Astroparticle Physics

352 17 SolutionsSolid angle: Ω =A4πR 2 ≈ 6.16 × 10−41 .Number of recorded photons: N R = ΩN E ≈ 1.15 × 10 −5 /s ≈ 364/a .Minimum flux: assumed 10/a, P min ≈ P Ω 10A 364 ≈ 6.35 × 10−19 J/(cm 2 s).5. (a) Assume isotropic emission, which leads to a total power of P = 4πr 2 P S ,wherer = 150 × 10 6 km is the astronomical unit (distance Sun–Earth). One getsP ≈ 3.96 × 10 26 W .(b) In a period of 10 6 years the emitted energy is E = P × t ≈ 1.25 × 10 40 J, whichcorresponds to a mass of m = E/c 2 ≈ 1.39 × 10 23 kg, which represents a relativefraction of the solar mass off = m M ⊙≈ 6.9 × 10 −8 .(c) The effective area of the Earth is A = πR 2 , so that the daily energy transport toEarth is worked out to be E = πR 2 P S t. This corresponds to a mass ofm = E c 2 ≈ 1.71 × 105 kg = 171 tons .Sect. 6.41. The power emitted in the frequency interval [ν, ν + dν] corresponds to the one emittedin the wavelength interval [λ, λ + dλ], λ = c/ν, i.e., P(ν)dν = P(λ)dλ, orP(λ) = P(ν) dνdλ ∼ ν 3 dνe hν/kT − 1 dλ ∼ 1λ 5 (e hc/λkT − 1) ,since dν/dλ =−c/λ 2 .2. The luminosity of a star is proportional to the integral over Planck’s radiation formula:L ∼∫ ∞0ϱ(ν, T ) dν =∫ ∞08πhν 3c 3 1e hν/kT − 1 dν ;use x =kThν ⇒L ∼ 8π ( ) kT 3 ∫ ∞c 3 h x 3 1 kT 8π k 4 T 4 ∫ ∞x 3 dxh 0 e x dx =− 1 h c 3 h 3 0 e x − 1= 8π k 4 π 4c 3 h 3 15 T 4 ∼ T 4 .In addition, the luminosity varies with the size of the surface (∼ R 2 ).