I think that space itself is not infinite. Space is also limited. Space does not stretch rather than spacetime fabric in which universe lie stretches.
We know mass bends spacetime fabric and after big bang everything is expanding. That means that here spacetime fabric is acting like a elastic substance.
As we know there is a law in physics called Hooke's law. It implies for all elastic objects until the point till which that object turns into plastic state.
According to Hooke's law:
F=K.x .........................................(I)
Where, F= force applied on that elastic object.
K=elasticity constant of that particular elastic substance.
x=distance covered by that elastic substance when force was applied to it.
Let us consider V as the volume of space which holds our universe.
As we know spacetime fabric can only be stretched till the space allows. Hooke's law can only be applicable till the object stays in elastic state. The point from which elastic changes into plastic is called maximum stretch limit or stretch limit. Stretch limit of spacetime is till the end of space.
Since, Hooke's law is only applicable till stretch limit we can take x as spacetime's stretch limit on one dimension.
We know that space forms three dimensions of spacetime fabric.
So, x*3 = volume of spacetime (V).
It means that any direction spacetime stretches it stretches till x*3 or V.
Then,
Cubing both sides of eqn(I), we get
F*3 = K*3 . x*3
Putting V=x*3 , we get
F*3= K*3 . V ...................................(II)
Then,
Force applied = Energy / Displacement
I.e. F= E/ s
Since the energy exerted on spacetime fabric is kinetic energy.
F= K.E. / s
Putting F= K.E. / s on eqn( II), we have
( K.E / s)*3 = K*3 . V
We know that K.E.= (m.v*2) / 2
{(m.v*2) / 2.s }*3 = K*3 . V
{( m*3 . v*6) / 8s*3} = K*3 . V
Since, v= s/ t
{( m*3 . s*6) / 8s*3 . t*6 } = K*3 . V
{( m*3 . s*3) / 8t*6} = K*3 . V
Cross multiplying, we get
{( m*3 . s*3) / 8K*3 . V } = t*6
Since x*3 = V,
{( m.s) / 2 K.x}*3 = ( t*2)*3
Cancelling powers we get
{( m.s.) / 2K.x}= t*2
Square rooting both sides, we have
t= √( m.s. / 2K.x) ............................(III)
Now let's move to eqn ( I ) again.
F= K.x
We know K= F / displaced distance( d).
F = F.x / d
By solving it, we get.
d = x ...........................................(IV)
Now let's go to eqn ( III)
t= √( m.s.d / 2F.x)
We know F= m.a
So,
t= √( m.s.d /2m.a.x)
t= √ ( s / 2a)
Thus
t = √ ( s / 2.a)
Where, t= time till which universe expands.
s= displacement from original place that is from big bang singularity.
a= acceleration of moving object.
This equation is dimensionally correct. It shows that universe expands only till certain time and then starts contracting again.
But this equation is only measured with the reference of one object. Hence to make it more informative, I add ∆ to both sides.
So, the final equation is:
∆t = √(∆s / 2.∆a)
I will someday prove this experimentally. But till then good bye. Explations about final equation will be there in next episode.
THANKS FOR READING. Stay tuned for more episodes of Black Hole.
