Plane and Spherical Pendulums

All spherical pendulums are similar to each other. Upon appropriate non-dimensionalization of time and action, they are described by one and the same dimensionless Lagrangian

θ˙2 ϕ˙ 2 L = + sin2 θ + cos θ , (1) 2 2 where θ and ϕ are the polar and azimuthal angles, respectively. Consistent with the axial rotational symmetry of the problem, the variable ϕ is cyclic and corresponding equation of motion yields the law of conservation of the z-component of the (non-dimesionalized) angular momentum, lz:

2 ϕ˙ sin θ = lz . (2)

This provides us with a simple relation between ϕ(t) and θ(t):

l Z dt ϕ˙ = z ⇔ ϕ(t) = l . (3) sin2 θ z sin2 θ(t)

This relation (and many other relations as well, as we will see in what follows) reveals a deep qualitative analogy between the problem of spherical pendulum and the problem of a two-dimensional particle performing a bound motion in a central potential. In both cases, we have a cyclic (azimuthal) angle associated with the rotational symmetry; the behavior of pendulums’s polar angle θ is qualitatively equivalent to that of the polar radius of the 2D particle trapped in a central potential.1 The Lagrange equation for θ is

θ¨ =ϕ ˙ 2 sin θ cos θ − sin θ . (4)

Relation (3) allows us to excludeϕ ˙ from (4), thus getting a closed equation of motion for θ(t). It is both convenient and instructive to directly arrive at the ﬁrst integral of the resulting closed equation of motion from the conservation of energy, in a direct analogy with our treatment of a 2D particle in a central potential. To this end we ﬁnd the expression for the (dimensionless) energy. We can do that directly from the Lagrangian:

∂L ∂L θ˙2 ϕ˙ 2 ε = θ˙ +ϕ ˙ − L ⇒ ε = + sin2 θ − cos θ . (5) ∂θ˙ ∂ϕ˙ 2 2 Using (3) to excludeϕ ˙, we get a ﬁrst-order closed diﬀerential equation for θ(t)

θ˙2 l2 dθ q + z − cos θ = ε ⇒ = ± 2ε + 2 cos θ − l2/ sin2 θ . (6) 2 2 sin2 θ dt z We see that our problem is equivalent to the problem of a one-dimensional particle of the unit mass moving in the dimensionless eﬀective potential

l2 U (θ) = z − cos θ . (7) eﬀ 2 sin2 θ We already developed a detailed theory of such motion and can readily write down the answers.

1Note, however, an interesting quantitative (or semi-qualitative, if you will) aspect of Eq. (3) taking place at θ > π/2, where the denominator decreases with increasing θ, and even diverges at θ → π.

1 Our first observation is that there are two quite different cases: a generic case lz 6= 0 and a special case lz = 0. The special case lz = 0 corresponds to the regime when a spherical pendulum behaves like a plane pendulum. We will discuss this case in a separate section dedicated to the plane pendulum. From now on we will be assuming that lz 6= 0 so that we necessarily have the centrifugal part of the effective potential:

l2 z > 0 (centrifugal term) . (8) 2 sin2 θ The centrifugal term diverges at θ → 0 and at θ → π, implying that the motion of θ is necessarily bound and thus necessarily periodic (the subscripts “p” and “a” for the two turning points allude to the perigee and apogee): θ ∈ [θp, θa] , 0 < θp ≤ θa < π , (9) where the angles θp and θa obey the condition Ueﬀ (θp,a) = ε. Hence,

2 lz 2 − cos θp,a = ε . (10) 2 sin θp,a For practical calculations, it is convenient to use the variable

χ = cos θ. (11)

From (6) we ﬁnd the equation of motion for χ(t) dχ q Z dχ = ± 2(ε + χ)(1 − χ2) − l2 ⇒ t = ± . (12) z p 2 2 dt 2(ε + χ)(1 − χ ) − lz

