In reply to  David Roberson's message of Sat, 22 Jun 2013 20:29:00 -0400 (EDT):
Hi,
[snip]
>
>I suppose He4 is a relatively tough little nut to crack, but it seems to 
>happen more often than not when D's fuse.  Isn't that the reason that T or He3 
>tend to remain alive after a free space event?  COM will always be achieved 
>when two D's collide and remain stuck together.  

That isn't the momentum that's in question. It's all about the momentum of the
final "bits" not the momentum of the initial bits.

When a gamma is formed, it shares momentum with the 4He nucleus, thus allowing
the 4He to remain in tact. However gamma production is usually a relatively slow
process, and the T, 3He branches are much faster. In all cases two particles are
involved. "4He + gamma" (counts as a particle because photons have momentum); "T
+ p"; "3He + n". Because there are two particles, momentum is conserved (i.e.
always zero in CM frame). 
When a spectator nucleus is available, it can take the place of one particle
resulting in "spectator + 4He".


D + D + e (slow) => 4He + e (fast)

here the "spectator" particle is an electron.

This would essentially produce a 24 MeV electron.

>COE suggests that they retain the amount of energy required to break apart 
>immediately unless some of the fusion energy is released.  I can not think of 
>a better trick to pull off than for a nearby electro magnetically coupled 
>"friend" to take away a portion of it.  In hot fusion there is no one nearby 
>to help out most of the time.  I am suggesting that perhaps on those rare 
>occasions when the He4 survives that a helper is close by.  In LENR help is 
>around the corner in most cases due to close quarters.

Perhaps something like:- 

D + D + e (slow) => 4He + e (fast) ?

here the "spectator" particle is an electron.

This would essentially produce a 24 MeV electron.

The problem with such an approach is that apparently a helper is almost *always*
at hand, since the product is almost always 4He.

If the de-energizing mechanism were independent of the Coulomb barrier
penetration mechanism, and reliant upon chance, then one would expect to see
much more T and 3He.

Regards,

Robin van Spaandonk

http://rvanspaa.freehostia.com/project.html

Reply via email to