounding static spheroids which form around the
central blue spheroid according to the closepacking laws for spheres. The
central blue spheroid represents the Betaparticle which Walter Kaufmann proved
to be the electron, i.e. from some Radium which he got from Marie Curie in
1901. The centretocentre distance from the centralblue staticelectron to the
surrounding 18 static electrons is 1.5 electron radii. The centralblue
betaparticle and the sixsurrounding blue staticelectrons are all in the
equatorial plane of the neutron (and proton). The orangeelectron spheroids lie
above the equatorial plane and the greenelectron spheroids lie below the
equatorial plane. As the static electrons are 1.5 radii apart, i.e. due to the closepacking laws for spheres, the 1st shell is 1.5 radii from the centre, the 2nd shell is twice that, the 3rd shell is thrice that, the 4th shell is quadruple that, the 5th shell is quintuple that and the 6th shell is 6 times this ratio. This can tend to explain how Balmer derived the formula for explaining the ratio between photons released from the 1st, 2nd, 3rd, 4th, and 5th shell electron captures and the inverse square law for distances (e.g. as explained earlier regarding photonlength changes and electron shell velocities).
The neutron structure follows
the Proton structure depicted in the Proton.pdf and BRITGRAV4 figures 3 and 4.
The incoming and outgoing graviton units attached to the static spheroids in
BRITGRAV4 Figure 3, i.e. in the equatorial plane of the proton, are proposed to
form a mechanical lock on the 1st shell, 2nd shell, 3rd shell, 4th shell, 5th
shell and 5th shell magnetons of the proton. This mechanical lock is proposed
to be the phenomenon which holds the 6 magneton shells in their positions and
maintains their radial distance from the proton centre. It is interesting to point out that the subprotonic muon particle still has Hydrogenlike shelldrop emissions, i.e. even after it is broken down from a protoncosmic ray collision. This means that the magneton shells of the muon are still at the same radial distance (from the muon centre) as the proton's magneton shells are from the proton's centre.
The absorption of an electron
by the proton (e.g. the white central spheroid in BRITGRAV4 Figure 3 or the
blue betaparticle in BRITGRAV4 Figure 5), is proposed to be the phenomenon
which causes the contraction of the proton magnetons to the fusionapproach
distance, i.e. 2.1879 x 10^{15} metre instead of the molar or atomic radius. The magnetoncollapse
phenomenon is proposed to be due to the central betaparticle being able to
reroute the incoming and outgoinggraviton units (in a manner which prevents
the equatorialplane gravitonunits from forming a mechanical lock on the
proton's six magneton shells).
The rerouting of the
proposed subatomic mattercorpuscles (see earlier section on the graviton
subunits) could also help to explain how a proton magneton could expand and
contract radially, (i.e. centrifugally from the proton's south magneticpole),
without making the proton heavier or lighter. If the proton centre could absorb
subatomic mattercorpuscles from the graviton flow, (through the staticspheroids surfacemembrane and
then into the magneton), then the phenomenon of mass commutation via subatomic
mattercorpuscle flow could explain experimental recording of magneton
characteristics. These subatomic matter corpuscletransfers could occur in
quantum units, i.e. in much the same manner as photons account for mass
commutation in superatomic corpuscleunits.
Using our model, e.g.
BRITGRAV4 Figures 1 and 4, we can attempt to explain what phenomena occur
during betaparticle capture. The incoming betaparticle ( BRITGRAV4 Figure 1 )
will have its gravitons penetrate the Northpole (the top) of the proton in
BRITGRAV4 figure 4 and travel inside the proton (along the North poleSouth
pole axis) until the gravitons exit via the South pole (the bottom of the
proton, e.g. BRITGRAV4 figure 4). As the betapartcle travels along the inside
(hollowed) northsouth polar axis, (see figure BRITGRAV4), the betaparticle
and the proton begin to react with one another, i.e. via their mutualgraviton interectionpotentials.
The 6 magnetons depicted in
BRITGRAV4 Figure 4 form the 6th magneton shell of the proton and there are 6
shells (5 not shown) where a further 6 magnetons (per proton shell) form the 6
shells. This makes 36 magnetons in all, which mathematically leaves 1800
magnetons to form the fusionapproach barrier at a distance of twice the protoncentre
radius (from the proton centre).
The 36 magnetons, those which form the 6 shells of the
proton, will contract their orbital radii upon Betaparticle capture by the proton's centre.
The magneton flow must be diverted by the phenomenon of Betaparticle capture.
The logical place for the flow to be diverted to would be the graviton/electron
surface area on the incoming Betaparticle. Since the graviton has been estimated to be
4/7 the mass of the electron, i.e. 9 x 32,444.608 eV, then if the graviton would be found to
be 4 times the mass of the magneton, then 4 x 9 Betaparticle gravitonunits, would equal 36 magneton units.
An experiment needs to be done to corroborate this hypothesis.
It is
hypothesised that when the betaparticle’s gravitons exit from the South pole
(as depicted above) they interact with these hypothesised 1800 magnetons and
are bent outward (so that they reverse direction and travel inside the orbit of
the 1800 magnetons at the fusionapproach distance). This would tend to force
the betaparticle’s gravitonsurface contact point to be reversed also and the
gravitonsurface contactpoints would tend to be forced backwards towards the
centralmagneton rim of the electron (see BRITGRAV4 Figure 2).
At the time of betaparticle
penetration (into the North pole of the proton in BRITGRAV4 Figure 4) the
gravitons of the static electrons in the northern half of BRITGRAV4 Figure 4
would tend to enter into the back of the electron (BRITGRAV4 Figure 2) and tend
to invert the innerelectron hemisphere. This hypothesised inversion of the
electron’s inner hemisphere would tend to force the incoming graviton’s surface
contact points out toward the electron’s magneton rim (i.e. where they would be
in confluence with the outgoinggraviton contactpoints of the electron
surface). At the time of betaparticle penetration (into the centre of the depicted
proton in BRITGRAV4 Figure 4) the gravitons of the static electrons in the
southern (lower) half of the depicted proton in BRITGRAV4 Figure 4 would tend to enter into the front of the
electron (BRITGRAV4 Figure 2) and tend to keep it hemispherical.
The rotatory flow of
subatomic corpuscles about the gravitoncontact points in the electron surface
will tend to find the lowest entropy (the mathematical point of focal
stability), i.e. the lowest entropy in classical thermodynamics. It was
Heaviside (Electrical Papers) who first had the idea that the magnetons might have
a rotatory flow as well as an orbital flow (so we use his word rotatory instead
of any other word). This convergence of subatomic corpuscular flow of the
gravitons (about the equator of the nowformed betaparticle spheroid) will
tend to disrupt the electronmagneton rim (BRITGRAV4 Figures 1 & 2). The
subatomic (circumferential) corpuscular flow in the betaparticle rim will be
forced temporarily inside the betaparticle spheroid by the rearrangement of
the gravitons, i.e. from a dipole flow in (BRITGRAV4 Figures 1 & 2) into an
isotropic flow in (BRITGRAV4 Figure 5).
The lowest entropy state for
the Hydrogen proton (in classical thermodynamics) is an isotropicgravitational
state (field) combined with a dipolarmagnetic state (field), i.e. as we know
exists in the ionised proton classically. The magneton of the betaparticle
would have to emanate out of the south pole of the neutron (i.e. parallel to
the other magnetons of the neutron) and would have to travel out around the
neutron. The betaparticle magneton would now have to be reabsorbed by the
neutron at its North pole and return to the betaparticle. The other manner
in which the betaparticle magneton could be viewed is in the following fashion.
The betaparticle magneton might sit just above the surface of the betaparticle at
the neutron centre, i.e. some sort of betaparticle restructuralisation of its surface must occur.
The manner in which to test between the two experimental hypotheses is to test for
anisotropic versus isotropicneutron spins, i.e. due to the presence or absence of
the betaparticle magneton at the neutron surface. This is the same
type of test which might be made to test for the gravitational isotropy of the proton
versus the gravitational isotropy of the neutron, i.e. due to the gravitising
restructuralisation of the proton/neutron change when the gravitising electron is
captured by a proton.
