From: logajan@anubis.network.com (John Logajan)
Subject: Re: Electrostatic fusion
Reply-To: logajan@anubis.network.com (John Logajan)
Organization: Sci.physics.fusion/Mail Gateway
Date: Thu, 31 Dec 1992 02:24:47 GMT

Lloyd G. Allred, Ph. D. writes:
>The sharp points produce enormous fields. 

It is true that the gradients can be over very short distances, but a
proton falling through, say, 20kv has the same final kinetic energy 
whether the 20kv gradient is spread over a meter or an angstrom.

>I made the conjecture that cold fusion could be induced by focusing
>electric fields using electrostatic devices, in particular, that electric
>fields at the corners of such devices could be made sufficient to induce
>thermonuclear fusion.

If the mode of fusion is kinetic energy induced coulomb barrier penetration,
then I'd be inclined to doubt it.  However,  if the applied gradient is
used to directly oppose and nullify the coulomb repulsion of the nucleus,
that might be something to consider.

>At any point in the interior of a conductive
>device, the electric field must therefore be zero, otherwise, any
>electron at that point would be accelerated, thus changing the
>distribution of the electric charges and the corresponding electric
>fields and the electric fields would not therefore be static.

This is a useful insight into basic physics.  Thanks.

>For an electrostatic object of a cone, we intend to show that the
>electric field approaches infinity as one approaches the tip of the
>cone. 

I'll have to beg off on understanding the calculations used to show this,
but on an intuative level it seems an overstatement.  You have a finite
charge (X electrons) and a non-infinitely small distance (Y meters) and
therefore the inverse product of the two is going to result in a finite
density or gradient.   To appeal to infinities in this case seems to be
a bit of marketing hyperbola.

>If the conductive metal becomes positively charged, the hydrogen nuclei
>will congregate at the sharp points of the surface, and be ejected
>from the surface.

I believe you just argued against this above where you said that the
electric field gradient would be zero at all points inside and on the
surface of the conductor.  Thus, rather than a congregation of nucleium
at the point, it would, due to the higher external charge gradient, see
the exact opposite, a lower population of nuclei than at other locations
on the charged surface.

You can, as you suggest, however, expect some ejection or ionization at the
point due to the geometry of the system generating a high field gradient at
the point.

>I fully understand that infinite electrical fields are an impossibility.
>As a consequence, static conditions cannot be achieved on an object with a
>tip because the charge leaks off into empty space.  The charge at such a
>tip will be less than infinite.

As pointed out above, even without leakage, the gradient would be finite.

>As a result, the charged particles within the device will be accelerated
>toward the tip these accelerations approach infinity as the charges move
>toward the tip.

Again, there will be no current unless it is to replace ejected charges
(or under startup transient conditions).  When an ejection deficit does
occur, the high gradient will accelerate the nuclie quicker, but over a
correspondingly shorter distance, the net result is that the final kinetic
energy is never more than that associated with the applied voltage (for
instance, you mentioned 20kv.)

I see that latter on you appeal to tidal analogues.  While this is an
interesting possibility, the exact mechanism by which this would occur is
not sufficiently spelled out.


- John Logajan MS010, Network Systems; 7600 Boone Ave; Brooklyn Park, MN 55428
- logajan@network.com, 612-424-4888, Fax 612-424-2853
Newsgroups: sci.physics.fusion