Shape Of Things To Come

Let's take a look at the cascade of collisions again,

We can identify a series of forward velocities after  $n$  number of collisions,

$v\left\{\cfrac { 2 }{ N_{ n }(\rho )+1 }\right\}^n$,  

$v\left\{\cfrac {N_{ n }(\rho )-1 }{ N_{ n }(\rho )+1 }\right\}^2\left\{\cfrac { 2 }{ N_{ n }(\rho )+1 }\right\}^n$,  

$v\left\{\cfrac {N_{ n }(\rho )-1 }{ N_{ n }(\rho )+1 }\right\}^4\left\{\cfrac { 2 }{ N_{ n }(\rho )+1 }\right\}^n$...

The total momentum of this forward particles is,

$P_{ps}=m_sv\left\{\cfrac { 2 }{ N_{ n }(\rho )+1 }\right\}^n lim_{m\rightarrow\infty}\left\{ 1+\left\{\cfrac {N_{ n }(\rho )-1 }{ N_{ n }(\rho )+1 }\right\}^2+\left\{\cfrac {N_{ n }(\rho )-1 }{ N_{ n }(\rho )+1 }\right\}^4+...\left\{\cfrac {N_{ n }(\rho )-1 }{ N_{ n }(\rho )+1 }\right\}^{2m}+... \right\}$

$=m_sv \left\{\cfrac { 2}{ N_{ n }(\rho )+1 }\right\}^n lim_{m\rightarrow\infty} \sum^m_0{\left\{\cfrac {N_{ n }(\rho )-1 }{ N_{ n }(\rho )+1 }\right\}^{2m}}$

$P_{ps}=m_sv \left\{\cfrac { 2}{ N_{ n }(\rho )+1 }\right\}^n\cfrac{1}{1-\left\{\cfrac {N_{ n }(\rho )-1 }{ N_{ n }(\rho )+1 }\right\}^2}$

$=m_{ s }v\left\{ \cfrac { 2 }{ N_{ n }(\rho )+1 }  \right\} ^{ n }\cfrac { (N_{ n }(\rho )+1)^{ 2 } }{ (N_{ n }(\rho )+1)^{ 2 }-(N_{ n }(\rho )-1)^{ 2 } }$

$=m_{ s }v\cfrac { 2^{ n } }{ (N_{ n }(\rho )+1)^{ n-2 } } \cfrac { 1 }{ (N_{ n }(\rho )+1)^{ 2 }-(N_{ n }(\rho )-1)^{ 2 } }$

$=m_{ s }v\cfrac { 1 }{ N_{ n }(\rho ) } \left\{ \cfrac { 2 }{ (N_{ n }(\rho )+1) }  \right\} ^{ n-2 }$

If  $N_{ n }(\rho )>1$  the forward momentum dies down eventually as  $n\rightarrow\infty$,  after many collisions.

And a series of backward velocities,

$v\left\{\cfrac {N_{ n }(\rho )-1 }{ N_{ n }(\rho )+1 }\right\}\left\{\cfrac { 2 }{ N_{ n }(\rho )+1 }\right\}^n$,  

$v\left\{\cfrac {N_{ n }(\rho )-1 }{ N_{ n }(\rho )+1 }\right\}^3\left\{\cfrac { 2 }{ N_{ n }(\rho )+1 }\right\}^n$,  

$v\left\{\cfrac {N_{ n }(\rho )-1 }{ N_{ n }(\rho )+1 }\right\}^5\left\{\cfrac { 2 }{ N_{ n }(\rho )+1 }\right\}^n$...

The total momentum of this backward particles is,

$P_{ps\,r}=m_sv\left\{\cfrac { 2 }{ N_{ n }(\rho )+1 }\right\}^n\left\{\cfrac {N_{ n }(\rho )-1 }{ N_{ n }(\rho )+1 }\right\} lim_{m\rightarrow\infty}\left\{ 1+\left\{\cfrac {N_{ n }(\rho )-1 }{ N_{ n }(\rho )+1 }\right\}^2+...\left\{\cfrac {N_{ n }(\rho )-1 }{ N_{ n }(\rho )+1 }\right\}^{2m}+... \right\}$

$=m_sv\left\{\cfrac { 2 }{ N_{ n }(\rho )+1 }\right\}^n\left\{\cfrac {N_{ n }(\rho )-1 }{ N_{ n }(\rho )+1 }\right\}  lim_{m\rightarrow\infty} \sum^m_0{\left\{\cfrac {N_{ n }(\rho )-1 }{ N_{ n }(\rho )+1 }\right\}^{2m}}$

$=m_{ s }v\left\{ \cfrac { 2 }{ N_{ n }(\rho )+1 }  \right\} ^{ n }\cfrac { N_{ n }(\rho )-1 }{ N_{ n }(\rho )+1 } \cfrac { (N_{ n }(\rho )+1)^{ 2 } }{ (N_{ n }(\rho )+1)^{ 2 }-(N_{ n }(\rho )-1)^{ 2 } }$

$=m_{ s }v\left\{ \cfrac { 2 }{ N_{ n }(\rho )+1 }  \right\} ^{ n-1 }\cfrac { N_{ n }(\rho )-1 }{ 2N_{ n }(\rho ) }$

$=m_{ s }v\left\{ \cfrac { 2 }{ N_{ n }(\rho )+1 }  \right\} ^{ n-1 }.\cfrac { 1 }{ 2 }\left\{ 1 -\cfrac { 1 }{ N_{ n }(\rho ) }  \right\}$

If  $N_{ n }(\rho )>1$ then the backward momentum also attenuate to zero.  When  $N_{ n }(\rho )=1$  there is no backward momentum, the particle has zero velocity.

From the post "Collisions And More Collisions" the last graph shows that when  $N_{ n }(\rho )=2$ velocity the attenuate very quickly within 10 collisions.  This would suggest that under normal circumstances  $N_{ n }(\rho )=1$.  That the forward momentum is carried from particle to particle without loss.