There is usually no emission of
energy (other than an electron neutrino) in betaparticle (KShell) capture, (i.e. if the atom is protonrich
and if there is an energy difference between the
new and old atoms of less than the restmass energy of the free electron)
[http://radiopaedia.org/articles/betadecay, http://en.wikipedia.org/wiki/Beta_decay],
so by Maxwell’s laws, the massenergy of the
betaparticle must be added to the proton. This is accounted for by the extra
mass of the neutron, e.g. the neutron has a mass of approximately 2 and 4/7
electron masses more than the proton. 2 of these electron masses are accounted
for by the groundstate mass of the electron and the unbound state of the free
electron, i.e. its classical mass increase due to classicalphoton capture
(Kaufmann 1906). The 4/7 times the mass of the groundstate electron would be
accounted for by the absorption of the betaparticle’s graviton units by the
rest of the static spheroids in the proton. Since, classically, the ability of
nucleons to absorb gravitons during fusion is due to the ability of nucleons to
emit radiation ( from within the 1800 magnetons ) during fusion, there would
have to be a 4/7 mass increase due to there being no radiation emission during
betaparticle capture. Radiation emission during normal betaparticle capture
will occur if the absorbed electron has a greater energy than twice the restmass energy,
i.e. if the energy difference between the old and new atoms (nucleons) is greater than
1.022 megaelectronVolts. The other way to account for the increased mass of the neutron,
i.e. when compared with the proton, is to state that
restructuralisation of the graviton matrix within the neutron (see BRITGRAV4 Fig. 5)
causes the extramass anomaly. This means that the gravitons within the neutron
are now penetrating the neutron matter, whereas before these specific neutron
gravitons were penetrating free space within the neutron.
The fusion of a proton to a neutron is slightly different from the "fusion" of an electron to a proton and progresses classically. A proton approaching a neutron in our Sun will have its forward (transverse) velocity and its rearward (incomingtransverse) velocity static, i.e. in a specific relation between the proton's travellinggravitons and the proton's surfacemembrane electrons. The proton's and neutron's magnetons, (i.e. at the fusionapproach distance), will be compressed by each other and by the (fixed) vector positions of the gravitons, (e.g. of each particle). The principle of lowest entropy will apply, (from Clausius) and the gravitons of each particle will have to cut through each other's magnetons to reach a newstable point of lowest entropy (Clausius). The gravitons cutting of the magnetons flight path (at this close range) will force the magnetons to release energy, (i.e. in the form of photons) and this energy will equal the massenergy of the radiation which is emitted, e.g. which is equal to the "massanomaly" (massenergy radiationvalue) commonly found in fusion experiments.
From BRITGRAV4 Figures 1 and 2; for the orbiting electron, from BRITGRAV4 Figures 3, 4 and 5, for the stationary Betaparticles and electrons within the atom, we can see that the inflowingoutflowing gravitons and magnetons must interact with the membranesurface, e.g. the electricconvection potential of G.F.C. Searle (Cavendish 1897). From Robert Turnbull (Glasgow 1979), we know that the proton binding energy is 1.072229 GeV. This equates to 32,444.608 eV per graviton unit in BRITGRAV4 Figures 2 and 5, e.g. 32,444.608 eV times 18 graviton units times 1836 electron masses equals 1.072229 GeV. 32,444.608 eV equates to 4/63 the restmass energy of the groundstate electron, i.e. 511,002 eV. Magnetons are known to lengthen and contract, i.e. in the ionised proton and in the solarmagnetic field. It is now proposed that the lengthening of magnetons can be explained by the conversion of gravitonunit cyclelengths into magnetonunit cyclelengths, i.e. through diversion of the Newtoniancorpuscular atomic subunits through the electricconvection potential of G.F.C. Searle.
Our figures show us what the atom could be like, i.e. for simple Hydrogen and the neutron. So what do Helium, Lithium and the rest of the nonradioactive atoms look like, i.e. in all their elemental Fire and Brimstone, as the good devil no doubt intended? In the 1920s and 1930s it was (classically) thought that the atom consisted of stacks of Helium atoms or Lithium atoms placed upon one another. If you can imagine 2 Deuterium atoms fused together so that the 4 nucleons are equidistant from one another, (all lying in the same plane), with their South poles pointing towards one another, then you can imagine Helium.
With Lithium, the protons are at the tips of the equilateral triangle and the neutrons are between the protons, all lying in the same plane.
Beryllium would consist of 2 Helium (Alpha particle) atoms stacked on top of one another, i.e. at right angles to one another.
Boron would consist of a Lithium atom fused to 2 protons, i.e. 1 proton would lie on top of the Lithium particle and the other proton would lie beneath the Lithiumparticle equatorialplane, towards the centre of the Lithiumparticle but on its northtosouth polaraxis.
Carbon would thus be 3 alpha particles stacked on top of one another, again at right angles to one another. This is evidenced by the glucose molecule where the socalled "carbon chair" molecule predicates that the 4electronic bonds of the organiccarbon molecule must come from the 4stationary protons which lie on the top and bottom of the triplealpha particlestack, as the central pair of carbon protons are known not to take part in molecularelectron bonding with organiccarbon molecules.
At this point we can see that Bohr was wrong in thinking that nucleons moved around within the nucleus as we can see that there is no proof of this, i.e. the proton (molecular) bonds in organic carbon do not move (or switch position at Absolute Zero). We can also see why Bohr was wrong, i.e. Bohr did his MSc on surface tension in water molecules and as a professor he also did experiments on water molecules, so the idea for moving nucleons came from this. We can see that by probing totallyionised elements with laser photons within the coherence length, (i.e. the length from the end of the laser to where the photons are still in known positions with respect to one another), we can attempt to detect the constantgraviton density between nucleons.
From Britgrav4 figures 3 to 5, we can hypothesise that the nuclearbinding energies
between the static nucleons are determined by multiples of 68 graviton units, For example the energy required
to remove the neutron from the proton in Deuterium is 68 times 32,444.608 eVolts. The energy required to remove
the neutron or the proton from Helium 3 is 3 times 68 x 32,444.608 eVolts. The energy required to remove the neutron
or the proton from Helium 4 is 9 times 68 x 32,444.608 eVolts. The total binding energy of the nucleons in Helium 4
is then 13 times 68 x 32,444.608 eVolts, (i.e. 28.681 MegaeVolts).
32,444.608 eVolts is the energy per metre cubed, of 1 graviton unit, which was derived classically
from "proof by construction" due to classical experiments on the breakup energy of the proton,
(e.g. due to hitting the proton with a photon of the Compton Length). We can now move on towards
discussing the electron's radius, together with its internal graviton.
The
electron radius is one of the most misunderstood, e.g. misinterpreted
measurements in Classical Physics. It was thought to be 2.8179 x 10^{15}
m from the equation "e^{2} x The Magnetic Constant per electron mass per
4Pi steradian". This would make it larger than the proton, i.e. an "impossible"
idea, since we have just seen from the earlier discussion that the electron
must fit inside the proton, (as a neutron (sub)particle). The electron was also
thought of as a point charge, i.e. a point charge has no metric width or diameter, much
less 2.8179 x 10^{15} m. In 1906 Planck was studying the
electronmass uptake, which increased the electron velocity as mass was absorbed by
the electron due to photon aborption (as Walter Kaufmann had
proved in 1906). Henri Poincaré pointed this out in 1910 (Poincaré "Dernier Pensées" 1910),
but the world refused to take note and went on to follow Bohr with his "Rutherfod memorandum" in 1913.
Planck wrote that it would be important to know the ratio of the electron mass times its
velocity to its radius. This would give the mass flow of the electron
rim and hence by the continuity equation, this would give the mass flow of the
electron surface (G.F.C. Searle 1897) and the electron graviton throughout the entire electron,
This would be due to the 1st Law of Thermodynamics. The mass
flow times the volumetric flow per metre would give the energy of the electron,
"kg s^{1} x m^{2} x s^{1} = Energy" in units of
Joules.
[ M_{e}c / Eq. 1 ] x h / M_{e}
= M_{e} c^{2} x [ 4Pi x 10^{3} / 17,275 x F.S.C.^{3}]

(Planck Massflow Equation) 
We may now see that the Planck
Equation can be set equal to the PoincaréEnergy Equation " energy in Joules = M_{e} x c^{2}
" from Henri Poincaré in 1898. His thoughts are a progression of the work of Lazare Carnot
in the 1790s, i.e. concerning "mass times velocity". If we multiply the Poincaré Equation by 2Pi
then the energy of the electron radius and spin can be set equal to the energy
of 1 graviton cycle, i.e. the force of the graviton (0.212 N) times the length
of one gravitoncycle unit
M_{e}c^{2} x 2Pi =
[ M_{e}^{2} x G_{0} / P_{Length}^{2}
] x 2.4263 x 10^{12}m 
(Poincaréenergy Equation) 
Equations concerning the
electron radius are mostly applied to reality by considering the electron
torque, i.e. due to its physical spin. When we consider the electron in the
groundstate orbit, e.g. as in Figure 1, we can see it depicted as spinning in
the same direction as the 1st shell magneton inside the orbit of the 1st shell
magneton. Hence the electron radius and torque are lost within the measurement
of the 1st shell radius, i.e. the far greater value of 5.291783 x 10^{11}
m.
However, when the electron is
in the inductionstate orbit of 9.1427 x 10^{7} m, it is spinning on
the outside of the 62,584 magnetons and it is spinning against the direction of
the magnetons. The electron radius and its spinning torque, i.e. due to its
selfgravitation, become very important here mathematically.
