Free from; http://www.reciprocalsystem.com/rs/satz/cover.htm

(PDF made on 24 July 2009)

RECIPROCAL SYSTEM

DYNAMICS

Ronald W. Satz

RECIPROCAL SYSTEM DYNAMICS

RONALD W. SATZ

Some Myths of Modern Physics

White Lies About Black Holes

A Tall Tale: A Review of Stephen W. Hawking‘s A Brief History of Time

Four Scientific Mysteries Unraveled

Reference Systems

Reference Systems and Speed Limits in the Reciprocal System

The Lorentz Transformation

The Two-Photon Problem

A Note on Scalar Motion

A Note on the Force of the Space-Time Progression

On the Nature of Undisplaced Space-Time

Clock Space, Coordinate Space, Clock Time, Coordinate Time

Particle Physics

Cosmic Rays and Elementary Particles

Identification of Cosmic Particles 3695 MeV/c² and 3105 MeV/c²

A Note on the Cosmic Proton

The Cohesive Energies of Crystals of the Elements

Progress on the Theoretical Calculation of the Cohesive Energy of the Elements

Equation of State of Solid Matter

Further Mathematics of the Reciprocal System

A New Derivation of Planck‘s Constant

Time Region Particle Dynamics

Calculation of the Dissociation Energy of Diatomic Molecules

The Liquid State in the Reciprocal System: The Volume/Pressure Relation, part I

The Liquid State in the Reciprocal System: The Volume/Pressure Relation, part II

Electricity and Magnetism

Permittivity, Permeability and the Speed of Light in the Reciprocal System

The Unit of Magnetic Charge

Photoionization and Photomagnetization

Theory of Electrons and Currents

Astrophysics

Hubble‘s Law and the Reciprocal System

Globular Cluster Mechanics in the Reciprocal System

Stellar Energy Generation in the Reciprocal System

The Gravitational Attraction of the Galaxy

Discussion of Larson‘s Gravitational Equation

The Gravitational Formula at High Velocities

A Crucial Experiment

The Interaction of Alpha Particles and Gold Atoms

Proving Rutherford Wrong

A Proposal for a Crucial Experiment

More Calculations with the R.S. Scattering Equation

Detailed Steps for the Design and Performance of the Proposed Crucial

Experiment More Details for the Proposed Crucial Experiment

WHITE LIES ABOUT BLACK HOLES

Kip Thorne introduces his article on black holes in Scientific American by stating:

Or all the conceptions of the human mind from unicorns to gargoyles to the hydrogen

bomb perhaps the most fantastic is the black hole: a hole in space with a definite edge

over which anything can fall and nothing can escape; a hole with a gravitational field so

strong that even light is caught and held in its grip; a hole that curves space and warps

time. Like the unicorn and the gargoyle, the black hole seems much more at home in

science fiction or in ancient myth than in the real universe. Nevertheless, the laws of

modern physics virtually demand that black holes exist. In our galaxy alone there may be

millions of them.

Thorn is saying that because the ―laws‖ of modern physics require them, black holes

must exist. However, it is more rational to conclude that those ―laws‖ which give rise to

the gargoyles, unicorns, and black holes of physics are wrong—that, as ordinarily

expected, deductions from false premises yield bizarre results. Let us now investigate

how such concepts as the black hole arose historically. The theory of the black hole stems

from the theories of general relativity, the nuclear atom, and the hydrogen-to-helium

conversion process in stars.

In the 1930s, Subrahmanyon Chandrasekhar‘s investigation of stellar evolution and

structure led him to conclude that, in the process of converting hydrogen to helium, most

stars lose energy and contract until internal pressures become great enough to cause

collapse of atomic structure. Back in 1924, Sir Arthur Eddington had suggested that the

high density of the white dwarf companion of the bright star Sirius was due to ―electron

degeneracy,‖ with all electrens stripped from individual atoms. Chandrasekhar seemed to

provide an explanation of how this could occur.

At this point someone might have pointed to a simpler solution: perhaps the nuclear atom

concept was incorrect because of the grave difficulty in explaining the high density of the

white dwarfs. Perhaps atoms do not have electrons circling around ihem at relatively

large distances. Perhaps the postulated hydrogento-helium conversion process in stars

was incorrect...

Chandrasekhar believed that a ―non-relativistic gas‖ at the center of a white dwarf could

always adjust itself until the gravitational forces compressing the star are countered.

liowever, according to the theory of general relativity, with a certain limiting mass, the

gravitational forces are net countered fully and so the star does not come into

equilibrium. The limiting mass, termed the Chandrasekhar limit, has been calculated to

be 1.2 solar masses.²

Oppenheimer and Volkoff considered what would happen to stars of mass larger than the

Chandrasekhar limit. As the central density increases, inverse beta decay would take

place, driving electrons into protons. Thus increasingly rich neutron elements would be

formed—giving rise to a ―neutron star.‖ Recently astronomers have concluded that the

pulsars are neutron stars.

However, it must be pointed out that there is no evidence that pulsars are neutron rich, in

the same way that there is no evidence that white dwarfs are electron degenerate. In order

to obtain such densities with the nuclear atom concept, those deductions might be correct.

But there is an atomic theory, developed by Mr. Larson, that explains such high densities

without the use of the nuclear atom...

According to current theory if the remaining mass exceeds two solar masses, it will

continue to contract to a ―Schwarzchild singularity,‖ a bottomless pit, a black hole. The

properties of a black hole are supposed to be:

1. a gravitational field so strong that not even light can escape, and thus no observer

can see any phenomena occurring within the Schwarzchild radius;

2. a curvature of the space-time whirlpool becoming infinite at the central

singularity;

3. a circumference of 19 kilometers multiplied by the mass of the hole and divided

by the mass of the sun;

4. a mass of between 3 and 50 solar masses;

5. a compositicn of matter compressed to near infinite density, losing in the process

every property of separate identity except mass, electric charge, and angular

momentum.

To say the least, such properties are astounding. It is a relief to know that the Reciprocal

System, developed by Mr. Larson, contains no such theoretical objects: Here is a brief

tabulation of the relevant points of the Reciprocal System:

1. Atoms are not composed of electrons, protons, and neutrons, but are whole units

comprised of various rotational motions; at equilibrium there is equality of inward

and outward forces on groups of atoms; great compression can take place without

―electron degeneracy‖ or ―neutron formation.‖

2. In the sector of the universe in which we live there are two regions. In the time-

space region, gravitation is inward, whereas the space-time pregression is

eutward. In the time region, which lies within unit space, the motions are

reversed: gravitation is outward, the progressicn inward. According to the theory,

during a type I supernova explosion, part of the material is dispersed outward in

space (to form a red giant star) and part dispersed outward in time (to form a

white dwarf). The expansion outward in time is equivalent to a contraction in

space—hence the extreme density of the white dwarfs. Type I supernovae occur

because a thermal limit is reached in the energy conversion process taking place.

Type II supernovae occur because of a stellar age limit. Here, instead of a white

dwarf being formed, a pulsar is formed. The type II process is what ultimately

produces the quasars. All of these high speed explosion products—white dwarfs,

pulsars, and quasars—originate from expansion in time.

3. A different mechanism of energy generation is postulated, which in turn produces

a different pattern of stellar evolution. In the Reciprocal System, stars slowly

increase in mass and temperature until the destructive thermal limit of the iron

group elements is reached. At this point, a type 1 supernova occurs, creating a red

giant and white dwarf. Gravity acts in both directions to bring the white dwarf and

red giant back to the main sequence. There is no stellar death into a black hole.

Simply, a succession of type 1 supernovae occur until the star reaches its upper

age limit and terminates in a type II supernova, producing pulsars which

eventually leave this sector for the space-time region.

What about claims of observation of a black hole? Kip Thorne states that he is 90%

certain of a black hole in Cygnus X-1. It seems that its mass is eight times that of the sun

—meaning that in current theory, a white dwarf and neutron star are ruled out. However,

in the Reciprocal System no such mass limit exists. In fact, it is apparent that the ―black

hole‖ in Cygnus X-l, because of its copious emissions of radio waves and X-rays, is

really a body that will eventually become a pulsar. It is a product of a type II supernova.

At present the periodicity of its radiation is not distinguishable from continuous radiation,

but as the high speed explosion product moves outward it will be. Thus here is a test

between current theory and Reciprocal System—we predict that this so-called black hole

will turn out to be a pulsar, but a pulsar that is more massive than any neutron star could

be.

Currently, it is postulated that black holes account for the great mass discrepancy in giant

elliptical galaxies. Mr. Larson provides the explanation from the Reciprocal System:

A star pressure is building up in the interiors of the older galaxies; that is, an increasing

proportion of the constituent stars are being accelerated to uttra high speeds by the energy

released in the explosion of stars that reach the destructive age limit. The cores of these

galaxies are thus in the same condition as the white dwarf stars and quasars; their densiiy

is abnormally high because the introduction of the time displacement of the ultra high

speeds reduces the equivalent space occupied by the central portion of the galaxy. In

brief, we may say that the reason for the abnormal relation between mass and luminosity

in the giant ellipticals is that these galaxies have white dwarf cores—not white dwarf

stars in the core, but white dwarf cores.

It seems that many individuals are intrigued with the term ―black hole.‖ Perhaps we could

retain this term in the Reciprocal System to denote the location at which mass has left this

sector for the inverse sector, the space-time sector:

References

1. K. S. Thorne, ―The Search for Black Holes,‖ Scientific American, December

1974, p. 32.

2. Martin Harwit, Astrophysical Concepts, (New York: John Wiley and Sons, 1973),

p. 359.

3. J R. Oppenheimer and G. M. Volkoff, ―On Massive Neutron Cores,‖ Phys. Rev.,

56, 455 (1939).

4. Dewey B. Larson, Quasars and Pulsars, (Portland, Oregon: North Pacific

Publishers, 1971), pp. 148-149.

A TALL TALE

A Review of Stephen W. Hawking‘s A Brief History of Time

This paper will critique the conventional physical theory espoused by Stephen Hawking

(hereafter abbreviated to SH) in his best-selling book. From the perspective of the

Reciprocal System (hereafter abbreviated to RS), the book is full of errors on practically

every page. Still I urge members of ISUS to read it-SH does a good job of presenting the

―establishments‖ viewpoint and its worth pondering his thinking. The book,s dust jacket

proclaims SH to be the ―most brilliant theoretical physicist since Einstein‖ yet he begins

his conclusion (p. 171) with the words ―We find ourselves in a bewildering world.‖ This

is a far cry from my Unmysterious Universe, published in 1971.

SH is confused as to whether the universe was created or not. He states that in ―imaginary

time‖ the universe has no beginning or end, no singularities or boundaries—it simply is.

But in ―real time‖ the universe does have a beginning and an end at singularities that

form a boundary to space-time and at which the laws of science break down (p. 139). The

beginning singularity is, of course, called the ―big bang‖, when the universe was

―infinitesimally small and infinitely dense‖(p. 8) and ―infinitely hot‖ (p. 117) and space-

time had infinite curvature [!]. Time, and by implication, space-time, had no meaning

prior to the beginning (p. 8). SH defines an event (p. 23) as ―something that happens at a

particular point in space and at a particular time. So one can specify it by four numbers or

coordinates.‖ This is the conventional four-dimensional space-time, 3 spatial coordinates

and 1 temporal coordinate, which is quite different from the 3-dimensional space-time of

the RS (where each dimension is a dimension of motion, not of space or time

individually). In the RS, space-time has an inbuilt expansion and hence there is no need

for a ―big bang‖ to explain the recession of the distant galaxies. SH states (p. 33) that

―Space and time are now dynamic quantities: when a body moves, or a force acts, it

affects the curvature of space and time—and in turn the structure of space-time affects

the way in which bodies move and forces act.‖ Certainly there is no physical evidence for

this; bare space-time is nonphysical since it cannot be changed into something else—it is

a simply a reference system for motion. SH does not specify a mechanism for producing

this alleged warping of space-time by the ―distribution of mass and energy in it‖ (p. 29).

SH treats space and time as both purely relative, not absolute (p. 21, 33); he says that

"each observer must have his own measure of time, as recorded by a clock carried with

him, and identical clocks carried by different observers would not necessarily agree.― Of

course, this subjectivist belief leads to numerous logical contradictions and is wrong;

space-time is the fundamental component of the universe and must thus be absolute

(there is nothing for it to be ―relative to‖ since it is itself the reference).

An electromagnetic field fills the space-time of SH. There can be (p. 18)―wavelike

disturbances in the combined electromagnetic field‖; (p, 38) ―..visible light consists of

fluctuations or waves, in the electromagnetic field‖. But (p, 54) ―Although light is made

up of waves, Planck‘s quantum hypothesis tells us that in some ways it behaves as if it

were composed of particles: it can be emitted or absorbed only in packets or quanta.‖

This is the old wave-particle duality. Compare this with the definition of photon in the

RS: a linear vibration within a same space-time unit, which itself progresses

perpendicularly, the combination thus generating a wave motion. SH states (p. 31) that

―Light rays too must follow geodesics in space-time‖and (p. 28) ―,nothing can travel

faster than light‖. In the RS, spatial motion is limited to speeds lower than that of light,

but temporal speeds are higher: the speed of light is the midpoint of the speed range of

the universe, not the upper limit. According to SH (p. 117) ―One second after the big

bang, (the temperature] would have fallen to about ten thousand million degrees... At this

time the universe would have contained mostly photons, electrons, and neutrinos... and

their antiparticles, together with some protons and neutrons.‖ SH does not provide a

fundamental definition of electrons or neutrinos (the quark hypothesis does not include

them; quarks are the undefined elementary particles of matter). He states (p. 65) that ―A

proton contains two up quarks and one down quark; a neutron contains two down and one

up.‖ (p. 73) ―One cannot have a single quark on its own... confinement prevents one from

observing an isolated quark or gluon‖ (which carries the ―strong nuclear force‖]. (p, 75)

―,the uncertainty principle means that the energy of the quarks inside the proton cannot

be fixed exactly. The proton would then decay.‖ (p, 73) ―another possibility is a pair

consisting of a quark and an antiquark... such combinations make up the particles known

as mesons, which are unstable because the quark and antiquark can annihilate each other,

producing electrons and other particles.‖ Each subatom has a spin, but (p. 66) ―quantum

mechanics tells us that the particles do not have any well-defined axes‖. (p. 67) ―Particles

of spin 1/2... make up the matter of the universe, and particles of spin 0, 1, and 2 give rise

to forces between the matter particles.‖ (p. 69) ―The electric repulsive force between two

electrons is due to the exchange of virtual photons (spin 1)‖ and (p. 70) ―the force

between two matter particles is pictured as being carried by a particle of spin 2 called the

graviton.‖ This is all in contrast with the RS, in which each subatom is a set of quantized

spins of a photon, with definite axes. The mesons are actually cosmic atoms in the

process of converting to the prevailing structures of our sector. The apparent force

interactions are not due to exchange of particles (virtual or otherwise), but rather

represent an interaction between the particle and the omnipresent space-time progression.

SH continues (p. 117): ―About one hundred seconds after the big bang, the temperature

would have fallen to one thousand million degrees, the temperature inside the hottest

stars. At this temperature protons and neutrons would no longer have sufficient energy to

escape the attraction of the strong nuclear force, and would start to combine together to

produce the nuclei of atoms of deuterium... The deuterium nuclei then would have

combined with more protons and neutrons to make the helium nuclei... and also small

amounts of a couple of heavier elements, lithium and beryllium... Within only a few

hours of the big bang, the production of helium and other elements would have stopped.‖

SH doesn‘t question the validity of the nuclear theory of the atom. No mention is made of

the instability of the neutron: how can an atom be stable if one of its main components is

known to be unstable? Also, why don‘t the alleged protons in the nucleus repel each

other? To explain the ―strong nuclear force‖ as due to the exchange of unobserved

―gluons‖ is mystical. Also, the alleged orbiting electrons are thought not to combine with

protons and neutralize their charges, whereas other pairs of oppositely charged particles

do. Finally, why should the production of new elements cease? It seems more likely that

over the course of thousands, or millions, or billions of years that atoms would continue

to combine to form heavier atoms, and these would join to form ever more complex

molecules. SH states (p. 60) that ―Since the structure of molecules and their reactions

with each other underlie all of chemistry and biology, quantum mechanics allows us in

principle to predict nearly everything we see around us, within the limits set by the

uncertainty principle. (In practice, however, the calculations required for systems

containing more than a few electrons are so complicated that we cannot do them.)‖ Well,

what good is a theory if we can only use it in principle?

SH continues (p. 117): ―The universe as a whole would have continued expanding and

cooling, but in regions that were slightly denser than average, the expansion would have

been slowed down by the extra gravitational attraction. This would eventually stop

expansion in some regions and cause them to start to recollapse... in this way disklike

rotating galaxies were born.‖ SH uses the uncertainty principle to explain the non-

uniform density. He says (p. 140) that ―... there must have have been some uncertainties

or fluctuations in the positions or velocities of the particles. Using the no boundary

condition, we find that the universe must in fact have started off with just the minimum

possible non-uniformity allowed by the uncertainty principle.‖ The RS explanation is

much better: the two main forces of the universe are the space-time progression and

gravitation; where gravitation is stronger, galaxies and stars are formed; where the

progression is stronger, the galaxies move away from each other. Also, the latest

evidence is that the formation of galaxies is not a one-time occurrence—observations

indicate that galaxy building is going on right now, just as the RS predicts.

An average galaxy has a hundred billion stars. SH states (p. 82) that ―A star is formed

when a large amount of gas (mostly hydrogen) starts to collapse in on itself due to

gravitational attraction. As it contracts the atoms of the gas collide with each other more

and more frequently and at greater and greater speeds—the gas heats up. Eventually, the

gas will be so hot that when the hydrogen atoms collide they no longer bounce off each

other, but instead coalesce to form helium. The heat released in this reaction, which is

like a controlled hydrogen bomb explosion, is what makes the star shine. This additional

heat also increases the pressure of the gas until it is sufficient to balance the gravitational

attraction, and the gas stops contracting... Stars will remain stable like this for a long

time... Eventually, however, the star will run out of its hydrogen and other fuels.

Paradoxically, the more fuel a star starts off with, the sooner it runs out.‖ This is

paradoxical indeed, for the larger stars should be the oldest and the smaller stars the

youngest. In fact, the observed evidence indicates that the astronomers have the

evolutionary sequence precisely upside down. SH continues (p. 83, 84, 87): ―When a star

runs out of fuel, it starts to cool off and so to contact... If a star‘s mass is less than the

Chandrasekhar limit, it can eventually stop contracting and settle down to a possible final

state as a white dwarf with a radius of a few thousand miles and a density of hundreds of

tons per cubic inch... [If a stars mass is greater than the Chandrasekhar limit, the star will

eventually collapse to a black hole] in which neither light nor anything else can escape.‖

In the RS, stars slowly accumulate mass, rather than losing mass, and there is no end in a

black hole. Rather at the mass limit, a supernova explosion occurs, and a red giant/white

dwarf pair is formed (or a planetary satellite system) and both stars eventually return to

the main sequence. Its interesting that SH proposes a radical theoretical change in the

black hole construct: he says that actually a black hole would emit particles and radiation

and (p. 115) ―..the black hole, along with any singularity inside it, would evaporate away

and eventually disappear.‖

SH continues (p. 46): ―The present evidence therefore suggests that the universe will

probably expand forever, but all we can really be sure of is that even if the universe is

going to recollapse, it wont do so for at least another ten thousand million years, since it

has already been expanding for at least that long.‖ Compare this with the RS: there are

two sectors, the material sector, and the cosmic sector. Outward spatial expansion in the

material sector is terminated with a galactic explosion which sends the matter over to the

cosmic (inverse) sector, where outward temporal expansion (the inverse of spatial

expansion) occurs. This expansion in turn is terminated with a cosmic galactic explosion

which sends the matter back to the material sector. Thus the main process in the universe

is cyclic, rather than a singular one-time expansion. Also the net total displacement in the

universe is zero, because the number of cosmic displacement units balances the number

of material displacement units (whereas SH states (p. 129) ―... the total energy of the

universe is exactly zero... this negative gravitational energy exactly cancels the positive

energy represented by the matter.‖)

Having carefully studied this book, I think the supporters of the RS have nothing to fear.

If this is the best popular rendition of conventional theoretical physics, the future looks

bright for the spread of the RS, which is a unified, general theoretical system. (Contrast

the RS with the current ―Grand Unified theories‖, which (p. 156) ―are not very

satisfactory because they do not include the force of gravity and because they contain a

number of quantities, like the relative masses of different particles, that cannot be

predicted from the theory but have to be chosen to fit observations‖. Further details of the

RS refutation of the ideas expressed by SH can be found in Dewey Larson‘s book The

Universe of Motion (particularly chapter 29) and his article ―The Mythical Universe of

Modern Astronomy,‖ Reciprocity, Vol. XII, No. 2, Autumn, 1982.

FOUR SCIENTIFIC MYSTERIES UNRAVELLED

I. Quarks

―Quarks are fantastic jive‖ :

James Joyce might have said if alive.

―We started with three

For atomic debris,

And now we find we have five.‖

(F. A. Moen)

(A wake for a fellow named Finnegan

Gave theorists quarks to begin again.

Having swallowed down four,

They keep asking for more

And a theory by which they can win again.)

(Ed., Science News, August 7, 1976)

The present scientific premise is that matter is ultimately composed of a set of matter

particles called quarks. Originally three quarks and three anti-quarks were postulated.

These were called ―up,‖ ―down ― and ―strange. To account for recently observed particle

behavior, theorists hsve postulated a fourth quark a ―charmed‖ quark as it‗s called. To get

around the Fermi-Dirac statistics, an additional quantum number has now been

postulated, ―color.‖ Then a class of intermediate particles called gluons was formulated to

carry the force between quarks. Finally other theorists began to feel that now quantum

numbers beyond charm and color are at work. One theorist said that beyond charm and

color lay truth and beauty. Thus the current hypothesis is a set of as many as seven quarks

and seven antiquarks, which are supposed to combine to form all hadrons. The leptons

(the electron, the muon, and the two kinds of neutrino) are considered different, showing

no intelnal quark structure.

To me such terms as ―strange,‖ ―color,‖ and ―charm‖ are mere names covering a great

deal or ignorance. They remind me of such terms as ―phlogiston‖ and ―levity.‖

Phlogiston, in early chemical theory, was the hypochetical principle of fire, of which

every combustible substance was in part composed.. Levity was a term employed by the

Montgolfier brothers to explain why smoke rose in a chimney. None of these terms really

advanced scientific understanding.

The Reciprocal System developed by Dewey Larson avoids any such mysterious terms.

The basic premise here is that the fundamental component of the universe is motion,

existing in discrete units. Each atom is composed of three types of motion or rotational

spin, rather than threa or more types of quarks. The difference between ―leptons‖ and

―hadrons‖ lies in the number of dimensions of the motion.

The quark theory gives no indication as to how matter can change into radiation and vice

versa. By contrast the Reciprocal System provides a simple answer: both atoms and

radiation are forms of motion and one can simply change one form of motion into

another.

II. Nuclear Atom

According to present theory, all forces between entities result from the exchange of

quanta. It has been theorized that the force between the supposed neutron and proton in

the atom results from an exchange of mesons. These mesons were supposed to be

charged, thus producing a current when in motion. Benson T. Chertok spent some ten

years of work at the Stanford Linear Accelerator . Center to find these meson-exchange

currents. As quoted in the Science News of May l0, 1975, Chertok said his experiment

found no evidence for these mesonexchange currents.‖Ten years of work shot down,‖ he

says.

This is typical of present practice, where theorists keep on postulating entities that

experimenters do not find, and experimenters keep on finding entities that theorists have

not postulated.

In the Reciprocal System, the atom is not compesed of neutrons, protons, and electrons,

and thus no force is needed to explain the attraction between protons and neutrons. The

atom is really one whole unit, composed of three different types of rotational spin.

III. Electrostatics

Present theory assumes that matter is composed of a large number of charged electrons

circling nucleuses of neutrons and protons. It is interesting to derive a consequence of

this hypothesis, showing its implausibility. This derivation comes from the

Encyclopaedia Britannica's article on electricity.

The normal density of solid matter is around five grams par cubic centimeter or 5000

kilograms per cubic meter and a gram molecule of a single compound occupies

something between five and 100 cubic centimeters. Thus the number of atoms per cubic

centimeter is in the region of 1022 to 1023 . The number of electrons pet atam varies from

one to more than 92 in the periodic table, but the number of relatively loosely bound

valence electrons per atom that can be appreciably influenced by electric fields is usually

small, in many cases there is only one per atom. ,As a general rule these are of the order

of 1023 effective electrons per cubic centimeter of 1029 per cubic meter. The total charge

associated with these electrons is thus some 1010 coulombs per cubic meter. The number

of electrons lying within one atomic diameter of the surface of a solid is 1029(2/3) per

square meter (i.e., about l020) a:nd so the available charge per square meter is of the order

of 10 coulombs per square mater. Now a surface charge density of 10 coulombs per

square meter is associated with an electric field normal to the surface of 10/ = 1012 volts

per meter. Even in extreme cases fields rarely exceed 108 volts per meter, and so, in

general the surface charges that appear in elec-troscatics correspond to fractional changes

in the internal charge distribution. Furthermore, a field of 1012 volts per meter

corresponds to a field of 100 volts per angstrom. A field of this strength would be

sufficient to disrupt and destroy the surface atoms. . . . !

It certainly wasn't the purpose of the author of the article to question the validity of

present atomic theory -- but he has surely provided an excellent disproof !

Thankfully, electric charges are not inherently in matter of the Reciprocal System.

IV. Gravitation

Present theories assume that the gravitational effect is propagated at the speed of light.

However, following LaPlace it is possible to show that if gravitational force is

propagated, the velocity must be at least 10°c!

(i) Assume that the gravitational force on the earth due to the sun is ―falling‖ upon the

earth with a velocity of ß.

(ii) Let r be the radius of the earth's orbit and let v be the earth's orbital valocity. Then the

component of solar acceleration in the direction of the earth's motion is

a = GM /r² / v/ß

(iii) By the method of perturbations, this acceleration can be shown to change the radius

of the earth's orbit at the rate

dr/dt = 2 · v²/ß

(iv). The celestial longitude 1(t) of the earth may by expanded in a power series:

1(t) = n t + A(n t)² + . . . (3)

where n is the present mean motion given by

no = (v/r) PRESENT (4)

and t is the time measured from the present. Now

n = v/r = dl/dt (5)

Differentiating eq. (3) twice gives

d²1/dt² = dn/dt = 2An ² (6)

From orbital mechanics,

n²r³ = const (7)

Using eqs. (7) and (2),

dn/dt = -3n² / v/ß (8)

Thus the coefficient of the second term of the celestial longitude equation is

A = -2/3 / v/ß

(v) In Che case of the earth the velocity is

v = 10-4 c (10)

If ß were equal to c, then A would have to be -1.5 x 10-4. In one century, the second

power term in equation (3) would become -60 radians or -1.2‖ x l07. Ant the largest

admissible variation is 2‖ . Therefore, ß must be greater than 6 x l06 c. This same result

occurs if other planets are considered. The assumption that gravitational force travels

with a speed of ß seems always to lead to a relation

ß > 106 c (11)

The relativists try to get around this problem by asserting that it is the gravitational

potential that is propagated. But they don't explain just how that is accomplished.

The Reciprocal System solves the problem by showing that the gravitational effect is not

propagated at all. Gravitational motion is simply an interaction between the inherent

rotational spin of the atoms and the translational progression of space-time.

One reason why Newton's Law of Gravitation is no longer considered fully tenable is tho

observation of the advance of the perihelion of the planet Mercury. The teory of General

Relativity came up with a rather complicated expression to explain the effect. However,

as I showed mathematically in a previous paper, this effect is really in the same class as

those other high velocity effects. And so the mathematical tretatment is of the same

nature as that of Special Relativity.

In the case of a planet orbiting the sun, the gravitational expression becomes

F = G m m2 / r² (1-v²/c²)½ (12)

In the case of an object moving toward another object, the expression is

F = G m m2 r² (1-v²/c²) (13)

These expressions provide an obvious unification of gravitational. and electrical effects,

and I'm amazed that these expressions haven't been stumbled on before The advance of

the perihelion of Mercury is fully explained by eq. 12. Note that eq. 13 shows that force

vanishes at the speed of light—this indicates that the natural datum of the universe is the

speed of light, rather than zero velocity.

Tn sum up some problems of modern physics:

1. Originally 3 quarks and 3 anti-quarks were postulated as the fundamental

components of matter. To account for recent experiments, this has been expanded

into 7 quarks and 7 anti-quarks—and the end is not in sight.

2. The meso-exchange currents supposed to carry the force between nucleons have

not bcen found.

3. The assump:ion of electric charges in matter leads to the conclusion that an

electric field 100 volts per angstrom is developed on surface atoms, an effect

certainly not observed.

4. The assumption a finite velocity the gravitational force leads to the conclusion

that this must be at least 106 c, rather than c.

Reciprocal System is. able solve all ef these problems in elegant fashion. It for such,

reasons I entitled my book Unmysterious Universe.

References

1. Science News, August 7, 1976.

2. Science News, June 26, 1976.

3. Science News, May 10, 1975.

4. ‖Electricity,‖ Encylopaedia Britannica, Fifteenth Edition, Vol. 6, pp. 550-551.

5. H. P. Robertson and Thomas W. Noonan, Relativity and Cosmology

(Philadelphia:W. B. Saunders Company, 1968), 401-403.

