[General] Potential energies of particles and photons

[Andrew Meulenberg]

Dear Richard and all,

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.

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.)

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?

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.

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. *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.*

*Something that I have not yet worked out:*
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.

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.

Can anyone argue against this conclusion or tell a better story?

Andrew
____________________

On Fri, May 15, 2015 at 7:40 PM, Richard Gauthier <richgauthier at gmail.com>
wrote:
Martin and John D and all,
   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).

    Richard
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