Consider a classical isolated relativistic elastic collision of two free scalar point masses in 1 D. The aim is to get the final velocities of the particles given

  1. initial rest masses and
  2. initial velocities

in some inertial reference frame.

The question is as follows: Is there any motivation to assume that the rest masses of each particle after the collision are same as before?

If not so,

  1. is it correct to say that the above problem becomes solvable only once the rest masses after collision are also given?
  2. For that matter, is it correct to further say that nothing (the no of particles produced or their masses) about the system after the collision is known except the 4 momentum of the system?

Please note that the fact that we are asking the collision to be elastic doesn't imply the assumption being asked about--it only says the total kinetic energy be conserved(or does it?) which is already taken care of in 4 momentum conservation. Also note that though rest mass of the system is necessarily $\ge$ sum of individual rest masses, it doesn't rule out the masses to stay the same across the collision.


1 Answer 1


Short answer to the title question: "No, masses are fixed in elastic collisions."

The issue here is that mass is energy, and changing the mass of one or more particles requires that the enegy come from (or go to) some channel.1 But in the microscopic realm many of the channels that confound us at the human scale are unavailable or too small to matter.

A perfectly workable (if rather casual) way of defining an elastic collision in the sub-atomic realm might be "the same set of particles come out as went in".

This differs a little from the usually way the issue is discussed in the macroscopic realm where energy gains and losses to changing internal state of the participants are too small to be noticed as changes in mass (they are changes in mass, but who can tell?).

In other words, in the macroscopc realm you will still say "that's my car" even when a great deal of energy has been trapped in permanant deformations of the bumper and body panels, so object identity is not a means of defining elasticity. But in the microscopic realm, when you stash some energy in a proton, you identify the particle that comes out as (say) a Delta-plus (a differnt particle).

However, you seem to have a misconception.

Please note that the fact that we are asking the collision to be elastic doesn't imply the assumption being asked about--it only says the total kinetic energy be conserved(or does it?) which is already taken care of in 4 momentum conservation.

This is not correct. Four-momentum can be (and is) conseved in collisions which excite some of the particles (inelastic) or create wholey new particles (also ineastic). Indeed an at-threshold particle creation reaction can result in zero three-momentum in the CoM (a maximally inelastic collision) while conserving four-meomentum.

Even in the intro class we expect that all colisions conserve energy. Elasticity is defined by the special property of conserving kinetic energy.

1 Quibble. There is the case where the participants exchange internal energy resulting in a no net change of kinetic energy. Say charged-current sacttering of electrons and neutrinos.

  • $\begingroup$ ok 4 momentum conservation doesn't imply elasticity. Is pair production elastic or inelastic? $\endgroup$
    – lineage
    Commented Oct 17, 2019 at 23:52
  • $\begingroup$ "or create wholey new particles (also ineastic)" $\endgroup$ Commented Oct 17, 2019 at 23:54
  • $\begingroup$ how does KE conservation imply particle identities stay same?or as you said, is it that we say the collision is elastic only when the particle identities stay same and therefore their KEs must be conserved? $\endgroup$
    – lineage
    Commented Oct 17, 2019 at 23:59
  • $\begingroup$ Let's start from the other side. A inelastic collision is one where kinetic energy is lost (or gained). Where can it go? In macroscopic collisions it can go lots of places. there are tons of channels available and some of them can hide a lot of energy. In the sub-atomic realm, however, you can (de-)excite a particle or you can create or destroy particles. And different excitation states are treated as different particles. So inelasticity means that you changed either the number or the identity of particles. Elastic collisions are left as the case where neither has changed. $\endgroup$ Commented Oct 18, 2019 at 0:04
  • $\begingroup$ Are you saying that pair production is inelastic? because in pair production (I am talking about a single photon near an atomic nucleus), the energy of the particles (photon before, electron positron after) is basically the same. I thought only the atomic nucleus (where nearby the pair production happens for conservation of momentum) gets a recoil (momentum gets transferred to the nucleus), but the energy (of the photon and electron positron pair) does not change (does not get transferred to the atomic nucleus). Thus I thought this is elastic. $\endgroup$ Commented Oct 18, 2019 at 1:36

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