Astroparticle Physics

350 17 Solutionsc) With the numbers used so far the mass of the human is 81 kg. Therefore, the equivalentannual dose comes out to beH ν = Em w R = 1.06 × 10 −14 Sv ,actually independent of the assumed human mass. The contribution of solar neutrinosto the normal natural dose rate is negligible, sinceH = H νH 0= 5.3 × 10 −12 .4. a) The time evolution of the electron neutrino is|ν e ; t〉 =cos θ e −iE ν 1 t |ν 1 〉+sin θ e −iE ν 2 t |ν 2 〉 .This leads to〈ν µ |ν e ; t〉 =sin θ cos θbecause(e −iE ν 2 t − e −iE ν 1 t ) ,(− sin θ|ν 1 〉+cos θ|ν 2 〉)(cos θ e −iE ν 1 t |ν 1 〉+sin θ e −iE ν 2 t |ν 2 〉)=−sin θ cos θ e −iE ν 1 t + sin θ cos θ e −iE ν 2 tsince |ν 1 〉; |ν 2 〉 are orthogonal states.Squaring the time-dependent part of this relation, i.e., multiplying the expression byits complex conjugate, yields as an intermediate step(e −iE ν 2 t − e −iE ν 1 t )( e +iE ν 2 t − e +iE ν 1 t ) = 1 − e i(E ν 2 −E ν1 )t − e i(E ν 1 −E ν2 )t + 1= 2 − (e i(E ν 2 −E ν1 )t + e −i(E ν 2 −E ν1 )t ) = 2 − 2cos[(E ν2 − E ν1 )t]= 2 (1 − cos [(E ν2 − E ν1 )t]) = 4sin 2 [(E ν2 − E ν1 )t/2] .Using 2 sin θ cos θ = sin 2θ one finally getsP νe →ν µ(t) =|〈ν µ |ν e ; t〉| 2 = sin 2 2θ sin 2 [(E ν1 − E ν2 )t/2] .Since the states |ν e 〉 and |ν µ 〉 are orthogonal one has P νe →ν e(t) = 1 − P νe →ν µ(t).In the approximation of small masses, E νi = p + m 2 i /2p + O(m4 i ), one gets E ν 1−E ν2 ≈ (m 2 1 − m2 2 )/2p.Since t = E νi x/p and because of m i ≪ E νi , the momentum p can be identifiedwith the neutrino energy E ν . If, finally, also the correct powers of ¯h and c areintroduced, one obtains( 1P νe →ν e(x) = 1 − sin 2 2θ sin 2 x )4¯hc δm2 E νwith 1/4¯hc = 1.27 × 10 9 eV −1 km −1 , which is the desired relation.