Griffiths Introduction to Elementary Particles · Problem 3.16

Symmetries — David J. Griffiths

Physics — QFT · two-body-decay-kinematics

Problem

Particle AA, at rest, decays into particles BB and CC (A→B+CA \to B + C). (a) Find the energy of the outgoing particles, in terms of the various masses. [Answer:EB=mA2+mB2−mC22mA c2]\left[Answer: \quad E_B = \frac{m_A^2 + m_B^2 - m_C^2}{2m_A}\,c^2\right] (b) Find the magnitudes of the outgoing momenta. [Answer:∣pB∣=∣pC∣=λ(mA2,mB2,mC2)2mA c,\left[Answer: \quad |\mathbf{p}_B| = |\mathbf{p}_C| = \frac{\sqrt{\lambda(m_A^2, m_B^2, m_C^2)}}{2m_A}\,c,\right. where λ\lambda is the so-called *triangle function*: λ(x,y,z)=x2+y2+z2−2xy−2xz−2yz.]\left.\lambda(x,y,z) = x^2 + y^2 + z^2 - 2xy - 2xz - 2yz.\right] (c) Note that λ\lambda factors: λ(a2,b2,c2)=(a+b+c)(a+b−c)(a−b+c)(a−b−c)\lambda(a^2, b^2, c^2) = (a+b+c)(a+b-c)(a-b+c)(a-b-c). Thus ∣pB∣|\mathbf{p}_B| goes to zero when mA=mB+mCm_A = m_B + m_C, and runs imaginary if mA<(mB+mC)m_A < (m_B + m_C). Explain.

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