Essentially you have asked two questions here.

• The overall spin can only be calculated by resolving the full set of couplings from a Schrödinger equation that involves the spin-something-coupling Hamiltonian contributions for all atoms involved. So you don't just need the overall particle parameters r, v, m, you need r, p, m for every atom.
• One additional thing that is always a "killer" in quantum mechanical calculations is the wavefunction phase which can never be measured absolutely and only be attributed for by relative values. Therefore, you have a differential equation with an underdefined initial value problem.

Dear Dr.Jürgen Weippert

Thank you for your reply.

Almost you have given pointed reply.

But Still I need clarity on 2nd point

"The overall spin can only be calculated by resolving the full set of couplings from a Schrödinger equation that involves the spin-something-coupling Hamiltonian contributions for all atoms involved. So you don't just need the overall particle parameters r, v, m, you need r, p, m for every atom".

My question is not about atom. It is about a particle without charge but having a mass. we know v, p and m only. v is equivalent linear velocity of rotation about its own axis.

please let me know whether it is possible or not

The reason why I asked this question is-

I theoretically found that Higgs boson is nothing but a massive photon. Its mass is same. But Higgs boson is a spin 0 particle and photon is mass less and with spin 1. I want to know the difference by studying further..

OK, now I understand your intention. No, for an elementary particle you can't just conclude the spin from mass and velocity, you actually have to solve the fundamental equation, let it be the Dirac equation or something more advanced, and see what sets come out. All quarks and leptons are spin 1/2, but their masses are completely different.

If there were simpler rules, you would have found them.

Jürgen Weippert ,

Thank you for your reply.

To answer the question “Is there any relation between a particles rotation(rotation on its own axis)and its quantum spin ?” is necessary before to understand – what is “particle”?; rotates it or not, if rotates, then how?, and not only, though.

The scientific answers can be, and are, given only in the Shevchenko-Tokarevsky’s Planck scale informational physical model, in this case it is enough to read

https://www.researchgate.net/publication/354418793_The_Informational_Conception_and_the_Base_of_Physics

Besides that was discussed on RG already, see, e.g., SS posts in the thread https://www.researchgate.net/post/No9_Is_the_spin_of_an_electron_really_spin2/1

, pages 1,2,3

To read SS post in

https://www.researchgate.net/post/H_denotes_the_constant_ratio_E_f_Ehf_is_it_possible_that_h_has_an_equation_both_without_E_and_f/2 it is useful as well.

Cheers

@ Dr. Sergey Shevchenko,

To answer the question “Is there any relation between a particles rotation(rotation on its own axis)and its quantum spin ?” is necessary before to understand – what is “particle”?; rotates it or not, if rotates, then how?, and not only, though.

1.This answer is irrelevant to my question. I found a particle with rotation around its own axis. I asked the question whether it is possible to interpret in quantum spin.

Somebody confirmed 'No'.

I agreed with it because quantum mechanics developed the spin concept in a specific manner. That we can not deny.

I will explain in my next paper how a massive particle can be described with rotation classically. Then how it can be treated as a quantum particle which even can obey relativity. It seems for me the same can be interpreted with quantum spin. For that reason I asked this question. Not to learn about particle and its rotation(which is classical)in quantum mechanics.

Thank you for your reply.

2. I have read the RG threads you have provided for reference.

In my view, as some body replied Planck constant h is a practical output.

It separates continuity to quantum. So wherever we apply quantum mechanics there h exists. As per my works this h is also not constant it depends on space time in which we are considering. Our four dimensional space time in which velocity of light is the signal for transformation there the value of this h is constant. Otherwise this h also will change. Further more in a particular case h will not exist also. There even time is also not quantized. It is a continuous flow of every thing. There quantum mechanics is not applicable(since there is no question of probability).

3. A theoretical explanation for reduced Planck constant(why it is required)has been provided in the following paper :

Please refer equation(46) and two three sentences below that.

It gives a theoretical explanation why reduced h is required in place of h

Plank Scale with Siva’s Constant “K”—A New Road to Grand Unification.

Journal of Modern Physics , 9, 1179-1194.

https://doi.org/10.4236/jmp.2018.96071

Thanking you.

The concept of "spin" was developed in the early stages of Quantum Mechanics development in an attempt to explain why electrons stabilize only in pairs in electronic orbitals in atoms. Once two electrons are paired in an atom, no more electrons can occupy this orbital.

Since the Schrödinger equation directly explains and measures only the momentum energy of an electron, they were then assumed to rotate perpendicularly to their expected direction of motion. Since only clockwise and counterclockwise rotations are possible, it was then assumed that only an electron rotating counterclockwise can pair up with a clockwise rotating electron to fill an orbital. Clockwise rotation was assigned value + ½, and counterclockwise rotation was assigned value - ½.

But we know better now that it is understood that electrons are electromagnetic in nature, and consequently each have a local magnetic field, meaning that they interact as small magnets.

So pairing is more logically explained by anti-parallel magnetic alignment of the magnetic fields of two electrons. Like poles repel (parallel orientation, corresponding to two + ½ particles or two – ½ particles), and unlike poles attract (antiparallel orientation, corresponding to a pair of + ½ - ½ particles).

My answer is probably not correct, but if I calculate the spin of the Earth with r=6400km, mass=6E24kg, Inertia=8E37kgm2 around the spinning axis (mass density distribution inside the earth is needed to get this value), then, s=I w~5E67 h/(2pi), where w is the angular velocity, h the Planck const., pi=3.14.