QuantumNet
Nov20-03, 03:14 AM
|---------------------|
s = ct
I claim that ptot = mc for all particles, and thereby:
p = mc = mv/(1 - (v/c)2)½
and thereby v = c/2½
If the Loretz-transformation is true for x, y, z and tcoord (and also for v) inside the referencesystems:
((1 - (v0/c)2)½) = ((1 - (v1/c)2)½)
The derivata is also true for dx, dy, dz and dtcoord (and also for dv) even if v is constant:
-v0/c2(1 - (v0/c)2)½ = -v1/c2(1 - (v1/c)2)½
(This is seen from a third referencesystem, the equation is alternatively 1 = -v1/c2(1 - (v1/c)2)½ seen from a second)
so:
dv0v/c2(1 - (v/c)2)½ = dv0v/c2(1 - (v/c)2)½
and (thereby)
a0v/c2(1 - (v/c)2)½ = a1v/c2(1 - (v/c)2)½
so m0a0v/c2(1 - (v/c)2)½ = m1a1v/c2(1 - (v/c)2)½ = Ftot
This means that the force m0 attracts m1 with is the force m1 attracts m0 with. It also means that if m0 x-doubles, a1 do to. if m0 thereafter y-doubles, the force has xy doubled.
We understand that we probably should look at this from the second referencesystem as in the case of de Broglie's instead of the third:
m0a0 = m1a1v/c2(1 - (v/c)2)½ = Ftot
and get that:
m0m1[/sub]v/c2(1 - (v/c)2)½ = F since v/c2(1 - (v/c)2)½ is what causes the acceleration.
But the force decrease sphearically and thereby we get that the total force between two bodies is m1m2v/c2(1 - (v1/c)2)½/(4\pi\r2)
But v/c2(1 - (v/c)2)½ = 1/(2c).
Thereby F = m1m2/(16\pi\r2c)
But c variates from medium to medium, therefor c in this case is the average lightspeed, ca, in the mediums between the masses.
G = 1/( 8 \pi\c_a)
I must add that the force variates with the form of the object, The force between two cylindershaped platinum weight at the distance one meter, both with the weight 1 kg is hard to calculate but seems to be more or less 1/( 8 \pi\c_a)
This means that nasa's problem with their satellite should not exist.
And that the force of gravity variates with length and temperature.
s = ct
I claim that ptot = mc for all particles, and thereby:
p = mc = mv/(1 - (v/c)2)½
and thereby v = c/2½
If the Loretz-transformation is true for x, y, z and tcoord (and also for v) inside the referencesystems:
((1 - (v0/c)2)½) = ((1 - (v1/c)2)½)
The derivata is also true for dx, dy, dz and dtcoord (and also for dv) even if v is constant:
-v0/c2(1 - (v0/c)2)½ = -v1/c2(1 - (v1/c)2)½
(This is seen from a third referencesystem, the equation is alternatively 1 = -v1/c2(1 - (v1/c)2)½ seen from a second)
so:
dv0v/c2(1 - (v/c)2)½ = dv0v/c2(1 - (v/c)2)½
and (thereby)
a0v/c2(1 - (v/c)2)½ = a1v/c2(1 - (v/c)2)½
so m0a0v/c2(1 - (v/c)2)½ = m1a1v/c2(1 - (v/c)2)½ = Ftot
This means that the force m0 attracts m1 with is the force m1 attracts m0 with. It also means that if m0 x-doubles, a1 do to. if m0 thereafter y-doubles, the force has xy doubled.
We understand that we probably should look at this from the second referencesystem as in the case of de Broglie's instead of the third:
m0a0 = m1a1v/c2(1 - (v/c)2)½ = Ftot
and get that:
m0m1[/sub]v/c2(1 - (v/c)2)½ = F since v/c2(1 - (v/c)2)½ is what causes the acceleration.
But the force decrease sphearically and thereby we get that the total force between two bodies is m1m2v/c2(1 - (v1/c)2)½/(4\pi\r2)
But v/c2(1 - (v/c)2)½ = 1/(2c).
Thereby F = m1m2/(16\pi\r2c)
But c variates from medium to medium, therefor c in this case is the average lightspeed, ca, in the mediums between the masses.
G = 1/( 8 \pi\c_a)
I must add that the force variates with the form of the object, The force between two cylindershaped platinum weight at the distance one meter, both with the weight 1 kg is hard to calculate but seems to be more or less 1/( 8 \pi\c_a)
This means that nasa's problem with their satellite should not exist.
And that the force of gravity variates with length and temperature.