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Conservation of momentum on relativistic collisions

  1. Sep 3, 2017 #1
    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. jcsd
  3. Sep 3, 2017 #2


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    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.
  4. Sep 3, 2017 #3


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    ##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.
  5. Sep 4, 2017 #4

    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.
  6. Sep 4, 2017 #5
    Ahh sorry, I figured it out just after posting.
  7. Sep 4, 2017 #6


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    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##.
  8. Sep 4, 2017 #7
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