Assignment 5 Solutions

Physics 106a: Assignment 5 (Solutions prepared by Alvin Li)

8 November, 2019; due 5pm Friday, 15 November in the “Ph106 In Box” in East Bridge mailbox.

One Dimensional Systems: Central Forces and the Kepler Problem Reading: Hand and Finch Chapter 4.

Problems 1. Binary-sun solar system: Consider a binary pair of identical suns of mass M orbiting in the x − y plane in an orbit centered at the origin. The gravitational constant is G. Now add a planet of small mass m with an initial condition on the z axis above the center of mass of the two suns and with a velocity in the z direction; by the symmetry of the system the small mass will remain on the z axis and suns will have equal z coordinates and their center of mass will also remain on the z axis. We can choose as coordinates describing the dynamics: the z coordinate of the planet z; the z coordinate of the two suns Z; and the polar coordinates (R, θ) giving the x, y coordinates of the suns (±R cos θ, ±R sin θ)— see ﬁgure.

z M Z R M R e

(a) What is the Lagrangian of the system in terms of these coordinates and their time derivatives? [Solution] For each of the large suns i, the kinetic energy is 1 1 1 T = MR˙ 2 + MZ˙ 2 + MR2θ˙2, (1) Mi 2 2 2 and for the small planet, the kinetic energy is 1 T = mz˙2. (2) m 2

The potential energy comes from three contributions: The gravitational potential between M1 and M2 (separated by a distance of 2R), that between M1 and m (separated by a distance of p 2 2 p 2 2 R + (z − Z) ), and lastly that between M2 and m (separated by a distance of R + (z − Z) ). Note that the contributions from the latter two are identical. Mathematically, we have GM 2 GMm V = − − 2 × (3) 2R pR2 + (z − Z)2 | {z } GPE between M1 and M2 | {z } GPE between Mi and m

Hence, the Lagrangian is given by

L = 2 × TMi + Tm − V (4) 1 GM 2 2GMm = MR˙ 2 + MZ˙ 2 + MR2θ˙2 + mz˙2 + + (5) 2 2R pR2 + (z − Z)2

1 • (0.5 pt) : Writing down the kinetic energy of each of the large suns. • (0.5 pt) : Writing down the kinetic energy of the small planet.

• (0.5 pt) : Writing down the potential between Mi and m. • (0.5 pt) : Writing down the potential between M1 and M2. • (1 pt) : Correct ﬁnal expression for the Lagrangian. Note that if the student directly writes down the Lagrangian without the above derivations, full credit should still be given. Total sub-points : 3

(b) Using constants of the motion implied by translational and rotational symmetries ﬁnd the coupled equations of motion forq, ¨ R¨ with q = z − Z. Your result will depend on the angular momentum l of the two suns as well as the parameters of the problem. [Solution] We notice that the Lagrangian: 1 GM 2 2GMm L = MR˙ 2 + MZ˙ 2 + MR2θ˙2 + mz˙2 + + (6) 2 2R pR2 + (z − Z)2 1 GM 2 2GMm = MR˙ 2 + MZ˙ 2 + MR2θ˙2 + mz˙2 + + (7) 2 2R pR2 + q2

does not depend on explicitly on the coordinate θ. This means that θ is an ignorable / cyclic coordinate, and hence the conjugate (generalized) momentum corresponding to θ, which, in this case, is the total angular momentum l of the system, is a conserved quantity. We have ∂L l = = 2MR2θ˙ = constant (8) ∂θ˙ The center of mass of the two suns lies on the z axis (and is of the same height as the two suns), which coincide with the planet. The overall center of mass of the system is given by m 2M Z = Q = z + Z. (9) com m + 2M m + 2M Together with the new coordinate q = z − Z, (10) we can solve for z and Z in terms of q and Q as

( 2M z = Q + m+2M q m . (11) Z = Q − m+2M q and hence we can write the Lagrangian as 1 GM 2 2GMm L = MR˙ 2 + MZ˙ 2 + MR2θ˙2 + mz˙2 + + (12) 2 2R pR2 + q2 m 2 1 2M 2 GM 2 2GMm = MR˙ 2 + M Q˙ − q˙ + MR2θ˙2 + m Q˙ + q˙ + + (13) m + 2M 2 m + 2M 2R pR2 + q2 m m 2 m 2M 2 GM 2 2GMm = MR˙ 2 + M + Q˙ 2 + M + q˙2 + MR2θ˙2 + + 2 m + 2M 2 m + 2M 2R pR2 + q2 (14) 1 Mm(m + 2M) GM 2 2GMm = MR˙ 2 + (m + 2M) Q˙ 2 + q˙2 + MR2θ˙2 + + (15) 2 p 2 | {z } (m + 2M) 2R R2 + q2 total mass 1 1 2Mm GM 2 2GMm = MR˙ 2 + (m + 2M)Q˙ 2 + q˙2 + MR2θ˙2 + + (16) 2 2 m + 2M 2R pR2 + q2

2 −1 1 1 1 1 GM 2 2GMm L = MR˙ 2 + (m + 2M)Q˙ 2 + + q˙2 + MR2θ˙2 + + (17) 2 2 2M m 2R pR2 + q2 | {z } 1 = µ 1 1 GM 2 2GMm = MR˙ 2 + (m + 2M)Q˙ 2 + µq˙2 + MR2θ˙2 + + (18) 2 2 2R pR2 + q2 where µ is the reduced mass of the system deﬁned by 1 1 1 = + (19) µ 2M m Now, since the Lagrangian does not depend explicitly on the coordinate Q, Q is therefore a cyclic / ignorable coordinate, and hence the conjugate momentum, which is the linear momentum PQ along the z direction, is a conserved quantity. Mathematically, we have ∂L PQ = = (m + 2M)Q˙ = constant, (20) ∂Q˙ Using Lagrange equation d ∂L ∂L − = 0, (21) dt ∂q˙i ∂qi we have the equation of motion for the R component as 2 d ˙ ˙2 GM 2GMmR (2MR) − 2MRθ − 2 − 3 = 0 (22) dt 2R (R2 + q2) 2 2 ¨ ˙2 GM 2GMmR 2MR − 2MRθ + 2 + 3 = 0 (23) 2R (R2 + q2) 2 2 2 ¨ l GM 2GMmR 2MR − 3 + 2 + 3 = 0 (24) 2MR 2R (R2 + q2) 2 and the equation of motion for the q component as d 2GMmq (µq˙) − − 3 = 0 (25) dt (R2 + q2) 2 2GMmq µq¨ + 3 = 0 (26) (R2 + q2) 2

