🚀 go-pugleaf

I thought I sent this message last Friday, Jan 22.  Maybe I didn't send it.
 Maybe it got lost.  If this is a duplicate, my apologies.

Terry Bollinger: terry@asl.dl.nec.com  writes

> 
>PLANCK'S CONSTANT GOES BYE-BYE
> 
>Firstly:  If you choose to use a Bohr-like approximation of atomic orbitals
>and then draw out any one dimension of the "motion" of a ground state
>hydrogen electron in "phase space" (x=distance, y=momentum), you will find
>that it always encloses an area equal in size and units to Planck's constant.
> 
Terry continues with some historical arguments of why Bohr-type orbits
cannot be used to describe the hydrogen atom electron.

What does this have to do with the Mills/Farrell theory?  We propose that a
bound electron is a two-dimensional *surface* (and a free electron is a
plane wave)--it is not a particle that moves in an orbit or a probility
density function.  For example, a n=1 electron in the hydrogen atom looks
like a soap bubble of zero thickness at the Bohr radius, a0. The mass and
charge of the electron are uniformly distributed on the surface.  (The mass
and charge do move, BTW, and give rise to angular momentum.  I will not
elaborate here.)  The electric field *inside* the bubble or cavity--we call
it an electron orbitsphere--is the electric field of a bare proton.  The
electric field *outside* of the orbitsphere (cavity) is zero.  The
spherical cavity is of a particular size (volume = (4/3) pi r^3)) and can
absorb photons of certain frequencies (quantization).  An absorbed photon
does not disappear--it is trapped inside the cavity.  The "standing wave"
of the trapped photon sets up surface charge on the surface (described by
spherical harmonics) or, if you prefer, the trapped photon creates an
electric field (described by spherical harmonics) inside of the cavity that
opposes the electric field of the proton. The electron density on the
surface may be uniform (kind of like s-orbitals) or non-umiform (like p, d,
f, ... orbitals)--the resultant surface is described by the spherical
harmonics.  For any hydrogen atom where the electron state is described by
"n" the effective nuclear charge is 1/n.  The effective nuclear charge =
the nuclear charge (+1 for hydrogen)  +  the trapped photon charge. For
example, for n = 2 the proton charge is + 1, the photon charge is -1/2, the
effective nuclear charge is +1/2, and the radius is 2 a0; for n = 3, the
proton charge is -2/3, the effective nuclear charge is +1/3, and the radius
is 3 a0; for n = infinity, the proton charge is +1, the photon charge is
-1, the effective nuclear charge is zero, the radius is infinity * a0, and
the electron is ionized. The electron orbitsphere (bubble) gets larger as
the effective nuclear charge decreases, radius(n) = n*a0.  The n = 1 state
is the only state where there is no trapped photon.  When the atom absorbs
a photon the effective nuclear charge decreases and the atom gets bigger;
when the atom emits a photon the effective nuclear charge increases and the
atom gets smaller.

I know that I have not directly addressed the angular momentum issue here. 
But that requires some time and some abilities that this electronic mail
system does not provide.  Let me assure you however that we do not throw
out Planck's constant--quite the contrary.   
Clearly, this is not a Bohr atom.
 
>DRASTIC SYSTEM MASS REDUCTION
> 
>Ever notice that the nominal masses of quarks can be larger than that of
>particles that they form?  That's because they release a _lot_ of energy
>(mass) when they combine, so that the resulting bound system (e.g., a
>proton or neutron) may actually be lighter than its constituent parts.
> 
>If you allow suborbitals the same sort of thing will happen.  The farther
>you drop, the more energetic will be the photons that are released, and
>the less the final bound system will weigh.
> 

This is true.  So? 
 

>PROTONS AREN'T POINT PARTICLES
> 
>Now the aforementioned shrinking assumes that the proton is a point particle.
>It isn't, of course.  Assuming that the electron could drop at all, I would
>guess that before it got to the point of not having enough mass left to jump
>anymore it would find itself immersed in a small cloud of three very upset
>quarks.
 
We do not assume that the proton is a point particle.  In fact, we
calculate the proton radius:

r(proton) =  1.3214 x 10(-15) m

Furthermore, as the electron orbitsphere radius decreases the electron
moves faster and *gains* mass.

Consider a bare proton.  The electric field, ef is given by

    ef = e/(4 pi epsilon0 r^2)

where epsilon0 is the permittivity of vacuum.

