Alain J. Brizard
Saint Michael's College
Motion in a Central-Force Field
1
Motion in the Center-of-Mass Frame
The Lagrangian for an isolated two-particle system is
L =
m1
m
j_r1j 2 + 2 j_r 2j2 ¡ U(r 1 ¡ r2);
2
2
where r1 and r2 represent the positions of the particles of mass m1 and m2 , respectively,
and U(r1 ; r 2) = U(jr1 ¡r 2j) is the potential energy for an isolated two-particle system (see
Figure below).
Let us now de¯ne the position R of the center-of-mass (CM)
R =
m1 r1 + m2 r2
;
m1 + m2
and de¯ne the inter-particle vector r = r1 ¡ r2 , so that the particle positions can be
expressed as
m
m
r1 = R + 2 r and r 2 = R ¡ 1 r;
M
M
where M = m1 + m2 is the total mass of the two-particle system. The Lagrangian of the
isolated two-particle system thus becomes
L =
M _ 2
¹ 2
_ ¡ U(r);
jRj + jrj
2
2
1
where
µ
m1 m2
1
1
¹ =
=
+
m1 + m2
m1
m2
denotes the reduced mass of the two-particle system.
¶ ¡1
For an isolated system, the CM canonical momentum
@L
_
= MR
_
@R
is a constant of the motion. The CM reference frame is de¯ned by the condition R = 0,
i.e., we move the origin of our coordinate system to the CM position (the Figure below
shows the case where m1 > m2 ).
P =
In this case, the Lagrangian for an isolated two-particle system in the CM reference
frame is
¹
L(r; r_ ) = j r_ j 2 ¡ U (r);
2
where
m2
m
r1 =
r and r2 = ¡ 1 r:
M
M
Hence, once the Euler-Lagrange equation for r
d
dt
Ã
@L
@ r_
!
=
@L
@r
!
¹ Är = ¡ rU(r)
is solved for r = r(t), the motion of m1 and m2 are determined through r1 (t) = (m2=M ) r(t)
and r2 (t) = ¡ (m1 =M) r(t).
2
Motion in a Central-Force Field
A particle moves under the in°uence of a central-force ¯eld F(r) = F (r) br if the force on
the particle is independent of the angular position of the particle about the center of force
2
and depends only on its distance r from the center of force. Here, the magnitude F (r)
(which is positive for a repulsive force and negative for an attractive force) is de¯ned in
terms of the central potential U (r) as F (r) = ¡ U 0 (r).
2.1
Lagrangian Formalism
The motion of two particles in an isolated system takes place on a two-dimensional plane.
When these particles move in a central-force ¯eld, the Lagrangian is simply
´
¹³ 2
L =
r_ + r 2 µ_2 ¡ U(r);
(1)
2
where polar coordinates (r; µ) are most conveniently used. Since the potential U is independent of µ, the canonical momentum
pµ =
@L
= ¹ r2 µ_ ´ `
@ µ_
(2)
is a constant of motion (here, labeled `). The Euler-Lagrange equation for r, therefore,
becomes the radial force equation
³
¹ Är ¡ r µ_2
´
= ¹ rÄ ¡
`2
= F (r):
¹ r3
(3)
In this description, the planar orbit is parametrized by time, i.e., once r(t) and µ(t) are
obtained, a path r(µ) onto the plane is de¯ned.
Since µ_ does not change sign on its path along the orbit, we may replace r_ and Är with
r0 (µ) and r 00 (µ) as follows. First,
µ ¶0
` r0
` 1
r_ = µ_ r0 =
=
¡
¹ r2
¹ r
:
Next, using Eq. (2)
2 ` r_
2 `2 r0
µÄ = ¡
=
¡
;
¹ r3
¹ 2 r5
we ¯nd
i
`2 h
rÄ = µÄ r0 + µ_ 2 r 00 = ¡ 2 5 2 (r0 )2 ¡ r r00 :
¹ r
Lastly, using the identity
µ ¶ 00
1
r
Ã
=
r0
¡ 2
r
!0
we ¯nd an expression for rÄ:
Är = ¡
=
i
1 h
0 2
00
2
(r
)
¡
r
r
;
r3
`2 µ 1 ¶00
;
¹ 2 r2 r
3
and the radial force equation (3) becomes
s00 + s = ¡
¹
`2 s2
F (1=s) = ¡
dU (s)
;
ds
(4)
where s(µ) = 1=r(µ) and U(s) = (¹=`2) U(1=s).