Supplementary box 5.1: Conserved quantities for which coordinate? Some readers may try to rewrite the Lagrangian as

P 2 1 l2 GM 2 2GMm L = MR˙ 2 + Q + µq˙2 + + + (27) 2(m + 2M) 2 4MR2 2R pR2 + q2

If they do so, when they apply the Euler-Lagrange equation they will ﬁnd that the sign of l2 the term 4MR2 will be incorrect. What is wrong here?

Readers should note that when we are going through the logic that “since the Lagrangian is not depending explicitly on a speciﬁc coordinate qi, and hence the conjugate momentum ∂L pi = corresponding to that coordinate is a conserved quantity”, what we really mean ∂q˙i is that the conjugate momentum is a conserved quantity in time (i.e. it is not changing with time). However, it does not immediately mean that the conjugate momentum is not changing with respect to the other coordinates! For example, consider the total angular momentum l = 2MR2θ.˙ (28)

3 It is obvious that the angular momentum is changing with respect to the coordinate R. In particular, ∂l 2l = 4MRθ˙ = 6= 0 (29) ∂R R And hence this explains why one will not get the correct answer if they substitute l into the Lagrangian before applying the Lagrange equation. Again, readers are reminded to be careful about the “conserved quantity” we are talking about here are only conserved in time.

• (1 pt) : Stating that L does not depend on θ explicitly, and hence θ is ignorable / cyclic, and its conjugate momentum (the total angular momentum) is a conserved quantity. • (1 pt) : Solving l in terms of M, R and θ. • (2 pt) : Introducing the coordinate Q, and solve z and Z in terms of q and Q. • (1 pt) : Replace z, Z,z ˙ and Z˙ in the Lagrangian by q, Q,q ˙ and Q˙ . Note that if the student does not express the expression beforeq ˙2 as the reduced mass µ, credit should still be given. • (1 pt) : Stating that L does not depend on Q exlicitly, and hence Q is ignorable / cyclic, and its conjugate momentum (the linear momentum along the z direction) is a conserved quantity. • (1 pt) : Finding and expressing the equation of motion for r in terms of l and the other parameters. Note that if the term involving l does not have a correct sign, credit SHOULD NOT be awarded. • (1 pt) : Finding and expressing the equation of motion for q in terms of the other parameters. Total sub-points : 8

(c) In the limit of small planetary mass m M we can ignore the eﬀect of the planet on the motion of the suns. Find the explicit solution for the motion of the suns R(t) for orbits with small eccentricity . Write your solution as circular motion plus a term proportional to . [Solution] m If M → 0, the equation of motion for R can be written as

2 2 ¨ l GM 2GMmR 2MR − 3 + 2 + 3 = 0 (30) 2MR 2R (R2 + q2) 2 2 2 ¨ l G 2GR m R − 3 3 + 2 + 3 = 0 (31) M 2M R 2R (R2 + q2) 2 M | {z } ≈0 2 l2 G R¨ − + = 0. (32) M 2M 3R3 2R2

For circular orbit, we have R = R0 =constant and R˙ = R¨ = 0, hence the equation of motion for R can be written as l2 G 3 3 + 2 = 0 (33) 2M R0 2R0 l2 R = (34) 0 GM 3 For a slightly non-circular orbit, we can write

R(t) = R0(1 + f(t)) (35)

where , the eccentricity, is a small quantity, and f(t) embeds the information of the time evolution of the orbit. We call f(t) as the perturbation term to the orbit.

4 Note that under the assumption that is small, we have

1 1 −3 1 2 1 3 = 3 (1 + g) = 3 (1 − 3f + O( )) ≈ 3 (1 − 3f) (36) R R0 R0 R0 and 1 1 −2 1 2 1 2 = 2 (1 + g) = 2 (1 − 2f + O( )) ≈ 2 (1 − 2f). (37) R R0 R0 R0 Putting R(t) back to the equation of motion for R (equation 34), we have

2 2R0 ¨ l G f − 3 3 (1 − 3f) + 2 (1 − 2f) = 0 (38) M 2M R0 2R0 | {z } 2 3 l =GM R0 2R0 ¨ G G f − 2 (1 − 3f) + 2 (1 − 2f) = 0 (39) M 2R0 2R0 2R0 ¨ Gf f − 2 = 0 (40) M 2R0 GM f¨ = − f (41) 4R3 √ 0 GM f = A cos √ t + φ (42) 2R0 R0 where A and φ are constants to be determined by initial conditions. The general solution is given by √ GM R(t) = R0 1 + A cos √ t + φ (43) 2R0 R0

Assuming the initial radial velocity of the orbit of the suns is R˙ = 0, we have φ = 0. Hence the solution for R can be written as √ GM R(t) = R0 1 + A cos √ t (44) 2R0 R0

If we further assume that the initial radius of the orbit is R = R0(1 + ), then we have A = 1 and hence the solution can be written as √ GM R(t) = R0 1 + cos √ t (45) 2R0 R0 • (1 pt) : Showing / Arguing that the fourth term in the equation of motion for R is negligible if m << M.