This electric field represents *stored electric energy*.

A free electron (plane wave) comes by.  It is negative and it is attracted
to the proton. The electron forms a sphere around it--a minimum and
constant potential energy surface in a central force field.  (BTW, we now
have a two-dimensional particle that is curved (positive)--we have curved
space-time, gravity.)  The electron surface is a wave (lambda = 2 pi r),
has velocity (velocity = h/(2 pi mass(electron) r)) and the forces balance
(coulombic and centrifugal) at the Bohr radius, a0.  Recall that the
electric field *outside* of the orbitsphere is zero. Thus, the *proton*
electric field between r = infinity and r = a0 has been destroyed
(superposition of the proton's positive field and the field of a sphere of
-1 negative charge).  The stored energy of the proton's field is given by

     E(elec) =  (1/2)*epsilon0* integral(from infinity to a0) of (ef)^2 * 4
pi r^2 dr

             =  e^2/(8 * pi* epsilon0 * a0)  =  13.6 eV

Of course, it is the *electron's* field that "destroys" the proton's field.
 Thus 13.6 eV of the electron's field has been destroyed.  A grand total is
27.2 eV of stored energy is annihilated. 

Conservation of energy requires that the stored electric energy = the
kinetic energy of the electron, 13.6 eV.  Because this is a central force
problem the potential energy = - 2*kinetic energy or -27.2 eV.

Thus, as the electon goes from infinity to a0, 27.2 eV of electric field is
destroyed.  The electron's KE goes from 0 to 13.6 eV (and its mass
increases by 13.6 eV) and 13.6 eV is emitted as a photon.  On balance, the
atom is 13.6 eV lighter; it is the *proton* that is lighter.  (This
calculation ought to be close to the correct answer.) 

Now, assume for a moment that n = 1/2, 1/3, 1/4 ... states are possible. 
Granted that this is a *big* assumption.  As the orbitsphere gets smaller
and smaller, more and more  electric field is annihilated.  One-half of
this annihilation energy is emitted as radiation and the atom gets lighter
and lighter.  Simultaneously, the electron is moving faster (getting
heavier).  The net effect, of course, is that the atom is lighter.

Now, when the electron goes from r = infinity to r = r(proton) the
potential energy is

        V = 2 * (1/2) *epsilon0* integral(from infinity to r(proton) of
(electric field)^2 * 4 pi r^2 dr

             =  e^2/(4 * pi* epsilon0 * r(proton))  =  1.09 M eV

Thus, a maximum of 1.09 MeV of stored energy can be annihilated. One-half
of this will be radiated, 0.545 eV. The whole atom is 0.545 MeV lighter.  I
haven't worked out the masses of the individual particles--the electron and
the proton--but there is sufficent energy in the system to allow this to
happen.

The electron cannot get any smaller, under these conditions (bound to a
proton), than the radius of the proton.  About the only thing that can
happen now is electron capture.  The two mechanisms--going to lower quantum
states and electron capture--are quite different.  Nevertheless, going to
smaller lower quantum states should enhance electron capture. 
      
> 
>CONCLUSION
> 
>Suborbitals do severe damages to a lot pretty solid (understatement) work
>in both the theory and practice of quantum mechanics in general, and particle
>physics in particular.  Not to mention chemistry, electronics, materials
>physics, astronomy, and just about any other QM related field of physics or
>modern technology.  Because it appears to either dismiss or drastically
>alter the idea of Planck's constant, the suborbital hypothesis should give
>rather huge deviations in the predictions of every QM based physics theory.
>It should show up just about everywhere, literally, and not just in one
>limited class of chemical reactions.
> 
>If you wish to follow this hypothesis, I'm certainly not one to object, but
>please don't take the implications too casually.  This is the kind of change
>that cannot be introduced locally into _one_ theory and be expected to give
>the same results everywhere else.
> 
>I think some further exploration and quantification of the consequences
>of the suborbital hypothesis would be useful for anyone interested in this
>line of thinking.
> 
>                                Cheers,
>                                Terry

We are trying.  We intend to do a lot of damage (but not to Planck).  So
far, we maintain that we have not violated special relativity, general
relativity, Maxwell's equations, or the deBroglie relationship.  We have
violated the Bohr atom and the Schrodinger's mechanics.  Frankly, my dear,
we don't give a damn.  

John Farrell
Franklin & Marshall College