Note that the form of the potential can be calculated from the solution s(µ) = 1=r(µ)
as follows. For example, consider the particle trajectory described in terms of the solution
r(µ) = r0 sec(® µ), where r0 and ® are constants, then
³
´
s00 + s = ¡ ®2 ¡ 1 s = ¡
dU(s)
;
ds
and thus
´
´
1 ³ 2
`2 ³ 2
® ¡ 1 s2 ! U(r) =
®
¡
1
:
2
2¹ r2
Note also that the function µ(t) is determined from the relation
U (s) =
µ_ =
`
¹r2 (µ)
!
¹ Z µ 2
t(µ) =
r (µ) dµ:
` 0
Returning to our example, we ¯nd
¹r02 Z ®µ
¹r 20
t(µ) =
sec2 Á dÁ =
tan(®µ)
®` 0
®`
!
r(t) = r0
v
u
u
t1
Ã
+
®` t
¹r20
!2
and the total energy
®2 `2
E =
;
2¹r20
is determined from the initial conditions r(0) = r0 and r(0)
_
= 0.
2.2
Hamiltonian Formalism
The Hamiltonian for the central-force problem is
p2r
`2
H =
+
+ U(r);
2¹
2 ¹ r2
where pr = ¹ r_ is the radial canonical momentum. Since energy is also conserved, we solve
E =
¹ r_ 2
`2
¹ r_ 2
+
+
U(r)
=
+ V (r);
2
2 ¹ r2
2
as
s
r_ = §
2
[ E ¡ V (r) ];
¹
4
(5)
where V (r) is known as the e®ective potential and the sign § depends on initial conditions.
This equation can then be used with Eq. (2) to yield
dµ =
` dr
¡ ds
`
dt =
= q
;
2
2
¹r
¹ r r_
² ¡ 2 U(s) ¡ s2
where ² = 2¹ E=`2, or
s 0(µ) = §
q
² ¡ 2 U (s) ¡ s 2:
(6)
(7)
We readily check that this equation is a proper solution of the radial force equation (4)
since
s0 [dU=ds + s]
dU
s00 = q
= ¡
¡ s
ds
² ¡ 2 U(s) ¡ s2
is indeed identical to Eq. (4). Hence, for a given central-force potential U (r), we can solve
for r(µ) = 1=s(µ) by integrating
µ(s) = ¡
Z s
s0
q
d¾
;
(8)
² ¡ 2 U(¾) ¡ ¾2
where s0 de¯nes µ(s0) = 0, and performing the inversion µ(s) ! s(µ).
2.3
Turning Points
Eq. (7) yields the following energy equation
E =
i
¹ 2
`2
`2 h 0 2
2
r_ +
+
U(r)
=
(s
)
+
s
+
2
U(s)
;
2
2¹ r 2
2¹
where s0 = ¡ ¹ r=`
_ = ¡ pr =pµ . Turning points are those special values of rn (or sn ) (n =
1; 2; :::) for which
"
#
`2
`2
s 2n
E = U(rn ) +
=
U (sn ) +
;
2¹ r2n
¹
2
i.e., r_ (or s0 ) vanishes at these points. If two non-vanishing turning points r2 < r1 < 1
(or 0 < s 1 < s2) exist, the motion is said to be bounded in the interval r2 < r < r1 (or
s1 < s < s2), otherwise the motion is unbounded.