The electron, named for Electra, the daughter of Agamemnon,
i.e. for her amber hair, and known from the ancients, onwards, to electrify men, their thoughts and science,
may be solved finally by us, e.g. by breaking the riddle of the Fine Structure Constant
and the proof by construction, (by experiment), of the electron radius, orbit as well as its thickness.
If we multiply the electron
radius by the electronselfgravitating force, we have Newtonmetres in Torque.
This value should be equal to the force of the 62,584 magnetons and their
electronorbit radius of 9.1427 x 10^{7} m. Instead of their 1:1
ratio, we find a coupling constant whose value is equal to the electron charge
divided by the 1stshell radiussquared per unit length, i.e. e/1st shell
radius^{2} times unit length. There is also a small (coupling)
coefficient involved, i.e. 1.000082877. This coupling coefficient is equal
to: 1 + ( 19 divided by the Fine Structure Constant divided by Pi, divided by
10^{7} ).
When attempting to insert the electron radius into any equation, i.e. in order to test what to see
can happen to atomic relationships, simply insert the
following term
e_{r} x 1.000082877
x 1st shell radius^{2} x unit length x e^{1}
= 1.29620937 x 10^{18} metre

(1) 
and proceed with your physics equation.
The derivation of equation one.pdf
This new term, (i.e. Eq. 1),
might be the term for describing the actualveritable thickness of the electron surface,
e.g. the thickness of the "electricconvection potential",
which G.F.C. Searle described in his 1897 published work;
On the Steady Motion of an Electrified Ellipsoid. This means that the surface thickness is the next
allimportant classicalatomic factor
which we muct discover. We can say this because if the electron is hollow
(as G.F.C. Searle states that both he and Heaviside agreed that the Electricfield vector
E and the Magneticfield vector H must be zero at the electron centre), then
the Electricfield vector
E and the Magneticfield vector H must also be orthogonal to each other,
i.e. for Maxwell's Laws to apply at the atomic level.
If the Electricfield vector
E and the Magneticfield vector H must be zero at the electron centre,
then the Electricfield vector
E and the Magneticfield vector H must flow through the electron surface.
This occurrence indicates that the Gravitationalfield vector must be orthogonal to both
the the Electricfield vector
E and the Magneticfield vector H and the Gravitationalfield vector
must flow through the electron centre. This occurrence would make a great deal of mechanical sense.
If we look at BRITGRAV4 Figure 2, we can see that the the Gravitationalfield vector must be orthogonal to both the the Electricfield vector E and the Magneticfield vector H, i.e. for the bound electron and the free electron. If we look at BRITGRAV4 Figure 3, then we can depict the magnetons in BRITGRAV4 Figure 4 and the gravitons in BRIRGRAV4 Figure 5 flowing into the constellation of electrons in BRITGRAV4 Figure 3. The magnetons must flow vertically downward into the electron surface and the gravitons must flow moreorless sideways into the constellation of electrons in BRITGRAV4 Figure 3. BRITGRAV4 Figure 5 depicts the neutron centre, but the static electron depicted at the centre of BRITGRAV4 Figure 5 could equally well depict any of the static electrons in BRITGRAV4 Figure 3. The arrows emanating outwards from (and going into) the central static electron depicted in BRITGRAV4 Figure 5 represent the binding energy of the gravitons' mechanical interaction with the surface of the electricconvection potentialmembrane, i.e. they constitute the binding energy of the proton (e.g. 32,444.608 eV x 18 graviton units times 1836 static electrons equals 1.07222 GeV.
The gravitons depicted in BRITGRAV4 Figure 5 must flow sideways, i.e. orthogonally into the electricconvection potentialmembrane as well as flowing orthogonally all the way into the centre and out again if the gravitationalfield vector is to remain orthogonal to both the Electricfield vector E and the Magneticfield vector H. The natural expansion cycles and contraction cycles of the proton's magnetons, (i.e. the proton ionisation cycles), predicate that some graviton flow must emanate from the electron surface and orbit/return to the same staticelectron surface, in order for conservation of mass to be upheld. The proton does not change its mass when it expands and contracts its magnetons, (e.g. only the loss/gain of a bound electron changes the proton mass), so we know the mass needed to expand a magneton must come from within the proton. Only the absorption of part of the graviton mass can account for this phenomenon according to our model, so that is why the gravitationalfield vector must flow partly through the electricconvection potential (i.e. that is why we say that the mass is convected through the surface) and part of the gravitationalfield vector must flow orthogonally through the static electrons and bind the proton together internally.
Since atoms (other than Hydrogen) have ionisation potentials which change according to the number of atomic protons which are actually ionised at any one time, the model depicts why the different atomic protons have sequentially higher and higher ionisation levels, i.e. the graviton flow between protons via their commonlybound neutrons allows magneton expansion in sequentiallyionised intraatomic protons to be accounted for. Graviton flow intraatomically is the only manner by which intraatomic magneton expansion can be accounted for, i.e. according to the conservationofmass law of the First Law of Thermodytnamics. The contracting magneton, e.g. imagine the six magnetons in BRITGRAV4 Figure 4 are contracting from the molar radius as the proton captures a free electron, must return their excess mass/length to the gravitons, which must in turn allow proportional expansion and contraction of intraatomic protons to return to their appropriate mass/length variations within the atom. Further to this phenomenon; the returning/contracting magneton must flow entirely into the outgoing gravitons for this appropriate magnetonmass/length variation within the atom. We can now state; that according to our model, mechanicalmagneton flow must integrate with outgoing mechanicalgraviton flow, i.e. in order to account for the phenomenon known as successiveproton ionisation.
For example, if one multiplies 4Pi times the electron radius squared, by Eq. 1, then one has the hypothesised volume of the electricconvection potential, i.e. which the graviton must flow into and out of again. If one then divides the volume of the hypothesised gravitonunit cycle, e.g. Pi times the Planck Length squared times 2.426316 x 10 ^{12} m, then one gets a value of 4.506313734 x 10 ^{31}. This value can be factorised to give: the electron mass, the velocity of light squared, the reciprocal of the electron radius, the electron charge, the reciprocal of the electron mass, 10 ^{7}, the Fine Structure Constant squared and the frequency of the proposed graviton (c/2.426316 x 10 ^{12} m). The only coefficient involved is the square of the term in Eq. 1, i.e. 1.000082877.
e_{r} = electron radius =
7.415649545 x 10^{17}metre
1st
Shell radius = 5.291785381 x 10^{11}metre
e
= electron charge volume = 1.6021917 x 10^{19}Coulumb
Let us look
at a few examples. Voltage = force divided by area. The force of 1 magneton
divided by the electron radius squared (e.g. in Figures 1, BRITGRAV3 and
BRITGRAV4) should equal the electron (rest mass) voltage of 511,002 eV (its
energy per electroncharge volume). The force of 1 magneton is found from the
classical equation:
Force of magneton = e x c x (number of magnetons in Amperes per metre) x
magnetic constant /4Pi
Force of 1 magneton = e x c x (1 Ampere per metre) x 4Pi x 10^{7}kg m^{5}/4Pi
= 4.803249 x 10^{18}N
Electron radius squared = eq. 1 squared
Electron (rest mass) voltage = 511,002.576 eV = M_{e} x c^{2}/e
Engineering Terms in Classical Physics.pdf
4.803249 x 10^{18}N/Eq. 1^{2}
= 511,002 eV x F.S.C.^{5} x 2 x 10^{6}/17,275 
(2) 
This compares well with the
proton restmass voltage equation
1836 magnetons x 4.803249 x 10^{18}N / [ 1.40897 x 10^{15}m ]^{2}
= 938.200 MegaeV x 17,275 x F.S.C.^{1} x 2
Proton radius = 1.40897341 x 10^{15}m
Proton rest mass = 938,200,727.7 eV
The value
given by Eq. 2 on the lefthand side equals 2.8588 x 10^{18} Volts.
This depicts how the magneton melts (i.e. flows) into the electron surface and
the electron surface becomes the "Electricconvection potential" of
G. F. C. Searle (a Cavendish Laboratory presenter at Cambridge University in
the 1890s). If we multiply this value by Eq. 1 and divide by the electroncharge
volume, then our units are Voltmetres per Coulomb. These are the reciprocal of
the units of the Electric Constant.
Electric Constant = ε_{0} = 8.854187816 x 10^{12}Coulomb Volt^{1}m^{1}
L.H.S. Eq. 2 x Eq. 1 =
[ e / Electric constant ] x 10^{3} x F.S.C.^{3} /4Pi 
(3) 
We can now
depict how the magneton might flow (solidify or condense) into the outgoing
graviton. The value given by Eq. 2 on the lefthand side equals 2.8588 x 10^{18}
Volts. If we multiply this by the electron radius and the electronradius
couplingcoefficient and the unit length, (but without using the
electronradius couplingconstant), we find the force of the graviton in
milliNewtons.
