[General] Potential energies of particles and photons

Sat May 16 07:24:02 PDT 2015

Dear Andrew,
I have just read your email again, but now not on my Iphone but on my laptop. It is much easier that way.

You are completely missing the fact that you are using the Born-Oppenheimer APPROXIMATION to the atom structure. This is the beginning of all the confusion you load on yourself (and the reader) as a consequence.
http://en.wikipedia.org/wiki/Born_Oppenheimer

The proton is not infinitely massive, and in the H-atom it is rotating around/oscillating against the electron just as much, in terms of its momentum as is the electron against the proton. As I said before, the momenta of electron and proton are exactly equal, and so are their de Broglie wavelengths. The electron and proton are quantum mechanically in tune! The 2-body problem can be translated into a 1-body problem using the reduced mass.
http://en.wikipedia.org/wiki/Reduced_mass

The electric potential of the proton is meaningless against a neutral object, but against a charged object, also having such a potential (let the word sink in... "potential"... it is nothing until...) the two may repel or attract. It is not one or the other, it is both.

Any loss of the energy of a closed system comes from the system as a whole, and the system finds a new balance in doing so.

The way you are talking about the hydrogen atom violates Newton's 3rd  law.
http://en.wikipedia.org/wiki/Newton%27s_laws_of_motion

I hope this helps. Please stop confusing the poor students.
Cheers, Martin

Dr. Martin B. van der Mark
Principal Scientist, Minimally Invasive Healthcare

Philips Research Europe - Eindhoven
High Tech Campus, Building 34 (WB2.025)
Prof. Holstlaan 4
5656 AE  Eindhoven, The Netherlands
Tel: +31 40 2747548

From: General [mailto:general-bounces+martin.van.der.mark=philips.com at lists.natureoflightandparticles.org] On Behalf Of Andrew Meulenberg
Sent: zaterdag 16 mei 2015 12:55
To: Nature of Light and Particles - General Discussion; Andrew Meulenberg
Subject: [General] Potential energies of particles and photons

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<mailto: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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