# Invariant Mass/Inelastic Collision Problem

1. Dec 9, 2012

### demoncore

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

Suppose two "small" particles of equal mass m collide, annihilate each other, and create another particle of mass M > 2m . (Note that the final state is just that one "big" particle, nothing else.) If one of the small particles is initially at rest, what must be the minimum total energy E of the other? Give your final result in terms of m and M.

2. Relevant equations

The problem also asks me to set c=1.

Invariant mass: (m)^2 = E^2 - p^2

Energy conservation:

E_initial = 2m + E_k,m = E_final = M + E_k,M

momentum conservation:

p_initial = p_final

3. The attempt at a solution

I've been wrestling with this problem in a lot of different ways (studying for finals), but cannot get the posted answer. I am confused about which quantities are conserved, since it seems there is no reference frame where all the particles are at rest. I'm not sure what values to equate in the energy four-vector. Here is one try, where I take the final momentum to be zero:

Energy conservation:

2m + E_k,m = M

=> E_k,m = M - 2m

momentum:

p_initial = p_final = 0

Invariant mass:

(m)^2 = E^2 - p^2
=> (2m +E_k,m)^2 - 0 = M^2 - 0
=> 4m^2 + 4mE_k,m + E_k,m = M^2

Already, I can tell that I've done something wrong. I also think that I'm solving for the wrong value here, since I don't have any idea of how to get the energy of the one 'moving' particle from E_k,m.

I chose this (center of mass?) frame because I can't think of any other way to write the momenta of the particles in terms of the kinetic energies without doing Lorentz gymnastics that don't seem at all appropriate to the problem. Since it doesn't seem to help me, I think I must be misunderstanding the situation in some serious way, if not completely misapplying these principles.

Thanks,
A.

2. Dec 9, 2012

### haruspex

Surely you can only do that if you take a reference frame in which the initial total momentum is zero. And having obtained an answer, you'd need to calculate the answer for the given reference frame (in which one particle was initially at rest).
So why not just stick with the given frame? What is the initial total momentum in that frame? What, then, is the momentum of the resulting particle, and thus what is its total energy?

3. Dec 9, 2012

### demoncore

Well, this is what i tried:

p_initial^2 = (2m + E_k,m)^2 - 4m^2 = E_k,m^2 + 2mE_k,m = p_final^2

=> E,M,tot^2 = E_k,m^2 + 2mE_k,m + M^2 = (M +E_k,M)^2 = M^2 + 2ME_k,M + E_k,M^2

=> E_k,m^2 + 2mE_k,m = E_k,M^2 + 2ME_k,M

E_initial = 2m + E_k,m = E_final = M + E_k,M
=> E_k,M = (2m - M) + E_k,m

=> E_k,m^2 + 2mE_k,m = ((2m - M) + E_k,m)^2 + 2M((2m - M) + E_k,m)
= (2m-M)^2 + 2(2m -M)E_k,m + E_k,m^2 + 2M(2m - M) + 2ME_k,m

=> 2mE_k,m = (2m-M)^2 + 2(2m -M)E_k,m + 2M(2m - M) + 2ME_k,m
= (2m-M)(2m-M + 2E_k,m + 2M) + 2ME_k,m
= (2m-M)((2m + M) + 2E_k,m) + 2ME_k,m
= (2m-M)(2m + M) + 2E_k,m(2m-M + M)
= (2m-M)(2m + M) + 4mE_k,m

=> E_k,m = -(2m-M)(2m + M)/(2m) = (M^2 - 4m^2)/2m

E_tot,m = (M^2 - 4m^2)/2m + m = (M^2 - 2m^2)/2m

This is the nearest I've gotten to the right answer, which should be:

(M^2 - m^2)/2m

I'm not sure where I picked up that factor of 2.

This also seems like it took way too much work to actually figure out, which will be pretty important in a test scenario. Just deciding whether to solve using the invariant mass, or the momentum/energy squared relations took me a while. Is there a faster way to figure out this kind of annihilation/particle creation problem?

4. Dec 9, 2012

### haruspex

It's easily shown that your answer is feasible while the supposedly correct answer is not. A valid solution should be M=2m, p=0.
I get the same answer as you, but in a way that may be less complicated. At the least, it looks less complicated, but perhaps only by using simpler notation.
Momentum before = momentum after = p.
Total energy before = m + √(m2+p2) = Total energy after = √(M2+p2)
Squaring and collecting up terms:
4M2m2+4m2p2-M4 = 0
Whence
m2+p2 = [M4-4M2m2+4m4]/(4m2)
So √(m2+p2) = [M2-2m2]/(2m)

5. Dec 9, 2012

### demoncore

Thanks a lot for that.