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9
Central Forces and Kepler’s Problem
Furnished with the most accurate observations of the position of Mars on the celestial
sphere Johannes Kepler devised his celebrated first law of planetary motion. The
observations of Tycho Brahe of the planet Mars were not consistent with circular
motion, nor with corrections to this using so called epicycles. He discovered (1605)
that the motion was very accurately described by motion along an ellipse with the
Sun at one focus. This law of planetary motion was followed in 1609 (published in
the Astronomia Nova and based on observations of Earth’s orbit) by the second law
that the radial vector from the Sun to the planet sweeps out equal area in equal time.
In 1618 Kepler published (the Harmonices Mundi) his third law of planetary motion
which states that the mean apsidal distance cubed relates to the square of the period
of the orbit of planets.
1. Motion of planets describe ellipses with the Sun at one focus.
2. The radial vector from the Sun to the planet sweeps out equal area in equal
time.
3. That a3 ∝ T 2 .
It is the first and last of these laws which lead Newton to his universal law of
gravitation which he published in his famous work, the Principia Mathematica (1687).
We will do things in the other logical direction by demonstrating how each of Kepler’s
laws follow from the dynamics of bodies being acted on by a central force and in
particular by an inverse square force law such as Newton’s universal gravity. This
application will again allow us to make use of Noether’s Theorem and the EulerLagrange equation.
9.1
Reduction of the 2-Body Problem to a 1-Body Problem
The first step is to realize that we can reduce the problem of the motion of two bodies
with masses m1 and m2 with a force acting between them to a 1-Body problem. Let
r1 and r2 be the position vectors from some origin to each of these masses. Let us
assume that V is any function of r ≡ r2 − r1 or of ṙ2 − ṙ1 or higher derivatives, not
restricting the problem any further for the time being. Let R be the position of the
center of mass of the system. We have 6 degrees of freedom implying 6 independent
generalized coordinates (R, r). The Lagrangian is then written,
L = T (Ṙ, ṙ) − V (r, ṙ, ...).
T can be written as a sum of the kinetic energy of the motion of the center of mass
plus the kinetic energy of motion about the center of mass, T 0 .
1
T = (m1 + m2 )Ṙ2 + T 0
2
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with,
1
1
T 0 = m1 ẋ21 + m2 ẋ22
2
2
where x1 and x2 are the radii relative to the center of mass,
x1 = −
m2
r
m1 + m2
x2 =
m1
r.
m1 + m2
Now expressed in terms of r,
T0 =
1 m1 m2 2
ṙ .
2 m1 + m2
The mass factor here is known as the reduced mass and is often written with µ. We
now write the Lagrangian out in full,
1
1
L = (m1 + m2 )Ṙ2 + µṙ2 − V (r, ṙ, ...).
2
2
1. You can see that the 3 coordinates R are cyclic (see page 32), so that the center
of mass is either at rest or moving with constant velocity (uniformly).
2. None of the equations of motion for r will contain R or Ṙ.
Therefore we simply drop the first term of the Lagrangian and write, without loss of
generality,
1
L0 = µṙ2 − V (r, ṙ, ...).
2
9.2
Equation of Motion and First Integrals
We will now use various symmetries of the system to obtain first integrals and thereby
restrict the motion. We want to find out as much about motion in central potentials
without solving the full equations of motion. This we will do later for the specific
case of the inverse square force law.
1. We will now restrict ourselves to conservative central forces, V (r) a function of
r ≡ |r| only, so that now the force is always along r (i.e., central).
2. We have shown that we only need to consider a single particle of reduced mass
µ or m (we will now switch to m for simplicity) moving about a fixed center of
force at the origin.
3. Since V (r) implies spherical symmetry, i.e., L is invariant under rotation about
any axis, an angle coordinate about a fixed axis must be cyclic.
Spherical symmetry gives us a conserved quantity of the motion, namely the angular
momentum vector, L = r × p. This is given to us by Noether’s Theorem and was
shown previously (page 59). This means that the radius vector, r, is always perpendicular to the fixed direction of L in space. Therefore the motion is restricted to the
plane perpendicular to L. What can we do about the case where L = 0?
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r
dθ
Figure 1: The shaded region has area dA = 12 r2 dθ. Since the small remaining region
has an area of order O(dθ2 ), dA gives us the area swept out by the radius vector to
first order in dθ.