L.H.S. Eq. 2 x e_{r}
x 1.000082877 x unit length = 10^{3}M_{e}^{2}
G_{0}P_{L}^{2} mN 
(4) 
Gravitational Constant = 6.662031411 x 10^{11}m^{3}kg^{1}s^{2}
Graviton force = M_{e}^{2}x G_{0} x P_{L}^{2}
= 0.2120166946 Newtons
P_{L} = Planck Length = 1.61478412 x 10^{35}metre
We have
depicted how the magneton flow might operate in the static electron, (i.e. the proton's
electron). This phenomenon would occur as the magneton flows through the electron surface
and onward out into the
proton's outgoing graviton. We can now depict how the inertial force of the orbitalelectron
magnetonrim equals the gravitational force of the orbitalelectron. Inertial
force is equal to mass multiplied by velocity squared divided by the radius of
spin.
Velocity of spin = c x F.S.C. ms^{1}
= 2,178,691 ms^{1}
Inertial force = M_{e} v^{2} /eq. 1
= M_{e}^{2}x G_{0} x P_{L}^{2}
x 2000 x F.S.C.^{1} x 17,275^{1}
M_{e} v^{2} /
[Eq. 1 x 2000 / F.S.C. x 17,275 N ] =
M_{e}^{2} G_{0} x P_{L}^{2}
x = 0.21201 N 
(5) 
The force of the electron can
now be related to the Coulomb Force of the 1st shell of Hydrogen, i.e. other
than by multiplying the electronselfgravitating force of 0.212 Newtons by the
cube of the Fine Structure Constant.
The electron radius (Eq. 1) multiplied by the
electronselfgravitating force of 0.212 Newtons, multiplied by 4Pi x 10^{3}
times the electron groundstate mass, times the velocity of light, divided by
Planck’s Constant and the ratio of the induction radius to the 1st shell
radius, (i.e. 17,275) equals the Coulomb Force.
Eq. 1 x 0.21201 N x 10^{3}4PiM_{e}c
/ 17,275 h = 8.2388 x 10^{8} N 
(6) 
The Coulomb force = 8.238837117 x 10^{8} Newtons.
The Coulomb force is the mechanical force (i.e. magneticelectronic screening force) due to
the interaction of the orbital electron at the 1stshell radius
of the Hydrogen Atom. It is the force which is needed to be overcome, by any particle,
in order for the Hydrogen atom to be ionised so that the Hydrogen atom will break
its simple electronproton bond and bind to a new atom or molecule.
The particle
might be another electron, proton, neutron, magneton, graviton or photon which can
contribute sufficient energy to initiate the protonelectron breakup interaction.
From the
earlier discussion on a particle subunit of matter which was a fibrelike
coilshaped particle, we can test if this unit of matter within the magneton
could contribute to the known force of the magneton, i.e. 4.803249 x 10^{18}Newtons
= (ec4Pi x 10^{7}kg m^{5}/4Pi steradian) Newtons. If we
multiply this unit of mass ( 7.3726 x 10^{51}kg = 1 Hz x h c^{2}
) by the velocity^{2} of the groundstate magneton (c x F.S.C)^{2}
and divided by the electron radius (Eq. 1), we get the force of one magneton
and some of our scaling factors (4.803249 x 10^{18}Newtons).
[ 7.3726 x 10^{51}kg ] x
[ c^{2} F.S.C.^{2} ] / Eq. 1 = 4.8032 x 10^{18} N x
[ 17,275 x F.S.C. x e / {M_{e} 4Pi x 10^{10} } ]^{1}

(7) 
All of the variables in
brackets on the righthandside of eq. 7 are dimensionless.
This may be because the term "e/{M_{e} 4Pi x 10^{10}}"
can be substituted by the electron charge to mass ratio times the magnetic constant,
times Unit Area times 1/1000. This dimensionless ratio can be found in many of these new equations,
i.e. especially when the ratio is a large number or involves gravitational phenomena.
It is interesting
to see that the electron charge to mass ratio (the specific volume of
ionisation of the hydrogen atom) again shows up as a scaling factor. It will
doubtlessly show up again, i.e. in other forthcoming equations.
The Coulomb Force of the 1st shell of Hydrogen divided by the
electronselfgravitating force of 0.212 Newtons equals the cube of the
Fine Structure Constant. This last equation shows how, mathematically speaking,
the electron radius has been lost, physically speaking, (in past history). The
electron radius multiplied by 4Pi x 10^{3} times the electron
groundstate mass, times the velocity of light, divided by Planck’s Constant,
divided by e/1st shell radius^{2} per unit distance and the ratio of
the induction radius to the 1st shell radius, equals the cube of the Fine
Structure Constant.
e_{Radius} x e_{mass} x c x 4Pi x 10^{3}
x 1st shell radius^{2} x Unit Distance/( h x e x 17,275)
= F.S.C.^{3}

e_{Radius} x c x 10^{3}
x 1st shell radius^{2} x Unit Distance/H x e
= F.S.C.^{3}
Since 4Pi times the electron groundstate mass,
divided by Planck’s Constant (times Unit Amperes per metre),
equals the ratio of the induction radius to the 1st shell radius,
the above electronradius equation simplifies to
where H is Maxwell's Magneticfield vector "unit amperes per metre"
and the electronradius coefficient 1.000082877 is multiplied by the electron radius, i.e. to make the equation balance.
So where do these last seven equations come from? The
first one is the electron radius. It is derived by "proof by
construction" from BRITGRAV4 Figure 3 and by the discovered coupling
constant/correction coefficient. Equations 2 and 4 depict how the force of the
magneton (returning to the electron) must equal the (outgoing) force of the
Graviton. The incominggraviton force must equal the force of the
staticelectron surface and the force of the outgoing magneton. This is due to
the volumetricflow law of the 1st Law of Thermodynamics, i.e. if no volumetric
flow is lost (from the magneton, staticelectron surface or the graviton), then
no force can be lost. This is also true because the magneton, staticelectron
surface and the graviton are all basically orthogonal to each other.
For the incoming magneton the force per metre is
onedimensional. For (half) the electron surface, the force per metre squared
is twodimensional. For the outgoing gravitons the force per metre cubed is
threedimensional. The force per metre cubed (in classical physics) equals the
Volts per metre. The Volts per metre value is the value of the Electric field
(E). From Maxwell's current displacement law, [J = Amperes m^{2} + ε_{0}E/d(t)],
we know that the electric field is a magnetic field with electrons in it. We
hypothesise (we can test for this) that the electrons
(travelling moreorless parallel to the magneton lines of force) travel in a magnetic
field due to the electron's gravitons
wrapping themselves around the protons' magnetons mechanically. This phenomenon
is due to at least two occurences; firstly, the helical nature of the electron's gravitons
(e.g. as depicted in BRITGRAV4 Figure 2), would make the gravitons mechanically wrap themselves
about the proton's straightparallel
magnetic field lines, (i.e. an electrical field is a magnetic field with a charged particle in it).
Secondly, the proton's straightparallel magnetons cause the electron's polarised
gravitons to be deflected and turn away from the magnetons in a
counterclockwisehelical spiralflight, i.e. when the electron orbit is orthogonal to
parallel megneton lines of force. This deflection causes Theta pinching, e.g. the phenomenon
whereby an electron orbiting within a magnetic field actually contracts the magnetons within
its orbit so that the magnetons are closer together (pinched).
4.8032 x 10^{18}N / Eq. 1^{3}
= [ 13.605eV / 1st Shell_{R} ] x [ 8 x 10^{9} / 17,275 ^{2}
x F.S.C.^{11} ] 
(8) 
Force of 1 magneton = e x c x (1 Ampere per metre) x 4Pi x 10^{7}kg m^{5}/4Pi
= 4.803249 x 10^{18}N
Electric Field = 13.60578693 Volts/5.291785381 x 10^{11}m
(see Table 2 in "Maxwell's 150 Years PDF" on the link above for the
tables on the application of Maxwell's Laws to the atom)
Eq. 1 can be considered as the electron
radius or the part of the electron radius which contains matter. Eq. 1 depicts the
mathematicalelectron radius which includes the real thickness of the electronsurface membrane
which the mathematical and virtualelectron radius passes through. We can see
from Eq. 8 that for every time that one has divided the force of the magneton
by the electronradius equation, one has multiplied the electron
radius by
F.S.C.^{3} x 10^{3} 
(9) 
By doing this one has the velocity of light reciprocal
in units of metres. To get the specific relation between the velocity of light
and the electron radius (Eq. 1), we divide Eq. 9 by Planck's Constant/[4Pi x M_{e}]
and by the scaling factor 17,275.
One can now see how Maxwell's
equations, e.g. the curl functions, are applied at the atomic level. The
curl function equation is when one makes a change in the function by
dividing the equation by an atomic length, i.e a length which is involved in
the equation. The opposite is Stokes' equations. In Stokes' equations one
multiplies (integrates) the equation by a length which is involved in the
equation. These multiplications and integrations change the dimension of the
equation field from a line to a square to a cube and back to a square and a
line, i.e. just as the magneton does when flowing into the electron surface,
then into the electron inner volume, then out to the electron surface again and
out into the outflowing magneton (or outflowing graviton).
