Inelastic Collision: Conservation Laws and Calculations

[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
?

 

[pervect]

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
?

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.

 

[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.

 

[pervect]

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).

 

[nrqed]

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.
.

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)

 

[nrqed]

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)

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

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.

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

 

[pervect]

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)

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

 

[terry]

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 + \gamma_1 \gamma_2 m_1 m_2 - \gamma_1 \gamma_2 m_1 m_2 u_1 u_2

Patrick

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

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

Thanks for the help

 

[nrqed]

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

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

 

[pervect]

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 occurring 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

https://www.physicsforums.com/showthread.php?t=117773

Another way of describing the issue here - the system m3 has relativistically significant amounts of heat energy, so one needs a theory of relativistic thermodyamics to model the system m3 in detail.

One simple paper that I've read recently that I like is
http://arxiv.org/PS_cache/physics/pdf/0505/0505004.pdf

but I really haven't read extensively about relativistic thermodynamics, there may be better approaches out there that I'm not aware of.

The bottom line is that the answer to this problem is OK, but the problem actually involves more advanced elements that are being glossed over to get a simple answer.