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I immediately have m 1 q¨ 1 = f (|q|)q/|q| and m 2 q¨ 2 = −f (|q|)q/|q|. Multiply- ing these equations by m 2 and m 1 (respectively) and then subtracting, I get m 1 m 2 (¨q 1 − ¨q 2 ) = (m 2 + m 1 )f (|q|)q/|q|. The desired equation follows after dividing by m 1 + m 2 , which is licit since I assume m 1 , m 2 > 0.
The kinetic energies of the particles are T 1 = 12 m 1 | q˙ 1 |^2 and T 2 = 12 m 2 | q˙ 2 |^2. Since m 1 q 1 +m 2 q 2 = 0, I have 0 = |m 1 q˙ 1 + m 2 q˙ 2 |^2 = m 12 | q˙ 1 |^2 +2m 1 m 2 q˙ 1 · q˙ 2 + m 22 | q˙ 2 |^2. Adding this to m 1 m 2 | q˙|^2 = m 1 m 2 | q˙ 1 |^2 − 2 m 1 m 2 q˙ 1 · q˙ 2 + m 1 m 2 | q˙ 2 |^2 yields m 1 m 2 | q˙|^2 = (m 1 + m 2 )(m 1 | q˙ 1 |^2 + m 2 | q˙ 2 |^2 ). Thus the total kinetic en- ergy is T 1 + T 2 = (^12) m^ m 11 +mm^22 | q˙|^2 , the first term of the desired formula. Meanwhile,
the particles’ potential energies are V 1 = 12 V (|q 1 − q 2 |) and V 2 = 12 V (|q 2 − q 1 |). Each of these is simply 12 V (|q|), so the total potential energy is the other term of the desired formula.
The angular momenta of the particles are J 1 = m 1 q 1 × q˙ 1 and J 2 = m 2 q 2 × q˙ 2. Since m 1 q 1 +m 2 q 2 = 0, I have 0 = (m 1 q 1 + m 2 q 2 )×(m 1 q˙ 1 + m 2 q˙ 2 ) = m 12 q 1 × q^ ˙ 1 + m 1 m 2 (q 1 × q˙ 2 + q 2 × q˙ 1 ) + m 22 q 2 × q˙ 2. Adding this to m 1 m 2 q × q˙ = m 1 m 2 q 1 × q˙ 1 − m 1 m 2 (q 1 × q˙ 2 + q 2 × q˙ 1 ) + m 1 m 2 q 2 × q˙ 2 yields m 1 m 2 q × q˙ = (m 1 + m 2 )(m 1 q 1 × q˙ 1 + m 2 q 2 × q˙ 2 ). Thus the total angular momentum is J 1 + J 2 = (^) mm 11 +mm^22 q × q˙, which is the desired formula. Since the z components of q 1 and q 2 are zero, so the z components of q and ˙q are also zero. Thus, the x and y components of q × q˙ are zero.
I have r^2 = |q|^2 and r cos θ = q· xˆ, where ˆx is the unit vector along the positive x axis. Differentiating these formulas, I get 2r r˙ = 2q· q˙ and ˙r cos θ−r θ˙ sin θ = ˙q·ˆx. Multiplying the first of these formulæ by 12 xˆ and subtracting the other formula
multiplied by q, I get ˙q × (ˆx × q) = r r˙xˆ − r˙q cos θ + r θ˙q sin θ, where I’ve used the vector identity that ˙q × (ˆx × q) = ( ˙q · q)ˆx − ( ˙q · xˆ)q. Now, ˆx × q = rˆz sin θ, where ˆz is the unit vector along the positive z axis, and | q˙ × ˆz| = | q˙| since ˙q has zero z component. Thus I have
r^2 | q˙|^2 sin^2 θ = |r r˙ˆx − r˙q cos θ + r θ˙q sin θ|^2 = r^2 r˙^2 + ˙r^2 |q|^2 cos^2 θ + r^2 θ˙^2 |q|^2 sin^2 θ − 2 r r˙^2 xˆ · q cos θ + 2r^2 r˙ θ˙xˆ · q sin θ − 2 r r˙ θ˙|q|^2 cos θ sin θ, (1)
where I’ve used that |ˆx| = 1. Substituting r^2 for |q|^2 and r cos θ for ˆx · q, this becomes
r^2 | q˙|^2 sin^2 θ = r^2 r˙^2 + r^2 r˙^2 cos^2 θ + r^4 θ˙^2 sin^2 θ − 2 r^2 r˙^2 cos^2 θ + 2r^3 r˙ θ˙ cos θ sin θ − 2 r^3 r˙ θ˙ cos θ sin θ = r^2 r˙^2 sin^2 θ + r^4 θ˙^2 sin^2 θ. (2)
