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Time Of Annihilation Of Form Of Matter In Time Dependent Matrix

Posted by Robindranath Haldar, 02 December 2012 · 3489 views

Proposed Calculation of Time of Annihilation of Form of Matter in Time Dependent Matrix.
PROPOSITION: From R.N.HALDAR,
47,GURUPADA HALDAR ROAD, KOLKATA-700026
The forms of matter change with time. The subatomic dissipates to take another form or goes to its original form due to internal and external work done under certain and various forces. I have proposed here that the time of such annihilation is inversely proportional to the total work done by the system or upon the system in 3-dimentional expansion or contraction.
Mathematically, TDEATH ∞ total work done by the system or upon the system in 3-dimentional expansion or contraction
Tdeath = 1/total work done by the system or upon the system in 3-dimentional expansion or contraction
or, Tdeath = K/ WTotal where, K= a constant ; WTOTAL= Total work done by the system
= K/ P x ds where P = Force, and
ds = the displacement in 3-dimentional expansion or contraction
= K/mg.ds where m = mass; g= gravitational force
We get, Tdeath = K/ mg.ds …………………( 1 )
Alternatively,
When, Tdeath = K/ WTotal = K/⅟2(mv2-mu2) =2K/m (v2-u2)……………….(2)
Where, v=final velocity & u=initial velocity
Now the term (v2-u2) can also be expressed as
(v2-u2) =(s2/t2)2-(s1/t1)2 where, s2=final spatial quantity & s1=initial spatial quantity.
=(s22t12-s12t22)/t22.t12 s1 & s2 are expressed in volume.
= μ (say) s2=volume of universe at time t2
s1 = volume of universe at time t1
t1 = time of beginning/birth,
t2 = time at which calculation is derived.
Here we can derive as follows:
When t1 = 0, s1 = 0 then (v2-u2) =∞ and thereby Td =2K/m (v2-u2) = 2K/m ∞ = 0,
So, t1 ≠ 0 It proves the universe was always present and there was no time when universe
was not there. It has no end and no beginning.
We know: t1 = 10-34 sec. Tpresent = 4.6x109x12x30x24x60x60 sec = 1.43 x 1017sec
Again
By putting the value of (v2-u2) in (2) we get, Td = 2K/mμ……………..(3)
Now from eq.1 & eq.3 we get, Td = K/ mg.ds =2K/mμ
Or K/ mg.ds =2K/mμ…………….(4)
Or 1/mg.ds =2/mμ
Or mg.ds =mμ/2
Therefore, a) ds = μ/2g (B) g= μ/2ds
Replacing μ by (s22t12-s12t22)/t22.t12 we get :
• ds =(s22t12-s12t22)/ 2g (t22.t12)
• g = (s22t12-s12t22)/ 2ds (t22.t12)





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