Astroparticle Physics

17.5 Chapter 5 347Consequences:The electrons are pressed into the protons,e − + p → n + ν e .The neutrons cannot decay, since in free neutron decay the maximum energy transfer tothe electron is only ≈ 780 keV, and all energy levels in the Fermi gas are occupied, evenif only 1% of the electrons are left: (n ∗ = 0.01 n ⇒ E F ≈ 148 MeV).3. Event horizon of a black hole with mass M = 10 6 M ⊙ :R S = 2GMc 2 ≈ 2.96 × 10 9 m ,∫ RSE =− G m pM∞ r 2 dr = G m pM= 1 R S 2 m pc 2 ≈ 469 MeV ≈ 7.5 × 10 −8 J .The result is independent of the mass of the black hole. This classical calculation, however,is not suitable for this problem and just yields an idea of the energy gain. In addition,the value of the energy depends on the frame of reference. In the vicinity of blackholes only formulae should be used which hold under general relativity.4. a) Conservation of angular momentumL = r × p ⇒ L ≈ mrv = mr 2 ωrequires r 2 ω to be constant (no mass loss):(R⊙ 2 ω ⊙ = RNS 2 ω R⊙) 2ω⊙NS ⇒ ω NS =≈ 588 Hz .R NSThen the rotational energy isE rot = 1 22 5 M NSRNS 2 ω2 NS ≈ 0.4 × 1030 kg × (5 × 10 4 ) 2 m 2 × 588 2 s −2≈ 0.35 × 10 45 J .b) Assume that the Sun consists of protons only (plus electrons, of course). Four protonseach are fused to He with an energy release of 26 MeV corresponding to amass–energy conversion efficiency of ≈ 0.7%:E = M ⊙ c 2 × 7 × 10 −3 = 2 × 10 30 kg × (3 × 10 8 ) 2 m 2 /s 2 × 7 × 10 −3= 1.26 × 10 45 J ,which is comparable to the rotational energy of the neutron star.5. A dipole field is needed to compensate the centrifugal forcemv 2R = evB guide . (∗)Equation (5.38) yields p = mv = 1 2eRB. Comparison with (∗)givesB guide = 1 B (Wideroe condition) .2