Let us look at one of Maxwell's 4
great equations, i.e. one of the equations which Heaviside has translated to us
from Maxwell's archaic Gothic script to the Clarendon font. The equation should
read: The Electric Field voltage per metre square equals the difference in
magnetic flux density (i.e. the difference between the maximum flux density at
the pole of the magnet passing at 90 degrees to the wire and zero flux density
when the magnetic pole is not passing the wire), divided by the difference in
time between the magnetic pole passing the wire the first and second times. The
negative (B) term means that the magnetic field ( i.e. the voltage
field ) induced into the conducting wire when the magnet passes close by it, is
in the opposite direction to the magnetic field direction coming out of the
magnet. This is explained by Newton's equal and oppositereaction law.
curl E = B/dt 
(10) 
The Electric Field of the atom = 13.60578693 Volts / 5.291785381 x 10^{11}m,
i.e. from Eq. 8. The curl of the Electric Field means the division of the
Electric field by the length of the 1st Shell radius again, i.e. by 5.2917 x 10^{11}m
again. Maxwell was trying to develop Stokes' Equations from some work that
Kelvin had given Maxwell.
When
Maxwell described what he was trying to explain, he wrote that by using the
word "curl", he did not mean swirl or twirl. So he did not mean
constant movement. However, he did mean a rotation of a geometrical plane about
one of the XYZ axes was occurring once only. His first equation was "curl
A = B". He was trying to make an electromagnetic alphabet
using the gothicscript letters A to K. One thing which I have found is that
when you are trying to introduce something to people, do not introduce a new alphabet
or a language as well. A is the magnetic field which always emanates and
returns to a conductor when you apply a voltage to the conducting wire. If you
curl a straight piece of wire into a circle, you have the magnetic field
emanating out of the wire into the circle, back around the outside of the wire
(and eventually going back into the proton which the magneton emanated out
from). The magnetic field going into the centre of the circle has a
magneticflux density which is labelled the "Bfield". It is measured
in units called Tesla and the dimensional metric dimensions are kg per metre
cubed per second (Volt seconds per metre square). The "Afield" is
measured in metric dimensions of kg per metre square per second. It is called
the "flux rate" in the 20th century and can measure any particles as
they pass through a square metre in a second. It is derived from the pressure
(e.g. the voltage) divided by the velocity. The pressure (e.g. the voltage)
divided by the frequency gives the flux. It is measured in Volt seconds and
gives the Voltage per cycle of a single particle, (e.g. the number of cycles on
a singlelinear particle stream). It is Planck's Constant divided by the
electron charge. It is measured in units of kg per metre per second, So, the
flux is used to measure the cycle number of a single particle passing down the
line that forms the edge of a cube ( at rightangles to the square base of the
metresquare cube ) and the curl of the flux measures the total number of
particles passing through the squaremetre base of the cube each second.
Why did Maxwell bother to try
to set up an alphabet for us? It is because Kelvin formulated the 1st Law of
Thermodynamics, i.e. the law which states that mass flowing into a junction
must come out of it at some point and time. Kelvin and Clausius were trying to
make a formal explanation for heat and energy. Clausius and Kelvin were
colleagues and Clausius was Planck's teacher. Clausius and Kelvin credited Sadi
Carnot with the 2nd Law of Thermodynamics, i.e. the law which states that you
will get a little friction and heat loss when you apply energy (heat) to any
working machinery. Clausius developed the 3rd Law of Thermodynamics, the law
which states that there is a maximum amount of work which you can get out of
any process (machinery). This law also means that you must apply energy (heat)
to any machinery to get the machinery to work. Carnot was trying to define the
amount of energy which must be applied to the machinery process (e.g. the
heating of water to make steam), so that French engineers could do what
Trevithick had done for Cornish mines. Together these three laws form the
grandunified field theory. Heaviside then decoded Maxwell's equations and
simplified them into 4 useful equations (the first of which is Eq. 10).
Maxwell's secondmost importantequation involved the Amperage per square metre
at the ends of a conductor. His term J, which symbolises "Amps per
square metre", symbolises the number of electrons per second emanating out
of the crosssection of a conducting wire (and a capacitor plate if you
discharge static electricity).
J = Amperes m^{2} + ε_{0}E/d(t) 
(11) 
If we
integrate the lefthandside of Eq. 10, (with respect to length in order to
think of the voltage per length of conducting wire instead of the voltage per
metre square), we have to integrate the R.H.S. (if we wish to keep Eq. 10
balanced), while we are making changes to show how Eq. 11 was derived. The
integration of Eq. 10 (deriving the integrated length) gives
E= ∫  [B/dt ] d (length )
This gives
E = V/m
Multiplying the E = V/m by the Electric Constant gives
ε_{0}E = ε_{0}V/m
This
gives us the ability to find the Amperage in a conductor, (e.g. because Eq. 3
shows that the Electric Constant is equal to the electron charge divided by
the product of the Voltage and the metric length).
The number
of Coulombs (of staticelectricity electrons) per Volt (applied to a capacitor
plate) per metre square of the capacitor plate (at rightangles to the wire
that it is soldered to), equals the number of Coulombs (of dynamicelectricity
electrons) per square metre of the crosssection of the currentcarrying
conducting wire (for every moment that we pass a magnet passed the conducting
wire). If we divide the equation by a difference of time, (e.g. 1 second from
now), we get
ε_{0}E/d(dt) = ε_{0}V/m(dt)
= Amperes per metre square 
(12) 
ε_{0}E/d(dt)
= D. D is Maxwell's term for the electric (electronic or
protonic) charge stored on a capacitor plate. The R.H.S. of Eq. 12 now becomes
Coulombs per metre square per second (or Amperes per metre square) per
crosssection of wire. We now have Maxwell's 2ndmost importantequation, i.e.
the total current is equal to the current being discharged by a capacitor plate
and the current flowing past a crosssection of the conducting wire.
J = D/(dt) + Coulombs/m^{2}(dt)
= Amperes per metre square 
(13) 
We can make Eq. 10 work for the Hydrogen atom now.
curl 13.605 Volts/ 5.2917 x 10^{11}m = 17,275Pi [13.605
T / 1.5198 x 10^{16} s] 
(14) 
1st Shell Orbit time = 1.519833811 x 10^{16}s
1st Shell radius = 5.291785381 x 10^{11}m
Electric Field = 13.60578693 Volts/5.291785381 x 10^{11}m
B = Flux density = 13.60578693 Tesla
Now we can try to apply the
HeavisideMaxwell equation to the electron. The Electric Field per metre square
on the L.H.S. of Eq. 14 is 511,002 Volts per electron radius squared (511,002
Volts/ Eq. 1 squared). The R.H.S. is 511,002 Tesla divided by the time it takes
1 magneton to move 1 electron radius at the velocity of (light times the Fine
Structure Constant).
curl 511,002 Volts/ [ Eq. 1 ] =  511,002 T / [5.92501
x 10^{25} s x 2 x 10^{3} x F.S.C.^{4}] 
(14a) 
Electron radius travel time = 5.925011083 x 10 ^{25}s
Electron radius = Eq. 1
curl Electric Field = 511,002.575 Volts/ [ Eq. 1^{2} ]
B = Flux density = 511,002.575 Tesla
If we look at Stokes'
Equations (Theoretical Concepts in Physics Longair), we can see how Maxwell
developed his equations, i.e. from the laws of massfluid flow. If we look at
Heaviside's books (Electrical Papers Heaviside), we can see how Heaviside applied
Maxwell's Equations to electricity, light, magnetism and gravitation. If we
look at G.F.C. Searle's 1897 paper on the "electric convection potential
of the electron surface", then we can see how mass flows into the electron
surface from the magneton current (which both Heaviside and G.F.C. Searle
mention) and out of it again. The important observation to note is that G.F.C.
Searle knew both Heaviside and J. J. Thomson and that Maxwell, Stokes, Joules
and Kelvin were all colleagues. Noone except us has bothered to put their work
together until now. We can see that we can now formulate gravitation according
to Maxwell's Laws and make a grandunified fieldtheory, i.e. one which will
allow space travel by classical magnetism and classical gravitation. We shall
see more about this later.
If we now
look at Heaviside's equation for gravitationalenergy dissipation, (i.e. Eq. 26
in Maxwell's 150 Years.pdf), we see 3 terms. The first term symbolises
"the divergence of the acceleration times gravitational flux times the
sine of the angle between the flux and its acceleration", i.e. the
inertial spin of the electron surface is at right angles to its forward
velocity (and its incoming graviton).
m s^{2} x kg m^{2} s^{1}
The second term symbolises "acceleration times the density of the electron
(the force per metre cubed), times the velocity of the gravitational
flux", i.e.
