Physics 105 (Fall 2013) Notes:The Two-Body Conservative Central Force Problem

A.E. Charman∗

Department of Physics, University of California, Berkeley

(Dated: 10/25/13; revised 12/12/13; revised 12/18/13)

Here we analyze the two body central force problem, or two-body problem for short, describingthe dynamics of two point particles subject to conservative, central (and reciprocal) forces. As animportant example, Kepler’s Laws are derived.

I. OVERVIEW

The Two-Body Conservative Central Force Problem (usually just abbreviated as the “Two-Body Prob-lem”) is one of the most important in classical mechanics. It concerns the motion of two point particlessubject to internal forces that are conservative (derivable from a time-independent potential energy),central (acting in a direction along the line joining the particles), and hence reciprocal (satisfying theThird Law). We refer to the two-body problem and not the two-particle problem because the resultingframework applies somewhat more generally, for example to the gravitational interactions between twouniform spherical masses.

The two-body problem undergirds the derivation of Kepler’s Laws directly from the Laws of Motionand of Universal Gravitation, one of the early triumph’s of Newtonian mechanics. It is one of the fewknown integrable systems in three spatial dimensions, illustrating the connections between symmetries,conservations laws, and reduction of the degrees of freedom for which we must simultaneously solve, andalso forms the starting point for a variety of perturbation and other approximation techniques used totackle more complicated and realistic problems in celestial mechanics, orbital dynamics, and atomic andmolecular physics. As such, this model, like the harmonic oscillator, is one of the primary “workhorses”of classical mechanics, and is a true cornerstone of physics and high water mark in intellectual history.

The general strategy for solving the two-body problem is illuminating, as it takes advantage of someof the most important ideas in classical mechanics: the decomposition of the dynamics into COM andrelative motion, use of symmetries, and mechanical analogic reasoning, whereby we reduce problems wedo not at first know how to solve to problems we already do know how to solve.

R1

R2m1

m2

Rcom

r = (R1 – R2) !F12

origin (in inertial frame)

FIG. 1 The two-body central force problem. The forces on each particle are equal in magnitude but opposite indirection, parallel or antiparallel to the relative displacement r = R1 −R2 between the particles. The relativepositions (npot shown) R′1 and R′2 with respect to the center of mass (COM) are also oppositely directed alongthis line.

∗Electronic address: acharman@physics.berkeley.edu

2

To see where we are headed, we outline our strategy and tactics:

• decompose the dynamics into motion of the center of mass and motion about the center of mass;

• use momentum conservation to deduce the COM motion;

• change variables to map dynamics of the two particles in the COM frame to the analogous dynamicsof one particle of reduced mass;

• use angular momentum conservation to restrict relative motion to one plane, and to decoupleangular and radial motion;

• map the radial motion to the analogous dynamics of a single particle moving in one-dimensionaleffective potential

• use energy conservation to solve for the radial dynamics;

• use angular momentum conservation to solve for the angular dynamics;

• map back to the original variables if needed.

Also, we will look at the important special case of attractive inverse-square force laws like the gravita-tional or Coloumb forces, and derive Kepler’s Law’s from Newton’s Laws.

II. DYNAMICAL SETTING

In some chosen inertial frame, suppose a point particle of mass m1, positionR1, and velocity R1 interactswith a second point particle of mass m2, position R2, and velocity R2, according to a conservative,central pair-interaction potential U(|R1 − R2|) which is a function only of the distance between theparticles (and of course possibly invariant parameters such as their masses or charges). This means theinstantaneous force F12 on particle 1 due to particle 2 (equal and opposite to the force F21 on particle2 due to particle 1) is

F12 = F12(R1,R2) = − ∂∂R1

U ′(|R1 −R2|) = − (R1−R2)|R1−R2| U

′(|R1 −R2|) = −F21, (1)

where U ′(R) = − ddRU(R) is the negative derivative of the potential with respect to distance between

the particles.

The corrresponding Lagrangian governing the dynamics is simply

L(R1,R2, R1, R2) = T (R1, R2)− U(|R1 −R2|), (2)

where the kinetic energy is given by the usual non-relativistic expression:

T (R1, R2) = 12m1|R1|2 + 1

2m1|R1|2, (3)

and we are implicitly assuming that any other forces on these particles, internal or external, are negligible.

III. CHANGE OF VARIABLES

We seek to change variables from those specifying the individual positions of the two particles to a pairof collective variables specifying, respectively, the position of the center of mass (COM) and the relativedisplacement of the particles. But before transforming kinematic variables, we first transform someparameters for subsequent convenience:

3

A. Total Mass and Reduced Mass

From the individual particle masses m1 and m2, we can define the total mass

M = m1 +m2, (4)

which is twice the arithmetic mean of the two particle masses, and also the reduced mass:

µ =( 1

m1+

1

m2

)−1

=m1m2

m1 +m2=m1m2

M, (5)

defined as half the harmonic mean of the two particles masses — the same rule as you use to addresistance or inductance in parallel, or to add capacitance in series, or to combine spring constants forideal springs connected in series, etc. Note that m1m2 = µM , implying the geometric mean of theindividual particle masses m1 and m2 equals the geometric mean of µ and M .

Because m1 ≥ 0, m2 ≥ 0, and (m1 ±m2)2 ≥ 0, it is easy to verify that

0 ≤ µ ≤ 12

√m1m2 ≤ 1

4M, (6)

where µ = 0 if and only if both m1 = 0 and m2 = 0, and µ = 14M if and only if m1 = m2 = 1

2M .

We see that in the limit as m1 → 0, we have M → m2 and µ→ 0, while as m1 →∞, we have M →∞and µ→ m2. Similarly, as m2 → 0, we see that M → m1 and µ→ 0, while as m2 →∞, it can be shownthat M →∞ and µ→ m1. In particular, in the limit as one of the particles possesses much more massthan the other, the reduced mass µ becomes numerically close to the mass of the lighter particle, whilethe total mass M will be dominated by the mass of the heavier particle.

Because M and µ are both symmetric functions of m1 and m2, they cannot be fully inverted to find m1

and m2 separately. The best we can do is:

max[m1,m2] = M[12 +

√14 −

µM

](7a)

min[m1,m2] = M[12 −

√14 −

µM

], (7b)

determining the two individual particle masses but not which is associated with which particle.

B. Center of Mass Position and Relative Displacement

We can transform in an invertible and coordinate-independent manner from a pair of individual particlepositions to the center-of-mass position and the relative displacement. We define the center of mass(COM) position as usual:

Rcom = m1R1+m2R2

m1+m2= m1

M R1 + m2

M R2, (8)

and then the relative displacement between particles:

r = R1 −R2 = R′1 −R′2, (9)

where the relative positions of each particle with respect to the COM are given by

R′1 = R1 −Rcom =(1− m1

M

)R1 − m2

M R2 = m2

M (R1 −R2) = +m2

M r (10a)

R′2 = R2 −Rcom = −m1

M R1 +(1− m2

M

)R2 = −m1

M (R1 −R2) = −m1

M r, (10b)

so each particle’s displacement relative to the COM is proportional to the full inter-particle displacementr but also to the other particle’s relative mass fraction.

