# Conservation of momentum on relativistic collisions

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1. Sep 3, 2017

### Lukanol

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

( This question is from the textbook of Introduction to Elementary Particle Physics, written by Griffiths, on the problem set of Chapter 3 )

Particle A (energy E) hits particle B (at rest), producing particles C1, C2, ...: A + B → C1 + C2 + ... + CN. Calculate the threshold (i.e. minimum E) for this reaction, in terms of the various particle masses.

[ Answer: ( M2 - m2A - m2B ) / ( 2mB ), where M = m1 + m2 + ... + mn ]

2. Relevant equations

Assuming that c = 1 and h = 1,

Before collision:
pμ i = ( EA + mB , pA )

After collision:
pμ f = ( M , 0 )
, where M = m1 + m2 + ... + mn
, and subsctibe i and f represents initial and final state of collision.

pμpμ = m2

3. The attempt at a solution

I calculated the answer but I do not understand why momentum pA seems to be not conserved if I want to get the answer.

(a). I assumed that the energy-momentum four vector is conserved.

(b). I simply calculate ( pμi )2 = ( pμf )2, then I solve for EA .
⇒ ( EA + mB )2 - pA2 = M 2
, then I solved for EA by subsituting pA2 with EA2 = mA2 + pA2

(c) My question is: why should I consider the final momentum state as:
pμf = ( M, 0 )
, but not something like:
pμf = ( Ej + M , pCj )
for M = M - mj
Shouldn't the momentum be conserved, just like a ball hitting a bunch of balls when playing snooker?

2. Sep 3, 2017

### Orodruin

Staff Emeritus
You are computing a Lorentz invariant. This has the same value in all inertial frames and you can therefore just as well compute it in the CoM frame.

3. Sep 3, 2017

### vela

Staff Emeritus
$p^\mu_\text{f} = (M,0)$ is wrong. You should say it's $p^\mu_\text{f} = (E,\vec{p}_\text{A})$. You got the right answer because your calculation involves only $p_\text{f}^2$, which is an invariant.

4. Sep 4, 2017

### Lukanol

Thanks!

Does that mean, for the equation below, $$\left( E_A + m_B \right)^2 - \vec p_A^2 = M^2,$$ I actually used the invariant property of $(p_f^\mu)^2 = M^2$ on the right-hand side, while on the left-hand side I used the "dot product" of $(p_i^\mu)^2$?

But then why:$$\frac {M^2- m_A^2 - m_B^2} {2m_B},$$ is the threshold (minimum E) for this reaction?
If the reaction is explosive, such that particle $C_1, C_2, ..., C_N$ has momentum $\vec p_1, \vec p_2, ..., \vec p_n$. If they adds up to $\vec p_A = \vec p_1 + \vec p_2 + ... + \vec p_n$, our result seems to be the same because the right-hand side is invariant.

5. Sep 4, 2017

### Lukanol

Ahh sorry, I figured it out just after posting.

6. Sep 4, 2017

### Orodruin

Staff Emeritus
The threshold occurs when the products are at relative rest. Therefore, in their rest frame $P_f = (M,0)$ and consequently $P_f^2 = M^2$.

7. Sep 4, 2017

Thanks!