• (2 pt) : Solving R0 in terms of l, G and M for the circular orbit case.

• (1 pt) : Expanding R(t) as the circular orbital radius R0 plus a perturbation term f. 1 1 • (2 pt) : Expanding R2 and R3 up to the ﬁrst order term of . • (3 pt) : Putting R(t) back into the equation of motion, solving the equation, and showing that f is a sinusoidal function. • (1 pt) : Correct ﬁnal expression for R(t). Note that students do not have to assume any initial conditions, but should keep the undetermined constants in the solution. Total sub-points : 10

5 (d) Hence show for these limits, keeping terms linear in , that the equation of motion for q is a nonlinear oscillator (oscillator with a non-quadratic potential) driven by a term f(q) cos Ωt with 2π/Ω the period of the solar motion, and ﬁnd the form of f(q). [Solution] In the regime of m << M, we have 1 1 1 1 = + ≈ (46) µ m 2M m

Hence we have the equation of motion for q as 2GMmq mq¨+ 3 = 0 (47) (R2 + q2) 2

Note that

√ 3 2 − 2 3 GM 2 2 − 2 2 2 (R + q ) = R0 1 + A cos √ t + φ + q (48) 2R0 R0 √ 3 − 2 2 2 2 GM 2 = (R0 + q ) + 2R0A cos √ t + φ + O( ) (49) 2R0 R0 √ 3 2 − 2 2 − 3 2R0 GM ≈ (R + q) 2 1 + A cos √ t + φ (50) 0 R2 + q2 2R R 0 0√ 0 2 2 − 3 3 2R0 GM 2 = (R + q) 2 1 − A cos √ t + φ + O( ) (51) 0 2 R2 + q2 2R R 0 √ 0 0 1 3R2 GM ≈ − 0 A cos √ t + φ . (52) 2 2 3 2 2 5 (R0 + q ) 2 (R0 + q ) 2 2R0 R0

Hence, the equation of motion for q can be written as √ 2GMm 6GMmR2 GM mq¨+ q − 0 qA cos √ t + φ = 0 (53) 2 2 3 2 2 5 (R0 + q ) 2 (R0 + q ) 2 2R0 R0

Using the initial conditions, we have A = 1 and φ = 0, which further simpliﬁes the equation of motion to the form √ 2GMm 6GMmR2 GM mq¨ + q − 0 q cos √ t = 0 (54) 2 2 3 2 2 5 (R0 + q ) 2 (R0 + q ) 2 2R0 R0 | {z } | {z } ≈Usual EOM for SHO non-linear force term

We can see that the equation of motion is composed of a usual “EOM for√ simple harnmonic oscillator” and a non-linear “driving force” in the form f(q) cos Ωt (Ω = GM√ ), where f(q) 2R0 R0 is given by 6GMmR2 0 q. (55) 2 2 5 (R0 + q ) 2 If you want to learn more about the complex dynamics (chaos) of this system, you can Google the “Slitnikov Problem”. One good reference is Hevia and Ranada, Eur. J. Phys. 17, 295 (1996). • (1 pt) : Arguing that for m << M, µ ≈ m. 2 2 − 3 • (2 pt) : Expanding (R + q ) 2 up to ﬁrst order of . • (1 pt) : Correct ﬁnal expression for the equation of motion of q. • (1 pt) : Correct expression for f(q).

Total sub-points : 5

6 For your interest only: The dynamics of this system given by the equation of motion you have calculated turns out to be complex even in the sensible limit m M. One way of illustrating the complexity is to consider the crossing times τ1,τ2 ... for which the planet crosses the plane of the orbit of the suns. It can be shown that for any chosen sequence of numbers (e.g. random numbers) an initial condition can be found such that the τi reproduce this sequence, with escape to infinity corresponding to the end of a finite sequence. Thus no matter how many τi are measured from observation of such a system, no prediction can be made for the next τ in the dynamics! These chaotic dynamics are a demonstration of the difficulty of analyzing the 3 body problem in gravitational physics. 2. The Yukawa potential: A particle of mass m moves in a force field described by the Yukawa potential k r V (r) = − r exp(− a ), where k and a are positive. In the limit a → ∞, this reduces to the Kepler problem. When a is finite, this results in an attenuation of the potential, and force, for r a; i.e., a short-range potential. It is commonly used to describe the potential between nucleons in the nucleus, with a ≈ 10−15 m.

(a) Write the equations of motion and reduce them to the equivalent one-dimensional problem, in terms of an eﬀective potential. [Solution] Assuming that the motion of the particle is restricted to a plane, we can use (r, θ) to describe the position of the particle. The Lagrangian for this problem is then given by 1 1 k r 1 1 L = mr˙2 + mr2θ˙2 + exp − = mr˙2 + mr2θ˙2 − V (r) (56) 2 2 r a 2 2

Since the Lagrangian does not depend on θ explicitly, θ is ignorable / cyclic and hence its conjugate momentum (which is the angular momentum l) is conserved. We have ∂L l = = mr2θ˙ (57) ∂θ˙ Using the Lagrange equation d ∂L ∂L − = 0, (58) dt ∂q˙i ∂qi the equation of motion for the r component is d ∂V (r) (mr˙) − mrθ˙2 − = 0 (59) dt ∂r ∂V (r) mr¨ − mrθ˙2 + = 0 (60) ∂r l2 ∂V (r) mr¨ − + = 0 (61) mr3 ∂r ∂ l2 ∂V (r) mr¨ − − − = 0 (62) ∂r 2mr2 ∂r ∂ l2 mr¨ − − − V (r) = 0 (63) ∂r 2mr2 | {z } −Veff ∂V mr¨ + eff = 0 (64) ∂r where the effective potential is given by l2 l2 k r V = + V (r) = − exp − (65) eff 2mr2 2mr2 r a • (2 pt) : Stating that the angular momentum l is a conserved quantity (either through the equation of motion, or by observing that θ is a cyclic / ignorable coordinate), and finding the correct expression for l.

