Inertial Motion Viewed from a Potential Well

JANUARY 2015 P H I L L I P S 409

NORMAN PHILLIPS Merrimack, New Hampshire

(Manuscript received 18 June 2014, in ﬁnal form 5 September 2014)

ABSTRACT

The motion of a mass undergoing free horizontal motion on a planet’s surface is analyzed in an inertial frame by using a potential well presentation. The following results from the conventional analysis in rotating coordinates are also found in inertial coordinates. Under typical circumstances the mass oscillates about the latitude where its angular momentum per unit mass matches that of the planet’s surface. For masses with nonzero angular momentum, conservation of angular momentum bounds the mass away from the pole. Motion with zero angular momentum is confined to a fixed meridian plane on which the mass may oscillate across the pole. Motion across the equator is possible only for masses with sufficient energy. Masses with angular momentum per unit mass greater than that of the planet at the equator oscillate about the equator with a limiting period determined by the excess angular momentum.

1. Presentation in the rotating frame called the ‘‘inertial circle.’’ The corresponding period is 2pr/C,or2p/(2v sinu). Free horizontal motion is a common topic when dis- Inertial motion is seen most clearly in the ocean, cussing dynamics of a point mass confined to the surface perhaps because the atmosphere does not experience of a planet. In a rotating reference frame, four forces act the external forcing such as surface drag, and the much in the absence of friction: Newtonian gravitation, cen- larger background velocities seem to mask any inertial trifugal force, surface reaction, and Coriolis force. The motion. Figure 1 shows a spectral analysis from 598N sum of the first two is ‘‘gravity,’’ which is perpendicular [taken from Weller (1982)]. to the surface. In the absence of friction, the surface This is the traditional analysis in the rotating reference reaction is also perpendicular to the surface. For free frame. However, when viewed from a nonrotating frame, horizontal motion, this leaves only the Coriolis force. the forces are quite different; they reveal the roles of cen- The horizontal component of this force on a mass m trifugal energy and conservation of angular momentum. moving horizontally at speed C is mC2v sinu, where v is The organization into a logical whole is greatly enhanced the angular speed of the planet’s rotation and u is the by the use of the potential well concept. Such an approach latitude of the mass (the angle between the surface leads naturally into a classification of several modes of normal and the normal to the equatorial plane). The free horizontal motion. force acts at right angles to the motion and thus pro- The appendix defines the potential well properties of the duces a centripetal acceleration C2/r, where r is the ra- trajectories pictured by Paldor and Killworth (1988).(Their dius of curvature of the path (clockwise in the Northern calculations are based in rotating coordinates.) The present Hemisphere and counterclockwise in the Southern article does not delve into the physics of the subject, but an Hemisphere, when viewed from above). excellent description of the physics behind the Coriolis If this is equated to the force per unit mass, we get force is found in the book by Stommel and Moore (1989). r 5 C/(2v sinu). (1)

When C is sufﬁciently small, and thus sinu is sensibly 2. A potential well formulation constant, the path approximates a reentrant circle, Marvin (1915) seems to have been one of the ﬁrst Corresponding author address: Norman Phillips, 18 Edward Lane, Americans after Ferrel to recognize that inertia motion Merrimack, NH 03054. is partly due to a component of Newtonian attraction E-mail: [email protected] that is parallel to the surface of the earth. He carried this

