Celletti A. Stability and Chaos in Celestial Mechanics
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214 10 Regularization theory
U −
C
J
2
=
1
2
(q
2
1
+ q
2
2
)+V −
C
J
2
=
1
2
(μ
1
r
2
1
+ μ
3
r
2
3
) −
1
2
μ
1
μ
3
+
μ
1
r
1
+
μ
3
r
3
−
C
J
2
=
1
2
μ
1
|w|
4
+ μ
3
|w
2
− 1|
2
−
1
2
μ
1
μ
3
+
μ
1
|w|
2
+
μ
3
|w
2
− 1|
−
C
J
2
.
Taking into account that D =4(Q
2
1
+ Q
2
2
)=4|w|
2
, we conclude that the term
D(U −
C
J
2
) does not contain singularities at P
1
.
We add a remark on the values of the velocities at the location of the primaries.
From the Jacobi integral in the physical space we get |˙z|
2
=2(U −
C
J
2
); since
z = w
2
−μ
3
,then ˙z =2w ˙w =
2
D
ww
or |w
|
2
=
D
2
4|w
2
|
|˙z|
2
, so that in the parametric
space we obtain
|w
|
2
=8|w|
2
U −
C
J
2
.
Therefore, we find
|w
|
2
=8μ
1
+ |w|
2
8μ
3
|w
2
− 1|
+4μ
1
|w|
4
+4μ
3
|w
2
− 1|
2
− 4C
J
− 4μ
1
μ
3
.
We remark that in P
1
one has r
1
= 0, namely w =0,sothat|w
|
2
=8μ
1
and the
velocity is finite. In P
3
one has r
3
= 0, namely w = ±1, so that |w
|
2
= ∞ and the
velocity is infinite.
10.2 The Kustaanheimo–Stiefel regularization
10.2.1 The restricted, spatial three–body problem
We consider the three–body gravitational interaction of P
1
, P
2
, P
3
in the spatial
case, in which P
2
is allowed to move in any direction. The masses of P
1
and P
3
are, respectively, μ
1
and μ
3
, while P
2
has infinitesimal mass. We assume that the
primaries move in the xy–plane around their common barycenter; their coordi-
nates in the synodic reference frame are given by P
1
(−μ
3
, 0, 0), P
3
(μ
1
, 0, 0). The
Hamiltonian function of this model is given by
H(p
1
,p
2
,p
3
,q
1
,q
2
,q
3
)=
1
2
(p
2
1
+ p
2
2
+ p
2
3
)+q
2
p
1
− q
1
p
2
− V (q
1
,q
2
,q
3
) , (10.3)
where V (q
1
,q
2
,q
3
) ≡
μ
1
r
1
+
μ
3
r
3
and r
2
1
≡ (q
1
+μ
3
)
2
+q
2
2
+q
2
3
, r
2
3
≡ (q
1
−μ
1
)
2
+q
2
2
+q
2
3
.
The equations of motion of P
2
are (see (4.12))
¨q
1
− 2˙q
2
= U
q
1
¨q
2
+2˙q
1
= U
q
2
¨q
3
= U
q
3
, (10.4)
where U =
1
2
(q
2
1
+ q
2
2
)+
μ
1
r
1
+
μ
3
r
3
and U
q
i
≡
∂U
∂q
i
for i =1, 2, 3.
10.2 The Kustaanheimo–Stiefel regularization 215
10.2.2 The KS–transformation
Starting from (10.4), which can be written as
¨q
1
− 2˙q
2
= q
1
−
μ
1
r
3
1
(q
1
+ μ
3
) −
μ
3
r
3
3
(q
1
− μ
1
)
¨q
2
+2˙q
1
= q
2
−
μ
1
r
3
1
q
2
−
μ
3
r
3
3
q
2
¨q
3
= −
μ
1
r
3
1
q
3
−
μ
3
r
3
3
q
3
,
we proceed to perform a time transformation and then a coordinate transformation.
We define the fictitious time s through the relation
dt = Dds,
namely
d
dt
=
1
D
d
ds
,whereD will be defined later. The second derivatives with
respect to t and s are related by
d
2
dt
2
=
d
dt
1
D
d
ds
=
1
D
d
ds
1
D
d
ds
=
1
D
2
d
2
ds
2
−
1
D
3
dD
ds
d
ds
.
