# Inelastic collision

1. Apr 20, 2006

### terry

Say we have 2 particles moving in the same direction, which undergo an elastic collision. Take the rest mass of particle to be $$m_1$$, the rest mass of particle 2 to be $$m_2$$, with velocities $$u_1,u_2$$, with $$u_1>u_2$$.

Assume they collidie inelastically to form a new particle m3

Are my conservation laws

$$m_1 \gamma_1+m_2 \gamma_2 = m_3 \gamma_3$$
$$m_1 \gamma_1 u_1 + m_2 \gamma_2 u_2 = m_3 \gamma_3 u_3$$
?

2. Apr 20, 2006

### pervect

Staff Emeritus
Yes, assuming the single occurence of "elastic" was a typo and that the collision is inelastic. The first expression is the result of the conservation of energy, the second expression is the result of the conservation of momentum.

This assumes that nothing is radiated away from the third particle (no other particles, no significant thermal energy).

The last assumption may be a bit difficult to achieve in practice, relativistic particle collisions often generate other particles, and macroscopic relativistic collisions would probably release a fair amount of energy causing the resulting fireball to radiate significantly.

3. Apr 20, 2006

### terry

Yeah, sorry inelastic collision.

Has anyone seen a general form for m3/u3?

I was asked to show that

$${m_3}^2=m_1^2+m_2^2+2m_1m_2(1-u_1u_1)$$

$$u_3=\frac{m_1 \gamma_1 u_1 + m_2 \gamma_2 u_2}{m_1 \gamma_1 + m_2 \gamma_2}$$

Its the expression for m3 that bugs me. I tried picking m1,m2,u1,u2, then using the formulas to get m3,u3, and it doesn't seem to satisfy my conservation equations.

Last edited: Apr 20, 2006
4. Apr 21, 2006

### pervect

Staff Emeritus
I also do not get that expression for m3.

I will give you what I compute for (m3^2 - m1^2 - m2^2) if you give me what you compute. (The result is simple, the computation is rather messy, it was greatly aided by using computer algebra).

5. Apr 21, 2006

### nrqed

But if it was inelastic, energy would not be conserved...Or am I missing something? (I know that in classical mechanics when two objects get stuck together the collision is inelastic but here the mass is not conserved so the situation is different)

6. Apr 21, 2006

### nrqed

Are you sure there is no factor $\gamma_1 \gamma_2$ multiplying the last term?

Using four-vectors, it is easy to calculate m_3^2. One has
$P_3 = P_1 + P_2 \rightarrow P_3^2 = P_1^2 + P_2^2 + 2 P_1 \cdot P_2 \rightarrow m_3^2 = m_1^2 +m_2^2 + 2 \gamma_1 \gamma_2 m_1 m_2 - 2 \gamma_1 \gamma_2 m_1 m_2 u_1 u_2$

Patrick

EDIT: I had forgotten the factor of 2 multiplying the last two terms. Sorry

Last edited: Apr 22, 2006
7. Apr 21, 2006

### pervect

Staff Emeritus
*IF* you include the energy in heat (molecular motion), energy is conserved even in an inelastic collision.

8. Apr 21, 2006

### terry

Thats what I got...albeit not so easily.

*sigh*...I guess the question was wrong.

Thanks for the help

9. Apr 21, 2006

### nrqed

Yes, this is absolutely right. I should have been more careful with the wording. I meant that kinetic energy is not conserved in an inelastic collision in classical mechanics.

Now going back to your comment that the collision in this problem is inelastic, what did you mean? You did not mean that the (relativistic) kinetic energy was not conserved since it is one of the two equations used. So what is your definition of inelastic collision in this context?

Regards

Patrick

10. Apr 22, 2006

### pervect

Staff Emeritus
Nice work on the 4-vector approach, the calculation was a lot simpler when you keep E1,E2,P1,P2 as symbols.

Anyway, onto inelastic collisions. An inelastic collision just means that v1=v2 after the collision.

The important facts about systems of particles are these

1) The total energy of all the particles involved add

2) The total momentum of all the particles involved add

3) The mass (invariant mass) m of any specific system or subsystem of particles is given by m^2 = E^2 - p^2 in geometric units, where E is the energy of the system or subsystem, and p is its momentum.

Masses don't necessarily add, as can be seen from the defintion.

4) As long as the system is isolated (not interacting with something else), the invariant mass of the system will be invariant. This can be shown with a particle-swarm model, if and only if every particle interacts with other particles only when they are at the same point. Particles that interact with each other via long-range forces need more attention.

If the particles interact with each other via fields, one has to include the energy and momentum of the fields into the system to get a coherent picture in which energy and momentum remain conserved. One might also be able to picture the field interactions as occuring by "virtual" particles to get the same result.

It's not particularly relevant to this problem, which is an isolated system but an example of a problem where a non-isolated system gives a non-invariant mass E^2-p^2 can be found