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# Re: HOT FUSION

I THINK THAT A PERSON WHOSE MESSAGE BECOMES
UGLY, FOR SOME REASON, SHOULD WORK ON IT TILL
OTHERS CAN READ IT AS INTENDED. WHO ELSE WILL
DO THIS FOR US? I APPOLOGIZE AGAIN.

1) In electrostatic you might ask students to estimate
the coulomb barrier preventing positive D ions to fuse
at low temperatures. Here is my suggestion. First tell
students that in addition to repulsive forces the ions
attract each other by very strong nuclear forces. But
these forces do not obey the 1/r^2 law. They are
negligibly small when r is above a distance R (called
range). For x<R the attractive nuclear forces are much
larger than repulsive electric forces. That would be a
good justification for defining the coulomb barrier, CB,
as the value of the electric potential at x=R.

The rest is trivial. Assume that R=3 F, for example, and
calculate CB. Note that F is the unit of length (femtometer
or 10^-15 m). My answers were 480 kV and 360 kV, at
R=3 F and 4 F, respectively. You may or may not link this
problem with the cold fusion controversy.

2) If you do then consider addressing the screening effect.
Some scientists say that screening is possible when D ions
are embedded in metals, such as Pd or Ti. Simply stated,
and without trying to argue about what causes screening
(local clouds of electrons at crystal's boundaries?) one
can simply declare: screening consists of lowering of the
coulomb by nearby electrons.

To illustrate screening do the following. Place one deuteron
at x=0 and treat the other deuteron, at x>0, as a probe charge.
That is what I did to calculate coulomb barriers. Then place
an electron at some negative value of x, for example, -2 F.
The CB is now V1 + V2 (where V1 is the positive part due
to the deuteron and negative V2 is the negative part due the
electron). You will see that CB approaches zero when electron
is approaching the origin, as it should be. My answer, for R=4 F,
was CB=120 kV for the electron placed at x=-2 F. And nothing
prevents you from introducing more than one screening electron.

Suppose a single electron at x=10 A =100000 F is replaced by
a negative particle of variable charge. Assuming R=4 F, how
does the magnitude of the negative point charge affect the
coulomb barrier? It turns out that the charge of only
215000*e is sufficient to eliminate the coulomb barrier.
Here my results:

charge in 10^5*e QB in kV
-------------------------------------
2.1 0.0015
2.0 72
1.0 216
0.5 288
0.1 346
0.01 358
0.00 ideal two-body barrier 360

I find these numbers shocking. First because the distance of
10 A is about ten times larger than the distance between
atoms in most metals. And second because the number of
electrons needed to eliminate the barrier (for that distance)
is a neglibible fraction free electrons in each cubic micron.

4) Yes, I know that three or more particles would usually not
be at rest on the x axis. And I know that the nuclear potential
is not a rectangular well. But my goal is to estimate the orders
of magnitude of CB, and to illustrate the idea of screening.

5) By the way, we usually think that the so-called "free
electrons" in metals are uniformly distributed, like in
ionized gases. What evidence do we have for this? Yes,
I know physics of surfaces if very complex, even for
something familiar, such as friction.

6) No, I am not trying to poison your minds with
heretical pseudo science. It is essentially an attempt
to show how a trivial electrostatic problem can be
made relevant, even in an introductory course. I hope
the message will not become ugly again.
Ludwik Kowalski