4

Equivalently, these relative positions R′1 and R′2 can be interpreted as the particle positions as measuredin the so-called barycentric or center-of-mass (COM) frame, i.e., a frame of reference co-moving with theCOM, whose origin coincides with the COM, but with unrotated axes remaining parallel to the originalinertial frame’s axes. This will be an inertial frame, because in the absence of external forces, we knowthat the COM must move at constant velocity.

Note that the direction of the displacements of the particles are in opposite directions in this frame, asmust be the case since the COM lies at the origin in the COM frame. As a quick check on our algebra,notice that

R′com =m1R

′1+m2R

′2

M = m1m2

M2 r − m2m1

M2 r = µM (r − r) = 0, (11)

while

R′1 −R′2 = m2

M r − (−m1

M r) = m2+m1

M r = r, (12)

as expected.

C. Velocities

The corresponding velocities are

vcom = Rcom = m1R1+m2R2

m1+m2= m1

M R1 + m2

M R2, (13)

for the COM, and

r = R1 − R2 = R′1 − R′2, (14)

for the relative motion, where

R′1 = R1 − Rcom = +m2

M r (15a)

R′2 = R2 − Rcom = −m1

M r (15b)

are the velocities of the particles relative to the COM, or equivalently, the velocities in the COM frame.

Notice that each particle’s motion in the COM frame is proportional to the relative velocity but also tothe other particle’s mass fraction — that is, the particles move in opposite directions in the COM frame,but the heavier particle moves more slowly in this frame, in order to maintain zero total momentum.

D. Linear Momenta

The total linear momentum is given as usual by

Ptot = Mvcom = MRcom = m1R1 +m2R2 = p1 + p2, (16)

where

p1 = m1R1 (17a)

p2 = m1R2 (17b)

are the usual linear momenta of each particle as measured in the original inertial frame. The corre-sponding momenta measured in the COM frame are:

p′1 = m1R′1 = m1(+m2

M r) = +m1m2

M r = +µ r (18a)

p′2 = m2R′2 = m2(−m1

M r) = −m2m1

M r = −µ r, (18b)

and we may define (without any prime, by convention) a momentum p = µ r = +p′1 = −p′2. Asexpected, the particles have opposite momenta in the COM frame, so that P ′tot = p′1 + p′2 = 0. Noticethat p = µr is equal to the linear momentum that a single point particle of mass µ would have if it weremoving in the COM frame with velocity r.

5

E. Kinetic and Potential Energies

The potential energy (which assumes the same value in either the original inertial frame or COM frame)can obviously be written as

U = U(r), (19)

where

r = |r| = |R1 −R2| = |R′1 −R′2| (20)

is the instantaneous distance between the particles, as measured in either (in fact any) frame.

The kinetic energy can be decomposed as

T = Tcom + T ′, (21)

where

Tcom = 12M |Rcom|2 = |Ptot|2

2M (22)

is the COM kinetic energy, and

T ′ = 12m1|R′1|2 + 1

2m2|R′2|2 = 12m1|m2

M r|2 + 12m2|−m1

M r|2

=m1m

22+m2m

21

M2 |r|2 = 12m1m2(m1+m2)

M2 |r|2 = 12µ |r|

2 = |p|22µ

(23)

is the kinetic energy relative to the COM, or equivalently the kinetic energy as measured in the COMframe. Notice that T ′ = 1

2µ|r|2 is equal to the kinetic energy that a single point particle of mass µ would

have if it were moving with moving with velocity r.

F. Angular Momenta

Similarly, we can decompose the total angular momentum about the origin as

Ltot = Lcom +L′, (24)

where

Lcom = Rcom × Ptot = Rcom ×MRcom (25)

is the angular momentum of the COM, and

L′ = R′1 ×m1R′1 +R′1 ×m2R

′2 = (m2

M r)×m1(m2

M r) + (−m1

M r)×m2(−m1

M r)

=m1m

22+m

21m2

M2 r × r = m1m2(m2+m1)M2 r × r = µ r × r = r × µr = r × p

(26)

is the angular momentum relative to the COM. Equivalently, L′ can be interpreted as the full angularmomentum as measured in the COM frame, with respect to the origin in that frame. Notice thatL′ = r × µr is equal to the angular momentum about that origin that a single point particle of mass µwould possess, were it to be located at r and moving with velocity r.

6

G. Transformed Lagrangian

Hence, the full Lagrangian can be written as:

L = L(Rcom, r, Rcom, r) = T − U = Tcom + T ′ − U = 12M |Rcom|2 + 1

2µ |r|2 − U(r) (27)

in terms of the COM and relative variables.

The canonical momentum conjugate to Rcom is just ∂L∂R com

= MRcom = Ptot, while the canonical momen-

tum conjugate to r is ∂L∂r = µ r = p.

The full Lagrangian can be decomposed as

L = Lcom + L′, (28)

where

Lcom = Lcom(Rcom) = 12M |Rcom|2 (29)

governs the COM motion, analgous to the free motion of a single fictitious particle of mass M locatedat Rcom; and

L′ = L′(r, r) = 12µ |r|

2 − U(r) (30)

governs the COM-frame relative motion, which we see is analogous to the motion of a single fictitiouspoint particle of mass µ and position r moving in a conservative, central potential U(r).

Note that in the context of the two-body problem, L′(r, r) can be interpreted as describing the COM-frame motion of a fictitious single particle. But the same Lagrangian will of course apply to the dynamicsof an actual single particle if it is subject to a conservative, central (radially-directed, radially symmetricin magnitude) force field in some inertial frame. The two problems are actually exactly equivalent inthe limit as m2 → ∞, such that particle 2 will just be stuck at what we can take to be the origin, inwhich case we we have µ→ m1, vcom → 0, and r → R′1 = R1.

IV. EULER-LAGRANGE EQUATIONS OF MOTION AND THEIR SOLUTION

We see that in (27) we have completely decoupled the COM and relative variables in terms of separateadditive contributions to the Lagrangian. We can solve separately for the respective COM and relativetrajectories, and moreover, both the COM and relative dynamics are separately analogous to those of asingle fictitious point particle, of mass M and µ, respectively.

That is, rather than having to solve directly for the coupled motion of two physical particles, we cansolve for the uncoupled effective motion of two fictitious point particles, one a free particle of mass M ,and the other a particle of mass µ subject to a spherically-symmetric force

F (r) = r F (r) = −rU ′(r), (31)

where we define F (r) = −U ′(r) = − ddrU(r) as the radial component of the force (which is in fact the

only non-vanishing component).

A. Choice of Coordinates for Rcom and r

So far we have not actually adopted any specific choice of coordinate systems with which to representRcom and r. A suitable choice will simplify the subsequent mathematics significantly. Because Rcom will

7

just follow a rectilinear trajectory at fixed velocity, the choice of coordinates does not matter too much,but Cartesian coordinates Xcom = (Xcom, Xcom, Zcom)T (with respect to some fixed orthogonal axes in theoriginal inertial frame) are perhaps the most convenient.

