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*From*: Ludwik Kowalski <kowalskil@MAIL.MONTCLAIR.EDU>*Date*: Sun, 15 Feb 2004 20:33:05 -0500

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

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