[Gyula Szász]

The comparison of the relativity theories (RT) with the Atomistic Theory of Matter (ATOM).

RT:
In the RT is the inertial mass mi set equal to the gravitational mass mg, mi = mg, and the gravitational mass is thrown away. Furthermore, it is only calculated in RT with inertial mass m and the following relations are valid: E = m∙c^2 and E^2 = (m0∙c^2)^2 +(p∙c)^2. The inertial mass m depends of the velocity v; it is the relativistic mass. The calculation of the “rest mass”, m0, cannot be done in RT; it remains open what the “rest mass” is. This situation is not only dissatisfying, but scientifically impossible. The physicists do not know what “rest mass” is and they cannot really calculate what relativistic mass is! The ambiguity of “mass” remains at the use of RT and nobody take notice from it. It remains the central unsolved problem of the whole accepted modern physics.

ATOM:
For the four elementary particles e,p,P, E are the gravitational masse mgj equal to the inertial masses mij: mgj = mij, j=e,p,P,E.
All masses composed bodies of Ni elementary particles can be calculated with the elementary masses mP and me. The gravitational masses of composed bodies are
mg(Ne, Np, NP, NE) = |(NP – NE)∙mP + (Np – Ne)∙me|,
At the gravitational mass is respected that the gravitational charges, gi, have different signs gi = ± g ∙mi; the gravitational constant is G and g = (4π∙G)^1/2 and further, in the Newtonian force equation the product of gravitational charges appears.
The inertial “rest masses” (at COM motion with v=0) of composed bodies are
mi(Ne, Np, NP, NE) = (NP + NE)∙mP + (Np + Ne)∙me – E(bound)/c^2.
E(bound) is the bound energy of the composed body; the inertial mass is greater or equal zero. The “rest mass” is unambiguously defined in ATOM and obviously the two masses of composed bodies are different. Alone this fact is an advantage in ATOM and is against RT!
The relativistic mass is defined only as approximation from the covariant equations of particle motions; the equations of motions contain also the two fundamental fields which propagate with c. But the bound energy is defined only at v=0 COM motion. Furthermore, we have also the uncertainty that we never know the velocity of particles exactly. The Lorentz factor contains the relativistic addition of velocities and it is only usable for exactly known velocities of particles.
The comparison of RT and ATOM shows that RT cannot be used as well at “rest masses” as at the relativistic masses correctly thus without approximation.
Gyula Szász

[Gyula Szász]

The Higgs bosons are not needed to understand what masses of particles are!

[Gyula Szász]

Prognoses of the Atomistic Theory of Matter (ATOM)

The ATOM based on the four stable elementary particles e, p, P and E connect the electromagnetism and gravitation in a unified way in Minkowski space and build in the gravitation in the particle physics. The particle number conservations, as subsidiary conditions of the particles at the variation, deliver Lagrange multipliers for the equations of particle motions. The Planck constant h is connected to a Lagrange multiplier, however, at the appearance of h a second condition should also be considered: the stationary solution of the variation must be also timely stationary. We want now consider these conditions at two-particle states. Because only at the appearance of attractive electric forces are stationary bound states awaited, it is evidently, we must only consider the following systems: (e,P), (e,p), (p,E) and (P,E). The first two, are well observed on the Earth; they are connected with the hydrogen atom and with the positronium. The other two contain elton (=“antiproton”) and they are seldom seen on the Earth because between elton and proton based matter a repulsive gravitational force exists.

For hydrogen atom H a phenomenological relation is known (Sommerfeld) for the Planck constant h (as Lagrange multiplier)

h =e^2/2c ∙(m’(H)∙c^2/2∙E(H;bound))^1/2 = 6.62607004 × 10^-34 m2 kg / s,

between the reduced mass m’(H) = me∙mP/(me+mP) and the energy of the ground state E(H;bound) = 13.6 eV. The same relation is also available for the positronium, however with the changed reduced mass m’(positronium) = me/2 and for the bound energy E(positronium;bound) = 6.8 eV.

Also for the sizes of bound states is a phenomenological relation is known

r(bound state) = h^2/(4π2m’e^2).