Our main focus will be on the relative motion in the COM frame, which in fact must be confined toa single plane because of angular momentum conservation. Even though the relative Lagrangian L′ isspherically symmetric, it therefore makes sense to actually choose cylindrical coordinates for r, with thez-axis taken to be perpendicular to the plane in which the relative motion is confined.

That is, because both the magnitude and direction of L′ = r× µr are conserved for all time, as long asL′ 6= 0, we can choose a COM-frame cylindrical coordinate system such that

z =r × r|r × r|

. (32)

Otherwise, if L′ = 0, meaning r × r = 0 and hence r and r are parallel (or else one or both vanish),then any acceleration r arising from U(r) will also be parallel to r. so we can choose as z any directionsuch that z · r = z · r = 0 initially, and these orthogonality conditions will continue to hold for allsubsequent time. In either case, the particle can only move transversely to z, and therefore if we choosez = 0 initially, the subsequent motion must remain confined to the z = 0 plane for all time.

In fact, since r ·z ∝ r ·(r × r) = 0, we naturally choose the origin of our coordinate system at r = 0, sothe local transverse radial direction can be taken as

r⊥ = r =r

|r|, (33)

and finally, the local azimuthal direction is then chosen to be

φ = z × r =|r|2r − (r ·r)r

|r| |r × r|. (34)

With these conventions, the transverse radial coordinate is just r⊥ = r = |r|, while the vertical coordinatealways vanishes: z = 0, since, once again, angular momentum conservation guarantees that the motionremains confined for all time to the z = 0 plane, i.e., the plane containing the initial displacement r(0)and initial relative velocity r(0).

For definiteness, we could then choose a particular x direction, say x = r(0) if it is non-zero, and thenmeasure the azimuthal coordinate φ counterclockwise from this axis. Alternatively, we can choose φ = 0to coincide with some privileged point on the r(t) trajectory — for example, a radial turning point.

The relative velocity, or equivalently COM-frame velocity of the fictitious particle,

r = r r + rφ φ+ z z = r r + rφ φ (35)

is in fact at most two-dimensional, because z = 0.

In these coordinates, the vertical component ` of relative angular momentum,

` = L′z = z ·L′ = z · [r × µr] = z · |r × µr| z = |r × µr| = |L′| (36)

is equal to the full magnitude of the COM-frame angular momentum. Of course, this angular momentumcan also be written explicitly as

` = z · [r × µr] = µ[z × r] · r = µ[z × (rr + zz)] · [rr + rφφ+ zz]

= µ[z × rr] · [rr + rφφ] = 0 + µr φ · [rφ φ] = µr2φ(37)

in terms of the radial distance r and angular velocity φ of the fictitious particle.

8

B. Transformed Lagrangian, Conservation Laws, and Equations of Motion

In these chosen coordinates, the Lagrangian can be re-written as

L = Lcom + L′ = 12M |Xcom|2 + 1

2µ[r2 + r2φ2 + z2

]− U(r) (38)

Physically, there are two point particles of mass m1 and m2 and positions R1 and R2, respectively. Dy-namically, we therefore should have a total of 3+3 = 6 position coordinates and 3+3 = 6 components ofvelocity. After transforming, we do in fact have 3 COM position coordinates and 3 relative displacementcoordinates, and well as 3 components of COM velocity and 3 components of relative velocity. Butmathematically, we can now think of the system as describing two fictitious uncoupled particles, one ofmass M at a position described by Xcom = (Xcom, Ycom, Zcom)T, and one of mass µ at r at a positiondescribed by r, φ, and z = 0.

Note this Lagrangian has several symmetries, and hence associated conservation laws: translationalinvariance, rotational invariance, and boost invariance (up to a gauge term) with respect to the COM,rotational invariance with respect to the relative displacement, and explicit time translation invariance.

In particular, the position Xcom is cyclic, so

ddt∂Lcom

∂Xcom= ∂Lcom

Xcom= 0⇒MXcom = 0. (39)

which can be immediately integrated once to yield

Xcom(t) = Xcom(0) (40)

and again to yield:

Xcom(t) = Xcom(0) + Xcom(0) t. (41)

So we turn to the less-trivial relative motion, as governed by L′, analogous to the motion of a singlefictitious particle of mass µ located at r.

Note that the z coordinate is cyclic, so

ddt∂L′∂z = ∂L′

∂z = 0⇒ µz = 0, (42)

and hence

z(t) = z(0) = 0, (43)

and

z(t) = z(0) + z(0) t = 0 + 0 = 0, (44)

confirming what we had already concluded based on angular momentum conservation, namely that themotion in the COM frame is confined to a single plane for all time. (Analogously, the motion of thefictitious particle is confined to the z = 0 plane for all time).

The φ coordinate is also cyclic, so

ddt∂L′∂φ

= ∂L′∂φ = 0⇒ d

dt [µr2φ] = d

dt` = 0, (45)

confirming that the angular momentum about the COM is conserved. (Analogously, the vertical com-ponent of angular momentum about the origin of the fictitious particle is conserved).

In addition, time-translation invariance guarantees that

H = 12M |Xcom|2 + 1

2µ[r2 + r2φ2 + z2

]+ U(r) (46)

9

is conserved, but we already know that z = 0 is separately conserved, and also Xcom and hence Tcom =12M |Xcom|2 are separately conserved, so in fact we can conclude that

E = 12µ[r2 + r2φ2

]+ U(r) (47)

is conserved, which can be interpreted as the total COM-frame energy, or equivalently as the total energyof the fictitious particle of mass µ.

Using the conservation of ` and E , we can now work out the dynamics for the radial coordinate r(t).The Euler-Lagrange equation for r is

ddt∂L′∂r −

∂L′∂r = 0⇒ µr − µr2φ2 = − d

drU(r), (48)

where as usual the transverse radial acceleration involves two terms, one reflecting changes in radialspeed and one reflecting centripetal effects. Using the conservation of angular momentum ` = µr2φ, thiscan be rewritten as

µr = µr2φ2 − dU(r)dr = `2

µr3 −dU(r)dr = − d

dr

[`2

2µr2 + U(r)]

(49)

which we recognize as being analogous to the motion of a single point particle in one dimension, subjectto the effective potential

Veff(r) = U(r) + `2

2µr2 (50)

which includes the original potential energy U(r) plus a so-called centrifugal barrier term `2

2µr2 , which

acts like a repulsive contribution to the effective radial potential energy. Why? If ` = µr2φ is to remainconstant, as r decreases, φ must increase, and since total energy in the COM frame is conserved, thisangular kinetic energy must grow at the expense of either true potential energy or radial kinetic energy(as there remains no vertical motion, and hence nowhere else from which to draw the energy).