The conservation of L gives us 3 independent constants of motion, two of which are
effectively specifying the direction (a unit vector needs 2 components) and restricting
the motion to the plane. So this means we still have one constant of motion left,
l = |L|, that we can use to further restrict the motion. We now have the following
Lagrangian,
1
L = T − V = m(ṙ2 + r2 θ̇2 ) − V (r).
2
As stated θ is seen to be cyclic, whose corresponding canonical momentum is the
angular momentum of the system:
pθ =
∂L
= mr2 θ̇.
∂ θ̇
One of the two equations of motion is then simply,
ṗθ =
d
(mr2 θ̇) = 0,
dt
(1)
with mr2 θ̇ = l and l ≡ |L|! If we rewrite this in a more suggestive way as,
d
dt
1 2
r θ̇ = 0,
2
where the factor of 12 has simply been inserted so that it looks more like Kepler’s
second law and is not important. The area swept out by the radius vector per unit
time is constant (see figure 1).
dA
1 dθ
= r2
dt
2 dt
1
dA = r(rdθ),
2
|L| is constant ≡ Kepler’s 2nd law and this is a general property of all central forces
and is not restricted to just the motion of the planets.
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Now we want to look in more detail at r-part of the equation of motion. We begin
by using the Euler-Lagrange equation on L,
d
∂V
(mṙ) − mrθ̇2 +
= 0,
dt
∂r
or since f (r) = −∇V · r̂,
mr̈ − mrθ̇2 = f (r).
Eliminate θ̇ using the relation l = mr2 θ̇,
mr̈ −
l2
= f (r).
mr3
(2)
There is another first integral of motion, namely the total energy since the forces are
conservative. Just a quick recap from the previous:
1. First write down,
dL X ∂L dqk X ∂L dq̇k
=
+
+
dt
∂qk dt
∂ q̇k dt
k
k
∂L
.
∂t
2. Since the Lagrangian is invariant under time transformations (conservative
forces), we know that the last term is zero.
3. We now subtract a term which is effectively the Euler-Lagrange equation multiplied by q˙k , this is subtracting a zero term so we are not changing anything
here.
dL X ∂L
∂L d
∂L
d ∂L
=
q̇k +
(q̇k ) −
−
q̇k .
dt
∂qk
∂ q̇k dt
∂qk
dt ∂ q̇k
k
4. The first and third terms clearly cancel and we can collect the derivatives of
the two second terms,
dL X d ∂L
=
q̇k ,
dt
dt ∂ q̇k
k
or
#
"
d X ∂L
q̇k − L = 0
dt k ∂ q̇k
5. Insert our coordinates into this problem to get,
d
1
1
mṙ2 + mr2 θ̇2 − mṙ2 − mr2 θ̇2 + V (r) = 0,
dt
2
2
and our constant of the motion (or first integral) is the total energy of course,
1
E = (mṙ2 + mr2 θ̇2 ) + V (r).
2
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9.2.1
Who needs Lagrange when you have Tricks!
Another way to proceed without using any fancy theorems is to start directly from
the equations of motion given by equations 1 and 2. Here are the tricks:
1. Rewrite equation 2 in the following way,
d
mr̈ = −
dr
1 l2
V +
2 mr2
!
2. Multiply this by ṙ!
d
mr̈ṙ =
dt
1 2
d
1 l2
mṙ = −
V +
2
dt
2 mr2
!
or,
d
dt
1 2 1 l2
mṙ +
+V
2
2 mr2
!
= 0.
Now which makes you feel better, the algebraic tricks or the theorems? Next time
you may not be so lucky to figure out the right tricks to extract a first integral of
motion!
9.2.2
How to solve, a quick sketch.
These 2 first integrals of motion give us 2 of 4 integrals we need (r, θ, ṙ, θ̇) to completely
solve the problem. We outline the way a solution can be obtained.
s
ṙ =
2
l2
E−V −
,
m
2mr2
Z r
dr
dt = √
,
−−−−−
t=
√
r0
(3)
dr
.
−−−−−
This gives t(r) which can, at least formally, be inverted to obtain r(t). Once we have
this we can easily get θ(t) by substituting into,
Z t
θ=l
0
dt
+ θ0 .
mr2 (t)
See Goldstein p.76 for a nice comment on the transition to Quantum mechanics
here. Clearly we could simply throw this at a computer and numerically integrate
our solutions, but we haven’t yet learned all we can about the general properties of
the motion. These can be understood without needing to solve for the equations of
motion.
