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Stars as Laboratories for Fundamental Physics - MPP Theory Group

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Oscillations of Trapped Neutrinos 331<br />

of freedom, and one may use the weak-damping limit where neutrino<br />

oscillations are much f<strong>as</strong>ter than their rates of collision or absorption.<br />

Then one finds <strong>for</strong> two flavors (Raffelt and Sigl 1993)<br />

˙ f x p = (1 − f x p) P x P − f x p A x P<br />

+ 1 4 s2 p[<br />

(2 − f<br />

x<br />

p − f e p)(P e P − P x P ) − (f x p + f e p)(A e P − A x P ) ] ,<br />

(9.48)<br />

where f e p and f x p are the occupation numbers <strong>for</strong> ν e and ν x <strong>as</strong> in<br />

Sect. 9.3.5. The corresponding equation <strong>for</strong> ν e is found by exchanging<br />

e ↔ x everywhere.<br />

For ν x = ν µ flavor conversion can build up a nonvanishing muon density<br />

in a SN core because they are light enough to be produced initially<br />

when the electron chemical potential is on the order of 200−300 MeV.<br />

For ν x = ν τ or some sterile flavor, the direct production or absorption<br />

is not possible, A x P = P x P = 0. This simplifies Eq. (9.48) considerably,<br />

˙ f x p = 1 4 s2 p[<br />

(2 − f<br />

x<br />

p − f e p) P e P − (f x p + f e p) A e P<br />

]<br />

. (9.49)<br />

If neither ν e nor ν τ are occupied because, <strong>for</strong> example, the medium<br />

is transparent to neutrinos so that they escape after production, one<br />

h<strong>as</strong> f ˙ p x = 1 2 s2 pPP<br />

e so that the production rate of ν x is that of ν e times<br />

1<br />

2 sin2 2θ p <strong>as</strong> one would have expected.<br />

In a SN core where normal neutrinos are trapped Eq. (9.49) is<br />

more complicated <strong>as</strong> backreaction and Pauli blocking effects must be<br />

included. It becomes simple again if s 2 p ≪ 1, because then the production<br />

of ν x causes only a small perturbation of β equilibrium. There<strong>for</strong>e,<br />

one may use the detailed-balance condition (1 − fp)P e P e − fpA e e P = 0.<br />

Inserting this into Eq. (9.49) leads to<br />

˙ f x p = 1 4 s2 p[<br />

(1 − f<br />

x<br />

p ) P e P − f x p A e P<br />

]<br />

, (9.50)<br />

so that now the ν x follow a Boltzmann collision equation with rates<br />

of gain and loss given by those of ν e times 1 4 sin2 2θ p . If the ν x are<br />

sterile they escape without building up so that their production rate is<br />

1<br />

4 sin2 2θ p that of ν e .

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