6. R.W. Satz, Question Box,‖ Reciprocity, Vol. IV, No. 2, July, 1974.

REFERENCE SYSTEMS AND SPEED LIMITS

IN THE RECIPROCAL SYSTEM: A REVIEW

Current theoretical physics views time as one-dimensional and constituting a kind of

quasispace which joins with the three dimensions of space to form a four-dimensional

space-time framework, within which physical objects move one-dimensionally. This view

has been formulated to help explain some of the new phenomena discovered in the

twentieth century, such as the very small, the very large, and the very fast. These

phenomena exist outside of our normal everyday world, where Newton‘s laws

predominate and where space seems to be totally separate from time. However, even with

this modern framework, most of the phenomena remain mysteries, in whole or in part.

In contrast, the Reciprocal System of theory, originated by D.B. Larson,¹ postulates that

both space and time have three-dimensional aspects and join together to form one entity,

space-time or motion, which itself is. three-dimensional. Space and time are the two

reciprocal aspects of motion and have no properties other than what they have in motion.

Here, space-time or motion is theorized to be the sole component of the universe, not the

framework or the background for particles of matter. Matter in the theory is itself a form

of rotational motion and may move translationally in more than one-dimension

coincidentally. To be sure, these ideas are novel and unfamiliar but they do overcome the

current difficulties in treating phenomena of the very small, the very large, and the very

fast, as will be demonstrated in this paper.

The conventional physical reference system is based on three dimensions of space and

one of time. The space is considered to be stationary and the time is considered to be

ffowing. Within this space, material objects move as a function of time in one dimension

in a specific vectorial direction. This one-dimension of motion may be resolved into three

components, one along each of the three orthogonal axes of the reference system (usually

denoted x, y, and z).

In the Reciprocal System, space and time each have the properties of the other. Time is

three-dimensional, like space, and space progresses, like time. Of course in a

gravitationally-bound material environment, space appears to be stationary and

threedimensional and time appears to be onedimensional and progressing, and so the

conventional reference system works for this situation. In a gravitationally-bound cosmic

(or inverse) environment, where space and time are interchanged, time would appear to

be stationary and three-dimensional, and space would appear to be one-dimensional and

progressing. The inverse of the conventional reference system would work in this

situation. So, in the Reciprocal System two types of reference systems exist: the first with

threedimensions of space and one of time, and the second with three dimensions of time

and one of space. The first is applicable to objects which are aggregated in space (as in

our ordinary material sector) and the second is applicable to objects which are aggregated

in time (as in the cosmic sector). Conventional physical science does recognize anti-

particles and anti-galaxies but does not stipulate an ―anti-reference system.‖

A common mistake of students of the Reciprocal System is to deduce from the above that

there are thus six dimensions of the universe, three of space and three of time. This is not

so, however. All that actually exists are three dimensions of motion, not three dimensions

of space or three dimensions of time individually. Of course, where convenient, we can

mentally fix one component, while allowing the other to move. This has the effect of

concentrating on one aspect of each of the components while ignoring the others. But it is

important not to forget that space and time do not exist separately; they are bound

together in units of motion, which are the actual reality.

Outside of our gravitationally-bound region, what happens? It is an observed fact in

astronomy that distant galaxies are moving away from our galaxy (and all others) at

speeds approaching that of light. The current explanation is that this is a result of a

hypothetical big bang some 16 billion years ago. But this is not, of course, the

explanation of the Reciprocal System. Here, the cause is the space progression, which

manifests itself when gravitation is attenuated. It is an effect brought about by the motion

of the natural reference system relative to our conventional spatial reference system. The

equation for this motion (at the full speed of light) in our sector is

x1² + x2² + x3² = c²t² (1)

(where c, of course, is the speed of light). Please note that because the motion is actually

scalarly outward only, the imputation of a specific vectorial direction for a particular

space unit is arbitrary.

Similarly there is an outward motion of cosmic galaxies relative to a cosmic observer‘s

galaxy, such motion also approaching the speed of light in the limit. Again, this is not due

to some sort of big bang in the cosmic sector; rather it is due to the motion of the natural

reference system relative to the three-dimensional temporal reference system. At the full

speed of light, the equation is:

t1² + t2² + t3² = x²/c² (2)

Again, the assigned vectorial directions are arbitrary, since this is actually a scalar motion

outward in all directions.

The two major sectors of the universe, the material sector and the cosmic sector, each

with their appropriate reference system, are stable. In between these two sectors is an

unstable transition zone, which cannot be represented properly by either reference

system.

In the material sector, the velocities of ordinary phenomena are below that of the speed of

light. In the cosmic sector the inverse velocities of ordinary cosmic phenomena are below

that of the speed of light. Hence the speed of light is the dividing line between the two

sectors, and, in fact, the photons are not actually part of either sector, but at the boundary.

They have no independent motion (other than their vibration) and so remain in the same

spacetime location, which is carried outward by the space-time progression (in a

perpendicular direction) with respect to the conventional space or time reference system.

From the standpoint of a gravitationally-bound cosmic aggregate, the photons appear to

be moving outward in all coordinate time directions. Actually, however, the photons

remain stationary in the natural reference system; the matter particles are moving inward

in space and the cosmic particles are moving inward in time, and so the opposite motions

are imputed to the photons.

Because motion can be three-dimensional, the actual separation between the two major

sectors of the universe is scalar unit speed in all three dimensions, or 3c. Each of the three

dimensions is limited to a maximum of one unit of speed and if motion were limited to

one dimension we would agree with current physics that nothing could travel faster than

the speed of light. However, because the motion is not so limited, the actual limit is 3c.

Notice that we are summing the value of speed of each dimension. These are, after all,

scalar quantities, which can only be added or subtracted. It is not possible to add

vectorially the motions in the different dimensions of multi-dimensional motion! The

sequence of additions of units of speed are: first one unit, in dimension l; then the second

unit, in dimension 2; and finally the third unit, in dimension 3.

In our sector, velocity is measured as s/t. In the cosmic sector it is t/s. To material

observers, photons appear to move through space; to cosmic observers, photons appear to

move through time. Both observers see photons as being the upper limit of speed (or

energy). Many beginning students of the Reciprocal System conclude that when we talk

about motion at speeds faster than that of light, we are referring to rate of change of

position in space. But this is not so. At speeds above the speed of light the rate of change

applies to change of position in time, which moves objects further apart in time, or

(equivalently) moves objects closer together in space.

According to the Reciprocal System, motion exists only in discrete units, so the question

arises, how can we have fractional units? Obviously the velocities on Earth are only a

tiny fraction of the speed of light, so how can this be? Recall that the theory requires that

we start with one unit, not from zero units. This one unit of motion is equal to one unit of

energy, because of the reciprocal relation between space and time. To achieve effective

translational speeds below unity we simply subtract the appropriate number of energy

units from one. The equation in natural units is

v = 1-1/x (3)

where x is the number of one-dimensional energy units (with dimensions t/s). Note that as

x is increased the speed is increased, and in the limit reaches 1 (or c). In the time region,

the region inside unit space, the numerical value of the energy term must be squared, for

reasons given by Larson.² So the equation actually is

v = 1-(1/x)² (4)

Suppose x has the value n. Motion at this speed often appears in combination with a

motion 1-1/m² that has the opposite vectorial direction. The net result is

vH = 1/n²-1/m)² (5)

This is clearly recognized as the Rydberg relation (in natural units) that defines the

spectral frequencies of atomic hydrogen—the possible speeds of the hydrogen atom.

Other elements have similar relations for their thermal motion. The important point is that

translational motion is quantized; it is not a continuum. To this extent, we can agree with

current theory.

Because of the ability of adding or subtracting energy units to the three basic speed

ranges, we can have speeds of 1 - 1/x, 2 - 1/x, and 3 - 1/x. Larson denotes the speed range

1 - 1/x ―low speed‖; the speed 2 - 1/x, ―intermediate speed‖; and the speed 3 - 1/x, ―high

speed.‖ Because of the one-dimensional nature of energy, it is not possible to go from

one speed range to the next by simply adding more energy. The only way to accomplish

this is by direct addition of units of speed. And the only way that can be accomplished is

by huge stellar or galactic explosions.

In a Type I supernova explosion (caused when the temperature limit of the iron group of

elements is reached) part of the material moves outward in space at speeds less than that

of light and part moves outward in time (or inward in space) at speeds in the intermediate

range. This results in a very dense compact star, a white dwarf, stationary in space,

surrounded by a cloud of dust and gas moving spatially outward. Eventually the speeds of

the particles of the white dwarf fall below that of unit speed (or c) and the white dwarf

begins to expand in space, eventually becoming a normal star on the main sequence. The

dust cloud recondenses into a red giant star, which also eventually returns to the main

sequence.

A Type II supernova is even more powerful. Here the explosion results in high speed

motion of the compact star away from the scene, together with the usual cloud of dust

and gas. This compact start, or pulsar, is similar to the white dwarf, except that it has an

additional translational motion in the high speed range (or third dimension of motion).

Once the effect of gravitation is attenuated (and the net speed goes above two units) the

pulsar will leave our sector and move into the cosmic sector, where the processes of that

sector will disintegrate the spatial aggregate and recondense it in time by means of

cosmic gravitation. It will thus disappear from our view.

In the central regions of the largest galaxies, the spheroidal galaxies, consisting of 1012 to

1013 stars, the matter is at the upper limit of age. Instead of isolated Type II supernova

explosions, a whole chain reaction of such explosions occurs, resulting in galactic

fragments (between 7 x 107 to 2 x 109 stars) being ejected. The fragments ejected at

upper-range speeds are the quasars; those at low or intermediate speeds are the radio

galaxies. Note that although the quasar itself is moving at high speed, the particles of

which it is composed are moving at intermediate speed, hence the compact structure.

The Reciprocal System of theory explains many of the puzzling characteristics of

quasars. One such characteristic is the observed double image of some quasars. Larson

explains this as follows:

Scalar motion does not distinguish between the direction AB and the direction BA. The

lateral recession outward from point X is therefore divided equally between a direction

XA and the opposite direction XB by the operation of probability. Matter moving

translationally at upper range speeds thus appears in the reference system in two locations

equidistant from the line of motion in the coincident dimension (the optical line of sight,

in most cases).³

Hence there is no need for such a hypothesis as a ―gravitational lens.‖

Initially the explosion speed of the quasar is applied to overcoming the effect of

gravitation, and thus there is a rapid change of position in the reference system. As

gravitation is gradually overcome, the net speed increases, but the rate of change of

position decreases, because the speed in the explosion dimension is not visible in the

reference system. However, the speed in the explosion (high speed) dimension can be

detected by the shift in frequency toward the red of the radiation coming from the quasar

and received on Earth. This Doppler shift is a measure of the scalar sum of the outward

motions of the quasar, both that due to the recession and that due to the explosion speed.

It is a direct speed measurement, and the relative adjustment factors do not apply to it.

Hence values greater than 1 actually do mean speeds greater than c, which is what the

Reciprocal System requires. Thus the quasars are not nearly as distant as the current

cosmological explanation of the quasars‘redshifts suggests.

Another unusual characteristic of the quasars is their seemingly impossible great energy

generation. But the conventional assumption is that this energy is carried away in all

three dimensions by radiation. In the Reciprocal System, the radiation that comes from an

object at the upper range of speed is distributed two-dimensionally. As Larson states:

If we find that we are receiving the same amount of radiation from a quasar as from a

nearby star, and the quasar is a billion (109) times as far away as the star, then if the

quasar radiation is distributed over three dimensions, as currently assumed, the quasar

must be emitting a billion billion (1018) times as much energy as the star. But on the basis

of the two-dimensional distribution that takes place in equivalent space, according to the

theory of the universe of motion, the quasar is only radiating a billion (109) times as

much energy as the star,... which is equivalent to no more than a rather small galaxy.4

Larson calculates5 that the explosion redshift is a function of recession redshift and

normally takes the value 3.5 z½. The quasar will begin converting to the cosmic status

when this speed reaches a value of 2.0. The corresponding redshift is then 0.3265, and the

total quasar redshift (the sum of the recession redshift and the explosion redshift) is

2.3265. According to the observers, there is a sudden cutoff in the distribution of quasars

above a Doppler shift of 2.2, which is consistent with the theory. (By probability there

will be a few quasars that linger on for higher redshifts).

The boundary between the two major sectors of the universe is thus quite unstable, as it is

filled with material quasars and pulsars and cosmic quasars and pulsars, all in the process

of moving to the opposite sector. It is also filled with material white dwarfs and cosmic

white dwarfs which will eventually return to being normal stars in their respective

sectors. The boundary here is a speed or energy boundary. There is another boundary

within each sector at unit distance or unit time. In the material sector, the region inside

unit distance represents and important subdivision; in the cosmic sector, the region inside

unit time also represents an important subdivision. Reversals of motion occur at these

unit distance or time boundaries just as they do at the unit velocity or energy boundaries.

Inside unit space, only motion in time is possible, because fractional space units do not

exist. But the motion in time may be expressed in equivalent space units by means of the

reciprocal relation. Similarly, inside unit time, only motion in space is possible, because

fractional time units do not exist. This motion in space may also be expressed in terms of

equivalent time units by means of the reciprocal relation.

Inside unit distance or time, the space-time progression and gravitation reverse directions.

The progression is always away from unity and gravitation always toward it. So outside

unit space or time, the progression moves the very large spheroidal galaxies apart; inside

unit space or time, the progression moves the atoms of matter or cosmic matter close

together, in opposition to gravitation, which in this region is a force of repulsion. Atoms

of matter or cosmic matter can thus reach equilibrium positions in the solid state at the

locations where the two forces reach equality. There is thus no need for the ad hoc

electrical forces of conventional theory, which, in any case, supply only one of the two

necessary forces for equilibrium.

I have thus shown, in brief, that the Reciprocal System of theory, based on the novel

concept of three dimensions of motion, with space and time being reciprocally related,

can handle some of the current problems in physics: those phenomena involving the very

small, the very large, and the very fast. The very small, the atoms and the subatoms, are

subject to the relations that apply inside unit space or unit time, where the roles of the

progression and gravitation are reversed from what they are in the time-space (material)

or space-time (cosmic) regions. The very large, the spheroidal galaxies, are moving away

from each other at speeds approaching that of light because of the space-time progression

and the attenuation of gravitation at great distances. After many billions of years of

aggregation, these galaxies are now in the process of discombobulating—emitting jets of

high speed gases, quasars, and radio galaxies—because their matter has reached the upper

limit of age. The very fast, the quasars, have extraordinarily high redshifts because they

are moving at speeds faster than light and they will ultimately disappear into the cosmic

sector.

References

1. D. Larson, Nothing But Motion, Portland, Oregon: North Pacific Publishers, 1979,

p. 30.

2. D. Larson, The Structure of the Physical Universe, Portland, Oregon: North

Pacific Publishers, 1959, p. 19.

3. D. Larson, The Universe of Motion, Portland, Oregon: North Pacific Publishers,

1984), p. 301.

4. D. Larson, Ibid., pp. 287-288.

5. D. Larson, Ibid., p. 210.

THE LORENTZ TRANSFORMATION

Question: Please provide a detailed rationale of how the RS theory produces the correct

answer to this ―Lorentz transform‖ problem:

M

L <———|———> R

When L and R travel at the speed of light relative to M, Larson says the speed of R

relative to L is 2 units of space divided by two units of time; thus, the velocity of R

relative to L is 2/2 = l. Now suppose that L and R both travel at C/2 relative to M. If we

seemingly follow the same procedure as above, it appears that the total distance involved

is (½+ ½) and the total time involved is (1 + 1), so that the velocity of R relative to L

should be distance/time - 1/2. Obviously somethirg is wrong. What?

Answer: Assume that the partIcles traveling with the speed C/2 are atoms. Then the rate

of motion of the atom toward R relative to the motion of the atom toward L is 0.8C. The

RS theory thus offers the same answer to your question as does the Lorentz

transformation equation. The mode of motion of a photon (vibration) is different from

that of an atom (a combination of vibration and rotation). Photons remain in the space-

time locations in which they originate; atoms do not. The space-time locations of photons

move at the unit rate C. Atoms do not remain in the space-time locations in which they

originate. Therefore, the procedure for calculating the rate of motion of two photons

going apart from each other is not applicable to calculate the motion rate of two atorns

moving apart from eacil other, each at v = C/2 with respect to M. The relative motion rate

of the two atorns in this case is

0.5C + 0.5C 1.0C

u = ————— = ——— = 0.8C (1)

1 +[(0.5C) (0.5C)/C²]

1.25

The procedure, equation (1), is called the Lorentz transformation equation. How does the

RS theory arrive at the Lorentz equation? How does the RS theory deduce this equation?

This question amounts to asking how does RS theory imply the transformation equation:

vdb + wba

uda = —————

1 +[(vdb) (wba)/C²]

This Lorentz equation or law about the composition of velocities follows from the RS

theory, because the latter assumes that a light photon remains in the space-time location

in which it originates and further assumes that the location progresses at unit speed or at

the uniform rate of C = 3 x 105 km/sec., independent of the motion of source or detector

of the photon. These assumptions are incompatible with the Newtonian-Galilean

transformation equation, the Newtonian law of the composition of velocities, uda = vdb +

wba ; uda = - uad.

The fact that the velocity of light is independent of the velocity of the source of the light

implies that any finite velocity of the source, when added to the velocity of light, yields a

resultant for the light whose magnitude equals that of the speed of light.

Now in Newtonian physics, when three particles A, B, D are moving in a straight line,

and if U is the velocity of A relative to D, V is the velocity of D relative to B and W is

the velocity of B relative to A, then uad + vdb + wba = 0.

However, the just stated fact and RS principle asserts that when v = C, then u = - C,

whatever value w may have. This implies that the equation u+v+w = 0 is not true when

velocities commensurable with that of light are involved: it works satisfactorily only

when all the velocities are small compared with C.

How then to deduce the correct form of tkre law ef composition of velocities for

velocities of any magnitude is now the task. Specify then that the exact relation between

the three velocities is F (u,v,w) = 0. Agree that wba = - wab etc.

By permuting the three particles A, B and D note that the function F has to be a

symmetric function of u, v and w.

Further, the function F has to be a linear function so that may yield a one-valued solution

when solved with respect to u, v or w. Consequently, the equation assumes the form

g + h (u+v+w) + k(vw + wv) + 1 (uvw) = 0

Since when w = 0, u = - v, then g - ku² = 0 for all values of u and so g and k are zero.

Thus, the equation takes the form h(u+v+w)+1(uvw)=D. Also, u= -C when v=C, no

matter what the value of w. Hence hw-1C²=0 and h=1C². Therefore,

1C²(u+v+w) +1(uvw)=0

or

u+v+w + (uvw/C²) = 0

This is the exact relation which replaces the Newtonian relation u+v+w=0. This exact

relation implies that

-u-uvw/C²=v+w

and

-u[1+(vw/C²)]=v+w

-u=v+w/[1+(vw/C²)]

-uad = uda

Therefore

vdb + wba

uda = —————

1 +[(vdb) (wba)/C²]

Q.E.D.

Thus the Lorentz Law of the composition of velocities is simly the mathematically

equivalent expressin of every physical theory which assumes that the speed of radiation

in vacuo is independent of the mtion of the radiation source.

—Reciprocity IV.1 (April 1974)

THE TWO-PHOTON PROBLEM

Question: How can two photons from separate sources meet if their space-time locations

are moving away from each other with the space-time progression?

Answer :

1. The essential point here is that if an object is in motion relative to a stationary

reference system, and acquires an additional motion, this new motion does not

replace the previously existing motions; it adds to them.

2. A completely free object is moving outward from all other such objects by reason

of the space-time progression (Motion I). Two such objects having no other

motions therefore cannot collide.

3. Gravitationally bound objects without independent motions are likewise moving

outward from all other similar objects (Motion I), but coincidentally are moving

inward toward all of these objects at the same rate of speed, by reason of

gravitation (Motion II). Two such objects maintain the same separation, and

therefore cannot collide.

4. An object A in a gravitationally bound system may acquire an independent

motion in any direction (Motion III). The sum of all three of the motions of this

object (equal to its independent motion) may then carry it to a point where it will

collide with a similar object B.

5. A photon released from object A participates in all three of the motions of that

object, and inasmuch as it is not under any restraint in the dimensions

perpendicular to the direction of Motions I and II, it is also moved outward at unit

speed in one of these dimensions by the space-time progression. (This motion can

be in any direction relative to the reference system, as the gravitational motion is

random) . The second progression is Motion IV; that is, it is an addition to all of

the other three motions. The net resultant of all four motions is a combination of

Motion III and Motion IV. If object A maintains the same speed and direction, the

motion of the photon, as seen in the context of a stationary reference system, is

directly outward from object A. The emitted photon may therefore collide with

any object B in the gravitational system, or with a photon emitted from object E.

—Reciprocity V. 3 (October 1975)

A Note on Scalar Motion

Beginning students of the Reciprocal System often have difficulty understanding scalar

motion, confusing it with vectorial motion. I will attempt here to clarify matters.

Assume as a thought experiment, a spherical light source in gravitational equilibrium

with us, the observers. In our ordinary 3-D spatial reference system, the source is

stationary, and photons are streaming away from it at the speed of light in all directions.

Without the knowledge of the Reciprocal System, you might conclude that the photons

have independent motion and are moving vectorially through coordinate space away from

the source. Iiowever, the theory says that this is not the case from the standpoint of the

true, natural reference system. The photons actually have no independent motion and thus

are stuck in the same space-time units in which they originate. It is the aioms of the

source that have independent motion and are moving against the space-time progression.

This resulting motion is termed gravitation and is always inward. Most importantly, this

motion is scalar: it is inward in all directions, with no one direction favored. It is because

the source is moving inward in all directions that makes it appear that the photons are

moving outward in all directions. Since the inward gravitational motion is taking place in

space, the motion imputed to the photons is outward in space.

Likewise for the cosmic sector: the cosmic atoms are moving inward in time, and so

cosmic observers would conclude that the photons from their light are moving outward in

coordinate time. In actuality, of course, the photons remain in the same absolute

spacetime locations and are not moving either in coordinate space or coordinate time!

Now suppose, as in the Einstein-Podolsky Rosen experiment, that two photons originate

at the same event and move in opposite directions. In the material sector, the motion

appears to be outward in space; in the cosmic sector, the motion appears to be outward in

time. In actuality, we are moving inward in space away from the photons, and cosmic

observers are moving inward in time away from photons. We have no independent

motion in coordinate time (at low vectorial speeds), and since the photons do not either,

we are able to effect a change in both photons by means of a change in one of them.

Likewise, the cosmic observers have no independent motion in coordinate space (at low

vectorial inverse speeds), and since the photons do not either, the cosmic observers are

able to effect a change in both photons by means of a change in one of them. (Existents

which are contiguous in either space or time may both be affected by application of a

suitable single force).

Because photons are stationary in the natural reference system, they are nat ―lost‖ from

either sector and are not ―disappearing over the time or space horizon‖; the universe is

not ―running down‖ toward a slow ―heat death‖.

A NOTE ON THE FORCE OF THE

SPACE-TIME PROGRESSION

In a previous paper of mine¹ I discussed Hubble‘s Law and the Reciprocal System. I

integrated Hubble‘s equation to obtain the following equation representing the distance of

a galaxy from our galaxy as a function of time:

r = ro + (vi - ro)e}Ht

(1)

where ro is the gravitational limit, H is Hubble‘s constant, and ri is the initial distance.

Now if we go back to Hubble‘s original equation,

V = Hr (2)

we can differentiate this, instead of integrating it as before.

dv/dt = H dr/dt = a = Hv = H²r (3)

Hence, the force of the space-time progression must be

Fp = mG² H²r + FG¹-G² (4)

where mG² is the mass of the galaxy moving away and FG¹-G² is the gravitational attraction.

This equation should have many applications in the Reciprocal System.

References

1. R. W. Satz, ―Hubble‘s Law and the Reciprocal System,‖ Reciprocity, Vol. VII, No. 3

ON THE NATURE OF

UNDISPLACED SPACE-TIME

Question: With respect to what is space-time moving? If there is not something

more fundamental than space-time with respect to which space-time, itself, is

moving, then space-time cannot properly be said to move (or progress) at all.

Answer: The term ―space-time,‖ as used in the Reciprocal System of theory, is

equivalent to, and interchangeable with, the term ―motion,‖ in the broadest sense of

the latter term, and the general nature of the answer to the foregoing question can

readily be seen if the equivalent term is substituted for ―space-time‖ in the wording

of the question. No one appears to have any difficulty in recognizing that the end of

a unit of time is later—more advanced—than the beginning of that unit; that is,

there is a progression from the beginning to the end. Furthermore, it is commonly

understood that this is simply a progression, not a progression relative to something

else, and hence a unit of time, a section of the progression, is a self-contained entity.

As the published expositions of the Reciprocal System have demonstrated, the

concept of a universe of motion requires that space be defined in exactly the same

terms as time, except that it is the inverse quantity. Thus the end of a unit of space

is also more advanced—that is, more distant (the spatial equivalent of later)—than

the beginning; not more distant from something but simply more advanced. Space,

too, is a progression, and since both of its components progress, notion (space-time)

is likewise a self-contained progression; it is not a ―something‖ that progresses

relative to something else.

Of course, a certain amount of mental effort is required in order to lift our thinking

out of the grooves in which it has been running so long, but obviously if it is

possible to conceive of time as a progression, independently of any hypothetical

background—a mental feat that seems to present no particular difficulty—then it is

also possible to conceive of an inverse quantity of exactly the same general nature.

If there is any difficulty in so doing, it does not arise from the nature of the concept

itself, but from an unwillingness, or inability, to let go of ideas that are derived from

premises that have no relevance in a universe of motion. When space and time are

viewed in terms of these concepts there should be no obstacle to recognizing that

motion (or space-time) is a similar self-contained progression, According to the

fundamental concept on which the new theoretical system is based, the unit of this

progression—the unit of motion—is the basic entity of the universe, that from

which all else is constructed. It cannot be related to anything ―more fundamental.‖

The idea of a background to which motion must be related belongs to some concept

such as that of a universe of matter; it has no place in a universe of motion, where

motion itself is the ultimate reality.

CLOCK SPACE, COORDINATE SPACE;

CLOCK TIME, COORDINATE TIME:

What is the difference? At last year‗s ISUS convention, a number of individuals expressed difficulty in

comprehending the difference between clock space and coordinate space and the

difference between clock time and coordinate time. This note will review these concepts

to aid the understanding of these individuals.

Larson states [1]: ―We begin with one-dimensional space s and one-dimensional time t....

Dividing space by time we obtain velocity s/t....‖ Space and time do not exist separately;

they exist only as aspects of motion. Motion in the most general sense is thus a relation of

space to time, and in the Reciprocal System space and time have no properties other than

what they have as aspects of motion. In defining motion, we can start with units of space

and time, as Larson did in the quotation, or with units of motion and define space and

time as the two aspects of that motion; the definitions are equivalent.

The basic space-time unit is thus one-dimensional and is a progression. We reject the

Relativity doctrine that space and time are joined in four-dimensional continuum and that

space and time magnitudes are purely relative. From the postulates of the Reciprocal

System we compute the absolute natural unit of space to be 4.558816x10-6 cm and the

absolute natural unit of time to be 1.520655x10-16 sec. Their ration is 2.997930x1010

cm/sec, the speed of light. The progression originates everywhere and is thus

omnipresent. Larson stats [2]: ―Unit velocity is a ... true physical datum with a finite

magnitude.‖ Thus we begin with clock space-time, rather than with coordinate space or

coordinate time (or a combination of coordinate space with clock time). Conventional

physicists (and individuals new to the Reciprocal System) keep trying to start with some

type of 3-D or higher metric; we reject this approach entirely.

Coordinate space and coordinate time result from clock space and clock time. Larson

explicitly states [3]: ―There is a general framework of the universe, an extension space,

generated by translational motion....‖ ; likewise, there is an ―extension‖ time, generated

by translational motion, the progression. This motion is scalarly outward in all directions

and thus the overall distribution of the 1-dimensional progressing units is 3-dimensional.

In The Unmysterious Uniuerse [4], I wrote ―...with respect to any particular progressing

unit, coordinate space and coordinate time include all other progressing units.... The

progression of a single unit of space is one-dimensional, but the progression of all space

units is distributed in three-dimensions.‖ Stationary coordinate space (or stationary

coordinate time) can arise only in the context of a gravitationally-bound material system

(or cosmic system), in which the atoms of matter (or c-matter) have neutralized the space

progression (or time progression).

When Larson states [5] that ―undisplaced space-time is the physical equivalent of

nothing‖ he means that a unit of space-time is not a photon, a subatom, an atom, or an

electric or magnetic charge. These other entities are interchangeable, either directly or

indirectly, but a unit of spacetime is not. It cannot be changed into something else, and it

cannot be used as an energy source. But this does not mean that space-time is ―nothing‖;

it is unit motion, not zero motion, and is every bit as much an existent as anything

physical. Larson says [6] ―In terms of [a] building analogy, [space and time] correspond

to the bricks of which the building [i.e., the universe] is to be constructed.‖

REFERENCES:

1. D. Larson, New Light on Space and Time (Portland Oregon: North Pacific

Publishers, 1965), p. 226.

