Newton Einstein ALEXANDER ABIAN,ARCHIMEDES PLUTONIUM a.k.a LUDWIG PLUTONIUM,OVE TEDENSTIG,NILS BÖRJESSON

Newton Einstein ALEXANDER ABIAN,ARCHIMEDES PLUTONIUM a.k.a LUDWIG
PLUTONIUM,OVE TEDENSTIG,NILS BÖRJESSON

M IS the mass of a particle at distance R
It's velocity is V.

It is a COSMIC particle, and Hence :it obey HUBBLE'S LAW.

(1) V=HR

We multiply


both sides by , M , giving :





(2) MV=MHR

MV is the MOTION p

giving :

(2) p=MHR

Newton's second law of force,: f=p'
"'" is the derivate.



, giving :

(3) f=M'HR+MH'R+MHR'

f is the COSMIC force, it has TWO terms:

(4) fgravity = -M.M1.G/R^2

Where M1 is the mass in a ball with radius R.
and
G is Newton's gravity constant, G;

The second force is the COMOLOGICAL FORCE

(5) fcosmologic = M.K.R

Where K is the COMOLOGICAL CONSTANT

giving : giving :

(6) -M.M1.G/R^2+ M.K.R = M'HR+MH'R+MHR'

ALEXANDER ABIAN GIVED THE FOLLOWING:
TIME HAS INERTIA and some energy is lost to move Time forward
E = mcc (Einstein) must be replaced by E = m(0) exp(-At) (Abian)

ARCHIMEDES PLUTONIUM a.k.a LUDWIG PLUTONIUM GIVED THE FOLLOWING:
RADIOACTIVE SPONTANEOUS ALPHA PARTICLE OR NEUTRON MATERIALIZATION

OVE TEDENSTIG GIVED THE FOLLOWING:
Our hypothesis is that a mass quantity of , dmt , is absorbed


by a body of mass , m , during time , dt. During time , T ,
the absorbed mass constitutes the mass of the by itself, hence
the time where the body has doubled its own mass.
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7) ==================





R = dma/(m.dt)





=====================

Out hypothesis now is that R corresponds to Hubble's cosmical
constant, H , having the same definition, namely the dimension
-1
of frequency 1/t or s .

NILS BÖRJESSON GIVED THE FOLLOWING:
BROUWER BROWN SHOUS!
EXPONETIAL: EXP(AT) A0 IS TIME FORWARD!
EXPONETIAL: EXP(-AT) A0 IS TIME BACKWARD!


Hence :


M' CAN'T bee 0!

EXPONETIAL: EXP(AT) A0 IS TIME FORWARD!
EXPONETIAL: EXP(-AT) A0 IS TIME BACKWARD!


E = m(0) exp(-At)
Substitute + for -
E = m(0) exp(+At)
substitute m for E
m = m(0) exp(+At)
m'=+A m(0) exp(+At)
m'=+A m
substitute H for A!
M'=+H.M.

giving :

(8) -M.M1.G/R^2+ M.K.R = H.M.HR+MH'R+MHR'

DIVISION BY M:

(9) -M1.G/R^2+ K.R = H.HR+H'R+HR'

R' is the velocity V from (1) V=HR

(10) -M1.G/R^2+ K.R = H.HR+H'R+HHR
(11) -M1.G/R^2+ K.R = 2H.HR+H'R

IF fgravity=0, then H'=0
giving :

(12) [-M1.G/R^2=0]+ K.R = 2H.HR+[H'R=0]
(13) + K.R = 2H.HR
(14) + K = 2H.H

giving :

(15) -M1.G/R^2+ 2H.H.R = 2H.HR+H'R
(16) -M1.G/R^2 = +H'R

IF G IS CONSTANT:
(17) G'=0



We make a summary of important formulae that have been
obtained, from which the , can be
calculated :


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R'=HR
+H'=-M1.G/R^3
M1'=+H.M1.
G'=0

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giving :



M1 ~ R
M1=C.R
+H'=- C.G/R^2
+H'H=- C.GR'/R^3

H^2/2+D/2= C.G/2/R^2

(R'/R)^2/2+D/2= C.G/2/R^2
(R')^2+D.R^2= C.G

I WILL SOLVE this for D=0 and for D0 and for D0

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D=0
(R')^2= C.G
R'=+SQRT(C.G)
or
R'=-SQRT(C.G)

R=+SQRT(C.G).T+E
or
R=-SQRT(C.G).T+F

THE SAME FOR THE MASS, BUT DIFFERENT CONSTANTS!
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D0
(R')^2+D.R^2= C.G
R''+D..R= 0

giving : Oscillation With Time-Period=2.PI.SQRT(1/D)
THE SAME FOR THE MASS, BUT DIFFERENT CONSTANTS!

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"""""""""""""""""""""""""""""""""""""""""""
D0
(R')^2+D.R^2= C.G
(R'')+DR= 0 D0
giving : Oscillation With IMAGINERY Time-Period=2.PI.SQRT(1/D)
THE SAME FOR THE MASS, BUT DIFFERENT CONSTANTS!

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ALL THE FORMULAS ARE IF G IS CONSTANT!
BUT BUT BUT BUT BUT!!!!!!!!!!!!!!!!!!!!!!