Astroparticle Physics

354 17 SolutionsThe solution of this differential equation is obtained by separation of variables,∫ EdE ′ ∫ tE 0E ′2 =−α 0dt ′ ⇒ 1 E 0− 1 E =−αt ⇒ E = E 01 + αE 0 t .It is only valid for E ≫ m 0 c 2 . (This limit is not respected for longer times.)The bending radius follows with E → pc toϱ =EceB = 1 E 0ceB 1 + αE 0 t = ϱ 01 + αceBϱ 0 t .b) In the general (relativistic) case there is p 2 = γ 2 m 2 0 v2 , i.e., γ 2 v 2 = p 2 /m 2 0 =(E 2 − m 2 0 c4 )/m 2 0 c2 , thusĖ =− 2 e 4 B 2 c 33 (m 0 c 2 ) 4 (E2 − m 2 0 c4 ) =−α(E 2 − m 2 0 c4 ), αas in (a) .The solution is calculated as in (a):∫ EdE ′ ∫ tE 0 E ′2 − m 2 =− dt ′ =−αt ⇒0 c4 0∫1 E()∣12m 0 c 2 E 0E ′ − m 0 c 2 − 1E ′ + m 0 c 2 dE ′ 1=2m 0 c 2 ln E′ − m 0 c 2 ∣∣∣∣ EE ′ + m 0 c 2⎛⎞=− 1 ⎝ 1m 0 c 2 2 ln 1 + m 0c 2E− 11 − m 0c 2 2 ln 1 + m 0c 2E 0 ⎠E1 − m 0c 2E 0()=− 1m 0 c 2 artanh m 0c 2E − artanh m 0c 2=−αt .E 0E ′ =E 0Solving this equation for E leads to()m 0 c 2E = () = m 0 c 2 coth αm 0 c 2 t + artanh m 0c 2.tanh αm 0 c 2 t + artanh m 0c 2E 0E 0For larger times (in the non-relativistic regime) the rest energy is appoached exponentially.From this the bending radius results in√ϱ = peB = E 2 − m 2 0 c4ceBm 0 c/eB=() =sinh αm 0 c 2 t + artanh m 0c 2E 0= m √0c ( )coth 2 αm 0 c 2 t + artanh m 0c 2eBE 0− 1m 0 c/eB(sinh αm 0 c 2 t + artanh) .√m 0 cϱ0 2e2 B 2 +m 2 0 c2