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

of find exact polynomial expression, we might use recursion relation, that is,∂ 2 p M +2∂p=−∂f2 1+f ∂f + mc 2 n c g 5/2(1+f) M+2 (1+g) N−3/2×∑[ (1+m)mpmn f m g n +m(m−M)+m−Mf(∂p 2 1∂f∂g = f − M +1 ) ∂p1+f ∂g + mc 2 n c g 3/2(1+f) M+1 (1+g) N−1/2×∑pmn f m g n m[n+ 5 ]2 +g(4+n−N) ,],(B.12)(∂ 2 p 3∂g = 2 2g − N − 1 21+g) ∂p∂g + mc 2 n c fg 3/2(1+f) M+1 (1+g) N−1/2×∑[ n(n+5pmn f m g n ) 2+n(4+n−N)+(4+n−N)g].From thermodynamic quantities, we can getn = 1 ( ) ∂p= 1 (∂f ∂pT ∂ψTT ∂ψ ∂f + ∂g )∂p∂f ∂g= 1 ( 1 f √ )1+f ∂pmc 2 g 1+f/a ∂f + f2 √ 1+f √ ∂p1+f/a∂g( ) √ ∂p 1+f ∂pns = −nψ =∂T mc 2 ∂g −nψ.ψThe derivatives of the above items w.r.t f and g are given by( )∂n 1∂f = f − 1n+ 1 f2a(1+f/a) mc 2∂n∂g = 1mc 2∂(ns)∂f∂(ns)∂g( ∂p∂f +2(1+f)∂2 pf√1+f/a(−= 1 √1 ∂p 1+f√2mc 2 1+f ∂g + mc√ 21+f ∂ 2 p=mc 2 ∂g − ∂n2 ∂g ψ.2g √ 1+f √ 1+f/a ×∂f − 1 )g ∂p2 21+f∂g +g ∂2 p,∂f∂g√ √ 1+f ∂p 1+fg 2 ∂f + g)∂ 2 p∂f∂g + 12 √ ∂ 2 p,1+f ∂g 2√∂ 2 p 1+f/a∂f∂g −ψ∂n ∂f −n ,f(B.13)(B.14)Finally we can get thermodynamic derivatives using the above formulae and constraints119