Griffiths Introduction to Elementary Particles · Problem 3.22

Symmetries — David J. Griffiths

Physics — QFT · mandelstam-variables

Problem

In a two-body scattering event, A+B→C+DA + B \to C + D, it is convenient to introduce the Mandelstam variables s≡(pA+pB)2/c2t≡(pA−pC)2/c2u≡(pA−pD)2/c2\begin{aligned} s &\equiv (p_A + p_B)^2/c^2\\ t &\equiv (p_A - p_C)^2/c^2\\ u &\equiv (p_A - p_D)^2/c^2 \end{aligned} (a) Show that s+t+u=mA2+mB2+mC2+mD2s + t + u = m_A^2 + m_B^2 + m_C^2 + m_D^2. The *theoretical* virtue of the Mandelstam variables is that they are Lorentz invariants, with the same value in any inertial system. *Experimentally*, though, the more accessible parameters are energies and scattering angles. (b) Find the CM energy of AA, in terms of ss, tt, uu and the masses. [AnswerAnswer: EACM=(s+mA2−mB2)c2/2sE_A^{\text{CM}} = (s + m_A^2 - m_B^2)c^2/2\sqrt{s}] (c) Find the Lab (BB at rest) energy of AA. [AnswerAnswer: EAlab=(s−mA2−mB2)c2/2mBE_A^{\text{lab}} = (s - m_A^2 - m_B^2)c^2/2m_B] (d) Find the total CM energy (Etot=EA+EB=EC+EDE_{\text{tot}} = E_A + E_B = E_C + E_D). [AnswerAnswer: EtotCM=sc2E_{\text{tot}}^{\text{CM}} = \sqrt{s}c^2]

Browse the problem library