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V0
l2
2mr2
E1
0
E2
r
E3
E4
−k
r
Figure 2: Plot of the effective potential for the attractive inverse square force law
showing the turning points for each of 4 different energy levels.
9.3
Effective Potential Again
Once again we note that we look at the motion in just the radial dimension since θ
is cyclic. We can consider the motion under an effective force or potential given by,
f0 = f +
l2
,
mr3
V0 =V +
1 l2
.
2 mr2
Consider the plot of V 0 (r) for an attractive inverse square law of force, V = −k/r,
shown in figure 2. The solid line is the effective potential. For this potential there
is a centrifugal barrier. A particle with energy E1 will never get closer than the
intersection with V 0 shown. This particle is also on an unbound orbit, it will come in
from infinity and “strike” the barrier and then leave to infinity. The vertical distance
between the E level and V’ curve is obviously the kinetic energy which means that
at the intersection the velocity comes to zero. This is called a turning point. For E3
we have a bound orbit with 2 turning points r1 and r2 known as the pericenter and
apocenter (or apsidal distances). This doesn’t mean that the orbit is closed in the
general case (for inverse square force law this happens to be the case), just that it
lives between r1 and r2 . E4 is the minimum of the potential and it corresponds to
circular motion. We will see that E1 , E2 and E3 correspond to hyperbolic, parabolic
and elliptical orbits respectively for an attractive inverse square force law. We can
of course study other potentials like the linear restoring force which was presented
previously when looking at the circular pendulum for small oscillations.
9.4
Differential Equation for the Orbit
Normally we want to solve for the equations of motion, r(t) and θ(t) with integration
constants E, l, θ0 , r0 . However, what if we want an equation for the shape of the
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orbit r(θ)? We can simplify the problem by removing the time dependence.
d
l d
=
dt
mr2 dθ
ldt = mr2 dθ,
means we can simplify the equation of motion,
mr̈ −
l2
= f (r)
mr3
to,
1 d
r2 dθ
1 dr
mr2 dθ
−
l2
= f (r)
mr3
and by substituting u = 1/r, we get a fairly reasonable looking differential equation,
d2 u
m d
1
+u=− 2
V ( ).
2
dθ
l du u
(4)
This equation implies that the resulting orbit is symmetrical about two adjacent
turning points. Let the origin for θ be at a turning point. If u(θ) is a solution to the
above equation then so is u(−θ) and the initial conditions at θ = 0 will also not be
affected,
du
u0 = u(0),
=0
dθ 0
(the last is the definition of a turning point, it is an extremum). We can state this in
another way, the orbit is invariant under reflection about apsidal vectors. See figure 3
9.5
Conditions for Closed Orbits
It is possible to derive a powerful theorem about what types of attractive central
forces lead to closed orbits. One kind of closed orbit, the circle, has already been
mentioned. These circular orbits will form the basis for the theorem. For any given l
we have a circular orbit if V 0 (r) has a minimum or maximum at some distance r0 and
that the energy be E = V 0 (r0 ). The maximum is unstable, any slight perturbation
will send the orbit to zero or to infinity. The minimum is stable and in general it is
the sign of the 2nd derivative of the effective potential which decides stability,
∂ 2 V 0 ∂f 3l2
=
−
+
> 0,
∂r2 r
∂r r0 mr04
0
f (r0 ) = −
l2
mr03
∂f f (r0 )
< −3
,
∂r r0
r0
∂ ln f > −3.
∂ ln r r0
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symmetric
rmin
symmetric
rmax
Figure 3: The orbit is invariant under reflection about apsidal vectors.
(in the above f (r0 )/r0 is assumed to be negative). If the force behaves like a power
law of r in the vicinity of the circular radius r0 ,
f = −krn
, then the stability condition for the orbit becomes,
−knrn−1 < 3krn−1
or, n > −3. If we perturb the particle away from the circular radius by a small
amount it will execute simple harmonic motion in u ≡ 1/r about u0 ,
u = u0 + a cos βθ.
We can see this by Taylor expanding the force law to 1st order. This gives us a
defining equation for β which is given by,
r ∂f β =3+
.
f ∂r r0
2
As the radius vector sweeps around the plane, u goes through β cycles of its oscillation.