For the gravitational masses of (e,P) and (e,p), taking the signs of the gravitational charges into account, we get
mg(H) = mP – me, and mg(positronium) = 0.
We calculate at this point also the gravitational masse of the two other two-particle systems
mg((p,E)-system) = mP – me and mg((P,E)-system) = 0.
On the other side, the inertial masses of the (e,P) and (e,p) systems are
mi(H) = mp + me – 13.6 eV/c^2 and mi(positronium) = 2∙me – 6.8 eV/c^2.
Obviously the two kinds of masses mg and mi are different for the hydrogen atom, (e,P), and for the positronium, (e,p) in the “ground states”.
As the inertial masses cannot be negative, we could ask for the case what happens if the bound energy of (e,p) is equal to 2∙me∙c^2, that is if mi((e,p)-system) = 0? The energy of the (e,p)-system cannot be lower as 2∙me∙c^2. We identify this state with the electron-neutrino νe at the bound energy E(νe, bound) = 2∙me∙c^2. With this bound energy and the reduced mass m’(e,p) = me/2 we get, according the formula for h, another value

h0 = e^2/2c∙(me/(2∙4∙me))^1/2= e2/2c∙(1/8)^1/2 = h/387.

This new constant h0 fixes as well the state of electron-neutrino νe =(e,p), as also the proton-neutrino νP =(P,E). For the neutrinos are as well the gravitational masses as the inertial masses zero, and we remark, the masse of the composing particles, e, p, respectively P and E, are not annihilated at the building of the neutrinos. This is a hint that masses cannot be converted in energy; masses are not equivalent to the energy.
If we take the value of h0 also for the (e,P)-system, we get a new bound energy

E((e,P)-system;bound) = 2.04 MeV,

and we identify this state with the stable neutron N0. The gravitational mass of N0 remains the same as for the hydrogen atom, however, the inertial mass is changed

mi(N0) = mP + me – 2.04 MeV/c^2.

The inertial mass mi(N0) is smaller than the inertial mass of the hydrogen atom and mi(N0) is smaller than the gravitational masses of N0, mg(N0), and smaller than that of the gravitational mass of the hydrogen atom. We notice, mg(N0) = mg(H).

We state, within the ATOM we can calculate as well the gravitational masses mg as the inertial masses mi of composed systems and these masses are different. Within the ATOM neither the weak equivalence principle, mi = mg, nor the energy-mass equivalence, E = m∙c^2 is valid. Einstein thrown away the gravitational masses of bodies and he could not calculate within his special relativity the “rest masses” mi(v=0). The advantages of ATOM are obviously in comparison to the special relativity.

Furthermore, within the ATOM we have definitions for the two kinds of neutrinos, νe and νP,. These definitions can be accounted as prognoses of the theory.
We can also calculate the sizes of N0, νe and νP
d(N0) = 2∙r(N0) = 0.702∙10^-13 cm, (two times the radius),
r(νe) = 0.703∙10^-13 cm,
r(νP) = 0.383∙10^-16 cm.

The size of the proton-neutrino, νP, is about 1936 times smaller, than the sizes of N0 and νe. The sizes of N0 and νe nearly the same and 10^-13 cm and it is in the size ranges of the nuclei. We conclude, the nuclei are composed of protons, stable neutrons N0 and electron-neutrinos νe. The instable neutron N which is coming out at nucleon decays and which is experimentally observed is obviously N =(P,e,p,e) which the decay mode

N =(P,e,p,e) → P + e + (e,p) = P + e + ve.

The instable neutron N is a four-particle system.
We can conclude that all sub-atomic systems composed of e, p, P and E with the new constant h0 and not with the Planck constant h. For instant, h0 is to be applied at the calculation of the energies and of the sizes of nuclei, and also at the calculation of other instable particles.

The ATOM is a very powerful theory to treat sub-atomic systems. We have not only the circumstance that the electromagnetism and the gravitation are treated in a unified level, but we can calculate all sub-atomic system. The prognoses of the Atomistic Theory of Matter are undoubtedly very valuable.

Gyula Szász

• This reply was modified 9 years, 2 months ago by Gyula Szász.
• This reply was modified 9 years, 2 months ago by Gyula Szász.
• This reply was modified 9 years, 2 months ago by Gyula Szász.

[Gyula Szász]

Additional we state:
With Newton’s law the gravitational masses of each body can be calculated with the elementary masses. The gravitational mass is conserved. The inertial masses follow from the equation of motion and we can state these both masses are obviously different.

[Gyula Szász]

The Atomistic Theory of Matter

After the fundamental knowledge that neither the positions, not the velocities of particles can ever be exactly known, we must look how many different kinds of elementary particles exist. The experimental observations tell us, there are four: the electron (e), the positron (p), the proton (P) end the elton (E). How many different kind of elementary properties have these particles? We count two: the electric charges and masses. We have observed two kinds of elementary electric charges qi = ± e and two different elementary masses me and mP with mP/me = 1936.152. For the characterization of each particles e, p, P and E is enough to know the electric charge qi and the mass mi. However, this is uncomfortable and is not on the same level. The electric charges have the same amount e, but different signs; the masses mP and me have different amounts and are > 0.

