Etudes des proprietes des neutrinos dans les contextes ...

tel-00450051, version 1 - 25 Jan 2010
In this case the system is equivalent to a harmonic oscillator with frequency κ
and the potential has a classical parabolic form whose minimum is reached when
ϕ = −2θV . The second derivative of this potential is:
d2V (ϕ)
= κ2
dϕ2 (5.39)
Therefore, ϕ = −2θV corresponds to a stable equilibrium position. In the approximation
of small angle, using Eqs.(5.31) and (5.34) the initial conditions yield
ϕ(0) ≈ −(ω/µQ) 2θV . Putting ϕ near the minimum of the potential V (ϕ) at
t = 0, with zero initial speed ( ˙ϕ(0) = 0), will only make ϕ oscillates around its
minimum of potential energy.
As discussed in chapter 1, there are two ways to describe an inverted hierarchy,
either we consider a positive cos 2θV and a negative ∆m 2 or the contrary. We
choose here the latter, so that a mixing angle close to zero corresponds to the
normal hierarchy 6 while θV near π/2 corresponds to the inverted hierarchy. We
can define
˜θV = π/2 − θV . (5.40)
for simplicity in inverted hierarchy, thus ˜ θV is small in this case. Therefore, the
potential (5.38) can be written as
V (ϕ) = κ 2
1 + cos(ϕ − 2˜
θV )
= − κ2
ϕ − 2
2
˜ 2 θV + . . .. (5.41)
Here ϕ = 2 ˜ θV corresponds to the maximum of the potential, and to an unstable
equilibrium position. The initial conditions being equivalent to the previous
ones, i.e ϕ(0) ≈ −(ω/µQ) 2 ˜ θV and ˙ϕ(0) = 0, here it will make ϕ go down to the
potential minimum where ϕmin ≈ −π 7 . Evaluating for Pz and ¯ Pz at ϕ = ϕmin,
one finds in the strong neutrino-neutrino coupling limit µ/ω ≫ 1 :
Pz|ϕmin = ¯ Pz|ϕmin
≈ −1 (5.42)
Therefore one can see that complete flavour conversion is possible. Nevertheless
this conversion takes a certain time to happen. Indeed, without calculation, one
can see that the smaller ˜ θV is, the closer ϕ(0) will be to the stable position, and
the longer it will take to go to the potential minimum. Consequently there will
be a certain time when the vector Q almost does not move. Pz and ¯ Pz will remain
6 We remind here that we consider a 2 flavour system where the relevant 3 flavour mixing
angle is θ13 which is at most, according tho Chooz results, equal to 9 ◦ .
7 It goes to −π because its initial value is negative, if it were positive ϕ would have gone to
π.
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