Conservation of momentum on relativistic collisions

[Lukanol]

1. Homework Statement

( 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 + ... + C_N. Calculate the threshold (i.e. minimum E) for this reaction, in terms of the various particle masses.

[ Answer: ( M² - m²_A - m²_B ) / ( 2m_B ), where M = m₁ + m₂ + ... + m_n ]

2. Homework Equations

Assuming that c = 1 and h = 1,

Before collision:
p^μ _i = ( E_A + m_B , p_A )

After collision:
p^μ _f = ( M , 0 )
, where M = m₁ + m₂ + ... + m_n
, and subsctibe i and f represents initial and final state of collision.

p^μp_μ = m²

3. The Attempt at a Solution

I calculated the answer but I do not understand why momentum p_A 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 ( p^μ_i )² = ( p^μ_f )², then I solve for E_A .
⇒ ( E_A + m_B )² - p_A² = M ²
, then I solved for E_A by subsituting p_A² with E_A² = m_A² + p_A²

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

 

[Orodruin]

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.

 

[vela]

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

$p^\mu_\text{f} = (M,0)$ 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.

 

[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.

 

[Lukanol]

Ahh sorry, I figured it out just after posting.

 

[Orodruin]

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

 

[Lukanol]

Thanks!