If β is an irreducible rational number given by p/q (p and q being integers), then after
q revolutions the orbit would begin to retrace itself again. This then mathematically
defines a closed small oscillatory orbit about the circular orbit r0 .
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Now each choice of l gives a different value of E and hence r0 , but the inequality
holds for any value of r0 for which a circular orbit is possible. This means it is not
enough for β to be rational, it must also be the same rational number at all
distances for which a circular orbit is possible (with given l). Otherwise β would
change discontinuously with r0 and orbits could not be closed since the number of
cycles would jump with small perturbations. The other way to look at this is that
with a small perturbation the orbit can also be thought of oscillating about another
very nearby r00 .
With β everywhere the same the definition of β actually becomes a differential
equation for f ,
∂ ln f
= β2 − 3
∂ ln r
over all r for which a stable circular orbit is possible. To solve this is easy,
f (r) = −
k
r3−β 2
.
All force laws of this form with β a rational number lead to closed stable orbits for
initial conditions that differ only slightly from circular. However, suppose the initial
conditions do deviate more than slightly from circular. This can be answered directly
by looking at the next term in the Taylor series expansion of the force law. J. Bertrand
(1873) did this and found for more than 1st order deviations from circularity that the
orbits are closed only for β = 1 and β = 2! The first of these is the inverse square
force law of Newton’s universal gravity and the second is Hooke’s law. The later of
these is ill suited to a universal law of gravity since it increases without bound as the
distance. The existence of closed orbits on planetary and stellar scales (binary stars
for example) leads to the conclusion that at least at these scales the only possible
form for the force law of gravity is to vary as 1/r2 !
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10
10.1
Kepler Problem: Solution of the Orbit
Kepler’s Laws 1 & 3
Now we turn to the solution for the shape of the orbit in an inverse square force law.
We have already shown that Kepler’s 2nd law actually applies to all central force laws
since it is just a restatement of conservation of angular momentum. We now want to
develop the remaining two of Kepler’s laws which are specific to the −k/r potential.
Recall from our sketch of the full solution that we can write,
s
ṙ =
l2
2
E−V −
.
m
2mr2
With the relation dt = (mr2 /l)dθ we can remove the time dependence as before and
write,
du
,
dθ = − q
2mk
2mE
2
+
u
−
u
2
2
l
l
where we have substituted u = 1/r, du = (−1/r2 )dr for clarity. This can be integrated
using the integral form,
Z
du
β − 2u
p
= cos−1 − p 2
2
α + βu − u
β + 4α
!
with α = 2mE/l2 and β = 2mk/l2 . If we put this all together we get,

θ = θ0 − cos
l2
−1  q
mk u
1+
−1
2El2
mk2

.
Finally solving for r = 1/u we get,

1
mk
= 2 1 +
r
l
s

2El2
1+
cos (θ − θ0 ) .
mk 2
Without loss of generality we can let θ0 = 0 to define the origin of the angular
coordinate to be at perihelion (or pericenter). In order to make the geometrical
nature of the above equation more obvious we will introduce,
l2
p=
mk
allowing us to write,
r=
s
e=
1+
p
.
1 + e cos(θ)
73
2El2
,
mk 2
(5)
(6)
a
a
b
g
F1 a
F2
ae
d
Figure 4: Some geometrical aspects of an ellipse.
This is the defining equation of a conic section in polar coordinates with focus at the
origin. Each of the well known conic sections are recovered with the following values.
2
e = 0 E = − mk
2l2
e<1
E<0
e=1
E=0
e>1
E>0
circle
bound
ellipse
bound
parabola
critical
hyperbola unbound
Here e is known as the eccentricity and p is often referred to as the parameter of the
ellipse. We have now shown that Kepler’s 1st law of planetary motion is a consequence
of Newton’s gravitational force.
