On Mar 14, 2013, at 7:05 PM, Jones Beene wrote:
From: Edmund Storms
Here is the mass change
D =
2.014101778
H=
1.00727647
n=
1.0086649
The gain in mass is D-n= p
You are making an incorrect assumption. The O-P effect (i.e.
“stripping”) is not thermonuclear, it is quantum mechanical - in
effect a tunneling reaction. Quantum tunneling is one of
Oppenheimer’s claims to fame.
OK Jones, then were does the mass come from? No matter what you
call the process, the energy MUST be conserved. This reaction
requires energy be added to create the mass of the product. Where
does this energy come from?
Yes, mass-energy is conserved but we are talking about deuterium
being converted into something else (tritium or He3)– so there is
NOT necessarily a non-conserved mass of anything, since there is
always the neutrino “wild card”. That, essentially, is the crux of
your incorrect assumption.
I'm making no assumption. I'm simply applying conservation of energy.
If instead of the D= n +P reaction, you propose the normal hot fusion
reaction, then of course the situation changes. When two D come
together with enough energy, the nuclei combine and then explodes into
tritium + p and He3 + n. This is the normal hot fusion reaction that
generates energy. That is not a neutron stripping reaction.
In the Fusor, the transmuted nucleus is left in an energy state as
if it had fused with a neutron of negative kinetic energy, so there
far less mass change than the thermonuclear reaction. The Fusor can
be called “warm fusion” not hot, since the threshold energy for
thermonuclear reaction is never attained.
The only issue here is how the barrier is overcome, because once
this happens, energy is created by the normal hot fusion reaction,
i.e. the combined nucleus fragments into the observed particles
which includes neutrons.
That is what you seem to be missing in all of this. It is not hot
fusion but CoE does apply. In the O-P reaction, the Coulomb barrier
is overcome when two deuterons approach each other with the neutron
end of each facing the other – i.e. being geometrically ahead of the
proton end. The 1.7 MeV barrier is effectively lowered to about 10
keV.
Yes, the barrier is lowered and the expected fusion reaction occurs.
That was not the original subject.
Why suggest some magic condition like negative energy.
Robert Oppenheimer and Melba Philips suggested this. Who am I, or
you, to suggest otherwise?
The process is very simple. The two D are given enough energy to
surmount the barrier. The Fusor simply does this in an efficient way.
No, the Fusor never gets close to doing this at all, without QM. The
energy to surmount the barrier is reduced by a similar amount to the
deficit in net energy transfer.
Once again, we appear to be seeing experts in one field who do not
understand the full implications of QM and nuclear tunneling - and
refuse to believe that energy on the quantum scale can be “borrowed”
for a few femtoseconds before it is repaid.
I know about tunneling. It is simply a way of saying that the expected
barrier is lowered by some process. You can describe the process using
QM if you want. Or you can propose that the orientation of the two d
is important or, if the d are in a material, the electron
concentration is important. Or you can imagine borrowed energy. These
are all assumptions used to explain what is observed. It has nothing
to do with the initial subject.
Ed
There is no 1.7 MeV threshold and there is corresponding mass
change. In QM tunneling, the energy barrier for fusion is reduced
and the excess energy is likewise reduced.
Jones
