Let me try this again.
On Nov 15, 2008, at 8:14 PM, Jones Beene wrote:
Hi Robin,
> How then is the electron bound to the proton at all?
..consider: given that a circular "orbit", in QM terms, is an
illusion,
the lack of an apparent binding electrostatic attraction (even if
one did
not agree with Mills) does not necessarily prelude magnetic (or other)
attraction.
The magnetic field of an electron at the Bohr radius is a least
a factor of 1000 times stronger than the electric field, no ?
The electron waveform is co-centered with the proton so one way to
look at it is there is on average no net Coulomb force or magnetic
force on an orbital electron, though there is some degree of spin
coupling between the electron and proton in a hydrogen atom. The
question then becomes where do we actually observe the electron when
we sample for it and what size is it there, what forces should it
have on it there? These are questions having random variables in the
answers.
Looking at this from a Bohr point of view, i.e. the electron
essentially being a point charge with a magnetic moment orbiting the
proton at the Bohr radius, the Coulomb force is 8.239x10^-8 N and the
magnetic force is 1.0010x10^-14 N, so the Coulomb force is about 2
million times larger than the magnetic. The dipole magnetic force is
a 1/r^4 force, so can exceed the 1/r^2 Coulomb force at a small
enough radius, but for high energy nucleus plunging electrons only.
The Lorentz force on the electron due to moving through the proton's
magnetic field at the Bohr radius at orbital speed, is about
1.048x10^-14 N, which is roughly on the order of the attraction due
to an assumed perfect coupling of the magnetic moments.
As always, I'm a bit prone to computational errors, but the gist of
what I'm saying above is right.
Best regards,
Horace Heffner
http://www.mtaonline.net/~hheffner/
