Homework Help: 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!