On Dec 27, 2011, at 9:05 AM, pagnu...@htdconnect.com wrote:
Horace,
Thanks for the comment.
What is needed are some toy models with some simple simulations.
I will check out your theory.
Do you believe any "new physics" is required
- or does standard QM suffice?
I am getting pretty boggled by the complexity of it all.
LP
I should have noted that my application of zero point energy to
nuclear reactions is possibly "new" physics, though the concepts
applied are not new at all, i.e. Casimir force, uncertainty energy,
Fermi pressure, etc. What is new is the concept of the energetic
trapping of electrons in heavy nuclei. This concept requires no new
physics I think, just an understanding of a simple mechanism by which
a net zero charge ensemble can enter the nucleus via tunneling and a
net magnetic energy gain. That this is feasible is to me self evident.
The basic concept behind the deflated state is simple conventional
physics - namely that the magnetic force, a 1/r^4 force, becomes
larger than the 1/r^2 Coulomb force at close radii.
Feynman and Wheeler computed there is enough energy in the vacuum of
a light bulb to boil the oceans of the earth.
The energy of the zero point field is vast. More specifically, in:
http://www.earthtech.org/publications/PRAv49_678.pdf
Haisch, Reuda, and Puthoff give the spectral energy density rho by
(virtual) photon angular frequency omega as:
rho(omega) d omega = (h_bar/(2 Pi^2 c^3)) omega^3 d omega
Integrating for omega = 0 to omega1 obtains energy density E_rho:
E_rho(omega1) = (h_bar/(8 Pi^2 c^3))) omega1^4
E_rho(omega1) = 4.95707 kg s m^-1 omega1^4
Using the Planck angular frequency, 1.85487x10^43 s^-1, for omega1 we
have energy density:
E_rho = 5.86784x10^111 J/m^3
Note that this also represents a pressure to the vacuum of
5.8678x10^111 Pa. Using m = E/c^2 that also represents
M_rho = E_rho/c^2 = 6.5289x10^94 kg/m^3
This, to me, is and indication that the photons of the zero point
field are virtual and have no mass.
Obtaining use of the vacuum pressure requires exclusion of the
virtual photons from a given volume, the source of the Casimir
force. Uncertainty energy and the Casimir force can be viewed as
different sides of the same coin.
There is a vast pool of energy all around us. All we have to do is
figure out a way to tap it. The system is then no longer closed, and
it is no longer a zero sum game. I think close electron-nucleus
interaction may be the way to tap this energy. Such close
interactions can be induced with low energies using various means,
including resonant effects.
In 2007 I first estimated the magnetic binding force on a stable
electron orbital on deuterium, one in which the electron deBroglie
wavelength was less than the orbital radius:
http://www.mtaonline.net/~hheffner/FusionSpreadDualRel.pdf
This was referenced in my first article on deflation fusion,
published in I.E. The magnetic force was 4 orders of magnitude
larger than the Coulomb force.
It seems to me that Jefimenko has it right when he shows magnetism is
just the relativistic effects, the effects of retardation of the
action of the Coulomb field. It is retardation effects due to
virtual photon velocity. If so, then this is a key link into the
energy sea of the zero point field. This link may provide some of
the energy of LENR. The Casimir force may even provide some insight
to the mechanisms of weak reactions.
The energy of a particle is constrained by Heisenberg to be
delta KE ~= k2 / (delta x)^2
The Coulomb potential Uc is:
Uc = k1 / x
so Heisenberg energy overcomes the Coulomb potential at some radius.
The magnetic potential Um is given by:
Um = k3 / x^3
Here k1, k2 and k3 are constants.
Therefore, given that k2 is positive and k1 and k3 are negative, as
distance between an electron and nucleus goes to zero, uncertainty
energy opposes and overcomes the Coulomb potential, but eventually
the magnetic potential opposes and overcomes the uncertainty energy.
