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Re: Eureka X -G v2:- Gravitational Force ensured that Mass of any given Entity, which is interacted with other Massive Entity or Entities / Particle(s), is greater than Zero

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Posted by kx21 on July 04, 2002 at 23:24:00:

In Reply to: Eureka X:- Newton's Gravitational forces ensured that the Mass of any given Photon is greater than Zero, with mathematical proof... posted by kx21 on July 03, 2002 at 04:06:40:

(A revised version of Eureka X-G)


Eureka X (Gravity):- Gravitational Force ensured that Mass of any given Entity, which is interacted with other Massive Entity or Entities / Particle(s), is greater than Zero,

Proof:-

(1) Let the Total number of Massive physical Entities which have the Interaction(s) / 'appropriate' relationship(s) with the given Entity (P) in the Cosmos is N. i.e. The mass of Entity i, m(i) > 0.

(1.1) m(p) the mass of the given Entity (P)

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 Entity (P), in the physical world.

(4.1) m(i): The Mass of massive Physical Entity i in the Cosmos

(4.2) G: Gravitational Constant

(5) r(i): The distance between the Entity P & 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)]/ G * N.


(8) Assumed that the 'Movement' of the Entity P is affected by gravitational field of one of the other Entities.

(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)] / G*N is > 0.

And

(10.1) Abs[m(p)] > 0

Thus,

(11) m(p) > 0 as the mass of any physical entity can't be negative, by definition.


i.e. The mass of the given Entity (P) is > 0.


This completes the proof.


Any comments on points (1) to (11) are welcome.


Copyright 2002 All Rights Reserved kx21.com



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