The equations of motion with respect to the new time s are
1
D
2
q
1
−
1
D
3
D
q
1
−
2
D
q
2
= U
q
1
1
D
2
q
2
−
1
D
3
D
q
2
+
2
D
q
1
= U
q
2
1
D
2
q
3
−
1
D
3
D
q
3
= U
q
3
and multiplying by D
3
one obtains
Dq
1
− D
q
1
− 2D
2
q
2
= D
3
U
q
1
Dq
2
− D
q
2
+2D
2
q
1
= D
3
U
q
2
Dq
3
− D
q
3
= D
3
U
q
3
. (10.5)
The singular terms are the right–hand sides of (10.5).
In the planar case we denoted by (q
1
,q
2
) the physical plane and by (Q
1
,Q
2
)the
parametric plane. The Levi–Civita transformation of the coordinates was written
in the form (compare with (10.1))
q
1
q
2
= A
0
(Q)
Q
1
Q
2
=
Q
2
1
− Q
2
2
2Q
1
Q
2
,
where, setting Q
≡ (Q
1
,Q
2
)wehaveA
0
(Q) ≡
Q
1
−Q
2
Q
2
Q
1
.ThematrixA
0
(Q)
has the property that every element is linear and that it is orthogonal. In the n–
dimensional case we investigate whether there exists a generalization A(u
)with
u
∈ R
n
of the regularizing matrix with the following properties:
216 10 Regularization theory
(i) the elements of A(u) are linear homogeneous functions of u
1
,...,u
n
;
(ii) the matrix is orthogonal, namely
(a) the scalar product of different rows vanishes,
(b) each row has norm u
2
1
+ ···+ u
2
n
.
A result by A. Hurwitz [167] states that such a matrix can only be produced
whenever n =1, 2, 4 or 8, but not in the spatial case n = 3. As a consequence it
is necessary to map the three–dimensional physical space into a four–dimensional
parametric space by defining the transformation matrix
A(u
)=
⎛
⎜
⎜
⎝
u
1
−u
2
−u
3
u
4
u
2
u
1
−u
4
−u
3
u
3
u
4
u
1
u
2
u
4
−u
3
u
2
−u
1
⎞
⎟
⎟
⎠
.
As for the coordinates, we extend the three–dimensional physical space to a four–
dimensional space by imposing that the fourth component is equal to zero, namely
(q
1
,q
2
,q
3
, 0).
The Kustaanheimo–Stiefel (KS) transformation [167] for the case of a collision
with P
1
is introduced by defining the following change of coordinates:
⎛
⎜
⎜
⎝
q
1
q
2
q
3
0
⎞
⎟
⎟
⎠
= A(u
)
⎛
⎜
⎜
⎝
u
1
u
2
u
3
u
4
⎞
⎟
⎟
⎠
−
⎛
⎜
⎜
⎝
μ
3
0
0
0
⎞
⎟
⎟
⎠
=
⎛
⎜
⎜
⎝
u
1
−u
2
−u
3
u
4
u
2
u
1
−u
4
−u
3
u
3
u
4
u
1
u
2
u
4
−u
3
u
2
−u
1
⎞
⎟
⎟
⎠
⎛
⎜
⎜
⎝
u
1
u
2
u
3
u
4
⎞
⎟
⎟
⎠
−
⎛
⎜
⎜
⎝
μ
3
0
0
0
⎞
⎟
⎟
⎠
, (10.6)
namely
q
1
= u
2
1
− u
2
2
− u
2
3
+ u
2
4
− μ
3
q
2
=2u
1
u
2
− 2u
3
u
4
q
3
=2u
1
u
3
+2u
2
u
4
.
Notice that the fourth component of the right–hand side of (10.6) is trivially zero.
Remarks.
(1) Setting u
3
= u
4
= 0 the KS–transformation (10.6) reduces to the Levi–Civita
transformation, which cannot be trivially extended to the spatial case introducing
only three parameters (say u
1
, u
2
, u
3
).
