Astroparticle Physics

17.4 Chapter 4 3455. From the expression for the change of the thermal energy, Q = c sp mT , one can findthe temperature rise:T = Qc sp m = 104 eV × 1.6 × 10 −19 J/eV8 × 10 −5 = 2 × 10 −11 K .J/(gK) × 1g6. q γ + q e = q ′ γ + q′ e ⇒ q γ − q ′ γ = q′ e − q e ⇒q 2 γ + q2 γ ′ − 2q γ q ′ γ =−2(E γ E ′ γ − p γ p ′ γ ) = q′2 e + q2 e − 2q′ e q e ⇒−2E γ E ′ γ (1 − cos θ) = 2m2 e − 2E′ e m e ; (p e = 0)=−2m e (E ′ e − m e) =−2m e E kine =−2m e (E γ − E ′ γ ) ⇒E γ − E ′ γE ′ γ= E γE ′ γ− 1 = E γm e(1 − cos θ)⇒E γ′ 1=E γ 1 + E γm e(1 − cos θ) = 11 + ε(1 − cos θ) .7. Start from (4.13). The maximum energy is transferred for backscattering, θ = π;E γ′ = 1E γ 1 + 2ε ,E maxe = E γ − E ′ γ = E γ − E γ1 + 2ε = E γwith numbers: Eemax = 478 keV (‘Compton edge’).For ε →∞(∗) yields Ee max = E γ .For θ γ = π one hasE ′ γ = E γ11 + 2 εand consequentlyE ′ γ = m ec 2 ε1 + 2 ε = m ec 22 + 1/ε .For ε ≫ 1 this fraction approaches m e c 2 /2.2ε1 + 2ε = 2ε21 + 2ε m ec 2 ; (∗)8. Momentum: p = mv; m = γm 0 ,wherem 0 is the rest mass and γ =p = γm 0 βc ⇒ γβ = p/m 0 c.1√1 − β 2 :9. In the X-ray region the index of refraction is n = 1, therefore there is no dispersion andconsequently no Cherenkov radiation.