Invariant Mass/Inelastic Collision Problem

In summary: IIn summary,The problem asks for the minimum total energy of two particles that collide and annihilate each other, creating another particle. The attempted solution states that the final momentum is zero and the invariant mass is (m)^2. When solving for the energy of the particles, the wrong value is obtained for the momentum and the equation is rewritten. The final result is that the energy of the new particle is (M+E_k,M)^2.
  • #1
demoncore
18
1

Homework Statement



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.

Homework 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

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.
 
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  • #2
demoncore said:
I take the final momentum to be zero:
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
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
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
Thanks a lot for that.
 

1. What is invariant mass?

Invariant mass is a concept in physics that refers to the mass of a system calculated using the equations of special relativity. It is the mass of a system as measured by an observer in any frame of reference, and it does not change regardless of the relative motion of the observer and the system.

2. How is invariant mass calculated?

The formula for calculating invariant mass is E^2 = (mc^2)^2 + (pc)^2, where E is energy, m is mass, c is the speed of light, and p is momentum. Invariant mass can also be calculated using the energy and momentum of the system in the center-of-mass frame.

3. What is an inelastic collision?

An inelastic collision is a collision in which kinetic energy is not conserved. This means that the total kinetic energy of the system before the collision is not equal to the total kinetic energy after the collision. In inelastic collisions, some of the kinetic energy is converted into other forms of energy, such as heat or sound.

4. How is invariant mass used in inelastic collision problems?

Invariant mass is often used in inelastic collision problems to calculate the total energy and momentum of the system before and after the collision. This allows for the conservation of energy and momentum to be applied, even if some kinetic energy is lost during the collision.

5. What is the importance of invariant mass in physics?

Invariant mass is important in physics because it is a fundamental concept in the theory of special relativity. It helps to explain the behavior of particles at high speeds and is essential for understanding concepts such as mass-energy equivalence. Invariant mass also plays a crucial role in the study of particle physics and cosmology.

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