Widdekind Posted August 8, 2011 Posted August 8, 2011 The Weizsaecker Formula, for the total nuclear binding-energy E_B, of a nucleus, of mass number A, and proton number Z, is (roughly): Now, viewing a NS, as a "single super-sized nucleus", amounts to taking the limit, as A \rightarrow \infty, with Z \approx 0. Accordingly, the binding-energy-per-nucleon, \frac{E_B}{A} \rightarrow 16 \, MeV \; - \; 24 MeV \approx -8 \, MeV where only the "volume" & "Pauli" terms have survived. That's a repulsion, of roughly 1% of the rest-mass-energy of the matter making up the NS. By way of comparison, the gravitational binding-energy, for NS of \approx 1.5 M_{\odot}, and \approx 7 \, km, is roughly 20%. It certainly seems like, in NS, both GR & QM play important roles, requiring some kind of "quantum gravity" theory to accurately model (??).
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