2. Ibid, p. 83.

3. Ibid, p. 242.

4. R. Satz, The Unmysterious Uniuerse (Troy, NY: Troy Printers, 1971), p.24, 27.

5. D. Larson, op. cit, p. 242.

6. D. Larson, op. cit, p. 240. 5

COSMIC RAYS AND ELEMENTARY PARTICLES

A View of the Reciprocal System

Introduction

Recently the existence of two new particles has been discovered at the Brookhaven

National Laboratory and the SLAC-Lawrence Berkeley Laboratory. They have been

named the psi [3695) and psi (3100) resonances (although their exact masses, in MeV/c²,

are still somewhat in dispute). A whole bumper crop of theories has sprung up to explain

these heavy resonances. In this endeavor physicists are postulating quantities such as

―charm, color, paracharge, gentleness and chimerity.‖ These terms are added to the

original term, ―strangeness.‖ In this paper an explanation of these particles, others like

them, and their natural origins, will be offered without the use of any quantity such as

―charm.‖

I. Critique of Present Theory

Around the turn of the present century physicists had just two particles to work with: the

electron and the proton. The theory then was that the atom was composed only of these

particles. In 1932 the neutron was discovered and added to the list of particles contained

within the atom. At the same time the positron was discovered, but not added to the list of

particles contained within the atom. Then during the 1940s the first two mesons, the

muon and the pion, were discovered. The physicists found that they could use the pi

meson to ―explain‖ the force of attraction between protons and neutrons within the

postulated atomic nucleus. But they knew of no function for the mu meson.

With the increase of energy in the linear accelerators and storage rings in the last two

decades, scores of new particles were found. Certainly Physicists could not assume that

all played a role in atomic structure. They decided that some particles are more

elementary than others: the present theory is that all the particles of the universe are made

up of three ―quarks‖ and three ―anti-quarks.‖. These somehow combine to make up every

other particle. (The original theory postulated an eight-fold way, but that hypothesis has

broken down).

There are two immediate criticisms to the quark hypothesis:

1. In all attempts to find quarks, no one has succeeded—it seems that every time an

attempt is made, new particles are discovered, but unfortunately, they aren‘t

quarks.

2. The quark theory gives no indication as to how matter can change into radiation

and vice versa.

Another aspect of present theory concerns the forces between particles. In essence the

theory states that all forces arise from the exchange of quanta, as follows:

1. Gravity results from the exchange of ―gravitons.‖

2. Electromagnetic forces result from the exchange of ―virtual photons.‖

3. The weak beta decay of atoms results from the exchange of ―the weak boson W.‖

4. The strong nuclear forces result from the ―exchange of mesons between

nucleons.‖

5. Chemical attraction between atoms results from the exchange of electrons.

All of these hypotheses suffer from one and the same defect: none of the exchange

particles has ever been observed in action doing what they are supposed to be doing

Despite major efforts no one has observed gravitons, weak bosons, exchange electrons or

exchange photons. (The last category is, by definition, not observable; hence there is no

way to test the latter theory.) Nor has the actual mechanism of coupling ever been

specified. It may be true that an exchange of gifts in the human realm promotes bonds of

friendship and it is true that our rate of effecting closer bonds is limited to the speed of

light. But the atomic realm is different. There the communication of forces is evidently

instantaneous; there is no time factor in Newton‘s law of gravitational force or

Coulomb‘s law of electrostatic force. To postulate an exchange of force and to limit the

propagation speed to that of light is, in a sense, anthropomorphic.

II. The Theory of the Reciprocal System

In the Reciprocal System as advanced by D. B. Larson in his books, beginning with the

Structure of the Physical Universe, no particle is posited as being actually elementary.

The fundamental component of the universe is not a set of matter particles (quarks).

Rather, the fundamental component is motion, existing in discrete units. Quantitatively,

the motion may be above or below unit value (one unit of space per unit of time). In the

sector we live in, the motion is ordinarily below unity. According to the theory, another

sector exists with motion above unity.

In our sector exists a series of chemical elements and subatomic particles, each a specific

quantity of motion. Likewise the other sector has a series of chemical elements and

subatomic particles: In our sector there exist stupendous galactic explosions resulting in

quasars, whose matter, the theory postulates, leaves our sector and enters the inverse

sector. Likewise, in the other sector stupendous galactic explosions occur, in which

inverse matter is dispersed into our sector. This dispersed, very energetic inverse matter

may be identified as cosmic rays.

In Mr. Larson‘s theory an atom, either in this sector or the other, is not composed of

protons, neutrons and electrons (or their ―antiparticles‖). Rather each atom and each

cosmic atom has a specific quantity of rotational motion about three perpendicular axes.

All of the particles, whether atoms, mesons, resonances or baryons, are units of motion,

some having more than others. The rotational motion making up the particles can, under

certain conditions, convert to linear motion or radiation. Forces arise from the interaction

of the particles‘ motion with that of the space-time progression. The latter is the general

translational outward motion of the universe, which arises from the equivalence of single

units of space and time. Since the particles and the progression are always interacting,

there is no Propagation of gravitational or electrostatic force and thus NO speed limit.

In our sector the most common element is hydrogen. According to theory, H² (deuteron)

is the natural atom, and H¹ is H² with a magnetic charge. Likewise in the other sector the

most common element is designated cosmic hydrogen, co-H² (co-deuteron). This co-H² is

the most common natural cosmic atom, and co-H¹ is co-H² with a cosmic magnetic

charge.

In the cosmic ray stream cosmic hydrogen should be much more abundant than other

cosmic elements. As elements constructed of motion above unity, these cosmic elements

should have properties the inverse of those ordinarily associated with corresponding

elements in our sector. this means that their masses must be the reciprocals of the masses

of their namesakes in this sector. Mathematically, the mass of a cosmic element is

m-co-element = 1/n natural mass units

= 2/n atomic mass units

= 364.66/n electron mass units

= 1862.95/n MeV/c²

where n is the cosmic atomic number. In the case of cosmic isotopes,

m-co-elements =1862.95/ (n +½G) MeV/c²,

where G is the number of cosmic gravitational charges. Once in our sector the co-

elements are subject to the same forces that produce material isotopes. Let I be the

number of material isotopic charges. Then the complete mass equation (ignoring

secondary mass effects) is

m-co-elements = (1862.95/ [n+½G) + I [931 .478) MeV/c²,

where I and G must have opposite signs for compatible motions.

Using the first equation above for cosmic deuteron, I find that it has the same mass as

deuteron in our sector. Once in our sector, with its lower translational velocity and its

change in space-time reference point, the cosmic deuteron transforms to the deuteron of

our sector. This is a true CPT transformation: an inversion of space-time coordinates

followed by an interchange of a particle and "antiparticle". Momentum and energy are

conserved in this transformation.

III. Critiquing the Reciprocal Theory

Old readers of Mr. Larson's books will note that here I am disagreeing with the original

presentation of the theory. There it was said that cosmic helium transformed to cosmic

krypton, emitting neutrons along the way, until finally the cosmic krypton converted into

a neutron (or equivalent). Although the present author agrees with Mr. Larson on the

framework of the theory, here we differ on details. Since the New Science Advocates is

not a religious sect, but a scientific body, members have the freedom, indeed the

obligation, to question the details of the theory. Cosmic krypton cannot decay to a

neutron because to do so Would violate the conservation law of momentum and energy.

One body decays, except in the case of hydrogen, are simply impossible. Furthermore, a

cosmic element in the middle of the cosmic periodic table of elements cannot emit a

neutron to move to the last column in the table, anymore than an element in the middle of

our periodic table could accept a neutron to move to the last column of the table.

I thus conclude that cosmic hydrogen transforms to our hydrogen, rather than that cosmic

krypton transforms to a neutron.

Once transformed, the new deuteron is unstable in our atmosphere and soon decays to a

proton, electron and neutrino. Thus, the cosmic ray stream is composed mostly of high

energy protons, precisely as observed.

Other cosmic elements and their isotopes exist in the cosmic ray stream and are produced

in laboratories on earth. Because of the variability in G and I, two different particles can

sometimes have the same mass. Thus identification of particles cannot be based solely on

mass. Some of the better known heavy mesons (hyperons) together with their probable

identifications are here listed:

SYMBOL OBSERVED MASS N G I IDENTIFICATION

3695 1 -1 0 co-H¹

1 0 2 co-H²+2

½ 0 0 co-n

751 2 1 0 co-He5

1672.5 3 -1 1 co-Li5+1

547 3 1 0 co-Li7

1304 5 0 1 co-B10+1

1236 6 0 1 co-C12+1

3100 6 0 3 co-C12+3

1198 7 0 1 co-N14+1

1118 10 0 1 co-Ne20+1

I think that the recently found psi [3695) particle is either cosmic hydrogen, isotope 1 or

cosmic deuteron with two material isotopic charges or cosmic neutron. A firm decision

will have to await experimental results.

On the basis of preliminary calculation I tentatively submit that the other recently found

particle psi (3100), is cosmic carbon with three material isotopic charges.

In addition to the heavy mesons two light mesons are commonly observed in the cosmic

ray streams and produced in laboratories on earth: the muon and the pion. I agree with D.

B. Larson that the muon is co-argon and the pion is co-silicon. On earth these mesons are

created from kinetic energy; an energetic proton strikes another proton, producing a third

particle, the pion. This pion decays to a muon and neutrino. I do not agree with Mr.

Larson that the muon then decays to co-cobalt and then to co-krypton. My reading of the

evidence indicates that the muon simply decays to positrons and neutrinos. The rotational

kinetic energy is converted to linear kinetic energy of simple rotational units, positrons or

electrons or neutrinos. The light mesons are created from kinetic energy and to kinetic

energy they return.

Before it decays, the pion is ―strongly interacting,‖ because it has both space and time

displacements. The muon is ―weakly interacting,‖ because it has only space

displacements. This is also the reason why muons are the ―hard‖ component of cosmic

rays—they can penetrate many meters into the ground. In the decay processes certain

conservation laws seem to hold true. Physicists are currently proposing one new

conservation law after another (strangeness number, hypercharge, paracharge, charm ,

baryon number, lepton number, etc.). In the Reciprocal System there is but one

conservation law: space-time displacements can be neither created nor destroyed. Energy,

t/s, and thus momentum, (t/s) , are conserved. In some reactions certain groups of space-

time units are conserved, for example, electric charge.

In many cases the heavy mesons decay to the lambda meson, which then ejects a neutron.

The remaining pion, if neutral, decays to two gamma rays. If charged, the pion decays to

a muon, which then transforms to a positron (or electron) and neutrino. The somewhat

lighter mesons, the eta, the rho, and the (small letter) omega, decay to two or more pions.

The theoretical explanation of decay process involves probability: smaller quantities are

more probable than larger ones. In the steps from rotational kinetic energy to translational

kinetic energy, the fewest number of particles are utilized. The decay pattern of the new

particles appears to be along these lines. The psi (3695) decays to the psi (3100), emitting

two pions, and then the psi (3100) decays to two muons. Many questions, however, still

remain to be answered.

CONCLUSION

The following are the essentials of the new theory:

1. A sector, the inverse of ours, is hypothesized, which provides a natural source for

scores of new particles, such as the mesons.

2. The cosmic rays are a stream of such particles coming from the inverse sector.

3. The masses of these particles are the inverses of their namesakes in our sector.

4. Inverse deuteron is equivalent to and converts to our deuteron. The most abundant

element in the cosmic sector converts to the most abundant element in our sector.

5. Cosmic atoms may have cosmic isotopic charges and/or material isotopic charges.

6. The cosmic elements in the cosmic ray stream, other than hydrogen, decay

eventually to the kinetic energy of simple rotational units, the electron, the

positron, the neutrino. In our laboratories mesons are created out of kinetic

energy. In our atmosphere the natural mesons return to kinetic energy.

7. All conservation laws relate to one: space-time displacements are neither created

nor destroyed.

Reciprocity Vol. V, No. 2 (May, 1975)

LETTER TO EDITOR

Dear Prof. Meyer,

In my opinion, the development of the consequences of the postulates of the Reciprocal

System of theory, and the correlation of these consequences with the results of

observation, have now been carried far enough to make it evident that the theoretical

system is basically correct. There are, however, many questions still remaining with

reference to the details, even in the areas that have already been studied, and, of course,

there are a great many other areas yet to be examined. I believe it is very desirable to

encourage free and open discussion of the theory and its application, so that we can have

the benefit of as many points of view as possible in extending and clarifying the

theoretical structure, and I want to avoid saying or doing anything that might give the

impression that I am trying to discourage dissenting opinions. For that reason I would

prefer not to comment at this time on Ronald Satz‘ article discussing the newly

discovered heavy ―resonances,‖ except to say that I agree with his conclusions l, 2, 3 and

7, and in part, with conclusion 5. I hope that readers of RECIPROCITY will give this

article careful consideration, and will not hesitate to express their opinions, pro and con,

in ―letters to the editor.‖

D. B. Larson

IDENTIFICATION OF COSMIC PARTICLES

3695 MeV/C² AND MeV/C²

In November, 1974, two teams, one at the Brookhaven National Laboratory and the other

at the Lawrence Livermore Laboratory, announced the discovery of a new particle with a

mass equivalent to 3,105 MeV/c² of energy. The lifetime of this particle is about 10-20

second, considered by some to be a remarkably long lifetime for a particle of this heavy

mass. This particle is named with the Greek letter, psi, and is referred to as a psi

resonance.

Shortly afterward, the two teams discovered a second psi resonance with a mass

equivalent to 3,695 MeV/c² of energy and lifetime of about 10-20 second. Cosmic decay of

the 3,695 MeV/c² particle apparently results in production of 3,105 MeV/c² particle.

Discovery of these two new physical entities is exciting news from the frontiers of

physics. How the psi resonances fit into the physical scheme of things has remained a

mystery until now. The discovery of the mere existence of these high-energy particles has

been deemed so important that the leaders of the two teams, Drs. Samuel Ting and

Burton Richter, were awarded the 1976 Nobel Prize in physics for this discovery.

In the Reciprocal System psi resonances and other related cosmic particles are identified

as specific isotopes of cosmic chemical elements.

The identification procedure depends on the convergence of several lines of approach,

including theoretical computation ot the mass and lifetime of each particle and also

examination whether and how ic can fit into the regular cosmic decay sequence after the

particle enters the material sector.

Cosmic element mass once the cosmic element enters the material sector is generally

made up ot its rotational mass,the inverse of the material element mass (Figures 1 and 2),

and of its material gravitational charges (Figure 3) acquired with entry into the material

sector (Larson, 1979).

Figure 1

COMPUTATION OF COSMIC ELEMENT MASS

1 atomic mass unit = 1.66 × 10-27 kg.

c = 2.99 x 108 m/ s ; c² = 8.94 × 1016 m²/ s²

Equivalent energy of 1 a.m.u. = mc²

1 a.m.u. = (1.66 x 10-27 kg) (8.94 × 1016 m ²/ s² = 14.9 x 10-11 J

1 electron volt = 1 ev = 1.6 x 10-19 J

Energy equivalent of 1 a.m.u. = 14.9 × 10-11 J / 1.6 X 10-19J

1 atomic mass unit = 931.15 MeV/c²

Mass of a material atom of atomic number Z:

m = 1862.30Z MeV/c² (1862.3 = 2 (931.15))

Mass of a cosmic atom is INVERSE mass

We observe cosmic mass as 1862.30/Z MeV/c²

Figure 2

COMPUTATION OF COSMIC ELEMENT MASS

Let Z = atomic number of cosmic element

cosmic mass = 1862.30/Z MeV/c²

Alternative Procedure

Instead of atomic number units (Z),

use atomic mass (or weight) units to express osmic mass.

Atomic weight units are half the size of units of atomic number.

Then cosmic mass = 3724.61/m MeV/c²

This is the mass of cosmic atom (isotope)

in the condition in which it enters material sector.

m here represents atomic weight units

Figure 3

COMPUTATION OF COSMIC ELEMENT MASS

after element enters material sector.

Mass of cosmic element in atomic weight units when it enters material

sector:

Cosmic mass = 3724.61/m MeV/c2, m here represents atomic weight units.

Superscripts for isotope symbols are atomic weight units.

After entering material sector cosmic atoms

may acquire gravitational charges of material type.

Mass of each gravitational charge is one atomic weight unit = 931.15 MeV.

The psi resonance with a mass equivalent to 3695 MeV/C² has been identified as the

isotope of cosmic hydrogen, c-H², cosmic deuteron with two material gravitational

charges (Figure 4). This is a deduction from the Reciprocal System theory and the

achievement of Ronald W. Satz (1975) and Larson (1979).

Figure 4

IDENTIFICATION OF 3695 MeV/c² PARTICLE

Identified by R. W. Satz as ―cosmic deuteron with two material isotope

charges‖ (c-H²).

Rotational mass of a material hydrogen (H²) atom is 1.007405 units of

atomic number scale.

Mass of a cosmic H² atom is the reciprocal of this number = 0.99265 units.

For hydrogen Z = 1, first portion of

Cosmic mass of c-H² = 1862.31 (0.99265/Z:

Rotational cosmic mass of c-H² = 1848 MeV/c2

After entry to material sector c-H² acquires two material gravitational

charges

2(931.15 MeV/c²) = 1862.3 MeV/c²

Total cosmic mass of c-H² =

1848 MeV/c² + 1862 MeV/c² = 3710 MeV/c²

Observed mass of c-H² reported as 3695 MeV/c²

The psi resonance with a mass equivalent of 3105 MeV/c² has bean identified as an

isotope of cosmic helium, c-He³ with two material gravitational charges (Figure 5). This

is an achievement of D.B. Larson (1979).

Figure 5

IDENTIFICATION OF 3105 MeV/c² PARTICLE

Identified by D. B. Larson as cosmic helium with two material gravitational

charges (c-He³).

The material He³ isotope is a He atom (mass = 4 atomic weight units) with a

one-unit negative gravitational charge (one negative atomic weight unit).

The mass of the isotope is then 3 atomic weight units.

The cosmic He³ isotope is a similar but inverse structure, with a net mass of

3 cosmic atomic weight units.

Since the c-He3 isotope has a mass of 3 cosmic atomic weight units, its

rotational mass as observed in the material sector is 3724.61/3 = 1242

MeV/c².

After entry to material sector the c-He³ isotope adds two material

gravitational charges mass 931.15 each making total mass 3104 MeV/c².

Observed mass reported as 3105 MeV/c² .

Some 20 years ago Larson (1959) already identified as isotopes of other cosmic chemical

elements the muon, the pion, the lambda, sigma, xi and omega particles (Table 1).

TABLE 1

SOME COSMIC ELEMENT ISOTOPES IDENTIFIED

Isotope

Cosmic

Mass

3724.61/ m

MeV/ c²

Gravitational

Number

Charges

mass

MeV/c²

Total

Mass

MeV/c²

Observed

Mass

MeV/c²

Name

c-H² 1848 2 1862 3710 3695 psi

c-He³ 1242 2 1862 3104 3105 psi

c-Li5 745 1 931 1676 1673 omega

c-B10 373 1 931 1304 1321 xi

c-N14 266 1 931 1197 1197 sigma

c-Ne20 185 1 931 1116 1116 lambda

c-Si27 138 0 0 138 140 pion

c-Ar35 106 0 0 106 106 muon

References

Dewey B. Larson, The Structure of The Physical Universe, North Pacific Publishers,

1959.

Dewey B. Larson, Nothing But Motion, Volume I of a Revised and Enlarged Edition of

The Structure of The Physical Universe, 1979. North Pacific Publishers.

Ronald W. Satz, Cosmic Rays and Elementary Particles: A View of the Reciprocal

System Reciprocity Vol. V, no. 2 (May 1975)

A NOTE ON THE COSMIC PROTON

The rotational displacements of the material proton (a single rotating system of a 2R

photon) are 2-1-(1); the rotational displacements of the cosmic proton (a single rotating

system of a ½ R photon) are (2)-(1)-1. Both protons can take a positive or negative

charge (because both have the necessary space and time displacements). In the material

sector, rotational time displacements are predominant, hence the opposing rotational

vibrational space displacements (positive charges) are more common; in the cosmic

sector, rotational space displacements are predominant, hence the opposing rotational

vibrational time displacements (negative charges) are more common.

So we would expect the observed material proton to carry a positive charge (usually) and

the observed cosmic or ―anti‖ proton to carry a negative charge. The first observed ―anti‖

proton (in 1955) was created by the bombardment of one proton on another at rest. Either

this proton and the subsequent ones created in this manner are actually material protons

with negative charges or they are cosmic protons. Certainly the ones that are used in the

annihilation experiments are cosmic protons; two material protons would combine to

form deuterium, not terminate rotational displacements. Because material deuterium and

cosmic deuterium have theoretically identical masses, each of the individual rotating

systems has identical mass.

Thus the mass of the material proton and the cosmic proton should be identical! And both

should be perfectly stable. The only way to tell them apart would be by their subsequent

reaetions. Two cosmic protons would combine to form cosmic deuterium; with two

gravitational charges, the mass of this particle comes to 3710.91 Mev, which is elose to

the observed mass of the psi particle. This particle then follows the decay sequence

described in Larson‗s book Nothing But Motion, chapter 15. So the forthcoming

experiment should see what happens when two ―anti‖ protons combine; our predictions

are clear.

—Cf. D. B. Larson’s comments in a letter to R. W. Satz, dated Sept. 22, 1988

THE COHESIVE ENERGY OF THE ELEMENTS AT

ZERO TEMPERATURE AND ZERO EXTERNAL

PRESSURE

The equation for the internal energy of a substance is

u = h - pv (1)

where h is enthalpy, p is pressure, and v is volwne. At zero absolute temperature, the

enthalpy is zero.

uo= -pv (2)

For a gas at zero temperature governed by the ideal gas law, the internal energy must also

be zero. This is not so with a solid. Larson has shown that the equivalent of an external

pressure exists which provides the cohesion of the solid state. This pressure arises from

the force of the space-time progression, which is inward directed within the time region.

With zero external pressure and zero temperature, the internal energy must equal the

cohesive energy. Letting * be the internal pressure in kN/m² and vo be the volume in m³

/mole, and dropping the sign convention, we obtain the cohesive energy in kJ/mole:

uo = po vo (3)

However, as shown in reference one, motion in the time region (whether inward or

outward) is effective only hatf the time. This reduces the cohesive energy given by

equation (3) by a factor of two.

uo = ½po vo (4)

This equation is directly applicable to the "rare gas" elements.

The equation for molar volume is

vo = GNso ³ (5)

where so is the nearest neighbor distance N is Avogadro's number, and G is a geometric

factor. For face-centered-cubic crystals,

Gfcc= .707 (6)

For body-centered-cubic crystals,

Gbcc = .770 (7)

For other crystals,

(GMW/density) x 10-6

G = ———————— (8)

so³N

where GMW is the gram molecular weight, density is in grams per cubic centimeter, and

so is in meters (10 Angstroms).

In Chapter 25 of reference one, Larson derives the equation for the internal pressure in

natural units:

aZR

Po = ———————— (9)

312.89(so/sut)³

where a is the effective displacement in the active dimension, Z is either the electric

displacement or the second magnetic disptacement (depending on the orientation of the

atom), and R is the number of rotational units. sut' is the time region natural unit of space,

given by sut = (1/156.44) su (10)

In kN/m² the value for po becomes

Po = 4.177 x 10-17 aZr kN

—— —— (11)

So³ m²

Then,

uo = 12.57GaZR kJ/mole (12)

The parameters a, Z, and R have been deduced by Larson for most of the elements, but

not yet for the rare gas elements. Pending this, the value of the internal pressure can be

determined as the reciprocal of twice the initial compressibility (equation 25-14,

reference one):

po = 1/2kT (13)

Table I gives the values for po , vo, and uo for the rare gas elements. Overall the values

compare within 8% of the experimental values.

Elements other than the rare gas elements have electric displacement and this must

obviously have an effect on cohesive energy. The additional energy is given by this

expression:

ut' = INEu (1/156.44)4 (14)

where I is an integer or half integer value, N is Avogadro's number, and Eu is the natural

unit of energy. Alternatively from the cohesive energy standpoint, the effective volume,

v, may be altered. The factor is the interregional ratio (applicabte to energy, as well as

force). I is one for most of the displacement one elements, one and one-half or two for

displacement two elements, three or more for displacement three elements, and from 3½

to 5½ for displacement four etements. I can be zero or negative for the electronegative

etements. An exact equation for I cannot as yet be given.

The final reduced equation for cohesive energy is

kJ

uo = 12.57 GaZR + 50.31 —— (15)

mole

Table II gives the values of G, a, Z R, I, and uo for most of the remaining elements,

together with the experimental values from reference two. Usually agreement is within a

few percent.

Present atomic theory has nothing comparable to equation (15). The so-called Lennard-

Jones potential commonly used is empirically based and has not been deduced from first

principles--and even then it has usually been applied only to the noble elements and a few

other elements of 1ow atomic weight. Thus we have here a definite advantage of the

Reciprocal System over current theory.

References

1. Dewey B. Larson, Nothing But Motion, Vol. 1 of the revised Structure of the

Physical Universe, presently in manuscript form.

2. C. Kittel, Introduction to Solid State Phisics, Fifth Edition (New York: John

Wiley & Sons, Inc., 1976 , p. 74.

Table 1

—————————————————————————–———————

kN m³ kJ kJ

Element — — — —

po m²

vo mole

uo mole

uexp mole

—————————————————————————–———————

Helium 8.56x104 1.950x10-5 .835 --

Neon 5.00x105 1.395x10-5 3.488 1.92

Argon 5.33x105 2.227x10-5 5.935 7.74

Krypton 8.93x105 2.806x10-5 12.53 11.2

Xenon 9.52x105* 3.528x10-5 16.79 15.9

Radon 12.30x105* 3.584x10-5* 22.04 19.5

*Estimated values based on trend line analysis or assumed specific rotational values.

Table II

—————————————————————————–—————

——

kJ

kJ

Element Form G a Z R I

——

——

uo mole

uexp mole

—————————————————————————–—————

——

Li bcc .770 4 1 1 2½ 164.5 158

Be hcp .752 4 4 1 3½ 327.3 320

C dia 1.554 4 6 1 5 720.3 711

Na bcc .770 4 1 1 1 89.0 107

Mg hcp .780 4 3 1 1 157.1 145

Al fcc .707 4 5 1 3 328.6 327

Si dia 1.543 4 5 2 -6½ 448.9 446

K bcc .770 4 1 1 1 89.0 90.1

Ca fcc .707 4 3 1 1½ 182.1 178

Ti hcp .731 4 8 1 3½ 470.1 468

V bcc .770 4 8 1 4 510.9 512

Cr bcc .770 4 8 1 2 410.3 395

Mn cu.com. 1.087 4 8 1 -3 286.3 282

Fe bcc .770 4 8 1 2 410.3 413

Co hcp .696 4 8 1 3 430.9 424

Ni fcc .707 4 8 1 3 435.3 428

Cu fcc .707 4 8 1 1½ 359.9 336

Zn hcp .809 4 4 1 -1 112.4 130

Ge dia 1.541 4 4 1 1 360.2 372

Rb bcc .770 4 1 1 1 89.0 82.2

Sr fcc .707 4 3 1 1 156.9 166

Zr hcp .731 4 6 1½ 5½ 607.5 603

Nb bcc .770 4 8 1½ 5 716.1 730

Mo bcc .770 4 8 2 1 669.7 658

Ru hcp .730 4 8 2 1½ 662.7 650

Rh fcc .707 4 8 2 0 568.8 554

Pd fcc .707 4 8 1½ -1 376.3 376

Ag fcc .707 4 8 1 0 284.4 284

Cd hcp .816 4 4 1 -1 113.8 112

In tet .762 4 4 1 2 253.9 243

Sn dia 1.543 4 4 1 0 310.3 303

Sb rho 1.227 4 4 1 0 246.8 265

Cs bcc .770 4 1 1 1 89.0 77.6

Ba bcc .770 4 2 1 2 178.0 183

La hex .721 4 4 1 5½ 421.7 431

Ce fcc .707 4 4 1 5½ 418.9 417

Pr hex .722 4 4 1 4 346.4 357

Nd hex .698 4 4 1 4 341.6 328

Sm com .716 4 4 1 1 194.3 206

Gd hcp .722 4 4 1 5 396.7 400

Dy hcp .732 4 4 1 3 298.1 294

Ho hcp .732 4 4 1 3 298.1 302

Er hcp .736 4 4 1 3½ 324.1 317

Tm hcp .679 4 4 1 2 237.2 233

Yb fcc .707 4 2 1 1½ 146.5 154

Lu hcp .732 4 4 1 5½ 423.9 428

Ta bcc .770 4 8 2 3 770.3 782

W bcc .770 4 8 3 -1½ 853.8 859

Ir fcc .770 4 8 3 -3½ 677.2 670

Pt fcc .770 4 8 2 0 568.8 564

Au fcc .770 4 8 1½ -3 275.7 284

Tl hcp .690 4 4 1 1 189.1 182

Pb fcc .770 4 4 1 2½ 268.0 265

Bi rho 1.224 4 3 1 1 234.9 210

Th fcc .770 4 8 1 6 586.2 598

U com .998 4 8 1 3 552.3 536

—————————————————————————–—————

——

THE EQUATION OF STATE OF SOLID MATTER

For many years scientists and engineers have had available an excellent equation of state

for gaseous matter. Now, at last, the Reciprocal System of Dewey B. Larson is able to

give us an exact equation of state for solid matter. This paper will present a unified

treatment of the subject, with Reference 1 as the starting point.

