Siacci's Resolution of the Acceleration Vector for a Space Curve

Meccanica (2011) 46: 471–476 DOI 10.1007/s11012-010-9296-x

BRIEF NOTE

Siacci’s resolution of the acceleration vector for a space curve

James Casey

Received: 17 January 2009 / Accepted: 16 March 2010 / Published online: 9 June 2010 © The Author(s) 2010. This article is published with open access at Springerlink.com

Abstract For motion of a material point along a space For spatial motions of a particle, Siacci succeeded curve, a kinematical decomposition, discovered by in showing that a similar resolution can be obtained [8]. Siacci, expresses the acceleration vector as the sum In this case, the two components are in the instanta- of two special oblique components in the osculating neous osculating plane to the path of the particle. One plane to the curve. A new proof of Siacci’s theorem is is tangent to the path, while the other is directed to- presented. wards the point where the perpendicular from a fixed origin meets the osculating plane. Although Siacci’s Keywords Siacci · Space curve · Kinematics · formulas are quite remarkable, his statement of the Classical mechanics · Central forces theorem is not altogether precise and his proof is cum- bersome.2 The purpose of this note is to present a proof of 1 Introduction Siacci’s theorem in space that is based on the Serret- Frenet formulae. Siacci’s theorem in the plane is in- For planar motions of a particle, a resolution of the cluded as a corollary. acceleration vector, due to Siacci [1], is well known and is particularly useful for problems in which angular momentum is constant [2–6]. This resolution com- 2 Preliminaries prises a radial component and a tangential component, which, in general, are not orthogonal to each other.1 Consider a particle P of mass m(> 0) moving in a 3-dimensional Euclidean space E3 under the influence of arbitrary forces. Choose an arbitrary fixed origin O 3 1Whittaker [2, p. 21] gives a geometrical proof of Siacci’s theo- in E , and let x be the position vector of P at time t. rem in the plane. Grossman [3] has a more modern proof (but the Let C be the oriented curve traced out by P, and let s accompanying diagram is misdrawn). Francesco Siacci (1839– be the arc length of C corresponding to time t. 1907) held professorships in ballistics at the military academy { } in Genoa, and in mechanics at the Universities of Turin and Let et , en, eb be a right-handed orthonormal basis Naples [7]. He played an active role in the movement for unifi- consisting of the unit tangent vector cation of Italy, and in his later years was elected a Senator for dx life. A street in Rome is named in his honor. e = x = , (1) t ds J. Casey () Department of Mechanical Engineering, University of California, Berkeley, CA 94720, USA 2Whittaker [2, p. 24, Problem 12] repeats Siacci’s statement of e-mail: [email protected] the theorem. 472 Meccanica (2011) 46: 471–476 the unit normal vector en, and the unit binormal vec- The foregoing resolution of x is indicated in Fig. 1. tor eb. Recall the Serret-Frenet formulae The vector beb is the position vector of the perpendicular OB from O to the osculating plane, and r is the = =− + =− et κen, en κet τeb, eb τen, (2) position vector of P relative to B. BZ is the perpendicular from B to the tangent line SPT , so that the where κ and τ are the curvature and torsion functions. position vector of Z relative to B is −pe . The posi- For κ = 0, ρ = 1/κ is the radius of curvature of C. n tion vector of P relative to Z is qe . The velocity v and acceleration a of P at time t are t given by Substituting (8)in(1) and invoking (2)1,2,3, we obtain = ˙ = =˙ = ˙ =˙ + 2 v x vet ,vs, a v vet κv en, (3) et = q et + qκen − p en − p(−κet + τeb) + b eb where the superposed dot denotes d/dt. Along C, v + b(−τe ), (12) may be expressed as a function of s, and the accelera- n tion may then be written as which yields

dv 2 a = v et + κv en, (4) q = 1 − κp, p = κq − τb, b = τp. (13) ds which always lies in the osculating plane Π to C at P. It follows from (11) and (13)1,2,3 that The angular velocity vector of the Serret-Frenet basis is rr = q − τpb = q − bb . (14)

ω =˙s(τet + κeb) (5) With the aid of (8), the angular momentum vector in (7) takes the form and lies in the rectifying plane to C. Clearly, O H = m(ben + peb)v. (15) e˙t = ω × et , e˙n = ω × en, e˙b = ω × eb. (6) Let The angular momentum of P about O is

