[Vo]:Re: HUP-spread-out electron "feels" (and thus Coulomb-screens?) like a point charge... - T.GIF

[Michel Jullian]

Robin,

Although it seems to make sense, something doesn't fit in your center of mass 
frame of reference and therefore equal De Broglie wavelengths (DBW) paradigm: 
in that frame, as you say, "when the particles are stationary relative to one 
another, the DBW is infinite, hence no longer relevant", whereas that distance 
r1 where the incident d has lost all its initial kinetic energy is precisely 
where the Li et al paper compares the distance by which it "missed" (r1-r0) 
with the DBW, which they don't find infinite but equal to 0.78 Å...

But on the other hand, how can the DBW not be infinite if momentum is zero??

On yet another hand, the DBW seems the right parameter to define the "spread" 
of a particle and therefore its capacity to tunnel or be tunneled to... if it's 
infinite, it's all over the place so tunneling should have 100% probability!

This point is definitely unclear to me, any enlightening welcome.

Michel

----- Original Message ----- 
From: "Robin van Spaandonk" <[EMAIL PROTECTED]>
To: <vortex-l@eskimo.com>
Sent: Monday, April 28, 2008 12:13 AM
Subject: Re: [Vo]:Re: HUP-spread-out electron "feels" (and thus 
Coulomb-screens?) like a point charge... - T.GIF


In reply to  Michel Jullian's message of Sun, 27 Apr 2008 18:05:33 +0200:
Hi,
[snip]
>Indeed two immobile d's wouldn't attempt much I don't think, but isn't it ok 
>if _one_ d in the pair, namely the incident one, makes the attempt, as in the 
>Desorbing vs Incident Excess Surface Electron Catalyzed Fusion (DIESECF) 
>scenario we are discussing?
[snip]
>I was thinking of permanently cooling the back (loading) side and exposing the 
>front side to more energetic incident deuterons. This way we would benefit 
>from the cooling-induced wider quantum spread (larger De Broglie wavelength 
>h/p) of the target deuteron, allowing tunneling to it from a larger distance, 
>and of the energy of the incident one to get as close as possible to the 
>target, does this make sense?

The De Broglie wavelength is based on the *relative* velocity between the
particles. In fact one should probably calculate this in the common centre of
mass frame of reference (in which case both particles have the same De Broglie
wavelength - my thanks to Charles Cagle).

What this means is that you can't have one fast particle and one slow one. They
both need to be slow. This also implies that tunneling probably only takes place
when the approaching particle has used up all its kinetic energy in overcoming
the electrostatic potential of the other particle, and is just on the verge of
reversing course. I.e. it's usually a "one shot" affair. Consequently, "glancing
blows" may not contribute, it may need to be a head on collision.
[snip]
>I appreciate, thanks also for your other posts. What do you mean by "initial 
>distance before tunneling", is this the distance at which the incident 
>(projectile) deuteron comes to a halt? 

Yes.

>I don't see where the De Broglie wavelength or the energy of the particle(s) 
>comes into play BTW, it should matter as we discussed above.

The De Broglie wavelength is only a guide - a rule of thumb if you will. Note
that when the particles are stationary relative to one another, the De Broglie
wavelength is infinite, hence no longer relevant. When two particles are within
the De Broglie wavelength of one another tunneling is possible, but not
guaranteed, and the chance that it will happen is strongly related to the
separation distance. That chance is what I have attempted to calculate in the
gif file I attached to my previous post.
Note also that even when tunneling does take place, fusion is not guaranteed.
Whether or not it happens, depends also on the nuclear cross section of the
reaction. Some reactions are more likely than others.
A good example of a poor reaction is the p-p reaction. Tunneling probably
happens quite often, yet a nuclear reaction seldom ensues.
OTOH, a reaction with a good cross section is the p-B reaction, but this is
limited by the reduced tunneling probability due to the high charge on the B
nucleus.

>
>Surely there must be a standard way to compute this, could the approach used 
>in the Chinese paper I quoted above be a standard one?

Yes, in fact the essence of it is the only method I have seen employed. If you
look closely, you will see it is also the method I used.
If you feel like working on this, and you come up with something that works
well, I would very much appreciate it, if you would pass it on.

>
>>See also the paper:- "Catalysis of Nuclear Reactions between Hydrogen Isotopes
>>by mu- Mesons" by J.D. Jackson, Physical Review, Vol. 106, Number 2, April 15
>>1957, page 330.
>
>Thanks, do you have a pdf version by any chance?

I'm afraid it's copyrighted, but you can purchase a version on line, as I did,
or visit your local technical/university library, and read it for free.
(BTW Jackson essentially uses the same basic concept).

Regards,

Robin van Spaandonk

The shrub is a plant.