On Dec 28, 2005, at 4:29 AM, [EMAIL PROTECTED] wrote:
The existence of zero point energy follows directly from the
uncertainty principle. I know Mills dismisses ZPE outright; so, I
gather he cares not for Heisenberg. I wonder what Randell thinks
happens at absolute zero?
It seems Heisenberg is indeed a problem with regard to scrunching the
electron wave packet into the tiny hydrino sized volumes, at least
from a Bohr perspective.
Summary from an old post:
I think this squeezing of the electron, i.e. rolling it up to fit
into a more compact volume, requires energy which is not available.
This energy is required by the Heisenberg principle. The smaller the
region a particle is confined to, the more the uncertainty on its
energy. If the uncertainty is large, that means you can sample such
particles and on average take a large amount of energy from them.
There must be an antecedent to the energy, or COE is violated. There
is no such antecedent, even on a statistical basis, in that the
energy is already all accounted for. The uncertainty grows to hugely
disproportionate values compared to the energy of the electron in its
orbital. The following calculation, from a simplified Bohr
perspective, hopefully demonstrates this fact.
The kinetic energy K of an electron in Bohr orbit radius r is given by:
K = q^2/((8Pi)(e0)(r)) = (1/2)(m)(v^2)
So speed v is:
v = (q^2/((4pi)(e0)(r)(m))^0.5
and momentum is thus ~ 1/r:
p = mv = ((m)(q^2)/(4(pi)e0(r))^0.5
Given that the radius is quantized to:
r = (n^2) ((h^2)(e0))/((pi)(q^2)(m)), for n = 1,2,3, ...
(or in Mills' case: n = 1/2,
1/3, ...)
we have:
v = [q^2/(2(e0)(h))] (1/n) = [q^2/(2(e0)(h)(n))]
instead of the Mills velocity:
v = [q^2/(2(e0)(h))] 1/(n^0.5), for n = 1/2,1/3,1/4, ...
Uncertainty of momentum (delta mv) for a particle (electron)
constrained by distance delta x is given by Heisenberg as:
delta mv = h/(2 Pi delta x)
but 2r acts as our delta x because the electron is contained within
the orbitsphere, so we have (substituting 2r for delta x in the above):
delta mv = h/(2 Pi [ 2 (n^2) ((h^2)(e0))/((pi)(q^2)(m)) ] )
delta mv = [h (q^2)(m)] / [4 (n^2)(h^2)(e0)]
delta mv = (q^2)(m) / [4 (n^2)(h)(e0)]
and an uncertainty in velocity delta v:
delta v = (q^2) / (4 (n^2)(h)(e0))
So now the question is how does delta v compare to v? That is to say
is the uncertainty on v small in comparison to v? To see, let's look
at the ratio:
(delta v)/v = [ (q^2) / (4 (n^2)(h)(e0)) ] / [q^2/(2(e0)(h)(n))]
(delta v)/v = [(2)(e0)(h)(n)] / [(4)(n^2)(h)(e0)]
(delta v)/v = [(2)(n)] / [4 (n^2)]
(delta v)/v = 1 / (2n)
However, with normal (non-Mills) orbitals, n is a whole number, so
delta v remains small with respect to v. There is not the large
imbalance which is the subject of discussion here, which occurs
because (1) n is a fraction, and (2) the exponents in the Mills
equations differ such that as n goes to increasingly lower values,
i.e. n = 1/x as x gets large, we have
delta v = (1/2) (x^0.5) v
for Mill's, and the resulting energy gets way out of whack in states
other than n=1.
The above relations for K, v, p remain valid for inner electrons in
the Bohr or Mills model (ignoring relativistic effects). With n = 1
in these inner states, it appears r is valid for either model, thus
either model works fine for the inner electrons. Neither then
violates Heisenberg. It is only the hypothesized (by Mills)
fractional quantum states that violate Heisenberg. At least that's
the way it appears to me.
I should also note that analysis from this perspective should have a
dramatic effect on feasible catalysts, energy availability, and
potential device design.
Horace Heffner