However, we can achieve the same kind of characterization of the elementary particle properties if we assign to the particles instead of the masses, the elementary gravitational charges gi = {- g me, + g me, + g mP, – g mE}, i=e,p,P,E. At the introduction of gravitational charges we have arbitrary decide that proton has a positive gravitational charge, as we decided that proton has positive electric charge. The relative sings of the other elementary particles follow then. Furthermore, we do not only want set up the elementary properties of particles, but we want use these also for the interactions between the particles. For this reason we use the observed static forces, in the shape of the Coulomb law and of the Newton law between two charges

F(Coulomb) = + qi∙qj/4πr^2 = ± e^2/4πr^2,

F(Newton) = – gi∙gj/4πr^2 = – (±) g^2∙mi ∙mj/4πr^2 = – (±) G∙mi ∙mj/r^2.

We notice, the two laws are completely equivalent only the overall signs are different. Under the assumption of universal gravitational forces expressed with the constant G we can express the factor g in the elementary gravitational charges as g =(G∙4∙π)^1/2. Now, we are ready with the complete characterization of elementary particles. Particles with the same signs of electric charges repulse each other; with different electric charges they are attractive. For the gravitational charges it is reserved: particles with the same signs of gravitational charges are attractive; they repulse each other if the signs are different. The observed electric force is by a factor of ca. 10^42 greater than the gravitational force.

We have to complete the description of particles and interactions with the observed properties that the fields propagate with c and the speeds of the fields do not depend of the motion of sources. The unique speed c allows constructing a space-time connection which we call the Minkowski space. In the Minkowski space the elementary charges qi and gi appear as invariants. The interacting time dependent fields can be described with four-vector potential A(e.m.)ν(x) and A(g.)ν(x) in finite ranges of Minkowski space {x}εΩ. With the two kinds of four-vector potentials and with the two kinds of four-charge probability currents

j(e.m.)ν(x) = Σ(i=e,p,P,E) qi∙ji(n)ν(x)

and

j(g.)ν(x) = Σ(i=e,p,P,E) gi∙ji(n)ν(x)

we can construct a Lorentz invariant Lagrange density with the elementary charges, with the four-particle number probability densities ji(n)ν(x) and with the two kinds of four-vector potentials. The integration runs about Ω and delivers a Lorentz invariant action integral for the variation calculus in order to get the covariant equations of motions. The action integral is not an expression for the energy.

However, appropriate subsidiary and boundary conditions must be applied. The boundary conditions are such that the physics within Ω cannot depend on the surface of Ω. Within Ω the following subsidiary conditions must be applied

∂νA(e.m.)ν(x) = 0, ∂νA(g.)ν(x) = 0 for the fields (Lorenz conditions) and

∂νji(n)ν(x) = 0, i=e,p,P,E for the particles.

The subsidiary conditions of the particles are equivalent with the particle number conservations. The particle number conservations cause the appearance of Lagrange multipliers in the equations of particles motions. For instant the Planck constant is connected with a Lagrange multiplier. This theory is also a quantum theory; however in this are only the sources of the fields quantized.

This theory is obviously a mathematic correct constucted Atomistic Theory of Matter based on the four kinds of elementary particles e, p, P, and E. This theory is quite the contrary to the energetic oriented physics developed in the 20th century. Again the energetic physics speak that the energy is none conserved and none quantized. The emission of light by atoms is a wave process and not a corpuscular phenomenon. The energy decreases continuously during the emission and not in discrete energy packages. The atomistic theory is able to replace completely the developed, but invalid energetic theory.

Gyula Szász

P.S. Unfortunately the text editor does not transfer the lower and upper indices.

[Gyula Szász]

Dear Gyula,
But I suspect that there may be a “baby” in the wash-water; maybe the “babies” are the ideas Einstein stole from Lorentz.

Picasso once said, I am told, “Good artists copy, great artists steal”

And it’s probably an unfortunate Nature of the Human Condition that the originators of ideas, like yourself, are dismissed by the simple fact that new ideas cannot be understood using the old conventions that folks have gotten accustomed to using. 🙁

Sincerely,
Bill Eshleman

Bill,
exact inertial conditions of positions x(body;t) and of velocities v(body;t) of bodies at an exact time t = 0, are physically not allowed simplifications. Also bodies cannot fall with the identical acceleration a(body1) = a(body i) = constant.
At physical description of motions, in particular for microscopic objects and for the stars in Universe firstly these conditions must be take in to consideration.

