HR/20190827T2143 Radius of neutron star p.1/2 4 Mass of homogeneous sphere: 푀 = 휌푉 = 휋휌푟3 [1] 3 2퐺푀 푟 2퐺푀 2퐺 4 8휋퐺휌 Schwarzschild radius: 푟 = ∴ 푆 = = ∙ 휋휌푟3 = 푟2 [2] 푆 푐2 푟 푟푐2 푟푐2 3 3푐2 2퐺푀 2퐺푀 푟 푐2 푟 푣2 Escape velocity: 푣 = √ ∴ 푣2 = = 푆 ∴ 푆 = 푒푠푐 [3] 푒푠푐 푟 푒푠푐 푟 푟 푟 푐2 1 Lorentz factor: 훾 = [4] 푣2 √1− 푐2 Relativistic mass: 푚 = 훾푚0 [5] 2 2 2 2 2 Kinetic energy: 퐸푘 = 푚푐 − 푚0푐 = 훾푚0푐 − 푚0푐 = (훾 − 1)푚0푐 [6] 2 퐸푘 퐸푘 푚0푐 +퐸푘 ∴ 2 = 훾 − 1 ∴ 1 + 2 = 훾 ∴ 훾 = 2 [7] 푚0푐 푚0푐 푚0푐 2 2 푣 푚0푐 substitute [4]: ∴ √1 − 2 = 2 [8] 푐 푚0푐 +퐸푘 2 2 푣푒푠푐 푚0푐 Escape energy vs. radius: √1 − 2 = 2 [9] 푐 푚0푐 +퐸푒푠푐 2 2 2 푟푆 푚0푐 푟푆 푚0푐 using [3]: ∴ √1 − = 2 ∴ 1 − = ( 2 ) [10] 푟푒푠푐 푚0푐 +퐸푒푠푐 푟푒푠푐 푚0푐 +퐸푒푠푐 2 2 푟푆 푚0푐 ∴ = 1 − ( 2 ) [11] 푟푒푠푐 푚0푐 +퐸푒푠푐 푟푆 8휋퐺휌 2 from [2]: = 2 푟푒푠푐 [12] 푟푒푠푐 3푐 2 2 2 2 3푐 푚0푐 ∴ 푟푒푠푐 = ∙ [1 − ( 2 ) ] [13] 8휋퐺휌 푚0푐 +퐸푒푠푐 2 2 3 √ 푚0푐 ∴ 푟푒푠푐 = 푐 ∙ √ ∙ 1 − ( 2 ) [14] 8휋퐺휌 푚0푐 +퐸푒푠푐 An object with rest mass 푚0 and kinetic energy 퐸푒푠푐 can/will escape if it is at a distance of at least 푟푒푠푐. 18 3 For a neutron star, we've got: 휌 = 휌푁 = 1.392134 × 10 kg/m [15] and of course: 푐 = 299792458 m/s [16] as well as: 퐺 = 6.67408 × 10−11 m3/kg/s3 [17] 3 3 so: 푐 ∙ √ = 299792458 ∙ √ = 10745 m [18] 8휋퐺휌 8휋∙6.67408×10−11∙1.392134×1018 This value equals the critical black hole radius as calculated in http://henk-reints.nl/astro/HR-on-the-universe.php. Suppose this object is an electron originating from a decaying neutron of a neutron star without obtaining energy from the star's binding energy. Then: −31 Electron mass: 푚0 = 푚푒 = 9.10938356 × 10 kg [19] 2 2 −14 so: 푚0푐 = 푚푒푐 = 8.18710565 × 10 J [20] 2 −13 Neutron decay energy: 퐸푑푒푐 = (푚푛 − 푚푝 − 푚푒)푐 = 0.78257 MeV = 1.2538 × 10 J [21] First assume the electron obtains all this energy. Then: 8.18710565×10−14 2 푟 = 10745 ∙ √1 − ( ) = 9871 m [22] 0 8.18710565×10−14+1.2538×10−13 4 Its corresponding mass equals: 푀 = 휋휌푟3 = 7.234 × 1030 kg ≈ 3.637푀 [23] 0 3 0 ⊙ So a neutron star with a diameter that exceeds 19.74 km and thus a mass above 3.637 times that of our sun cannot lose any electrons by neutron decay because gravitation definitely wins that game. Above this size a ball of neutronium would not be able to disintegrate. That's why I do still not really understand what the big bang actually was. The IniAll must have had a radius of nearly 3 astronomical units. Copyright © 2019, Henk Reints HR/20190827T2143 Radius of neutron star p.2/2 In practice, the electron antineutrino takes away roughly 50% of the neutron decay energy on average, 2 −14 √ 8.18710565×10 yielding: 푟1 = 10745 ∙ 1 − ( 1.2538×10−13) = 8856 m [24] 8.18710565×10−14+ 2 corresponding to a diameter of: 푑1 = 17.71 km [25] 4 and its corresponding mass is: 푀 = 휋휌푟3 = 4.050 × 1030 kg ≈ 2.036푀 [26] 1 3 1 ⊙ To me it is remarkable that this mass is within the tolerance of the lower bound of the TOV-limit as stated on https://en.wikipedia.org/wiki/Tolman%E2%80%93Oppenheimer%E2%80%93Volkoff_limit#History (as of 2019-08-27): The mass of the pulsar PSR J0348+0432, at 2.01±0.04 solar masses, puts an empirical lower bound on the TOV limit. Below this mass, a neutron star would be prone to evaporation by neutron decay. This could mean twice the solar mass is a lower mass limit for neutron stars to be able to stay alive at all. But the neutrons in a neutron star are definitely not free neutrons, so they may have a far longer decay time, which would also avoid this evaporation. Copyright © 2019, Henk Reints .