# [Special Relativity] Energy-momentum invariant question

1. Feb 3, 2008

### Inquisitus

I was just wondering why what I've done in a spec rel question is wrong.

1. The problem statement, all variables and given/known data

A particle of mass m is travelling at 0.8c with respect to the lab frame towards an identical particle that is stationary with respect to the lab frame. If the particles undergo an inelastic collision and coalesce, what is the fractional change in the system's kinetic energy as measured from the lab frame?

http://img85.imageshack.us/img85/6922/croppercapture1io7.png [Broken]

2. Relevant equations

$$p=\gamma m v$$
$$E^2 = (mc^2)^2 + (pc)^2$$

3. The attempt at a solution

http://img183.imageshack.us/img183/6559/croppercapture3vr8.png [Broken]

My lecturer marked the first line here as being wrong; he said that for the momentum term in the denominator I should have put 2m rather than just m. However, my thinking here was that since the momentum of a system is conserved within any given frame, the momentum prior to the collision should be the same as following the collision, which is why I used $$\gamma m v c$$ rather than $$\gamma m_1 v_1 c$$. This makes things vastly simpler as we know the velocity of the moving particle prior to the collision, but we don't know the velocity of two particles after the collision.

Has my lecturer just misunderstood what I was doing, or am I doing something wrong?

Last edited by a moderator: May 3, 2017
2. Feb 3, 2008

### StatusX

Your logic is right, but you've computed the ratio of the total energies, not the kinetic energies. Just subtract out the total rest mass in the numerator and denominator and it should work.

3. Feb 4, 2008

### Inquisitus

Whoops, I actually meant total energies rather than kinetic energies, sorry

I'll see my lecturer tomorrow and see what he says. Thanks for the help!

4. Feb 4, 2008

### Ich

The total energy of the system cannot change. And the system's mass is definitely not 2m. The problem seems well stated, but neither your approach nor what your tutor told you seem correct.

5. Feb 4, 2008

### Inquisitus

Well the question asked what the fractional change in the systems total energy is upon the particle's collision; it was just a type in my post. The question, everything I've learnt, my solution to the problem, and my lecturer all say that this is possible

6. Feb 4, 2008

### StatusX

Sorry, I was assuming that since this is an inelastic collision, we shouldn't expect energy to be conserved (ie, some of it would go into heat, sound, etc). But assuming these are the only two particles around, and there is no radiation, the only way for energy to be conserved is if the rest mass of the particles increases (as would happen, for example, if some of the energy is converted into heat). In this case we have, both before and after the collision:

$$p^{tot}_\mu= ( \gamma m c^2, \gamma m v, 0, 0) + (mc^2,0,0,0)$$

$$= ( (\gamma + 1 ) m c^2, \gamma m v, 0, 0)$$

The velocity of a single particle with four momentum $p_\mu$ can be read off as $v_i = c^2 p_i/p_0$, giving the resulting particle a velocity of:

$$v' = c^2 \frac{\gamma m v}{ (\gamma + 1 ) m c^2} = \frac{\gamma}{\gamma+1} v$$

and a rest mass of

$$m' = \frac{1}{c^2}\sqrt{p_0^2 - c^2p_1^2} = \sqrt{ (\gamma+1)^2 m^2 c^4 - \gamma^2 m^2 v^2 c^2 } = m \sqrt{ (\gamma+1)^2 - \gamma^2 \frac{v^2}{c^2}}$$

$$= m \sqrt{ (\gamma+1)^2 - (\gamma^2 -1)} = m \sqrt{2(\gamma+1)}$$

Last edited: Feb 4, 2008