see

Article Ground state of hydrogen as a zero-point-fluctuation-determined state

Dear Hadi Jabbar Alagealy Thank you for such a quick, positive response and posting H. Puthoff's paper.

I fully expected to be jumped on by QM's when I posted such a heretical question.

regards

Barry

No, since the hydrogen atom isn’t stable in the classical approximation. Zero point fluctuations don't have anything to do with it, since they describe fluctuations. They are, also, quantum effects. In fact one way to understand the stability of the hydrogen atom in quantum mechanics is that it's an integrable system (that was noticed by Pauli in 1926), whereas the classical system is not, when radiation is taken into account-which can be shown to be absent in the quantum system.

There is a physical explanation, in fact: The states that describe the radiation of a bound electron, are classical, they don’t exist, when quantum effects are taken into account. The classical potential is unbounded from below, the quantum potential isn’t. This was understood by Vladimir Fock in 1935: The Schrödinger equation on R^3, for the Coulomb potential, is equivalent to the Laplace equation, in momentum space, on the 3-sphere. It’s the property that the 3-sphere is a compact manifold, that ensures that the spectrum is discrete and bounded from below.

It’s a standard exercise in any course on electrodynamics to compute the time it would take for an electron rotating about a center, at the distance of a hydrogen atom, to lose energy due to radiation and fall in the center. The reason is that there exist states that can detect the radiation, in the classical aproximation, when Maxwell's equations are valid, that's what radiation means.

The reason for the stability is the Heisenberg principle, i.e. the electron reacts to confinement by increasing its kinetic energy. If you plot then the potential energy, this blows up negatively as -1/r when r->0 and goes to 0 from below when r->infinity. The kinetic energy goes as p^2->(h/r)^2 which goes to 0 as1/r^2 for r->infinity and blows up stronger than 1/r for r>0. The net result is a total energy function that:

-blow up positively for r->0

-goes to 0 from below (i.e. negative values) for r->infinity.

Therefore, it MUST have a minimum which is the ground state. But that requires the Heisenberg principle, or equivalently that the kinetic energy increases faster than 1/r as r->0

Dear Barry Gilbert,

There is a planetary model of the atom proposed by W. Weber in the second half of the nineteenth century.

See:

Book Assis, Andre Koch Torres; Wiederkehr, Karl Heinrich and Gudr...

and

Book "Wilhelm Weber Main Works on Electrodynamics Translated into...

regards

JP Chaib

Stam Nicolis i have tried to explain the ground state and other states of hydrogen using the Spacetime Wave theory. The electron is considered as a looped wave encircling the nucleus many times. Within each wavelength there is a small positive charge followed by a slightly larger negative charge so that successive loops tend to align with a half wavelength displacement.

Then the ground state is the lowest possible energy state for the electron.

Preprint The Unification of Physics (June 2021)

Richard

These details change the sources of Maxwell’s equations, not the final result, which is the transfer of energy from the sources to infinity and, therefore, imply that there can’t exist any state of finite energy, because all the energy will be radiated away

(Coils radiate and require power sources to keep them emitting radiation.)

The solution now is known so it suffices to read Fock’s paper, in German, https://link.springer.com/article/10.1007%2FBF01336904

or the two papers, in English, by Bander and Itzykson, that appeared in Rev. Mod. Phys. in 1966: https://inspirehep.net/literature/48744 and https://inspirehep.net/literature/48601

Dear Stam Nicolis I also am very familiar with Maxwell and will attempt to put my interpretation to the test.

A half wave dipole or infinitesimal dipole is a standard tool for us antenna folk. and as you may or may not know? in free space dipoles (half wave) at resonance have a predominantly resistive component to their complex impedance. This is simply a result of energy leaving the dipole and expanding into space. the 75 Ohm real impedance of the dipole is transformed into the 377 Ohm impedance of free space. If we prevent energy leaving the dipole by some means then the complex impedance becomes purely reactive. A six sided copper cavity will be considered: A half wave dipole inserted in such a cavity will indicate a purely reactive impedance and will accumulate energy until It forms a plasma and causes arcing or physical breakdown or reaches some form of equilibrium by dissipating energy into the small Ohmic loss in the copper cavity. The copper cavity effectively is a six faced mirror that appears to the dipole as a three dimensional infinite array of dipoles. If we model a hydrogen atom as a dipole in a universe with 10^80 hydrogen atom in it ( a 3D array), with each atom modelled as a dipole . Then all 10^80 dipoles are in equilibrium with their input impedances being pure reactive indicating zero energy leaving any dipole. We now have an EM field permeating the vacuum of space preventing orbiting electrons (modelled as dipoles) from crashing into their respective protons. We have classical atomic stability, what is more, we have a fluctuating field with an average energy = to half Planck's constant per mode as well as a classical explanation for Heisenbergs uncertainty principle and a real source of the mysterious zero point radiation, energy. fluctuations, quantum noise etc, etc.

Just a thought

Barry

Barry Gilbert See the paper titled The Hydrogen Bond. This provides a classical explanation for the energy states of the hydrogen atom.

Starting with the Balmer formula and following the Spacetime wave theory in which the entire mass/energy of the electron is wave energy given by E=hf the formula for the number of wavelengths in the loop is developed and this must be an integer for all values of n.

The nature of electric charge means that each wavelength of the electron is a small positive charge followed by a slightly larger negative charge. This means that successive passes around the nucleus will align with half a wavelength displacement. The is the cause of the stable states of the electron and explains the quantisation of the electron and the quantisation of light using only the classical ideas of the Spacetime Wave theory.

Preprint The Hydrogen Bond (June 2022)

Richard

NO, there is none. Classically an electron moving in a circular orbit has acceleration, which means it should radiate, lose energy and eventually spiral into the nucleous. The reason for this stability is the uncertainity principle, the electron reacts to confinement by increasing the spread of its velocities, hence momenta, hence energies. If you do a rough calculation and substitute h/r for the momentum in the kinetic energy term, you will get an expression E(r)=h^2/2mr^2-Ze^2/r. For small r, the first term dominates and gives infinity. For large r, the second term dominates and gives 0 from the negative side. Hence there must be a negative energy minimum which you can get by taking the derivative.