All that is needed for a small magnetically bound hydrogen state to
be entered is (1) wavefunction collapse, (2) tunneling of the
electron to the magnetically bound state, or (3) the electron to act
like a point particle in the wavefunction.
If itinerant electrons approach an absorbed hydrogen nucleus, with
zero angular momentum, then a direct pass through of the nucleus will
occur. It is notable here that angular momentum is quantized, so an
approaching electron either will have some quantum of orbital angular
momentum, or exactly zero orbital angular momentum, thus directly
passing through the nucleus. Further, the electron, if t has zero
orbital angular momentum, and if not magnetically captured, or
otherwise perturbed, can oscillate back and forth through the
nucleus, increasing the exposure rate to magnetic capture.
When considering a zero angular momentum orbital one dimensional
values can be applied for uncertainty energy. The zero point energy
of a particle in a box with sides of length L is given by:
Ezp = (h^2/(8 m)) (1/L^2)
The zero point energy of a particle confined to a box with radius r
is thus approximated by:
Ezp ~= (h^2/(8 m)) (1/(2*r)^2)
Ezp ~= (h^2/(32 m)) (1/r^2)
An alternate view on uncertainty energy can be obtained by starting
with the uncertainty on momentum, given a constraint x in position
location, provided by Heisenberg as
(delta p) = h/(2 Pi (x))
but since:
KE = (1/2) m v^2 = (1/(2 m)) (mv)^2
delta KE = (1/(2 m)) (h/ (2 Pi (delta x)))^2
delta KE = (h^2/(8 Pi^2 m)) (1/(delta x)^2)
In motion constrained to (delta x) = 2 * r we have:
delta KE = (h^2/(8 Pi^2 m)) (1/(delta x)^2)
delta KE = (h^2/(32 Pi^2 m)) (1/r^2)
To be conservative in computing an electron potential Ezp will be
used, because it is larger by a factor Pi^2. We now are using
Ezp ~= k2 / x^2
Uc = k1 / x
Um = k3 / x^3
where, for the electron:
K2 = h^2/(32 Pi^2 m) = 1.5261x10^-39 J m^2
and for the electron-deuteron:
K1 = -2*(1/(4*Pi*e0))*q^2 = -4.61416x10^-28 J m
K3 = -mu0*muD*muB/(2*Pi) = -8.03267x10^-57 J m^3
when using constants:
e0 = 8.85419x10^-12 F/m
q = 1.602177x10^-19 coul
mu0 = 1.25664x10^-6 m/A^2
muB = 9.27402x10^-24 A m^2
muD = 4.33074x10^-27 A m^2
We now have the potential function:
Ut = k1/x + k2/x^2 + k3/x^3
where x is the distance between electron and deuteron.
Some numbers of interest are the points at which the uncertainty
potential matches the Coulomb plus magnetic potential. In other
words, where:
-k1/x - k3/x^3 = k2/x^2
(4.61416x10^-28 J m)/x +
(8.03267x10^-57 J m^3)/x^3 =
(1.5261x10^-39 J m^2)/x^2
The solutions are x=3.3074x10^-12 m, and x=5.2635x10^-18 m. This
means to find a correct solution the relativistic mass must be used
for the electron in k2. This will result in a much larger value for
the smaller solution for x.
This so far resolves that the tunneling distance for the electron
from normal state to deflated state is limited to no more than
3.3x10^-12 m, far less than the tunneling distance in Josephson
junctions. Given the potential plus kinetic energy of the system is
constant, i.e that no radiation is involved, the deflated state is a
degenerate state of the orbital, thus it can be expected the electron
wavefunction can have a dual existence embodying a normal state with
probability p, and a magnetically bound small state, with probability
1-p.
For some miscellaneous thoughts on nuclear zero point energy (ZPE)
tapping, and some excellent references, see:
http://mtaonline.net/~hheffner/NuclearZPEtapping.pdf
Best regards,
Horace Heffner
http://www.mtaonline.net/~hheffner/
