Re: Proof for E=mc2

From: Androcles (dummy_at_dummy.net)
Date: 01/16/05 

Date: Sun, 16 Jan 2005 01:05:38 GMT

"Zigoteau" <zigoteau@yahoo.com> wrote in message
news:1105825173.790959.128690@f14g2000cwb.googlegroups.com...
> Hi, Greg,
>
>> > In fact what
>> > he derived was Delta E = Delta m.c^2
>>
>> If we accept the additional postulate that E=0 when m=0, then
>>
>> E(m) - E(0) = (m - 0) c^2
>
> Sure, once you realize that energy has something to do with mass, it
> is
> just as reasonable to define the energy origin to be zero mass as zero
> velocity. However it's a definition thing, and there's no way to prove
> that either is "correct", because in Newtonian physics all energy has
> to be is a constant of motion of a certain kind.

I cannot agree with that. Newtonian physics readily accepts potential
energy, given up as kinetic. The cork bobbing on water or the mass on
a spring are examplars. We accept the sun is constantly pouring away
enormous quantities of energy, the potential form of which is it's
mass, and mass is measured as force. The potential energy (mass)
concentrates into a small space by some force we call gravity,
(again in NM) and the resultant pressure at the core gives rise to
the release of the mass as radiation. Finding an equation relating
the two was not something Newton did, that is agreed, but Newton's
physics didn't end with Newton. They continued once experiments
with electricity and magnetism were made, and diverged when
Einstein began guessing with his "thought" experiments.
Any equations you write are intended to represent the world
in which we live, not a wild imagination where time, previously
invariant, and distance, previously invariant, give up those
properties in favour of light's speed being invariant, and gives
up the PoR as well.
Einstein states:
 "But this result comes into conflict with the principle of
relativity set forth in Section V. "
ref. http://www.bartleby.com/173/7.html

He is in fact stating that the velocty of A with respect to B
differs from the velocity of B with respect to A.

How can you reconcile the collision energy? Does the stationary
car do less damage to the moving car than the moving car
does to the stationary?
Androcles.

 I think that a
> stronger argument for taking the origin at zero mass is that
> energy-momentum is then a 4-vector. Again, this is an elegance thing -
> there is no a priori requirement for energy to be the component of a
> 4-vector. It is only in hindsight that it appears natural.
>
> Are you on a postdoc now? Still sloshing that liquid helium about?
> Cheers,
>
> Zigoteau.
>

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