Thus, | q˙|^2 = ˙r^2 + r^2 θ˙^2 , and the formula for the total energy is proved. (The conclusion is valid even when sin θ = 0, by continuity, since sin θ can’t be con- stantly 0 in the physically relevant situation where an orbit exists. Similarly, I’ve been assuming r > 0 all along.) As for the angular momentum,
|q × q˙|^2 = |xˆ × (q × q˙)|^2 = |(ˆx · q˙)q − (ˆx · q) ˙q|^2 = | r˙q cos θ − r θ˙q sin θ − r q˙ cos θ|^2 = r˙^2 |q|^2 cos^2 θ + r^2 θ˙^2 |q|^2 sin^2 θ + r^2 | q˙|^2 cos^2 θ − 2 r r˙ θ˙|q|^2 cos θ sin θ − 2 r r˙(q · q˙) cos^2 θ + 2r^2 θ˙(q · q˙) sin θ cos θ = r^2 r˙^2 cos^2 θ + r^4 θ˙^2 sin^2 θ + r^2 ( ˙r^2 + r^2 θ˙^2 ) cos^2 θ − 2 r^3 r˙ θ˙ cos θ sin θ − 2 r^2 r˙^2 cos^2 θ + 2r^3 r˙ θ˙ sin θ cos θ = r^4 θ˙^2 , (3)
so |j| = m|q × q˙| = mr^2 | θ˙|. I in fact have j = mr^2 θ˙, since both sides are positive for counterclockwise motion, although ultimately I won’t need the sign.
Note that the sign of θ˙ always equals the sign of the constant j in this for- mula, and the physical application requires that θ˙ not be constantly 0, so I can conclude that j 6 = 0, which will be needed when I divide by j later on. To continue,
E =
m(r^2 θ˙^2 + ˙r^2 )+V (r) =
m(r^2
j^2 m^2 r^4
m r˙^2 +(V (r) +
j^2 2 mr^2
as desired. By the way, note that the centrifugal force is −(d/dr)(j^2 / 2 mr^2 ) = j^2 /mr^3 , not j^2 /mr as stated.
Solving for ˙r in the formula for E, I get E −U (r) = 12 m r˙^2 , or ˙r^2 = (^) m^2 (E − U (r)),
so ˙r = ±
2 m (E^ −^ U^ (r)). However, ˙r^ may or may not be nonnegative.
If ˙r 6 = 0, θ/˙ r˙ = ± (^) mrj 2 /
2 m (E^ −^ U^ (r)), or θ˙ = ± j mr^2 r/˙
2 m (E^ −^ U^ (r)), when
r˙ 6 = 0. Integrating this with respect to time, θ = θ(0)±
0 (^
j mr^2 r dt/˙
2 m (E^ −^ U^ (r))).
To avoid picking branch cuts for √^ and arccos (or even for θ!), let me revert to
a differential equation and look at θ˙^2. Since θ˙ 6 = 0 but I can’t be so sure about r˙, I’ll use
r˙^2 θ^ ˙^2 =
2 m (E^ −^ U^ (r)) j^2 /m^2 r^4
2 mr^4 (E − V (r) − j^2 / 2 mr^2 ) j^2
r^2 (2mEr^2 + 2mkr − j^2 ) j^2
Let me simplify the constants by introducing p := j^2 /mk and e :=
1 + 2Ej^2 /mk^2 now, so j^2 = mkp and mk^2 e^2 = mk^2 +2Ej^2 = mk^2 +2Emkp, or 2Ep = ke^2 −k. Note that since E ≥ −mk^2 / 2 j^2 , or 2Ej^2 /mk^2 ≥ −1, the square root defining e is real. Then
r˙^2 θ^ ˙^2 =^
r^2 (2mEr^2 + 2mkr − mkp) mkp
r^2 (2Epr^2 + 2kpr − kp^2 ) kp^2 =
r^2 (ke^2 r^2 − kr^2 + 2kpr − kp^2 ) kp^2
r^2 (e^2 r^2 − r^2 + 2pr − p^2 ) p^2
The numerator now contains the subtraction of a square, which suggests the use of trigonometry. Since the left side of this equation must be nonnegative, I can conclude that e^2 r^2 ≥ r^2 − 2 pr + p^2 = (p − r)^2. Thus, there must always be an angle α such that p − r = er cos α. In terms of this angle, then, I have r˙^2 / θ˙^2 = r^2 (e^2 r^2 − e^2 r^2 cos^2 α)/p^2 = e^2 r^4 sin^2 α/p^2. If ˙r 6 = 0, then, I can now write this as θ/˙ r˙ = ±p/er^2 sin α, or θ˙ = ±p r/er˙^2 sin α. In order to integrate this, I’ll want to understand ˙α, so differentiate the equation p − r = er cos α to get − r˙ = e r˙ cos α − er α˙ sin α; since e cos α = p/r − 1, this becomes − r˙ = p r/r˙ − r ˙ − er α˙ sin α, or ˙α = p r/er˙^2 sin α. I need analyse this no further, for now I see that θ˙ = ± α˙, so θ = ±α + θ 0 for some constant θ 0. In particular, cos(θ − θ 0 ) = cos(±α) = cos α = ( p r − 1)/e, so I may write θ = θ 0 + arccos
