Reply To: Einstein

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Gyula Szász

New Mass Concept in Physics

Einstein’s imagination about mass was disastrously; he could not clear up what mass in physics is. He has created the weak equivalence principle then he believed on the equivalence of the inertial and gravitational mass. He did not recognize that UFF is violated. Einstein has connected the relativistic mass with energy and proposed that mass can be annihilated and created. He stated the energy-mass-equivalence principle E = m∙c^2. At the end in his general relativity, he has thrown away the gravitational mass. The modern physics takes Einstein’s concept about mass as fundament, but it could not cleared up in the last 100 years wherefrom the masses of particles are coming. Nowadays, the Higgs-particle is taken to explain the mass of particles. The question of quantum gravity is further on unsolved.

If have broken with Einstein’s mass-concept.

First of all, I have recognized that the gravitational mass and the inertial mass are different. This is confirmed in the violation of the UFF. The gravitational mass is derived from the conserved gravitational charges, gi ={± g∙mi} with help of the invariant masses mP and me of proton and electron. The universal gravitational constant is G = g^2/4π. The gravitational charges of the stable elementary particles, e, p, P and E, generate the time dependent gravitation field. In the expression of inertial masses appears the bound energy of particles, beside the elementary masses, mP and me. While the gravitational mass of a system, mg, remain always unchanged, the inertial mass, mi, is changing. We weight gravitational masses with balances; however, the inertial masses appear in the equation of motion under the influence of interactions. If the interaction is pure gravitation, in the equation of motion appear the relation of both masses

mg(material)/mi(material) = 1 + Delta(material).

The mass defect, Delta(material), is between – 0.109% (hydrogen atom) and + 0.784% (56Fe isotope).

Supposed, besides the gravitation only the electromagnetism exist as interaction between particles and both interactions propagate with c, an action integral is constructed in finite ranges of Minkowski space and with Lorentz-invariant Lagrange density, from which the Lorentz-invariant equation of the fields and the particles could be derived. The electromagnetism is also caused through conserved elementary charges. This formulation is valid for all possible velocities of particles. The elementary masses mP and me can be neither annihilated, nor created. For the most physicists is unusual that the elementary gravitational charges have two signs; that means repulsive gravitational interaction also exists. It is the case if two gravitational charges, Gi and Gj, have different signs. The static gravitational force is

F(rij) = – Gi∙Gj∙rij/4∙π∙rij^3.

Newton’s force equation

F(rij) = – G∙Mj∙mj rij/4∙π∙rij^3,

is valid only if the masse Mj and mj are connected to gravitational charges with the same sign and in this case the masses are connected to the conserved gravitational masses of bodies.

I’m going on to explain the new mass concept.
The gravitational mass mg of a proton-electron (P,e) system is

mg(P,e) = mP – me = proton mass – electron mass = gravitational mass of hydrogen atom.

The proton-electron system could have different bound energies, 13.8 eV, 2.04 MeV and even E(bound:P,e) = (mP+me)∙c^2 depending on the values of Lagrange multipliers. These bound energies correspond to states which have different inertial masses

mi(P,e) = mP + me – E(bound:P,e)/c^2.

The electron-positron, νe = (e,p) system has the gravitational mass zero,

mg(e,p) = 0.

This system contains the positronium and the electron-neutrino, νe = (e,p), depending on the bound energy. Electron and positron pair can neither annihilate, nor can be created.

An instable neutron N =(P,e,p,e) has the calculated gravitational mass

mg(N) = mP – me = 937.761 0821 MeV/c^2,

which is the same as that of the hydrogen atom. The inertial mass of instable neutron is measured in nuclear physics

mi(N) = 939.565 4133(58) MeV/c^2.

The gravitational mass of an electric neutral isotope with the mass number A is A time the hydrogen atoms gravitational mass

mg(isotope;A) = A∙( mP – me) = A∙937.761 0821 MeV/c^2.

The natural mass unit of electric neutral matter is

Mass Unit = 937.761 0821 MeV/c^2.

If we weight matter with a balance, we weight a multiple of this Mass Unit.