I. Volume of Solid as a Function of Temperature with Pressure Constant

From the material presented in Chapter 8 of Ref. 1, I have drawn a generalized plot of

thermal expansion coefficient versus temperature, Figure l. The symbols are defined as

follows:

ß = thermal expansion coefficient

T = temperature

TM = temperature of solid end point (at or close to melting point)

VM = volume at solid end point

T1 = first transition temperature

ß0 = initial value of thercnal expansion coefficient at absolute zero temperature

V0 = initial volume at absolute zero temperature

V1 = volume at tranSition temperature

ß0‘ = initial vatue of thermal expansion coefficient based on second segment of curve

V0‘ = initial volume based on second segment of curve

With the initial votume of the first segment of the curve included, eq. (8-4) of Ref. 1

becomes

V = V0 + K/n³ T² (1)

where K is a constant and n is the number of rotational units that are themally

vibrating.This equation can be put into a more usable form involving T, T1, ß1, end ß0 —

all of which can be determined from theory. The thermal coefficient of expansion at

temperature T is

ß = 1/V dV/dT + ß0

= 2KT/n³ [1/V0 + KT²/n³ + ß0 (2)

= 2KT/n³V0 + KT² + ß0

At T1,

ß1 = 2KT1/n³V0 + KT1² + ß0

Then,

ß1ß0/2T1 = K/n³V0 + T1²

So,

2T1/ß1-ß0 = n³V0/K + t1²

K/n³ = V0/2T1/ß1-ß0 - T1² (3)

Therefore,

V = V0 + V0T²/2T1/ß1-ß0 - T1² (4)

This equation holds from T = 0 to T = T1. Larson has deduced the following values of ß1,

ß0, and T1:

ß0 = 5.17 * 10-6/°K for one unit

= 10.3 * 10-6/°K for two units

= 15.5 * 10-6/°K for three units

= 20.7 * 10-6/°K for four units units

= units = [3576/TM]

ß0 = -2/7 * ß1 for electropositive elements

= -1/7 * ß1 for some 2lectronegative alements

Tl = 8.98 (a + z + y)°K

a, z, y are from Table 22 of Ref. 1

Thus given the volume of the solid at zero temperature, che rotational factcrs of the

element, and the solid and point temperature, the volume V at any other temperature, (up

to T1) can be easily determined.

The equation for the volume for temperatures above T1 has the same form as eq. (4):

V0 T²

V = V01 + ———

2T1 (5)

——— - T²

ß1-ß0 For this equation to be of use,V0‘ and 0‘ must be expressed in terms of known quantities

such as VM and TM. Now,

V0‘ TM²

VM = V0‘ + ———–

2T1

———– - T1²

ß1-ß0‘

V0‘ T1²

V1 = V0‘ + ———

–

2T1

———

– - T1²

ß1-ß0‘

ln the equation for V1, solve for V0‘ and put in equation for VM:

V1

(

TM²

)

VM = —————— 1 + —————

1 + T1² 2T1

——––—– ——– - T1²

2T1 ß1ß0‘

———– - T1²

ß1-ß0‘

Or,

TM²

1 + —————–

2T1

——–– - T1²

ß1-ß0‘

VM/V1 = ———————

T1²

1 + —————–

2T1

——–– - T1²

ß1-ß0‘

Let,

1

C2 = —————

2T1 (6)

——– - T1²

ß1-ß0

Then,

VM 1 + C2TM²

—– = ————–

V1 1 - C2T1²

Solve for C2:

VM - V1

C2 = ——————— (7)

V1TM² - VMT1²

From eqs. (6) and (7), ß0‘, can be found:

2T1

ß0‘ = ß1 -

V1TM² - VMT1² (8)

+ T1²

VM - V1

This value of ß0‘ can then be substituted into the equation for V0‘:

V1

VO‘ = ———————–

T1²

1 + —————— (9)

2T1

———– - T1²

ß1 - ßo‘

With ß0‘ and V0‘ known, eq. (5) is ready for use. Larson has deduced the following

values of TM and VM:

TM = 1.80 * T1 for one rot. unit vibrating

= 4.56 * T1 for two rot. units

= 9.32 * T1 for three rot. units

= 17.87 * T1 for four rot. units

GM

VM = 1.0625 V0 —–

G0

where G0 is the initial crystal geometric constant and GM is the final one (some solids

cnange crystalline structure as they expand).

The ratio VM/V1 can be generalized to any pair of final to initial volumes:

Vf 1 + C2Tf²

— = ———— (10)

Vi 1 + C2Ti²

Compare this with the equation for a gas:

Vf Tf

— = — (11)

Vi Ti

II. Volume of Solid as a Function of Pressure with Temperature Constant

The comoression of a solid by hydrostatic pressure is discontinuous at certain aressures

nere denoted as P1, P2, P3, P4 etc. At these pressures the internal pressure P0 can change to

P01, P02, P03, P04, etc., thus altering the slope of the compression curve. Larson has shown

that between the transitions, the volume naries as the inverse square root of the pressure.

The most general way to express this is with the following aquation:

V P0 + Pref

—— = ———— (12)

Vref P0 + P

For the given value of P, the vatues of Pref, Vref, and P0 must be detennined by theory (or

empiricnlly if necessary) before V can be found. For a four transition solid we have the

following:

0 < P < P1 Vref = V0 Pref = 0 P0 = P0

P1 < P < P2 Vref = V1 Pref = P1 P0 P01

P2 < P < P3 Vref = V2 Pref = P2 P0 = P02

P3 < P < P4 Vref = V3 Pref = P3 P0 = P03

Now, in the MKS system,

aZy KN

P0 = 4.177 * 10-23 —– —– (13)

S0³ m²

where a, Z, y are the rotational compression values (simitar to the thermat values) and so

is the base interatomic spacing. At each transition a, Z, y can change (and possible S0),

thus causing P0 to change.

Before continuing the discussion of the equation of state I will discuss some subsidiary

properties of matter: the bulk modulus, the modulus of elasticity, and Poisson‘s ratio.

Larson has derived the equation for compressibility; the solid bulk modulus is the inverse

of this:

B = 2 * P0 (14)

(at zero external pressure and zero temperature for a pure substance). I witl not derive the

equation for the modulus of elasticity, E. In eq. (13) let the constants of the equation be

written as J and generalize s0 (for the moment) to s. Then the initial internal stress is

= -P = -Js-3 (15)

By definition,

d

E = —–

d

And,

d d ds

— = — —

d ds d

where is the strain:

s - s0

= ——–

s0

So,

s = s0 + s0

thus,

ds

— = s0

d

Since

d

— = 3Js-4

ds

then

d

— = 3Js-4 s0

d

and when s = s0 and T = 0 °K for a pure substance,

d

— s0,T = 0 = 3Js0-3

= -3P0

d

E = -3P0 (16)

(stress and pressure are in opposite directions)

Poisson‘s ratio can be determined from the well-known equation

3P0

= .5 - ——– (17)

6 2P0

Thus at zero temperature and pressure for a pure substance,

3P0

n = .5 - ——– = .25 (18)

6 2P0

This is in the ―ball park‖ for most solids; however, most substances used in construction

are impure and at other than zero temperature dnd, in addition, may contain a proportion

of tiquid molecules — thus drastically changing the values of Poisson‘s ratio and the

modulus of elasticity. These considerations will be left to another paper.

Going back to eq. (12) we can generalize to the ratio of final to initial volume within a

segment:

Vf P0 + Pi

— = ———– (19)

Vi P0 + Pf

This compares with the equation for a gas:

Vf Pi

— = — (20)

Vi Pf

III.Volume of Solid as a Function of Both Temperature and Pressure

The solid can be considered to undergo a aressure change at zero temperature and then a

temperature change from the new volume. Let P < P1 Then

P0

V0NEW = V0 ——––

P0 + P

Let T < T1. Then,

V0NEW T²

V = V0NEW + ————————

2T1NEW (21)

———— - T1NEW2

ß1 - ß0

The value of T1 is not the same as before. To get to the original value of V1 the new value

of T1 must be higher

2

T1NEW = ———————

ß1 - ß0

——— + ß1 - ß0 (22)

V1

—————–

V0 (P0/ P0 + P)½

where V1 is calculated from the original T1. I am assuming here that. is as before.

If P < P1 and T > T1, then the value of V0‘ has to be modified, since T1 and TM are

different. I assume that ß0‘ is the same. Then the second term on the right in eq. (8) is the

same and the new value of TM can be found:

( {

T1NEW

[

V1TM2 - VMT12

]

}

VM-V1 +VM

) ½

TMNEW = ———– ———————

+ T12 - T1NEW² — T1NEW

T1 VM - V1 V1

(23)

Eq. (5) becomes

V0NEWT²

V = V0NEW + ———–

2T1NEW (24)

———– - T1NEW2

ß1 - ß0

Equations (21) and (24j (combined) represent the complete equation of the solid state.

IV. Exampte Calculations

As an example, consider one volume unit of silver at zero degrets K and zero external

pressure. Whnt is the volume at temperature T and pressure P?

First the thermal rotational factors, a-Z-y, from Tabte 22 of Ref, 1 are found; they are 4-

3-l. With these, the temperature of the first transition point, T1, can be calcutated:

T1 = 8.98 (a+Z+y) = 8.98 (8) = 71.84oK

Silver has a maximum of four magnetit rotational units vibrating, so the solid end point is

TM = 17.87 * T1 = 17.87 * 71.84 = 1283.78oK

In this case the endgoint appears to be somewhat higher than the empirical melting point,

1234 oK Thus it would seem that the thermal factors at the end point are towered by one

to 3-3-1, so that

TM = 17.87 * 8.98 (3+3+1) = 1123.31oK

Now the number of units to use in selecting ß1 is

[3576 / 1123.31] = 3

and therefore

ß1 = 15.5 * 10-6 / oK

Since siiver is etectronegative,

s0 = - 1/7 ß1 = 1/7 * 15.5 * 10-6 = -2.214 * 10-6

Then from eq. (4),

V T²

— = 1 + ————————

V0 2 * 71.84

———————– - 71.84²

(15.5 + 2.214) * 10-6

V

— = 1 + 1.234 * 10-7 T² T < T1, P = 0

V0

This equation holds good up to T = T1 at which point

V1

— = 1 + 1.234 * 10-7 T1² = 1.0006369

V0

For temperatures above T1, the values of V0‘ and ß0‘ are needed. To calculate ß0‘ I am

going to use the empirical value of TM pending theoretical clarification.

From eqs. (8) and (9),

2 * 71.84

ß0‘ = 15.5 * 10-6 - ————————————

1.0006369 * 1234² - 1.0625 * 71.84²

——————————————— + 7184²

1.0625 - 1.0006369

= 9.647 * 10-6

1.0006369

V0‘ = —————–

1 + 71.84²

———— = 1.0004265

2 * 71.84²

———————– - 71.84²

(15.5 - 9.647) * 10-6

Thus from eq. (5),

1.0004265 T²

V = 1.0004265 + ——————

2 * 7184

———————— - 71.84²

(15.5 - 9.647) * 10-6

V = 1.0004265 + 4.07623 * 10-8 T² T > T1, P = 0, V0 = 1

(Note: no crystal change from FCC is assumed here).

Now we‘11 go on to look at the pressure relations. Assume that P is less than the first

transition pressure P1 (which is approx. 107 KN/m² ) so that the initial compressibility

factors from Table 14 of Ref. 1 can be used: a-Z-y = 4-8-1. From Table 4 of Ref. 1,

s0 = 2.87 x 10-10 m. Then from eq. (13),

P0 = 4.177 * 10-23 (4 * 8 * 1) / (2.87 * 10-10)³

= 5.654 * 107 KN/m²

Since P c p , P = 0, dnd V = V . Then eq. (12) is

V

(

5.554 * 107

)

— = ——————– ½ P < 1.0 * 107 KN/m²

V0 5.554 * 107 + P T = 0

If P = .001 P ,

V/V0 = (P0/1.001 P0)½ = .999500

The bulk modulus B, modulus of elasticity E, and Poisson‘s ratio can now be calculated

for a pure sample of silver at zero temperature:

B = 2 * P0 = 2 * 5.654 * 107 = 1.1305 * 108 KN/m²

E = 13 * P01 = 3 * 5.654 * 107 = 1.1692 * 108 KN/m²

V = .25

For the combined pressure and temperature loading, eq. (21) yields

V0 5.654 * 107

——————– T²

V = V0 5.654 * 107 5.654 * 107 + P

——————– + ————————

5.654 * 107 + P

2 T1

——————— - T12

(15.5 + 2.214) * 10-6

The value of T1 to be used here comes from eq. (22):

T1 = 2

————————————————————

15.5 * 10-6 + 2.214 * 10-6 + (15.5 + 2.214) * 10-6

———————————

1.0006369

————–

If P = .001 P0, then

5.654 * 10T - 1

——————–

5.65 * 107 1P

T1 = 128.24 °K

Putting this value of T1 into the above gives:

V

— = .99950 + 6.91095 * 10-8 T² T < T1

V0 P = .001 P0

The new value of T1 gives the new value of V0‘:

V0‘ = 1.0006369

————–

1 + 128.224

—————————— = 1.0002614

2 * 128.24

—————— - 128.242

(15.5 - 9.647) * 10-6

Thus, for temperatures above T1

V = 1.0002614 + 1,0002614 T²

——————––

2 * 128.24

————–— - 128.24²

(15.5 - 9.647) * 10-6

V = 1.0002614 + 2.28350 * 10-8 T² T > T1, P = .001P0

V0 = 1

Finally, from eq. (23) I find that the new melting temperature is:

TMNEW = 1650.88 °K

(I have assumed, however, that this does not affect the original value of ß1) .

Basically the same procedure could be used with other elements, atloys, and compounds.

Corresponding equations do not exist in quantum mechanics. A solution in ―principle

only‖ is not a true solution. A true solution is based on principle and works in practice.

**********************

Reference

1. Dewey B. Larson, Solid Matter, prepublication version of second volume of the

revised edition of The Structure of the Physical Universe (Portland, Oregon: North

Pacific Publishers, 1980).

FURTHER MATHEMATICS OF THE RECIPROCAL

SYSTEM

This paper will present in the most concrete, explicit manner the mathematics of space–

time, radiation, and matter of the Reciprocal System. Readers without special knowledge

of the Reciprocal System are first urged to study Larson‘s books, especially Nothing But

Motion¹ before undertaking the study of this paper.

I. Mathematics of Space–Time

A. Rectangular Coordinates

Starting from any reference point x0, y0, z0, t0 in the 0–system, the space–time

progression is a spherical expansion. In rectangular coordinates the equation

is

(x–x0)² + (y–y0)

² + (z–z0)

² = c

² t

² (1)

where c is the speed of light. If we choose the reference point to be x0 = 0, y0 = 0, z0 = 0,

t0 = 0, then the equation is simply

x² + y

² + z

² = c

² t

²

Now consider a second system, the 0´–system, moving translationally with respect to the

0´–system in the x–direction. What is the equation for the progression in the 0´–system?

From the inverse Lortentz transformations,

[x´+vt´]² 1

x²= ————— = ———— (x´²+v²t´²+2v x´t´) (2)

[ v²/c²)]² 1-(v²/c²)

[t´+(v/c²)x´]² 1

t²= ————— = ———— (v²/c4)x´²+t´²+(2v/c²)x´t´) (3)

[ v²/c²)]² 1-(v²/c²)

y² = y´² (4)

z² = z´² (5)

Upon substitution, we obtain

x² + y

² + z

² – c

² t² = x´

² + y´² + z´

² – c

²t´² (6)

But since the left side of the equation equals zero, so must the right side:

x´² + y´

² + z´

² = c

²t´

² (7)

Thus the progression as determined by 0' is also spherical. And so the equation for the

progression is invariant under a Lorentz transformation.

B. Polar Coordinates

In polar coordinates the equation is simply

r – r0 = c(t – t0) (8)

Or, letting r0 = 0, t0 = 0, (9)

r/t = c

In the Reciprocal System the speed of light is the natural unit of velocity and so r and t

must take equal natural values. The space–time progression is thus ¹/1,

²/2,

3/3, etc. Thus

one unit of space is equivalent to one unit of time. If there are an infinite number of

space units, there must be an infinite number of time units; if there are a finite number of

space units, there must be a finite number of time units.

II. Mathematics of Radiation

In the Reciprocal System radiation is the combined motion of a simple harmonic

oscillation in one dimension and a uniform translation in a perpendicular direction.

A. Simple Harmonic Oscillation

The equation for a simple harmonic oscillation in one dimension (say the y direction) is

y = A*SIN(–2 fost) (10)

where A is the amplitude and fos is the frequency. Since the oscillation takes place over

one natural space unit, the amplitude must be one–half a natural space unit and thus is

A = .5*Snat = .5 x 4.558816 x 10–8

m = 2.279408 x 10–8

m (11)

for all photons. In observation from the time–space region this value is reduced by the

interregional ratio142.222 to 1.6027 x 10–10

m = 1.6027 Å

The other variable to be determined in eq.(10) is the frequency, fos. In one cycle the

oscillation travels one space unit up and one space unit down, for a total of two units.

The average velocity of the oscillation is th

vos = (2*snat/cycle) * fos (cycles/sec) (12)

The natural unit of frequency must occur when the average velocity is

c.

vos = c = 2*snat * fos nat (13)

But c = snat/tnat´

so snat/tnat = 2*snat * fos nat

Solving for fos nat we have

fos nat = 1/(2*tnat) = 1/(2*1.520655 * 10–16

sec) = 3.2880575 * 1015

cycles/sec (14)

the Rydberg frequency. (Actually, Larson derived the natural unit of time from the

Rydberg frequency, but I think it was instructive to do the reverse, and this method will

be used to calculate rotational and rotational vibration frequencies as well. Of course,

this method assumes that the natural unit of time can be found by some other means.)

Because of the discrete nature of the Reciprocal System, it is only possible to have

integer multiples or reciprocal integer multiples of the Rydberg frequency.

Putting the values of A and fos in eq.(10) we have

y = 1.6027*SIN(–2 n * 3.2880575 * 1015

* t) Å (15)

where

n = 1, 2, 3, ...or (16)

n = ½, ¹/3,

¼, ....

B. Perpendicular Translation

Perpendicular to the oscillation is a translation at unit velocity (the speed of light). Let x

be perpendicular to y.

Then

x = c * t (17)

C. Combined Motion

From eq.(17) t can be found in terms of x and c and put in eq.(15). The result

is

y = 1.6027 * SIN(–2 n * 3.2880575 * 1015

* x/2.997930 * 108) (18)

or y = 1.6027 * SIN (–6.8912465 * 107 * nx)Å

if x is given in meters. This is the equation for a monochromatic wave of radiation in the

Reciprocal System.

III. Mathematics of Matter

Particles of matter consist of rotating photons. Subatoms have one rotating photon;

atoms have two rotating photons (both photons rotate about the same central point). The

rotational motion has a translational effect, which will be discussed after the mathematics

of the rotation has been worked out.

A. Rotation

1. single systems–particles

A photon can rotate around either of two horizontal axes passing through its midpoint,

and also around itself. In the Reciprocal System the true physical zero is motion at unit

speed. Anything physical must have a motion either greater or less than unit speed. This

deviation is called a speed displacement by Larson. The first particle has 1 speed

displacement around one horizontal axis of the photon and is called the rotational base.

Actually there are two rotational bases: one with one speed displacement above unity,

the other with one speed displacement below unity. As will be discussed later, the one

displacement unit neutralizes the translational motion of the photon in the original

dimension, but the progression now continues in the remaining dimension, so the

effective displacement is zero. In the ground state condition, the photon that is being

rotated is one vibrational displacement away from unity (either 2R or (1/2)R). Here is a

table giving the photon frequency, the rotational displacement, the effective rotational

displacement, and rotational speed of the cosmic rotational base and the material

rotational base:

Photon

frequency

Rotational

displacement

Effective

rotational

displacement

Rotational

speed

———— ————— ———————— ——————

Cosmic rotational

base ½R (1)-0-0 0-0-0 2-1-1

Material rotational

base 2R 1-0-0 0-0-0 ½-1-1

In the above table the speeds are calculated from the displacements as follows. For

displace

ments of np, ns, and nE, the speeds are (np+1), (ns + 1), and (nE + 1) for a cosmic particle,

and 1/(np + 1), 1/(ns + 1), and 1/(nE + 1) for a material particle. (Of course material

particles could have high speed electric displacement, and cosmic particles could have

low speed electric displacement).

These speeds can be converted to conventional units, such as revolutions per second, as

follows. In one rotation of a photon about a horizontal axis the tip of the photon coverse

a distance of * snat´ a circumference. The speed of the rotation is

then

Vrot = [( * snat) /rev] * frot (rev/sec) (19)

The natural frequency of rotation must occur when the speed is c.

c = ¹ * snat * frot nat

But c = snat/tnat´

so solving for frot nat, we get

frot nat = 1/(¹tnat)

But tnat = 1/(2R)

so frot nat = 2R

/¹ (20)

where R is the Rydberg frequency, as before. In these terms, then, the cosmic rotational

base is a photon that has a vibrational oscillation of 1.6440288 x 1015

cycles/sec and is

rotating at 4.1864848 x 1015

revolutions/sec around one axis, and 2.0932424 x 1015

revolutions/sec around the other two axes. Likewise, the material rotational base is a

photon that has a vibrational oscillation of 6.576115 x 1015

cycles/sec and is rotating at

1.0466212 x 10 15

revolutions/sec around one axis, and 2.093242 x 1015

revolutions/sec

around the other two axes.

All the other particles have photon vibrational frequencies, rotational displacements,

effective rotational displacements, rotational speeds, and rotational frequencies. Here is a

complete tabulation:

Photon

frequency

Rotational

displacement

Effective

rot.

displacement

Rotational

speed

Rotational

frequency

———— ————— ————— ———————— ——————

M–

positron 2R 1–0–1 0–0–1

½-1–

½ R

/2R

/R/

C–positron ½R (1)–0–(1) 0–0–(1) 2–1–2 4R/

2R/

4R/

M–

electron 2R 1–0–(1) 0–0–(1)

½-1–2 R

/-2R

/4R

/

C–electron ½R (1)–0–1 0–0–1 2–1–½ 4R

/2R

/R/

Photon

frequency

Rotational

displacement

Effective rot.

displacement

Rotational

speed

Rotational

frequency

———— ————— ————— ———————— ————————

M-

massless

neutron

2R 1–1–0 ½ –½–0 ½–

½-1 R

/R/

²R/

C-

massless

neutron

½R (1)–(1)–0 (½)–(½)–

0 2–2–1 4R

/4R

/²R

/

M-

neutrino 2R 1–1–(1) ½–½–(1)

½-½–2 R

/-R

/4R

/

C–neutrino ½R (1)–(1)–1 (½)–(½)–1 2–2–½ 4R

/4R

/R/

Many more permutations appear to be possible, but the probability principles keep

eccentricity to a minimum. Since none of the above particles has an effective

displacement of 1 or more, they are all massless (aside from the mass contribution of an

electric charge). The diameter of all the particles is one natural space unit, reduced by the

(one–photon) interregional ratio, or 3.2054 Å. However, because these particles do not

exert any force in the uncharged state, a particle–measuring probe would not be able to

detect any size of these particles at all.

2. intermediate systems

Intermediate particles have two rotating photons, but one of the two sets has no

effective displacement and thus contributes no primary mass. The two intermediate

particles are the neutron and the mass one hydrogen isotope (and their cosmic analogs).

There are only two kinds of rotations that can combine to form this kind of particle, the

proton type and the neutrino type We identify the combination of the material proton

rotation and the material neutrino rotation as the mass one hydrogen atom; the

combination of the material proton rotation and the cosmic neutrino rotation as the

neutron; the combination of the cosmic proton rotation and the cosmic neutrino rotation

as the mass one atom of cosmic hydrogen; and the combination of the cosmic proton

rotation and the material neutrino rotation as the cosmic neutron. The proton is a single

system with displacements 2–1–(1), effective displacements 1–1–(1), speeds ½-½

-2, and

rotational frequencies 2R/3 –R/ –4R/ . Then we would have the following table for the

neutron and mass one hydrogen.

Photon

frequency

Rotational

displacement

Effective rot.

displacement

Rotational

speed

Rotational

frequency

———— ————— ————— ———————— ————————

Neutron {

2R 2–1–(1) 1–1–(1) ¹/² –½-2

2R/³

R/

R/

½R (1)–(1)–1 (½)–(½)–

1 2–2–½ 4R

/4R

/R/

H¹ {

2R 2–1

> (1) 1–1

> (1) ¹/³-½

> 2

2R/³

-R/

>4R/ 2R 1–1 ½–½ ½ –½ R

/R/

The new notation makes clear the two photons involved and the five rotations (to be

discussed next).

3. Atomic cycles

Atoms have two rotating photons, but here both systems have effective displacements

and both systems ordinarily have the same velocities. Let the first photon be called A

and the second be called B. A and B are mutually perpendicular. We have the following

five rotations:

(i) the rotation of A about B produces disk a;

(ii) the rotation of B about A produces disk b;

(iii) then disk a can be rotated about A;

(iv) then disk b can be rotated about B;

(v) finally the whole structure can be rotated in the electric dimension.

This last rotation is in the scalar direction opposite to that of the previous rotations.

Cosmic atoms have speeds above unity for the first four types of rotations, whereas

material atoms have speeds below unity for the first four types. The electric rotation may

be above or below unity for both cosmic and material atoms.

The first particle with two effective rotating systems is deuterium, the second is helium,

etc. A table similar to that for the intermediate particles can be made.

Photon

frequency

Rotational

displacement

Rotational

speed

Rotational

frequency

———— ————— ——————— ————————

Deuterium {

2R 2–1

> (1) ¹/³ –

½

> 2

2R/³

R/

>4R/ 2R 2–1 2³–½ 2R

/³R/

Helium { 2R 2–1 >

0 ¹/³-½ > 1 2R

/³-R

/ >2R/

2R 2–1 ½³ –½ R

/³R/

All other atoms can be given appropriate values in the same manner. In the solid state,

however, the values that govern the physical properties are not the actual rotations, but

the relative rotations, and the different values there are not due to inherent differences in

the rotational speeds, but to differences in the orientations of the interacting atoms, and

this will be discussed further later.

4. Electric charges and magnetic charges

According to the Reciprocal System an electric charge is a rotational vibration about the

electric axis, and the magnetic charge is a rotational vibration about one of the magnetic

axes. Both charges have the same natural frequency, calculated as follows. In one cycle

the motion covers a distance of ¹ * snat one way and * snat back, for a total of 2 * snat.

So we have

vch = (2 * snat/cycle) * fch (cycles/sec) (21)

At the unit level, vch = c = snat/tnat, so

snat/tnat = (2 * snat/cycle) * fch nat (22)

Solving for fch nat and recalling that tnat = 1/(2R),

fch nat = R/ (23)

This frequency is one–half that of a full rotation and can thus be considered to be

effective in one direction only half the time. One negative electric charge is a rotational

vibration of R/2 = 5.233106 x 1014

cycles/sec. One positive electric charge is a

rotational vibration of 2R/ = 2.093242 x 1015

cycles/sec. Similarly one unit of magnetic

charge is a rotational vibration of 2R/ = 5.233106 x 1014

cycles/sec, whereas one unit of

isotopic charge is a rotational vibration of R/2 = 5.233106 * 1014

cycles/sec. The

isotopes of atoms result from the addition of isotopic charges.

B. Translation

The rotational motion of particles has a translational effect. The maximum inward

translation is two full units, giving one net inward unit. In terms of rotation we can have

2³ = 8 one–dimensional rotational electric displacements or 4 two–dimensional rotational

magnetic displacements. Note that since 1³ = 1, the first magnetic rotational

displacement, which is ½ unit rotational speed, produces one unit of inward translation

and thus neutralizes the original translational motion of the photon, but the progression

still continues in the third dimension. Thus the rotational base and all the single system

massless particles previously discussed move at the speed of light. Additional magnetic

and electric displacements produce a net inward motion, and the inward motion of a

group of atoms is termed gravitation.

For atoms with magnetic displacements of less than 4 and electric displacements of less

than 8, the frequency of the rotating photons is normally one displacement above unity,

or 2R (the frequency of photons in cosmic atoms is (½)R). When the magnetic

displacement reaches 4 or the electric displacement reaches 8, the rotation must be

extended to a second vibrational displacement unit–which means that the frequencies of

the photons are now 3R (or (¹/3)R for cosmic atoms). As Larson points out, though, it is

possible to have these higher frequency photons even when the rotational displacements

are less than 4 or 8, in which case we can say that the atom is ―excited‖.

After the change to vibration two, two units of vibrational displacement exist to be

rotated, and so each added unit of rotational displacement corresponds to only one–half

unit of added specific speed. Thus the speeds corresponding to magnetic displacements

can be listed as follows:

Magnetic Displacement Magnetic Speeds

1 ½ ½ ½

2 ¹/3 ²/5 ¹/3

3 ¹/4 ¹/3 ¹/4

4 ¹/9 ²/7 ¹/5

5 ²/5 ¹/4 in one

displacement

axis only

And the speeds corresponding to electric displacements can be listed as follows:

Electric Displacement Electric Speeds Electric Displacement Electric Speeds

1 ½ ½ 9 ¹/9 ¹/6

2 ¹/3 ²/5 10 ²/19 ²/13

3 ¹/4 ¹/3 11 ¹/10 ¹/7

4 ¹/5 ²/7 12 ²/21 ²/15

5 ¹/6 ¹/4 13 ¹/11 ¹/8

6 ¹/7 ²/9 14 ²/23 ²/17

7 ¹/8 ¹/5 15 ¹/12 ¹/9

8 ²/17 ²/11 16 ²/25 ²/19

In the solid state, the values for electric rotation can be further altered. Larson states

that a combination of one atom of electric displacement x with another atom of electric

displacement 8–x results in a neutral bond. This bond gives rise to an electric speed of ¹/10 for vibration one, and

¹/5 for vibration two. Also there can be a combination of two 8–

x atoms, which Larson calls a secondary positive bond. In this case the rotational speed

comes to ¹/(18–2x).

One final set of complications involves the lower group elements. Here there is just one

subordinate magnetic displacement unit and thus these elements have less rotational force

and thus are closer together in the solid state. The force is proportional to ln t, where t is

the inverse of the magnetic speed, and since ln 2 is less than 1, atoms that have magnetic

speed greater than ¹/3 in any dimension have no effective force in that dimension. The

number of ―active‖ dimensions is given in the following tabulation.

Lower Group Atomic Table in Solid Stat

Atom.

No Elem.

Oscillation

Frequency

Rotational

Displ. Bond

Rotational

Speed

Rotational

Frequency

Active

Dim.