The effective dynamics of this fictitious one-dimensional particle are therefore governed by

µr = ddtT′ = − d

drVeff(r), (51)

which has as an obvious constant of the motion

Heff = 12µr

2 + Veff(r) = 12µr

2 + `2

2µr2 + U(r) = 12µr

2 + 12µr

2φ2 + U(r) = E , (52)

equal to the actual COM-frame energy of the original particles.

1. Formal solution for the Radial Motion as a Function of Time

In the equivalent one-dimensional problem, note that r is acting like the one-dimensional position of thefictitious particle, not the distance from the origin. So in order to avoid possible jumps in angle φ by πif r reaches r = 0, it is convenient to allow for the possibility that r may take on positive or negativevalues, while being careful to now write

Veff(r) = U(|r|) + `2

2µr2 (53)

to ensure the correct physical behavior of the potential energy. However, as we will see momentarily,usually r cannot ever reach r = 0.

Consider the evolution of r(t) for given values of the constants of motion E and `. Because of theirconservation, the radial coordinate r must be confined to a classically allowed region, consisting of acontinuous, connected interval bracketing r(0) and for which Veff(r) ≤ E . This interval may be finite,half-infinite, or doubly-infinite, or even reduce to a single point in the case of a circular orbit. Generically,solutions to E = Veff(r) will lead to radial turning points r± between the radial motion will be confined.

10

Actually, we can always assume −∞ ≤ r− ≤ r(t) ≤ r+ ≤ +∞, where we agree to set r− = −∞ ifthere is no finite turning point for r ≤ r(0), and set r+ = +∞ if there is no finite turning point for anyr ≥ r(0). The motion will be bounded if |r±| <∞ strictly, and unbounded otherwise.

Typically — specifically, unless ` = 0 exactly or else U(r) has a singularity as r → 0 that cancels that

in the centrifugal barrier term `2

2µr2 — we will have Veff(r) → +∞ as r → 0, leading to an inner radial

turning point at some r− ≥ 0. Smaller values of |r| are not possible, because in order to conserveangular momentum the system would require more rotational kinetic energy than the total energyavailable. In exceptional cases without a sufficient centrifugal barrier, there will be a regimes for whichr− = −r+ = +|r+| for certain values of E , but there may be other classically allowed radial ranges aswell.

Given the radial turning points r± defining the boundaries of the relevant classically allowed region, wecan calculate the time intervals from the most recently encountered turning point via an integral of theform:

t− t− = +

r(t)∫r−

dr√2µ(E − Veff(r))

(54)

or

t− t+ = −r(t)∫r+

dr√2µ(E − Veff(r))

= +

r+∫r(t)

dr√2µ(E − Veff(r))

(55)

where t− is the most recent time for which r(t−) = r−, and t+ is the most recent time for whichr(t+) = r+. In principle, these relations can then be inverted over each branch of the motion to obtainthe radial trajectory r(t).

2. Circular Orbits

If r− = r+ = r0, then the classically allowed region shrinks to a single point, and the trajectorycorresponds to a circular orbit r(t) = r0, requiring r = 0 and r = 0. But r = 0 is equivalent toV ′eff(r0) = d

dtVeff(r)|r=r0 = 0, meaning r0 is a local extrema (or at least critical point) , while r = 0implies that energy is chosen just right, so that E = Veff(r0). The circular orbit r = r0 is stable withrespect to small radial perturbations if V ′′eff(r0) > 0 and unstable if V ′′eff(r0) < 0. Note however, thatthis only tests for stability with respect to small displacements in radial position or radial velocity.Other perturbations could change the functional form of Veff(r) or shift the plane in which the motionis confined, and cannot be assessed by examining the effective radial potential alone.

3. Formal solution for the Angular Motion

Once r(t) is determined, the angular motion can in principle be determined using angular momentumconservation:

φ(t)− φ(0) =

t∫0

dt′ φ(t′) =

t∫0

dt′ `µ r(t′)2 = `

µ

t∫0

dt′ [r(t′)]−2, (56)

so we formally have “reduced to quadratures” the problem of finding both r(t)and φ(t).

11

4. Kepler’s Second Law

Note that, regardless of the functional form of U(r), the angular motion satisfies Kepler’s Second Law,which says the displacement vector r(t) sweeps out equal areas in equal times. indeed, Kepler’s SecondLaw is just a restatement of the conservation of angular momentum in this context. Recall that themagnitude of the vector a×b can be interpreted as the area of the parallelogram whose sides are specifiedby the three-dimensional vectors a and b, while the vector itself points in a direction normal to thisparallelogram, which is the usual convention for ascribing a direction to a small area. So, neglectingO(dt2) terms, the vector area dA swept out by r(t) between t and t+ dt is just one-half the area of theparallelogram defined by r(t) and r(t+ dt):

dA = 12r(t)× r(t+ dt) = r(t)× [r(t) + dt r(t)] = 1

2r(t)× r(t) + 12dt r(t)× r(t) = 1

2dt r(t)× r(t), (57)

so

dAdt = 1

2 r × r = 12µL

′, (58)

which is a constant vector.

Perhaps a bit more simply, in cylindrical coordinates, we see that

` = µr2 dφdt , (59)

but 12r rdφ = 1

2r2φ dt can be interpreted as the scalar area dA of an infinitesimal triangle of “height” r

and “base” r dφ = rφ dt which is swept out by r(t) between t and t+ dt, so that

` = 2µdAdt . (60)

Again, note that this holds for any central force, not just for gravitational attraction.

5. Caveat

Notice that the equation of motion (51) could be derived from an effective Lagrangian

Leff(r, r) = 12µr

2 − Veff(r) = 12µr

2 − `2

2µr2 − U(|r|). (61)

However, this is not equal to the Lagrangian L′ —in fact the centrifugal term enters with opposite sign.This is because what we have reinterpreted as a contribution to an effective radial potential energy whatstarted out as an angular contribution to the kinetic energy, and kinetic and potential energies appearin a Lagrangian with opposite sign.

We can also think in terms of the nature of the constraint µr2φ − ` = 0. This is a non-holonomicconstraint, so we cannot simply substitute it into the Lagrangian L′ before taking variational derivatives.

When we study Hamiltonian methods, this subtlety regarding the sign of the centrifugal term in theradial dynamics can largely be avoided.

6. Binet Method and Shape of Orbits

If ` = 0, the the motion is confined to a single line through the origin. so φ = 0 and φ(t) = φ0 does not

advance, Otherwise, if ` 6= 0, then φ(t) 6= 0, and we can use the azimuthal angle φ as an independentvariable instead of the time t. In fact, it is often more useful to look at the spatial shape of the orbit,that is, at r(φ) regarded as a function of the angle variable φ.

12

In order for the angle φ(t) to be an invertible function of time t, we need to allow φ to grow continuouslyand to assume arbitrary values as t increases, rather than wrapping back to 0 at ±2π. An orbit is closedif and only if r(φ) is a periodic function of φ for some integral multiple of 2π (and multiples thereof).

