Since you think this paper is relevant, perhaps you can suggest where
all the BEC are in PdD and why great effort is made to achieve
temperatures near 0 K to study BEC?
Ed
On Feb 14, 2013, at 12:31 PM, Axil Axil wrote:
Experimens have shown that a BEC can form at
a temperature of 2640 K.
arxiv.org/pdf/1210.7086
I have posted on this elsewhere.Cheers: Axil
On Thu, Feb 14, 2013 at 2:23 PM, Edmund Storms
<[email protected]> wrote:
On Feb 14, 2013, at 12:19 AM, Chuck Sites wrote:
I think you are being very dismissive of the way quantum mechanics
works with in the nuclear realm.
I have no problem applying QM if it is applied to realistic
conditions. Simply assuming a condition that has no reality and
then applying QM to justify the assumption means nothing. This is
only a dog chasing its tail. You can use any vocabulary you want,
but Gibbs energy determines the basic behavior of atoms. The
temperature must be low because the bonding energy, obtained from
the process you describe, is very low. The entropy * T will
overwhelm the enthalpy if the value for T is large, thereby causing
the BEC structure to decompose. Or do you think BEC formation
violates the Laws of Thermodynamics?
Ed
It all boils down to PSI and if the nuclear force is point charge
with a probability of interacting defined by PSI, or that PSI is
blurred motion where the nuclear force is spread over space
describing PSI. Is it a wave or is a particle probability? It's
a very fundamental question with respect to BECs. The BEC comes
about by the overlapping wave functions of integral spin. By it's
nature bose particles when chilled they like to fall towards ground
states and as they do, their PSI's will completely overlap making
one big PSI(n) where PSI(n) describes all of the properties of that
mix. The PSI is the matter wave, and with the matter wave all of
the other attributes of a particle are carried along, so the
electric force and the nuclear force(s) are just aspects of that
PSI(n). The overlap of the PSI is where there is a probability of
interaction. That's why I mentioned the Gamow factor is that it
describes perfectly what the collision of two PSI's with nuclear
interactions looks like. At very high energies, it looks like
CERN, but at very low energies it looks like solid state.
Eventually you have to have PSI(x) describing the model. If that
wave function overlap doesn't occur, there is no probability for
interaction and nothing will occur.
I think maybe a hybrid of Chubbs' and Kim's theory could be very
compelling. Specifically with the Nano scale BECs or 100 atom Bose-
band states.
Best Regards,
Chuck
On Wed, Feb 13, 2013 at 10:15 AM, Edmund Storms <[email protected]
> wrote:
Eric, the details do not matter. The basic idea is wrong. The
details are just a series of arbitrary assumptions to avoid
dropping the initial premise. We are simply playing whack-a-mole.
He strings a collection of words together that have no logical
relationship, but because the vocabulary of QM mathematics is used,
no one questions the statements. If Ron wants to make a
contribution, he needs to apply his ideas to what actually exists
in the real world based on what material science has agreed is real
based on much study. Simply making up concepts to which math is
applied is not useful except as a game. Also, we are describing a
mechanism. Describing one part in isolation is not useful. This is
like saying an automobile works by turning the key in the ignition
and then go on to describe the key in great detail.
On Feb 12, 2013, at 10:42 PM, Eric Walker wrote:
On Tue, Feb 12, 2013 at 7:13 AM, Edmund Storms <[email protected]
> wrote:
There is no alpha. The helium CAN NOT MOVE spontaneously. The
helium contains extra energy as mass. This mass must be converted
to energy before it can appear as reaction energy. The He is fixed
in space. Normally the He nucleus explodes into fragments
producing hot fusion. Or it emits a gamma which releases the mass-
energy. This conversion CAN NOT OCCUR outside of the nucleus
simply by being near a Pd.
I suspect that you are very busy and haven't had time to read
Ron's writeup closely. Here is what he says about the production
of the alpha:
The fusion of deuterons always happens through unstable
intermediate states, and the cross section to alpha particle is
only small because of the same non-relativistic issue. To get an
alpha, you need to emit a gamma-ray photon, and emissions of
photons are suppressed by 1/c factors.
Yes, this is why the hot fusion products occur rather than helium.