(2) The norms of each row of the matrix A are equal to the square of the norm of
the vector u
: u
2
1
+ u
2
2
+ u
2
3
+ u
2
4
.
(3) To regularize collisions with P
3
, one needs to replace the constant vector
(−μ
3
, 0, 0, 0) with (μ
1
, 0, 0, 0).
(4) The matrix A is orthogonal: A
T
(u)A(u)=(u,u) · I
4
(where I
4
is the 4 × 4–
dimensional identity matrix). Denoting by X
≡ (q
1
+ μ
3
,q
2
,q
3
, 0) it follows that
under the transformation X
= A(u)u,itis
r
2
1
=(X,X)=X
T
X = u
T
A
T
(u)A(u)u = u
T
u(u,u)=(u,u)
2
,
namely r
1
=(u,u)=|u|
2
= u
2
1
+ u
2
2
+ u
2
3
+ u
2
4
.
10.2 The Kustaanheimo–Stiefel regularization 217
(5) A direct computation shows that A(u)
= A(u
). From X = A(u)u, it follows
that X
=2A(u)u
.Infact,
q
1
=2u
1
u
1
− 2u
2
u
2
− 2u
3
u
3
+2u
4
u
4
q
2
=2u
2
u
1
+2u
1
u
2
− 2u
4
u
3
− 2u
3
u
4
q
3
=2u
3
u
1
+2u
1
u
3
+2u
4
u
2
+2u
2
u
4
, (10.7)
which can be written as
⎛
⎜
⎜
⎝
q
1
q
2
q
3
0
⎞
⎟
⎟
⎠
=2A(u
)u
=2
⎛
⎜
⎜
⎝
u
1
u
1
− u
2
u
2
− u
3
u
3
+ u
4
u
4
u
2
u
1
+ u
1
u
2
− u
4
u
3
− u
3
u
4
u
3
u
1
+ u
4
u
2
+ u
1
u
3
+ u
2
u
4
u
4
u
1
− u
3
u
2
+ u
2
u
3
− u
1
u
4
⎞
⎟
⎟
⎠
,
provided the last row, which is named the bilinear relation, is identically zero:
u
4
u
1
− u
3
u
2
+ u
2
u
3
− u
1
u
4
=0.
(6) The second derivative is given by
X
=2A(u)u
+2A(u
)u
,
namely
q
1
=2(u
1
u
1
− u
2
u
2
− u
3
u
3
+ u
4
u
4
)+2(u
2
1
− u
2
2
− u
3
3
+ u
3
4
)
q
2
=2(u
2
u
1
+ u
1
u
2
− u
4
u
3
− u
3
u
4
)+4(u
1
u
2
− u
3
u
4
)
q
3
=2(u
3
u
1
+ u
2
u
4
+ u
1
u
3
+ u
2
u
4
)+4(u
1
u
3
+ u
2
u
4
)
0=2(u
4
u
1
− u
3
u
2
+ u
2
u
3
− u
1
u
4
) (10.8)
(last equation follows from the bilinear relation).
Let us now conclude by showing that the KS–transformation (10.6) provides the
desired regularization. We select the scale factor D as
D ≡ r
1
=(u
,u)=u
2
1
+ u
2
2
+ u
2
3
+ u
2
4
;
one finds that D
= r
1
=2(u
1
u
1
+ u
2
u
2
+ u
3
u
3
+ u
4
u
4
). As a consequence the
equations of motion are given by (10.5), where q
1
, q
2
, q
3
,aswellastheirfirst
and second derivatives, are expressed in terms of u
, u
, u
through (10.6), (10.7),
(10.8). The singular parts of equations (10.5) are given by D
3
U
q
1
(or equivalently
by D
3
U
q
2
, D
3
U
q
3
). Since U
q
1
is proportional to
1
r
3
1
and D is proportional to r
1
,
it follows that D
3
Ω
q
1
does not contain singularities; a similar result holds for the
terms D
3
Ω
q
2
, D
3
Ω
q
3
, thus yielding the required regularization of the equations of
motion at P
1
.
