Etudes des proprietes des neutrinos dans les contextes ...

tel-00450051, version 1 - 25 Jan 2010
We can always redefine the Ce(t) and Cµ(t) functions modulo a phase, thus for
practical purpose we actually solve the following equation:
i d
Cee
dt
−i H1(t)dt
Cµe−i
−
H2(t)dt =
∆m2
4E cos 2θV + √ 2
2 GFNe
∆m2 sin 2θV 4E
∆m2 ∆m sin 2θV 4E 2
4E cos 2θV − √ 2
2 GFNe
Cee−i
H1(t)dt
. (B.25)
Cµe −i H2(t)dt
This relation reduces to two simultaneous first order differential equations for
Ce(t) and Cµ(t):
H12 Ce e i (H2(t)−H1(t))dt = i ˙ Cµ (B.26)
H12 Cµ e −i (H2(t)−H1(t))dt = i ˙ Ce
We impose the initial conditions we are interested in: Ce(ti) = 0 and |Cµ(ti)| = 1
(i.e a νµ created initially). Combining those two equations and defining H12 = f
we find:
¨Ce(t) − i αt ˙ Ce(t) + f 2 Ce(t) = 0 (B.27)
The substitution
Ce(t) = U1 e −i (H2(t)−H1(t))dt
reduces Eq.(B.27) to the Weber 3 equation:
A last redefinition of the variables:
(B.28)
d2 2
U1 α
+
dt2 4 t2 − iα
+ f2 U1(t) = 0 (B.29)
2
z = α 1
2e −iπ/4 t (B.30)
n = if 2 /α
allows us to put Eq.(B.29) in the standard form:
d2
U1
+ −
dz2 z2
4
1
+ n + U1(z) = 0 (B.31)
2
The Weber function D−n−1(iz) is a particular solution of this equation which
vanishes for infinite z along the directions ∞ exp(−i3π ) and ∞ exp(−iπ ). Hence
4 4
the solution satisfies the first boundary condition:
U1(z) = A±D−n−1(∓iz), α ≷ 0 (B.32)
3 d
A definition of a general Weber equation is 2 f
dz2
2
z + 4 − a f = 0
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