Even this statement is ambiguous - what does 1/c factors mean? In
fact, the explanation is much easier to understand simply by noting
that energy can be lost by the nucleus exploding into its parts
faster than it can be released by gamma emission. The issue is
based on relative rates. Why is gamma emission slow? It is slow for
the same reason it is slow when photons are emitted from any
energetic nucleus. Many explanations have been suggested including
the need to assemble the required energy and spin in the nucleus
before the photon can be emitted. The statement of 1/c factors has
no relationship to this process.
When there is a nucleus nearby, it can be kicked
electrostatically, and this process is easier than kicking out a
photon, because it is nonrelativistic (the same holds for an
electron, but with much smaller cross section due to the smaller
charge, and there is no reason to suspect concentration of
wavefunction around electron density, as there is for a nucleus).
Here Ron makes an assumption that has no justification. The nearest
nuclei is many Å away and surrounded by electrons. Any nuclear-
nuclear interaction is impossible. That is why spontaneous nuclear
reactions are so rare.
The time-scale for kicking a nucleus is the lifetime of the two-
deuteron resonance, which is not very long, in terms of distance,
it is about 100 fermis, this is about the same size as the inner
shell. If the deuterons are kicking about at random, this
coincidence is not significant, but if the deuteron-hole
excitations are banded, it is plausible that nearly all the
energetic deuteron-deuteron collisions take place very close to a
nucleus, as explained above.
This is word salad without meaning in the real world. He makes up a
number and then assumes it applies it to an imagined process. Yes,
the d must be bonded (or as he says banded), but how?
There are conservation laws broken when a nucleus is nearby. The
nucleus breaks parity, so it might open up a fusion channel, by
allowing deuteron pairs to decay to an alpha from a parity odd
state. Such a transition would never be observed in a dilute beam
fusion, because these fusions happen far away from anything else.
This hypothesis is not excluded by alpha particle spectroscopy
(there are a lot of relevant levels of different parities), but it
is not predicted either.
This is word salad. His statement about beams reveals an ignorance
about how beams are used. They are used to bombard a solid in which
many interactions take place resulting in hot fission.
Here there is a concept of a "two-deuteron resonance," i.e., the
metastable 4He you're talking about following upon the d+d fusion,
which will not last long and must shed some energy. Ron states or
alludes to the following in the above paragraph:
There is a metastable "two-deuteron resonance" that will decay.
This is the energetic 4He you're referring to, which will then go
and do something else.
There are three channels for the decay of the two-deuteron
resonance: (a) d+d → [2d]* → 3He+n, (b) d+d → [2d]* → t+p,
(c) d+d → [2d]* → 4He+ɣ. Normally (a) and (b) predominate and
(c) is rare. But the reason that (c) is rare is that it takes a
while for the photon to be produced (my reading, anyway, of
"emission of photons are suppressed by 1/c factors").
This is a restatement of the earlier comment, which is correct.
When there is a palladium nucleus (not atom) nearby, however, the
energy that would have been dumped as a photon will instead be
kicked to a proton in the palladium nucleus, a process that occurs
quickly rather than slowly. Because this occurs quickly, branch
(c) is enhanced and branches (a) and (b) are suppressed in direct
proportion.
This is an impossible assumption.
When the mass deficit of the two-deuteron resonance is
electrostatically dumped into the proton in the nearby palladium
nucleus on the order of 24 MeV, you will get a palladium nucleus
with additional kinetic energy an energetically stable recoil
alpha, moving quite quickly.
In his original description Ron has touched on points that address
nearly every objection you have raised so far. His description
may well be incorrect, but I suspect it is not incorrect for the
reasons you have mentioned so far.
Eric, this discussion is a waste of time simply because the concept
has no relationship to reality. Clever people can create all kinds
of personal realities that are useful as games or as a guide for
their lives. But in science, the reality has to be shared based on
centuries of hard work by millions of people. New ideas have to fit
into what is known and must be described using words that have
common meaning. People seem free to imagine anything about CF that
would be laughable if applied to any other field of study.
I don't mean to press this issue. I just think Ron's theory
should be read closely before objections are raised; some very
good objections have already been raised in earlier threads. I
understand if you're too busy or if this lead does not seem to
merit your time. There may be interest among others here. It is
also entirely possible that while Ron knows something about the
math involved, he knows nothing about what happens with these
things in real-life. I am wary of drawing this conclusion myself
without further evidence.
You are on the right tract. Just have more courage to call a spade
a spade, or more exactly call nonsense what it is.
Ed
Eric
