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

Solving, we haveα = 9 5 + 5B + 1 ( 3T o ǫ 5 + 5B − K )o= 18 T o 3T o 5 + 20B − K o,T o T oβ ′′ = 5K/T o −3−9ǫ−45(1+ǫ)B/T o10−15ǫη = 18−10K/T o +150B/T o15ǫ(2−3ǫ)= K oT o− 6 5 − 12BT o,= 18 5 + 30BT o− 2K T o,K o ′ = 143 K o −15B − 3 (5 T o +ǫ K o −15B − 9 )5 T o(3.20)= 5K o −20B − 6 5 T o.The last expressions are obtained with ǫ = 1/3. We also find, for ǫ = 1/3,[ 1S v = T o3 + α 2 −α L + 5 3 β′′ L − 5 9 β′′ +η L − η ],2[ 2S v ′ = T o9 + α 2 −α L + 259 β′′ L − 2527 β′′ + 4 3 η L − 2 ]3 η .Pure neutron matter has pressure[ 2p N = T o ρ o u5 (2u)2/3 −α L u+ 5 ]3 22/3 β Lu ′′ 5/3 +(1+ǫ)η L u 4/3(3.21)(3.22)(3.23)At ρ o , p N has a value about half of the non-interacting Fermi pressure. Combining this resultwith reasonable values S v = 30 MeV and S v ′ = 15 MeV, and assuming ǫ = 1/3, one findsα L ≃ −4.4, β L ′′ ≃ 3.6, and η L ≃ −10.7. In the case S v ′ = 30 MeV, one finds α L ≃ −8.4,β L ′′ ≃ 6.7, and η L ≃ −19.9. In both cases, since β L ′′ > 0, the neutron matter pressure willcontinuously rise with u.However, we now observe that[ ( 3Q = T o5 u2/3 2 2/3 − 14 ) [ β′′+(29 18 + 2/3 − 5 ) ]β L]u ′′ 5/3 , (3.24)3evaluated at the saturation density and using ǫ = 1/3, S v = 30 MeV and S v ′ = 15 MeV, isQ 1 ≃ −10.3 MeV. Obviously, the neutron matter properties and the symmetry propertiescannot be separately adjusted.Most Skyrme forces have the properties that both Q 1 and Q ′ 1 ≡ (dQ(u)/dlnu) 1 = [p n −p s −S v ′ ] have small magnitudes: if they don’t, their neutron matter has problems. Setting44