DOI: 10.1175/JAS-D-14-0185.1

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FIG. 2. Gravity components acting on a mass at point P. The solid line t–t represents the horizontal tangent plane, normal to Pg (gravity). PN is Newtonian gravitation, with a component PG in the plane t–t. Ng is the centrifugal force per unit mass. PZ is the local vertical at an angle u (geocentric latitude) from the plane of the equator. PG has a magnitude equal to the negative component of Ng in the horizontal plane. The force of reaction needed to FIG. 1. Spectrum of clockwise motion near the inertia frequency P PZ (solid line) and counterclockwise motion (dashed line) in the At- support a mass at would be directed upward along . lantic Ocean near 598N. Copied from Fig. 7 in Weller (1982). dV/dt 1 UU sinu/r 5 v2r sinu and (3a) idea far enough that in a later paper (Marvin 1920)he introduced the phrase ‘‘the law of the geoidal slope’’ as dU/dt 2 UV sinu/r 5 0, (3b) part of the Coriolis force. More recently, Durran (1993) has revived this consideration. where v2r sinu is the horizontal component of Newtonian Let u denote latitude (increasing northward from the attraction. It is directed poleward and arises from the equator) and l denote longitude (increasing eastward) in a rotational flattening of the planet’s figure, as pictured in nonrotating frame. For a mass confined to the surface of the Fig. 2. This term vanishes at the pole and equator and is planet the northward and eastward velocity components are independent of the sign of v. A mass at rest with respect to the rotating planetary V 5 Rdu/dt and (2a) surface will have V 5 0 and U 5 rv. Under these con- U 5 rdl/dt, (2b) ditions (3) shows that the mass will remain at rest, since then the right-hand side of (3a) is balanced by the U2 where r denotes the perpendicular distance of the mass term on the left-hand side. For convenience, the symbol from the axis of the planet and R denotes the radius of m for mass will be omitted, inasmuch as no trans- the planet. (For planets of small oblateness, such as formations involving mass are involved (i.e., as if m 5 1). Earth, r can be replaced by cosu times R.) Changes in Equations (3a) and (3b) imply energy invariance r are related to changes in latitude by dE/dt 5 0, where

Rdu/dt 5 (dr/dt)/sinu. (2c) E 5 (1/2)(U2 1 V2 1 v2r 2), (4)

The inertial (nonrotating) equations of motion on the where (1/2)v2r2 is the ‘‘centrifugal energy,’’ or the po- planet’s surface are tential energy associated with the poleward Newtonian

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invert the relation to express r2 in terms of V. Our principal interest will be in the turning points of the path taken by the mass—that is, the points where and when V 5 0andE 5 W. For these points, the roots of the quadratic equation are

2 2 2 2 2 1/2 v r2 5 E 2 (E 2 v M ) and (8a)

2 2 2 2 2 1/2 v r1 5 E 1 (E 2 v M ) , (8b)

where the minus sign and plus sign subscripts respectively denote the turning points closest to the pole and equator. Since E is nonnegative, the right-hand sides of (8) are nonnegative, as required by the left-hand side. However, the left-hand side of (8b) must not be greater than v2R2 since R is the maximum possible value of r.Wewillfind that if E is sufficiently large, the latter restriction makes it 0 FIG. 3. The nondimensional effective potential W for four values necessary to exclude the root (8b)—or in other words, the of the nondimensional angular momentum: M0 5 0, 0.5, 1.0, and 1.5 0 02 turning point at low latitudes is ignored when E is suffi- [see (8)]. The curves satisfy the relation W 5 x 1 M /x, where x 5 (r/R)2 is the abscissa. (The abscissa is shown for both hemispheres.) ciently large. attraction. In addition, (3b) implies angular momentum 3. Paths limited to one hemisphere invariance 0 5 dM/dt 5 0, (5a) The curve for M 0.5 in Fig. 3 shows that a fixed value of E can have two intersections with that curve within one where hemisphere, so the mass’s path can be confined entirely to one hemisphere. (Since the potential is symmetric with 5 M rU (5b) respect to the equator, the same statement also applies to each hemisphere.) This requires both (8a) and (8b) to be is the angular momentum about the planet’s axis. admissible, and therefore in (8b) we must have E , v2R2. U After eliminating between (4) and (5b), we can write Further, in any case the radical requires jMj , E/jvj,so 2 V2 5 2(E 2 W) and (6a) we must have jMj , jvjR as an additional condition for this case. Still another condition is that E must be greater W 5 (1/2)(M2/r2 1 v2r 2), (6b) than the minimum potential in the well, since according to (6a) there is otherwise no solution. where W is the effective potential—in this case, the The effective potential is a function of v and the mass current sum of azimuthal kinetic energy and gravita- property M, plus the radius r. The r location of the tional potential energy. Since the left-hand side of (6a) is minimum potential occurs at r 5 r0, where nonnegative, motion with V different from zero can 2 5 j vj occur only when there is a range of r in which E happens r 0 M/ (9) to exceed W.IfM is nonzero, the leading term in W is infinite at the poles, serving to bound the motion away and from either pole. The second term bounds the motion 5 v2 2 away from the equator unless E is sufficiently large. W0 r 0 . (10) Curves of constant W are shown in Fig. 3 for a series of M values using the primed nondimensional equivalents: At this point the mass has an azimuthal velocity of 0 magnitude jvjr , and, as shown by (6a), its maximum E 5 E/[(v2R2)/2], (7a) 0 meridional velocity value Vmax. The azimuthal velocity 0 u M r 2 rv W 5 W/[(v2R2)/2], and (7b) relative to Earth is equal to / and in this sit- uation is zero: 0 2 5 M 5 jMj/(jvjR ). (7c) u0 0. (11) Equation (6a) expresses V as a function of r 2.By Instead of the energy E, the meridional velocity at any solving the implied quadratic equation for r 2, we can reference point can be used as a proxy, by means of (6a).

