In reply to Horace Heffner's message of Wed, 13 Apr 2011 06:12:17 -0800:
Hi,
[snip]
>>> to the interaction of the strong force. This loss of potential
>>> energy does not prevent electron capture of the now energetically
>>> trapped electron, if capture occurs very fast, because that electron
>>> initially has the kinetic energy necessary.
^^^^^^^^^
>>
>> It isn't clear why kinetic energy is "necessary".
>
>Uh ... why do you say "necessary".
...because that's the word you used yourself here above.
>It *is* there. As the potential
>energy of the system, the field energy, is lost, the kinetic energy +
>relativistic mass of the electron (and hydrogen nucleus, because in
>relativistic orbitals they rotate about each other) makes up for it.
[snip]
>> I am left wondering (literally) where the energy comes from during
>> normal beta
>> decay for the electron to escape the nucleus. The only conclusion I
>> can come to
>> it that conversion of a neutron to a proton results in 28 MeV more,
>
>The amount is stochastic, it isn't a fixed "28 MeV".
>
...yes I realize that, but I used it because you used it, and I didn't want to
further complicate matters.
>Well, then, you can't buy the trapping part either
...which as I have previously suggested, I'm not sure I do, though I must admit
to finding it somewhat confusing.
>- so you are left
>having to explain why heavy element LENR provides neither the
>enthalpy nor the high energy signatures that would be expected, and
>why branching ratios change for cold fusion D+D-->He, etc, etc., etc.
>You can no longer explain LENR. It is an experimental fact that
>stable nuclei have heat. See: E. Melby et al, Observation of
>Thermodynamical Properties in the 162Dy, 166Er, and 172Yb Nuclei ,
>Phys. Rev. Lett. 83/16 (October 1999): 3150 - p. 3153, the basis for
[snip]
This paper pertains to excited nuclei. I see no implication that heat exists
when the nucleus is not excited, or that the temperature exceeds that which
might be explained by the excitation energy. (IOW no sign of ZPE energy).
Quote: "The experiments were carried out with 45 MeV 3He-projectiles at the Oslo
Cyclotron Laboratory (OCL)."
(Which projectiles supplied the excitation energy in this case).
[snip]
>C. A. Chatzidimitriou-Dreismann et al,Anomalous Deep Inelastic Neutron
>Scattering from Liquid H2O-D2O: Evidence of Nuclear Quantum
>Entanglement,Phys. Rev. Lett. 79, 2839 - 2842 (1997),
>http://prola.aps.org/abstract/PRL/v79/i15/p2839_1
>
>The fact that, when H2O is looked at fast enough, it is H1.5O,
>meaning 1/4 of the time a proton "disappears", is just too much
>linked to cold fusion to ignore! Even though the disappearance is
>only attoseconds in duration, it must be repeated frequently i a
>water molecule to have such a probability.
If time spent tunneling were the reason for disappearing H (which I presume is
what you are implying here), then it would imply that tunneling is not
instantaneous, i.e. that your previous statement about wavefunction collapse
just "being there" is in doubt.
[snip]
>You can look the energetics another way. When, in hot fusion, a
>naked proton is forced into a Ni nucleus to the radius where 28 MeV
>(for example) is released upon its departure if fissioned, then that
>energy was in the form of kinetic energy a priori. Once the proton
>binds to the nucleus by the strong force, that a priori kinetic
>energy is gone - it is now in the form of potential energy, the field
>energy of the new nucleus. It is locked in, no longer available,
>lost, in either hot or cold fusion.
Agreed. However the implication is that the binding nuclear force is *large
enough to prevent the proton from leaving again*, which in turn implies that it
actually supplied 28+6 MeV (e.g.) when binding the proton. The extra six MeV
being the difference between the total binding energy of the initial nucleus and
that of the final one. Now when an electron and a proton enter the nucleus
together, the work done in overcoming the Coulomb barrier presented to the
proton is done by the electron, since its attractive electrical force to the
target nucleus exactly balances the repulsive electrical force exerted on the
proton (IOW they enter as a neutral ensemble). However the nuclear binding
energy of the reaction remains the same (i.e. 28+6 MeV), so there is 28+6 MeV
available to the electron, of which 28 MeV is needed to overcome the Coulomb
force, and escape, with 6 MeV left over once it is free.
(If this is wrong, then at least you may see the cause of my confusion (see also
below) ;)
(Another possibility might be that in hot fusion one shoots a proton up a high
hill till it falls over the top and lands in a shallow crater at high elevation
with a slight "plop" as it binds with the nucleus. However if this were so, then
it would only require the return of that slight "plop" energy to a nucleus in
order to free the proton and retrieve the full 28 MeV in repulsive Coulomb
energy, and I doubt nuclei would be very stable in that case, not to mention the
huge potential as an energy source, which I suspect would have long been
exploited, i.e. 28 MeV - "plop" ;).
Regards,
Robin van Spaandonk
http://rvanspaa.freehostia.com/Project.html