218 10 Regularization theory
10.2.3 Canonicity of the KS–transformation
Following [167], we proceed to prove that the KS–transformation is canonical. In
the planar case the KS–transformation reduces to the Levi–Civita transformation
(hereweuse(u
1
,u
2
) instead of (Q
1
,Q
2
)):
q
1
q
2
=
u
1
−u
2
u
2
u
1
u
1
u
2
=
u
2
1
− u
2
2
2u
1
u
2
≡ A
0
(u)u
with A
0
(u)=
u
1
−u
2
u
2
u
1
, such that if U
=(U
1
,U
2
) is conjugated to u =(u
1
,u
2
),
then
p
1
p
2
=
1
2
(A
T
0
(u))
−1
U =
1
2(u
2
1
+ u
2
2
)
A
0
(u)
U
1
U
2
. (10.9)
In the spatial case let us define
Λ(u
) ≡
⎛
⎝
u
1
−u
2
−u
3
u
4
u
2
u
1
−u
4
−u
3
u
3
u
4
u
1
u
2
⎞
⎠
,
obtained from A(u
) eliminating the last row. Let
⎛
⎝
q
1
q
2
q
3
⎞
⎠
= Λ(u
)
⎛
⎜
⎜
⎝
u
1
u
2
u
3
u
4
⎞
⎟
⎟
⎠
−
⎛
⎝
μ
3
0
0
⎞
⎠
. (10.10)
By analogy with (10.9) we define
⎛
⎝
p
1
p
2
p
3
⎞
⎠
≡
1
2(u
2
1
+ u
2
2
+ u
2
3
+ u
2
4
)
Λ(u
)
⎛
⎜
⎜
⎝
U
1
U
2
U
3
U
4
⎞
⎟
⎟
⎠
. (10.11)
Setting u
0
≡ t, U
0
≡ T ,whereT is the variable conjugated to t, we proceed to
verify that (10.10), (10.11) define a canonical transformation. Let
r
1
≡
(q
1
+ μ
3
)
2
+ q
2
2
+ q
2
3
= u
2
1
+ u
2
2
+ u
2
3
+ u
2
4
= |u|
2
;
introducing the scalar product
σ(U
,u) ≡ (U,u
∗
)
with u
∗
≡ (u
4
, −u
3
,u
2
, −u
1
) (coinciding with the last row of A(u)), one obtains
⎛
⎜
⎜
⎝
p
1
p
2
p
3
ρ
⎞
⎟
⎟
⎠
=
1
2r
1
A(u)
⎛
⎜
⎜
⎝
U
1
U
2
U
3
U
4
⎞
⎟
⎟
⎠
,
10.2 The Kustaanheimo–Stiefel regularization 219
where the auxiliary variable ρ is defined as
ρ ≡
1
2r
1
(U
1
u
4
− U
2
u
3
+ U
3
u
2
− U
4
u
1
)=
1
2r
1
σ(U,u) .
From
(p
1
,p
2
,p
3
,ρ)
⎛
⎜
⎜
⎝
p
1
p
2
p
3
ρ
⎞
⎟
⎟
⎠
=(U
1
,U
2
,U
3
,U
4
)
1
2r
1
A(u)
T
·
1
2r
1
A(u)
⎛
⎜
⎜
⎝
U
1
U
2
U
3
U
4
⎞
⎟
⎟
⎠
=
1
4r
2
1
r
1
|U|
2
=
1
4r
1
|U|
2
,
it follows that (p
2
1
+ p
2
2
+ p
2
3
+ ρ
2
)r
1
=
1
4
|U|
2
, namely
(p
2
1
+ p
2
2
+ p
2
3
)r
1
=
1
4
|U
|
2
−
1
4r
1
σ(U,u)
2
. (10.12)
Putting into evidence the term
1
r
1
corresponding to the interaction with P
1
,letus
write the Hamiltonian describing the spatial case in a fixed reference frame within
the framework of the extended phase space in the form
H
0
=
1
2
(p
2
1
+ p
2
2
+ p
2
3
)+T −
1
r
1
− V
1
(q
1
,q
2
,q
3
)
for a suitable function V
1
(q
1
,q
2
,q
3
,t). The introduction of the fictitious time defined
through dt = r
1
ds provides the Hamiltonian
H
1
=
1
2
(p
2
1
+ p
2
2
+ p
2
3
)r
1
+ Tr
1
− 1 − r
1
V
1
(q
1
,q
2
,q
3
) .