Now instead of solving for r(t) then φ(t), we can solve directly for r(φ) by using a clever transformationintroduced by Jacques Philippe Marie Binet, a 19th century French mathematician and astronomer.Introducing the change of variable

u = 1r , (62)

and using ` = µr2φ along with the chain rule, we can rewrite the differential operator ddt in terms of d

dφ :

ddt = dφ

dtddφ = `

µr2ddφ = `

µ u2 ddφ . (63)

Using this relation, we can write

r = ddtr = `

µ u2 ddφ

1u = `

µu2 (−1)u2

ddφu = − `

µddφu, (64)

and then

r = ddt r = `

µ u2 ddφ

[− `µ

ddφu

]= − `2

µ2 u2 d2

dφ2u, (65)

so the radial equation of motion

µr = `2

µr3 − U′(r) = `2

µr3 + F (r) (66)

can be written equivalently as

− `2

µ u2 d2udφ2 = `2

µ u3 + F (1/u) (67)

or

d2

dφ2u = −u− µ`2

1u2 F

(1u

)(68)

where F (r) = −U ′(r) = − ddrU(r). Provided ` 6= 0, this second-order differential equation can be solved

for u(φ) given appropriate initial conditions, thereby determining r(φ) = 1/u(φ).

If we can solve this differential equation for appropriate initial conditions, then we can deduce the shapeof the orbit. Conversely, if given the shape of the orbit, we can deduce the force law without having toknow anything about the velocity along the orbit.

V. INVERSE SQUARE FORCE LAWS AND KEPLER’S LAWS

An very important special case of conservative central force motion, applicable in gravitational andCoulomb problems, is that of the inverse-square force law, corresponding to the interaction potentialenergy

U(r) = −κr

(69)

for all r > 0, or equivalently, to the force

F (r) = −r κr2, (70)

for some constant parameter κ, where the force is attractive if κ > 0 and repulsive if κ < 0. For example,for the gravitational interaction between two point masses, κ = Gm1m2 = GMµ ≥ 0.

13

c

f

ϕ

FIG. 2 The bounded orbits of the attractive Kepler (inverse square law) problem are ellipses in the COM frame,with the center of force at one focus, specified in polar cylindrical coordinates r and φ. Shown are the lengths ofthe semimajor axis (a), the semiminor axis b, the semilatus rectum (c), and also the length f from the geometriccenter to a focus, as well as the minimal radius rmin (achieved at the periapsis) and the maximal radius rmax

(reached at the apoapsis). All these lengths are determined by the COM-frame energy E and magnitude ofangular momentum `.

A. Kepler’s First Law

We can now establish Kepler’s First Law, which its most general form says that all relative trajectoriesr(t) for an inverse-square force law are conic sections —- circles, ellipses, parabolas, or hyperbolas.

If ` = 0, then r and r remain parallel, and the motion is confined to a straight line (or line segment)through the origin, which can be thought of as a degenerate (limiting) case of an ellipse (for motionalong a segment) or a hyperbola (for unbounded motion along a line).

Otherwise, if ` 6= 0, then we can use Binet’s transformation. For the inverse square force law,

F (r) = − ddrU(r) = − κ

r2 = −κu2, (71)

for all r > 0, and the Binet equation (68) becomes

d2

dφ2u = −u− µ`2

1u2 F

(1u

)= −u+ µ

`21u2 κu

2 = −u+ µκ`2 , (72)

which is a linear differential equation in u which can be integrated to obtain:

u(φ) = 1r(φ) = α cos(φ− φ0) + µκ

`2 (73)

for some integration constants α and φ0. For later convenience, we define new constants

c = `2

µκ (74a)

ε = αc = α`2

µκ , (74b)

so that

r(φ) =c

1 + ε cos(φ− φ0), (75)

where c is fully determined once ` is fixed, and ε and φ0 are dimensionless constants to be chosen basedon further initial conditions.

14

1. Bounded Kepler Orbits

First, assume an attractive force such that κ > 0, and hence c > 0. So bounded (and in fact, closed)orbits are possible. As long as we are interested in the shape of single orbits, and not, say, transfersbetween orbits, we can always orient the in-plane (x and y) axes so that: (i) r(φ) is at a radial turningpoint when φ = 0, and (ii) ε ≥ 0. This means that we are choosing φ0 = nπ for some integer n, so theorbit can be written as

r(φ) =c

1 + ε cos(φ), (76)

for some ε ≥ 0,

The radial turning points must be

rmin = c1+ε = `2

µκ1

1+ε (77a)

rmax = c1−ε = 1+ε

1−ε rmin, (77b)

and so we can relate the parameter ε to the energy E by equating

E = Veff(rmin) = `2

2µ1

r2min− κ

rmin= `2

2µ1

rmin

[1

rmin− 2

c

]= `2

2µ1+εc

[1+εc −

2c

]= µκ2

2`2 (ε2 − 1), (78)

and we see that ε is a function of the conserved quantities ` and E :

ε = +√

1 + 2µκ2 E `2, (79)

which also establishes the non-obvious inequality E `2 > −µκ2

2 for bounded Kepler orbits.

This orbit will be truly bounded if rmax <∞ strictly, meaning the system does not have enough energyto achieve arbitrarily large separations r. Since lim

r→∞U(r) = 0, this is equivalent to saying the energy is

negative: E < 0, which from (78) implies 0 ≤ ε < 1.

But it is not difficult to show that for c > 0 and 0 ≤ ε < 1, (89) describes an ellipse with one focus atthe origin r = 0, with eccentricity ε, and with a length c of the so-called semilatus rectum, but we willfirst need to review some geometric facts about the ellipse.

The eccentricity ε of the ellipse is a measure of the aspect ratio of the ellipse, and is equal to:

ε =rmax − rmin

rmax + rmin

. (80)

so clearly satisfies 0 ≤ ε < 1, where ε = 0 corresponds to a circle of constant radius rmin = rmax.

The ellipse is defined such that the sum of the distances between any point on the ellipse to each oftwo specified foci is some constant, which we will denote 2a. By center of the ellipse we mean theintersection of the two mutually perpendicular axes of reflection symmetry, the longer of which is calledthe major axis and contains the foci located symmetrically about the center, and the shorter is the minoraxis. Except for the special case of a circle, the center does not coincide with the origin r = 0 of ourcoordinates; rather the origin is taken to be one of the foci (specifically the one on the right lookingdown at a diagram of the ellipse on the page, according to our usual conventions for relating Cartesianand polar coordinates).

The line segment between the center and either vertex on the ellipse, that is, between points of maximaldistance from the center, is a semimajor axis, and is of length a, which is the arithmetic mean of theminimal and maximal distances between the ellipse and the origin/focus:

a = 12 (rmin + rmax). (81)

15

The length a of the semimajor axis is also equal to the mean radius weighted by arc length around theellipse:

∫ds r =

2π∫0

dφ√

( drdφ )2 + r2 r(φ) = a. (82)

Both foci of the ellipse lie along the major axis, equidistant from the center. The length f is defined asthe distance between one focus and the center of the ellipse.