{ 2R 2-1 >

(1) ¹/³–½ >

¹/10 2R

/³R/ >

R/

1 H

Neutral

1

2R 2-1 ¹/³–

½ 2R

/³R/

{

2R 2-1

> 0

¹/³–½

> 1

2R/³

R/

> 2R/

2 He

Zero

1

2R 2-1 ¹/³–

½ 2R

/³R/

{

3R 2-1

> 1

²/5–²/5

> ½

4R/5

R/5

> R/

3 Li

Positive

2

3R 2-1 ²/5–

²/5

4R/5

R/5

{

3R 2-1

> 2

¹/³–½

> ²/5

2R/³

R/

> 4R/5

4 Be

Positive

2

3R 2-1 ¹/³–

½ 2R

/³R/

{

3R 2-1

> 3

¹/³–½

> ¹/5

2R/³

R/

> 2R/5

5 B

Neutral

1

3R 2-1 ¹/³–

½ 2R

/³R/

{

2R 2-2

> (5)

¹/³–¹/³

> ¹/10

2R/³

2R/³

> R/5

or B

Neutral

1

2R 2-2 ¹/³–¹/³

2R/³

2R/³

{

2R 2-1

> 4

¹/³–½

> ¹/10

2R/³

R/

> R/5

6 C

Neutral

2

2R 2-1 ¹/³–

½ 2R

/³R/

{

2R 2-2

>

¹/³–¹/³

> 1

2R/³

2R/³

> 2R/

or C

Zero

3

2R 2-2 ¹/³–¹/³

2R/³

2R/³

{

2R 2-2

> (3)

¹/³–¹/³

> ¹/10

2R/³

2R/³

> R/5

7 N

Neutral

1½

2R 2-2 ¹/³–¹/³

2R/³

2R/³

{

2R 2-2

> (3)

¹/³–¹/³

> 1

2R/³

2R/³

> 2R/

or N

Zero

3

2R 2-2 ¹/³–¹/³

2R/³

2R/³

{

2R 2-2

> (2)

¹/³–¹/³

> ¹/10

2R/³

2R/³

> R/5

8 O

Neutral

1½

2R 2-2 ¹/³–¹/³

2R/³

2R/³

{

2R 2-2

> (2)

¹/³–¹/³

> 1

2R/³

2R/³

> 2R/5

or O

Zero

3

2R 2-2 ¹/³–¹/³

2R/³

2R/³

{

2R 2-2

> (1)

¹/³–¹/³

> ¹/10

2R/³

2R/³

> R/5

9 F

Neutral

2

2R 2-2 ¹/³–¹/³

2R/³

2R/³

The reader can continue the table all the way to element 118. Again, one must first

determine the kind of bond involved before the electric rotational speed can be

determined.

Since different atoms have different rotational speeds and thus different rotational

forces, a particle probe of equal energy shot at atoms of different elements would

―penetrate‖ to different depths. Thus experimenters have concluded that ―nuclear‖ size is

proportional to atomic weight. Actually what they are measuring is atomic size, and

according to the Reciprocal System this is constant (2.914 Å diameter)–but the force is

proportional to the atomic weight of the atom. Also, where interatomic distances are less

than 2.914 Å, the atoms are partially merged; where distances are greater than 2.914Å,

the atoms are separate.

Reference

1. Dewey B. Larson, Nothing But Motion (North Pacific Publishers: Portland, Oregon,

1979)

A NEW DERIVATION OF PLANCK‘S CONSTANT

To present-day physical science the numerical value of Planck‘s constant is a mystery:

quantum mechanica does not have a theoretical method for its calculation. By contrast the

Reciprocal System of theory derives the value of all physical constants, including

Planck‘s constant, from its fundamental postulates. However, because of errors in the

previous derivations, this paper presents a new, dimensionally sound method for the

calculation.

Larson¹ was the first to attempt to derive Planck‘s constant from the Reciprocal System.

Because of the change in the calculated natural values of mass and energy in the second

edition of his work², the original derivation has been invalidated. The factor of three that

was used is dimensionally incorrect since the photon is a one-dimensional vibration. And

the use of the cgs gravitational constant in such an equation is wrong since the result

cannot be converted to a different system of units such as the Sl (mks) system. The

remainder of Larson‘s original equation (including the use of the interregional ratio and

the square of the natural unit of time) will be shown to be correct.

Nehru³ made the second attempt to calculate the constant. However, he started by setting

the dimensions of energy to be space divided by time, which is, of course, the reverse of

what they are. The rest of the derivation was very tortuous, although he ended up with a

good numerical result (with faulty dimensions).

Let us now proceed with the new derivation. First, consider conceptually the linear

vibration of the photon. The oscillation takes place over one space unitwhich,

simultaneously, is also one time unit. In the material sector of the universe, we define

frequency to be cycles/sec, because here it is time that appears to have a uniform

progression; in the cosmic sector of the universe, hypothetical cosmic observers would

define frequency to be cycles/cm (or some such length unit), because there it is space that

would appear to have a uniform progression. Actually, the photon exists at the boundary

between the two sectors, where both space and time progress uniformly. Here the correct,

nAtural definition of frequency must be cycles/(cm-sec) (or equivalent units). To put it

another way, frequency in the natural sense is the number of cycles per space-time unit.

Photons of all frequencies can be observed in both sectors, and the only way that this

could be possible is if the denominator of the natural definition contains both a space unit

and a time unit. This then causes Planck's constant to have the actual dimensions of erg-

cm-sec. However, if the dimensions of frequency are assumed to be cycles/sec, rather

than cycles/(cm-sec), then the dimensions of Planck‘s constant are erg-sec. Let E be

photon energy, h be Planck‘s constant, and be photon frequency. Then, as usual, we

have

E=h* (1)

In space-time terms, equation (1) is

[

t²

]

t ——– 1

— =

( t/s

) ——

(2)

s ——– s * t

t/s

In the cgs system of units, equation (1) is

[

sec²

]

——– 1

erg =

( sec/cm

) ——–—

(3)

——– cm * sec

erg

Observe, in both cases, the dimensional consistency. Since the oscillation of the photon

takes place with in a unit of space-time, the interregional ratio must be contained within

Planck‘s constant. With this factor and the dimensional information from above, Planck‘s

constant is

1 t²0

h= ———— * ———–

156.4444

( sec/cm

)

———–

erg

where t is the natural unit of time (1.620666 * 10'18 sec).

Ref. 3 states that the ratio of (sec/cm)/erg is 2.236066 x 10-8

. This figure is deduced as

follows. Dimensionally unit mass is t³/s³ , or 3.711381 * 10-32

sec³/cm³. Avogadro‘s

constant is the number of atoms per gram atomic weight. 6.02486 * 10-23

. The reciprocal

of this number, 1.66979 * 10-24

, in grams, is therefore the mass equivalent of unit atomic

weightz². Thus to convert from the unit sec³/cm³ to grams we must divide by 2.236066 *

10-8

. From the euprsssion E = mc² we see that the sa.me conversion factor must apply to

energy (in ergs) to keep the equation balanced. (Nehru³ modi ied his equation to include

secondary mass; however, his rasulting equation is dimensionally incorrsct. Furthermore,

secondary mass varies between the subatoms and atoms and so cannot be a part of the

conversion factor). Thus the numerical value of Planck‘s constant is

h = 6.6102662 g 10-27

erg-sec (5)

(when frequency is assumed to have the dimensions cycles/sec).

This is 99.77% of the egperimental value of 6.6266 * 10-27

erg-sec. Given the

uncertainties involved in the determination of Avogadro‘s constant and the natural unit of

time, the result is satisfaetory. Any improvement in the accuracy of these values would be

reflected in an improvement in the accuracy of the calculation of Planck‘s constant.

References

1. Dewey B. Laraon, The Structure of the Physical Universe (Portland, Oregon:

North Pacific Pub- lishers, 1969), pp. 117-118.

2. Dewey B. Larson, Nothing But Motion (Port- land, Oregon: North Pacific

Publishers,1979), pp. 157-168.

3. K.V.K. Nehru, ―Theoretical Evaluation of Planck’s Constant,‖ Reciprocity, Vol.

XII, No. 3.

TIME REGION PARTICLE DYNAMICS

Mr. Larson has worked. out the static relations between particles in the time region;

specifically, he has calculated the equilibrium interatomic distances for all the elements

and many compounds (see pages 27-49 of The Structure of the Physlcal Universe). This

paper will explore he dynamic relations between particles in the time region.

Consider a particle (say an alpha particle) moving directly towards a stationary atom (say

a gold atom fixed in thin foil). Initially the particle has a velocity vo. Once it enters the

tlme region, that is, when its distance is less than one natural unit of space, two forces are

encountered: the progression and gravitation. In the time region, the progression acts to

bring particles closer together, whereas gravitation acts to repel particles — the reverse of

gravitation in the time-space region. The progression is stronger until the equilibrium

distance is reached, then the gravitational force becomes stronger. I believe that the

equation of motion is

Fp - KG / (xu- x)4 = m d²x / dt² (1)

where

Fp = unit force of the progression

KG = magnitude of the rotatianal motion of the partlcles

xu = natural unit of space

m = mass of the moving particle

x = distance-measured from start of time region

Dividing by m gives

Fp / m - KG / m (xu - x4 = d²x / dt²

The right hand side reduces to

d²x/dt² = dv/dt = dv/dx dx/dt = v dv/dx (2)

Thus

Fp/m - KG/m (xu - x)4 = v dv/dx

Separating variables and integrating, we have

xf

xf

vf

Fp/m dx - KG dx/m (xu -x)4 = vdv

o o vo

or

Fp/m xf - KG /3m [(xu - xf)-3 - (xu)

-3] = ½ (vf² - vo²) (3)

There are two cases of interest with this equation.

Case 1: Suppose we want to know the initial velocity required to bring the particles to a

certain distance apart from each other. Equation (3) is solved for vo, letting vf be zero.

Vo = (2[KG/3m{(xu - xf)-3 - (xu)

-3}-Fp/m xf] )½ (4)

Case 2: Suppose we want to know the final separation between two particles, given vo.

Let

xsep = xu -x

Equation (3) becomes

Fp/m (xu - xsep) - KG/3m xsep-3 - xu

-3] = ½ (vf² - vo²)

With vf = 0, and putting the terms irevolving xsep on one side af the equation, we have

Fp/m xu + KG xu-3/3m + 1/2 vo² = Fp/m xsep + KG/3m xsep

-3

Define the following coefficients:

C2 = Fp/m xu + KG xu-3/3m + ½ vo²

C2 = Fp/m

C3 = KG/3m

C4 = -C1/C2

C5 = C3/C2

With these coefficients, the result is a quartic equation:

xsep4 + C4 xsep

3 + C5 = 0 (5)

This equation can then be solved by the usual means.

Now, going back to equation (3) we can solve for v as a function of x:

v = dx/dt = {2[ Fp/m x - KG/3m[(xu - x)-3 -xu

-3]] + vo²}½

Separating variables and integrating we have

x

t = dx/{2[Fp/m x -KG/3m((xu - x)-3 -xu

-3)] + vo²}½

o

The integral can be evaluated numerically by Romberg‘s method.

Example

Consider an alpha particle moving directly towards a gold atom in a foil, at an initial

velocity of 2.06x107 meters/sec. What is the distance of closest approach? How long does

it take to get there? What happens afterward.?

Here we have

vo = 2.06 x 107 m/sec

m = 6.64 x 10-27 kg

xu = .455884 x 10-7 m

Fp = 1.09699 x 10-3 n

Now,

KG = 1.09699 * 10-3 * (.455884 * 10-7 )4 /(156.44) * ln² tA ln² tB

For gold, tA =4.5; for helium, tB = 3. But helium has only one active dimension so the

force is multiplied by 1/3. Thus

KG = 1.09699 * 10-3 * (.455884 * 10-7 )4 /(156.44)4 * ln² (4.5) ln² (3)

* 1/3 = 7.20006 * 10-42 N-m4

(This assumes that since helium is inert, the electric displacements of gold have no

bearing on the motion). The coefficients are next calculated:

C2 = 2.7438094 * 1015

C2 = 1.6520934 * 1023

C3 = 3.614488 * 10-16

C4 = 4.6872709 * 10-8

C5 = 2.197923 * 10-39

The quartic equation is

xsep4 - 4.6972709 * 10-8 xsep

3 + 2.187823 * 10-39 = 0

The only physical solution is

xsep- = 3.6014328 * 10-11 m .36 Å 3.6226287 * -11 m with revised Fp]

Note that his is cons:lderably greater than that predicted by use of classical atomic theory

and Coulomb‘s law: 2.581 x 1014 m.

Using equation (6) and Romberg‘s method I find that

t = 6.3041312 * 10-16 sec

The average velocity of the particle to the point of closest approach is

4.552396 * 10-8 /6.3041312 * 10-16 = 7.2212742 * 107 m/sec

The initial velocity having been dissipated, the particle goes back to the equilibrium

point. Of course, at room temperature, helium is a gas, and so the particle would not

remain in the time region!

Situations in which the particle is not moving directly towards the atom will be treated in

a future paper.

CALCULATION OF THE DISSOCIATION ENERGY

OF DIATOMIC MOLECULES

This paper presents the first rational calculation of the dissociation energy of diatomic

molecules. Quantum mechanics does not have such a calculation, even in principle. The

importance of this calculation is that it provides additional quantitative verification of the

molecular force and energy concepts of the Reciprocal System.

Dissociation energy is the change in energy (usually expressed in kcal per mole) at

absolute zero temperature in the ideal gas state for the reaction

A-B —> A + B (1)

the products (atoms A and B) being in their ground states and the reactant (molecule A-

B) in the zeroth vibrational level. Note that dissociation energy is slightly different from

bond energy, which is defined as the standard enthalpy change at 25º C for the ideal gas

reaction given above. Calculating dissociation energy rather than bond energy frees us

from having to consider molecular thermal energy.

Now let us proceed to the derivation of the expression for bond dissociation energy from

the principles of the Reciprocal System. A diatomic molecule, as a unit, exists in the

time-space region. However, the two individual atoms of the molecule, relative to one

another, exist in the time region because the interatomic distance is less than one space

unit; hence, time region expressions apply to the attributes of the bond. To quote Larson,

The motion in time which can take place inside the space unit is equivalent to a motion in

space because of the inverse relation between space and time. An increase in the time

aspect of a motion in this inside region (the time region, where space remains constant at

unity) from 1 to t is equivalent to a decrease in the space aspect from 1 to 1/t. Where the

time is t, the speed in this region is equivalent space 1/t divided by time t, or 1/t²[Ref. 1].

Thus,

v = 1/t² (2)

In the Reciprocal System, energy is the reciprocal of speed. Hence, in the time region,

E = t² (3)

This energy equation gives the proper dimensional form of the expression for dissociation

energy. It can be generalized to

E = ta * tz (4)

In application to the problem at hand, ta and tz refer to the rotational time

displacements of the atoms of the molecule, where ta is the primary magnetic

displacement or the secondary magnetic displacement and tz is the second magnetic

displacement or the electric displacement. To justify this intepretation, let us recall

that the two atoms of the molecule are in translational equilibrium; in the Reciprocal

system this means that the scalar translational repulsion effect of the rotational force of

the atoms is equal and opposite to the cohesive translational force of the space-time

progression; the magnitude of the force is thus equal to the translational equivalent force

of the rotational force of the atoms and so the required dissociation energy must equal the

rotational energy. Because of the discrete unit postulate, less than this amount of energy

would be ineffective.

As it stands, equation (4) expresses the energy in natural units of the time region. We

have to convert the equation to an equivalent expression for the time-space region so that

we can compare calculated to observed results. First of all we must apply the fourth

power of the interregional ratio, 1/156.44, to the equation, just as is done in the atomic

force equation.

E = (1/156.44)4 * ta * tz (5)

This is the energy in natural units as would be observed in the time-space region. To

convert this to conventional units of measurement we multiply by the value of the natural

unit of energy expressed in conventional units, Eu.

E = (1/156.44)4 * ta * tz * Eu (6)

The experimental values are expressed as kcal/mole so we must multiply the right side of

the equation (6) by a conversion factor, k, and by Avogadro‘s number, N.

E = (1/156.44)4 * ta * tz * Eu * k * N (7)

Next we must append a factor of ½ to the expression to account for the inherent

vibrational nature of the time region motions and a factor of 1/3 to the expression to

reduce the energy to one dimension. So now we have

E = (1/156.44)4 * ta * tz * Eu * k * N * (1/6) (8)

From Ref. 1, Eu is 1.49175 x 10-3

ergs and N is 6.02486 x 1023

. k is 2.389 x 10-11

kcal/erg. The final working equation is

E = 5.9747 * ta * tz kcal/mole (9)

Displacement ta can range from 1 to 4 and displacement tz can range from 1 to 8. Table I

lists the possible values of E for the various combinations of ta and tz.

I have applied equation (9) to 18 diatomic molecules of the elements. The theoretical and

experimental results are given in table II. Let t1 symbolize the primary magnetic

displacement of an element, t2 the secondary magnetic displacement, and t3 the electric

displacement. It is clear from the table that

ta = t1, or ta = t1 + 1, or ta = t1 - 1, or ta = t2, or ta = t2 + 1, or ta = t2 - 1 (10)

And

tz = t2, or tz = t2 + 1, or tz = t2 - 1, or tz = t3 (11)

For electronegative elements, the 8-t3 rule applies:

tz = 8 - t3, or tz = 8 - t3 + 1 (12)

Generally, only one (if any) of the two displacements has to be incremented or

decremented by 1 to obtain a good fit with the experimental data; the other displacement

equals the rotational displacement (or 8 minus the rotational space displacement) as the

theory requires. Elements that require an increment of displacement usually have low

atomic number; elements that require a decrement of displacement usually have high

atomic number.

The values of ta and tz thus fit the normal variations in the elements that have appeared

in other Reciprocal System calculations. This, together with allowance for experimental

error, allows us to conclude that we have good agreement between theory and reality.

A future paper will apply equation (9) to diatomic molecules of unlike atoms.

References

1. Dewey B. Larson, Nothing But Motion (Portland, Oregon: North Pacific

Publishers, 1979), p. 155.

2. John A. Dean, ed., Lange’s Handbood of Chemistry, Eleventh Edition (New

York: McGraw-Hill Book Company, 1973), pp. 3-123 to 3-127.

Table I: Allowed Values of Dissociation Energy

E kcal/mole ta tz ta tz ta tz

5.9747 1 1

11.9494 1 2 2 1

17.9241 1 3 3 1

23.8988 1 4 2 2 4 1

29.8735 1 5

35.8482 1 6 2 3 3 2

41.8229 1 7

47.7976 1 8 2 4 4 2

53.7723 3 3

59.7470 2 5

71.6964 2 6 3 4 4 3

83.6458 2 7

89.6205 3 5

95.5952 2 8 4 4

107.5446 3 6

119.4940 4 5

125.4687 3 7

143.3928 3 8 4 6

167.2916 4 7

191.1904 4 8

Table II: Calculated and Observed Values of Dissociation Energy

Molecule Displacement Method ta tz Ecalc. Eobs.

As-As 3-3-(3) t2 8-t3 3 5 89.62 91

Cs-Cs 4-3-1 t2-1 t3 2 1 11.95 10.4

Cl-Cl 3-2-(1) t1 t2+1 3 3 53.77 57.07

Cu-Cu 3-3-(7) t1 t2 3 3 53.77 48

F-F 3-2-8 t1 t2 3 2 35.85 36

Ga-Ga 3-3-(5) t1 t2-1 3 2 35.85 35

Au-Au 4-4-(7) t2 8-t3+1 4 2 47.80 52

D-D 2-1-(1) t1 8-t3+1 2 8 95.60 105

I-I 4-3-(1) t1-1 t2-1 3 2 35.85 35.55

Li-Li 2-1-1 t1 t2+1 2 2 23.90 25

P-P 3-2-(3) t1+1 8-t3 4 5 119.49 116.0

K-K 3-2-1 t2 t3 2 1 11.95 11.8

Se-Se 3-3-(2) t2-1 8-t3 2 6 71.70 65

Ag-Ag 4-3-(7) t2 8-t3+1 3 2 35.85 39

Na-Na 2-2-1 t2+1 t3 3 1 17.92 17.3

S-S 3-2-(2) t2 8-t3+1 2 7 83.65 83

Te-Te 4-3-(2) t1-1 t2 3 3 53.77 53

Sn-Sn 4-3-(4) t2-1 8-t3 2 4 47.80 46

Note: the observed values, Eobs., come from Reference 2.

THE LIQUID STATE IN THE RECIPROCAL SYSTEM:

THE VOLUME/TEMPERATURE RELATION,

A CONTEMPORARY MATHEMATICAL TREATMENT

This paper provides a step-by-step procedure for the calculation of liquid specific volume

as a function of composition and temperature, based on the Reciprocal System of D. B.

Larson1. In this theory, each individual molecule may be in the solid, liquid, or gaseous

(or vapor) state, regardless of the state of the majority of molecules of the substance.

First let's define some terms:

= overall specific volume of liquid (cm3/g) (total volume/total mass)

= specific volume increment at 0 oK and that due to the solid molecules in

solution of the liquid (solid volume/total mass)

= specific volume increment due to the liquid molecules of the substance,

temperature above 0 oK (liquid volume/total mass)

= specific volume increment due to the critical (gaseous or vapor) molecules in

solution of the liquid (gaseous volume/total mass)

Then,

(1)

The initial values of these three components are designated . These differ only

by a geometric factor (designated ) applied to a base initial value, ,

determined as follows.

Just as the volume of a gas is determined by the number of molecules, so the

volume of a liquid is determined by the number of volumetric groups which it contains.

In an organic compound, for instance, each of the common interior groups, such as CH2,

CH, or CO, constitutes one volumetric group. The CH3 groups in the end positions of the

aliphatic chains occupy two units each. So hexane, represented as

CH3CH2CH2CH2CH2CH3, has 8 volumetric groups. Let be the number of volumetric

groups and recall that the factor .707 expresses the geometric reduction obtained by the

close-packed arrangement of the liquid groups because of their flexibility of movement.

Then, in natural units, the base initial volume is directly proportional to the number of

volumetric units, reduced by close-packing:

(2)

Let be the molecular weight (non-dimensional) of the molecule of the substance,

be the value of the natural unit of atomic mass in g, and be the value of the natural

unit of liquid volume expressed in cm3. Then, in conventional units, the basic initial value

is

cm3/g (3)

is not the cube of the natural unit of space in the time-space region, which is

applicable only to the gaseous state. Rather, is the cube of the natural unit of space in

the time region, which is 1/156.45 (the inter-regional ratio) of that in time-space region,

or 2.9139 x 10-8

cm. Cubing this we get

cm3

The natural unit of mass is 1 atomic mass unit, so is 1.65979x10-24

g. Putting these

values in eq. 3, we get

cm3/g (4)

For hexane, is 8 and the molecular weight is 86.18. Therefore,

cm3/g

For the critical (gaseous or vapor) specific volume increment, the geometric factor is

always 1.00. For the solid specific volume increment, the geometric factor is .891

(the cube root of .707) where close-packing in the solid state can be achieved. Where

such packing cannot be achieved, the geometric factor is 1.000. The same applies to

the geometric factor for the liquid specific volume increment, . Therefore, the initial

values of the three volume components may be expressed as

(6)

(7)

(8)

In a multi-group molecule, the value of the geometric factors and represent

averages, since some groups may be at .891 while others at 1.000. Let = the number

of close-packed groups per molecule in the solid state, and let = the number of close-

packed groups per molecule in the liquid state. Then

(9)

(10)

For hexane, for instance, is .9864 (with 1 group at .891 and 7 groups at 1.0000, the

average is 7.891/8 or .9864) and is .9728 (with 2 groups at .891 and 6 groups at

1.0000). Therefore, for hexane, the initial values of the specific volume increments are

cm3/g

cm3/g

cm3/g

From eq. 10 it's clear that ordinarily . However, for lower group

elements, hyrdrogen through fluorine, closer packing than normal can be achieved

because of inactive dimensions of the gravitational repulsion force. This means that, in

effect, for lower group elements the geometric factors can be less than .891. We can still

use eq. 10, though, if we allow the value of the number of solid groups to exceed the

number of volumetric units.

Now that we have the initial values as a function of composition, we can determine the

values of the three components as a function of temperature. The solid specific volume

increment not only includes the initial volume at 0 oK but also a factor proportional to the

number of solid molecules in the substance at any temperature, , which can be

determined by probability considerations.

(11)

To use the normal probability function or table we need to know the value of the normal

random variable, zs, applicable. It should be proportional to the difference between the

liquid temperature and the melting point , in degrees K, divided by the melting

point. The coefficient and the intercept have unfortunately not been worked out

theoretically, but are given empirically by Larson (Ref. 1) as follows:

(12)

We want the right tail of the distribution, so we subtract the value of the normal function,

denoted by erf(zs), from 1 and then multiply by the average difference in specific volume

between solid and liquid molecules, denoted by :

(13)

Larson uses an average value of of .080 for paraffin hydrocarbons (C14 and below)

and .084 for paraffins above C14 (rather than computing the individual values). For

hexane, = 178 K (-95 oC). At = -50

oC, zs=1.41 and from the normal probability

table, erf(zs) .9207. Subtracting this from 1.0000, we get .0793, which means that 7.93

% of the molecules in the liquid hexane aggregate at -50 oC are in the solid state.

Multiplying this figure by the approximate difference in specific volume between solid

and liquid molecules, .080, we get .0063 cm3/g for the value of .

The thermal motion beyond the initial point of the liquid (considered as starting at

0 oK) is the one-dimensional equivalent of the thermal motion of a gas, and thus the

volume generated is directly proportional to the temperature, . Let be the natural

unit of temperature in the time region (for the condensed states of matter) and be the

temperature factor. Then

(14)

In Ref. 2, Larson derived the value of to be 510.8 K. For simple substances, is 1.

More complex or more electropositive substances have values of of 2 up to 16.

Hexane has a value of 1; water, 2; silver, 16. Compounds of electropositive and

electronegative elements have intermediate values (some with half-integral values, which

are averages), as would be expected.

The gaseous or vapor increment of specific volume depends on the proportion of

critical molecules existing in the aggregate at each temperature, which can be computed

from probability considerations. Larson uses two random variables for this computation,

both a function of the critical temperature, :

(15)

(16)

Then the specific volume increment due to critical molecules in the substance is

(17)

For hexane, = 508 K. At = 210 oC, = .2947 and =.8106. The corresponding

values of the normal probability function are .6144 and .8109. Then, from eq. 17,

cm3/g

The .5747 factor means that 57.47% of the molecules at this temperature are in the

critical state.

Having determined we can now calculate from eq. 1.

To automate the task of comparing the theoretical values with those observed, I've

prepared a computer program and run it on most of the same liquids Larson used in the

original series of papers: hexane, hexadecane, benzene, acetic acid, ethyl acetate, ethyl

chloride, ethanethiol, fluorine, hydrochloric acid, sulfur dioxide, carbon tetrachloride, and

water. Printouts from the program for all of these liquids follow. The observed values

come from the same sources Larson used: Timmermans' Physico-chemical Constants of

Pure Organic Compounds, the American Petroleum Institute, and the International

Critical Tables.

Most of the computer results are in harmony with Larson's manual calculations.

The two seeming exceptions are for acetic acid and water. For acetic acid, Larson used a

value of initial liquid specific volume of .5469, which is .7795 that of his base initial

volume, .7016; but .891 is supposedly the smallest allowed fraction. For water, Larson

used a value of .7640 for both the initial solid and liquid specific volumes, but this is only

.8713 that of his base initial volume, .8769, not .891. Actually. these differences are due

to "hydrogen bonding", which can allow closer packing than normal. In a second

calculation for water, I input 1.78 for and so as to get the initial volumes to be

.7640. The theoretical results computed came out to be much closer to the experimental

ones than the previous run.

To compute the specific volume for any liquid of your choice, follow these steps:

1. Determine the formula of the compound and its molecular weight.

2. From the formula, determine the number of volumetric units and number of

temperature units.

3. Use equation 4 to obtain the base initial volume.

4. Use equations 9 and 10 to compute the geometric factors; some iteration here may be

required to the get the right values.

5. Compute the initial volumes with equations 6, 7, and 8.

6. Using equations 12 and 13, compute the solid specific volume increment, equation 11.

7. Use equation 14 to compute the liquid specific volume increment.

8. Using equations 15 and 16, compute the critical specific volume increment, equation

17.

9. Sum the results to get the final value, from equation 1.

References:

1. D. Larson, The Liquid State , privately circulated series of papers on the liquid state,

circa. 1960-1964. Note: I made use of the papers numbered I, II, II-supplement, and III.

I've reorganized all of the equations and changed some of the symbols for the sake of

clarity. I've also used the latest values of the conversion constants. The computer

program is entirely original.

2. D. Larson, Basic Properties of Matter (Salt Lake City, UT: International Society of

Unified Science, 1959-1988), pp. 59-60.

Appendix: The Computer Program

The following pages show the input screens of the program. The data base

language is filePro Plus and the computation language is TrueBasic. This is the first of

what will be a comprehensive series of programs for the calculation of all properties of

matter based on the Reciprocal System of theory. Eventually the programs will be made

available for purchase.

THE LIQUID STATE IN THE RECIPROCAL SYSTEM:

THE VOLUME/PRESSURE RELATION,

A CONTEMPORARY MATHEMATICAL TREATMENT, PART II

From thermodynamics,¹ the general equation of state of a pure substance is

(1)

where

volume expansivity (2)

and

isothermal compressibility (3)

(Of course, V = volume, P = pressure, T = temperature.)