Using r
1
= |u|
2
and (10.12) one gets the Hamiltonian
H
2
=
1
8
|U
|
2
+ T |u|
2
− 1 −|u|
2
V
1
(u) −
1
8|u|
2
σ(U,u)
2
. (10.13)
We now show that σ = σ(U
,u) is constant along the solutions of the equations of
motion.
Lemma. The bilinear quantity
σ(U
,u)=U
1
u
4
− U
2
u
3
+ U
3
u
2
− U
4
u
1
is a first integral of the canonical equations
du
k
ds
=
∂H
2
∂U
k
,
dU
k
ds
= −
∂H
2
∂u
k
,k=0, 1, 2, 3, 4 ,
where u
0
≡ t, U
0
≡ T .
220 10 Regularization theory
Proof. One finds
dσ
ds
=
4
j=1
∂σ
∂U
j
dU
j
ds
+
∂σ
∂u
j
du
j
ds
=
u
∗
,
dU
ds
−
U
∗
,
du
ds
.
The new canonical equations are
du
ds
=
∂H
2
∂U
=
1
4
U
−
1
4|u|
2
u
∗
σ(U,u)
dU
ds
= −
∂H
2
∂u
= −2U
0
u +
∂
∂u
(|u|
2
V
1
) −
1
4|u|
4
u σ(U,u)
2
−
1
4|u|
2
U
∗
σ(U,u) . (10.14)
Since (u
∗
,u)=(U
∗
,U) = 0, one has
dσ
ds
= −2U
0
(u
∗
,u)+
u
∗
,
∂
∂u
(|u|
2
V
1
)
−
1
4|u|
4
(u
∗
,u)σ(U,u)
2
−
1
4|u|
2
(u
∗
,U
∗
)σ(U,u) −
1
4
(U
∗
,U)+
1
4|u|
2
(U
∗
,u
∗
)σ(U,u)
=
u
∗
,
∂
∂u
[(u,u)V
1
]
=2(u
∗
,u)V
1
+ |u|
2
u
∗
,
∂V
1
∂u
= |u
|
2
u
∗
,
∂V
1
∂u
.
Direct computations show that
∂V
1
∂u
=2A
T
(u)
∂V
1
∂q
(since
∂V
1
∂u
=
∂V
1
∂q
∂q
∂u
). Moreover
one has that
u
∗
, 2|u|
2
A
T
(u)
∂V
1
∂q
=0. (10.15)
In fact, if y
≡
∂V
1
∂q
and v ≡ 2|u|
2
A
T
(u)y,then(u
∗
,v)=(v,u
∗
)=0,since(v,u
∗
)=
σ(v
,u)=u
4
v
1
− u
3
v
2
+ u
2
v
3
− u
1
v
4
, which is zero. In fact, the latter expression
is equal to the fourth component of A(u
)v =2A(u)A
T
(u)y |u|
2
=2(u,u)y |u|
2
,
where the fourth component of y
≡
∂V
1
∂q
(which represents the derivative with
respect to the time in the physical space) is zero by definition. Finally from (10.14)
and (u
∗
,
∂V
1
∂u
) = 0, we obtain that
dσ
ds
= 0 along the equations of motion.
Lemma. Assume that the initial values u
(0), U (0) of u(s), U(s) satisfy at s =0
the bilinear relation
σ(U
(0),u(0)) = U
1
(0)u
4
(0) − U
2
(0)u
3
(0) + U
3
(0)u
2
(0) − U
4
(0)u
1
(0) = 0 .
Then one has:
(a) the value of the first integral σ is zero: σ(U
,u)=U
1
u
4
−U
2
u
3
+U
3
u
2
−U
4
u
1
=0;
(b) the Hamiltonian (10.13) is equivalent to the reduced Hamiltonian
H
3
=
1
8
|U
|
2
+ U
0
|u|
2
− 1 −|u|
2
V
1
(u,u
0
) .
10.2 The Kustaanheimo–Stiefel regularization 221
Proof.
(a) It is a consequence of the fact that σ(U
,u) is a first integral.