A semiminor axis is the line segment between the ellipse and its center which is perpendicular to thesemimajor axes; its length b is the geometric mean of the the minimal and maximal distances betweenthe ellipse and the origin/focus:

b =√rmin rmax (83)

The point on the ellipse closest to the origin, at the edge of one semimajor axis and correspondingto r = rmin, is called the periapsis (from the Greek, via Latin, meaning ”near arch”). The point onthe ellipse furthest from the origin, which must lie on the far edge of the other semimajor axis, andcorresponding to r = rmax, is called the apoapsis (“far arch”). Together these points are called theapsides. (For the particular case of a body orbiting the sun, they are called the perihelion and aphelion,respectively; for a body orbiting the Earth, the perigee and apogee). The area enclosed by the ellipse isjust A = πab.

The semilatus rectum (Latin for “half of a straight side”) is a line segment between one focus and apoint on the ellipse, oriented transverse to the major axis (and therefore parallel to the minor axis). Itslength, denoted by c (some books will use d, or p, or various other symbols) is the harmonic mean ofthe minimal and maximal radii measured from the origin/focus:

1c = 1

2 ( 1rmin

+ 1rmax

). (84)

For an ellipse, the various characteristic distances satisfy a number of additional elementary relationships,including:

f =√a2 − b2 = aε, (85a)

c = a(1− ε2) = b√

1− ε2 = rmin(1 + ε) = rmax(1− ε) = b2

a . (85b)

For f = 0, or equivalently ε = 0, the ellipse is in fact just a circle. As f →∞ or equivalently ε→ 1, theellipse becomes a parabola.

In polar coordinates centered on the right focus, with angle measured counterclockwise from the x axis,the ellipse must satisfy the defining relation specifying that the distances from the foci to any point onthe ellipse sum to exactly 2a:

|r r|+ |r r + 2f x| = 2a (86)

or

(r − 2a)2 = |r r + 2f x|2 = r2 + (2f)2 + 4fr cosφ (87)

which is equivalent to:

r =a2 − f2

a+ f cosφ=

a2 − ε2 a2

a+ εa cosφ=

a(1− ε2)

1 + ε cosφ=

c

1 + ε cosφ, (88)

as claimed.

Note that for a given energy E < 0, the eccentricity decreases as ` increases. Conversely, the circularorbit has the most angular momentum for a given energy.

16

2. Unbounded (Scattering) Orbits for Attractive Potentials

Supposing still that κ > 1 but that ε > 1, we see that

r(φ) =c

1 + ε cos(φ), (89)

will still have an inner radial turning point at

rmin =c

1 + ε=

`2

µκ

1

1 + ε, (90)

so this inner turning point still allows us to connect ε to the other conserved quantities:

E = Veff(rmin) = µκ2

2`2 (ε2 − 1) > 0, (91)

but here the energy is positive for ε > 1, meaning that the kinetic energy is sufficiently large so that theorbit is said to be unbounded, or equivalently to be a scattering orbit. In fact, we see that r → ∞ ascosφ → −ε−1 or equivalently as φ → ± arccos(−ε−1). Thus we expect a pair of asymptotes at certainfinite angles with respect to the origin, and we will find hyperbolic trajectories.

Although only one branch of the hyperbola wil be associated with a single trajectory, it is convenientto visualize both branches, and the center of reflection symmetry between them. The hyperbola canbe defined as the locus of all points such that the difference in distance between two foci (locatedsymmetrically along the major axis) is equal to the constant 2a, where f = εa is the distance betweenthe center and one of the foci, and it turns out a is also the distance between the center and a vertex,which is defined as a point where the hyperbola crosses the major axis, meaning it is the closest pointon a branch of the hyperbola to the center. The semilatus rectum is again defined as a line segmentfrom the focus to the ellipse, transverse to the major axis; but now has length c = a(ε2 − 1). Centeringour origin (r = 0) on the left focus, the defining relation

|2f x− r r| − |r r| = 2a (92)

after rearranging and squaring becomes

(r + 2a)2 = |2f x− r r|2 = (2f)2 + r2 − 4fr cosφ, (93)

which is equivalent to:

r =f2 − a2

a+ f cosφ=

ε2 a2 − a2

a+ εa cosφ=

a(ε2 − 1)

1 + ε cosφ=

c

1 + ε cosφ, (94)

as claimed.

If ε = 1 exactly, meaning E = 0 exactly, the particles have just enough energy to achieve arbitrary largeseparation (albeit in infinite time), and we claim the relative trajectory is then a parabola, which can bethought of as the limiting case of an ellipse where the foci become infinitely separated, or as the limitingcase of an hyperbola where the angle between the asymptotes approaches π.

In this case, by rearranging and squaring

r = c1+cosφ , (95)

we find

r2 = (c− r cosφ)2 = c2 + r2 cos2 φ− 2cr cosφ = c2 + r2 − r2 sin2 φ− 2cr cosφ, (96)

which is equivalent to

r2 sin2 φ = c2 − 2cr cosφ, (97)

or in Cartesian coordinates (x = r cosφ, y = r sinφ) centered at the same focus,

y2 = c2 − 2cx, (98)

which indeed describes a parabola.

17

3. Unbounded (Scattering) Orbits for Repulsive Potentials

In the case of repulsive potentials (corresponding to κ < 1), obviously no bound orbits are possible sinceE ≥ 0. The easiest way to work out the orbits is to choose ε ≤ −1 (it is then the negative eccentricity)but insist that r ≥ 0, leading to hyperbolic (or in the limiting case, parabolic) orbits, but with thefocus/origin outside rather than inside the convex hull of the orbit. That is, the incoming fictitiousparticle gets turned around before crossing the plane which passes through the center of force and istransverse to the initial velocity.

B. Kepler’s Second Law and Orbital Speeds

We have already proved Kepler’s Second Law for the general two-body central force problem. Forclosed Kepler orbits, it confirms that the angular speed is maximal at the periapsis and minimal at theapoapsis. Energy conservation additionally tells us that the total speed is also maximal at the periapsisand minimal at the apoapsis. However, at these apsides, where the total speed is extremal, the radialvelocity must vanish, because they are radial turning points. The radial speed is maximal in between,at points where the effective potential Veff(r) is minimal — specifically at each end of the latus rectum.

C. Kepler’s Third Law

Closed (and therefore elliptical) orbits are necessarily periodic. Integrating Kepler’s Second Law overone period τ , we find

A =

τ∫0

dt dAdt =

τ∫0

dt`

2µ=τ`

2µ(99)

but for an ellipse,

A = πab, (100)

so

τ =2πabµ

`(101)

and hence

τ2 =4π2a2b2µ2

`2= 4π2µ

2

`2a2a2(1− ε2) = 4π2µ

κa3a(1− ε2)

`2

µκ

= 4π2µ

κa3 cc = 4π2µ

κa3, (102)

so the square of the period is proportional to the the cube of the length a of the semimajor axis. Notethat a is a natural measure of the average radius of the obit, since it is both the arithmetic mean of theminimal and maximal radii, and also the arclength-weighted average of the orbital radius. This lengtha, and therefore the temporal period of the orbit, depend only on the COM-frame energy E , and not onthe COM-frame angular momentum:

a =c

1− ε2=

`2

µκ

2`2

µκ2 (−E )=

κ

2|E |, (103)

which itself a beautiful and useful result for the inverse-square central force problem, related to theintegrability properties of this system (see below) and the fact that in the quantum mechanics of the

18

hydrogen atom, the energy eigenvalues depend (in the absence of relativistic and spin effects) only onthe principal quantum number and not on the angular momentum quantum number.

