<div dir="ltr"><div><br><br><div><div><div><div><div>Dear Richard and all,<br><br></div>Richard, you have
made the comment several times now about the total energy TE of the
decaying atomic electron decreasing. I have been trying to convince
people that it is the proton in a H atom (or a hypothetical source of
the potential energy, PE) that is losing total energy. Perhaps you all
can help me make an acceptable story. But first I have to convince you.<br><br></div>The
non-relativistic story begins with energy conservation: TE1 = KE1 +
PE1 = KE2 + PE2 + photon => TE2 = KE2 + PE2. The virial theorem for
stable orbits in a 1/r central potential gives delta KE = -delta PE/2.
Thus, delta TE = - photon energy = -delta KE = delta PE/2. => TE1
> TE2. So it seems clear that the electron total energy decreases but
the system energy (including the photon) is constant and energy is
conserved. However, let us do the same thing for the proton. (Nobody ever thinks to do this.)</div></div></div><br>The non-relativistic story for the proton is the same,
except now no photon is released and we can assume that KE1 = KE2 = 0:
TE1 = KE1 + PE1 and TE2 = KE2 + PE2 => deltaTE = - delta
PE. (The virial theorem is not applicable here at this level of
approximation.) The change in total proton energy is twice that of the
electron. Thus, the change in system TE (including the proton, electron and photon) is twice
times that of the photon energy. Energy is not conserved. Is this why no one ever includes the proton?<br><br>Well,
mathematical physicists don't include the proton because they assume a
central potential that has no features except for a single charge and an infinite energy and
mass. Since infinity minus any finite value is still infinity, the
change in potential energy of the central potential is zero and
conservation of energy is maintained. Thus, they never have to answer
the embarrassing question (which I asked as a freshman), "which body
provides the potential energy, the electron or the proton?" If pressed,
they could simply say, "what proton? The Hamiltonian does not include
one! Besides, if you did include one, the Coulomb potential stands
alone; it belongs to neither." Now if any of you believe this, you are
not the independent thinkers I had assumed.<br><br>Practical
classical physics provides an answer. It is based on freshman physics
definitions: "potential energy is the ability to do work;" and "work is
force times distance." Since to 1st order the proton does not move, the
electron does no work. Therefore, the potential energy must come from
the proton. <b>The known mass decrease of a radiating hydrogen atom
(from the loss of a photon) must come from the proton's energy, not that
of the electron.</b><br><br></div><div><b>Something that I have not yet worked out:</b><br></div><div>A potential problem with this story is that the potential energy of the electron can change, even tho it has done no work. How does one rationalize this in terms of the energy conservation expression TE = KE+ PE? If PE is only the ability to do work, that ability can change without actually doing work (of course, doing work can also change that ability). If PE is the ability and not actual energy, how does it fit into the energy-balance equation? It seems that relativity provides part of the answer in including the mass of the particle(s). It is necessary to make the mass be potential dependent to complete the story. Then the potential is not needed in the equation at all, except as part of the mass.and as a means of determining forces. However, the mass that changes is the proton's and that is not in the equation unless the whole system is defined.<br><br></div><div>It seems that the abbreviated version of the energy accounting has become imbedded in the method and people have even forgotten the fine print exists. The proper understanding is required to understand the physical means of electron-positron formation and annihilation.<br></div><div><br>Can anyone argue against this conclusion or tell a better story?<br><br></div>Andrew<br><div>____________________<br><br>On Fri, May 15, 2015 at 7:40 PM, Richard Gauthier <span dir="ltr"><<a href="mailto:richgauthier@gmail.com" target="_blank">richgauthier@gmail.com</a>></span> wrote:<br>Martin and John D and all,<div>
When an electron (charged photon) “falls" into a previously ionized
atom, the total energy of the electron decreases as it becomes bound to
the atom and gives off one or more photons as it drops from one atomic
energy level to another, but the charged photon’s (electron’s) average
kinetic energy and momentum increase as it goes into the negative
potential energy well of the atom. The charged photon’s (electron’s)
average de Broglie wavelength decreases as its momentum and kinetic
energy increase. The charged photon (electron) gives off an uncharged
photon each time it drops from one energy level of the atom to another,
as described by QM. With each new lower total energy, increased average
kinetic energy and decreased average de Broglie wavelength, the charged
photon creates a kind of resonance state (quantum wave eigenfunction)
throughout the atom corresponding to its particular energy eigenvalue.
The charged photon as it circulates is continually generating plane
waves corresponding to its energy. These plane waves from the charged
photon generate the charged photon’s (electron’s) de Broglie wavelength
and corresponding quantum wave functions along the helically circulating
charged photon's longitudinal direction of motion. The probability
density for detecting the electron (charged photon) is given by Psi*Psi
of its particular eigenfunction in the atom. The charged photon appears
to be spread out but when detected it is more localized (the resonant
eigenstate produced by its de Broglie wavelength is destroyed) and the
electron (charged photon) is back to being a non-resonant charged photon
(electron), until it creates a new resonant state (new eigenfunction).<br><br></div><div> Richard</div><br></div></div>