(b) The canonical equations associated to the Hamiltonian H
2
and H
3
have the
same solutions as a consequence of (a), due to the fact that the term σ
2
(U,u)can
be factored out in H
2
−H
3
and that taking the derivatives there is always a factor
σ which is zero along the given solutions.
Next we need to associate to each set of initial conditions in the physical space,
say q
1
(0), q
2
(0), q
3
(0), p
1
(0), p
2
(0), p
3
(0), a new set of initial conditions in the
parametric space, say u
(0), u
0
(0), U(0), U
0
(0), which are obtained by means of
(10.10), (10.11), so that
(i) u
0
(0) = 0, U
0
(0) = −H(q
1
(0),q
2
(0),q
3
(0),p
1
(0),p
2
(0),p
3
(0));
(ii) σ(U
(0),u(0)) = U
1
(0)u
4
(0) − U
2
(0)u
3
(0) + U
3
(0)u
2
(0) − U
4
(0)u
1
(0) = 0.
We remark that to obtain (ii) one can proceed as in (10.15) by setting v
= A
T
(u)y
with y = q
(0) and defining U(0) ≡
1
2|u(0)|
2
A
T
(u(0))q
(0).
Theorem. Assume that u
(0), u
0
(0), U(0), U
0
(0) satisfy (i) and (ii); then, the
solutions of
du
k
ds
=
∂H
3
∂U
k
,
dU
k
ds
= −
∂H
3
∂u
k
,k=0, 1, 2, 3, 4 ,
give by (10.10) and (10.11) the solution of Hamilton’s equations associated to H
defined in (10.3) with the corresponding initial data, namely if q
=(q
1
,q
2
,q
3
),
p
=(p
1
,p
2
,p
3
), one has
dq
dt
=
∂H
∂p
,
dp
dt
= −
∂H
∂q
.
Remark. The above theorem implies that the transformation from (q
,p)to(u,U)
is canonical, since it preserves the Hamiltonian structure.
Proof. The canonicity is proved using the criterion based on Poisson brackets,
namely that
(a) {q
k
,q
l
} =
4
j=0
∂q
k
∂U
j
∂q
l
∂u
j
−
∂q
k
∂u
j
∂q
l
∂U
j
=0
(b) {p
k
,q
l
} =
4
j=0
∂p
k
∂U
j
∂q
l
∂u
j
−
∂p
k
∂u
j
∂q
l
∂U
j
= δ
kl
(c) {p
k
,p
l
} =
4
j=0
∂p
k
∂U
j
∂p
l
∂u
j
−
∂p
k
∂u
j
∂p
l
∂U
j
=0,
for k =0, 1, 2, 3, l =0, 1, 2, 3. Cases k =0orl = 0 are trivial; therefore the
sum can be restricted to j =1, 2, 3, 4. Since the variables U
do not enter in the
KS–transformation (10.10), one has
∂q
k
∂U
j
= 0 providing (a). Moreover, (b) becomes
222 10 Regularization theory
{p
k
,q
l
} =
4
j=1
∂p
k
∂U
j
∂q
l
∂u
j
= δ
kl
,k,l=1, 2, 3 . (10.16)
Using matrix notation one obtains
∂q
l
∂u
j
=2Λ(u
) ,
∂p
k
∂U
j
=
1
2|u|
2
Λ(u) ,
∂p
k
∂u
j
= −
1
|u|
4
Λ(u)
⎛
⎜
⎜
⎝
U
1
U
2
U
3
U
4
⎞
⎟
⎟
⎠
(u
1
,u
2
,u
3
,u
4
)+
1
2|u|
2
Λ(U)
for l, k =1, 2, 3, j =1, 2, 3, 4. Therefore, due to the orthogonality of Λ,therelation
(10.16) is equivalent to
{p
k
,q
l
} =
1
|u|
2
[Λ(u)Λ
T
(u)]
kl
=
1
|u|
2
(u,u)[I
4
]
kl
= δ
kl
,
where [A]
kl
denotes the element at the crossing of the kth row and of the lth column
of the matrix A and I
4
is the 4 × 4–dimensional identity matrix. To prove (c)we
use the bilinear relation, which is valid along the solution due to (a) of the previous
lemma. Let A, B be the matrices defined as
A =
1
2
|u
|
2
Λ(U)Λ
T
(U)
B = Λ(U
)
⎛
⎜
⎜
⎝
U
1
U
2
U
3
U
4
⎞
⎟
⎟
⎠
⎡
⎢
⎢
⎣
Λ(U
)
⎛
⎜
⎜
⎝
u
1
u
2
u
3
u
4
⎞
⎟
⎟
⎠
⎤
⎥
⎥
⎦
T
,
so that
A − B =
4
j=1
∂p
k
∂u
j
∂p
l
∂U
j
2|u
|
6
.
