Goldstein 939a (3
rd
ed. 9.39a)
(1) USING CARTESIAN COORDINATES
Using Cartesian canonical variables, the Hamiltonian for the Kepler problem
is
H
=
p
2
2
m
−
k
r
,
(1)
where
p
2
=
p
2
x
+
p
2
y
+
p
2
z
,
(2)
and
r
=
p
x
2
+
y
2
+
z
2
.
(3)
The LaplaceRungeLenz vector is
A
=
p
×
L
−
mk
r
r
.
(4)
Let
x
,
y
,and
z
represent the Cartesian basis vectors.
We know that
L
is
conserved, and we are free to orient the coordinate system so that the
z
axis is
coincident with
L
,
L
=
L
z
z
.
(5)
We also know that
r
and
p
lie in the orbital
x

y
plane, so we may conveniently
set
z
=0
and
p
z
=0
.
It follows that
p
×
L
is a vector that also lies in the
orbital plane. Thus,
A
has only
x
and
y
components,
A
=
A
x
x
+
A
y
y
.
(6)
Since we now have
p
=
p
x
x
+
p
y
y
,
(7)
and
r
=
x
x
+
y
y
,
(8)
it follows that
A
x
=
p
y
L
z
−
mk
r
x,
(9)
and
A
y
=
−
p
x
L
z
−
mk
r
y.
(10)
1
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentNow, because
A
has no explicit time dependence, to prove that
A
is a
constant of the motion, we need to show only that
[
A
x
,H
]=0
,
(11)
and
[
A
y
,H
]=0
.
(12)
Actually, by taking advantage of the
x

y
symmetry of this problem, all we have
to prove is either Eq.(11) or Eq.(12) because the other will then automatically
be true.
To see this, just consider what happens to
A
y
,de
f
nedinEq
This is the end of the preview. Sign up
to
access the rest of the document.
 Fall '10
 Wilemski
 mechanics, Cartesian Coordinate System, René Descartes, Standard basis, Lz

Click to edit the document details