7 • (1 pt) : Using the Lagrange equation to ﬁnd an equation of motion for r in terms of r and θ (and V (r)). • (1 pt) : Replacing the term involving θ˙ by l.

• (2 pt) : Writing the equation of motion for r in terms of the effective potential Veff, and writing down the correct expression for the effective potential. Total sub-points : 6

(b) Use the effective potential to identify and discuss the qualitative nature of orbits for different values of the energy E and the angular momentum `. [Solution] The effective potential is

l2 l2 k r V = + V (r) = − exp − (66) eﬀ 2mr2 2mr2 r a and the total energy is

1 l2 1 l2 k r E = mr˙2 + + V (r) = mr˙2 + − exp − (67) 2 2mr2 2 2mr2 r a

As usual, we compare the total energy E with the effective potential Veff. The difference between E 1 2 and Veff (at different r) gives the radial kinetic energy, i.e. 2 mr˙ , of the particle (at that particular position). If there is a minimum point for Veff, there will be bound states for the particle. Otherwise, there will be no bound states!

We look for solutions such that Veﬀ has local extremum points: dV l2 k r k r eﬀ = − + exp − + exp − = 0 (68) dr mr3 r2 a ar a l2k 1 m m r − r + r2 exp − = 0 (69) mr3 k l2 al2 a 1 m m r − r + r2 exp − = 0 (70) k l2 al2 a

r l2 If we write a = x and b = ma , we have 1 x x2 − + e−x = 0 (71) k b b b l2 x(x + 1)e−x = = (72) k mka From the graph of f(x) = x(x + 1)e−x from x = 0,

8 2 l r1 r2 we can see that if mka ≥ 0 is small enough, we will have two solutions, x1 = a and x2 = a (where ∂Veff we take for convenience x1 < x2), for ∂r = 0. The slope at these points on the curve f(x) denotes 0 the nature of these local extremum points for Veff. In particular, we see that f (x1) > 0, meaning 0 that x1 is a local minimum, and f (x2) < 0, meaning that x2 is a local maximum point. In this case there can be bound states:

(1) Firstly, we note that for 0 < E < Veff(x2), both bound states and unbound states can exist. For unbound states, they are either hyperbolic (if E < Veff(x2)) or parabolic (if E = Veff(x2)). See the figure below for a graphical explanation. For items (2) to (4), we will focus on the bound states, but readers are asked not to forget that unbound states can also happen. min (2) If E = Veff(x1) = Veff , the radial kinetic energy of the particle will be 0 at all times, and r can only be a fixed value at r = ax1 = r1. This corresponds to a bound circular orbit.

(3) If Veff(x1) < E < Veff(x2), the particle will stay in a bound elliptical orbit with aphelion and perihelion distance equal to intersection points of y = E and y = Veff.

(4) If E = Veﬀ(x2), there will be an unstable bound state (Still a circular orbit, but unstable) for the particle at r = ax2 = r2. A slight deviation of the particle from the position r = r2 will lead to an unbound state, which corresponds to a hyperbolic orbit.

(5) Finally, if E > Veﬀ(x2), we will have an unbound orbit, i.e. a hyperbolic orbit.

l2 As mka gets larger, the two solutions x1 and x2 will come closer and closer to each other until 0 they finally merge into a single point with slope f (x1 = x2) = 0. This corresponds to the point of inflexion for Veff. At this point, there will only an unstable bound state, and at other energies E we will have unbound states only.

9 l2 When mka is larger than the maximum value of f(x) (for x ≥ 0), there will be no solutions for ∂Veff ∂r = 0. This means that only unbound states can exist. In other words, there will be no bound states.

• (1 pt) : Correct expression for the total energy E. This is important since we are comparing E and Veff. • (1 pt) : Differentiating Veff with respect to r to find the local extremum points of Veff. comparing E and Veff. l2 • (2 pt) : Finding the relationship between the value of mka (if students do not use the parametrization suggested in the solution, credit should still be given with plausible com- ∂Veff parisons between r and l, k, a and m in the equation for ∂r = 0) and the number of local extremum points for Veff. • (4 pt) : For the two local extremum points case, explaining the nature of orbits for (1) E = Veff(x1), (2) Veff(x1) < E < Veff(x2), (3) E = Veff(x2) and (4) E > Veff(x2) AND 0 < E < Veff(x2). One point for each item. If the student does not state that there could be bound state even if E < Veff(x2), awards only 0.5 point for the last item. • (2 pt) : Explaining that when the two local extremum points merge into one, the point becomes a point of inflexion and hence only one unstable bound state can exist, otherwise the orbit will be unstable. ∂Veff • (1 pt) : Stating there will only be unbound states if there is no solution to ∂r = 0. Total sub-points : 11

(c) In the limit a → ∞ (the Kepler problem), give expressions for the radius of the circular orbit rc, the orbital frequency (inverse of the orbital period), and the energy, in terms of `. k, and m. [Solution] As a → ∞, we have k V ≈ − (73) r and l2 k V ≈ − (74) eﬀ 2mr2 r

10 The equation of motion for r becomes l2 ∂ k mr¨ − + − = 0 (75) mr3 ∂r r l2 k mr¨ − + = 0 (76) mr3 r2

For circular orbit, r = rc =Constant, and hencer ˙ =r ¨ = 0. We have l2 k − 3 + 2 = 0 (77) mrc rc l2 r = (78) c mk Recall that the angular momentum of the particle is l = mr2θ˙ = mr2ω (79) where l l mk2 ω = 2 = 2 = 3 (80) mr mrc l is the orbital angular velocity. From this, the orbital frequency is given by ω mk2 f = = (81) 2π 2πl3 The energy is given by 2 2 2 2 1 2 l k mk mk mk E = r˙ +Veﬀ(r = rc) = 2 − = 2 − 2 = − 2 (82) 2 2mr rc 2l l 2l |{z} c =0 as expected.