Gyula Szász

[Gyula Szász]

Dear Gyula,

I cannot continue; my think about a fixed coordinate
system, an independent “stage”, is hopelessly crushed
by your statement, “Coordinate are never conserved”;
If I was not clear(I think I wasn’t), I was meaning that I
am considering that the whole coordinate system and
its shape, remain always the same, I think. And I am
still troubled by the inability of space to warp.

Einstein said this about “space”:

“When a smaller box s is situated, relatively at rest, inside the hollow space of a larger box S, then the hollow space of s is a part of the hollow space of S, and the same “space”, which contains both of them, belongs to each of the boxes. When s is in motion with respect to S, however, the concept is less simple. One is then inclined to think that s encloses always the same space, but a variable part of the space S. It then becomes necessary to apportion to each box its particular space, not thought of as bounded, and to assume that these two spaces are in motion with respect to each other.

Before one has become aware of this complication, space appears as an unbounded medium or container in which material objects swim around. But it must now be remembered that there is an infinite number of spaces, which are in motion with respect to each other. The concept of space as something existing objectively and independent of things belongs to pre- scientific thought, but not so the idea of the existence of an infinite number of spaces in motion relatively to each other. This latter idea is indeed logically unavoidable, but is far from having played a considerable role even in scientific thought.”

Albert Einstein

Is this quotation disagreeable to you?

I will post after you help me collect my think.

Sincerely,

Bill Eshleman

Einstein did not understand how electromagnetism work, he did not understand how gravity works, but, he thought that the gravitational mass is equal the inertial mass without prove! Einstein was not a good physicists and he was a wrong mathematician. He did also not understand what space and time are.
Szász Gyula

[Gyula Szász]

Do you not understand that “space” cannot warp and as well the positions, as the velocities of bodies are uncertain?

[Gyula Szász]

Einstein did not really understand how particle physics work. Quite the contrary to Einstein’s opinion, photons do not exist in Nature.
Szász Gyula

• This reply was modified 9 years, 2 months ago by Gyula Szász.

[Gyula Szász]

Bill, you must understand, even if you want use the Lorentz factor, you must know the exact velocity of a particle because of the “relativistic addition of velocities”.

[Gyula Szász]

“And I am still under the (stupid maybe) impression that a Lorentz invariant derivation of motion is identical to using the Lorentz factor to correct the Lorentz transformation matrix.”
How do you want transform an elementary particle from which you do not know the exact velocity?

[Gyula Szász]

Conservation of Coordinates

Dear Gyula,

Or the invariance of coordinate translations
(or transformations).

I thought I might run this past you to get my
customary slap of your ruler on my wrist.
Maybe even a thread on the Gravitation
forum once it’s cleaned-up.

So, here goes. By demanding the conservation
of “KonserviertMasse”, it also looks like you have
coincidentally or purposefully demanded the
conservation of the Cartesian Coordinate System;
or for that matter, the conservation of ANY
coordinate system that may be chosen; so the
conservation really implies an independence of
the coordinate system chosen; as you suggest.

Under these circumstances, so-called space-time
looks more like a three-dimensional space with
an additional, but imaginary, forth axis. A “stage”
quite independent (it does not warp), of the actors;
if it were not for that minus sign(s) in the invariant.

That minus sign(s) made Einstein think the spacetime
“stage” warps from the weight of the actors and therefore
even partakes in the performance we observe.

I’m considering that it was actually BECAUSE Einstein
was unsure of what so-called “rest mass” is in his
theory, that caused his theory to warp spacetime in
response; and that your treatment conserves
“KonserviertMasse”,and in response demands
KonserviertKoordinaten.

Is this just my mind focusing on trivialities? Or can
we repair my reasoning and discuss this from the
standpoint of physics instead “raw” mathematics?

Sincerely,
Bill Eshleman

Dear Bill,
Minkowski create the correct space-time connection as a 3+1 dimension manifold for the interaction fields which propagate with c. Within this manifold you can define several coordinate systems. The invariants which are given in the Minkowski space, or which you define as invariant, are independent from each coordinate system and invariants under Lorentz transformations: they do not change (are invariant) their values. Such invariant is for instant the distance between two points; the invariant distances connect space and time. Yor can define also an invariant action integral to determine the covariant equations of motions.
Invariant elementary gravitational charges are connected with elementary masses. The elementary masses are conserved. Coordinate are never conserved.
The gravitational masses of bodies, as addition of elementary masses regarding the signs of their elementary gravitational charges, are conserved. Einstein threw away the conserved gravitational masses. Furthermore, Einstein did not understand the “rest masses”, the inertial masses at the impulse p = 0. He could not interpret; he could not calculate the “rest masses”. Einstein thought that the gravitational mass is the same as the inertial mass for each bodies, but this is not true. Einstein did not understand how gravity works.
However, a difficulty arises: you can indeed assign invariants to elementary particles (such invariants are the elementary electric and gravitational charges, thus also elementary masses), but, you cannot assign a discrete point of the Minkowski space, with a precise velocity, to the elementary particles. That is the mean reason from the standpoint of physics.
Please Bill, continue our discussion in the Gravitation forum. It is interesting for other people too.
Sincerely,
Gyula Szász

