Geography Reference
In-Depth Information
n
L
(
S
)=
L
0
+
SL
0
+
1
1
m
!
S
m
L
(
m
0
,
2!
S
2
L
0
+ lim
n→∞
m
=3
n
B
(
S
)=
B
0
+
SB
0
+
1
1
m
!
S
m
B
(
m
0
,
2!
S
2
B
0
+ lim
(20.72)
n→∞
m
=3
n
α
(
S
)=
α
0
+
Sα
0
+
1
1
m
!
S
m
α
(
m
0
.
2!
S
2
α
0
+ lim
n→∞
m
=3
Question: “But how to
effectively
compute the higher
derivatives of
L
(
m
0
,B
(
m
0
,α
(
m
)
with respect to an initial
point
{L
0
,B
0
,α
0
}
and subject to the differential equa-
tions which govern a
geodesic
in the Hamilton portrait,
a submanifold in
{
}
0
A
1
,A
2
?” Answer: “A proper answer is
immediately given by the
Lie recurrence
(“Lie serie
s”)
of
{
E
in
terms
of
x
=
r
cos
α
0
=
S
√
G
11
L
0
L
(
m
0
,B
(
m
0
,α
(
m
)
}
and
y
=
r
sin
α
0
=
S
G
22
B
0
summarized in Box
20.4
.”
0
L
(
m
0
,B
(
m
0
,α
(
m
)
If we replace
in the Taylor expansion (
20.72
)bymeansofthe
Lie recurrence
of Box
20.4
,wehavesolvedthe
initial value problem
of the geodesic (
20.69
)and(
20.71
)inthe
Hamilton portrait in terms of power series
x, y, x
2
, xy,y
2
etc., in particular
{
}
0
L
=
L
0
+ [10]
x
+ [11]
xy
+ [12]
xy
2
+ [30]
x
3
+O
4
L
,
(20.73)
B
=
B
0
+ [01]
y
+ [20]
x
2
+ [02]
y
2
+ [03]
y
3
+O
4
B
,
(20.74)
α
=
α
0
+ [10]
α
x
+ [11]
α
xy
+ [12]
α
xy
2
+ [30]
α
x
3
+O
4
α
.
(20.75)
2
A
1
,A
2
The coecients [
μν
]aregiveninBox
20.5
. Solving the initial value problem for
E
with
L
0
=0
.
0003,
b
:=
semiaxes
A
1
and
A
2
of Earth dimension up to an accuracy of
l
:=
L
−
α
0
=0
.
001, we are limited to distances up to 100km for series
expansion
up to order five
(
Boltz 1942
).
B
0
=0
.
0002, and
α
B
−
−
Box 20.4 (The Lie recurrence (“Lie series”) of
{
L
(
m
)
,B
(
m
)
}
in terms of
x
=
r
cos
α
0
=
SN
(
B
0
)cos
B
0
L
0
and
y
=
r
sin
α
0
=
SM
(
B
0
)
B
0
,
E
2
=(
A
1
−
A
2
)
/A
1
,
N
0
=
N
(
B
0
),
M
0
=
M
(
B
0
)).
cos
α
sin
α
tan
B
N
(
B
)
cos
α,
L
=
N
(
B
)cos
B
, B
=
M
(
B
)
, α
=
−
(20.76)
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