Conservation of mass in relativistic collisions?

Question

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?

Answer 1

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.

Answer 2

It's stated in my textbook that relativistic mass is conserved in collisions, even in inelastic ones.

$\let\g=\gamma$ It's true simply because relativistic mass is nothing but energy (a factor $c^2$ without) and energy is always conserved in SR.

I would be curious to know how your book defines an inelastic collision. I try a guess: a collision in which mass transforms into energy or vice versa. I say this because I have seen several books proceding this way:

• defining relativistic mass
• stating that in an inelastic collision mass converts in energy or vice versa.

I hope you see the contradiction: you can't assert that (relativistic) mass is always conserved, and at the same time define an inelastic collision as one in which that doesn't happen.

A correct definition of inelastic collision is as one in which the sum of rest (i.e. invariant) masses is not conserved. Your instance would be inelastic if $M\ne2m$.

But there is more (and this is why I wrote "would be"). For generic values of $m$, $M$, $u$ that process is impossible. The reason is momentum conservation. Surely you know that momentum is $p=m\g(u)u$. Let's write both conservation equations, one after the other: $$m\,\g(u) + m = M\,\g(v) \tag1$$ $$m\,\g(u)\,u = M\,\g(v)\,v.$$ Multiply (1) by $c$, square both equations and subtract. Recalling definition of $\g$ after some algebra you will arrive at $$2 m^2\,[1 + \g(u)] = M^2.$$ So your reaction may only happen if $m$, $u$, $M$ are accurately chosen. If $m$ and $M$ are the masses of two real particles you had to adjust $u$ to an exact value, which is experimentally impossible. The least inaccuracy would mean failure.

Answer 3

I think the non-relativistic definition that
an "elastic collision" conserves the total kinetic-energy can be generalized to the relativistic case by saying that
an "elastic collision" conserves the "total relativistic KINETIC-energy".

Note that "total relativistic energy" (being the time-component of the total 4-momentum) is always conserved (since the total 4-momentum is conserved). Thus, $$\gamma_{1} m_1 +\gamma_{2} m_2 = \gamma_{3} m_3 +\gamma_{4} m_4 $$ or (better) in terms of rapidities $$m_1 \cosh\theta_{1}+m_2 \cosh\theta_{2} = m_3 \cosh\theta_{3}+m_4 \cosh\theta_{4}$$

If additionally total relativistic kinetic energy is conserved (an "elastic collision"), then $$m_1 (\cosh\theta_{1}-1)+m_2 (\cosh\theta_{2}-1) = m_3 (\cosh\theta_{3}-1)+m_4 (\cosh\theta_{4}-1).$$ By subtracting, we have $$m_1+m_2=m_3+m_4,$$ which says that total-rest-mass is conserved for an "elastic collision".
(If $m_3=m_1$ and $m_4=m_2$, then the particles retain their identities after the collision.
I'm not sure, but maybe this condition is somehow enforced by another condition (maybe having to do with momentum transfers). )

Answer 4

I think M could be thought as a new particle. Since two particle cannot be "stick". There cant be proton-electron but maybe a new particle with their masses.

Answer 5

The word "elastic" is used in different ways in different parts of physics. Sorry, but there it is. In the present example it is being used to refer to collisions where the rest mass of entities does not change. This is different from its usage in Newtonian collision physics where typically it means that kinetic energyis conserved.

Note that total energy is always conserved, so there is not much point in having a special term to mean "energy-conserving".

On a separate note, it is widely considered, and it is my opinion, that it does not help our understanding to refer to $\gamma m$ as "relativistic mass". Rather, $\gamma m c^2$ is energy and $\gamma m v$ is momentum and $\gamma m$ is not sufficiently important to earn its own name (except that it can and should be called energy when the units are such that $c=1$). It is better to avoid the notion of "conservation of mass"; it only leads to confusion. Better to stick to conservation of energy and conservation of momentum.