Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Suppose I have an inelastic head on collision between two idential particles of mass $m$ that come to rest in the centre of momentum frame where relativistic momentum is obviously conserved. If I now switch to the proper frame of one of the particles, the velocity of the other before the collision is $2v\gamma^2$, and after they both travel with velocity $v$.

collision

This would appear to suggest that relativistic momentum isn't conserved in the proper frame of either particle if I write the initial relativistic momentum as $(\gamma_{2v\gamma^2}) (m)(2v\gamma^2)$ and the final as $(\gamma_v) (2m)(v)$. So where is the error in this reasoning?

share|improve this question

1 Answer 1

up vote 1 down vote accepted

The collision is inelastic. In either frame some of the initial kinetic energy is lost.

Where can it go?

  1. It could be carried off by massless ejecta (photons).
  2. It could remain as a binding energy of the combined system or in internal modes of the combined system. Either way the mass, $M$, of the combined system is greater than $2m$.
  3. Or some combination of the above.

Your calculation assumes the final mass is $2m$, which is not true unless you have failed to tell us about some relevant momenta in the ejecta.

Assuming case 2, let's figure the kinematics in your second frame (one mass stopped); I'm going to use $c=1$ units and every mass I write is a rest mass:

$$ E_f = \sqrt{M^2 + P^2} = E_i = m + \sqrt{m^2 + p^2} $$

Applying conservation of momentum we can set the $P = p$, so we solve for $M$

$$ M^2 = m^2 + 2m\sqrt{m^2 + p^2} + m^2 + p^2 - p^2 $$

or

$$ M^2 = 2m^2 \left[ 1 + \sqrt{1 + \left(\frac{p}{m}\right) ^2} \right] $$

which is a bit more than $2m$.

share|improve this answer
    
Briefer: conservation of energy implies that the rest energy of the two particle system after the collision is the same as before (ie. $= 2m\gamma$, not 2m, and conservation of momentum clearly holds). –  James Jun 10 '12 at 12:32
    
They still both come to rest in the com frame, so the final velocity is still $v$ in the other frame. –  Larry Harson Jun 10 '12 at 13:00
    
Er...yes. Of course you are right about the velocity. –  dmckee Jun 10 '12 at 16:22

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.