Chemistry Reference
In-Depth Information
where
m
1
/
2 for
the minimization of the energy. Equation (26) can be straightforwardly obtained
from Eqs. (21), (23), and (24). Simple algebra leads to
=
j/ h
T
for the time minimum problem and
m
=
j
with
h
E
=
(
m
2
1) cos
2
θ
sin
2
θ
1
−
+
θ
2
=
(27)
We obtain for the positive branch that
θ
1
sin
θdθ
1
T
=
(28)
−
(1
+
m
2
) cos
2
θ
θ
0
where
θ
0
and
θ
1
are the initial and final values of
θ
. Using the fact that
sin
θdθ
√
a
arcsin(
√
a
cos
θ
)
1
√
1
a
cos
2
θ
=−
(29)
−
one arrives at
m
2
arcsin(
1
m
2
cos
θ
1
)
arcsin(
1
1
T
=
√
1
+
m
2
cos
θ
0
)
−
+
(30)
+
The inversion of this relation gives
arcsin
K
)
sin(
1
π
2
+
1
m
2
t
θ
(
t
)
=
√
1
+
+
(31)
+
m
2
in the case where
θ
is an increasing function of
t
. Here,
K
is a constant that can be
determined from
θ
(0). The evolution of
ϕ
is obtained from the differential equation
m
cot
2
θ
ϕ
=
(32)
and Eq. (31).
The extremal trajectories, which are the same for
H
E
and
H
T
, are represented
in Fig. 11 for a given value of
j
and different values of
h
. These trajectories
have three symmetries. The figure is first symmetric with respect to translation
in
ϕ
. In addition, two trajectories corresponding, respectively, to
p
ϕ
and
−
p
ϕ
are
symmetric with respect to the meridian
ϕ
cst
. Finally, two extremals starting
from the same point (
ϕ
(0)
,θ
(0)), with the same value of
p
ϕ
but with opposite initial
values of
p
θ
(0) intersect on the antipodal parallel
θ
=
θ
(0) with the same value
of the cost. As shown in Fig. 11, we notice that the trajectory with
p
θ
(0)
=
π
−
0is
tangent to the antipodal parallel. Figure 12 illustrates these two symmetries that
=
