Constants or Proportionality?

The forum i refered to you was not as simple as i could put it since i was trying to explain to others my point. Let me try to simplify it:

As you know the particle graviton is not yet discovered or no one knows yet if it exists, many physicists and other people like me support the idea it does. Please see:http://hyperphysics.phy-astr.gsu.edu/hbase/particles/expar.html#c6

By studying the electromagnetic force one can predict the properties or behavior of an electromagentic particle, the same way for gravity force one can predict the properties or behavior of graviton...

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

In within graviton there are two forces, one force pushing outward and the other one is pushing inward or counter force. Force pushing outward we name it F<font size="1">o</font id="size1"> and force pushing inward we name it F<font size="1">i</font id="size1">. These forces are equal in magnitude yet opposite in dirrection. Therefore the Net Force or sum force of the graviton is zero. Fo+(-Fi)=F=0. Therefore the graviton is stable.

The velocity both forces balance out i postulated to be "c" then the graviton runs out of mass. Then F=F or c=c. Using the equation E=mc^2 I solved for c^2 to show that both foces hold this speed true in order for the graviton to be stable.

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

therefore c^2= c^2
therefore (Fo)r/m1= (Fi)r/m1

Notice the two forces are within the graviton therefore sharing equal amount of mass inorder to remain stable. Therefore it can be said the graviton has a set amount of energy.


Notice only in within the system or within the graviton is there any action. Here is an example from, hyperphysics.phy-astr.gsu.edu/hbase/thermo/inteng.html#c2 :
""For example, a room temperature glass of water sitting on a table has no apparent energy, either potential or kinetic . But on the microscopic scale it is a seething mass of high speed molecules traveling at hundreds of meters per second. If the water were tossed across the room, this microscopic energy would not necessarily be changed when we superimpose an ordered large scale motion on the water as a whole.""
Since it is said no 'isolated' force or matter-graviton in our case- exists in our universe, then a graviton naturaly is in contact with other matter basically gravitons. These other forces do not affect the two forces(Fo and Fi) within the graviton, but it affects the Net Force of the whole graviton, just like the glass can be tossed across the room the velocity is not enough to affect the water microscopic energy, only until the water experience extreme increase of velocity then the microscopic energy is affected. Only when the Net Force increases to extreme or net velocity or velocity of graviton as a whole experiences speed close to 'c' affects the forces within the graviton, and only when that speed passes 'c' the forces within the graviton become unbalance, which it does not occur naturally.

To explain the relation of a Graviton overall force to another Graviton, we start talking about mechanical forces, this relation is shown as:
<font color="red">(Fo)r/m1= (Fi)r/m1
Or Fr/m1=Frm1
Fr/m1=m2ar/m1 ----where F=m2a </font id="red">(<font color="red">IMPORTANT</font id="red">: graviton in contact with another graviton experiencing acceleration in relation to the collision)
<font color="red">Fr/m1m2=ar/m1 ----where a=F/m2
therefore Fr/m1m2=Fr/m1m2</font id="red">

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



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.







xeno

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by xenocidere</i>
<br />The forum i refered to you was not as simple as i could put it since i was trying to explain to others my point. Let me try to simplify it:

As you know the particle graviton is not yet discovered or no one knows yet if it exists, many physicists and other people like me support the idea it does. Please see:http://hyperphysics.phy-astr.gsu.edu/hbase/particles/expar.html#c6

By studying the electromagnetic force one can predict the properties or behavior of an electromagentic particle, the same way for gravity force one can predict the properties or behavior of graviton...

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

In within graviton there are two forces, one force pushing outward and the other one is pushing inward or counter force. Force pushing outward we name it F<font size="1">o</font id="size1"> and force pushing inward we name it F<font size="1">i</font id="size1">. These forces are equal in magnitude yet opposite in dirrection. Therefore the Net Force or sum force of the graviton is zero. Fo+(-Fi)=F=0. Therefore the graviton is stable.



The velocity both forces balance out i postulated to be "c" then the graviton runs out of mass. Then F=F or c=c. Using the equation E=mc^2 I solved for c^2 to show that both foces hold this speed true in order for the graviton to be stable.



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

therefore c^2= c^2
therefore (Fo)r/m1= (Fi)r/m1

Notice the two forces are within the graviton therefore sharing equal amount of mass inorder to remain stable. Therefore it can be said the graviton has a set amount of energy.


Notice only in within the system or within the graviton is there any action. Here is an example from, hyperphysics.phy-astr.gsu.edu/hbase/thermo/inteng.html#c2 :
""For example, a room temperature glass of water sitting on a table has no apparent energy, either potential or kinetic . But on the microscopic scale it is a seething mass of high speed molecules traveling at hundreds of meters per second. If the water were tossed across the room, this microscopic energy would not necessarily be changed when we superimpose an ordered large scale motion on the water as a whole.""
Since it is said no 'isolated' force or matter-graviton in our case- exists in our universe, then a graviton naturaly is in contact with other matter basically gravitons. These other forces do not affect the two forces(Fo and Fi) within the graviton, but it affects the Net Force of the whole graviton, just like the glass can be tossed across the room the velocity is not enough to affect the water microscopic energy, only until the water experience extreme increase of velocity then the microscopic energy is affected. Only when the Net Force increases to extreme or net velocity or velocity of graviton as a whole experiences speed close to 'c' affects the forces within the graviton, and only when that speed passes 'c' the forces within the graviton become unbalance, which it does not occur naturally.

<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">

So how do the gravitons differ in outward and inward force? As
a function of their velocity? Like if it is greater than c,
it has a net inward force and if less than c, it has a net
outward force?? How do they interact with matter to result
in gravitational effects or are they part of matter?


<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">

---

To explain the relation of a Graviton overall force to another Graviton, we start talking about mechanical forces, this relation is shown as:
<font color="red">(Fo)r/m1= (Fi)r/m1
Or Fr/m1=Frm1
Fr/m1=m2ar/m1 ----where F=m2a </font id="red">(<font color="red">IMPORTANT</font id="red">: graviton in contact with another graviton experiencing acceleration in relation to the collision)
<font color="red">Fr/m1m2=ar/m1 ----where a=F/m2
therefore Fr/m1m2=Fr/m1m2</font id="red">

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


<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">

Shouldn't there be a negative sign since the forces are
in opposite directions? If so, wouldn't c^2 = a negative
number for one of the forces? I'm lost.

Maybe you are onto something with the c^2 = c^2. It's
like if you started with pi = pi, the equations that
relate to pi all have circle properties?

Shouldn't there be a negative sign since the forces are
in opposite directions? If so, wouldn't c^2 = a negative
number for one of the forces? I'm lost.


It depends on the point of view of the force. Yes the dirrection is important.
For example if you push on a wall, the wall pushes back with equal force, to our point of view you are applying positive force and the force of the wall is negative, to the point of view of the wall its force is positive while yours is negative and so on...
The point is the (negative/positive) dirrection will be "important" when calculating the Net Force of your force and the walls force, but it won't be important when comparing the magnitude of your force and the walls. Because the negative can be either sides, so when I said Fo=Fi that is comparing the magnitude, c^2=c^2 is comparing the magnitude of the constant velocity of the forces, It is important to put dirrection of force when calculating the Net Force(F=-Fo+Fi OR F=Fo-Fi).




xeno

Quote:
How do they interact with matter to result
in gravitational effects or are they part of matter?



This is how I explained it on the other forum ( www.space-talk.com/ForumE/showthread.php3?threadid=1797 ):

"The tendency of Gravity force to push or pull is due to the tendency of graviton push and pull forces(Fo,Fi). 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...."
Yes, graviton is matter.




xeno