On most general exact solution for Vaidya-Tikekar isentropicsuperdense star

The energy density of Vaidya-Tikekar isentropic superdense star is found to be decreasing away from the center, only if the parameter K is negative. The most general exact solution for the star is derived for all negative values of K in terms of circular and inverse circular functions. Which can further be expressed in terms of algebraic functions for K = 2-(n/δ)² < 0 (n being integer andδ = 1,2,3 4). The energy conditions 0 ≤ p ≤ αρc ², (α = 1 or 1/3) and adiabatic sound speed conditiondp dρ ≤ c ², when applied at the center and at the boundary, restricted the parameters K and α such that .18 < −K −2287 and.004 ≤ α ≤ .86. The maximum mass of the star satisfying the strong energy condition (SEC), (α = 1/3) is found to be3.82 Mq_· at K=−2/3, while the same for the weak energy condition (WEC), (α =1) is 4.57 M_ _⊙ atK=−>5/2. In each case the surface density is assumed to be 2 × 10¹⁴ gm cm^-3. The solutions corresponding to K>0 (in fact K>1) are also made meaningful by considering the hypersurfaces t= constant as 3-hyperboloid by replacing the parameter R ² by −R² in Vaidya-Tikekar formalism. The solutions for the later case are also expressible in terms of algebraic functions for K=2-(n/δ² > 1 (n being integer or zero and δ =1,2,3 4). The cases for which 0 < K < 1 do not possess negative energy density gradient and therefore are incapable of representing any physically plausible star model. In totality the article provides all the physically plausible exact solutions for the Buchdahl static perfect fluid spheres. This revised version was published online in July 2006 with corrections to the Cover Date.