From my previous paper² (and Larson‘s original work8),

(4)

where

VL = overall specific volume of liquid (cm³/g) (total volume/total mass)

V1 = specific volume increment at 0ºK and that due to the solid molecules in

solution of the liquid (solid volume/total mass)

V2 = specific volume increment due to the liquid molecules of the substance,

temperature above 0ºK (liquid volume/total mass)

V3 = specific volume increment due to the critical (gaseous or vapor) molecules in

solution of the liquid (gaseous volume/total mass)

In this paper we will consider the effect of pressure on a liquid at temperatures below the

liquid natural temperature unit, 510.8ºK. At low temperature, . Pressure has a

different effect on V3 than it has on V2. Also, pressure has a different effect on a liquid at

a temperature above, rather than below, 510.8ºK. These differences will be handled in

another paper.

For a solid under pressure³, the volume is multiplied by , where Po is the internal

pressure and P is the external pressure. For a liquid under pressure, the volume is

multipled by the square of the solid factor, or simply . So,

(5)

It follows that isothermal compressibility is

(6)

It's often easier to work with the bulk modulus, B, which is the inverse of .

(7)

From my previous paper,

cm3/g (8)

since is negligible for most liquids above the melting point.

cm3/g (9)

cm3/g (10)

where nv is the number of volumetric units.

The internal pressure of a liquid is obviously different from that of a solid. The natural

unit of pressure in the Reciprocal System is4

15,538,642 atm. To calculate the internal

pressure of a solid we divide this number by the interregional ratio, 156.45. For a liquid,

we divide by the square of the interregional ratio. Because liquid cohesion is two-

dimensional rather than three-dimensional we must also multiply the expression by 2/3.

Therefore,

atm (11)

This expression is then multiplied by the number of pressure units, np, and divided by the

ratio of the base volume to 1, raised to the 2/3 power. (The solid expression just uses

volume, or so3.) Therefore,

atm (12)

Substituting eq. 10 in eq. 12, we get

atm (13)

np is the number of atoms effectively acting against the external pressure. It is

sometimes, but not usually, equal to the number of volumetric units, nv. Using eq. 8, 9,

and 10, B can be expressed as

atm (14)

Now let's turn to calculating the volume expansivity.

K-1 (15)

where is the value of the expansivity at the end point of the solid.

One could plug (or 1/B) and into eq. 1 and integrate, but the resulting equation is more

complex than eq. 5 and thus not useful.

In summary, to calculate bulk modulus and volume expansivity of a liquid, it is

necessary to determine

m, the molecular weight

nv, the number of volumetric units

s1, geometric factor

s2, geometric factor

nt, the number of temperature units

np, the number of pressure units

Example Calculations and Comparisons with Experiment5,6

I selected four important liquids: acetic acid, carbon tetrachloride, ethyl acetate, and

water. Here are the results, in table format.

Chemical Formula M

Nv Nt Np P atm

T ok Balc atm Bobs atm

Acetic Acit CH3CO2H 60.05 .9046 .7820 4 1.0 7 1 288.16 11441.503 11279.014

Carbon Tetrachloride

CCl4 153.81 1.0 .9183 6 1.0 5 1 250.26 12334.317 11878.218

Ethyl Acetate CH3CO2C2H5 88.10 .9818 .9818 6 1.0 6 1 293.16 8687.0274 8733.6283

Water H2O 18.0153 .8707 .8707 1.5 2.0 9 1 273.16 19697.992 19698.877

Chemical calc k1 obsK-1

acetit acid 1.1377x10-3 1.269x10-3

Carbon Tetrachloride

1.240x10-3 1.2987x10-3

Ethyl Acetate 1.24398x10-3 1.304x10-3

Water 7.72383x10-4 7.992x10-4

(The values of have not yet been determined, which explains the descrepancy between

calc and obs.)

The np values are easy to understand. In acetic acid, the CH3 contributes 3 units and the

CO2H contributes 4. In carbon tetrachloride, each atom contributes 1 unit. In ethyl

acetate, each volumetric group contributes a unit. In water, 3 molecules of 3 atoms each

act against the external pressure, for a total of 9. All values of nv, nt, and np are integral

or half-integral, as required by the nature of the Reciprocal System. This is very different

from the empirical correlations used by other investigators.7

In the coming years I hope some member of ISUS will calculate the results for thousands

of liquids following the equations given here.

References:

1. M. Abbott, H. Van Ness, Thermodynamics (New York: McGraw-Hill Book

Company, 1972), p. 105.

2. R. Satz, "The Liquid State in the Reciprocal System: The Volume/Temperature

Relation, a Contemporary Mathematical Treatement," Reciprocity, Vol. XXIII, No. 2,

Autumn 1994. Incidentally, the normal function should have been denoted by , not

erf.

(The numerical results of the paper do not change, because was actually used.)

3. R. Satz, "The Equation of State of Solid Matter," Reciprocity, Vol. X, No. 2, Spring-

Summer 1980.

4. D. Larson, Nothing But Motion (Portland, Oregon: North Pacific Publishers, 1979), p.

160.

5. Handbook of Chemistry and Physics, 72nd Edition (Cleveland: The Chemical Rubber

Company, 1991-1992), pp. 6-108 to 6-110.

6. American Institute of Physics Handbook (New York: McGraw-Hill Book Company,

1972). values are difficult to find. If you know the volume at temperature i and

temperature f (and the pressure is constant), then from equation 1, .

7. R. Reid, J. Prausnitz, B. Poling, The Properties of Gases and Liquids, 4th Edition

(New York: McGraw-Hill, Inc., 1987).

8. D. Larson, The Liquid State, privately circulated series of papers on the liquid state,

circa. 1960-1964. Note: for this work, I made use of his paper IV. Larson used the

semi-empirical value 415.84 atm for the liquid natural pressure unit. My derivation of

Plnu is unique.

PERMITTIVITY, PERMEABILITY AND THE SPEED

OF LIGHT IN THE RECIPROCAL SYSTEM

Introduction

Physics textbooks do not provide a theoretical derivation of Newton‗s law of gravitation

or Coulomb‘s laws of electrostatics and magnetostatics; they are simply stated as

empirical truths. By contrast, books on the Reciprocal System, such as Ref. [1], [2], [3],

do provide a theoretical derivation of these laws. This paper will take a closer look at the

terms of the electrostatic and magnetostatic equations--the terms of the gravitational

equation have already been discussed in detail, most recently in Ref. [3]. Unlike the force

of gravitation the forces of electrostatics and magnetostatics can be reduced by the

intervening media between the charges. The question to be answered is: what are the

dimensions of all the terms of Coulomb‘s laws?

Permittivity

Coulomb‘s law of electrostatic attraction is expressed as

Q1 Q²

Fe = ke ——— (1)

r²

where Q1 and Q² are the electric charges, r is the distance between them, Fe is the force,

and ke is the proportionality constant. In the Reciprocal System all physical quantities are

expressed in terms of space and time only; there are no separate dimensions for mass or

charge. So, electric charge is not taken as a fundamental entity and given an independent

unit. Larson has deduced that the dimensions of force and charge are

Fe = [t/s²]

Q = [t/s]

In the gravitational expression, the second mass and the distance are considered to be

dimensionless ratios. Such a procedure could be used in analyzing the electrostatic

expression; however I find it more fruitful to treat the second charge and the distance as

having dimensions. In this case Coulomb‗s law expressed in dimensions is

[t/s] [t/s]

[t/s²] = ke ———– (2)

[s²]

For this equation to be dimensionally correct, the dimensions of ke must be

ke = [s²/t] (3)

These are the dimensions of permittivity (in the Reciprocal System). However, the

conventional expression for the coefficient of the law is

ke = 1/(4 ) (4)

where is the permittivity, expressed in farads/meter. Thus the derivation gives a result

that is the inverse of the usual coefficient. But in the Reciprocal System the farad is

reducible to a length. So in the conventional units that are used, permittivity turns out to

be dimensionless whereas physically it is not. It should not then be surprising that the

numerical values of permittivity are inverted. Instead of saying that the permittivity of air

is 1.0006 times that of free space, I would say that it is 1/1.0006 = .9994 times that of free

space. This actually sounds better! (The other part of the coefficient (1/4 ) of the

conventional expression was put in for practical reasons having nothing to do with basic

physics; there is no point to keeping it in the Reciprocal System). Of course, the end

result-the calculated force--must be the same in both systems.

Permeability

Coulomb‗s law of magnetostatics is

M1 M²

Fm = km ——— (5)

r²

where M1 and M² are the magnetic charges, Fm is the force, r the separation distance, and

km the proportionaiity constant. Larson has deduced that the dimensions of magnetic

charge are

M = [t²/s²] (6)

Then, expressed dimensionally, Coulomb‗s law for magnetostatics is

[ t²/s² ] [ t² /s² ]

[t/s²] km ——————— (7)

[s²]

For this equation to be dimensionally correct, the dimensions of km must be

km = [s4/t³] (8)

But in the Reciprocal System magnetic permeability (symbol ) has the dimensions t /s .

Thus km = 1/ . This time the derivation from the Reciprocal System is in exact accord

with the conventional Kennelly system, with magnetic charges or poles expressed in

webers (volt-sec). The only thing awkward here is the name ―permeability‖ . On the basis

of the equation, a better name for this quantity would be ―impermeability‖ . Besides, as

Larson has pointed out ―permeability‖ is the magnetic analog of electric resistance

(t² /s³ * t/s). Perhaps for parallelism with the revised electrostatic expression we should

put the reciprocal of what is now called permeability into the numerator of the law but

call it by the same name. Thus the higher the electric force between charges, the higher

the (reciprocal) ―permittivity‖ ; and the higher the magnetic force between magnetic

charges, the higher the (reciprocal) ―permeability‖ .

Permittivity, Permeability, and the Speed of Light

One way to confirm the identification of the dimensions of permittivity and permeability

is to use them in the same expression. One such expression is Maxwell‗s famous result

from electromagnetic theory:

c = 1/( o o )½ (9)

where c is the speed of light, o is the permittivity of free space, and o is the

permeability of free space. In the dimensional terms of the Reciprocal System, the

equation is

[s/t] = 1/[(s²/t)(t³/s4)]½ (10)

As expected, the dimensions check out fine. These new results should help clarify

electrostatics and magnetostatics for both students and working scientists and engineers.

References

1. Dewey B. Larson, The Structure of the Physical Unieerse (Portland, Oregon: North

Pacific Publishers, 1959). Note: in this, the first presentation of the Reciprocal System,

the permittivity and permeability were treated as dimensionless (p. 82).

2. Dewey B. Larson, Nothing But Motion (Portland, Oregon: North Pacific Publishers,

1979).

3. Dewey B. Larson, Basic Properties of Matter (Salt Lake City, Utah: International

Society of Unified Science2 1988). Note: the dimensions of permittivity are stated as

[s²/t] on p. 172; the dimensions of permeability are stated as [t³ /s4] on p. 222.

APPENDIX: SAMPLE CALCULATIONS

1. Electrostatics

What is the force exerted by a charge of one coulomb on another charge of one coulomb

one km away, in air?

From eq. 3, the value of the permittivity in free space is

snat ²/tnat = (4.558816*10-6

)²/1.520655*10-16

= 136670.11 cm²/sec.

In air, the value is .9994 times this, or 136588.11 cm²/sec. Now a coulomb is defined as

the electrostatic charge which when placed at a distance of 1 meter from an equal charge

of the same sign produces a repulsive force of 8.98755*10-9

N. In space-time terms, this

force is

8.98755*109 N * 10

5 dynes/N * (7.316889*10

-6 sec/cm²) / (3.27223*10² dynes)

= 20096664 sec/cm² .

Then from Coulomb‗s law (for a vacuum) we have

20096664 sec/cm² = 136670.11 cm²/sec * Q² sec²/cm² / 10000 cm².

Solving for Q gives 1212.6213 see/em per coulomb. So for the problem at hand we have

Fe - 136588.11*1212.6213²/100000² = 20.084604 sec/cm².

Converting back to conventional units we have

20.084604 sec/cm² * 3.27223*10² dynes/7.316889*10-6

sec/cm² * 10-5

N/dynes

= 8982.1567 N.

This agrees with the value from experiment.

2. Magnetostatics

What is the force exerted by a magnetic pole with a strenath of one weber against another

magnetic pole of equal strength one km away, in vacuum?

The value of the permeability of free space is

t³/s4 = (1.520655*10

-16)³/(4.558816*10

-6)4 = 8.1411073*10

-27 sec /cm

4 .

Now a weber may be defined as the strength of a magnetic pole which exerts in a vacuum

a force of 63325.74 N upon another magnetic pole of the same strength one meter away.

In space-time terms this force is

63325.74 N * 105 dynes/N 2 (7.316889*10

-6 sec/cm²)/(3.27223*10² dynes) = 141.59989

sec/cm .

From Coulomb‘s magnestatic law we get

141.59989 sec/cm² = (1/8.1411073*10 -27 sec3/cm4) * M²sec4/cm4/10000 cm².

Solving for M gives 1.0736759*10-10

sec²/cm² per weber. So for the problem at hand we

have

Fm = (1/8·1411073*10-27

sec³/cm4) * (1.0736759*10-10 sec /cm ) / (100000 cm) =

1.41599890 sec/cm.

Converting back to conventional units we have

1.4159989*10-4 sec/cm² * 3.27223*10² dynes/(7.316889*10-6

sec/cm ) * 10-5 N dyne =

0633257 N.

This agrees with the value from experiment.

SUMMARY:

1. permittivity of free space = 136670.11 cm²/sec

2. permeabilit of free s ace = 8.1411073 10 sec /cm

3. one coulomb = 1212.6213 sec/cm

4. one weber = 1.0736759-10 10 sec²/cm²

THE UNIT OF MAGNETIC CHARGE

In terms of the egs system, the unit of electron charge (and quantity) is calculated by

multiplying the Faraday constant by the mass equivalent of unit atomic weight:

2.89366x1014 esu/g-equiv * 1.65979x10-24 g = 4.80287x10-10 esu

(ref. [1]). Of course, 4.80287x10-10 esu is equal to 1.602062x10-19 coulombs.

In space-time terms, the dimensions of electron charge (or electric charge in general) are

t/s. The magnetic charge is a two-dimensional form of the electric charge; its space-time

dimensions are t2 /s2 . The numeric value of the magnetic charge must therefore be the

value of the electric charge divided by the the natural value of s/t, or the speed of light. In

the egs system, this results in

4.80287x10-10 esu / 2.99793x1010 cm/sec = 1.602062x10-20 ―esu‖

(ref. [2]). The ―esu‖ here are the magnetic units of the electrostatic system. According to

ref. [3], 1 ―esu‖ of magnetic flux (equivalent to charge in the Reciprocal System) equals

299.8 webers. Thus one unit of magnetic charge equals 4.802982x10-18 webers.

Each atom has two rotating systems; if one system acquires a magnetic charge, the other

system must also acquire a charge if there is to be stability and permanence. Henee each

atom has two poles, or centers of magnetie effect--it is dipolar, not monopolar.

Consider a simple ―bar magnet‖ of four iron atoms. The poles would be arranged in this

manner: N-S - N-S - N-S - N-S. Only the end atoms are not neutralized; therefore, in

general only the surface atoms of a bar magnet contribute to its effective magnetic charge

(one unit of charge per surface atom). This means that it would take 2.0820399x1017

magnetically charged surface atoms to generate one weber of magnetic flux. Since iron

has a mean atomic weight of 55.847, the total mass of these surface atoms would come to

1.9301763x10-5 grams.

References:

1. Dewey B. Larson, Basic Properties of Matter (Salt Lake City, Utah: International

Society of Unified Science, 1988), p. 110.

2. Dewey B. Larson, The Structure of the Physical Universe (Portland, Oregon:

North Pacific Publishers, 1959), p. 211 (except that the numerical value has been

updated in ref.1).

3. Robert Resnick and David Halliday, Physics (New York: John Wiley & Sons,

Ine., 1966), p. 33, Appendix G, of the supplement.

PHOTOIONIZATION AND PHOTOMAGNETIZATION

Introduction: the Reciprocal System vs. Present Theory

Consider a group of atoms in an electric field and bombarded with ultra-violet photons or

a group of atoms in a magnetic field and bombarded with radio photons. What happens?

Two theories exist that can give an answer: Quantum Mechanics and the Reciprocal

system. Both are quantized, but the first is a matter-structure theory, whereas the second

is a motion-process theory. Quantum Mechanics considers atoms to consist of various

subatoms which have intrinsic charge, magnetic moment, and angular momentum; the

atom‘s charge, moment, and momentum are derived from that of its subatoms. The

Reciprocal system views atoms as composed of two photons, each having rotational

motion in three dimensions; the atom has no intrinsic electric charge or magnetic

moment—electric and magnetic effects result from additions of rotational vibratory

motions to the base rotational motions.

Quantum Mechanics‘ explanation of electric ionization is that previously 0existing

charged particles, the protons and electrons, are separated; the Reciprocal System‘s

explanation is that the positive and negative charges 0are created in the process, and thus

have no prior existence. Quantum Mechanics‘ explanation of the magnetic resonance

experiments is that the experimenters have found intrinsic magnetic moments of nuclei;

the Reciprocal System‘s explanation is that the experimenters have induced temporary

magnetic charges in their material. Quantitative details of both theories will now be

examined. (Full comprehension of this paper requires previous reading of two of D. B.

Larson‗s books, Refs. 1 and 2, and one of my papers, Ref. 3.)

I. Photoionization

A. Subatoms

1. Present theory.

According to present thought, the electric charge is unanalyzable and undefined, except

operationally. Either a particle has or does not have an intrinsic electric charge—there is

no possibility of ionizing an uncharged subatom.

Present photoelectric theory states that, upon absorption of a sufficiently energetic

photon, a pre-existing charged electron is ejected from its atom to move in an external

circuit. [4,5] The energy necessary to tear 0the electron loose is called the work function

of the material. No commonly accepted equation for the work function, based on

Quantum Mechanics, exists.

2. Reciprocal System

For details of subatom and atom motions, see Refs. 1, 2, and 3. In the Reciprocal System,

electric charge is not an intrinsic feature of a subatom; rather, charges may be created or

destroyed, not necessarily in pairs, and 0thus charge conservation in a process does not

always hold true. However, total motion displacement is conserved in each process.

As with all other phenomena in the Reciprocal system, electric charge is a motion, in this

case a simple harmonic rotational vibration, as shown in Figure I.

FIGURE I: ELECTRIC CHARGE

An equation for this motion will now be derived. Let q be the rotation angle in radians,

be its frequency in Hz, and t be the time in seconds. From the figure, the amplitude of the

motion is radians and the angular distance traveled each cycle is 4 radians. Hence the

equation is

q = cos(4 t) (1)

As shown in a previous paper of mine [3], a negative electric charge has the frequency

-elec. = R/2 (2)

where R is the Rydberg frequency (3.288 * 1015 Hz).

Electrons exist within matter, but not as intrinsic features of atoms. 0Also, these electrons

are ordinarily uncharged. To travel outside of matter 0the electrons must become charged

or ionized. The energy for the charge and 0the kinetic energy of the charged electron

come from absorption of a photon, 0thus producing the photoelectric effect. A rigorous

equation for this 0effect, slightly modified from Ref. 6, can now be given. Let

h = Planck‗s constant phot.

phot= photon frequency

v = electron velocity (outside of matter) 0

m = electron mass

eV = electric potential surrounding the matter

Wo. = work function of the matter

Uk. = energy of the electron before the process begins

Ul = energy lost by charged electron in moving to surface

The equation for the electron‗s kinetic energy outside of matter is then

1/2 mv² = h phot. + eV - Wo. + Uk - Ue (3)

According to the Reciprocal System the work function is the energy necessary to charge a

uncharged electron.Since the rotational vibration is scalar, like the linear vibration,

Planck‘s law holds for electric charges 0as well as for photons:

EI,e. = h -elec. = h * R/2 = 2.17 eV (4)

Note: any observed values of work function less than 2.17 eV emply previous electron

energy, Uk). The value of EI,e given in (4) is modified by the environment of the electron,

i.e., by the atom in which it currently exists. The electron‗s charge may be in the same

dimension as one of the atom‗s magnetic rotations or in the same dimension as the

electric rotation. In the Reciprocal system, charge is energy, t/s, the inverse of velocity.

The atom‗s magnetic rotation velocity is vmag, its electric rotation velocity is v; the

inverse of these in natural units is c/vmag and c/velec , where c is the speed of light. If the

atom has only one electric time displacement unit (velec= 1/2c), the ionization energy of

the electron is not increased, hence a 1 must be subtracted from c/velec. Finally, the atomic

motions take place in the time region, whereas we want the energy as measured in the

time-space region—so the square root of the inverse velocity expressions 0must be taken.

Thus

wo. = 2.17 * [c/vmag.]½ eV. -and/or (5a)

wo. = 2.17 * [c/velec. -1]½ eV

These equations, theoretically derived, are nearly identical to the ―empirical‖ equation

given by Larson in Ref 1 (p. 118), eq. (142)). The set of 0values of Wo. is

Wo={2.1 7, 3.07, 3.76, 4.34, 4.82} <5b>

Table I compares the theoretical results with those observed.

TABLE I

WORK FUNCTION AND IONIZATION ENERGY

Wo EI

Element c/vmag c/velec calc. obs calc. obs.

Li

2 2.17 2.28 4.34 5.39

Be 3

3.76 3.92 7.52 9.32

B 3

3.76 4.4 8.68 8.296

C

5 4.34 4.341 4.34 11.264

Na

2 2.17 2.25 7.52 5.138

Mg 3

3.76 3.78 7.52 7.644

Al

4 3.76 3.43 8.68 5.984

Si

5 4.34 4.2 4.34 8.149

K

2 2.17. 2.12 6.14 4.39

Ca

3 3.07 3.20 7.52 6.111

Ti 3

3.76 3.95 7.52 6.83

V 3

3.76 3.95 8.68 6.83

Cr 4

4.34 4.37 7.52 6.764

Mn 3

3.76 3.76 7.52 7.432

Fe 3

3.76 3.91 7.52 7.90

Co 3

3.76 3.9 7.52 7.86

Ni 3

3.76 3.67 7.52 7.633

Cu 3

3.76 3.85 7.52 7.724

Zn 3

3.76 3.89 7.52 9.391

Ga

4 3.76 3.80 7.52 6.00

Ge

5 4.34 4.29 8.68 7.88

As

6 4.82 5.11 9.64 9.81

Se 4

4.34 4.42 8.68 9.75

Rb

2 2.17 2.16 4.34 4.176

Sr

3 3.07 2.74 6.14 5.692

Zr 3

3.76 3.73 7.52 6.835

Nb 3

3.76 3.96 7.52 6.88

Mo 3

3.76 4.08 7.52 7.131

Ru 4

4.34 4.52 8.68 7.36

Rh 4

4.34 4.57 8.68 7.46

Pd 4

4.34 4.49 8.68 8.33

Ag 4

4.34 4.33 8.68 7.574

Cd 3

3.76 3.73 7.52 8.991

Sn 3

3.76 3.62 7.52 7.332

Sb 4

4.34 4.14 8.68 8.64

Te 4

4.34 4.70 8.68 9.09

Cs

2 2.17 1.96 4.34 3.893

Ba*

2 2.17 2.11 4.34 5.810

La* 2

3.07 3.3 6.14 5.61

Ce* 2

3.07 2.84 6.14 6.91

Pr* 2

3.07 2.7 6.14 5.76

Nd* 2

3.07 3.3 6.14 6.31

Sm* 2

3.07 3.2 6.14 5.6

Hf* 2

3.07 3.53 6.14 5.5

Ta 3

3.76 3.96 7.52 7.7

W 4

4.34 4.35 8.68 7.98

Re 4

4.34 5.0 8.68 7.87

Os 4

4.34 4.55 8.68 8.7

Ir 4

4.34 4.5 8.68 9.2

Pt 4

4.34 4.09 8.68 8.96

Au 4

4.34 4.46 8.68 9.223

Hg 4

4.34 4.5 8.68 10.434

Tl 3

3.76 3.84 7.52 6.106

Pb 3

3.76 3.94 7.52 7.415

Bi 4

4.34 4.31 8.68 7.287

The agreement is excellent (the correlation coefficient rcorr = .964).

As discussed by Ref. 7, the number of electrons emitted per incident pho-ton depends on

both the nature of the emitter and the frequency of the incident radiation.At frequencies

lower than that at which maximum yield is obtained, reflectivity of incident photons is so

high that only a few photons take part in the emission process.At frequencies higher than

that at which maximum yield is obtained, the photons penetrate to such a depth that the

electrons, newly charged at that depth, lose too much energy in coming to the surface.

Of the remaining subatoms only the positron, proton and H can take an electric charge.As

shown in Ref 3, a positive electric charge has the frequency

+elec. = 2R/ (6)

*Asterisk denotes entry of rotation into second space-time unit in the 4a and 4b groups;

also where value 3 appears in magnetic rotation, this is the inverse of actual rotation.

Values of c/vmag'

c/velec and Wo obs. are taken from Table.

The ionization energy of a free positron or proton (not including force-coupling effects)

is then

EI,p. = h +elec. = h * 2R/ = 8.68 eV (7)

Now consider the ionization of the intermediate particle, H. Here everything is unity: H

has one natural unit of primary mass and one unit of electric space displacement; one

positive charge is created and one negative charge. So the required photon energy must

also be unity by the Principle *Asterisk denotes entry of rotation into second space-time

unit in the 4A and 4B groups; also where value 3 appears in magnetic rotation, this is the

inverse of actual rotation. Values of c/vmag, c/velec., and Wo obs. are taken from Table CIX

of Ref. 1. of Equivalence (see Ref. 1, p. 21):

EI,H1 = h * R = 13.595 eV (8)

as observed.

B. Atoms

1. Present Theory

In present theory, ionization is thought to be the ejection of an elec-tron from its orbital,

leaving a net positive charge on the atom. No generally accepted equation has been

developed to calculate the energy of ionization from Quantum Mechanics.One might

expect the ionization energy to be practically the same as the work function— but they

are not.Indeed, there is no evidence that matter becomes ionized in the photoelectric

effect.

Gas and liquid ionization is currently thought to be the break-up of previously existing

oppositely-charged units.

2. Reciprocal System

In the photoelectric effect, only charged electrons are created, not charged atoms.But

generally in atomic ionization both positive and negative electric charges are created

(they do not exist previously).Where negative ions can form, as with the electronegative

elements, they do so. But usually negatively-charged electrons serve as the second

component of the force couple creating the charges. Thus in most cases a positively-

charged ion and a negatively-charged electron are the results of ionization. So, rather than

the ionization energy being the binding energy of an elec-tron to a nucleus, it is the

energy required to create two charges, one 1 -positive and one negative.

Equation (5) gave the energy necessary to create a negative charge on an electron. From

mechanical considerations it is obvious that the energy necessary to create a positive-

negative charge pair is twice that needed to create the negative charge on the

electron.Hence for the first ionization level the energy is

EI,atom = 4.34 * [c/vmag.]½ ev (9a)

EI,atom = 4.34 * [c/velec.-1]½ ev

The set of values of EI,atom is

EI,atom = {4.34, 6.14, 7.52, 8.68, 9.64} (9b)

Table I compares calculated values with observed values of ionization [8]. 0Agreement is

very good. (Avg. Wo calc. =

3.706; avg. Wo. obs.=3.779. Avg. EI. calc. =7.412; avg. EI.

obs=7.366. Avg. EI. calc./avg. Wo. calc.=2.000; avg. EI.

obs./avg. Wo. obs.=1.95. For EI. obs. and EI.calc. rcorr.=.803.

Individual discrepancies are most likely due to experimental error.)

Ordinarily, solids are not ionized; the resulting forces would overcome the cohesive

energy and break apart the solid.Liquids and gases are more readily ionized, with the

energy often being supplied by photons.An electric field is of course necessary to prevent

the charges from recombining. For a good discussion of liquid ionization, see Ref. 1.

Gaseous ionization depends on both electric field strength and gas pressure. An equation

for the saturation electric field will now be derived. Let

I = primary current (ions created by photons/time)

Ephot./t = photon energy absorbed per unit time

EI. = ionization energy

€f. = electric field strength

€fnat.= natural unit of electric field strength

P = gas pressure Pnat. = natural unit of pressure

The equation for the primary ion current is then

I =(Ephot./t) * (1/EI.) * ( €f./€ fnat.) * (Pnat./P) (10)

Clearly if

Ephot. = EI.,€ f =€ f nat, and P = Pnat., one ion

per time t will be created. For a fixed energy input, I can be increased either by

0increasing €f. or decreasing P until

( €f. / €fnat.) * (Pnat./P) = 1 (11a)

Solving for €f we have

f. = €fnat. * (P/Pnat.) (11b)

With €fnat. = 2.04133 * 1016 v/m and Pnat. = 1.5539 * 107 atm.

and letting P = 1 * 10-5 atm., then

€fsaturated = 13137 v/m

The newly created ions and charged electrons can themselves cause further ionization,

which is called secondary ionization.

The reverse of ionization is the addition of a negatively charged electron to a singly

charged positive ion, which results in a neutral atom. Thus the ―electron affinity‖ of a

singly charged positive ion is just the negative of the ionization energy of the

corresponding neutral atom.[9] Where current theory gets into difficulty is in

understanding the ―electron affinity‖ of neutral atoms. According to the Reciprocal

System, the electron loses its charge upon absorption in matter, resulting in a reverse

photoelectric effect.

II. Photomagnetization

A. Subatoms

1. Present Theory

According to present thought, the magnetic resonance experiments detect the magnetic

moments intrinsic to subatoms and atoms (nuclei).The magnetic moment is considered to

result from the angular spin of the electric charge of a particle.It is given in units of the

Bohr magneton or the nuclear magneton, as derived from Dirac‗s theory.