Supplementary Box 5.2 : Why “as expected”? The energy E for the circular orbit we calculated is

mk2 1 E = − = −T = V (83) 2l2 2 Let’s consider the situation when an object m is orbiting around a planet of mass M (with M >> m in a circular orbit with radius R. We have the gravitational force acting on the object by the planet equals to the centripetal force, i.e.

mv2 GMm = (84) R R2 r GM v = (85) R The kinetic energy of the particle is 1 GMm T = m|v|2 = (86) 2 2R The potential energy of the particle is GMm V = − = −2T (87) R Hence, we see that T = −2V (88)

11 This is actually a special case of the Virial theorem. We interpret virial theorem for our discussion as: If the orbit of the particle under a central force is bound and periodic (this is not a necessary condition, but for simplicity we assume this is true), then we will always have 1 ∂V < T >= − < r > (89) 2 ∂r where the expression < ... > means the mean of the quantity. If the potential V is a power-law function of r, i.e. V = arn+1, we will have 1 1 n + 1 < T >= − < a(n + 1)rn × r >= − (n + 1) < arn+1 >= − < V > (90) 2 2 2 Furthermore, if the central force is an inverse square law, i.e. V = ar−1 (or n = 0), we have 1 < T >= − < V > (91) 2 which is the result we obtained here. If readers wish to learn more about Virial theorem, they may refer to Goldstein section 3.4.

k • (1 pt) : Reducing the Yukawa potential to V = − r . • (2 pt) : Solving rc from the equation of motion for r. • (1 pt) : Correct expression for f. • (1 pt) : Correct expression for E. Total sub-points : 5

(d) For the Yukawa potential, give the equation whose solution yields the radius of the circular orbit rc,a in terms of `. k, m and a. [Solution] For the Yukawa potential, the equation of motion for r is l2 ∂V (r) mr¨ − + = 0 (92) mr3 ∂r l2 ∂ k r mr¨ − + − exp − = 0 (93) mr3 ∂r r a l2 k r k r mr¨ − + exp − + exp − = 0 (94) mr3 r2 a ar a

For circular orbit, we have r = rc,a =Constant andr ˙ =r ¨ = 0, so the equation of motion for r becomes 2 l k rc,a k rc,a − 3 + 2 exp − + exp − = 0 (95) mrc,a rc,a a arc,a a l2 r r = kr 1 + c,a exp − c,a (96) m c,a a a

Solving the above equation will give the radius of the circular orbit rc,a. • (1 pt) : Putting in the Yukawa potential into the equation of motion for r. Full credit should also be given.

• (2 pt) : Implementing r = rc,a andr ¨ = 0 into the equation, yielding the correct ﬁnal form of the equation. Note that students may be have the same form of equation as the solution. As long as they have implemented the conditions stated, credit should be given. Also note that ∂Veff students may also approach the problem through evaluating ∂r |r=rc,a . Full credit should also be given.

12 Total sub-points : 3

(e) In the limit rc/a 1, solve the equation from part (d) by approximation, to give rc,a in terms of rc and a. You can assume rc,a = rcg, where g is a function of rc/a (and which should go to 1 in the limit a → ∞). Find g to leading order in rc/a. Also ﬁnd the orbital frequency in this case. [Solution] l2 Students are reminded that rc = mk is not equal to rc,a. We have the equation as

l2 r r = kr 1 + c,a exp − c,a (97) m c,a a a r r kr = kr 1 + c,a exp − c,a (98) c c,a a a r r r = r 1 + c,a exp − c,a (99) c c,a a a

If we assume rc,a = rcg, we have the equation as

gr gr r = gr 1 + c exp − c (100) c c a a gr gr 1 = g 1 + c exp − c (101) a a

rc In the regime that a << 1, we have gr gr g2r2 g3r3 1 = g 1 + c 1 − c + c + O c (102) a a 2a2 a3 gr gr g2r2 1 ≈ g 1 + c 1 − c + c (103) a a 2a2 g2r2 g3r3 1 = g 1 − c + O c (104) 2a2 a3 g2r2 1 ≈ g 1 − c (105) 2a2 g2r2 −1 g = 1 − c (106) 2a2 g2r2 g4r4 g = 1 + c + O c (107) 2a2 4a4 g2r2 g ≈ 1 + c (108) 2a2 Hence we have g2r2 c − g + 1 = 0 (109) 2a2 q 2 2rc 1 ± 1 − a2 g = 2 (110) rc a2 2 2 r 2 a a 2rc g = 2 ± 2 1 − 2 (111) rc rc a 2 2 2 4 6 a a 1 2rc 1 4rc 8rc g = 2 ± 2 1 − 2 − 4 O 6 (112) rc rc 2 a 8 a a

13 2 2 2 4 a a rc rc g ≈ 2 ± 2 1 − 2 − 4 (113) rc rc a 2a 2 2 2 rc 2a rc g = 1 + 2 or 2 − 1 − 2 (rejected) (114) 2a rc 2a Hence, we have r2 r = r g = r 1 + c (115) c,a c c 2a2 Again, recall that the angular momentum is given by

l = mr2θ˙ = mr2ω (116)

where l l ω = 2 = 2 (117) mr mrc,a is the orbital angular velocity. From this, the orbital frequency is given by ω l l f = = = (118) 2 r2 2 2π 2πmrc,a 2 c 2πmrc 1 + 2a2

2 • (1 pt) : Replacing l by mkrc, and substituting rc,a by grc in the equation obtained in part (d). grc • (1 pt) : Expanding exp(− a ) up to the second order term. If students keep only up to the ﬁrst order term, award only 0.5 point. grc • (2 pt) : Simplifying the equation, and keeping terms only up to the second order of a . grc • (2 pt) : Solving for g, keeping terms only up to the second order of a . Note that students should not keep the negative solution for g since we require g → 1 when a → ∞.

