It's stated in my textbook that relativistic mass is conserved in collisions, even in inelastic ones. So if you have a particle with rest mass $m$ moving with speed $u$ (considerable fraction of the speed of light) in the lab frame and it collides with a stationary particle (as seen in lab frame) also of rest mass $m$ and it is given that the two particles coalesce into a new particle with rest mass $M$ (that moves with speed $v$ in the lab frame), then we can say that:

$\gamma(u)m+m=\gamma(v)M$

This is pretty much exactly what's written in my textbook. However, this inevitably leads to:

$\gamma(u)mc^2+mc^2=\gamma(v)Mc^2$

Therefore, this seems to show that the collision is elastic, as no energy is lost.

This has left me very confused, especially since it's shown here as well: http://www.feynmanlectures.caltech.edu/info/solutions/inelastic_relativistic_collision_sol_1.pdf

Could someone please help me to make sense of this?

## Best Answer

Elastic / inelastic refers not to conservation of energy but conservation of kinetic energy. In this case we have a incident particle with speed $u$ and energy $E$, mass $m$. We also have a stationary particle of mass $m$ and after the collision we will have a single particle of mass $M$. If the rest mass energy changes (i.e. $M \neq m$) then the kinetic energy must also have changed

because total energy is always conserved.(In fact in relativity an increase in thermal energy which in Newtonian mechanics would just be described as energy lost to heat will be described as an increase in rest mass.)

To be explicit about your example one can show that the composite particle has mass:

$$M = \sqrt{2(1+\gamma)} m > 2m$$

And so the total rest mass energy is increasing, the total energy is constant and thus the kinetic energy has gone down.