To show the last of the three laws it is useful here (and also for later) to deduce
some geometrical relations for the ellipse (see figure 4). We note that distance d
corresponds to θ = 0, so that d = p/(1 + e). Likewise g corresponds to θ = π, so
that g = p/(1 − e). Adding these two quantities gives us twice the semimajor axis,
d + g = 2a, so that p = a(1 − e2 ). Now clearly c = a − d = ae from which it is easy to
√
solve for b, b2 = a2 − c2 = a2 (1 − e2 ) = ap, giving b = ap. We also note that from
equations 5 we can write,
k
a=−
,
2E
which shows that the semimajor axis of the ellipse depends only on the energy and
force constant. Recall equation 1 from which we showed Kepler’s second law,
dA
1
1 l
= r2 θ̇ =
.
dt
2
2m
Which when integrated over an entire orbit gives,
A=
1 l
T,
2m
74
where A is the area of the ellipse and T is the period of the orbit. The area of the
ellipse is easily shown to be πab since the ellipse is just a projection of the circle. This
means we can solve for the period of the orbit,
1 l
√
T = πa ap,
2m
2m
2
T = (2π)
k
a3 ,
which is Kepler’s 3rd law. Note that if the units in the problem are astronomical
units (1 AU = a for the Earth) and solar masses (M ) and years, then the term in
square brackets is exactly 1.
10.2
The Motion in Time
Although we have now derived all of Kepler’s laws we would still like to be able to
find the exact position of the planet at a given time. This task becomes somewhat
analytically cumbersome, but its solution leads us to a very famous equation known
appropriately as Kepler’s Equation. We can proceed by once again considering the
equation,
Z r
dr
r t=
r0
2
k
l2
m E + r − 2mr2
and substitute in r = p/(1 + e cos(θ)) and after some algebra obtain,
t=
l3
mk 2
Z θ
0
dθ
.
[1 + e cos(θ)]2
Inversion of this equation to obtain a solution for θ(t) (and hence r(t) from the conic
section equation) is not so easy and in general there is no closed form solution. To
get a flavour of the solution we should try an easy case, namely that with e = 1 which
is the parabolic orbit. Here we use the half-angle formula 1 + cos θ = 2 cos2 (θ/2) to
obtain,
Z θ
l3
θ
t=
sec4 dθ.
2
4mk 0
2
Making the substitution, x = tan(θ/2) we get,
l3
t=
2mk 2
Z tan θ
2
0
l3
θ 1
θ
(1 + x )dx =
tan + tan3
.
2
2mk
2 3
2
2
To solve for θ(t) we have to first solve a cubic (which we can do analytically) for
tan(θ/2) and then simply take 2 tan−1 of this to get θ(t).
To solve the general case it is much easier to take a slight step back to look at the
geometry of the problem again. There is a very useful angle substitution which allows
us to perform the integral in the general case. This angle is known as the eccentric
75
y0
P
b
ah
h
a
y
F1
E
ae
θ
x
x0
Figure 5: The relationship between the eccentric anomaly and the true anomaly.
These are related to each other since the ellipse is simply an orthogonal projection of
the circle as is show by the rescaling of any height, h, onto the ellipse.
anomaly and is shown in figure 5 along with the angle θ which is also known as the
true anomaly. An unfortunate collision of notation occurs here since the eccentric
anomaly is usually denoted by E which should not be confused with the energy!
The context (used as an angle) usually makes this clear. The position in Cartesian
coordinate measured from the focus is given by,
√
x = a(cos E − e)
y = b√
sin E = a 1 − e2 sin E e < 1
(7)
x = a(cosh E − e)
y = a e2 − 1 sinh E
e>1
√
1
2
x = 2 (p − E )
pE
e=1
y =
The two angles are related by,
 q
1+e
E


 q 1−e tan 2
θ
tan =
2 


e+1
e−1
tanh E2
E
p
e<1
e>1
e=1
Let us look at just the elliptical orbit case to show how this substitution makes life
easy. Since a and e are constants of the motion,
√
ẋ = −a sin E Ė
ẏ = a 1 − e2 cos E Ė
Use of the angular momentum equation (equation 1) yields,
l
= |r × ṙ| = xẏ − y ẋ.
m
76
since r = (x, y, 0) for coordinates in the plane of motion. Substituting the time
derivatives of x and y given above gives,
p
xẏ − y ẋ = a2 1 − e2 (1 − e cos E)Ė =
l
.
m
This is easy to integrate! We get as a result Kepler’s Equation,
nt = E − e sin E,
(8)
where the quantity n is called the mean anomaly and is given by,
s
n=
2π
k/m
=
.