For the specific case of the gravitational two-body problem, note that

µ

κ=

µ

Gm1m2=

µ

GµM=

1

GM, (104)

so

τ2 = 4π2

GM a3, (105)

where Kepler’s constant 4π2

GM is not quite constant across all planets, since it depends on the total massM and not just the solar mass.

VI. DYNAMICAL INVARIANTS AND INTEGRABILITY

Because the bound orbits for the Kepler problem are all closed, with the shape of the orbit described bya single-valued mapping between angle and radius, it must be the case that in cylindrical coordinate theperiod of the radial motion and the the period of the angular motion are equal. Moreover, the periodof motion depends only on the energy and not the angular momentum. For a general problem, even fora completely integrable one, there is no reason that either of these properties would necessarily be thecase, unless some sort of additional constraint is present. Indeed, such a constraint arises from anotherconservation law, connected to a so-called hidden symmetry of the Kepler problem.

1. Laplace-Runge-Lenz Vector

In addition to E and L′, it can be shown directly from the equations of motion for the Kepler problemthat an additional vector-valued quantity is also conserved, the so-called Laplace-Runge-Lenz Vector :

A = p×L′ − µκ r, (106)

which includes a term which is bilinear in the angular and linear momenta.

Since A is a vector, this might seem to provide an additional three conserved quantities, but really onlyone additional independent constant of the motion has been found, not three, since A, L′, and E arerelated by two additional equations:

A·L′ = 0 (107a)

A·A = µ2κ2 + 2µE `2, (107b)

but the vector A does necessarily lie in the plane of motion, and its direction A in this plane doesprovide an additional, genuinely independent constant of the motion — always pointing in the directionfrom the COM-frame origin (center of force) to the periapsis of the orbit. Therefore its conservation iswhat guarantees that the bounded orbits must be closed — non-inverse-square-law central forces, forexample, will have non-closed trajectories, where the apsides precess over each successive orbit.

In fact, we can derive the Kepler orbits quite simply from the conservation of A. We have

A·r = Ar cosφ = r · (p×L′)− µκr = (r × p) ·L′ − µκr = L′ ·L′ − µκr = `2 − µκr, (108)

where we are measuring φ from the periapsis, and A = |A| ≡ µκε is the constant magnitude of A.Rearranging, this becomes:

r =

`2

µκ

1 + Aµκ cosφ

=c

1 + ε cosφ(109)

19

which is the equation for a conic section, with eccentricity ε = Aµκ and length of semilatus rectum c = `2

µκ .

From the above expression for A·A we can rewrite the eccentricity in terms of the energy and angularmomentum:

ε2 − 1 = 2`2

µκ2 E , (110)

which is equivalent to our earlier result.

It turns out that this additional conservation law is not associated with a symmetry under a transfor-mation of the configuration space coordinates by functions of these coordinates (and/or time) alone,but is associated with a more general sort of symmetry under transformations which mix the gener-alized positions and their conjugate momenta. Equivalently, the symmetry can be interpreted as ahigher-dimensional geometric symmetry, where the usual motion in three dimensions is regarded as theprojection of a motion in four dimensions. (For group theory aficionados: this “hidden” symmetry isassociated with SO(4) for bound orbits and SO(3, 1) for unbound orbits).

As is often the case in the history of science, the accepted name does not reflect the actual history ofdiscovery, but more the history of popularization. The conservation of A for a inverse-square centralforce was apparently first discovered for a special case by Jakob Hermann, then generalized by JohannBernoulli in 1710, then rediscovered by Laplace, then re-derived by Hamilton, then by Gibbs. The latter’sderivation was used in a popular textbook in Germany by Carl Runge, which was then referenced byWilhelm Lenz in a widely-read paper on the quantum mechanics of the hydrogen atom, and gainedfurther notoriety when used by Pauli to solve for the energy spectrum of hydrogen by algebraic means.

The Laplace-Runge-Lenz (LRL) vector also reveals an interesting fact about the momentum p in thecase of Kepler motion, namely that its tip traces out an offset circle in momentum space. We defineCartesian axes x ∝ A and y = z × x in the plane of motion (transverse to L′ ∝ z). Then by equatingthe squared-magnitude of both sides of

µκr = p×L′ −A, (111)

we can deduce that

p2x + (py −A/`)2 = (µκ/`)2, (112)

which is the equation for an offset circle of radius |µκ/`| in momentum space.

A. Subintegrability, Integrability, Superintegrability, and Maximal Superintegrability

Recall that we say a system is integrable when the problem of specifying any trajectory given the initialconditions can be “reduced to quadratures,” meaning we can in principle express the trajectories usingone-dimensional definite integrals as well as various purely algebraic relations involving root-finding orfunction inversion (even if, in practice, we cannot necessarily perform the needed integrations or algebraanalytically).

This notion of integrability is closely related to whether the system possesses a sufficient number ofautonomous constants of the motion that are independent (in ways that we will make more precise laterin the course).

A single-degree-of-freedom conservative problem is always integrable, because formally we can expressthe trajectories in terms of one-dimensional integrals involving the conserved value of the energy andthe potential energy function. Conservation of a single function of the (generalized) position and (gen-eralized) velocity determines the shape of the trajectory in the two-dimensional position/momentumphase space, and further initial conditions specify where on this curve the system lies at a specified time.Even though we have two dynamical variables (say q and q), we only need one conserved quantity to adetermine a (2 − 1) = 1 dimensional curve (trajectory), since we still must have the freedom to chooseinitial conditions, and therefore to specify where along this curve we are to start at time t = 0.

20

Similarly, an n-DOF holonomic system has 2n dynamical variables (n generalized coordinates and ngeneralized velocities), but it turns out that we only need n independent, autonomous (i.e., explicitlytime independent) constants of the motion for the system to be integrable. The conserved quantitiesrestrict the trajectory to an n-dimensional surface in the 2n-dimensional space of dynamical variables,and the remaining initial conditions pick out a particular starting point and subsequent curve on thissurface. In principle, the solution at any time can be determined using algebra at performing n definiteintegrals. (We will explore this in more detail after introducing the Hamiltonian formalism).

