Astroparticle Physics

17.11 Chapter 11 3694. Planck distributionϱ(ν, T ) = 8πh∫ ∞1c 3 ν3 e hν/kT − 1 , 〈ν〉 = 0νϱ(ν,T)dν∫ ∞0ϱ(ν, T ) dν ;substitution hνkT = x:∫ ∞0∫ ∞0∫ ∞0∫ ∞0ϱ(ν, T ) dν = 8πhc 3= 8πhc 3νϱ(ν,T)dν = 8πhc 3( kTh( kTh( kTh) 3 ∫ ∞0x 3e x − 1kTh dx) 4 ∫ ∞x 30 e x − 1 dx,) 5 ∫ ∞x 4e x − 1 dx ;0x 3e x − 1 dx = 3! ζ(4) , ζ(4) = π 4 /90 , ζ – Riemann’s zeta function,x 4e x dx = 4! ζ(5) , ζ(5) = 1.036 927 7551 ... ;− 1〈hν〉 = h 8πh ( kTh) 5c × 4! ζ(5)3(8πh kTh) 4= h kTc × 3! ζ(4) h × 4 × ζ(5)ζ(4)3= kT × 4 × ζ(5) 360× 90 = kT ζ(5) ≈ 900 µeV .π 4 π 45. a) At present: number density of bb photons: 410/cm 3 , average energy 〈E〉 =900 µeV(from the previous problem). This leads to a first estimate of the present energydensity ofϱ 0 ≈ 0.37 eV/cm 3 .Since, however, the temperature enters with the fourth power into the energy densityone has to useϱ 0 = π 215 T 4for photons (number of degrees of freedom g = 2, see (9.10)). One has to includethe adequate factors of k and ¯hc to get the correct numerical result:ϱ 0 = π 215 T 4 k 4≈ 0.26 eV/cm3(¯hc) 3(k = 8.617 × 10 −5 eV K −1 , ¯hc = 0.197 GeV fm).