A number of problems exist with this theory.The neutron has a magnetic moment, but no

electric charge. Pions and alpha particles do have electric charges but no magnetic

moment; present theory claims that these particles have zero spin— but that seems

equally strange.The magnetic moment of the proton has been obtained from experiments

with ice and water; thus the magnetic moment could actually be that of H itself,

regardless of what theory says.The magnetic moment of the anti-proton has been found

not to equal the magnetic moment of the proton (or rather, H1 ).

More fundamentally, the value of the magnetic moment is not measured + directly; it is

inferred from the data. To see this, let

o= resonant photon frequency

h o. = absorbed photon energy

B = magnetic field strength

µ = magnetic moment in field direction

L = angular spin no.

The equation (from Ref. 10) is

h o = [µ /L] * B (12)

The energy h o and field B are measured, L is inferred from other data (usually

spectroscopic), and then is calculated. If L is bogus, then µ is bogus. All that the

experiments tell us is that for a given B there exists a certain photon frequency at which

great amounts of energy are absorbed by he sub-atoms and atoms. The conclusion that

the relation between h o and B is µ /L is purely hypothetical.

2. Reciprocal System

My interpretation of the magnetic resonance experiments, on the basis of the Reciprocal

System, is radically different.

Here the sub-atoms and atoms have no intrinsic magnetic moments or magnetic charges.

(Note: the isotopic charges in atoms cancel the magnetic effect of the magnetic charges of

the neutrinos contained within). But under certain circumstances, such as in the magnetic

resonance experiments, temporary magnetic charges can be induced. A magnetic charge

is a rotational vibration of one of the inner magnetic rotations of the sub-atom or atom.

As given in Ref. 3, the vibrational frequency of a unit magnetic charge is

vmag. = 2R/ (13)

The required energy to produce this vibration depends on the environment: the magnetic

field and the velocity of the principal or subordinate magnetic rotation which is modified

by the charge. The resulting equation is related to, but different from, the equation for

energy of electric ionization, eq. (5). In the Reciprocal system, magnetic effects are the

square of electric effects—so the square root of eq. (5) is eliminated. Also, as in the

equations for force and distance in the Reciprocal System, the effect of the magnetic

velocity is inverse to that of the electric velocity. Thus, instead of the energy being

proportional to c/vmag., the energy is proportional to vmag./c. The complete

expression is

h o = [h * (2R/ * (vmag./c * 1/Bnat.] * B (14a)

Larson [1] has previously identified magnetic susceptibility as proportional to vmag./c.

This provides additional support for that term in the above equation.

FIGURE II: MAGNETIC CHARGE

As shown in Figure II, the sole difference between a magnetic charge and an electric

charge is that the magnetic charge is an electric charge that is given an extra angular spin

by either the subordinate magnetic rotation or the electric rotation (depending on whether

the charge is placed on the principal or subordinate magnetic rotation).

If no magnetic field is present, and the photons are not reflected, the photon energy is

simply transformed to thermal energy of the particle. For a given magnetic field B, o can

be calculated for each kind of sub-atom and atom.As discussed in Ref. 3, vmag./c can take

on the following values:

vmag./c = {.20, .22, .25, .29, .33, .40, .50} (14b)

Setting B equal to 1 Tesla and knowing that Bnat = 6.813*107 Tesla, the following

resonance frequencies are obtained:

o. in MHz = {6.14, 6.83, 7.68, 8.78, 10.24, 12.29, 15.36} (14c)

The calculation assumes that no other energy is present that can be utilized in the creation

of the magnetic charge.Unfortunately, magnetic resonance experiments have been done at

room temperature rather than at temperatures close to degrees K, so the absorption of

thermal energy is a definite possibility.

Nearly all atoms have magnetic resonance frequencies at or below 15.6 MHz (with B = 1

Tesla), in accord with eq.(14). But the intermediate parti- cles, the neutron and H ,have

higher observed frequencies,29.16 MHz and 42,57 MHz, respectively. One theoretical

explanation is that these particles require multiple magnetic charges if they are to have

any at all.The neutron is comprised of two rotational systems:a proton rotational system

and a cosmic neutrino rotational system. In terms of total rotational speed (in natural

units) the notation is

{

1/3 - 1/2

neutron

1

2 - 2

(See refs. 1 and 3 for details). Suppose each rotational system takes a magnetic charge on

its subordinate magnetic rotation. For the proton rotational system, the energy required is

h * (2R/ ) * (1/2) * (B/Bnat.)

For the cosmic neutrino rotation (which takes an inverse charge) the energy required is

h * (2R/ ) * 2 * (B/Bnat.)

The combined energy is

h * (2R/ ) * (B/Bnat.)

and the resonance frequency is 30.70 MHz (for B and Bnat.) as before. The small

discrepancy between the observed and cal- culated values may be due to the absorption of

thermal energy.

The notation for H1 is

{

1/3 - 1/2

H1

2

1/2 - 1/2

In this case, though, each rotational system apparently takes two charges, so that the

frequency is 3R/ rather than 2R/ . The energy required is then

[h * (3R/ ) * (1/2) + h * (3R/ ) * (1/2)] * (B/Bnat.)

giving a resonance frequency of 46.05 MHz. Again the discrepancy between that

observed and that calculated may be due to the utilization of thermal energy.

Both the real proton and anti-proton (inverse proton) and the material and cosmic

neutrinos and the material and cosmic massless neutrons should have resonance

frequencies of 15.36 MHz, unless they take multiple charges.

The electron and positron have no subordinate magnetic displacements at all and thus

cannot take magnetic charges. All magnetic effects of these particles (and also the muon),

whether uncharged or charged, result from their being in translational motion. To quote

Larson [2]:

As we have seen, the electric charge is a one-dimensional modification of the rotational

motion of an atom or sub-atomic particle and the magnetic charge is a similar two-

dimensional modification. The characteristic effects of the magnetic charge originate

because the one-dimensional forces are distributed over two dimensions by the second

rotation. But for this purpose it is not necessary that the motion in the second dimension

be rotational. We can see why this is true if we examined the behavior of the axes of

rotation. The axis of the electric rotation of an atom is a line: a one-dimensional figure. A

stationary electric charge thus has no two-dimensional rotational effects. For a magnetic

charge the locus of all positions of either axis is a disk: a two-dimensional figure and the

magnetic charge has two-dimensional properties. But if we move the electric charge

translationally, the locus of all positions of the axis is again a two-dimensional figure, and

hence a moving electric charge has a two-dimensional distribution of forces comparable

to that of a magnetic chargeÉ If an uncharged electron or positron is given a translational

motion, this again is motion in two dimensions and it produces electromagnetism, a

magnetic effect.

Particles heavier than the electron and positron would show a similar magnetic effect if

they could be accelerated to the same high velocities.

B. Atoms

1. Present Theory

Present theory regards the magnetic moment of nuclei to result from a combination of the

moments of its constituent sub-atoms. Even-even nuclei are regarded as having zero net

spin and thus zero moment. According to Segre‗s account of current theory [11], adding a

neutron to an even-even nucleus is supposed to yield (* + 1/2)[ -3.826/(2* + 1)] nuclear

magnetons for the magnetic moment, where* is the spin angular momentum of the added

neutron; subtracting a neutron is supposed to yield (* - 1/2)[3.826/(2* +1)] nuclear

magnetons. Adding a proton is supposed to yield (* + 1/2)[1+ 4.586/(2* +1)] nuclear

magnetons, whereas subtracting a proton is supposed to yield (* - 1/2)[1-4.586/2 *+ 1)]

nuclear magnetons. These expressions do bracket the data, as Segre points out, but that is

all: they do not work in detailed application.

The sign given for the moment is an inference, not a result from experiment (which

measures only photon energy and magnetic field strength at resonance). In present theory,

both the magnetic moment and angular spin are vectors.If they are aligned the magnetic

moment is said to be positive; if anti-aligned, negative.

2. Reciprocal System

In the Reciprocal System all basic motions — including the electric and magnetic

charges— are scalar. In addition, the magnetic charge is intrinsically dipolar: the

magnetic rotation of the one-dimensional rotational vibration can be viewed both

clockwise and counter-clockwise (see Figure 2). Generally in the magnetic resonance

experiments, the atoms are induced to take only a single magnetic charge and thus have

resonance frequencies of 15.36 MHz or less. The exceptions, such as F and T1, are

apparently induced to take multiple charges

Table II compares the observed resonance frequencies [12] (of stable isotopes) with the

theoretical results from eq. (14). (Note: Thermal energy and cohesive energy are not

taken into account; numerous isotopes (usually unstable) given in Ref. 12 have resonance

frequencies less than 6.14 MHz—these are not given in the table below. Further

theoretical work is necessary to include thermal energy, cohesive energy, and instability

effects in magnetic resonance).

TABLE II

MAGNETIC RESONANCE FREQUENCIES

vmag./c [B=1 Tesla]

Isotope Displ. Mag. Speed calc.(MHz) obs.(MHz)

5 B11 2-1-3 1/2 15.36 13.66

21.Sc45 3-2-3 1/3 10.24 10.36

25.Mn55 3-2-7 1/3 10.24 10.57

27.Co59 3-2-9 1/3 10.24 10.12

29.Cu61 3-3-(7) 1/3 10.24 10.12

31.Ga69 3-3-(5) 1/3 10.24 10.24

33.As75 3-3-(3) 1/4 7.68 7.31

34.Se77 3-3-(2) 2/7 8.78 8.14

35.Br79 3-3-(1) 1/3 10.24 10.70

41.Nb93 3-3-5 1/3 10.24 10.45

47.Ag104 4-3-(7) 1/5 6.14 6.10

48.Cd111 4-3-(6) 2/7 8.78 9.07

49.In113 4-3-(5) 2/7 8.78 9.35

51.Sb121 4-3-(3) 1/3 10.24 10.24

63.Eu151 4-3-9 1/3 10.24 10.56

65.Tb159 4-3-11 2/7 8.78 8.64

67.Ho165 4-3-13 2/7 8.78 8.93

70.Yb171 4-3-16 1/4 7.68 7.52

73.Ta181 4-4-(13) 1/5 6.14 5.14

78.Pt195 4-4-(8) 2/7 8.78 9.24

82.Pb207 4-4-(4) 2/7 8.78 8.99

The equation appears to work well for the majority of atoms studied (rcorr = .956 for the

table). In those cases where the equation does poorly, the thermal effects may be the

culprit. Ideally, the experiments should be repeated with the atoms widely separated and

at close to 0 degrees K in temperature — then the effects of thermal energy and cohesive

energy would be eliminated. Under such conditions all frequencies for the induction of a

single unit of magnetic charge in a stable isotope should be between 6.14 MHz and 15.36

MHz when the field B is 1 Tesla.

Atoms with an even number of neutrino-induced isotopic charges (half on each rotational

system) have no need to acquire another charge for balance. Hence in the current jargon

these atoms do not have a ‗magnetic moment.‘

In the series of isotopes of an element, the placement of the magnetic charge appears to

alternate with the placement of the isotopic charge. A quantitative check on this is

difficult to do because in most such series there are unstable isotopes —this instability

adds another variable to the problem. Apparently if the number of isotopic charges is

greater than that allowed by the magnetic ionizat ion level, the energy required to induce

a magnetic charge is decreased. Once the magnetic field is turned off and the photon

bombardment ceases, the magnetic charges are transformed to radio photons and lost.

Only a few elements, such as Co, Ni, and Fe, are able to retain a magnetic charge.

Summary. Electric and magnetic charges are not unanalyzable; they are one and two-

dimensional rotational vibrations of a sub-atom or atom. They are also not permanent and

inviolate; they may be created or destroyed, so long as overall motion displacement is

conserved.

1a. The work function of a material is not the energy required to remove the least tightly

bound electron; it is the energy required to charge an uncharged electron in that material.

1b. The ionization energy of an atom is not the energy required to separate pre-existing

charged protons and electrons; it is the energy required to create a negative charge (on an

electron) and a positive charge (on an atom) and is equal to twice the work function.

2. The magnetic resonance energy of a sub-atom or atom does not indicate a previously

existing intrinsic magnetic moment; it is the energy required to induce one or more

dipolar magnetic charges in sub-atoms or atoms.

References

1. D. B. Larson, The Structure of the Physical Universe (Portland, Oregon: North

Pacific Publishers, 1959) pp. 21, 61-89, 116-131. Larson‗s Nothing But Motion

(Portland,Oregon: North Pacific Publishers, 1979) is the first of a series of

volumes of the second edition of The Structure of the Physical Universe.

2. D. B. Larson, New Light on Space and Time (Portland, Oregon: North Pacific

Publishers, 1965) pp. 165-196.

3. R. W. Satz, ―Further Mathematics of the Reciprocal System,‖ Reciprocity, Vol. X,

No. 3, Autumn 1980.

4. F. Bueche, Introduction to Physics for Scientists and Engineers (New York:

McGraw-Hill, 1969) pp. 541-559, 741-745, 813-815.

5. R. Cautreau and W. Savin, Modern Physics (New York: McGraw-Hill, 1978) pp.

56-61.

6. E. L. Chaffee, ―Electronics‖ in Fundamental Formulas of Physics, ed. D. H.

Menzel (New York: Dover Publ., 1960) p. 353.

7. R. Rose, L. Shepard, J. Wulff, The Structure and Properties of Materials:

Electronic Properties (New York: John Wiley & Sons, 1966) pp. 26-31.

8. V. W. Finkelnburg and W. Humbach, ―Ionisierunggenergien von Atomen und

Atomionen,‖ Die Naturwissenschaften, Heft 2, Jg. 42, 1955, pp. 35-37.

9. T. L. Brown and H. E. Lemay, Jr., Chemistry: The Central Science (Englewood

Cliffs, New Jersey: Prentice-Hall, 1977) pp. 196-198.

10. E. R. Andrew, ―Nuclear Magnetic Resonance,‖ Encyclopaedic Dictionary of

Physics, Vol. 5, ed. J. Thewlis (New York: Macmillan, 1962) pp. 70-73.

11. E. Segre, Nuclei and Particles, second ed. (Reading Massachusetts: W. A.

Benjamin, 1977) pp. 274-279.

12. 12. E. U. Condon and H. Odishaw, eds., Handbook of Physics (New York:

McGraw-Hill, 1967) pp. 9-93 to 9-101. (The table presented gives and L, from

which the resonance frequency can be retrocalculated

THEORY OF ELECTRONS AND CURRENTS

This paper will present the Reciprocal System theory of electrons and currents and

compare it with the conventional theory

1. The Electron

a. conventional theory

According to present theory1 electrons are classified (along with muons and neutrinos) as

leptons, meaning that they are not affected by the strong interaction of nuclear forces but

suffer the weak interaction that causes beta decay. These subatoms are all considered to

be fermions: they obey Fermi-Dirac statistics, have spin s =½, and have spinor-wave

functions that satisfy the Dirac equation. The present theory does not yield equations

enabling the calculation of electron mass, charge, and magnetic moment. The empirical

values are:

mass: m = 9.109*10-31 kg (1)

charge: e = -1.601*10-19 coulombs (2)

magnetic moment: ue = 9.28*10-24 joule/tesla (3)

Also no size or shape is definitely specified. The closest we have is the following:

It is obviously tempting to picture an electon as a spinning sphere of electric charge

whose radius is determined by the dimensional relation e2/a = mc2 at which the

electrostatic self-energy of the charge distribution is comparable with the relativistic

energy of the rest mass. This classical electron radius, a = 2.81785*10-15 m, is an

important scale parameter in physics; but the uniqueness of e, the arbitrariness of the

quantization rules, and the difficulty of making it properly relativistic, forbid such a

purely classical model.

Note that for this radius, and for a spin angular momentum of ½ Ã3h, the angular

velocity of the electron must be 2*1025 rad/sec — giving an equatorial speed of about

200c!

b. Reciprocal System

The Reciprocal System is much more specific on the details of electron attributes than

conventional theory. My previous papers3 4have described the shape, size, and all

motions constituting the electron.

The electron is a spherical particle resulting from the rotation of a single photon. The

frequency of the photon is

phot = 2R = 6.576115*1015 cycles/sec (4)

(Here R is the Rydberg frequency). The rotational speeds in revolutions per second

around the three axes are r/ - 2R/ - 4R/ or in terms of rev/sec

elec= 1.0466212*1015rev/sec. - 2.0932424*1015 rev./sec -4.1864848*1015 (5)

The electron may be charged or uncharged. If charged, the electron has an added

rotational vibratory motion of

-elec = R/2 = 5.233106*1014 cycles/sec (6)

The diameter d of the electron is one natural space unit, reduced by the appropriate inter-

regional ratio (142.22 here). Thus,

d = 4.55884*10-8/142.22 = 3.2054 Å (7)

2. Electron Flow

a. conventional theory

According to present theory, conduction in metals takes place by movement of the

electrons in the outermost shells of the atoms making up the crystalline structure of the

solid. These electrons reach an average drift velocity which is directly proportional to the

electric field intensity

vd = E (8)

where µ, the mobility, has the units m2/V*s. For a conductor of length l, conductivity

ó(siemans per meter), and cross-sectional area A, eq. (8) may be rewritten as

vd = ( *1/( *A))*I m/s (9)

EXAMPLE: For a copper conductor 100 mm long and 3 mm in diameter, what is the

average drift velocity of the electrons if the current is 10 amps?

For copper,

= 5.8*107 S/M -

= 0.0032 m2/V*s

Here

A = ¼ (3*10-3)2 = 7.0686*10-6 m²

Thus,

vd = (.0032*.1/(5.8*107*7.0686*10-6))*10

= 7.805*10-6 m/s

b. Reciprocal System

In the Reciprocal System, the natural unit of velocity is 2.99793*108 m/s (the speed of

light) and the natural unit of current, which is also a velocity, is 1.0535*10-3 amperes.

The conversion is thus

2.99793*108 m/s/1.05353*10-3 amps = 2.8456048*1011 m/s/amps.

Hence the ―drift‖ velocity of electrons (here uncharged and massless) in the Reciprocal

System is

vd = 2.846*1011*I m/s (10)

EXAMPLE: For the case of the previous example,

vd = 2.846*1011*10 = 2.846*1012 m/s (11)

The answer of the Reciprocal System is 3.646*1017 times the answer of conventional

theory!

Of course, the number of electrons passing a given point per second must be the same in

both theories.

In the conventional theory,

N = (10 C/s)(1 electron/1.6*10-19C) = 6.25*1019 elec/s

In the Reciprocal System,

N = 3.15842*106 esu/s*1 electron/4.80287*10-10esu *10 amps/1.05353*10-3 amps

= 6.24*1019 elec/s

The difference in ―drift‖ velocities must therefore be due to vastly different numbers of

electrons in the matter of the two theories. More about this in another paper.

References

1. Encyclopedia Britannica, Vol. 6, pp. 665-672.

2. Ibid., p. 667.

3. R. Satz, ―Further Mathematics of the Reciprocal System,‖ Reciprocity, Vol. X,

No. 3.

4. R. Satz, ―Photoionization and Photomagnetization in the Reciprocal

System,‖Reciprocity, Vol. XII, No. 1.

HUBBLE‘S LAW AND THE RECIPROCAL SYSTEM

The conceptual basis for Hubble‘s Law in the Reciprocal System has been discussed by

Mr. Larson in a number of his works. This paper will present some additional

mathematical details.

Hubble‘s Law is commonly written as

v = Hr (1)

where v is the velocity of a distant galaxy, in km/sec, r is the distance to the galaxy, in

Mpc, and H is Hubble‘s constant, in km sec-1 Mpc-1 . In differential form, the equation is

dr/dt = Hr (2)

However, as shown in Larson‘s The Structure of the Physical Universe,¹ the recession

starts at the gravitational limit of our galaxy, denoted by r . Thus the correct expression is

dr/dt = H (r - ro) (3)

Clearly the velocity is zero when r = ro.

The equation is a first order linear differential equation² and can be easily solved for r.

The result if

r = ro + (ri - ro)eHt

(4)

where ri is the initial position of the external galaxy.

Of great interest is the determination of Hubble‘s constant from first principles.

According to the Structure, the ratio of effective to total gravitational units is 1/156.4444.

At the distance a galaxy recedes at the speed of light, the effective gravitational force

drops betow the value of unity and vanishes. Thus

1/156.4444 · M

G/r1² =1 (5)

(equation 159 of Structure). Solving for r1, the limiting distance, yields

r1 = MG½ /12.51 (6)

(equation 160 of Structure).

Putting this vatue of r1 in Hubble‘s Law, one can solve for the constant:

H = c/r1-ro c/r1 (7)

where c is the velocity of light.

The value of the constant thus depends on the value of the mass of the Galaxy, MG.

According to reference three, this is

MG = 2.587 1041 kg

(8)

With this value the constant is

h = 114.522 · km/ sec Mpc (9)

However, according to Sandage the value of H is

H = 75.0 · km/sec Mpc (10)

This implies that the actual value of the mass of the Galaxy is

MG = 6.032 * 1041 (11)

or 2.33 times that estimated.

It seems to me that Hubble‘s constant has been more accurately determined than the mass

of the Galaxy. Thus the analysis leads to the conclusion that the mass of our Galaxy is

greater than supposed—probably because of a white dwarf galactic core that still remains

difficult to observe.

References

1. Dewey B. Larson The Structure of the Physical Universe, First Edition (Portland,

Oregon: North Pacific Publishers, 1959.

2. Richard Bronson , Modern Introductor Differential Equations (New York:

McGraw-Hill Book Company, 1973.

3. Martin Harwit, Astrophysical Concepts (New York: John Wiley & Sons, 1973).

GLOBULAR CLUSTER MECHANICS IN THE

RECIPROCAL SYSTEM

This paper discusses the forces on stars in a globular cluster. Consider Figure 1; the

symbols are defined as follows:

Mg = mass of the stars of a globular ciuster Internal to g that of a particular star

m = mass of that particular star

mp = mass of the nearest neigboring stars

xg = distance of the star from the mass center of the globular g cluster

xp = distance of the star from the mass center of the nearest neighboring stars

xpg = distance of the mass center of the nearest neighboring stars from the mass center of

the globular cluster

xpo = equilibrium distance of the star from the mass center of the nearest neigboring stars

x = distance of the star from the mass center of the nearest neighboring stars, relative to

the equilibrlum distance

Recall that in the Reciprocal System two forces are acting on the star:

1. Gravitation of the star by the cluster as a whole—this produces an inward motion.

2. Progression of the star away from its nearest neighbors—this produces an

outward motion.

My goal in this paper is to derive the expression for the net force acting on the star, to

find the equilibrium position (xpo) of the star, and to determine whether or not this

position is stable.

Nehru‘s recent paper [1] provides the starting point. Some additional symbols are needed:

dog = gravitational limit of the globular cluster

dop = gravitational limit of nearest neigboring star:

yg = non-dimenslonal distance of the star from the mass center of the globular cluster

yp = non-dimensional distance of the star from the mass center of the nearest neighboring

stars

vog = ―zero-point speed‖ of the star relatlve to the globular cluster

vop = ―zero-point speed‖ of the star relative to the nearest neighboring stars

vng = net inward gravitational speed of the star

vnp - net outward progression speed of the star

vn = net speed of the star

G = ―universal‖ gravitational constant

Mo = mass of the sun

ag = acceleration from gravitation of the globular ctuster

ap = acceleration from progression away from the nearest neigbors

an = net acceleration of the star

In this notation,

dog = 3.77 * (Mg / Mo)½

(ly) (1)

yg = xg /dog = (x + xpo + xpg ) / dog (2)

vog = (2 * G * Mg / dog )½

(3)

vng = vog * (1 / yg ½ - yg½) (Inward) (4)

dop = 3.77*(mp / Mo)½ (ly) (5)

yp = xp / dop = (x + xpo) / dop (6)

vop = (2 * G * mp /dop )½ (7)

vnp = (½) * vop * (yp - 1/yp ) (outward) (8)

vn = vnp - vng (9)

Differentiating the velocity expressions with respect to time gives the accelerations:

ag = G * Mg * (1/xg² - 1/dog ²) (inward) (10)

ap = G * mp * (½) * (xp / dop³ - dop / xp³) (outward) (11)

an = ap - ag (12)

At equilibrium,

an = 0 (13)

Let

h = mp / (2 * dop³) (14)

i = Mg * (1/dog² - 1/xg²) (15)

j = (1/2) * mp * dop (16)

Then, in terms of xpo, at equilibrium,

h * xpo 4 + i * xpo ³ - 1 = 0 (17)

a quartic equation.

The appendix gives a simple computer program written in BASICA to solve equation 17

numerically. (An attempt to solve the equation analytically using the MU MATH AI

program failed). A sample run with Mg = 200*Mo , mp = 2*Mo , xg = 40 ly, dog = 53.32

ly, and dop = 5.33 ly produced xpo = 9.29 ly.. Another sample run with Mg = 30000*Mo ,

mp = 200*Mo ,xg = 400 ly, dog = 652.98 ly, and dop = 33.32 ly produced xpo = 178.94 ly.

Input parameters that are physically impossible produce negative distances.

Now let‘s turn to the question of the stability of this positlon, xpo The net force acting on

the star in terms of the distance from equilibrium, x, is

F = m * G * ((1/2) * mp * ((xpo + x)/dop³ - dop / (xpo + x)³)

- Mg * (1 / (xpo + x + xpg )² - 1 / dog²)) (18)

Differentiating F with respect to x gives

dF / dx = m *G * ((1/2) * mp * (1/dop³ + 3 * dop / (xpo + x)4)

+ 2 * Mg / (xpo + x + xpg )³) (19)

If x is positive, dF / dx is positive and hence F increases with x.

If x is negative, dF / dx is still positive. Thus

- dF / dx < 0 (20)

This is the definition of instability. Hence, xpo is a point of unstable equilibrium. But

there is one saving grace: the forces near this point are quite small, so sudden changes in

position are precluded.

Globular clusters continually grow by accretion until eventually being absorbed into

galaxies. The stars in the clusters must keep changing their temporary equilibrium

positions.

Reference

1. K. V. K. Nehru, ―The Gravitational Limit and the Hubble‘s Law,‖ presented at the

1986 Convention of ISUS.

STELLAR ENERGY GENERATION

IN THE RECIPROCAL SYSTEM

The theory of stellar energy generation in the Reciprocal System is stated qualitatively in

various works by Mr. Larson, such as Quasars and Pulsars. For the benefit of new

readers of Reciprocity, I quote Mr. Larson in full:

Inasmuch as a charge is a modification of the basic rotation, the number of charges that

an atom can acquire, the degree of ionization, as it is called, is limited by the number of

rotational units of the appropriate space-time direction that exist in the atomic structure;

the number of units available for modification. Negative ionization is confined to low

levels, as the effective negative rotation is never more than a few units. The limit of

positive ionization is the atomic number, which represents the net total nurnber of units

of rotational time displacement in the atom.

Electric ionization may be produced by any one of a number of agencies, inasmuch as the

requirement for this process is essentially nothing more than the availability of sufficient

energy under appropriate conditions. In the universe at large the predominant process is

thermal ionization. Thermal or heat energy is linear motion of material particles, and it is

therefore space displacement. In the ionization process this linear space displacement is

transformed into rotational space displacement: positive charge. As the temperature

increases, more and more space displacement becomes available for ionization, and the

degree of ionization rises until the atom finally reaches the point where it is fully ionized;

that is, each of its units of time displacement has acquired a positive charge.

If the temperature of the fully ionized atom continues to rise, a destructive limit is

ultimately reached at the point where the total space displacement, the sum of the

ionization and the thermal energy, is equal to the time displacement of one of the

magnetic rotational units. Here the oppositely directed rotational displacements neutralize

each other, and both revert to the linear basis, destroying this portion of the atomic

structure. Since the maximum ionization increases with the atomic number, the amount of

thermal energy required to bring the total space displacement of a fully ionized atom up

to the destructive limit is less for heavier atoms, and the effect is to establish a

temperature limit for each element that is inversely related to atomic number. As the

temperature of an aggregate rises the heaviest elements are therefore the first to

disintegrate.1

To sum up, when the destructive thermat limit is reached, the following word equation

holds true:

ionization energy of atom + thermal energy of atom = energy

equivalent of one unit of magnetic time displacement (la)

Let EI be the ionization energy, ET be the thermal energy, and EM be the oppositely

directed magnetic rotational energy. Then in symbols,

EI + ET = EM (lb)

Each of the terms in the equation will now be discussed.

Equivalent energy of one unit of magnetic time displacement

Before we can find the energy equivalent of one unit of magnetic time displacement, we

must find the mass equivalent. According to deductions previously made from the

postulates of the Reciprocal System the electric equivalent of a magnetic displacement n

is 2n²; this does not refer to the total from zero to n—it is the equivalent of the nth term

alone. Each electrical unit is equal to two atomic mass units, and each atomic mass unit is

equivalent to 931.48 MeV. For n equal to 3 and 4, the following table results:

n 2n² amu EM MeV

3 18 36 33533.28

4 32 64 59614.72

Thus, the third magnetic time displacement is equivatent to 33533.28 MeV, and the

fourth unit to 59614.72 MeV.

Ionization energy

At the present stage of development of the Reciprocal System we do not have a

theoretical equation giving the energy needed to completely ionize at atom—but then

neither does quantum mechanics. An empirical equation will have to do for now.