[acceleration x mass m^{3}] x m s^{1}
The third term symbolises "acceleration of the groundstate electron
times the acceleration of the upperstate electron divided by the Gravitational
Constant and by the cycletime of the Hydrogenminimum photonlength", i.e.
m s^{2} x m s^{2} x G_{0}^{1} s^{1}
Heaviside is explaining matter accretion (e.g. on a star) using Maxwell's
equations. We can attempt to show how photon accretion by an electron can be
partially explained by using this equation, (i.e. as we did with Eq. 22 in
Maxwell's 150 Years.pdf, when describing the power per metre squared of the
Hydrogen atom with the electron in its ground state). The third term in Eq. 26
is the inertial acceleration of the groundstate electron times the inertial
acceleration of the electron at the molar radius, divided by the Gravitational
Constant and the cycletime of one photon length of the Hydrogenmaximum
frequency.
a x d(a)/ [ G_{0} d(t) ] =  [ c^{2}
x F.S.C.^{2}/ 1st Shell_{Radius} ] x [ c^{2} / Molar_{Radius}
] / [ G_{0} x (3.03966 x 10^{16} s) ] 
(15) 
The answer
to Eq. 15 is 5.63125737 x 10^{74} Watts m^{3}. The negative
answer symbolises that the photonemission direction is opposite to the
electron forwardorbital direction, i.e. as the cosine of 180 degrees is = 1.
The second term in Eq. 26 symbolises the gravitational force of the
groundstate electron times the graviton's velocity
F x v = [ M_{e}^{2} x G_{0}
/ Planck_{Radius}^{2} x Eq. 1^{3} ] x c

(16) 
The answer
to Eq. 16 is 2.918537616 x 10^{61} Watts m^{3}. The ratio
between the third term and the second term is 8 x 10^{8}/17,275.94389^{3}x
F.S.C.^{8}.
The first term in Eq. 26 involves the divergence of gravitationalinertial acceleration times the gravitational flux, i.e. the phenomenon of electron selfgravitation (when the graviton enters the orbital electron). This means that one half of the incominggraviton massflow gains centrifgalspin momentum while losing one half of its component of forward momentum as half the graviton splits off from the origial incoming, (i.e. forwardmoving) graviton half. This is hypothesised because the graviton must flow into the electron hemisphere and form the electron hemisphere (due to the continuity equation of the 1st Law of Thermodynamics). The spin velocity goes down to c x F.S.C. by the time the graviton half reaches the electron rim and the forward velocity goes down to zero (though the electron momentum still carries the electron itself forward at c x F.S.C.. The graviton diverges orthogonally from itself (temporarily), i.e. at the point of contact where the graviton touches the electron surface. The graviton consequently moves backwards up the hemispherical rim until the graviton loses all of its centrifugal, (e.g. radial) momentum. The graviton half must now change its form as it flows through the magneton rim of the electron, i.e. the graviton subunits must now change into magneton subunits. The graviton/magneton subunits continue to rotate around the hemispherical rim (before returning down the front of the electron hemisphere until the graviton converges with the outgoinggraviton half). This latterhypothetical phenomenon is due to surfacetension forces within the electron hemisphere, (i.e. those forces which must change the graviton/magneton subunits back into graviton/electronhemisphere subunits and finally back into the outgoing graviton subunits).
The divergence of gravitation
times the cosine of the angle that it makes with the cross product of the
electron's inertial acceleration and gravitational flux gives Eq. 17

(17) 
The answer to Eq. 17 is
8.694760963 x 10^{66} Watts m^{3}. This forms a ratio of 4 x 10^{5}/17,275.94389^{2}x
F.S.C.^{5} with the third term in Eq. 26.
The electron magneton must
have some mathematical equivalence to the proton magnetons. The protons’
magnetons, i.e. the 62,584 magnetons which ionise the Hydrogen Atom (and the H_{2}
molecule), are easily found. 13.605 Tesla ionise the Hydrogen proton. 13.605
Tesla are defined as 13.605 Webers per metre square. 13.605 Webers per metre
square are defined as 13.605 x 10^{8} Maxwells per metre square. A
Maxwell is 1 magneton. 13.605 Webers per metre square x 4.5998 x 10^{5}
m^{2} = 6.2584 x 10^{4} Weber. A Weber equals 10^{8}
Maxwells or 10^{8} magnetons so 6.2584 x 10^{4} Weber x 10^{8}
magnetons = 62,584 magnetons.
A Coulomb is a unit of static
electricity (electrostatics), e.g. 1 Cb per second = 1 Ampere. The current
flowing past a Hydrogen proton when it is ionised (not the groundstate
boundelectron current), orbits the magnetons made up of several ionised
Hydrogen atoms (see Figs. 2 and 3). A Coulomb is the volume which contains the
62,584 magnetons together with the ionisedHydrogen proton and the
orbitingionised electron. It is the 62,584 magnetons flowing past the
1stshell area, i.e. the proton charge per metre square, (e/1stshell radius^{2}).
If we get this value (57.215 metres), which is our electron radius coefficient,
divided by the induction orbit radius of 9.1427 x 10^{7} metre, we get
62,584 magnetons (with a scaling factor of 10^{3}). These 62,584
parallel magnetons (travelling at the velocity of light), constitute the proton
charge per metre cubed e/ [1stshell radius^{2} x 9.1427 x 10^{7}
m], i.e. the phenomenon which occurs when we ionise the Hydrogen proton.
Fig. 3 depicts the induction
of the electron from the 1stshell orbit in the Hydrogen atom into the
inducedstate orbit, i.e. when 62,584 parallel magnetons (13.605 Volts), are
applied antiparallel to the northsouth polar axis of the Hydrogen atom. The magnetic
current of the electron rim, multiplied by the magnetic force of the 62,484
magnetons, equals the gravitational force of the selfgravitating electron,
multiplied by the electric current of the electron in the inducedstate orbit.
Fig. 3 depicts the famous experiment, i.e.
"Bev = mv^{2}/r"
where B symbolises the 13.605786 Volts per square metre applied to
an atom when a magneticflux density of 13.605786 Tesla is applied to a conductor
in a cyclic manner, "e" is the electroncharge volume of 1.6021917 x 10^{19}
m^{3}, "v" is
the velocity of the electron orbiting within the applied voltage field at 2.18761 x 10^{6}
metres per second, "m"
symbolises the electronground statemass of 9.109534 x 10^{31} kg and
"r" symbolises the orbit radius of the electron within the applied
Bfield at 9.142058 x 10^{7} m.
From figures 1, 2 and 3 one can see the depicted differences between an electron
orbiting the proton in the “ground sate” (Fig. 1) and an electron
which has just been pulled from the proton in the ground state
(Figs 2 and 3), i.e. in order to orbit the 62,584 magnetons in the
"inductionstate" orbit.
Fig. 1 depicts an electron spinning and orbiting on the inside of a
magneton “shell” (e.g. a virtual tube of force), while Figs, 2 and 3
depict an electron spinning and orbiting on he outside of a magneton
shell. The difference between the two states, e.g. the insidespin
state and the outsidespin state, is that in the inside state (Fig. 1),
the electron’s electrical and magnetic forces would cause the proton’s
magnetons to shorten their overall length of travel and become wound in,
thus building up the protonmagneton internalpotential energy,
(i.e. its springconstant energy). In Figs. 2 and 3, the electron
spin is depicted going against the spin direction of many protons’
magnetons, e.g. thus forcing the protons’ magnetons inward, due to
the electron’s spinningmagnetic rim as well as the electron’s
hypothetical graviton. This latter phenomenon is the socalled
wellknown "Thetapinching", e.g. the "Larmorfrequency"
phenomenon.
Since both the electron’s "groundstate" orbit and
"inductionstate" orbit are stable, i.e. both orbits are at
lowest entropy and permanent, the forces and volumetric flows are stable,
permanent and counterbalance one another. It is obvious from observing
the "Thetapinching" phenomenon, that their must be a circular
force which is involved in “pinching in” the Bfield magnetons
(see Fig, 2). It is not obvious what this force is, however our
hypothetical graviton offers up the numerical solution.
The force of the graviton, i.e. the electron’s selfgravitating force,
" m^{2}G_{0}/P_{L}^{2} ",
divided by the force of the "inductionstate" magnetons,
62,584 Amperes per metre x e c x µ_{0}/4π = 3.006084657 x 10^{12} Newtons
must equal the ratio of the electron’s magneticrim volumetric flow
to the volumetric flow of the electron in the "inducedstate"
orbit, i.e.
(6.102061 x 10^{8} Amperes).
The magnetic current is defined by the
electroncharge volume multiplied by the frequency of spin of the effectiveelectron radius, i.e. e x [c x F.S.C./2Pi x r_{e.e.}]. The magnetic force
is defined by the 62,584 magnetons x Unit Amperes per metre x the
Magnetic constant x e x c/4Pi. The Gravitational force is defined by
the electron mass squared x Newton's Gravitational constant divided by the Planck length squared.