[Gyula Szász]

Dear Bill,
my intention is mainly to state that gravity is caused by invariant elementary gravitational charges, similar to the cause of electromagnetism by invariant elementary gravitational charges. Both interaction fields propagate with c. The equations of motions of the fields and of the particles/bodies are to be derived with the use of invariants from a Lorentz-invariant action integral, integrated about a finite space-time domain. However, appropriate subsidiary conditions for the fields and for the particles must be considered. For the particles the subsidiary conditions are in accordance with particle number conservations.
Sincerely,
Gyula Szász

[Gyula Szász]

Dear Bill,
I can precisely say what the atomistic treatment of “reality” is:
The atomistic treatment describe matter and interacting fields correctly (for instant at each velocities of particles; no matter how large the velocities are) under the circumstance that neither the positions, nor the velocities are ever exactly known. For the description, only invariants are used (for instant invariant elementary charges, invariant terms in the Lagrange density, invariant action integral, etc.) in order to get covariant equations of motions; that means: equations of motions which are independent from coordinate systems (and are valid for all velocities of particles). That is exactly the dynamics of the universe. The laws of nature are non-deterministic, however causal.

You said “Invariance, covariance, form invariance… these are “buzz-words” to me…..”. I’m unhappy to hear this. You try to use treatments, either such within you don’t use invariants (entropic treatment), or you tries to use exactly known velocities (Lorentz factor with a mass, but which mass?). If you want, there is little connection between our both theoretical settings.

[Gyula Szász]

Dear Gyula,

Thank you for pasting the picture. I am still trying
to recover from your statement that the Lorentz factor
is an incorrect, special relativistic correction for
gravitational and electric charges. The most important
problem with the Lorentz factor is its infinity at the
speed of light. An infinity that is a curtain both to the
hypothetical black hole and to the microscopic quantum
world. Worlds that may just become visible by a
Feynman-type renormalization….maybe.

I am hoping that my splitting the Lorentz factor into a form
consisting of a purely additive part,

(addition of both ideal scalars and vectors),

and a purely, but speculated “relativistic” part………..will
come to my aid so that we can agree; always with my
primary intent to agree with your theory, “come hell or
high water.” And as of now, the extra factors of my
mathematical objects, I am interpreting as trying to
describe what the energies of bonds must approximate
to achieve longevity and stability. And it is those factors
that contribute to the infinities, whereas the first conservative
factor has no infinity at all.

My effort will usually have the goal that we agree in
some way that I can justify by manipulation of the counters
on my infinite sum and product identities. That is, it’s
treatment is flexible or even correctible on my part.
Sincerely,
Bill Eshleman

Dear Bill,
you must think on the equation of particles motion under the condition that neither the positions, no the velocities of particles are ever exactly known. We can assume that for stable particles the gravitational mass is equal to the inertial mass (they are not composed of any other particles), but the Lorentz factor contain the term v/c and which velocity will you input for v? Furthermore, the interactions between particles are non-conservative interactions. Anyhow, as the equations of motions are derived from an invariant action integral, the validity is given at each velocities smaller than c.
I state once more again the equation of stable particles motion is a covariant equation and it is composed of at least three covariant terms (the interaction term appears two times, the radiation term is suppressed). Covariant means “form invariant” again passive Lorentz transformations, and passive Lorentz transformations mean transformations of the coordinate system. The mass appears in two terms, in the kinetics part and in the interaction part through the gravitational interaction; in both cases the masses of stable particles enter as invariant masses.
If somebody tries to describe composed particles/bodies, in the kinetics part, the so called inertial mass occurs (which is not an invariant) in the interaction term appears the sum of gravitational charges as sum of gravitational masses. The inertial mass incorporates the bound energy. The uncertainty of velocities and the non-conservative interactions remain the same also for composed particles/bodies.
In my microscopic quantum word renormalization a la Feynman is not needed anymore and the particles cannot approach each other infinitely close and they cannot move with the velocity c. Furthermore, black holes do not exist; the space-time is not “deformed”.
Sincerely,

• This reply was modified 9 years, 2 months ago by Gyula Szász.

[Author]

Posts