• (2 pt) : Obtaining the correct expression for the orbital frequency in terms of l and rc. Total sub-points : 8

(f) If the orbit is nearly circular, i.e. r(t) = rc,a + δr(t)(δr(t) r0), derive the differential equation for δr(t). Show that it is stable, and corresponds to oscillations about the circular orbit, and find the frequency of oscillation. compare to the orbital frequency. [Solution] Note that the equation of motion for r is given by ∂V mr¨ = − eff (119) ∂r

In part (c) and (d), we evaluate the above equation at r = rc or r = rc,a such that either terms in the above equation is 0 (since the radius is fixed,r ¨ = 0; also, Veff is minimized at r = rc or r = rc,a, and hence its first derivative evaluated at these radii will vanish). Now, for a slightly non-circular orbit, its radius is given by

r(t) = rc,a + δr. (120)

We can also evaluate the equation of motion for r at this radius. The left hand side is simply d2 mr¨ = m (r + δr) = mδr¨ (121) dt2 c,a The right hand side becomes

∂Veﬀ ∂Veﬀ = (122) ∂r ∂r r=rc,a+δr

14 We can approximate this term by doing a Taylor expansion about r = rc,a as

2 ∂Veff ∂Veff ∂ Veff ≈ + (r − ra,c) (123) ∂r ∂r ∂r2 r=rc,a+δr r=rc,a r=rc,a | {z } =0 2 ∂ l k r k r = − + exp − + exp − δr (124) ∂r mr3 r2 a ar a r=rc,a 2 3l 2k r 2k r k r = − exp − − exp − − exp − δr (125) mr4 r3 a ar2 a a2r a r=rc,a 2 3l k 2 2 rc,a = 4 − 2 3 (2a + 2arc,a + rc,a) exp − δr (126) mrc,a a rc,a a

Since Veﬀ is minimized at r = rc,a, the second derivative of Veﬀ evaluated at r = rc (as well as its neighbouring region) should be > 0. Hence, we can see that the solution for δr is a stable solution. Hence, the required equation is

2 ¨ 3l k 2 2 rc,a mδr = − 4 − 2 3 (2a + 2arc,a + rc,a) exp − δr (127) mrc,a a rc,a a 2 ¨ 3l k 2 2 rc,a δr = − 2 4 − 2 3 (2a + 2arc,a + rc,a) exp − δr (128) m rc,a ma rc,a a

From the above equation, we can obtain the “angular velocity” of the changing of the orbital radius as

s 2 3l k 2 2 rc,a Ω = 2 4 − 2 3 (2a + 2arc,a + rc,a) exp − (129) m rc,a ma rc,a a s l mkrc,a 2 2 rc,a = 2 3 − 2 2 (2a + 2arc,a + rc,a) exp − (130) mrc,a a l a s mkr r = ω 3 − c,a (2a2 + 2ar + r2 ) exp − c,a (131) a2l2 c,a c,a a and hence the frequency of oscillation is s Ω mkr r F = = f 3 − c,a (2a2 + 2ar + r2 ) exp − c,a (132) 2π a2l2 c,a c,a a

• (1 pt) : Showing that mr¨ = mδr¨ . ∂Veff • (1 pt) : Arguing / stating that ∂r is 0 when evaluated at r = rc,a 2 ∂Veff ∂ Veff • (2 pt) : Expanding ∂r around r = rc,a up to ∂r2 . Note that if students uses the result that ∂Veff ∂r is 0 when evaluated at r = rc,a here directly without stating, the credit for the previous item should still be awarded. 2 ∂ Veff • (1 pt) : Arguing that since ∂r2 < 0 around r = rc,a, the solution to δr is stable. • (1 pt) : Obtaining the “angular velocity” from the resulting equation. • (1 pt) : Expressing the frequency of oscillation in terms of the orbital frequency for a circular orbit f. Note that the students’ ﬁnal expression inside the square root might be diﬀerent from the solution. As long as they have the “f” dependence in their solutions, credit should be awarded. Total sub-points : 7

15 (g) Think about how you might use observations of planetary motion to measure or limit the value of a. You might also want to think about what physics might result in a Yukawa potential. [Solution] Note that the frequency of oscillation of the orbit radius is NOT EQUAL to the orbital frequency. This implies that when the particle ﬁnishes a complete cycle around the center (focus) (i.e. when θ(Torbital) = 2π, where Torbital is the orbital period), it will not be at the same position as it was initially (i.e. r(Torbital 6= r(0)). Such a diﬀerence will cause the orbit to move (rotate) in time, a phenomenon known as apsidal precession.

Since the orbit is moving (rotating), the perihelion is also moving in time, which we call “advancing”. Perihelion advancement is observed for mercury and other celestial objects. Indeed, physicists have been investigating the deviation of short-range gravitational force from the usual Newtonian gravitational force by adding a assumed-Yukawa-like potential1 to the usual Newtonian potential. By measuring the apsidal precession rate of celestial objects, and comparing with the theoretically calculated results, one can provide somewhat a constraint to the value of a in the Yukawa potential. In fact, some physicists also postulated the existence of a “ﬁfth force”, aside from the weak, strong, electromagnetic and gravitational forces, in the form of the Yukawa potential 2. • (1 pt) : Arguing / Stating / Explaining that the diﬀerence in orbital frequency and frequency of oscillation of the orbital radius will lead to apsidal precession. • (1 pt) : Suggesting that measuring precession rates of orbits and comparing with theoretical results may provide constraints on a, or other reasonable suggestions.

