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

The symmetry energy in nuclear matter is defined asS v = 1 8d 2 (E/ρ)dx 2∣∣∣∣ρ=ρ0,x=1/2= T 0[ 13 + v 1L −v 1U2= 28 MeV.+(1+ǫ)((1+2ǫ)v 2L −v 2U )+ 5 18 v 3 − 1 16 t′ 3](4.23)Another parameter, which is related to symmetry energy, is given byL = 3ρ 08d 3 (E/ρ)dρdx 2 ∣∣∣∣ρ=ρ0,x=1/2[ 2= T 03 + 3 2 (v 1L −v 1U )+3(1+2ǫ)(1+ǫ)((1+2ǫ)v 2L −v 2U )+ 2518 v 3 − 3 ]8 t′ 3= 54 MeV.(4.24)We choose the effective mass at the saturation density as 0.78m b and use this number in theEq. (4.12),m ∗ m=1+2mρ 0 V 3 / = 0.78m b. (4.25)2Thus we can easily recover v 3 from Eq. (4.19). From Eqs. (4.20), (4.21), and (4.22), we canhave v 1 = v 1L +v 2L , v 2 = v 2L +v 2U and t ′ 3 :v 1 = 5K 0/T 0 +5v 3 (1−3ǫ)−72ǫ−90B 0 /T 0 (1+2ǫ)−1245ǫv 2 = 12+90B 0/T 0 −5K 0 /T 0 −5v 390ǫ(1−2ǫ)t ′ 3 = −8v 1 −16v 2 −8v 3 − 16B 0T 0− 485 . (4.26)Then we can manipulate Eq. (4.23), and (4.24) to get v 1L and v 2L ,v 1L = 1 2 v 1 + 1 [ 5(1−3ǫ)v 3 + 2ǫ−12ǫ 27 16 t′ 3 + (1+2ǫ)S vT 0v 2L =12(1+ǫ) v 2 −[1 54ǫ(1+ǫ) 2 27 v 3 − t′ 316 + S v− L − 1 T 0 3T 0 9− L − 1+6ǫ3T 0 9],](4.27)and we can have v 1U = v 1 −v 1L and v 2U = v 2 −v 2L .57