The electric current in the inducedelectron orbit is defined by e x c x F.S.C./2Pi x
Induction_{Orbit Radius}.
This makes classical sense as
the electron’s magnetic current density, i.e. the graviton’s classical
mechanical force which is turned into the electric force as the electron’s
graviton pulls the electron around the proton, must equal the proton’s magnetic
density (the magnetic force). This is proposed to be due to the graviton’s
subunits (discussed earlier) being converted into the electron’s magnetic rim
via the electron’s membrane in a classicalmechanical manner. This is why the
Coulomb force is twice the electric force and twice the magnetic force in the
simpleharmonic oscillatorequation.
It is hypothesised that when
the graviton enters the electron surface membrane, (which it is proposed to do
as a helical coil, e.g. a spring under tension), that (onehalf of) the
graviton, (e.g. the outer graviton core), loses one of its three degrees of
freedom, (i.e. the zaxis of forwardaxial velocity). This outer core then is
proposed to spread itself and its subunits, out in an expanding
twodimensional spiral, i.e. expanding and spiralling outwards (within the
electron's surface) in the same direction as the electron spin (see Fig. 1).
This is Stokes first equation, i.e. the integration (multiplication of) 1/m^{3}
by length to give 1/m^{2}, It means that we can tell what is going on
(i.e. what forces are occurring) inside an electron (a sphere) by studying its
outer surface,
The expanding and spiralling
twodimensional subunits, will eventually interact with their neighbouring,
(i.e. 9) graviton units. These 9 graviton units are proposed to be 40 degrees
apart and have their centres located at approximately 4.0139 x 10^{17}
metre out from the centre of the electron. The 9 graviton units define their
position within a steradian mathematically by being equidistant from each
other and the electron centre. The pressure of the collision between the
neighbouring gravitons' subunits would force the graviton subunits out
towards the electron’s magnetic rim. The subunits would join the electron’s
magneticrim subunits by losing another of the three degrees of freedom, (i.e.
the xaxis degree of radial velocity). The surface tension between the outgoing
innergraviton units’ cores, (i.e. the half of the graviton which did not lose
its axial degree of freedom and join the electronmembrane surface), would now
have its effect. Surface tension between the outgoinggraviton's subunits
would pull the subunits from within the electron's outersurface membrane into
the outgoing gravitons. Surface tension from the outgoing subunits within the
electron's outersurface membrane would pull subunits from within the magnetic
rim of the electron into the outer surface of the electron's surface membrane
and thence into the outgoing gravitons. The two lost degrees of freedom would be
regained as the subunits moved from the electron rim into the outerelectron
surface (i.e. first the negativeradial velocitycomponent and secondly the
outgoing axialdegree of freedom). This is proposed to occur as the subunits
rejoin and merge with the
outgoing graviton. This process would then be cyclic, i.e. it would follow the
continuity equation of the First Law of Thermodynamics.
The graviton would now have all of its mass and volume
back, i.e. as they were before the graviton split into an inner and outer core,
whence the outer core entered the electronsurface membrane.
This fascinating world of the
subatomic units leads us back to the exciting world of atomiclevel
physicsequations, (which is the whole point of the study). With the rigorous
proofs of the Laws of Thermodynamics to guide us, we can explore hypotheses and
posit possible hypotheses, (i.e. hypotheses without experiments to constrain
their positing). The purpose of such actions is to find experiments which can
uphold some of our hypotheses, (e.g. by a chisquared test), so that our use of
hypotheses without experiments can lead us onto new hypotheses which can have
experiments which will lead us further on.
For example, we know well
that Volts x Amperes equals Watts, so we can posit that Joules per metre cubed
x volumetric flow equals Watts.
The power of the
inductionstate electron, (e.g. the volumetric flow of the graviton x the
13.605 Volts of the electron in the inductionstate orbit where B Tesla x e Cbs
x v m/s = m kg x (v m/s) ^{2} /r _{inductionstate orbitradius}),
should equal the power of the free electron when it is orbiting within the last
magneton shell at the molar radius, i.e. 1.022005 x 10 ^{6} electron
Volts multiplied by e x c/ 2Pi x 7.347 x 10^{10} metre = 10633.37685
Watts.
The power
of the inductionstate electron is derived from the product of the volumetric
flow of the graviton multiplied by the pressure of 13.605 Volts of the
electron. The volumetric flow of the graviton can be defined by the crosssection
of the graviton,
Crosssection_{Graviton} = Pi x P_{L}^{2}
This multiplied by 2Pi x the gravitonorbit radius of one graviton unit to give the volume of one graviton unit.
Cosmic orbit volume_{Graviton}
= Pi x P_{L}^{2} x 2Pi x 1.178497606 x 10 ^{56} metres,
to give one graviton unit a volume of 6.06579 x 10 ^{13} m^{3}.
This is now multiplied by the graviton frequency.
Frequency_{Graviton} =
c/2.426316079 x 10 ^{12} metres = 1.235586989 x 10 ^{20} Hz.
This equals the volumetric flow of the graviton.
The Volumetric Flow_{Graviton}=
7.494811459 x 10 ^{7} m ^{3} s^{1}
The volumetric flow of the graviton, as we said earlier,
should equal the volumetric flow of the magneton,
i.e. as it was postulated that the graviton flows continually into the magneton
via the electron surface membrane (the electricconvection potential of G.F.C. Searle),
The volumetric flow of the magneton for both the orbiting electron and the proton's magneton,
is known to be 3.3118316 x 10 ^{3} Amperes. If we divided 7.494811459 x 10
^{7} m ^{3} s^{1} by (e/M_{e} x 10 ^{7}) and
multiplied it by the cube of the Fine Structure Constant and 2, we do in fact get
3.3118316 x 10 ^{3} Amperes (see the "G.F.C. Searle pdf" and the last 2
paragraphs of the "Engineering Terms in Classical Physics.pdf" for more equations depicting
3.3118316 x 10 ^{3} Amperes).
The volumetric flow of the
gravitonhelical coil multiplied by the ionisation voltage of the electron in the groundstate orbit,
( i.e. 1.019728078 x 10 ^{9} Watts ) divided by ( the volumetric flow of the free electron
at the molar radius multiplied by the voltage of the free electron ), e.g. 10633.37685 Watts,
gives a ratio of
191,797.5996.
This ratio, when multiplied by 8Pi x F.S.C. gives the electron chargetomass
ratio times 10 ^{7} in dimensionless units, i.e. 17588.07535. This
last ratio, when divided by 20Pi and multiplied by the 62,584 magneton number
(i.e. which create the inductionstate orbit of the electron), gives us the
exact ratio between the natural area of the 62,584 magnetons (when they emanate
from a magnet of 13.605 Teslaflux density) and the area occupied by the 62,584
magnetons when they are orbited by inductionstate electrons.
4.599836136 x 10 ^{5} m^{2}/2.625656364 x 10 ^{12} m ^{2}
= 1.75188048 x 10 ^{8}
If we look at the discussion which
immediately precedes the Planck Massflow Equation and the Poincaréenergy Equation,
we can see that
the volumetric flow of the graviton subunit, ( e.g. Newton's corpuscles ) per metre
(e.g. the gravitonunit cyclelength), multiplied by the mass flow of the graviton subunit,
i.e. Planck's constant (divided by c squared and multiplied by the Graviton frequency squared)
should equal the energy of the electron, M_{e}c^{2}, i.e. the Poincaré Equation. Amperes per metre = 0.25 m^{2}/s (18) Mass _{Flow} = 1.125562171 x 10^{10} kg s^{1} (19) = m_{e}c^{2} (20)
From the preceding paragraphs
the Volumetric Flow _{Graviton subunit} per gravitonunit cyclelength
=
[Planck _{L}^{2} x Pi/ Graviton _{Frequency} as a dimensionless number]
x 2Pi x Graviton
_{Orbit Radius} x Graviton _{Frequency} per gravitonunit
cyclelength
= [P_{L}^{2} x Pi/ 1.235586 x 10 ^{20}]
x 2Pi x 1.178497 x 10 ^{56} m x 1.235586 x 10 ^{20} Hz/ 2.426316 x 10 ^{12} m
Note that this is 1/4 of Maxwell's unit of magnetism, H,
i.e. H = the metric definition of 1 Magneton or the areal velocity of 1 square metre per second.
One can use the scaling terms "16Pi per steradian" to make Eq. 18 work, as H = 1 Magneton per steradian
and then use one's scaling factor of 17,275, (i.e. the quotient of the electron's inductionorbit radius to
the electron's grouindstate orbitradius), to derive mc^{2} (See Eq. 20).
This means that we can now derive a couple of equations, e.g. Maxwell's term for
magnetism, H (the magneticfield vector) in mathematics and one can derive the specific enthalpy term of the
second law of thermodynamics as well, i.e. a change in the magnetonareal flow per unit time change gives the specific enthalpy of matter.