3
Kepler Problem
We now solve the Kepler problem where U(r) = ¡ k=r, where k is a constant, so that
U(s) = ¡ s 0s, where s 0 = ¹k=`2. The turning points for the Kepler problem are solutions
of the quadratic equation
s2 ¡ 2 s0 s ¡ ² = 0;
5
q
s20 + ²
which can be written as s = s 0 §
s1 = s 0 (1 ¡ e) and s2 = s 0 (1 + e);
where
e =
q
1 + ²=s20 =
q
1 + 2 E`2=¹k2:
We note that motion is bounded when E < 0 (0 < e < 1) and unbounded when E ¸ 0
(e > 1).
3.1
Bounded Keplerian Orbits
We will now look at the bounded case (e < 1). We de¯ne µ(s2 ) = 0, so that for the Kepler
problem, Eq. (8) becomes
µ(s) = ¡
Z s
d¾
q
s20 e2 ¡ (¾ ¡ s 0)2
s 0 (1+e)
;
(9)
which can easily be integrated by using the identity
µ ¶
¡ dx
x
p
= d arccos
;
2
2
a
a ¡x
so that Eq. (9) yields
µ
¶
s ¡ s0
µ(s) = arccos
:
s0 e
This equation can easily be inverted to yield
s(µ) = s 0 (1 + e cos µ):
(10)
We can readily check that this solution also satis¯es the radial force equation (4).
3.1.1
Kepler's First Law
The solution for r(µ) is now trivially obtained from s(µ) as
r(µ) =
r0
;
1 + e cos µ
(11)
where r0 = 1=s0 denotes the position of the minimum of the e®ective potential
Ã
1
`2
V 0 (r0) = 2 k ¡
r0
¹ r0
6
!
= 0:
Eq. (11) generates an ellipse of major radius
a =
r0
k
=
2
1 ¡e
2 jEj
and minor radius
p
b = a 1 ¡ e2 =
v
u
u
t
`2
2¹ jEj
and, therefore, yields Kepler's First Law.
3.1.2
Kepler's Second Law
Using Eq. (2), we ¯nd
¹ 2
2¹
r dµ =
dA(µ);
`
`
R
where dA(µ) = r dr dµ = 12 [r(µ)] 2 dµ denotes an in¯nitesimal area swept by dµ. When
integrated, this relation yields Kepler's Second law
dt =
¢t =
2¹
¢A;
`
i.e., equal areas are swept in equal times since ¹ and ` are constants.
3.1.3
Kepler's Third Law
The orbital period T of a bound system is de¯ned as
T =
Z 2¼
0
dµ
¹ Z 2¼ 2
2¹
2¼ ¹
=
r dµ =
A =
ab
` 0
`
`
µ_
7
(12)
where A = ¼ ab denotes the area of an ellipse with major radius a and minor radius b.
Using the expressions for a and b found above, we ¯nd
v
u
v
u
u `2
u ¹ k2
k
2¼ ¹
¢
¢ t
= 2¼ t
:
T =
`
2jEj
2¹ jEj
(2 jEj)3
If we now substitute the expression for a = k=2jEj and square both sides of this equation,
we obtain Kepler's Third Law
(2¼)2¹ 3
T2 =
a :
(13)
k
Note that in Newtonian gravitational theory, k=¹ = G (m1 + m2 ); although Kepler's Third
Law states that T 2=a3 is a constant for all planets in the solar system, we ¯nd that this is
only an approximation that holds for m1 À m2.
3.2
Unbounded Keplerian Orbits
We now look at the case where the total energy is positive or zero (i.e., e ¸ 1). Eq. (11)
yields r (1 + e cos µ) = r 0 or
Ã
p
er
e2 ¡ 1 x ¡ p 2 0
e ¡1
!2
¡ y2 =
r20
:
e2 ¡ 1
For e = 1, the particle orbit is a parabola x = (r02 ¡ y2 )=2r0 , with distance of closest
approach at x(0) = r0=2, while for e > 1, the particle orbit is a hyperbola.
