Widdekind Posted August 8, 2011 Share Posted August 8, 2011 The Weizsaecker Formula, for the total nuclear binding-energy [math]E_B[/math], of a nucleus, of mass number [math]A[/math], and proton number [math]Z[/math], is (roughly): Now, viewing a NS, as a "single super-sized nucleus", amounts to taking the limit, as [math]A \rightarrow \infty[/math], with [math]Z \approx 0[/math]. Accordingly, the binding-energy-per-nucleon, [math]\frac{E_B}{A} \rightarrow 16 \, MeV \; - \; 24 MeV \approx -8 \, MeV[/math] 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 [math]\approx 1.5 M_{\odot}[/math], and [math]\approx 7 \, km[/math], 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 (??). Link to comment Share on other sites More sharing options...
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