Welcome › Forums › Gravitation › Atomistic Theory of Matter › Reply To: Atomistic Theory of Matter

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 4 years, 7 months ago by Gyula Szász.
- This reply was modified 4 years, 7 months ago by Gyula Szász.
- This reply was modified 4 years, 7 months ago by Gyula Szász.