Equation (10) for the turning points χp,a = cos θp,a now reads 2 2 2(ε + χp,a)(1 − χp,a) − lz = 0 . (13) This is a cubic equation, so that the number of roots in a general case is three. The two physical roots satisfy the condition −1 < χp,a < 1 , see the deﬁnition (11) and the inequalities (9). Representing Eq. (14) as 1 = (2/l )(ε + χ) 1 − χ2 z and “solving” it graphically within the interval |χ| < 1, we clearly see that there are no more than two physical solutions. For the half-period of the motion of χ we have [see Eq. (12)]

Z χa dχ t = . (14) half-period p 2 2 χp 2(ε + χ)(1 − χ ) − lz In terms of the variable χ, equation (3) reads dϕ l = z . (15) dt 1 − χ2 Excluding the time variable between the diﬀerential equations (12) and (15), we get the diﬀerential equation for the “orbit” ϕ(χ) dϕ l = ± z . (16) 2 p 2 2 dχ (1 − χ ) 2(ε + χ)(1 − χ ) − lz

2 The solution to this equation is the integral Z dχ ϕ = ± l . (17) z 2 p 2 2 (1 − χ ) 2(ε + χ)(1 − χ ) − lz

An important quantity is the angle ϕ0, the change of the azimuthal angle during the half-period of the motion of χ: Z χa dχ ϕ = ± l . (18) 0 z 2 p 2 2 χp (1 − χ ) 2(ε + χ)(1 − χ ) − lz In a general case, the angle ϕ0 is incommensurate with 2π, so that the motion of the spherical pendulum is quasi-periodic rather that genuinely periodic. As a result, the phenomenon of ergodicity takes place: In the asymptotic limit t → ∞, the system explores all possible pairs (θ, ϕ) consistent with given lz and ε. For a visualization of the ergodic motion of the spherical pendulum, Google “Spherical Pendulum Animation” and enjoy a nice video on YouTube; make sure you are watching an animation that shows the “orbit.”

Regime of uniform rotation. The special regime of uniform rotation,

lz θ(t) ≡ θ0 , ϕ(t) = 2 t (uniform rotation), (19) sin θ0 takes place if the energy is exactly equal to the minimal value of the effective potential, ε = ε0, 2 lz ε0 = Ueff (θ0) = 2 − cos θ0, (20) 2 sin θ0 4 0 sin θ0 2 Ueff (θ0) = 0 ⇒ = lz. (21) cos θ0

Perturbed uniform rotation. If the energy is only slightly larger than ε0, that is (in our dimensionless units) ε − ε0 1 , then we have a regime of perturbed uniform rotation: The angle θ(t) performs small-amplitude2 harmonic oscillations about the equilibrium value θ0.

Regime of small oscillations. This regime takes place when

θ(t) 1. (22)

Conﬁning our description to the leading order in the small parameter (22), we Taylor-expand La- grangian (1) in powers of θ, keeping only the leading relevant terms. Omitting the irrelevant constant, we get θ˙2 ϕ˙ 2 1 L = + θ2 − θ2 . (23) small 2 2 2 With the same level of accuracy, for the dimensionless Cartesian coordinates x and y we have

x = θ cos ϕ, y = θ sin ϕ.

In terms of the coordinates x and y, our Lagrangian is x˙ 2 y˙2 x2 + y2 L = + − . (24) small 2 2 2

2 Such that |θ(t) − θ0| 1.

3 We see that our problem is equivalent to the problem of a two-dimensional particle in the rotationally symmetric harmonic potential. We remember that this problem is special: The orbit is closed and thus there is no ergodicity. The ergodicity of the original problem here develops only as a small perturbation on top of the Lagrangian (24), resulting in a slow precession of the elliptic orbit.

