🚀 go-pugleaf

On Sep 30, 10:19 am, Igor <thoov...@excite.com> wrote:
> On Sep 30, 1:43 am, frankli...@yahoo.com wrote:
>
>
>
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> > On Sep 29, 1:16 pm, Loui...@edu.herlufsholm.dk wrote:
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> > >        Interpretation of E = m*c^2 as kinetic energies of sub-
> > > particles.
>
> > > By Louis Nielsen,   Treatise:http://www.rostra.dk/louis
>
> > > What is the physical interpretation of Einstein's equation E = m*c^2
> > > where E is the latent energy inside a portion of matter with the mass
> > > m and where c is the velocity of light?
>
> > > In my treatise I show that m*c^2 is equal to the total kinetic
> > > energies of the sub-particles I call Unitons and of which all matter
> > > consists.
>
> > > The Uniton Postulate:
> > > The unitons are the smallest and most fundamental energy-/matter
> > > quanta in the Universe.
> > > All matter consists of unitons and all interactions in the Universe
> > > are consequences of
> > > interacting unitons.
>
> > > In my treatise I show that the total latent energy E inside a given
> > > amount of matter with mass m is given by E = m*c^2 where c is the
> > > velocity of light. And the expression m*c^2 is a measure of the TOTAL
> > > KINETIC ENERGY OF THE UNITONS of which the matter consists.
> > > Kinetic energy is the most fundamental form of energy.
>
> > > By the above recognition of our physical world it is not difficult to
> > > understand that all matter,
> > > regardless of its chemical appearance contains energy as given by
> > > Einstein's famous equation.
>
> > > Read more in section 13 and section 14 of my treatise:  http://www.rostra.dk/louis
>
> > > Best regards
> > > Louis Nielsen
> > > Denmark
>
> > It's actually a lot easier than that. E=mc^2 because it is the sum of
> > the kinetic energies of the particles ejected during pair production
> > of a positron and electron. E = 1/2 mv^2 + 1/2 mv^2 = mv^2 where v=c
> > and m = mass of positron and electron.
>
> > See: What Does E=mc^2 Really Mean?
>
> >http://www.geocities.com/franklinhu/emc.html
>
> >fhuemc- Hide quoted text -
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> > - Show quoted text -
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> You do understand that you are using the wrong relativistic formula
> for kinetic energy?- Hide quoted text -
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> - Show quoted text -

Actually, one of the key aspects of how I calculate E=mc^2 is that
relatavistic formula for kinetic energy does not apply at this level.
The relatavistic formula cannot be correct since you get the correct
answer using the non-relatavistic equation.

The reason for this is that I believe that relatavistic effects are
due to masses moving THROUGH the aether (my sea of positron/electron
pairs). For reactions involving the aether itself, these rules do not
apply since the aether cannot move within itself and pair production
doesn't involve particles moving through or past any aether particles.
At this level, it breaks down into simple newtonian mechanics and E=
1/2mv^2 is completely applicable.

The proof is in the fact that the experimentally, the non-relatavistic
formula works to describe the energy required to do pair production,
not the relatavistic one which would assign much larger masses to the
positron/electron.