Etudes des proprietes des neutrinos dans les contextes ...

tel-00450051, version 1 - 25 Jan 2010
Consequently, the vector Q in flavour space plays the role of a spherical pendulum
in that its length is conserved so that it can move only on a sphere of radius
Q. Actually, looking at Eqs.(5.27) and (5.31), we realize that the pendulum’s
subsequent oscillations, represented by the vector Q, are confined in a plane
defined by B and the z-axis. Indeed, in vacuum the vectors P and ¯ P evolve
in the same way except that they rotate in opposite directions, therefore, their
respective y−component cancels each other. Consequently, the vector Q has only
two evolving components on the x− and z− axis. Therefore, we can define Q
such as:
⎛
Q = ⎝
sin ϕx
0y
cosϕz
⎞
⎠ (5.34)
where ϕ is the tilt angle of Q relative to the z-axis. Finally, the problem variables
are:
˙ϕ = µD ,
˙D = −ωQ sin(ϕ + 2θV ) . (5.35)
Noticing from Eqs.(5.35), that ϕ is a coordinate and D its canonically conjugate
momentum, we can obtain the form of the corresponding Hamiltonian, namely:
where
H(ϕ, D) = κ2
1 − cos(ϕ + 2θV )
µ
+ 1
2 µ D2 = V + T
= ωB · Q + µ
2 D2 , (5.36)
κ 2 = ωµQ . (5.37)
It corresponds to the motion of a simple pendulum where the first term V =
ωB · Q is the potential energy in a homogeneous field and the second term T =
µ2D 2 /2 is the kinetic energy. We now study the influence of the hierarchy on the
pendulum behaviour.
The importance of the hierarchy
We first consider the normal hierarchy case and then investigate the inverted
hierarchy case. Assuming a small vacuum mixing angle θ0 and a small excursion
angle ϕ of the pendulum, the potential can be expanded to the second order:
V (ϕ) = κ 2 [1 − cos(ϕ + 2θV )]
= κ2
2 (ϕ + 2θV ) 2 + . . . . (5.38)
100