But generically, a conservative n-DOF system will have one conserved quantity, namely the energy,but perhaps no more. If it has more than 1 but less than n independent (and autonomous) conservedquantities, it is said to be subintegrable or partially integrable. These count as non-integrable, and suchsystems can exhibit chaotic motion which we will discuss in more detail at the end of the course. If thesystem has n independent (and autonomous) constants of motion, it is (completely) integrable, leadingto regular trajectories which are quasi-periodic if bounded. And if it has more than n independent(autonomous) constants of the motion, it is said to be superintegrable, with the additional dynamicalinvariants further constraining the possible motion. How many independent constant of motion can ann-DOF system have? It can have up to 2n− 1, and their intersecting level surfaces would then define aone dimensional curve in the space of dynamical variables. Any more, and no actual dynamics would bepossible — we need the freedom to specify initial conditions which then determine where on this curvethe system is located at a given time.

For a two-body central force problem, in the relative coordinates in the COM frame, we have 3 degreesof freedom and 3 + 3 = 6 dynamical variables (the components of r and r), and also 4 independent,autonomous conserved quantities: the energy E , and 3 components of the relative angular momentumL′. Generically, these systems are therefore superintegrable, since 4 > 3. But for the inverse squareforce law, we have in addition 1 independent component of the Laplace-Runge-Lenz vector, say A·x inour notation. So the Kepler problem is in fact maximally superintegrable, because 5 = 2 · 3− 1 = 6− 1.

We have seen above that we could deduce the shape of the Kepler orbits algebraically, without any actualintegration — just by manipulating the Laplace-Runge-Lenz Vector. This is a consequence, and sign, ofmaximal superintegrability, in which 1-dimensional trajectories are determined entirely by conservationlaws, and not by calculus. The intersection of the level hypersurfaces associated with the conservedquantities determines the one-dimensional shape of the trajectory.

Specifically, in the full two-body problem we have 6 degrees of freedom for the relative motion, meaning6 generalized coordinates and 6 corresponding generalized velocities. A total of 6 initial conditionsspecifying the COM position and COM velocity determine the trivial (rectilinear) motion of the COM.

Then, 6 additional initial conditions are needed to specify the relative position and relative velocityshould determine the motion in the COM frame. Indeed, we need: 2 numbers to specify the orientationof the plane to which the relative motion is confined (for example, a unit normal vector); 2 parametersto specify the intrinsic shape of the ellipse (for example, the length of the semimajor axis and theeccentricity); another 1 parameter to specify the angular orientation of the ellipse (or hyperbola orparabola) in that plane; and finally 1 number to specify the location along the ellipse at some referencetime. For the Kepler problem, the first 5 of these are related to time-independent constants of themotion, respectively: the direction of the angular momentum; the COM-frame energy and magnitude ofCOM-frame angular momentum; and the in-plane direction of the Laplace-Runge-Lenz vector, leavingonly the freedom of choosing the time of arrival at periapsis.

What happens when we add more particles? The full three-body problem has 9 degrees of freedom, butinvestigations by Bruns, Poincare, Painleve, and others have established that there are no well-behavedextra autonomous constants of the motion beyond energy, linear momentum, and angular momentum.So there are 9 degrees of freedom but only 7 autonomous invariants, so the three-body problem is notintegrable. By adding just one more particle, we go from a maximally superintegrable system (in therelative coordinates) to a subintegrable, chaotic system.

21

VII. SOME ADDITIONAL THEOREMS

A. Virial Theorem

The Virial Theorem (coined by Rudolph Clausius, from the Latin for “strength”) is a useful relationshipbetween time-averaged kinetic and potential energies for certain systems undergoing bounded stablemotion. We can establish it as follows. Integrating with respect to time the COM-frame kinetic energy(denoted here by K ′, to avoid confusion with the duration T ), we have

T∫0

dtK ′ =

T∫0

dt 12µ r ·r = 1

2 r · µ r∣∣T0− 1

2

T∫0

dt r ·µr = 12 r ·p

∣∣T0− 1

2

T∫0

dt r ·F , (113)

or

r(T )·p(T )− r(0)·p(0)

T=

1

T

T∫0

dtK ′ +1

2

1

T

T∫0

dt r ·F . (114)

If the trajectory remains bounded, so that |r(t)| ≤ rmax < ∞, and additionally the kinetic energyrermains bounded, so |p(t)| ≤ pmax =

√2µKmax < ∞ (both of which hold true for the bound orbits of

the inverse square force law in the COM frame), then in the limit as T → ∞, the left hand side of (114)vanishes, so in these cases,

2 limT→∞

1

T

T∫0

dtK ′ = − limT→∞

1

T

T∫0

dt r ·F . (115)

If U(r) is homogeneous of degree n, meaning U(λr) = λnU(r) for all λ > 0, then Euler’s HomogeneousFunction Theorem guarantees that

−r · ∂∂rF = r · ∂∂rU(r) = nU(r), (116)

and so

2 limT→∞

1

T

T∫0

dtK ′ = n limT→∞

1

T

T∫0

dtU(r) , (117)

meaning 2 〈K ′(t)〉t = n 〈U(t)〉t in the COM frame, where here the brackets represent infinite-time av-erages. For central potentials, homogeneity in fact requires that potential energy be a power law:U(r) ∝ rn. For the case of the inverse-square force law, we have n = −1, so we could conclude thatfor all bound Kepler orbits, the time-averaged kinetic energy is one half of the absolute value of thetime-averaged potential energy. In the case of closed (i.e., periodic) orbits, the infinite time averagescan be replaced with time averages over one complete period (because the boundary term automaticallyvanishes). So for closed Kepler orbits, we find

2 〈K ′〉t = 21

τ

τ∫0

dt 12µ|r|

2 = −1

τ

τ∫0

dtU(r) =1

τ

τ∫0

dtκ

r= −〈U〉t. (118)

Relating as it does the time-averaged kinetic and potential energies in the case of bounded motionin a homogenous central force field, the Virial Theorem finds applications in statistical mechanics,astrophysics, quantum mechanics, and other fields.

22

B. Bertrand’s Theorem

We shall not prove this theorem, because it is a bit involved, but we mention a beautiful and non-obvious result by 19th century French mathematician Joseph Bertrand. Many conservative central forceproblems may have isolated circular orbits or other closed (i.e., periodic) orbits. But there are onlytwo conservative central potentials for which all bounded orbits are necessarily closed: the attractiveinverse square law, U(r) = −κ/r, and the harmonic oscillator potential, U(r) = 1

2kr2. For both of these

potentials, the closed orbits are all ellipses, only in the central force case, they force is directed towardone focus of the ellipse, while in the harmonic oscillator case, the force is directed toward the geometriccenter of the ellipse.

It is no coincidence that these two potentials picked out by Bertrand’s Theorem — the isotropic har-monic oscillator and the inverse square law — are also the only maximally superintegrable centralpotentials. (The three-dimensional harmonic oscillator has an additional conserved quantity, analogousto the Laplace-Runge-Lenz vector). For the same reason, in quantum mechanics, you can solve for theenergy spectrum of the hydrogen atom and the isotropic harmonic oscillator algebraically, using onlycommutation relations, without ever having to actually integrate the Schrodinger equation.