Vilasi G. Hamiltonian Dynamics
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Appendix
B
Concerning
the
Lie Derivative
Let
us
observe that
from
it
follows
that
or
equivalently,
Then,
we
may write
404
Concerning
the Lie Derivative
h=l
axk
0x6
‘
=
--
In this way, multiplying
by
8xi/8xk
and summing
over
k,
we
obtain
and
then
Therefore,
405
Appendix C
Concerning
the
Kepler
Action
Variables
The two closed curves of integration in the integrals
are fixed by requiring the vanishing
of
the corresponding velocities
or,
better,
of
the corresponding momenta
p8
and
pr
expressed,
of
course, in terms
of
variables
re
and
rv.
In
this way, the integration limits are fixed by
Therefore, the
“6”
integration must be performed between the
limits
$1
and
192
given by the solutions of
where
Eq.
(4.25)
has been used. Since
19
itself always lies between
0
and
r,
where sin8
>
0,
we have
sin291
=
sin&
=
COSLY.
Thus, the integration goes
406
Concerning
the
Kepler
Action
Variables
407
from
81
=
n/2
-
a
to
7r/2
to
82
=
n/2
+a
and again back to
61;
the sin6 goes
from
cosa
to
1,
then to
COSQ.
In this way, we obtain
a
-
cos2
8d8
*/2 cos2r
dr,
1
1
-
sin2
a
sin2
r
sin2
a
2x8
--
-
IT
with
r
defined by
cos
8
=
sin
a
sin
r
Therefore, with x
5
tanr, we have
1
+
22
dx
cos2
a
1
Jo=y+m[,,-
7r
1+x2 cos2
a
)
=
-
(z-
-cosa
27r#
7r
7r
7r
2
=
7rfi(l
-
cos
a)
,
and then, by using again Eq.
(4.25),
The
"r"
integration requires the application
of
the method
of
residues.
The roots
r1
and
r2
of the equation
2mk
7ri
--o
2mE+
-
-
-
-
r
T2
are positive if
E
<
0
and correspond to the radii
of
turning points.
In the complex
z
plane, the function
has two branch points at
408
Concerning
the
KepEer
Action
Variables
and
a
simple
pole
at
z
=
0,
so
that
J,
=
i(R(2
=
0)
+
R(z
=
+03)).
Since
R(z
=
0)
=
a
and
R(z
=
+m)
=
rnk/e,
we
finally
obtain
mlc
JZGiE.
Jr
--Z$
+
Appendix
D
Concerning the Reduced
Phase
Space
Appendix
E
On
the Canonical Differential
I-Form
Let
M
be
a
differentiable manifold and
T"M
its cotangent bundle. The
map
T:T*M-+M,
which associates with every differential 1-form
on
TqM
the point
q
E
M,
is
a
surjective differentiable map.
Let
Va,
E
Ta,(T*M)
be
a
tangent vector on the cotangent bundle at the
point
ag
E
TPM;
the derivative
r*
:
T(T*M>
-+
TM
of
the natural projection
I
maps the vector
V,,
to the vector
Tm*,(~a~~,
which
is
tangent to
M
at the point
q.
The map
defined by
is
called the
cunon~eul
1-form
on the cotangent bundle
T*M.
410
On the
Canonical
Differential
1-Form
411
If the manifold
M
is supposed to be a Lie group
G,
the diagram
@;-I
T*G
___)
T*G
.l
IT
G+G
@g
is a commutative diagram (here
ag
=
Lg
and, for every
g
E
G,
is the
symplectic diffeomorphism
of
the induced action of
G
on the cotangent bundle
T*G).
Indeed,
T(@;-,
(ah))
=
r(agh)
=
gh,
Vah
E
T*G
and
ag(7(ah))
=
Qg(h)
=gh,
Vah
E
T*G.
The
diagram
7
T*G
3
G
ti-G
1
T(T*G)
3
TG
T*
where
(G
and
<TOG
are given
by
d
<G
:
E
G
+
(G(g)
=
z@ett(g)lt=o
E
TgG
v<
E
6
and
d*
ET*G
:
Og
E
T*G
+
fT*G(ag)
=
-ae-t€
(ag)lt=o
E
Tag
(T*G)
,
dt
is
a
commutative diagram too.
Of
course, by using the commutative diagram
(E.2),
we have
d
dt
d
dt
T*cz,,
(<T*G(Qh))
=
-T(@:-t€
(%))lt=O
=
-@€?C(T(ah))(t=O
412
On
the
Canonical Differential
I-Form
d
- -
-&a&
(h)lt=o
=
kdh)
=
~G(T(~~)).
From
Eq.
(10.24), we have
J€(ag)
=
(i€pG@)(%)
=
@ag
(b*C(ag))
=
ag(T*ag
(<T*G(ag)))
=
ag(tG(g))
*
Let us
go
back to the general case.
It
is
not difficult to prove that
p*s
=
8
p
:
q
E
M
-+
p(q)
=
pq
E
T;M.
for every differential l-form
on
T*M,
(E.3)
We start by observing that the derivative of
Eq.
fE.3)
defines a map
p*
:
TM
+
T(T*M)
,
so
that,
if
d
dt
v,
=
-q(t)lt=o
7
(~s)q(~)
=
s~9(~*q(~))
=
pq(T*~9(p*q(~)))
*
i.e.
V,
E
TqM
with an integral curve
q(t),
where
q(0)
=
q,
then
On the other hand,
d
d
T*P,(P*q(l/P))
=
g'("q(t)))lt=o
=
-q(t)It=o
dt
=
v,
7
so
that
(P*%(Vq)
=P*(V*)r
vvq
E
TqM.
If
w
=
-d0
is the canonical symplectic form on
T*M,
we have
P*w
=
-p*&
=
-dp*s
=
-@
for every
differentia^
f-form
P
on
T*M.
Appendix
F
~oncerni~~
Rigid
ody
Equations
Let
us
observe that given the
map
Adra
:
g
E
G
-+
(A~a)(g)
=
Adia
E
g;*
CY
E
g;*
,
its deri~tive, at identity, gives the
map
d
dt
(dd:a)*e(~)
=
-Ad:tcblt=O
=
Qdga
I
E
0
*
Similarly, the derivative
of
the
map
Ad:AdiCY
:
h
E
G
-+
Ad~Ad;TCY
E
8%
is
given
by
d
dt
(Ad:Adla)..(()
==
-Ad~~~A~~al~~o
=z
u~;(Ad~b)
.
On
the other hand,
we
also
have
d
d
dt
dt
d
dt
(Ad~Adib)*~(e)
=
--Ad$cAd;Ttu)t=o
=
-dd&t<alt=a
=
-Adtpet~alt=o
==
(A~a)*g~(~g)*=(t~)
t
SO
that
(Ad:a)*g
((
Lg)*e(t))
=
ad; (A~ia~
-
If
g
=
~(t)
is
a
curve
in
G
and
a(t)
=I
A~~~~~-~~
is
a
curve
in
Q",
the above
relation
is
equivalent to Eq.
(10.49).
413