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Stars as Laboratories for Fundamental Physics - MPP Theory Group

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Oscillations of Trapped Neutrinos 335<br />

For l.h. electrons the coupling strengths are different, and they are<br />

relativistic so that recoil effects are not small. However, they are degenerate<br />

so that their contribution is expected to be smaller than the ep<br />

process. It is not entirely negligible, however, especially if the nucleon<br />

contribution is partly suppressed by many-body effects. The transition<br />

rates were worked out in detail by Raffelt and Sigl (1993) <strong>for</strong> entirely<br />

degenerate leptons. They found<br />

W R EE ′<br />

W L EE ′<br />

≈ G2 Fµ 2 e<br />

3π<br />

≈<br />

G2 Fµ 2 e<br />

3π<br />

(E − E ′ )E ′<br />

,<br />

E 2<br />

(E − E ′ )E<br />

, (9.62)<br />

E ′2<br />

a result which is the lowest-order term of an expansion in powers of<br />

E/µ e (electron chemical potential µ e ).<br />

9.5.3 Time Scale <strong>for</strong> Flavor Conversion<br />

Given enough time the ν τ ’s will reach the same number density <strong>as</strong> the<br />

ν e ’s. There<strong>for</strong>e, it is most practical to discuss the approach to chemical<br />

equilibrium in terms of a time scale<br />

τ −1 ≡ − d ( )<br />

dt ln nνe − n νx<br />

. (9.63)<br />

n νe<br />

If τ were a constant independent of n νx /n νe the difference between the<br />

number densities would be damped exponentially.<br />

Collecting the results of the previous section one then finds e<strong>as</strong>ily<br />

<strong>for</strong> the c<strong>as</strong>e of a “large” ∆m 2 of Eq. (9.59)<br />

τ −1 = θ 2 0<br />

3G 2 F<br />

5π n pµ 2 ν e<br />

[<br />

(C 2 V + 3C 2 A) F p + µ ν e<br />

µ e<br />

F e<br />

]<br />

, (9.64)<br />

where F p and F e give the contributions of protons (CC process) and<br />

electrons (effective NC process). They are functions of µ νx which is<br />

parametrized by<br />

η ≡ µ νx /µ νe . (9.65)<br />

One finds (Fig. 9.2, left panel)<br />

F p = (1 − η 5 )/(1 − η 3 ),<br />

F e = ( 5 6 + 1 2 η + 1 4 η2 + 1 12 η3 ) (1 − η) 3 /(1 − η 3 ), (9.66)

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