O h = pv, w = bv. (16) H = x × mv = x × mvet (7) The two components, h and w, of angular momentum and lies in the plane perpendicular to et (i.e., the normal plane to C). per unit mass may be interpreted as twice the rates at which the orthogonal projections of x sweep out area on the osculating plane Π and on the rectifying plane, 3 3 Siacci’s theorem respectively. Siacci’s idea is to resolve the acceleration vector Let the position vector of P be resolved on the Serret- a in (4) along the tangential direction SPT and the Frenet basis: radial direction BP in the osculating plane. To do this, we need to be able to express en in terms of r and et , = − + x qet pen beb, (8) whichinviewof(10), is possible if and only if p = 0. We may ensure that p never vanishes by making where the physical assumption that the binormal component of angular momentum is nonzero. Then, h = pv = 0. q = x · et , −p = x · en,b= x · eb. (9) Hence, in view of (11), r = 0 and we may deﬁne a unit Also, deﬁne a vector r in the osculating plane Π by vector er by r =−pe + qe . (10) 1 n t e = r. (17) r r Note that r2 = r · r = p2 + q2. (11) 3See [9] for a discussion of the areal velocity vector in space. Meccanica (2011) 46: 471–476 473

Fig. 1 A particle P travels along a space curve C. SPT is the to Π. BZ is perpendicular to SPT ,andr is the position vec- tangent line to C at P and Π is the osculating plane at P . OB is tor of P relative to B the perpendicular, from an arbitrary ﬁxed origin O in space,

Equation (10) then yields the desired resolution The tangential Siacci component may be put into a number of different forms. First note that q satisﬁes 1 the equations en = (−rer + qet ). (18) p κq = p + τb, q = rr + τpb = rr + bb , (21) Substituting (18)in(4), we obtain the fundamental form of Siacci’s resolution of the acceleration vector: which follow from (13)2 and (14). With the aid of (21)1, (16)1 and (13)3, we ﬁnd that 2 2 κv r dv κv q 2 a =− er + v + et = Sr er + St et . 1 2 κv q p ds p St = (v ) + 2 p (19) 1 v2p τv2b = (v2) + + p p In general, the unit vectors er and et are not orthogonal 2 to each other and the two components of a in (19)are 2 2 = (h ) + τbh not equal to the orthogonal projections of a onto er 2p2 p3 and e . We call S and S , respectively, the radial and t r t 2 1 h tangential Siacci components of the acceleration. = (h2) + (b2) . (22) p2 p2 Invoking (16)1 with p = 0, we may express the ra- 2 dial Siacci component as Likewise, employing (21)2, we obtain 2 2 κrh 1 2 2 (r ) S =− . (20) St = (v ) + κv + τb . (23) r p3 2 2p 474 Meccanica (2011) 46: 471–476

If instead, we use the second expression for q in (21)2, The expression (19) now holds in the plane of the we may deduce that motion. Equations (25) and (26) follow immediately from (22) and (23). 2 1 2 κv 2 2 St = (v ) + (r + b ) 2 2p It may be noted that in the planar case, (14) and 2 (13)2 yield 1 κv = (v2) + (x · x) , (24) 2 2p rr = q, p = κrr . (28) where (10) and (8) have also been utilized. Summarizing the above derivation, we state Corollary 2 Suppose that the fixed plane in Corol- lary 1 passes through O, and that the angular momen- Siacci’s Theorem Let a particle P of mass m(> 0) tum of P never vanishes. Then, Siacci’s resolution (19) travel along a curve C in space, and suppose that the holds, with St being given by (25) and (26). binormal component of its angular momentum vector never vanishes. Then, the acceleration vector of P may Proof This is a special case of Corollary 1 with = be expressed as the sum of two oblique components, b 0. the Siacci components, as in (19). One Siacci component lies along the tangent to C, while the other one is directed from P towards the point where the perpen- 4 Illustrative examples dicular from an arbitrary fixed origin meets the osculating plane. The radial component may also be writ- Example 1: Motion along a helix ten as in (20), and alternative expressions for the tangential component are given by (22), (23), and (24). Let the curve C be a circular helix, along which the particle P is traveling. The helix lies on a cylinder of Corollary 1 Suppose that the motion of P is confined radius A. to a fixed plane, not necessarily containing the origin Using a cylindrical coordinate system (R,θ,z),let O, and that the component of angular momentum that the position vector of P be given by is perpendicular to the plane never vanishes. Then, the acceleration of P can be expressed in the form (19), x = AeR + zk,z= Bθ, (29) where er is the radial basis vector of a polar coordinate system in the plane. The tangential Siacci compo- where eR = cos θ i + sin θ j, eθ =−sin θ i + cos θ j, nent reduces to {i, j, k} is a fixed right-handed orthonormal basis, and A(> 0) and B(= 0) are constants. Let ω = θ˙.Theve- (h2) locity and acceleration vectors of P are St = , (25) 2p2 v = ω(Aeθ + Bk), (30) or equivalently to 2 a =−Aω eR +˙ω(Aeθ + Bk). 2 √ 1 2 κv 2 St = (v ) + (r ) . (26) Let C = A2 + B2. The speed of P and its time deriv- 2 p ative are