Reference three has the most comprehensive table of ionization values available, giving

the complete ionization energy for the first twenty elements. From the values, I have

derived the following empirical equation:

EI - 13.595 + 52.148(Z-1)² 10-6 MeV (2)

where Z is the atomic number. Of course, other equations are possible.4 Extrapolating any

empirical equation to high values of Z is risky, but this will have to do. For thorium, at

no. 90, eq. 2 gives

EITh = .413 MeV

Thermal energy

Let k be Boltzmann‘s constant in MeV/°K and T be the temperature of an atom in °K

Then the standard equation for the thermal energy (based on the ideal gas taw) is

ET = 3/2 kt (mv²/2) = 1.292 x 10-10 (3)

Calculation of critical temperature and velocity

From eq. lb,

ET = EM - EI

Then,

TCRIT = EM - EI / 1.292 x 10-10 °K (4)

For thorium, EM is 59614.72 MeV and EI is .413 MeV, so

TCRITTh = 4.614110449 x 1014 °K

This is fantastically high from our view as spectators on the earth, but in terms of natural

units, the temperature is ―only‖ 127.44.

With k in J/°K, equation 3 can be solved for the velocity at the critical temperature.

vCRIT = (3kTCRIT/m)½ (5)

For thorium, this amounts to

vCRITTh = 2.5289 x 108 m/sec

This is 84% of the speed of light:

For iron, the critical temperature is

TCRITFe = 4.61413989 x 1014 o

K

and the critical velocity is

vCRITFe = 2.7165 x 108 m/sec

This is 91% of the speed of light! No wonder atoms are accelerated to velocities above

the speed of light during a supernova explosion:

Most likely the motion of the atoms in the core of a star is circular. The greater the

temperature, the higher the velocity—thus as theoretically expected O and B type stars

have much greater rotational velocities than G and K type stars.

Rate of energy generation

Since both the unit of magnetic time displacement and the opposing space displacement

revert to linear motion, the total energy radiated per critical atom is

ERAD = 2xEM =119229.44 MeV (6)

for n = 4.

The rate of energy generation depends on the number of atoms at the critical temperature,

NCRIT. This in turn depends on the total mass of the star M, the average mass per critical

atom, m, and on the fraction FCRIT of the mass M that is critical. Thus

NCRIT = FCRITxM /mCRIT (7)

Let PST be the total power output of a star. Then assuming no accretion whatever, the

litetime of a star can be calculated as follows:

LST = NCRITERAD/ PST sec (8)

For the sun,

M = 2 x 1030 kg

PST = 2.43x1039 Mev/ sec

Taking thorium as representative of the critical elements,

m = 2.988 x 10-25 kg/atom

Assuming various values of FCRIT we can calculate the lifetime of a star with no

accretion. The following table results.

FCRIT LST (no accretion) years

.01 1.0389 x 1011

.001 1.0389 x 1010

.0001 1.0389 x 1009

.00001 1.0389 x 1008

.000001 1.0389 x 1007

According to the Reciprocal System, net accretion does occur over the life of a star, but

there may be periods where there is a net loss. Since such a period may last as long as a

billion years, I believe we are on good ground assuming that FCRIT is equal to .0001. At

present we have no way of deducing theoretically the fraction of the mass of a star that is

critical. Certainly, observation is no help; observation can only indicate the composition

of the stellar atmosphere, not that of the central core.

Rate of accretion

The sun appears to be one-third along its way on the Herzsprung-Russell diagram. Since

the sun has been estimated to be in existence for about 5 billion years, we can roughly

assume that the average lifetime of a star is 15 billion years. According to the theory, a

star slowly increases in temperature until the critical temperature of the iron group of

elements is reached, at which point the life of the star is terminated in a supernova1

explosion. From the equations presented in this paper, the critical temperature of iron is

3,091,400,000 oK above that of uranium. Thus the rate of change of temperature with

time can be roughly expressed as follows:

dT/dt / T/ t = 3,091,400,000/15 x 109 = .206 °K (10)

(Even if L were only 7.5x109 years, the increase in T per year would be less than .5 °K).

Thus stars are for most of their lives very stable energy generators. From this we can

conclude that the rate of accretion is just slightly greater than the rate of mass lost

through burning. For calculating the rate of accretion we can assume that for the short

term they are identical.

Let RACC be the rate of accretion in kg/sec. Then

RACC = PSTmCRIT / ERADFCRIT (10)

Using previous values of PST, ERAD, mCRIT and FCRIT equal to .0001,

RACC = 6.099 x 1013 kg/ sec = 1.925 x 1021 kq/ yr

This amounts to .000000096% of the mass of the star per year. It would take over 3108

years for the accretion to amount to the mass of the earth:

Clearly we cannot observe this small rate of accretion. Observation cannot tell us whether

the mass of the sun is remaining constant or slowly increasing, as we believe, or whether

the mass of the sun is slowly decreasing, as present theory suggests.

Conclusion

The current theory of stellar energy generation has been criticized elsewhere, and a

summary of that criticism is presented in reference five. The basic differences between

the new theory and the current one are as follows:

1. In the new theory, energy is generated by disintegration of heavy elements; in the

current theory, energy is generated by fusion of light etements.

2. In the new theory the temperature of the stellar core is of the order of 4.6 x 1014

°K in current theory, it is 3x107 °K for the first phase, and 109 for later phases.

3. In the new theory, ordinary stellar energy generating processes do not give rise to

neutrino emission, but in current theory they do. So far, no neutrinos have been

found to emanate from the sun.

4. In the new theory, one method for energy generation serves all types of stars;

current theory proposes that various stars have different energy schemes: proton-

proton reaction: the CNO bi-cycle; helium burning; (y, ) reaction of C12 , O16 ,

Ne20 nuclei; e-process; r-process.

Thus, though observation (other than neutrino counts) cannot at present decide in favor of

one theory over the other, Occam‘s Razer can: the new theory wins hands down.

References

1. Dewey B. Larson, Quasars and Pulsars (Portland, Oregon: North Pacific

Publishers, 1971), p. 60.

2. Dewey B. Larson, New Light on Space and Time (Portland, Oregon: North Pacific

Publishers 1965) p. 234

3. Von W. Finkelnburg and W. Humbach, ―Ionisierungsenergien von Atomen and

Atomionen,‖ Die Naturwissenschaften, Heft 2, Jg. 42, 1955, pp. 35-37.

4. For instance a Polynomial equation in Z has been worked out by computer by

Frank V. Meyer: EI = 78.6411 - 72.8213 x Z + 33.675 x Z² + .801221 x Z³ .

5. Ronald W. Satz, The Unmysterious Universe (Troy, NY: Troy Printers, 1971),

p. 11.

6. R. Davis, Jr., D. S. Harmer, K. C. Hoffman, ―Search for Neutrinos from the Sun,‖

Phys. Rev. Let., 20, 1205 (1968).

Author‘s Note: This paper is not meant to be the last word on subject of stellar energy.

Rather it is meant only to be the second word. Constructive criticism would be welcome.

THE GRAVITATIONAL ATTRACTION OF

THE GALAXY

In a previous paper1 I worked out the general form of Newton‘s Law of Gravitation and

applied it to the special case of a planet orbiting the sun. In this case Newton‘s Law was

modified by the factor

1/(1 - v²/c²) ½

For the case of an object moving directly toward another object rather than orbiting, the

genetal equation reduces to Newton‘s Law multiplied by the factor

(1 - v²/c²)

This is exactly of the same form as Lorentz‘s modification of Coulomb‘s Law.

Before applying the new factor, it is important to realize that the galaxy cannot be

represented as a pofnt mass; rather it should be represented as a flat disk. The Newtonian

actraetion of a flat disk for a point mass has been worked out before², but will be repeated

here.

In Figure I let the radius of the disk be r and let its surface density be . I aim to find the

attraction of the disk for a polnt mass located at P on the perpendicular line passing

through the center of the disk. Let 0 be the origin of a system of polar ccordinates p and

, and let z be the distance along the line to the attracted location P.

Since pdp is the area of an element in polar coordinates, the mass of such an element is

dm = pdpd (1)

The dietance of the element dm from P is

R = (p² + z²)½ (2)

and the attraction of the mass dm for the mass at P is

- G dm = G pdpd

— ———— (3)

R² p² + z²

and the component of the attraction along the exis is

- G pdpd . z G pdpd

———— – = - ———— (4)

p² + z² R (p² + z²)3/2

The total intensity of attraction of the disk for the point P mass is

n

2 pdpd

I = G z ————–

o o (p² + z²)3/2

r pdp

I = 2 G z ————–

o (p² + z²)3/2

[

z - z

]

I = 2 G ———— —— (5)

(z² + r²)½

(z²)½

Assuming z positive,

[

z

]

I = 2 G ———— - 1 (5¹)

(z² + r²)½

Now with the modifying factor included, the acceleration of the point masa toward the

disk is

dv

[

z

]

[

1 - v²

]

— = 2 G ———— - 1 —

dt (z² + r²)½

c²

dv 2 G

[

z

]

[ c² - v² ]

— = —— ———— - 1 (6)

dt c² (z² + r²)½

dv = dv dz = v dv

— — — — (7)

dt dz dt dz

The crucial deduccion in Larson‘s gravitational theory is that the gravitational force of

any mass extends outwatd only a finite amount the gravitational force does not extend out

to ―infinity‖, as commonly assumed. At the gravitational limit of the galaxy, which will

be denoted by do, the attracted velocity of a mass is zero. This velocity becomes larger to

the degree that the mass is loeated closer to the galaxy. Let the velocity be v at distance z.

Then, separating the variables in equation 6 and integrating between the limits, the result

is

v vdv

z

2 G zdz

z 2 G dz

————— = ———– ———— ———— (8)

o c² - v² do c² (z² + r²)½

do c²

The outcome of this result is that

v

[

- 4 G [(z² + r²)

½ - (do² + r²)

½ + do - z]

½

]

– = 1 - e ——–– ½ (9)

c c²

For our galaxy the constants in the equation are as follows:

G = 6.67 x 10-11 N - m²/kg²

c² = (3 x 108)² m²/s²

= .2975 kg/m²

r = 4.626 x 1020 m

do = 2.177 x 1022 m

With these values equation 9 becomes

v

[

- 2.771 x 10-27[(z² + 2,140 x 10 41)½ - 9.087 x 1013 - z]

]

–

= 1 - e ½ (10)

c

Speed in km/sec vs. distance ia kiloparsecs is plotted in the graph (Fig. 2). Great caution

must be used in applying equation 10 to real masses:

1. A globular cluster or a small galaxy associated with the Milky Way galaxy is not

really a point mass; in fact, observation shows that the near side stars of such

objects are attracted at the expense of the farside stars.

2. Globular clusters are not falling directly toward the galactic center; rather they are

orbiting.

3. Small local galaxies are at a distance close to the gravitational limit of the galaxy

— their veloclties aze difficult to measure and compare with theory.

Even so, the calculated velocities, neverthaless, agree in a very general vay with those

observed for the local group of objects.

References:

1. Satz, R.W. REClPROCITY Vol. IV, No. 2, p. 25, July 1974,

2. MacMillan, W., The Theory of the Potential (New York: McGraw Hill Book

Company, 1930),pp. 15-16.

DISCUSSION OF LARSON‘S

GRAVITATIONAL EQUATION

As brought out at the recent convention, some confusion has arisen over Larson‘s

gravitational equation, eq. (2) of the original edition of the Structure of the Physical

Universe:

F = mm‗/s² (1)

The correct expanded version of this equation is

(m × 3.7115 x 10-32) × (m‘ × 3.7115 x 10-32)

F = ————————————————— (2)

1521 × 10-15 × (s/1)²

where 3.7115 x 10-32 sec³/cm³ is the value of the natural unit of mass and m and m‘ are

simply numbers. The number .1521 x 10-15 sec is the natural unit of time. From equation

(2), the natural value of the gravitational constant can be determined:

Gn.u.= (3.7115 × 10-32)²/.1521 × 10-15 = 9.0567 x 10-48 (3)

Thus equation (1) might better be written as

Fn.u = 9.0567 × 10-48 mm‘/s² (4)

where all values are in natural units. The expression for G in equation (3) can be

converted to conventional units. First, the cgs system:

Fn.u.sn.u.² dynes

Gcgs = 9.0567 ×10-48 ———— × 109.7 ——–

mn.u² Fn.u

(.456 x 10-5 cm mn.u²

× —————— × ——————

sn.u² (.5565 x10-24 g

= 6.67 x 10-8 dynes cm²/g² (5)

The MKS system

Fn.usn.u² N

GMKS = 9.0567 ×10-48 ———– 109.7 10-5 ——

mn.u² Fn.u

(.456 × 10-7 m)² mn.u²

× —————— × ———————

sn.u² (.5565 × 10-27 kg)²

= 6.67 × 10-11 N-m²/kg² (6)

Both check. The importance of this cannot be overestimated. These equations completely

confirm Larson‘s identification of all the fundamental units.

Note that if in equation (2), the value 3.7115 × 10-26 sec³/m³

is used then the correct time value must be .1521 × 10-3 sec for the denominator (or 1012

units of time).

THE INTERACTION OF ALPHA PARTICLES

AND GOLDS ATOMS

A New Explanation of Rutherford Scattering

Introduction

Nearly all present-day physicists are convinced of the truth of the assertion in the

following quotation from Weidner and Sells‘ Elementary Modern Physics(1)

:

It was by the alphaparticle scattering experiments, suggested by Rutherford, that the

existence of atomic nuclei was established.

However, when we study the literature of Rutherford‘s era, we find that he and his

associates, Geiger and Marsden, did not in fact discover the atomic nucleus. Geiger and

Marsden‘s paper, ‖The Laws of Deflexion of a Particles through Large Angles, ‖(2)

does

present strong experimental evidence of a central repulsive force originating from atoms,

but the paper does not prove that this force is electrical in nature. What their experiments

did prove is that the number of particles scattered through an angle q is proportional to

1/sin4 (q/2) and to the inverse of the square of the kinetic energy of the particles, 1/Ek

2.

Of course, the experiments did disprove the Thomson ‖plum pudding ‖ atom model,

which did not predict a strong central repulsive force; but it is one thing to disprove a

theory; it is quite another to prove one. If the Rutherford model were the only alternative

left, we might have to conclude that it is correctÑbut there are always other alternatives.

This paper will present one such alternative: the Reciprocal System of physical theory. A

new scattering equation will be derived and compared with the experimental facts as

found in an uptodate version of the original experiment, that conducted by Prof. Adrian

C. Melissinos and his students.(3)

The originator of the Reciprocal System is Dewey B.

Larson; for full comprehension of this paper, the reader should first study Larson‘s

books.(4,5,6)

I. Theory

A. The Repulsion Force: F

1. The Reciprocal System

In the Reciprocal System, nonionized and nonmagnetized matter is subject to only two

primary forces: the spacetime progression and gravitation. In the timespace region, the

progression is outward and gravitation inward, whereas in the time region (inside unit

space) the progression is inward and gravitation outward -- a repulsion. Right at the

boundary the progression is zero, but the net gravitation is not zero. Compared with the

repulsive gravitational force, the attractive gravitational force is negligible. Thus at the

boundary only the repulsive gravitational force is effective. Now consider what happens

when an atom A, which is moving towards an atom B, reaches this boundary. According

to Larson,(6)

When atom A reaches point X, one unit of space distant from B, it cannot move any

closer to B in space. It is, however, free to change its position in time relative to the time

location occupied by atom B. The reciprocal relation between space and time makes an

increase in time separation equivalent to a decrease in space separation, and while atom

A cannot move closer to atom B in space, it can move to the equivalent of a spatial

position that is closer to B by moving outward in coordinate time. . . . No matter what the

spatial direction of the motion of the atom may have been before unit distance was

reached, the temporal direction of the motion after it makes the transition to motion in

time is determined purely by chance.

A previous paper of mine, ‖Time Region Particle Dynamics, ‖(7)

dealt with the situation

in which atom A (say an alpha particle) is assumed to continue to move directly toward

atom B (say a gold atom) in the time region. This paper will consider the general case in

which no assumption is made as to the actual motion that takes place in the equivalent

space of the time region. All that will be considered here is the equivalent motion that

takes place at the boundary, i.e., in actual space. Here, with the atoms separated by so, the

repulsive gravitational force F is (from Ref. 4)

F = KG/s4 = KG/so

4 (if s = so) (1)

The repulsion coefficient KG is expressed by

KG = [Fpso4/(156.44)

4] * [ln

4 teff/ln

²t‘eff] (2)

where Fp is the natural unit of force and the number 156.44 is the interregional ratio. The

dimensionless variables teff and t‘eff are material constants determined by the

characteristics of the interacting particles, and will be discussed further later.

2. Conventional Theory

In present theory, the alpha particles somehow avoid interacting with the cloud of

electrons supposedly surrounding each gold nucleus. The only force involved comes from

the presumed nucleus. At low energies this is a Coulombic force, given by

F = zZe²/4 os

² (3)

where z is the atomic number of the alpha particule (helium), Z is the atomic number of

gold, e is the value of electric charge, eo is the permittivity constant of free space, and s is

the separation distance. This is an inverse square force, rather than an inverse quartic

force as in the Reciprocal System. Why this Coulombic force should act between

particles but not within nuclei is a question completely unanswered by current theory.

B. The Impact Parameter: b

MOTION IN ACTUAL SPACE: so remains constant, but angle f changes from o to -

where is the deflection angle observed in the time-space region.

The figure shows a typical collision process. The impact parameter is the distance that the

alpha particle would have passed the gold atom if there had been no force between them.

1. Reciprocal System

Let m be the mass of the alpha particle and vo be its initial velocity. Referring to the

figure, we have

impulse = (mv)y

Fdt = mvo sin( )

(KG/so4) sin(f)dt = mvosin( )

(4)

The alpha particle passes from the timespace region, through the time region, and back

into the timespace region. For the general case, we cannot write an equation for the actual

motion in the equivalent space of the time region, but we can write an equation for the

equivalent motion in the actual space of the timespace region. Throughout this motion,

neither the angular momentum, nor the actual spatial separation, changes. But the angle

does change. Therefore,

mvob = mso²(d /dt) (5)

Separating dt from d in eq. (5) and substituting in eq. (4), we have

(KG/so4) * (so

²/vob) sin( ) d( ) = mvo sin( )

(6a)

or

(KG/so²mvo

²b) sin( ) d( ) = sin( )

Since the kinetic energy, Ek, is (1

/2) mvo², eq. (6a) can be rewritten as

[KG/(so² * 2 * Ek * b)] sin( ) d( ) = sin( ) (6b)

Before the collision, = 0 and after the collision, = , so these are the limits on the

integral.

sin( ) df = -cos( ) o o

= cos ( ) + cos(o)

= cos( ) + 1

Thus, [KG/(so² * 2 * Ek * b)] * (cos( ) + 1) = sin( ) (7a)

But sin( )/(cos( ) + 1) = tan ( /2)

so KG/(so2 * 2 * Ek * b) = tan( /2) (7b)

Solving for b we finally obtain

b = [KG/(2 * so² * Ek)] * cot( /2) (7c)

2. Conventional Theory

Refs. 1, 3, and 8 all have derivations for b in current theory. The result is

b = [zZe²/(8 oEk)] * cot ( /2) (8)

C. Target Cross-Section:

The target Cross — Section is defined as = b².

1. The Reciprocal System

From eq. (7c), using the above, we obtain

= [ KG²/(4 * so4 * Ek

²)] * cot

² ( /2) (9)

2. Conventional Theory

With b from eq. (8),

= [z²Z

²e

4/(64 o

²Ek

²)] * cot

²( /2) (10)

D. Differential Cross-Section: d

The differential cross-section is

d = 2 bdb

1. Reciprocal System

Here, db = [-KG/(so² * 4 * Ek)] * [d /sin

²( /2)] (11)

Thus d = 2 * [KG/(2 * so² * Ek)] * [cos( /2)/sin( /2)]

* [-KG/(4 * so² * Ek)] * [d /sin

²( /2)]

= [-2 KG² cos( /2) d ] / [8 * so

4 * Ek

² * sin

³( /2)]

= [ KG² cos( /2) d ]/[4so

4 * Ek

² * sin

³( /2)] (12)

(with the minus sign dropped).

The angles and + d define two cones with the horizontal line through the gold atoms

as their axis. The differential solid angle d between the two cones is

d = 2 sin( ) d

Since the cosine term in eq. (12) can be expressed as

cos( /2) = sin( )/[2 * sin( /2)]

eq. (12) becomes

d = [ KG² (sin ( )) d ]/[8so

4 * Ek

² * sin

4( /2)]

= [KG² d ]/[16 * so

4 * Ek

² * sin

4( /2)]

or d /d = KG²/[16 * so

4 * Ek

² * sin

4( /2)] (13)

The units of d /d are meter squared per steradian, m²/sr.

2. Conventional Theory

Similarly, for conventional theory,

d /d = z²Z

²e

4/[256 *

² * o

2 * Ek

² * sin

4( /2)] (14)

E. The Scattering Constant: Ks

It is immediately seen from eqs. (13) and (14) that for both

theories,

(d /d ) * sin4( /2) = Ks

a constant.

1. Reciprocal System

Here,

K = KG²/(16 * so

4 * Ek

²) (15)

2. Conventional Theory

Here,

Ks = z²Z

²e

4/(256 *

²* o

² * Ek

²) (16)

F. The Number of Particles Scattered per Minute at the angle : Is

So far we have looked at the situation involving only one alpha particle and only one gold

atom. For the situation in which a beam of alpha particles strikes a gold foil, we would

like to compute the number of particles scattered through a certain angle q. Let

Io = number of incident alpha particles/minute

d = detector solid angle

N = area density of scatterers (number of gold atoms/m2)

The value of N is determined from this equation:

N = * * [No/A) * (104 cm

²/m

²)] (17)

where

d = thickness of foil (cm)

r = density of scatterer (g/cm3)

No = 6.02 * 1023

, Avogadro‘s number

A = atomic weight of scatterer

Then the number per minute, Is, of alpha particles scattered into the detector at the angle

is

Is = Io * N (d /d ) * d

= Io * * * (No/A) * 104 * (d /d ) * d

1. Reciprocal System

Here,

Is = Io * * * (No/A) * 104 * KG

²* d /[16 * so

4 * Ek

2 * sin

4( /2)] (18)

2. Conventional Theory

Here,

Is = Io * * * No/A * 104

* (z²Z

²e

4 * d )/[256 *

² * o² * Ek

² * sin

4( /2)] (19)

Note: Both equations are based on the assumption that E is sufficiently low such that

‖relativistic ‖ effects can be neglected and such that the gold atoms remain stationary

during the interaction -- this would not be the case at very high alpha particle energies.

II. Experiment

A. The Experimental Set-up

Professor Adrian C. Melissinos carried out a modern version of the original

GeigerMarsden experiment and described his findings in Ref. 3. In this experiment

Io = 1.1 * 105 incident alpha particles/minute

d = da/L2 = .8786/(6.67)

² = 0.197 sr

[da is differential area of detector and L is distance of detector from foil]

= 19.3 g/cm3 (gold)

= .00025 cm

A = 197 (gold)

Thus,

N = * * (No/A) 104

= (.00025)(19.3)(6.02 * 1023

/197)*104

N = 1.4744 * 1023

gold atoms/m2

So the equation for Is is

Is = Io * N * (d /d ) * d

= 1.1 * 105 * 1.4744 * 10

23 * (d /d ) * .0197

Is = 3.1950 * 1026

* d /d

Is is measured, and then (d /d )is computed. For each angle q the product (d /d ) *

sin4( /2) can be found and the results plotted. From leastsquares error analysis, the best

fit experimental value of Ks can be obtained.

Melissinos states that the above value of Io, and hence also Is, is subject to at least a ± 20

percent error in view of the approximations used and the nonuniformities in beam density

and direction. One other uncertainty is the energy of the alpha particles. The incoming

energy is 5.2 MeV, but since the particles lose a considerable amount of energy in

traversing the target, Melissinos believes that it is more appropriate to use a mean value

of Ek for the calculations. He calculates the mean value to be

Ek = 4.39 MeV = 7.03 * 10-13

J

B. Calculation of Ks for the Experimental Set-up

1. Reciprocal System

Here,

Ks = KG²/(16 * so

4 * Ek

²)

KG = [Fp * so4/(156.44)

4] * [ln

4 teff/ln

² t‘eff]

Fp = 3.27223 * 10-3

Newtons

so = 4.558816 * 10-8

meters

ln² t‘eff = 1, since helium has no electric dispacement -- it is ‖inert. ‖

The difficult part in calculating Keff for gold helium. The (tentative) method used here

will be different from that used in my previous paper, Ref. 7. Gold has 3 active rotational

dimensions with t = 4.5 in all (see Ref. 4 for more details). Helium has only 1 active

dimension, with t = 3. The other two dimensions have t = 1. In the first dimension the

mean value for gold and helium is

t = (4.5 * 3)1/2

= 3.67

In the other two dimensions, the full rotational force of the gold atom is present, so

instead of (4.5 * 1)1/2

= 2.1, we simply have t = 4.5. The mean over all three dimensions

is

teff = (3.67 * 4.5 * 4.5)1/3

= 4.2

Thus (tentatively),

KG = [3.27223 * 103 * (4.558816 * 10

8)4/(156.44)

4][ln

4(4.2)]

= 1.00 * 1040

nm4

The (tentative) value of Ks is then Ks = (1.00 * 1040)2/[16 * (4.558816 * 10

8)4 * (7.03 *

1013)2] Ks = (2.93 * 10

28 (m

2/sr)

2. Conventional Theory

Here,

Ks = z²Z²e

4/(256 *

² * o

² * Ek

²)

z = 2

Z = 79 e = 1.602 * 1019 coulombs

eo = 8.85 * 1012 coulombs/N-m²

Ek = 7.03 * 1013 J

Thus,

Ks = [(2²)(79

²)(1.602 * 10

19)4] / [256 *

2 * (8.85 * 10

-12)² * (7.03 * 10

-13)²] Ks =

1.68 * 10-28

(m²/sr)

3. Experiment

The best fit experimental value, according to Melissinos, is

Ks = 2.70 * 1028 (m²/sr)

Thus the theoretical and experimental results can be summarized as follows:

Ks(m²/sr)

Experiment: 2.70*10-28

Reciprocal Systems 2.93*10-28 (tentatively)

Conventional Theory 1.68*10-28

Appraisal

It is well known that Rutherford‘s theory of scattering fails at high energy. On the basis

of Melissinos‘ experiment, we must also reject this theory at low energy. But, given the

present climate of thought, Melissinos himself could not come to this conclusion. He says

(Ref. 3, p.250):

The difference between the observed and theoretical constants, while at first sight large,

can be traced to the limited sensitivity of the apparatus and mainly to

(a) Uncertainty in incoming flux

(b) Uncertainty in foil thickness [and, to a lesser extent, to]

(c) Extended size of the beam and lack of parallelism

(d) Extended angular size of the detector

(e) Plural scattering in the foil (for the data at small angles)

(f) Background (for the data at large angles)

However, given the close result of the Reciprocal System, it now appears that Melissinos

is too modest. Even with some uncertainty in the theoretical value of KG and the

experimental values of Io and Ek, it appears that the Reciprocal System is consistent with

the observed result, whereas conventional theory is not.

Can an appeal to more sophisticated mathematics rescue the current theory? It cannot.

From Ref. 1 (p. 223) we have this statement:

It is a remarkable fact that a thoroughgoing wavemechanical treatment of scattering by an

inversesquare force yields precisely the same result as that yielded by the strictly classical

particle analysis discussed here.

Thus, despite the thousands of books, the thousands of papers, and the thousands of

lectures on the nuclear theory of the atom, the physicists are going to have to discard their

cherished concept.

The scattering equation of the Reciprocal System will next have to be applied to other

pairs of incident and target particles and to other energy levels.

References

1. R. Weidner and R. Sells, Elementary Modern Physics (Boston: Allyn & Bacon,

1968), pp. 222-223.

2. H. Geiger and E. Marsden, ―The Laws of Deflexion of a Particles through Large

Angles, ‖ Philosophical Magazine, 25 (1913), 604-628.

3. A. Melissinos, Experiments in Modern Physics (New York: Academic Press,

1966), pp. 231-252.

4. D. Larson, The Structure of the Physical Universe (Portland, Ore.: North Pacific

Publishers, 1959). (The revised edition is being published in several volumes, the

first of which is Nothing But Motion, 1979).

5. D. Larson, The Case Against the Nuclear Atom (Portland, Ore.:North Pacific

Publishers, 1963).

6. D. Larson, New Light on Space and Time (Portland, Ore.: North Pacific

Publishers, 1965), pp. 115-116.

7. R. Satz, ―Time Region Particle Dynamics, ‖ Reciprocity IX.2 (Summer, 1979).

8. H. Enge, M. Wehr, J. Richards, Introduction to Atomic Physics (Reading, Mass.:

Addison Wesley Publishing Co., 1972).

A PROPOSAL FOR A CRUCIAL EXPERIMENT

Rutherford‘s nuclear theory of the atom has held sway in the scientific community for 70

years. I now propose a test which may disprove it.

In the original scattering experiment, charged helium atoms (alpha particles) were

beamed at a gold foil; the resultant scattering was claimed to be due to Coulombic

repulsion by charged nuclei. Now suppose that that non-charged helium atoms are

beamed at a gold foil. The Reciprocal System of Theory predicts the same scattering in

this case. The Rutherford theory predicts no scattering.

Of course, experimenting with non-charged helium atoms is more difficult than with

charged ones. The scattering apparatus will have to incorporate a means to inject

electrons to neutralize the alpha particles and a different material to detect the particles

after scattering.

Numerous physics laboratories in the world have the capability to carry out this

experiment. Which will be the first?

See

THE INTERNATIONAL SOCIETY OF UNIFIED SCIENCE

LINKS

The Collected Works of Dewey B. Larson

Reciprocal System Dynamics by Ronald W. Satz

Glimpses of a New Paradigm by K.V.K. Nehru

Toward a Unified Cosmological Physics by Arnold Studtmann

Kinetons and Quanta by Jan Sammer

Special Thanks to the folks who set up the web site at;

http://www.reciprocalsystem.com/rs/links.htm