Lesson 5 of 15

Two-Body Decay Kinematics

Two-Body Decay Kinematics

When a particle AA of mass MM decays at rest into two daughters BB and CC, conservation of 4-momentum uniquely fixes the kinematics. Working in natural units (c=1c = 1, all quantities in GeV):

Momentum of the Daughters

Both daughters recoil back-to-back with equal and opposite momenta of magnitude pp^*:

p=[M2(mB+mC)2][M2(mBmC)2]2Mp^* = \frac{\sqrt{\bigl[M^2 - (m_B + m_C)^2\bigr]\bigl[M^2 - (m_B - m_C)^2\bigr]}}{2M}

This is real only when MmB+mCM \geq m_B + m_C (energy conservation requires the parent to be heavier than the sum of daughters).

Energies of the Daughters

EB=M2+mB2mC22M,EC=M2+mC2mB22ME_B^* = \frac{M^2 + m_B^2 - m_C^2}{2M}, \qquad E_C^* = \frac{M^2 + m_C^2 - m_B^2}{2M}

Note that EB+EC=ME_B^* + E_C^* = M (total energy equals parent rest mass) and EB2p2=mB2E_B^{*2} - p^{*2} = m_B^2 as required.

Example: π0γγ\pi^0 \to \gamma\gamma

With M=0.135M = 0.135 GeV and mB=mC=0m_B = m_C = 0 (massless photons):

p=EB=EC=M2=0.0675 GeVp^* = E_B^* = E_C^* = \frac{M}{2} = 0.0675 \text{ GeV}

Your Task

Implement (all masses and energies in GeV, natural units):

  • decay_momentum(M_GeV, mB_GeV, mC_GeV) — returns pp^*
  • decay_energy_B(M_GeV, mB_GeV, mC_GeV) — returns EBE_B^*
  • decay_energy_C(M_GeV, mB_GeV, mC_GeV) — returns ECE_C^*
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