One can now say that the energy of the electron is proven by its internal energy plus
the product of the electron pressure and volume.
The mass flow of the graviton subunit is derived from Planck's Constant
divided by the energy of matter (c^{2} Joules per kilogram), multiplied by the
Graviton _{Frequency}^{2}
the Mass _{Flow} =
[ h / c^{2}] x 1.235586 x 10 ^{20} Hz ^{2}
The volumetric flow of the graviton subunit per gravitonunit cyclelength, Eq. 18,
multiplied by the mass flow of the graviton subunit, Eq. 19, times 16Pi/17,275
How to classically describe conjectures.pdf
From the 6 preceding equations, (Eqs. 15, 16, and 17
concerning photon emission and absorption by the electron, as well as Eqs. 18, 19 and 20
concerning magneton and graviton emission and absorption by the electron), we can hypothesise,
(i.e. classically one can say), that from Isaac Newton's famous 'Query',
"Is it not obvious that matter and light are interconvertible?", that as soon as the electron emits light,
it must have an equal and opposite reaction. The electron, we have hypothesised,
emits light because when the free electron is decelerated into the ground state orbit, the free electron is
at that point at twice its normal groundstate restmass, i.e. so it cannot accept any more mass from
the incoming graviton (which is still at the velocity of light). The free electron is decelerated
from the velocity of light to the velocity of the Fine Structure Constant times the velocity of light,
and the incoming graviton is still travelling at the velocity of light, so the energyconversion rate
must be changed.
If the incoming graviton is still travelling at the velocity of light and it is now forced by the
decelerated electron to travel through the G.F.C. Searle electricconvection potentialsurface
of the electron hemisphere, i.e. instead of travelling straight
through the decelerated electron, then this change (of direction) in the incominggraviton half would cause
the electron to spin proportionately faster
(as the electron is proportionately decelerated). This is hypothesised to be where the application of shear forces
overrides the application of surfacetension forces. The interaction between electronsurface membrane surfacetension forces
(shear forces) and viscosity forces will make the outergraviton half shear off from the incoming graviton,
e.g. in a halfmass times velocitysquared energyreaction.
The central core of the incoming graviton might now travel straight through
the electronsurface membranes and emanate out of the front of the decelerated electron.
The central core of the (halfmass) incoming graviton will now have the capacity to absorb matter
from the surface membrane at the front of the electron, i.e. as it is now at this point in space
an outgoing graviton. The former (as we call it) incoming (nowoutgoing) graviton will pull
the Newtonian corpuscles out of the front of the decelerated electron, i.e. due to viscosity forces
overriding shear forces. This viscosityforce
phenomenon will now cause matter, i.e. the socalled Newtonian corpuscles, (from the back of the electron where the incoming photon was absorbed)
to be pulled through to the front of the decelerated electron and into the outgoing graviton.
This classical phenomenon, i.e. involving the forces of shear, viscosity, equal and opposite reactions
and the Newtonian Law of the reversibility of light, will now cause the excess mass of the free electron
to flow towards the centre of the decelerated electron inside its inner hemisphere, i.e. where the
forces of shear stress will dominate over the forces of viscosity. The excess mass will now be forced
to leave the decelerated electron in the form of a photon (with its direction and internal spin already
well established). The decelerated electron will now remain in the groundstate orbit with half of its
full mass, e.g. as we now know it as the neutral electronproton pair, i.e. The electronproton pair will now follow the
Coulombforce equation of G.F.C. Searle (see Fig. 1). The electronproton pair will also follow the
first law of thermodynamics, i.e. in regards to lowest entropy, as the electronproton pair is now at Absolute Zero, so
the divergence of material flow will also be zero, (after J.C. Maxwell).
So we see that Newton's and Kepler's ancientclassical laws can be applied to the atomicground state.
Together with the modernclassical laws of Kelvin, Carnot and Clausius (which influenced Maxwell and Heaviside,
causing them to draw up their mathematics classically), the laboratory experiments remain to be proven so that the modernday
Hydrogenatom no longer remains lost in the mythical mists of Time. (Please see the portabledocuments file on The Experiments if you have time).
Perpetual Motion (GB2410770), Antigravitational Force Engines (GB2368910), Free Energy and all that remains for future studies
So where does all of our
classical mechanics and classical physics lead us? It leads us to experiments
to test for an antigravitational forceengine
(British Patent number GB2368910). British Patent number GB2368910
is a spinning gyroscope with spiralhoriontal arms. By D'Alembert's Principle,
a stationary object which spins at the Earth's escape velocity (11,181 metres
per second), will decouple from the Earth's gravitational acceleration,
(Newton's Second Law),
just tha same as will an object which travels past the Earth with a velocity
which is greater than the Earth's escape velocity (Newton's First Law).
What would occur is the following; the Earth's gravitationalacceleration field
would be moved outward from the interior of the spinning gyroscope, compressed
at the exterior perimeter of the spinning gyroscope causing increaseddownward
gravitationalacceleration and would permit an object to have decreaseddownward
gravitationalacceleration, (i.e. if the object were placed over the centre of the
spinning gyroscope). This is probably due to the Devil, (various gnomes) and the
Devil and the gnomes would probably say "It would have to be energised by
a perpetualmotion machine".
This study leads us to near100%
efficient electricitygenerators. Patent GB2410770, (by the author), describes
how the Maximum Power Transfer Theorem can be applied to a lessthan
atmosphericpressure liquidboiler, i.e. water at room temperature in a
nearvacuum. The Maximum Power Transfer Theorem states that to have maximum
efficiency of electricalenergy transfer between an electricity generator and
its load (the output, e.g. a lightbulb), the resistance of the
(generatorwinding) wire that the magnet passes across must equal the
resistance of the generatoroutput load, e.g. a wire filament in a light bulb.
At room temperature,
resistance in a wire that the magnet passes across, will always produce heat.
The heat produced is always measured in Watts as I squared times R, (where I is
the current in Amperes and R is the resistance in Ohms). The heating is known
as Joules heating and the power of this heating is always equal to the power
produced by the electricity generator, (i.e. which the light bulb uses) and
power is also commonly known as energy dissipation per second.
The heat produced by an
electricity generator is normally very low, e.g. it is slightly warmer than
room temperature, hence the power is dissipated as low energy dissipation per second,
compared to the power of a 100 Watt light bulb (a light bulb is highenergy
dissipation at a high temperature). The trick then to produce useful electrical
energy is to lower the atmospheric pressure around the magnet passing across
the wire (the magnet and the wire are now placed underwater). Water will start to boil
at 20 degrees Centigrade if the atmospheric pressure is reduced to
approximately 1% of atmosphere pressure.
If a heated fluid at 30  35 degrees Centigrade flows under the boiler base and the working fluid inside the boiler is at 20 degrees Centigrade, then the water will boil and the water vapour
will push the water up a tube, e.g. much like a geyser. A propeller placed at
the top of a shaft will rotate due to the geyser pressure and a magnet placed at the bottom of the propeller shaft will rotate across the generatorwinding wire (at the bottom of the
shaft within the boiler tube) and produce electrical power, i.e. 100 Watts. The generatorwinding wire will also
produce 100 Watts of Joules heating at ~30  35 degrees Centigrade and this heating will produce extra water vapour, which in turn will produce extra geyser
motion. This extra geyser power will produce extra dynamo power, i.e. at 100 Watts (considering the efficiency of the dynamo). The propeller will rotate the shaft ever faster (until the cooling ability of the coolant fluid at the top of the propellershaft tube is exhausted).
The coolant fluid must be able to
condense all of the vapour which is pushing up the water or
the pressure within the propellershaft tube will rise and hence the
temperature that the water will boil at, will rise above 20 degrees Centigrade.
If the pressure rises within the propellershaft tube,
then a higher and highertemperature heatsource will be required, until the pressure within
the propellershaft tube is at normal (outside) ambient pressure and
we will have lost the ability to boil water at less than 100degrees Centigrade.
If we can maintain the pressure within the propellershaft tube at ~1% of normalambient pressure,
then the efficiency of the
electricity generator will be raised to near 100% and if waste heating is
available from other appliances, (i.e. from other heat sources above 20 degrees Centigrade),
then perpetual motion can be achieved and perpetualfree electricitygeneration can be achieved.
A 100Watt tungstenfilament lghtbulb produces 10 Watts of visiblewhite light and 90 Watts of
infrared heat. If you place the 100Watt lightbulb, i.e. a motorcycle/bicycle headlight bulb,
back into the inside of the vacuum container, then you wll get 90% OF YOUR POWER BACK!!!
This patent is obtainable at £20,000,000 per licence (from The Hague). For maximising optimal profit from developing a nanotechnology product
to obtain cheapelectrical energy from renewables, I will provide a consultancy for up to £2 million a year, i.e. for a guaranteed return from a patentable
item.
Please feel free to send questions to Dunstan Dunstan at dunstand123@gmail.com