( p r − 1)/e
for some branch of θ and arccos. In preparation for writing this in terms of the original parameters, let me multiply both sides of the fraction in the arccosine by k/j. Then
θ = θ 0 + arccos(
kp jr −^
k j ke j
) = θ 0 + arccos(
kj^2 kmjr −^
k j k j
1 + 2 Ej 2 mk^2
= θ 0 + arccos(
j mr −^
k √ j k^2 j^2 +^
2 E m
which is the desired formula. Note that this θ 0 may not be the θ(0) from above, but that could be fixed by adjusting the indefinite integral.
I already reduced the clutter in the previous problem. However, to make sure of the formula for r, I should check also the situation when ˙r = 0. I know from
the qualitative analysis that this occurs only when E = −mk^2 / 2 j^2 and that r = j^2 /mk then. In terms of e and p, these are equations are, irrespectively, r = p and mk^2 (e^2 − 1)/ 2 j^2 = −mk^2 / 2 j^2 , or e = 0. Thus, the equation r = p/(1 + e cos(θ − θ 0 )) holds then as well.
You used the translation and Galilean symmetries by setting the origin at the centre of mass for all time, and you used some of the rotational symmetry by placing the problem within the x, y plane. But there remains a rotational symmetry within that plane, so I use it to set θ 0 to 0. (There is also a symmetry of reflection through the z axis, which could be set by requiring j > 0, and a scaling symmetry, which could be set by requiring p = 1. But I will not do this yet.) Then the equation for the orbit becomes p = r + er cos θ. Now, r cos θ = x, so I get r = ex − p, or r^2 = e^2 x^2 − 2 epx + p^2. Since r^2 = x^2 + y^2 , this becomes x^2 −e^2 x^2 +2epx+y^2 = p^2 , as desired. If e = 0, then this equation is x^2 +y^2 = p^2 , a circle of radius p centred at the origin. If 0 < e < 1, then prepare to complete the square by writing (1 − e^2 )^2 x^2 + 2ep(1 − e^2 )x + (1 − e^2 )y^2 = p^2 (1 − e^2 ), or (1 − e^2 )^2 x^2 + 2ep(1 − e^2 )x + e^2 p^2 + (1 − e^2 )y^2 = p^2 (1 − e^2 ) + e^2 p^2 ,
so
(1 − e^2 )x + ep
- (1 − e^2 )y^2 = p^2 , which becomes (x + (^1) −epe 2 )^2 / p
2 (1−e^2 )^2 + y^2 / p
2 1 −e^2 = 1, which is an ellipse centred at (−ep/(1^ −^ e
(^2) ), 0) with minor radius
p/(1 − e^2 ) along the x axis and major radius p/
1 − e^2 along the y axis. If e = 1, then the equation is 2px + y^2 = p^2 , or x = p/ 2 − y^2 / 2 p, which is a parabola centred along the x axis, opening to the left, and with a vertex at (p/ 2 , 0). Finally, if e > 1, then I can complete the square as in the elliptic case,
only now the equation should be written (x − (^) e 2 ep− 1 )^2 / p
2 (e^2 −1)^2 −^ y
(^2) / p^2 e^2 − 1 = 1, which is a hyperbola centred at (ep/(e^2 − 1), 0) with minimal radius p/(e^2 − 1) along the x axis and asymptotes x = ±(e^2 − 1)y.
