[AntonioLao]

What are the properties that make two given elementary particles identical? Classical mechanics gives mass and charge. In addition, quantum mechanics gives energy and spin. Energy and spin are both quantized in theory and in practice (verified by experiments). However, the extension of mass and charge into the quantum domain caused insurmountable problems for theorists but not for experimentalist. There are still no theoretical bases for the quantization of mass and electric charge from first principles. Although the minimum experimental mass value of the lightest fermion is still undetermined when talking about the electron neutrino, the experimental mass of the electron (supposedly 2nd lightest fermion) is experimentally determined as 9.11x10^{-31} kilogram. The electric charge of the electron is experimentally determined as -1.6x10^{-19} coulomb. As already mentioned these values have no theoretical bases for their validity. Their theoretical subtleties have yet to be discovered.

On the other hand, bosons likewise possess quantized energy and spin but no quantized mass or charge. However, the minimum mass value is set at zero for many bosons for example: photon, gluon, and graviton. All three also have zero electric charge, although gluons do have color charge. To be continued..

[Guille]

Can electric charge have motion? That is, can it leave a particle and go to another particle/place? And mass? No. This is what I see important. Energy can move everywhere, and leave. And so can spin, change. Does this mean something?

[AntonioLao]

Originally Posted by GUILLE Can electric charge have motion?

Yes. The electron jumps from orbital to orbitals in the chemical formations of elements. Their displacements in conductors are also responsible for the transmission of electrical currents and their vacancy in solid states was the cause of semiconductivity for transitors.

[AntonioLao]

All the elementary particles share one common relativistic equation relating their square of energy to their square of linear momentum and their square of rest mass given by E=cp+mc. For the case of the photon, since its rest mass is theoretically zero, the equation becomes E=cp or p=E/c which says that the linear momentum of a photon is its energy divided by its speed in vacuum. This gives two lines of questioning which hopefully will allow the conceptualization for time independence. more to follow..

[dleviwing]

Antonio;
It may be easier to simply say that since the rest mass is assumed as zero, the momentum factor (mass*velocity) "p" becomes zero and thus the cp potion becomes zero leaving E=mc. Take the squareroot and E=mc

[AntonioLao]

Originally Posted by dleviwing the momentum factor (mass*velocity) "p" becomes zero

Strictly speaking, the Hamiltonian formalism does not allow the separation of mass and velocity from the momentum. On the other hand the mass in p is really relativistic mass or mass of motion. Furthermore in the Lagrangian formalism, momentum is replaced by velocity which imply mass unity. I don't fully understand them therefore I could be mistaken.

[AntonioLao]

First, some preliminary questions: the linear momentum, is it a scalar or a vector? Under what conditions, it is appropriately a scalar, a vector? In differential geometry, [∂p]= ∂p▪∂p=[∂p]. But this does not indicate what the partial differentiation with respect of, time or space? If the partial is taken with respect to time the result is a differential force ∂F, with respect to a metric space ∂r the result is the infinitesimal change in mass density per unit volume or gradient of a mass density per unit volume. But what is the physical meaning of a volume, V, such that V=1, quantum of volume? Or the harder question V=0? For the harder problem, there is the divergence theorem of vector calculus. For the easier there still no satisfactory answer. If only the absolute value of p is considered then p=E/c can have two possible real solutions p=±E/c. If both the right-hand side (RHS) and the left-hand side (LHS) multiplied by E then the equation does not changed, pE=±E/c. This is just a simple scalar form of Einstein’s field equations. Where pE represents the scalar of the energy momentum tensor divided by c, and =±E/c represents the scalar form of positive and negative tensor curvature divided by the speed of light. Nevertheless if the momentum remains on the LHS by itself, while the numerator and denominator of the RHS are both multiplied by E then p=E/cE. This is a plausible redefinition for the absolute value of linear momentum. The numerator E is responsible for time independence. The term cE leads to a definition of continuous space. The term E is defined as the double integral of space-time quanta whose vector differential form is [∂E]=[∂F(α)×∂r(α)]▪[∂F(β)×∂r(β)], where α and β are indices for the binary interaction of two neighboring space-time points. The terms ∂F(α) and ∂F(β) are differential primary forces. The terms ∂r(α) and ∂r(β) are differential metric spaces. Then E=∫∫[∂E]=∫∫[∂F(α)×∂r(α)]▪[∂F(β)×∂r(β)]. However, expanding by Lagrange’s identity gives two distinct integrands as {[∂F(α)▪∂r(β)][∂F(β)▪∂r(α)]- [∂F(α)▪∂F(β)][∂r(α)▪∂r(β)]} and {[∂F(α)▪∂F(β)][∂r(α)▪∂r(β)]- [∂F(α)▪∂r(β)][∂F(β)▪∂r(α)]} for ∂H(+) and ∂H(-). Therefore E=∫∫∂H(+) + ∫∫∂H(-). If space-time have equal number of ∂H(+) and ∂H(-) then E=0. This is necessary and sufficient condition for vanishing linear momentum of the vacuum. Furthermore it does not depend on the energy of the vacuum or the value of the speed of light. That is to say that space-time can still have energy content but zero total linear momentum not necessarily zero total angular momentum although quantized angular momenta are localized by primary forces and previously demonstrated to be mass independence. However if the numbers of ∂H(+) and ∂H(-) are not globally symmetric then the density of ∂H(+) and ∂H(-) cannot be homogeneous nor isotropic. Nevertheless the enumerable groupings of ∂H(+) and ∂H(-) allow emergence of local symmetry for the formation of elementary particles, fermions (quarks and leptons) and bosons. These developments can be extended by using Hadamard matrices for algebraic computations such as determining the mass ratio of proton and electron.

Since the time parameter does not appear anywhere in the differential form [∂E(α,β)], its double integral is time independent. However, at the infinitesimal domain the local symmetry is subjected to the uncertainty principle such that the product of conjugate variables for virtual energy variation ∆E and virtual time variation ∆t is always greater or equal to Planck’s constant ћ, ∆E∆t≥ћ, as limit of measurability.

[Guille]

By 'qantized angular momenta' do you mean that angular momentum has qaunta? If so, what do you mean by this?

[AntonioLao]

Originally Posted by GUILLE If so, what do you mean by this?

Planck's constant is the quantum of action which is really angular momentum. Angular momentum is the product of mass, velocity, and distance. In Hamiltonian formalism, mass and velocity cannot be separated. In Lagrangian Formalism, the mass is given the value of 1. The generalized coordinates are momenta and position for Hamiltonian and velocity and position for Lagrangian.
But for mass independence of primary forces only a2▪r2=c makes any sense so that the quantization is derived when there is an integer n such that the next a2 is just na1 but the r2 will becomes r2/n.

[Guille]

Originally Posted by AntonioLao Planck's constant is the quantum of action which is really angular momentum. Angular momentum is the product of mass, velocity, and distance. In Hamiltonian formalism, mass and velocity cannot be separated. In Lagrangian Formalism, the mass is given the value of 1. The generalized coordinates are momenta and position for Hamiltonian and velocity and position for Lagrangian. But for mass independence of primary forces only a2▪r2=c makes any sense so that the quantization is derived when there is an integer n such that the next a2 is just na1 but the r2 will becomes r2/n.

In lagrangian formalism, given that there is independence of primary forces, could you substitude 'angular momentum is the product of velocity and distance' (as mass is 1) by '____________ is the product of E/M and E'? If you can't, why? If you can, then what is the ________ part, is it still angular momentum? Moreover, could the lagrangian ever describe gravity (I believe not as in it mass can't be used in mv=p therefore also there is no ma=F for this formalism)?

More comments comming.