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From: John Polasek on 29 Jul 2010 14:39
On Fri, 16 Jul 2010 12:16:41 -0700 (PDT), valls(a)icmf.inf.cu wrote:

>I suppose we are all in agreement about the increase of U(r) with an
>increase of r. Has U(r) a finite maximal limit value when r tends to
>infinite? In case of positive answer, which is that maximal value?
I am copying your first post here so people know what you're talking
about:
>"The conversion of a pair positron-electron into photons (and vice
>versa) is a common experimental fact of today Physics. If the
>particles are considered initially at rest infinitely separated, the
>final photons total energy results equal to the initial total rest
>energy. Then, to where the initial electrostatic field energy goes
>after the conversion?"

We have to examine what happens during this process and we will get
some help from the use of potential theory as used in gravity which is
also an inverse square field.
Unfortunately we don't have a radius of the electron but we can say
that initially the centers of the pair are separated by A. The pair
has a potential energy given by
W = -e^2/4pieps0*A = k/A (1)
This energy is the integral of force = k/r^2 from infinity down to A.
That energy had to be installed by someone dragging those two apart
and the effort was very real and so we have to ascribe the energy
either to the field or to the charges. As we know the energy can be
recovered by letting the two fly back toward each other.
It should be evident from equation (1) that its energy depends
sharply on A, with no contribution from infinity. So your concerns
about energy at infinity are unfounded.
With gravitational fields, the escape velocity is one measure of the
strength of the gravitational field. The Earth e.g. has a surface
acceleration of 9.8m/ss with an escape velocity Ves of 11,180 m/s. If
we drop anything from infinity to the Earth's radius R, Ves will be
its terminal velocity (in vacuum). The kinetic energy is equal to the
potential energy
.5mV^2 = mMG/R, therefore for unit mass
V = sqrt(2MG/R) = 11,180 m per second at earth
Similarly here with the electrons, except for their equal mass ( where
Earth was massive) they will meet and crash with a velocity that
depends on their (unknown) minimum separation which we denoted as A.
Each will share half the Newtonian energy .5mv2, as they race toward
their center of mass
.5*2*0.5me*v^2 = k/2A from (1)
would be the energy for each particle so there velocity would be
.5mev2 = k/2A
v = sqrt(K/A*me)
I'll work it out for you. Lets take A as the classical radius of the
electron which is computed as.
A = rcl = e^2/4pimc^2eps0 = 2.818e-15m
(which is 2 pi times the size of the dual space cell in my dual space
theory or 3.514e-14m)
Insert radius A into equation 1 and get nearly total cancellation:
W = mc^2 = 2*mv^2/2 for two charges
That's the Newtonian energy as they crash.
The initial electrostatic energy came from dragging the pair apart to
a great distance and the energy reappeared as they smashed together,
which could have been a thermal event.
As far as automatically assuming that two photons are thereby emitted,
you first have to try to visualize how an observer would see that;
it's easy to say and easy to believe but I don't think any laboratory
is prepared to take an electron and a positron and separate them by
say 1 m and then to measure the effect of their subsequent collision,
not even to mention as to how one would steer them toward each other
in the first place since the initial attraction is near zero.
The question is always what are you going to do with the answer?
I hope you at least see that there's no contribution to the energy at
the range infinity.

John Polasek
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