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In particular, if the location of the minimum potential is nonzero, since the right side of (13a) cannot be zero. used as the reference point, then we have However, under these conditions, the mass can move in a path that encircles the pole. A more interesting result E 5 (1/2)(V )2 1 W . (12) max 0 is for the case of M 5 0 since then U is zero and the mass 2 Thus, (Vmax) /2 determines the difference between E oscillates along a meridian ﬁxed in the absolute frame. and the minimum W. If E in (8a) and (8b) is replaced The energy is from (12), and W from (10), the expressions for the 0 5 2 1 v2 2 turning points can be rewritten E (1/2)(V r ), (16)

5 1 d2 1/2 2 d 5 v2 2 r2 r0[(1 ) ] and (13a) and the effective potential is W (1/2) r , with a minimum at the pole. The maximum value of r occurs 5 1 d2 1/2 1 d where V 5 0, so E is equal to (1/2)(v2r2 ). It follows r1 r0[(1 ) ], (13b) max from (6a) and (2a) that where d 5 Vmax/2jvjr0. For small d, the two values for r (dr/dt)2 5 v2(r 2 2 r 2) sin2u. (17) are symmetric about r0. By taking half the difference max between them, we find the following expression for the displacement as projected onto the equatorial plane: In the special case when the flattening of the planet is slight (e.g., Earth), this problem is mathematically 2 5 d 5 j vj (r1 r2)/2 r0 Vmax/2 . (13c) equivalent to that for a (nonlinear) plane pendulum, and the solution can be stated as an elliptic integral of the Division by sinu converts this to a local distance on the first kind. planet’s surface and recovers (1).(Expansiontonextorder A more interesting example is a mass with zero M in d shows that the motion is elongated toward the equator.) that crosses the pole with a small velocity directed To determine the azimuthal speed relative to Earth, along a fixed meridian—say, l 5 0. The motion will be we take advantage of (5) and (11) and write confined to small values of p/2 2 u,sothatsinu ; 1. But when the period is calculated from (17) according to 2v 5 v 1 2v 5 v 1 r0 r2(r2 u2) and r0 r1(r1 u1). the procedure used to obtain (15b), the result is T 5 2p/ v, or 1 day. This is twice the period calculated for (15) From which it follows from (13a) and (13b) that with sinu 5 1. The resolution of this seeming paradox lies in the fact u2 512r vd 51V (14a) 0 max that, when the mass with zero angular momentum has l 5 and moved along 0 for a half period of 12 h, it has re- turned to the pole and is moving in the opposite di- 52 vd 52 u1 2r0 Vmax . (14b) rection from its starting direction. At this time the original longitude in the rotating frame has rotated 8 l 5 Thus, Vmax is the speed relative to Earth experienced by counterclockwise 180 from 0 and coincides with the the mass in its motion; that is, it is the speed C referred to 1808 meridian in the fixed frame. Since the mass spends in section 1 of this article. And the motion is clockwise in the second 12 h on this 1808 fixed meridian, it will repeat the Northern Hemisphere, as shown by the signs here in the next 12 h the same path in the rotating frame that for the u and v values. it traversed during the first 12 h. The period as seen in Intheinertialframe,itisalsopossibletodeducethe the rotating frame is thus 12 h, agreeing with the con- period of free horizontal motion when the excursion from ventional polar result. r0 is not large. The circumference of the path on Earth’s surface is p(r1 2 r2)/sinu.From(13c), we find that the speed around the path V is equal to (r1 2 r2)v. Thus, the 5. Paths that cross the equator time to go around once is as recorded in section 1: If the mass starts at the pole with V0 equal to vR, p/v sinu 5 2p/2v sinu. (15) the rotational speed of the planet at the equator, then it has just enough energy to reach the equator. However, the time necessary to arrive at the equator 4. Polar latitudes from the pole with this critical starting speed is infinite. Equation (13a) confirms that the poleward limit of A matter of more interest is the question of when a a path with nonzero r0 cannot reach the pole when M is mass in free horizontal motion can cross from one