3
a
T
The solution of this equation has seen a lot of attention over the last several
hundred years, with of the order of one hundred published methods to solve it since
Kepler posed the problem early in the seventeenth century. The practical need to solve
the transcendental Kepler equation to accuracies of a second of arc over the whole
range of eccentricity has fathered many of the developments in numerical mathematics
in the eighteenth and nineteenth centuries. In those days a “computer” was a person
gifted at performing calculations in her head! Although there are series expansions
in powers of e or in series of Bessel functions, the fastest modern methods are based
on iterative numerical solution. A simple iterative scheme starts with a trial value of
E0 and substitutes this into equation 8 as follows,
Ei+1 = nt + e sin Ei ,
with possible trial solutions E0 = 0, or E0 = M , or by experimenting with combinations like E0 = M +sign(M )0.85e. This is a very slowly converging method compared
to what is used in practice! To obtain a quadratically convergent method (the number
of decimal places doubles with each iteration) we can use Netwon’s method to find
the root of the equation,
f (E) = E − nt − e sin E = 0
as long as we start with an initial guess that is not too far from the root. The following
iteration converges on the root of the equation, and hence on the solution to Kepler’s
equation for given nt,
Ei+1 = Ei −
f (Ei )
nt + e sin Ei
=
0
f (Ei )
1 − e cos Ei
The iteration can be stopped when the new value of Ei+1 differs from the old value
Ei by a small enough amount.
77
10.2.1
The Laplace-Runge-Lenz Vector
The Kepler problem is also special because it conserves another vector besides the
angular momentum. For a general force Newton’s second law of motion can be written,
r
ṗ = f (r) .
r
The cross product of ṗ with constant L is then,
mf (r)
[r × (r × ṙ)]
r
mf (r)
[r(r · ṙ) − r2 ṙ].
r
ṗ × L =
=
(9)
(10)
This can be further simplified by using,
r · ṙ =
1d
(r · r) = rṙ,
2 dt
and noting that L is constant so that we can make the first term a total derivative,
d
(p × L) =
dt
mf (r)
[rrṙ − r2 ṙ]
r
rṙ
2 ṙ
= −mf (r)r
−
r r2
dr̂
= −mf (r)r2
dt
(11)
(12)
(13)
At this point we should specify the force law to be f (r) = −k/r2 so that the above
equation becomes,
d
dr̂
(p × L) = mk ,
dt
dt
or simply,
d
(p × L − mkr̂) = 0.
dt
This means that A = p × L − mkr̂ is a further conserved quantity! A is called the
Laplace-Runge-Lenz vector. From the definition of A we have,
A · L = 0,
meaning that the vector A is in the orbital plane. It is shown in figure 6. Note that
A = mker̂1 .
If θ is used to denote the angle between r and the fixed direction in the orbital
plane of A then ,
A · r = Ar cos θ = r · (p × L) − mkr.
Now permute the terms in the triple product,
r · (p × L) = L · (r × p) = l2 ,
78
A
mk
p×L
p
p
r2
p×L A
r3
r1
F1
mk
p×L
mk
A
p
Figure 6: The vector diagram for the Laplace-Runge-Lenz vector in the orbital plane.
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so that the above equation becomes,
Ar cos θ = l2 − mkr,
or,
1
mk
A
= 2 1+
cos θ .
r
l
mk
So we see that this further conserved quantity gives us a really easy way to arrive at
the orbit equation for the Kepler problem. One final note on the conserved quantities.
The number of conserved quantities implied by A, L and E seems to be 7. However,
we have 6 degrees of freedom in the problem, and an angular variable is not set by
these conservation laws, which means we need only 5 quantities to constrain the orbit.
This can only mean that there must exist two other relations among A, L and E.
One of these is A · L = 0 and the other is obtained when we express e in terms of E
and l in the value of A,
2
2
2
A = (mke) = (mk)
2El2
1+
mk 2
!
= (mk)2 + 2mEl2 .
This conserved quantity does not seem to come from an obvious symmetry such as
time or spherical symmetry. However, it turns out that the Kepler problem has
an O(4) symmetry since it can be reduced to a harmonic oscillator problem in 4
dimensions. This over-riding symmetry provides both conserved vectors L and A.
For details on how this reduction is done, see ”Interpreting the Kustaanheimo-Stiefel
transform in gravitational dynamics”, Prasenjit Saha, Sept. 2009 (Monthly Notices
of the Royal Astronomical Society).
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