Total sub-points : 2

3. Precessing ellipses: Hand and Finch Problem 4-24 (slightly changed). Discuss the motion of a particle in a central force potential k β V (r) = − + . r r2 In particular, show that the equation of the orbit has an exact solution in the form p = 1 + cos αφ . r This is an ellipse for α = 1, but it is a precessing ellipse if α 6= 1. The precessing motion may be described in terms of the rate of precession of the apsides (turning points). Derive an approximate expression for the rate of precession when α is close to 1. Calculate the precession angle for one period of the (almost)

1See https://link.springer.com/article/10.1134/S1063772917050092 for example. 2See https://physicsworld.com/a/microscope-focuses-tighter-limits-on-a-fifth-force/ for example.

16 elliptical motion. If β is increased to the point where it is no longer small compared to the centrifugal term, how does this aﬀect the orbit? [Solution] Here, we assume the mass M of the source of the central force is much larger than the particle’s mass m. In general, one should replace the m in the following discussion by the reduced mass µ. As usual, we ﬁrst write down the Lagrangian of the particle as 1 1 k β L = mr˙2 + mr2φ˙2 + − (133) 2 2 r r2

Again, since the Lagrangian does not depend explicitly on the coordinate θ, its conjugate momentum (which is the total angular momentum l) is a time-conserved quantity. The angular momentum can be found by ∂L l = = mr2φ˙ (134) ∂φ˙ Using the Lagrange equation d ∂L ∂L − = 0, (135) dt ∂q˙i ∂qi the equation of motion for r can be calculated as d k 2β (mr˙) − mrφ˙2 − + = 0 (136) dt r2 r3 l2 k 2β mr¨ − + − = 0 (137) mr3 r2 r3

Note that dφ l = mr2φ˙ = mr2 (138) dt mr2 dt = dφ (139) l d l d = (140) dt mr2 dφ

Hence we can write the equation of motion for r as l2 k 2β mr¨ − + − = 0 (141) mr3 r2 r3 d d l2 k 2β m (r) − + − = 0 (142) dt dt mr3 r2 r3 l d l d l2 k 2β m (r) − + − = 0 (143) mr2 dφ mr2 dφ mr3 r2 r3 l2 d r0 l2 k 2β − + − = 0 (144) mr2 dφ r2 mr3 r2 r3 l2 (r2)(r00) − (r0)(2rr0) l2 k 2β − + − = 0 (145) mr2 r4 mr3 r2 r3 l2 r00 2l2 r02 l2 k 2β − − + − = 0 (146) m r4 m r5 mr3 r2 r3 where the “prime” denotes a diﬀerentiation with respect to φ. 1 If we let u = r , we have

( 0 u0 r = − u2 2 00 0 0 (147) 00 (u )(u )−u (2uu ) u00 2u02 r = − u4 = − u2 + u3

17 And hence the equation of motion for r can be written as

l2 u00 2u02 2l2 u0 2 l2 − + u4 − − u5 − u3 + ku2 − 2βu3 = 0 (148) m u2 u3 m u2 m l2 2l2 2l2 l2 − u2u00 + uu02 − uu02 − u3 + ku2 − 2βu3 = 0 (149) m m m m l2 − (u2u00 + u3) + ku2 − 2βu3 = 0 (150) m l2 − (u00 + u) + k − 2βu = 0 (151) m l2 l2 u00 + + 2β u = k (152) m m

The equation we obtained is a second-order inhomogeneous diﬀerential equation in u. First we solve for the homogeneous solution for the equation l2 l2 u00 + + 2β u = 0 (153) m m by using the trial solution u = esφ. We have l2 l2 s2 + + 2β = 0 (154) m m 2βm s2 = − 1 + (155) l2 r 2βm s = ±i 1 + (156) l2 which is imaginary. This corresponds to sinusoidal functions. Therefore, the homogeneous solution can be written as r 2βm u = A cos 1 + φ + γ (157) h l2 where A and γ are constants to be determined by initial conditions. Now, we observe that the right hand side of the inhomogeneous equation is a constant k. We are hence prompted to guess a particular solution in the form

up = C = constant (158)

Substituting up into the diﬀerential equation, we get l2 + 2β C = k (159) m 2βm + l2 C = k (160) m km C = (161) 2βm + l2

Hence, the general solution can be written as r 2βm km u = u + u = A cos 1 + φ + γ + (162) h p l2 2βm + l2 | {z } | {z } α = km α2l2 km = A cos αφ + γ + (163) α2l2

18 where we have denoted r 2βm 1 + = α (164) l2 If we assume that the initial change in u with respect to φ is 0, i.e.

∂u = 0 (165) ∂φ φ=0 Then we have

0 = −Aα sin(γ) (166) γ = 0 (167) for non-trivial solution. Hence, we can write 1 km u = = A cos(αφ) + (168) r α2l2 α2l2 1 Aα2l2 = cos(αφ) + 1 (169) km r km | {z } | {z } p p = 1 + cos(αφ) (170) r

Let’s consider the apsides (the perihelion) of a sinusoidal function. We have, from the previous equation, that p r = (171) 1 + cos(αφ)

Apisides are posiitons when the particle is furthest away from the “center” (focus), i.e. r is maximized. We can easily see that the maximum r happens when the denominator of the above equation is the smallest, i.e. cos(αφ) = −1, (172) which corresponds to αφ = (2n + 1)π (n=0, 1, 2...) (173)

Hence, the positions of the apisides are given by 2n + 1 φ(n) = π (174) α

The precession rate is then given by

2(n + 1) + 1 2n + 1 2π δφ = φ(n + 1) − φ(n) = π − π = (175) α α α

When α → 1, we have 2βm → 0 (176) l2 and hence the precession rate can be approximated as

− 1 2βm 2 δφ = 2π 1 + (177) l2 1 2βm 4β2m2 = 2π 1 − + O (178) 2 l2 l4 βm 2βm ≈ 2π 1 − = − π (mod 2π) (179) l2 l2

19 When β is no longer small, i.e. α is no longer close to 1, then the precession rate δφ (mod 2π) will not be regular anymore. The ﬁgures below show the plot of r(φ) with various value of α.

