I can't get anything right! Sorry, sorry, another correction!
I previously mentioned the Heisenberg traps article has been updated
with a computation of minimum uncertainty of quark size based on the
measured temperature inside the Dy nucleus. The result was 1.10
x10^-14 m. The article has been further corrected to show there is a
more reasonable conclusion than the quark radius uncertainty estimate
given. Article is at:
<http://mtaonline.net/~hheffner/HeisenbergTraps.pdf>
The corrected part follows.
An example of a Heisenberg trap, and a powerful conduit to the zero
point field, may be the nucleus itself. Stable nuclei, like stable
atoms, do not radiate. However, they sustain a large kinetic energy
density. Consider the following article from the AIP:
"PHYSICS NEWS UPDATE
The American Institute of Physics Bulletin of Physics News
Number 443 August 16, 1999 by Phillip F. Schewe and Ben
Stein
NUCLEAR THERMOMETER. How hot is it inside the nucleus
of a dysprosium atom (element 66, abbreviated Dy) Temperature
is a statistical concept that normally applies to an ensemble of
many particles, such as air molecules or a gas of atoms kept in a
bottle. Inside a heavy nucleus, swarming with protons and
neutrons (collectively called nucleons) it's not so easy to define
temperature, owing to the many pairing and other inter-nucleon
interactions that take place, but it can be done. The nuclear
environment can be sampled by colliding nuclei together and then
carefully measuring the photons that fly out: high energy gamma
rays, in this case, rather than the visible and infrared photons that
come out of heated-up atomic gases. In this way, physicists at the
University of Oslo have deduced the temperature inside a Dy
nucleus (in effect, a gas of 162 nucleons) to be 6 billion K. It can
be said, therefore, that even in winter parts of Norway (very small
parts) remain quite warm. This is the first time a nuclear
temperature has been measured strictly on the basis of the spectrum
of gammas emitted. (E. Melby et al., Physical Review Letters,
tent. 30 August 1999; contact Magne Guttormsen,
[EMAIL PROTECTED], 011-47-2285-6460.)"
The nucleus itself may be an endless repository of kinetic energy
which can be tapped if a means to repetitively and frequently
kinetically interact with it can be found. Such a means might
include photon stimulation, interaction with energetic electrons, or
coupling and jiggling via a lattice. It seems reasonable to
conjecture that the jiggle of rapidly diffusing Li, D or T nuclei
might siphon off some of their nuclear heat energy via spin coupling
or EM coupling with close energetic free electrons, or possibly even
the lattice. Further, in electron catalyzed fusion, the nucleus
itself may assist the escape of the catalysing electron via coupling
to the nuclear heat, the nuclear kinetic energy. Once tapped, the
nucleus can subsequently replenish its heat from the ZPF.
The radius of a Dy nucleus can be estimated at
R_Dy = 1.4x10^-13 cm * A^(1/3)
= 1.4x10^-13 cm * (162)^(1/3)
= 7.6x10^-15 m.
The atomic weight of Dy is 162.5 AMU, which, at 1.66x10^-27 kg/AMU
gives a mass m of 2.7x10^-25 kg. Letting gives
delta KE = h^2 /(8 Pi^2 (2.7x10^-25 kg) (delta x)^2)
= h^2 / ((8 Pi^2 (2.7x10^-25 kg)) (7.6x10^-15
m)^2)
= 3.6x10^-16 J = 2200 eV.
At 11,600 Deg. K per eV, we have a minimum Dy temperature of 26
million Deg. K. This is way short of 6 billion degrees.
Given there are 162 nucleons in Dy, that is 486 quarks. The quarks
have an average mass of 5.58x10^-28 kg. Now we can aske the
question, how big do the quarks have to be to account for thhe
nuclear temperature? The temperature of 6 billion degrees is
equivalent to 5.15x10^5 eV per quark. We have:
5.15x10^5 eV = h^2 /(8 Pi^2 (5.58x10^-28 kg) (delta x)^2)
(8 Pi^2 (5.58x10^-28 kg) (delta x)^2) = h^2 /(5.15x10^5 eV)
(delta x)^2 = h^2 /((5.15x10^5 eV) (8 Pi^2 (5.58x10^-28 kg)))
(delta x)^2 = 1.208 m^2
delta x = 1.10 x10^-14 m
We thus have the minimum diameter of the quark as being about 10^-14
m. The proton diameter is given by some references as about 10^-15
meters. See:
<http://hypertextbook.com/facts/1999/YelenaMeskina.shtml>.
If the quarks have an uncertainty in position of 10^-14 m then that
fits OK with the quarks staying in the 10^-23 diameter of the Dy
atom, but does not fit with some estimates of the diameter of the
proton. The quarks jammed into protons may thus be even hotter than
6 billion degrees.
Alternatively, we can assume the unit of mass involved in sustaining
the heat is not the quarks but rather the 162 nucleons. We thus
have a mean mass of (2.7x10^-25 kg)/162 = 1.667x10^-27 kg, and:
5.15x10^5 eV = h^2 /(8 Pi^2 (1.667x10^-27 kg) (delta x)^2)
delta x = 6.36x10^-15 m
And this fits very nicely into the Dy or radius of 7.6x10^-15 m. It
thus may be reasonable to expect that the nucleons are the basic
unit of interaction with the zero point field.
Assuming the nucleons assume a temperature depending on the confining
radius, we can predict, given A the mass number, the nuclear
temperature T in kelvin for large nuclei by:
T = (11,600 degrees K/eV) (h^2 ) /
(8 Pi^2 (1.667x10^-27 kg) (1.4x10^-13 cm *
A^(1/3) )^2)
We can see T is proportional to A^(3/2), thus using Dy as a baseline
we have a rule of thumb:
T = K_temp * A^(3/2)
where:
K_temp = 6x10^9/(162^(3/2)) deg.K = 2.91 x 10^6 deg. K
and thus
T = (2.91 x 10^6 deg. K) * A^(3/2)
