**0**

# Time Of Annihilation Of Form Of Matter In Time Dependent Matrix

Posted by

**Robindranath Haldar**, 02 December 2012 · 3711 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 ( g= μ/2ds

Replacing μ by (s22t12-s12t22)/t22.t12 we get :

• ds =(s22t12-s12t22)/ 2g (t22.t12)

• g = (s22t12-s12t22)/ 2ds (t22.t12)

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 ( g= μ/2ds

Replacing μ by (s22t12-s12t22)/t22.t12 we get :

• ds =(s22t12-s12t22)/ 2g (t22.t12)

• g = (s22t12-s12t22)/ 2ds (t22.t12)