[General] Potential energies of particles and photons

[Andrew Meulenberg]

Dear Martin,

Comments below:

On Sat, May 16, 2015 at 6:09 PM, Mark, Martin van der <
martin.van.der.mark at philips.com> wrote:

>  Dear andrew, you have a point but your reasoning is very crippled and
> hence wrong: in a closed, bound system it is not possible to say where the
> energy came from other than the system, balance must be maintained at all
> instant. Would you not be looking at a closed system, any bullshit can
> happen.
>

I am not looking at a closed system, I am looking into a closed system. How
many QM Hamiltonians include the photon? Without that, the system is not
closed. Are the answers bullshit? How many QM models, which use reduced
mass to tie the nucleus into the starting conditions, attempt to do the
same for the potential. Is it impossible to say where the mass resides in a
closed system? That mass includes most of the energy. Is the Coulomb
potential energy *in* the vacuum? In the *E*-field that fills the vacuum?
For the H atom, is it not reasonable to associate it with the 2 masses and
the 2 charges? Since the charges are equal, one might guess that the
potential energies are equally divided? But, the masses are not equal, so
could not the potential energies be unequally divided? Just because the
potential energy is measurable as a system quantity does not mean that it
cannot be partitioned. Because we can weigh an atom, that does not mean
that we cannot weight the nucleus. Just because photons have been
discovered doesn't mean that non-photonic Maxwellian radiation does not
exist. Physics is too often close-minded about options other than the
things that 'work'.


> I am writing a very detailed paper right now that a.o. contains this
> issue, it should be ready this week. It is one of the papers i will be
> presenting at the conference.
>

I look forward to it.

>
>  And, by the way, it is perhaps not wise to suggest that nobody has
> thought about whatever. You may simply not know about it, they may have
> come to the wrong conclusion, but very very likely someone somewhere has
> thought about it already. Nonetheless, i like to give you that you are
> thinking about relevant things indeed!
>

I thought about that statement and how it might be misinterpreted. I
decided that I meant that they never think to consider the potential energy
of the proton in their publications on the hydrogen atom. I should have
changed the statement. I am sure that some people have thought about it and
decided not to waste paper in adding the extra equations that do not change
their results. How often do you see the proton introduced by including its
mass in a reduced-mass form AND the radius changed to include both the
distance between charges and distance to the COM? The potential energy does
not change by introducing the proton if you use the distance between them
to calculate it. However, I believe it would, if instead you use the
electron distance to the COM. Which distance is used in the calculation for
the energies and which for the orbital radii?

When Richard commented that an electron lost total energy in its decay to a
lower level. I decided that I needed to raise the point. He had thought
about it; but, I believe he came to a wrong conclusion. Do you think not?
>From your comments, I would assume you think that he cannot predict either
way and so we are both wrong. Is this a valid assumption?

Andrew

>
>  Draft paper will be forwarded this week....comments welcome, final
> version before conference deadline.
>
>  Cheers, Martin
>
> Verstuurd vanaf mijn iPhone
>
> Op 16 mei 2015 om 12:55 heeft Andrew Meulenberg <mules333 at gmail.com> het
> volgende geschreven:
>
>
>
>   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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