Consider the elastic scattering reaction a + b rightarrow a + b in the lab frame (b initially at rest) and assume that the initial energy e_1 of the incoming a particle satisfies e_1 m_b so that the recoil of the target can be neglected. use the golden rule for scattering to show that the differential cross section is given by: d sigma/d ohm = |m|^2/d(8 pi m_b)^2 write down the lowest order diagram(s) for this scattering process in abc theory. calculate the scattering amplitude using the feynman rules for abc theory (express your result using the mandelstam variables s, t and/or u as relevant). combine the results from (a) and (c) to obtain the differential cross section (in the limit e_1 m_b and assuming that ma and m_c are tiny compared to mb). show that the total cross-section is sigma = g^4/4 pi m^6_b under the conditions stated in part (d).