Discussion:
Fusion Reaction Rates Between Deuterino Molecule Nuclei
Itzhak Shechtman
2002-03-20 11:09:27 UTC
Hi,

In a deuterino molecule (i.e. a molecule consisting of 2 deuterino atoms)
with reduced distance between the nuclei, the probability of fusion
reaction between them due to their closer proximity can be many orders of
magnitude larger than in the normal deuterium molecule.

Results are given below of a calculation using the expression given by
J.D.Jackson [Phys. Rev.,106(1957),330] for the fusion reaction rate, P
[sec**-1], between the nuclei of a deuterium molecule:

P = A/(4 pi a**3) * alpha/(2 pi) * Exp(-lambda)

where A is the reaction constant(in cm**3/sec), proportional to the
reaction cross section,sigma; a is the "ground-state" radius of the atom;
alpha is the harmonic oscillator parameter in units of a**-2(appearing in
the vibrational ground-state wave function as Exp[-1/2alpha(x-xsub0)**2],
where (x-xsub0) is the separation from equilibrium in units of a); lambda
is the Coulomb barrier penetration factor for the deuteron pair.

If we denote by B the ratio of the Bohr radius to the radius of the
electron in the dehydrino atom, the above expression raduces to:

P ~ 2.27*10**8 * B**5/2 * Exp[-109.34/B**1/2] .

On the basis of the deuteron-deuteron penetration factor and reaction rate
calculated by Jackson [loc.cit.,Tables 1 and 2]for a muon-deuterium
molecule, with the sum of nuclear radii taken to be 5*10**-13 cm, the
results are given for the range B = 1 - 207.

B = 1 P = 7.41*10**-40 sec**-1
B = 2 P = 3.40*10**-25 "
B = 3 P = 1.36*10**-18 "
B = 4 P = 1.31*10**-14 "
B = 5 P = 7.36*10**-12 "
B = 6 P = 8.23*10**-10 "
B = 7 P = 3.32*10**-8 "
B = 8 P = 6.68*10**-7 "
B = 9 P = 8.18*10**-6 "
B = 10 P = 6.91*10**-5 "
B = 20 P = 9.78*10**0 "
B = 50 P = 7.72*10**+5 "
B = 100 P = 4.05*10**+8 "
B = 207 P = 7.00*10**+10 "

Itzhak Shechtman
luthersetzer
2002-03-20 18:14:49 UTC
As I understand it, the fractions in Mills' model refer to
fractional Rydberg levels. So, doesn't that mean that a
hypothesized n=1/2 hydrino would have a radius of 1/2 the standard
Only if we can show that the Fourier Transform of a radial Dirac
delta function really is a sinc function. So far, that has not
happened. Given how much this whole theory hinges on that one
statement, it looks like we should suspend confidence in the theory
until Dr. Mills can post his derivation of this single, critical
mathematical proposition. If he can't, then this list will have to
move in the direction of analyzing and explaining the data in
conventional terms -- or substantially revising the math of CQM.

Luke Setzer
John A. Kassebaum
2002-03-20 19:51:13 UTC
Post by luthersetzer
Only if we can show that the Fourier Transform of a radial Dirac
delta function really is a sinc function. So far, that has not
happened. Given how much this whole theory hinges on that one
statement, it looks like we should suspend confidence in the theory
until Dr. Mills can post his derivation of this single, critical
mathematical proposition. If he can't, then this list will have to
move in the direction of analyzing and explaining the data in
conventional terms -- or substantially revising the math of CQM.
I must agree that a mathematical delta function has a constant fourier
transform. It is sinc if you use the limit as width->0 on a square
pulse. If you say that the limit as width->0 of a square pulse of
density concentrated at a radius is appropriate, then the sinc is not
improper is it?

[I've proposed this in an earlier message, yes. But this does not account for the 1/r^2 term in the *radial* Dirac delta function. --LS]

The delta function is after all a mathematical construction,
perhaps the charge need not be absolutely two dimensional, but only
infinitesimally close to two dimensional. Any deviation from exactly
two dimensional in the orbitsphere would lead to a sinc, right?

[Maybe. Dr. Mills will have to explain that one. --LS]

-John
--
John A. Kassebaum
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