Monte Carlo Particle Transport Methods: Neutron and Photon - gnssn

49from Equation (3.13) asE = -—- (cosft + Vcos 2 ff + A 2 - I) 2 (3.14)(A + I) 2Substituting the two extreme values: COS'S = - 1 and cost) = 1, respectively, one getsthe restrictionE„a 2 =£ E =S E 0for the new energy, where a = (A — 1)/(A + 1).For a wide range of energies (especially for light target nuclei) the elastic scattering isisotropic in the center-of-mass system. Then its cosine can be simply selected as(cos-d) cm= 1 - 2p (3.15)and the cosine of the scattering angle in the laboratory system iscv1 + A(cosd) c mCOStJ = — = =-=== (j.16)Vl -FA 2 + 2A(cosft) cmHereafter the subscript cm indicates the variable measured in the center-of-mass system,angles in the laboratory system have no subscripts.The new energy is to be computed from Equation (3.14), or directly fromE = V 2EJ(I - a 2 )(cosfl) cm+ 1 + a 2 ! (3.17)where a = (A - 1)/(A + 1).The selection procedure can further be simplified if the target is hydrogen: A = 1. Bypreserving the isotropic scattering assumption in the center-of-mass system (which is validup to about 10 MeV) and substituting Equation (3.15) into Equation (3,16) one getscos-d = VpThe new energy isE = E 0(cos-f>) 2= E„pWhen scatterings are simulated for such incident energies and target nuclei that scatteringis not isotropic, theoretically the Equationcr(p,)dp, (—1 5 S fx «£ i) (3.18)has to be solved for the actual selection, where p, denotes the cosine of the scattering angle— either in the laboratory or in the center-of-mass system. If the differential cross-sectionis given in the center-of-mass system, then there are two ways to proceed:1. One can first transform the cross-section according totr(p,)dp, = CT tm(p. cJdp. om(3.1 1 ))