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