3.3
Laplace-Runge-Lenz Vector
Let us now investigate an additional constant of the motion for the Kepler problem. First,
we consider the time derivative of the vector p £ L, where the linear momentum p and
angular momentum L are
³
p = ¹ r_ rb + rµ_ µb
´
and L = ` zb = ¹r2 µ_ zb:
The time derivative of the linear momentum is p_ = ¡ rU(r) = ¡ U 0 (r) rb while the angular
momentum L = r £ p is itself a constant of the motion so that
d
dp
(p £ L) =
£ L = ¡ ¹_r ¢ rU r + ¹r ¢ rU r_
dt
dt
d
= ¡ (¹ U r) + ¹ (r ¢ rU + U) r_ ;
dt
8
hence the vector A = p £ L + ¹ U(r) r is a constant of the motion if the potential U(r)
satis¯es the condition r ¢ rU(r) + U(r) = 0. For the Kepler problem, with central potential
U(r) = ¡ k=r, the Laplace-Runge-Lenz (LRL) vector
Ã
A = p £ L ¡ k¹ rb =
!
`2
¡ k¹ rb ¡ ` ¹r_ µb
r
is, therefore, a constant of the motion since r ¢ rU = ¡ U.
Since the vector A is constant in both magnitude and direction, we choose its direction
to be along the x-axis and its amplitude is determined at the distance of closest approach
rmin = r0 =(1 + e) and we can easily show that A ¢ rb = A cos µ leads to the Kepler solution
r0
r(µ) =
;
1 + e cos µ
where r0 = `2=k¹ and e = A=k¹.
Note that if the Keplerian orbital motion is perturbed by the introduction of an additional potential term ±U(r), we can show that the LRL vector is no longer conserved
(i.e., dA=dt 6= 0) and that the direction of the Keplerian elliptical orbit precesses with a
precession frequency
A
dA
!p (µ) = zb ¢ 2 £
;
A
dt
where the unperturbed Kepler solution r(µ) is to be used.
4
Isotropic Simple Harmonic Oscillator
We now investigate the case when the central potential is of the form
k 2
¹k
U (r) =
r
! U (s) =
:
2
2`2 s2
The turning points for this problem are expressed as
r1 = r0
µ
(14)
µ
1 ¡ e ¶4
1
1 + e ¶4
1
=
and r2 = r0
=
;
1+e
s2
1¡e
s1
1
1
where r0 = (`2=¹k)1=4 = 1=s 0 is the radial position at which the e®ective potential has a
minimum V0 = k r02 and
v
e =
u
u
t
Ã
1 ¡
kr02
E
!2
:
Here, we see that orbits exist and are always bounded for E > V0 (and thus 0
Next, using the change of coordinate q = s2 in Eq. (8), we obtain
¡1 Z q
dq
q
µ =
;
2 q2
(¹=`2 ) [2E q ¡ k] ¡ q2
9
e
1).
(15)
where q2 = (1 + e) ¹E=`2. We now substitute q(') = (1 + e cos ') ¹E=`2 in Eq. (15) to
obtain
" Ã
!#
1
1 `2 q
arccos
¡1
µ =
2
e ¹E
or
p
r0 1 ¡ e2
r(µ) = p
:
(16)
1 + e cos 2µ
This equation describes the ellipse
x2
y2
+
= r02
(1 ¡ e)
(1 + e)
p
p
of semi-major axis a = r0 1 + e and semi-minor axis b = r0 1 ¡ e. Lastly, we note that
one revolution along the orbit r(µ) corresponds to an angular period of ¼, i.e., r(µ + ¼) =
r(µ), and not 2¼ as found in the Kepler problem. The area of the ellipse A = ¼ ab =
¼ (`2=¹E) while the period is
T (E; `) =
Z ¼
0
dµ
¹A
¼`
_µ = ` = E :
If we introduce the angular frequency ! = ¼=T , then we ¯nd the important relation between
energy and angular momentum E = ` !.
10