In nuclei there are also positrons present. An electric neutral isotope consists of A = NP protons, Np positrons and (NP + Np) electrons. The inertial mass of an isotope, mi(A,Z), which is measured in mass spectrometers, is different from its gravitational mass

mi(isotope;A,Z) = A∙( mP + me) + 2∙Np∙me – E(bound;A,Z)/c^2,

E(bound;A,Z) is the bound energy of all particles in the isotope. From the (NP + Np) electrons are Z bound in the atomic shells and (A – Z + Np) in the nuclei.

Notice: The nuclear physicists calculate the bound energy of nuclei incorrect. They take (bound;A,Z)/c2 = Z∙ mP + N∙ mi(N) – mi(isotope;A,Z).

In the nucleus of an isotope, eltons (E) are obviously are not present then free eltons are not detected at the decays of nuclei. In particle physics elton is also called “antiproton”. The proton-neutrino, νP = (P,E), has the gravitational mass zero. For both neutrinos the inertial mass are also zero mi(neutrino) = 0, this condition define the neutrino states. At the neutrino decays of nuclei, we must pay attention; the neutrino could be electron-neutrino of proton-neutrino.

In particle physics the calculation of gravitational masses of particles is simple. I give some examples. The charged muons, μ+ and μ-, are composed of five elementary particles

μ+ = (p,P,e,p,E), μ- = (e,P,e,p,E).

The decays are

μ+ → p + νe + νP, μ- → e + νe + νP.

The gravitational masses of μ± are that of the (invariant) mass of an electron

mg(μ+) = mg(μ-) = me = 0.510998910(13) MeV/c^2.

The electric neutral four-particle system, (P,e,p,E), is most possible classified as a third kind of neutrino in particle physics, as the tau-neutrino ντ then the gravitational mass of this particle system is zero. The name electric neutral muon, μ0, would also fit to this particle. The assigned names, tau-neutrino or electric neutral muon, depends of their bound energy. The inertial mass of the charged muon is measured to be

mi(μ+) = mi(μ-) = 2∙mP + 3∙me – E(bound; μ±)/c^2 = 105.6583715(35) MeV/c^2.

The bound energy is remarkable great, compared to the bound energies of the nuclei

E(bound; μ±)/c^2 = 2∙mP + 3∙me – mi(μ±) = 1772.418787 MeV/c^2.

We continue the discussion of the masses of particles with those of the charged and neutral pions. The pions have the compositions

π+ = (p, P,2e,2p,E), π- = (e, P,2e,2p,E),
π0 = (P,2e,2p,E).

The gravitational masses of charged pions are again equal to the mass of electron, and that of the neutral pion is zero. It is not surprising that the inertial masses of charged pions

mi(π±) = 2∙mP + 5∙me – E(bound;π±)/c^2
= 139.57018(35) MeV/c^2

are noticeable different from the inertial mass of the neutral pion

mi(π0) = 2∙mP + 4∙me – E(bound;π0)/c^2
= 134.9766(6)MeV/c2,

since the bound energies and the compositions are different. The decays of charged pions are known

π± → μ ± + νe.

However, the often discussed decay of neutral pion

π0 → 2 γ

is physically impossible. Also events with the production and subsequent decay of “a new particle pair”:

p + e → τ+ + τ− → p + μ- + 4 neutrino or → e + μ+ + 4 neutrino

are also physically impossible.

I could continue the discussion of the masses of additional particles, those of kaons and baryons, Λ, Σ, Ξ, etc., but, at any time I would have the problem that I must have exact knowledge of the number of (e,p) and (P,E) in the considered particles. For this, model calculations could help, then I have a variation principle for the calculation of bound energies with the Lagrange multipliers, as h0 = h/387 and h, with the invariant masses, mP and me and the elementary electric charge, e, of the stable elementary particles e, p, P and E. The variation principle is able to determine stable and instable particle states. At instable states the simultaneous determination of the bound energy and life time is possible.

Obviously, we don’t need the weak- and the strong-interactions, we have alone the electromagnetic interaction. Furthermore, we don’t need gluons, partons, or whatever quarks. The stable elementary particles are not composed of quarks. And we don’t need obviously the Higgs-particles for the explanation of the masses of particles. The problem of quantum gravitation is also solved with the implementation of elementary gravitational charges.

Gyula I. Szász

  • This reply was modified 7 years, 11 months ago by Gyula Szász.