Astroparticle Physics

17.6 Chapter 6 351b) On both sides of the Earth a number of N muon neutrinos are created. The ones fromabove, being produced at an altitude of 20 km almost have no chance to oscillate,while the ones crossing the whole diameter of the Earth have passed a distance ofmore than 12 000 km, so that from below only NP νµ →ν µ(2R E ) muon neutrinos willarrive, leading to a ratio ofS = 0.54 = NP ν µ →ν µ(2R E )= 1 − sin 2 (1.27 × 12 700 × δm 2 ).NSolving for δm 2 one obtains δm 2 = 4.6 × 10 −5 eV 2 and m ντ ≈ 6.8 × 10 −3 eV.c) No. The masses of the lepton flavour eigenstates are quantum-mechanical expectationvalues of the mass operator M = √ H 2 − p 2 . For an assumed (ν µ ↔ ν τ )mixing with a similar definition of the mixing angle as for the (ν e ↔ ν µ ) mixing,one getsm νµ =〈ν µ |M|ν µ 〉=m 1 cos 2 θ + m 2 sin 2 θ,m ντ =〈ν τ |M|ν τ 〉=m 1 sin 2 θ + m 2 cos 2 θ.For maximum mixing (θ = 45 ◦ ,cos 2 θ = sin 2 θ = 1/2) one obtainsm νµ = m ντ = (m 1 + m 2 )/2 .Sect. 6.31. I = I 0 e −µx photons survive, I 0 − I = I 0 (1 − e −µx ) get absorbed;from Fig. 6.37 one reads µ = 0.2cm −1 ;detection efficiency: η = I 0(1 − e −µx )= 1 − e −µx ≈ 0.45 = 45% .I 02. The duration of the brightness excursion cannot be shorter than the time span for thelight to cross the cosmological object. Figure 6.48 shows t = 1s⇒ size ≈ ct =300 000 km.3. cos ϑ = 1nβ ; threshold at β>1 ⇒ v> c nn ;E µ = γm 0 c 2 =1√1 − β 2 m 0c 2 =1√1 − 1 n 2 m 0 c 2 =n√n 2 − 1 m 0c 2 ,{ {4.5 GeV in air4.4 GeV in airE µ ≈160.3 MeV in water , Ekin µ ≈ 54.6 MeV in water .4. Number of emitted photons:N E = P hν ≈3 × 10 27 W10 11 eV × 1.602 × 10 −19 J/eV ≈ 1.873 × 1035 s −1 .