The proof of the symmetry of the matrix A − B is not trivial and we refer the
reader to [167] for complete details. This latter result proves (c).
10.3 The Birkhoff regularization
The problem of regularizing both singularities simultaneously, namely to find a
transformation to regularize at once both collisions with the primaries, is the con-
tent of the Birkhoff transformation. Such global regularization is obtained as fol-
lows.
In the framework of the restricted, planar, circular, three–body problem, we
consider a synodic reference frame with origin at the barycenter of the primaries.
Let the primaries be P
1
, P
3
, while the small body is denoted by P
2
. We normalize
10.3 The Birkhoff regularization 223
to unity the sum of the masses of the primaries, which we denote as μ
1
=1− μ
and μ
3
= μ. Moreover we set to unity the distance between the primaries as well
as the gravitational constant. Let (x
1
,x
2
) be the synodic coordinates of P
2
and let
(p
1
,p
2
) be the corresponding momenta. The Hamiltonian function is given by
H(p
1
,p
2
,q
1
,q
2
)=
1
2
(p
2
1
+ p
2
2
)+q
2
p
1
− q
1
p
2
−
μ
1
r
1
−
μ
3
r
3
, (10.17)
where
r
1
=
(q
1
+ μ
3
)
2
+ q
2
2
,r
3
=
(q
1
− μ
1
)
2
+ q
2
2
.
Birkhoff regularization transforms the physical plane into the parametric plane with
the aim of removing the singularities in (10.17). To this end we shift the origin of
the reference frame by choosing the midpoint between the primaries through a
transformation (q
1
,q
2
)to(q
x
,q
y
) such that q
x
+ iq
y
= q
1
+ iq
2
−
1
2
+ μ.Asa
consequence P
1
is located at (−
1
2
, 0), while P
3
is at (
1
2
, 0). Denoting by (p
x
,p
y
)the
corresponding momenta, the transformed Hamiltonian takes the form
H
1
(p
x
,p
y
,q
x
,q
y
)=
1
2
(p
x
2
+ p
y
2
)+q
y
p
x
−
q
x
+
1
2
− μ
p
y
−
μ
1
r
1
−
μ
3
r
3
,
where
r
1
=
q
x
+
1
2
2
+ q
y
2
,r
3
=
q
x
−
1
2
2
+ q
y
2
.
Denoting by q = q
x
+ iq
y
and by ∇
q
the gradient with respect to q, we can write
the equations of motion in complex form as
¨q +2i ˙q = ∇
q
U(q) , (10.18)
where U (q) ≡
1
2
[(q
x
+
1
2
−μ)
2
+ q
2
y
]+U
c
(q), being U
c
(q)=
μ
1
r
1
+
μ
3
r
3
the critical part
of U. Setting w = w
1
+ iw
2
, we define the regularizing transformation through the
complex change of variables q = h(w)=αw +
β
w
for suitable constants α and β.
We introduce also a time transformation through the expression
dt
dτ
= g(w) ≡|k(w)|
2
for a suitable complex function k = k(w). In order to write the transformed equa-
tions of motion, let us compute the first and second derivatives as
˙q =
dh
dw
dw
dτ
dτ
dt
= h
w
˙τ
¨q = h
w
¨τ +(h
w
2
+ h
w
)˙τ
2
,
where the prime denotes the derivative with respect to the argument. The gradient
operator is transformed as
h
∇
q
U = ∇
w
U,
where
h is the complex conjugate of h. Finally, the equation of motion (10.18) takes
the form