Eureka X:- Newton's Gravitational forces ensured that the Mass of any given Photon is greater than Zero, with mathematical proof...
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Posted by kx21 on July 03, 2002 at 04:06:40:
In Reply to: DickT, concerning Photon mass posted by Haelfix on July 02, 2002 at 09:14:08:
Eureka X:- Newton's Gravitational forces ensured that the Mass of any given Photon is greater than Zero Proof:- (1) Let the Total number of Massive physical Entities which have the 'appropriate' relationship(s) with the given Photon (e.g. Star, Plant, other Massive particle floating outside the Matter (e.g. Cosmic Dust), etc), in the Cosmos is N. i.e. The mass of Entity i, m(i) > 0. Newton's Graviational force:- (2) f(i) = Gm(p)*m(i) / r(i)^2 Thus, (3) Abs[f(i)] = G*Abs[m(p]*m(i) / (r(i)^2, i =1,2,...,N Note: (4) m(p): The Mass of the given photon (4.1) m(i): The Mass of massive Physical Entity i in the Cosmos (5) r(i): The distance between the given photon & the Entity i. (6) Meta Laws 3 of the Cosmos dynamics implies that r(i) > 0. Thus, (7) Abs[m(p)] = Sum of [ Abs[f(i)]* r(i)^ 2) / m(i)] / N.
(8) Assumed that the speed of the Photon is c and its movement is affected by some of the Entities (e.g. Plant / Atom: The Photon absorber).
(9) For instance, Entity j. Then Abs[f(j)] > 0. This (7) implies that (10) the Sum of [ Abs[f(i)]* r(i)^ 2) / m(i)] / N is > 0. And (10.1) Abs[m(p)] > 0 Thus, (11) m(p) > 0 as the mass of any physical entity can be negative, by definition.
i.e. The mass of the given Photon is > 0.
This completes the proof. Any comments on points (1) to (11) are welcome.
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