Bouncy Balls, Sticky Balls, Transfer Of Momentum

Consider this,

$m_{ \rho \, p }c^{ 2 }=m_{ \rho  }c^{ 2 }-\int _{ 0 }^{ 2x_{ z} }{ \psi  } dx$

If the particle  is to deform along $x_z$,

$\cfrac { \partial \, m_{ \rho \, p } }{ \partial \, x_{ z } } c^{ 2 }=-2\psi (x_{ z })$

where $\psi(x_z)$ is a function in $x_z$.  $x_z$ marks the maximum of $\psi$.

$\cfrac { \partial \, m_{ \rho \, p } }{ \partial \, x_{ z } } \cfrac { \partial \, x_{ z } }{ \partial \, t } c^{ 2 }=-2\psi (x_{ z }) \cfrac { \partial \, x_{ z } }{ \partial \, t }$

$\cfrac { \partial \, m_{ \rho \, p } }{ \partial \, t } c^{ 2 }=-2\psi (x_{ z }) \cfrac { \partial \, x_{ z } }{ \partial \, t }$

The rate of change of momentum is force,

$F=\cfrac { \partial \, m_{ \rho \, p } }{ \partial \, t } c=-\cfrac{2}{c}\psi(x_{ z }) \cfrac { \partial \, x_{ z } }{ \partial \, t }$  per unit volume

the negative sign indicates that this force opposes the change in $x_z$.  The impulse over time of the deformation is,

$\Delta\,p= \int _{ 0 }^{ t_{ p } }{ F }\, d\, t=-\cfrac{2}{c}\int _{ 0 }^{ t_{ p } }{ \psi (x_{ z }) } \cfrac { \partial \, x_{z } }{ \partial \, t } d\, t=-\cfrac{2}{c}\int _{ x_{zo} }^{ x_{zf} }{ \psi (x_{ z })  } d\, x_{ z }$

So, the change in momentum per unit volume is the

$\Delta\,p=-\cfrac{2}{c}\int _{ x_{zo} }^{ x_{zf} }{ \psi } d\, x$

where $x_{zf}$ marks the deformation in $\psi$ from $x_{zo}$ and $\Delta p$ is in the opposite direction to the change $x_{zf}-x_{zo}$.

When $\cfrac{\partial\,x_z}{\partial\,t}=0$, $F=0$, thus the particle does not recover from the deformation.  Momentum is loss from the particle, the agent that cases the deformation $x_{zo}\rightarrow x_{zf}$ gains the momentum.

If the particle disintegrate completely,

$\Delta\,p=-\cfrac{2}{c}\int _{ x_{zo} }^{ 0 }{ \psi } d\, x=\cfrac{2}{c}\int ^{ x_{zo} }_{ 0 }{ \psi } d\, x=\cfrac{2}{c}\int ^{ x_{z} }_{ 0 }{ \psi } d\, x$

If the particle has mass of radius $a$,

$\Delta\,p=-\cfrac{2}{c}\int _{ x_{zo} }^{ a }{ \psi } d\, x=\cfrac{2}{c}\int ^{ x_{zo} }_{a }{ \psi } d\, x=\cfrac{2}{c}\int ^{ x_{z} }_{a }{ \psi } d\, x$

As such $\psi$ around the particle can be transferred as momentum.