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http://hyperphysics.phy-astr.gsu.edu/hbase/particles/expar.html#c6

THE MISSING LINK

We are all familiar to Newton 3rd law that says, for every force there is an opposite force in direction and equal in magnitude. He was mainly applying it to external force. Yet this can be applied to internal force, a force in one part of system will be countered by reaction force on another part of system. Please see:

http://hyperphysics.phy-astr.gsu.edu/hbase/newt.html#nt2cn

As we know matter has a tendency of (expanding)taking space over time, we explain it as velocity (v) where v=r/t (r-distance, t-time). Therefore a graviton in space has tendency of taking space over time, or expanding from in within, in order this to occur it has to experience acceleration (a) which comes from basic tendency of matter to take up space over time. This creates a force pushing outward from within a graviton (we will call this Fo-o stand for outward-where Fo=ma). But since every force in a system has an opposite force in direction and equal in magnitude, in within the graviton there is an opposite force in direction and equal in magnitude to force Fo. We will call this force pushing inward or Fi where Fi=ma. This force (Fi) is due to the tendency of matter contracting, in order to remain stable, or Fo will keep expanding forever if it is not balanced by Fi.

Since they share same mass in order to create the two equal forces and opposite in direction, the forces will balance each other when they run out of mass. Therefore Fo=Fi. Therefore Net Force of the Graviton is Fo plus 'negative' Fi(negative since it is opposite in direction). Therefore the Net Force F=0. The graviton is stable.

At what speed did the graviton forces reach before they balance each other? The conclusion I came up with is the speed 'c'. Also I noted that the graviton has a set amount of mass, therefore it has a set amount of energy. We all know the famous equation E=mc^2, I wondered how I can use that to convert or relate to the graviton with a net force of zero, where all its energy is used. Since I postulated Fo=Fi, therefore the velocity of both forces will be c=c. Then solving for c^2 in E=mc^2, c^2=E/m, in order to show c=c or Fo=Fi.

So c^2=E/m where E=W(work)=Fr(since we are comparing the two forces energy in the given mass the F stands for Fo or Fi and the distance they can travel with given mass which is infinite, and to solve for that it is likely to intergrate 'r').

So we substitute for E, c^2=Fr/m.

To compare the two forces(Fo=Fi or c=c) we substitute in the variables:

c^2=c^2.

therefore (Fo)r/m= (Fi)r/m (note the F or forces share mass, or they are within the graviton.)

Lets back up to where we were discussing the Net Force(or Sum Force) of the graviton when the forces cancel therefore it equals zero. In our universe it is said to be no isolated force, so there is no isolated matter, in our case a graviton. Therefore the graviton will always be in contact with other gravitons or matter. Since the forces within graviton stabilized and canceled each other the graviton "overall"(the whole graviton) velocity is zero, the Sum Force is zero.

When it experiences an overall acceleration or collides with another graviton, it therefore increases its overall velocity in relation to the contact or collision. This causes an increase of the Sum Force, but it does not affect the internal forces (Fo, Fi) of the graviton. Only until the graviton overall velocity is "greater" than the velocity "c" does the forces within graviton become unstable. (As we know matter is said to keep the velocity 'c' if it doesn't things will go wrong...)

To understand the above explanations please check:

http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/inteng.html#c2

To explain the relation of a Graviton overall force to another Graviton, we start talking about mechanical forces, this relation is shown as:

(Fo)r/m= (Fi)r/m

Fr/m1=m2ar/m1 ----where F=m2a (graviton in contact with another graviton experiencing acceleration in relation to the collision)

Fr/m1m2=ar/m1 ----where a=F/m2

therefore Fr/m1m2=Fr/m1m2

And so on...., until we end up with F=Gm1m2/r2. The whole derivation is at the bottom of the post.

The tendency of Gravity force to push or pull is due to the tendency of graviton push and pull forces. When another graviton collides to another graviton it will either bounce or attach to the other graviton depending on the collision impact(velocity) and direction. If they attach(due to pull and push force in within the graviton) the two gravitons net force or sum force form a relation sharing equal net force since they will travel same net velocity or overall velocity, this is where central tendency of gravity force comes from. The more gravitons attach to form one Net Force the more graviton force increases therefore gravity increases due to matter increasing and so on....

E=mc^2

c^2=E/m where E=W=Fr (r stands for radius or distance or ‘s’)

therefore c^2= Fr/m

therefore c^2= c^2----when forces interact the forces hold true the constant c^2 in order to remain stable

therefore Fr/m= Fr/m

Fr/m=mar/m ----where F=ma

Fr/mm=ar/m ----where a=F/m

therefore Fr/mm=Fr/mm

Fr/mm=Fvt/mm ----v=r/t where r=vt

Fr/mm=Frt/tmm ----where v=r/t

therefore Frt/mm=Frt/mm

Frt/mm=Frr/vmm ----where t=r/v

Frtv/mm=Frr/mm ----where v=r/t

therefore Frr/mm=Frr/mm

where F=(Frr/mm)mm/r^2 ----(Frr/mm) is proportionality for example: F=Kq1q2/r^2-electromagnetic force

The "G" (=Frr/mm) is a way of expressing how the constant c is kept true mathematically when forces interact just the way c^2 is a mathematical expression expressing that c should always remain true in order for the matter to remain stable.