Science > Physics > Newton ALEXANDER ABIAN,ARCHIMEDES PLUTONIUM a.k.a LUDWIG PLUTONIUM,OVE TEDENSTIG,NILS BÖRJESSON
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Science > Physics |
| User: |
"NILS BÖRJESSON" |
| Date: |
04 Feb 2006 10:52:29 AM |
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Newton ALEXANDER ABIAN,ARCHIMEDES PLUTONIUM a.k.a LUDWIG PLUTONIUM,OVE TEDENSTIG,NILS BÖRJESSON |
Newton 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.
""""""""""""""""""""""""""""""""
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) A>0 IS TIME FORWARD!
EXPONETIAL: EXP(-AT) A>0 IS TIME BACKWARD!
Hence :
M' CAN'T bee 0!
EXPONETIAL: EXP(AT) A>0 IS TIME FORWARD!
EXPONETIAL: EXP(-AT) A>0 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 :
""""""""""""""""""""""""""""""
R'=HR
+H'=-M1.G/R^3
M1'=+H.M1.
G'=0
"""""""""""""""""""""""""""""""""""2
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 D>0 and for D<0
"""""""""""""""""""""""""""""""""
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!
""""""""""""""""""""""""""""""""""""
"""""""""""""""""""""""""""""""""""""""
D>0
(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!
"""""""""""""""""""""""""""""""""""""""""""""""""""""""
"""""""""""""""""""""""""""""""""""""""""""
D<0
(R')^2+D.R^2= C.G
(R'')+DR= 0 D<0
giving : Oscillation With IMAGINERY Time-Period=2.PI.SQRT(1/D)
THE SAME FOR THE MASS, BUT DIFFERENT CONSTANTS!
""""""""""""""""""""""""""""""""""""""""""""""""""""""""""""""""
ALL THE FORMULAS ARE IF G IS CONSTANT!
BUT BUT BUT BUT BUT!!!!!!!!!!!!!!!!!!!!!!
.
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