Astroparticle Physics

346 17 Solutions17.5 Chapter 51. a) In the classical non-relativistic case (v ≪ c) the kinetic energy is justE kin = 1 2 m 0v 2 = 1 2 m 0( ) eRB 2= 1 e 2 R 2 B 2,m 0 2 m 0where the velocity v was found from the usual requirement that the centrifugal forcebe balanced by the Lorentz force:m 0 v 2R= evB ⇒ v =eBRm 0.b) In the relativistic case⎛⎞√E kin = γm 0 c 2 − m 0 c 2 = m 0 c 2 1⎝√− 1⎠ = c p 2 + m 2 0 c2 − m 0 c 21 − v2c 2 √√= c e 2 R 2 B 2 + m 2 0 c2 − m 0 c 2 = ecRB 1 + m2 0 c2e 2 R 2 B 2 − m 0c 2≈ 5.95 × 10 7 eV .Alternatively, with an early relativistic approximation√E kin = c p 2 + m 2 0 c2 − m 0 c 2 ≈ cp = ecRB ≈ 6 × 10 7 eV .2. Let us first find the number of electrons N e in the star:N e = M star /m p = 10 M ⊙ /m p ≈2 × 1034 g1.67 × 10 −24 g ≈ 1.2 × 1058 .Here it is assumed that the star consists mainly of hydrogen (m p ≈ 1.67 × 10 −24 g) andis electrically neutral, so that N e = N p .The pulsar volume isV = 4 3 πr3 ≈ 4.19 × 10 18 cm 3and the electron density is n = 0.5 N e /V ≈ 1.43 × 10 39 /cm 3 . Then the Fermi energyisE F = ¯hc(3π 2 n) 1/3≈ 6.582 × 10 −22 MeV s × 3 × 10 10 cm s(3 × 3.14162 × 1.43 × 10 39 ) 1/3 cm −1≈ 688 MeV .