If The Universe Is A Mochi...

If the Big Bang is really BIG, we would expect the system to be driven forward towards greater entropy that,

$m_p=\lim_{n\rightarrow\infty}\left\{{\cfrac{m}{2^n}}\right\}$

each of velocity,

$v=\sqrt{2}c$

as we have seen that under both conservation laws of energy and momentum the split masses do not lose velocity and have equal velocity.  The process does not cost energy and is expected to go on util all particles are of mass  $m_p$.

What happens after the big bang has gained maximum entropy?  It begins to reverse itself.  The energy released are slowly being nullified.  (Both forward and reverse process are simultaneous.)  From the post  "If The Universe Is A Banana...",  in equation  (1) we find that for each particle  $m_p$  there is another of the reverse velocity.  Pairing them up for collisions we have,

$m_pv-m_pv=0=2m_pu$

$u=0$

and

$m_pv+m_pv=2m_pu$

$u=v$

The direct collisions head-on, destroys velocity and incur a kinetic energy loss.  The resulting mass coalesce and has twice the initial mass.  We define the energy loss as,

$\Delta L=\cfrac{1}{2}m_pv^2+\cfrac{1}{2}m_pv^2=m_pv^2$

For the side collisions where both masses travel in parallel and coalesce , there is no energy cost.

$\Delta L_s=0$

Both type of collision are equally likely and we expect half of the moving  $m_p$  to be involved in each type of collisions.

Hypothetically, the  $2^{n-1}$  pair of particles will give rise to  $2^{n-2}$  collisions of each type with a total energy loss of,

$Loss_1 = 2^{n-2}\Delta L$

and  produce $2^{n-1}$  particles of mass  $2*m_p$,  ($2^{n-2}$  pairs).   Half of this with non zero velocity  ($2^{n-3}$  pairs) in turn can similarly collide,

$2m_pv-2m_pv=0=4m_pu$

$u=0$

and

$2m_pv+2m_pv=4m_pu$

$u=v$

The associated loss per collision is given by,

$\Delta L_2=\cfrac{1}{2}2m_pv^2+\cfrac{1}{2}2m_pv^2=2m_pv^2=2\Delta L$

Since only half of the pairs of particles of mass $2m_p$  actually incur energy cost ( ($2^{n-4}$  pairs), we have the total loss as,

$Loss_2 = 2^{n-4}2\Delta L= 2^{n-3}\Delta L$

A total of $2^{n-3}$  masses,  $4m_p$ are produced from the previous collisions, of which half  ($2^{n-4}$) have non zero velocity, of which there are  $2^{n-5}$  pairs  colliding to produce $8m_p$,  and half of this,  $2^{n-6}$  collide at a cost of

$\Delta L_4=\cfrac{1}{2}.4m_pv^2+\cfrac{1}{2}.4m_pv^2=4\Delta L$

The total loss as a result of this type of collision is,

$Loss_4 = 2^{n-6}4\Delta L= 2^{n-4}\Delta L$

The total process loss is thus given by,

$LP=Loss_1+Loss_2+Loss_4...$

$LP=2^{n-2}\Delta L+2^{n-3}\Delta L+ 2^{n-4}\Delta L...$

$LP=2^{n-1}\sum^n_1{2^{-i}\Delta L}$

$LP=2^{n-1}m_p2c^2\sum^n_1{\cfrac{1}{2^i}}=2^{n}m_pc^2\sum^n_1{\cfrac{1}{2^i}}$

Since,  $m_p=\cfrac{m}{2^n}$,

$LP=mc^2\sum^n_1{\cfrac{1}{2^i}}$

Taking the limit  $n\rightarrow\infty$,

$LP=\lim_{n\rightarrow\infty}{mc^2\sum^n_1{\cfrac{1}{2^i}}}=mc^2$

which is the amount of energy we started with.

So, the big bang as a whole considering both forward entropy gain (particles split) and reverse entropy loss (particle coalescence) is stable.  The net energy sum is zero.

Feeling mochi safer already.