Plane Pendulum

The non-dimensionalized Lagrangian of the plane pendulum reads

θ˙2 L = + cos θ , (25) 2 where θ is the angle with respect to the vertical axis having the same direction as the vector of the free-fall acceleration. Note that the Lagrangian (25) can be also obtained from the Lagrangian of the spherical pendulum, Eq. (1), by settingϕ ˙ ≡ 0, meaning that the plane pendulum model also describes the motion of the spherical pendulum in the special case of lz = 0. As opposed to the case of the spherical pendulum with lz 6= 0, where the domain of deﬁnition of the angle θ was restricted to the open interval (0, π), here it makes a perfect sense to extend the domain of deﬁnition of θ to the full number axis:

θ ∈ (−∞, +∞) . (26)

This way we explicitly capture the reﬂectional symmetry of the system, θ → −θ, and also have the most natural way of describing the rotational motion of the pendulum. In such a setup, our problem becomes explicitly equivalent to that of a one-dimensional particle (of the unit mass) with the Cartesian coordinate x ≡ θ moving in a periodic external potential U(x) = − cos(x). We have the motion, θ¨ = − sin θ , (27) and its ﬁrst integral in the form of energy conservation law:

θ˙2 dθ √ √ − cos θ = ε ⇒ = ± 2 ε + cos θ . (28) 2 dt The solution to this equation can be written in the integral form 1 Z dθ t = ±√ √ . (29) 2 ε + cos θ The integral in the r.h.s. is a standard special function. At ε > 1, the motion is unbound. Here Eq. (28) tells us that θ˙ cannot nullify, meaning that the pendulum keeps rotating in one and the same direction. At ε < 1, the motion is bound. The angle θ oscillates between the two values (turning points), ±θ∗, corresponding to the condition θ˙ = 0:

cos θ∗ = −ε . (30)

With Eq. (30), the integral (29) can be parameterized as

1 Z dθ t = ±√ √ (oscillatory regime) . (31) 2 cos θ − cos θ∗

4 The expression for the quarter of the period of oscillations then is3

1 Z θ∗ dθ 1 Z θ∗ dθ tquarter-period = √ √ = √ √ . (32) 2 0 ε + cos θ 2 0 cos θ − cos θ∗

The period diverges at → 1 (θ∗ → π). The case ε = 1 is special. Here we have the unstable equilibrium solution θ(t) = π and two similar—related to each other by time-reversal transformation—time-dependent solutions approach- ing ±π (or ∓π) at t → ±∞. At ε = 1, the integral (29) can be performed analytically. With the useful substitution y = sin(θ/2) ⇒ dy = (dθ/2) cos(θ/2) we reduce this integral to a standard one: 1 Z dθ Z dy 1 1 + y √ √ = = ln + C . 2 ε + cos θ 1 − y2 2 1 − y Absorbing the free integration constant C into the time variable, we get 1 + y ln = ± 2t ⇒ y = ± tanh t , 1 − y

θ(t) = ± 2 arcsin(tanh t). (33) The moment t = 0 corresponds to the angle θ = 0. The sign controls the sense of rotation. Of special interest is the behavior of the solution (33) in the limit |t| → ∞ when we have |θ| → π. To explore this limit, we represent Eq. (33) as sin(θ/2) = ± tanh t. (34) For the sake of deﬁniteness, we then chose the sign “+” and explore the limit t → +∞, in which case θ → +π. Introducing small positive δθ, such that

θ = π − δθ ,

We expand the l.h.s. of (34): (δθ)2 sin(θ/2) ≈ 1 − . 8 Finally we take into account that

tanh t ≈ 1 − 2e−2t at t 1, and arrive at the following asymptotic form of our solution

θ(t) → π − 4e−t at t → +∞ . (35)

This result is perfectly consistent with our general analysis of the motion of a one-dimensional particle in an external potential in the vicinity of the point of unstable equilibrium, where the Taylor expansion of the potential at the point of its maximum allows us to readily solve the equation of motion.

3The combination of time-reversal and reﬂectional symmetries implies that the piece of the trajectory from θ = 0 to θ = θ∗ contains all the information about the motion.