Unauthenticated | Downloaded 10/01/21 10:00 AM UTC JANUARY 2015 P H I L L I P S 413 hemisphere to the other. When jMj , jvjR2,thiscan be answered from (13b) by setting r1 equal to the equatorial radius R and solving for the critical value of jV0j. The answer is surprisingly simple:

j j 5 jvj 2 2 2 V0 CRIT R(1 r 0/R ). (18)

When r0 5 0, this agrees with the value obtained above for M 5 0. If jV0j exceeds the critical value, the path crosses the equator and lies symmetrically between the polar turning points given by (13a) (which is applicable to both hemispheres). j j . jvj 2 FIG. 4. Plot showing how the values of nondimensional angular For M R , W is a minimum at the equator and momentum M0 (abscissa) and energy E0 (ordinate) determine the motion is only bounded away from the poles. Under the character of free horizontal motion. The critical curves . jv j 0 5 0 these conditions, assuming E M (i.e., the necessary are defined by the nondimensional variables WMIN 2M ,and 2 2 0 5 1 condition for a solution), we have E . v R so the WEQ 1 M. equatorial turning point disappears. The exact polar turning point is given by (8a), where, if desired, E can be 6. Summary expressed in terms of V as a proxy by using the equator Figure 4 is the inertial frame equivalent of Fig. 3 of as a reference point; that is, Ripa (1997). It shows how the behavior of a mass in free horizontal motion is organized according to its angular 5 2 1 momentum (the abscissa) and energy (the ordinate). E (1/2)(VEQ) WEQ and (19a) Nondimensional values are most convenient for this 5 2 2 1 v2 2 purpose as defined earlier in (8). The conventional WEQ (1/2)(M /R R ), (19b) picture of inertial motion occurs in region A, in which the north–south excursions are confined to one hemi- in accordance with (6a) and (6b). sphere. (This is illustrated in Fig. 2 by the potential well An approximate expression for the turning point for M0 5 0.5.) The lower boundary of region A is the (when jMj . jvjR2) can be found from (19a) by as- straight line defined by E0 5 2M0—that is, when the suming VEQ is small and the turning point is corre- mass has a zero value for V. Below this curve, in region spondingly close to the equator. If a positive reference C, the mass has no energy. The upper limit of region V V5j j 2 . jvj 0 5 0 5 1 02 rotation rate is defined by M /R , then the A is bounded by the curve E WEQ 1 M , corre- result can be expressed as sponding to the critical value of E needed to reach the equator. 2 2 2 2 2 In region B the mass has sufficient energy to oscillate R 2 r2 5 (V ) /(V 2 v ), (20) EQ symmetrically across the equator. If the mass lies on the curve W0 , with 1 , E0 , 2, the mass stays in one and the approximate period of oscillation is EQ hemisphere but takes infinite time to reach the equator. 0 5 0 . If E WEQ 2, then the mass moves along the equator. 2 2 1=2 T 5 2p/(V 2 v ) . (21) Finally, the E0 axis corresponds to the case of zero angular momentum, when the mass oscillates on a fixed meridian plane. For E0 , 1, this motion is confined between the pole As jMj increases, T approaches the revolution period and the parallel where (r/R)2 5 E0.ForE0 . 1, the mass 2p/V associated with V. crosses both poles. Paldor and Killworth (1988) and Ripa (1997) have analyzed details of the two-dimensional aspects of free horizontal motion near the equator in rotating co- Acknowledgments. I am grateful for the improvement ordinates. Interesting patterns emerge for oscillations wrought by the late Professor G. Platzman on early that cross the equator. Paldor and Killworth illustrate forms of this manuscript and for the help provided by asymptotic approach to the equator in their Fig. 1f. Drs. Anders Persson and Robert Weller together with Values of M0 and E0 relating their figures to Fig. 4 in this Professors J. Fleming and J. Pedlosky in providing paper are provided in the appendix. copies of several references.

