Theory of Nuclear Matter for Neutron Stars and ... - Graduate Physics

This equation can be calculated by iteration scheme.u − i+1 = 12a= 12a∫ zi+1−∞∫ zi−∞=e −∆/a u − i + 12au(z ′ )e (z′ −z i+1 )/a dz ′u(z ′ )e (z′ −z i )/a e −∆/a dz ′ + 12a∫ zi+1z iu(z)e (z−z i+1)/a dz.∫ zi+1z iu(z ′ )e (z′ −z i+1 )/a dz ′(A.8)For the integration at the end, we expandthen we havewheref =e −∆/aI0 − = 12aI1 − = 12au(z) = u i + u i+1 −u i∆ (z −z i)+O(∆ 2 ) (A.9)∫ ∆0∫ ∆0u − i+1 = fu− i +I − 0 u i +I − 1 u i+1 ,(1− z )e z/a e −∆/a dz = a−(a+∆)e−∆/a∆2∆z∆ ez/a e −∆/a dz = ∆−a+ae−∆/a .2∆In the same manner, we have an equation for u + i ,u + i = 12a= 12a= 12a∫ ∞z iu(z ′ )e (z i−z ′ )/a dz ′∫ zi+1∫ ∞(A.10)(A.11)u(z ′ )e (z i−z ′)/a dz ′ + 1 u(z ′ )e (z i+1−z ′)/a e −∆/a dz ′ (A.12)z i2a z i+1u(z)e (zi−z)/a dz +e −∆/a u + i+1 .z i∫ zi+1With the first order Taylor expansion before, we havewhereu + i = fu + i+1 +I+ 0 u i +I + 1 u i+1 ,I 0 + = 12aI 1 + = 12aAs a second order expansion, we have∫ ∆0∫ ∆0(1− z )e −z/a dz ′ = I1− ∆z∆ e−z/a dz ′ = I − 0 .(A.13)(A.14)u(z) = u i + u i+1 −u i−1(z −z i )+ u i+1 −2u i +u i−1(z −z2∆ 2∆ 2 i ) 2 +O(∆ 3 ) (A.15)108