• (2 pt) : Writing down the equation of motion for r in terms of r, l, k and β. • (2 pt) : Rewrite the equation of motion in terms of r0. • (4 pt) : Solving r in terms of θ. • (2 pt) : Identifying α and in the solution for r. • (2 pt) : Correct derivation and expression for the position of the apisides. • (2 pt) : Correct precession rate for α → 1 • (1 pt) : Explaining what will happen if β is no longer small.

Total sub-points : 15

4. Hyperbolic orbits - asymptotes and impact parameters Hand and Finch Problem 4-27. A two-body system with reduced mass µ and orbital angular momentum l in a potential V (r) = ±k/r with k > 0, with eccentricity > 1, the (unbound) orbit is described in Cartesian coordinates as: p p = − x2 + y2 + x , where p ≡ l2/(µk) is ﬁxed by the initial conditions. The asymptotes (for both attractive and repulsive forces) are straight lines. (a) Prove that the equations for those lines in Cartesian coordinates for a repulsive force will have the general form p p y = ± 2 − 1 x − √ . 2 − 1 The asymptotes for an attractive inverse-square force obey a similar equation, but with → −. [Solution] We start from the orbit equation: p p = − x2 + y2 + x (180) x2 + y2 = (x − p)2 (181) x2 + y2 = 2x2 − 2px + p2 (182) [(2 − 1)x2 − 2px] − y2 = −p2 (183) 2p y2 p2 x2 − x − = − (184) 2 − 1 2 − 1 2 − 1 p 2 p22 y2 p2 x − − − = − (185) 2 − 1 (2 − 1)2 2 − 1 2 − 1 p 2 y2 p2 x − − = (186) 2 − 1 2 − 1 (2 − 1)2 2 p 2 x − 2 y −1 − = 1 (187) p2 p2 (2−1)2 2−1

20 Comparing to the hyperbola equation

(x − x )2 (y − y )2 0 − 0 = 1, (188) a2 b2 we see that the “center” of the hyperbolas is

p (x, y) = , 0 (189) 2 − 1

The asymptotes of hyperbolas are straight lines passing through the center. To ﬁnd the slope(s) of the asymptotes, ﬁrst note that the orbit equation can be written in terms of polar coordinates as

p = −r + r cos θ (190) p = −1 + cos θ (191) r As r → ∞, we have

0 = −1 + cos θ (192) 1 cos θ = (193) sec θ = (194) sec2 θ = 2 (195) 1 + tan2 θ = 2 (196) tan2 θ = 2 − 1 (197) p tan θ = ± 2 − 1 (198)

Hence, the equation of asymptotes are

y − 0 p 2 p = ± − 1 (199) x − 2−1 p p y = ± 2 − 1x ∓ √ (200) 2 − 1 • (1 pt) : Deriving the slope of the asymptotes. • (2 pt) : Deriving the center of the hyperbolas. • (1 pt) : Deriving the equations of the asymptotes. Total sub-points : 4

(b) The impact parameter b is defined as the distance of closest approach to the origin along the incoming asymptote. Prove that the angular momentum l is given in terms of the reduced mass µ and the center of mass relative velocity v∞ infiniely far away from the origin by the formula l = µv∞b. [Solution] As we have seen for many times, since under the proposed central force, the Lagrangian does not depend explicitly on the angle θ, its conjugate momentum, which is the angular momentum, is a time-conserved quantity. Hence, wherever we evaluate the angular momentum does not affect its value, so we can calculate its angular momentum at infinity, which is simply given by

l = µv∞b. (201)

• (1 pt) : Arguing that the angular momentum is a conserved quantity. • (1 pt) : Explaining why the angular momentum can be given by the expression suggested in the question.

21 Total sub-points : 2

(c) Show that the impact parameter can be written as a function of for constant total energy in the center of mass, E: k p b() = 2 − 1 . 2E [Solution] Initially, the particle is at inﬁnity and hence the only energy in the system is the kinetic energy of the particle. The energy at any position r is given by

1 l2 k E = µr˙2 − ± . (202) 2 2mr2 r Since energy is conserved, we have 1 E = µv2 (203) 2 ∞ s 2E v = . (204) ∞ µ

Hence, the angular momentum can be written as s 2E l = µv b = µb = bp2µE. (205) ∞ µ

Now, we wish to ﬁnd b in terms of p and . Recall that b is the closest approach to the origin along the incoming asympote. We ﬁnd the shortest distance√ from the origin (0, 0) to the asymptote. Any point on the asymptote can be denoted by (x, 2 − 1x − √ p . The distance s between the origin 2−1 and the line is given by

s 2 r p p 2p2 s = x2 + 2 − 1x − √ = 2x2 − 2px + = pf(x) (206) 2 − 1 2 − 1

To ﬁnd the closest distance, we simply have to ﬁnd the minimum value of f(x), i.e.

df = 0 (207) dx p 22x − 2p = 0x = (208) Putting this value of x back into s, we will obtain b as s p2 p 2p2 s = b = 2 − 2p + (209) 2 − 1 r 2p2 = p2 − 2p2 + (210) 2 − 1 r p2 = (211) 2 − 1 r 1 = p (212) 2 − 1 l2 r 1 = (213) µk 2 − 1

22 √ With l = b 2µE, we have √ (b 2µE)2 r 1 b = (214) µk 2 − 1 2b2E r 1 b = (215) k 2 − 1 k p b = 2 − 1 (216) 2E

• (2 pt) : Expressing v∞ in terms of E and µ. • (3 pt) : Expressing b in terms of l, µ and . • (2 pt) : Deriving the ﬁnal expression. Total sub-points : 7

Maximum points obtainable for this problem set: 81 pt