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TABLE A1. References for trajectories displayed in the ﬁgures in REFERENCES Paldor and Killworth (1988).

0 0 Durran, D., 1993: Is the Coriolis force really responsible for the Figure No. Lat Initial speed M E inertial oscillation? Bull. Amer. Meteor. Soc., 74, 2179–2184, 1a 158 0.1 1.030 1.035 doi:10.1175/1520-0477(1993)074,2179:ITCFRR.2.0.CO;2. 1b 268 0.1 0.898 0.903 Marvin, C., 1915: Deflection of bodies moving freely under gravity 1c 308 0.1 0.837 0.842 on a rotating sphere. Mon. Wea. Rev., 43, 503–506, doi:10.1175/ 1d 33.38 0.1 0.782 0.787 1520-0493(1915)43,503:DOBMFU.2.0.CO;2. 1e 358 0.1 0.753 0.758 ——, 1920: The law of the geoidal slope and fallacies in dynamic 1f 36.98 0.1 0.719 0.724 meteorology. Mon. Wea. Rev., 48, 565–582, doi:10.1175/ 1g 458 0.1 0.570 0.576 1520-0493(1920)48,565:TLOTGS.2.0.CO;2. Paldor, N., and P. Killworth, 1988: Inertial trajectories on a rotating Earth. J. Atmos. Sci., 45, 4013–4019, doi:10.1175/ APPENDIX 1520-0469(1988)045,4013:ITOARE.2.0.CO;2. b This section supplies convenient reference data for Ripa, P., 1997: ‘‘Inertial’’ oscillations and the -plane approxi- mation(s). J. Phys. Oceanogr., 27, 633–647, doi:10.1175/ the figures in Paldor and Killworth (1988); see Table A1. 1520-0485(1997)027,0633:IOATPA.2.0.CO;2. Nondimensional symbols are used, following (7a)–(7c), Stommel, H., and D. Moore, 1989: Huygens’ rotating oblate earth. 2 with the value 2Rv (5940 m s 1 for Earth) for scaling An Introduction to the Coriolis Force, Columbia University the velocity components U0 and V0. All examples have Press, 173–257. zero for the initial V, so that W0 is initially equal to E0. Weller, R., 1982: The relation of near-inertial motions observed in the 0 5 mixed layer during the JASIN (1978) experiment to the local All initial speeds are to the east. Thus, in this table, E wind stress and to the quasi-geostrophic flow field. J. Phys. 0 2 2 0 0 (1/2)[(cosu 1 u ) 1 cos u] and M 5 cosu(cosu 1 u ), in Oceanogr., 12, 1122–1135, doi:10.1175/1520-0485(1982)012,1122: 0 both of which u 5 0.1. TRONIM.2.0.CO;2.

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