Proof In the planar case, τ = 0, eb is constant and or- dv v =˙s = Cω, v = Cω.˙ (31) thogonal to the plane, and b is constant. Also, the com- ds ponent mpv eb of angular momentum in (15) is orthogonal to the plane, and, by assumption, never vanishes. The arc length increases linearly with θ. By virtue of (8), (10) and (11) the position vector of P The Serret-Frenet basis for the helix is is et = cos α eθ + sin α k, en =−eR, (32) =− + x = rer + beb. (27) eb sin α eθ cos α k, Meccanica (2011) 46: 471–476 475 where the helix angle α is given by tan α = B/A.The If f(r)is integrable, we obtain curvature and torsion are constant: 1 h2 2 + f(r)dr= const., A cos α B 2 (41) κ = = ,τ= = κ tan α. (33) 2 p C2 A C2 which is equivalent to energy conservation for the par- The angular velocity of the Serret-Frenet basis is ticle P . Additional material on central forces may be found ω = ωk. (34) in [2–6]. The components of x on the Serret-Frenet basis Acknowledgements I would like to thank Professor Charles may be calculated from (9)1,2,3, (29)1, and (32)1,2,3: E. Smith, of Oregon State University, and Professor Nathaniel Grossman of the University of California, Los Angeles, for their q = z sin α, p = A, b = z cos α. (35) comments on a draft of this note.

The two components of the angular momentum per Open Access This article is distributed under the terms of the unit mass, given by (16)1,2,are Creative Commons Attribution Noncommercial License which permits any noncommercial use, distribution, and reproduction h = ACω, w = Azω. (36) in any medium, provided the original author(s) and source are credited. Applying Siacci’s theorem in space, we find that 2 2 2 2 2 Sr =−rω =−ω A + z sin α, Appendix: The radial-perpendicular (37) 2 representation of planar curves St = Cω˙ + ω z sin α. In this appendix, some useful background material on Example 2: Central forces the radial-perpendicular, or (r, p), representation of Suppose that the hypotheses of Corollary 2 are satis- planar curves is provided. C fied. Suppose also that the angular momentum of P is Consider a particle P moving along a curve in a { } a nonzero constant and that the radial component S of fixed plane through O (Fig. 2). i, j is a fixed ortho- r = × acceleration is a function of r: normal basis and k i j. Choose polar coordinates and let {er , eθ } be the associated basis. Clearly, Sr =−f(r). (38) der deθ x = r = rer , = eθ , =−er . (42) The tangential Siacci component of acceleration in dθ dθ (25) now vanishes. From (20) and (38), it follows that Suppose that C is specified in polar form r = r(θ).The unit tangent vector to C is5 κrh2 = f(r). (39) p3 dr dr dθ e = = e + re . (43) t ds dθ r θ ds It is shown in the Appendix that at every point of an orbit, except possibly at its apses, the curvature is given Hence, by the formula in (50). Thus, for any portion of the dr 2 dθ 2 orbit not containing an apse, (39) may be written as a r2 + = 1. (44) first-order differential equation4 dθ ds h2 dp Observing that the triangle OZP in Fig. 2 is right- = f(r). (40) p3 dr angled, and ∠OPZ = β, we note that et can also be expressed as

4The key formula (40) is attributed to Abraham de Moivre, al- 1 e = cos β e + sin β e = (qe + pe ). (45) though a similar result may have been previously known to t r θ r r θ Newton. De Moivre communicated it in a letter to Johann Bernoulli in 1705; Bernoulli responded with a proof in 1706, 5 which he later published. For more details, see the historical In the plane (Fig. 2), et may also be expressed as et = cos φ i + study [10]. sin φ j and κ = dφ/ds. 476 Meccanica (2011) 46: 471–476

Fig. 2 Motion of a particle P along a curve C in a ﬁxed plane

It follows from (43) and (45) that dr/dθ = 0, and dp/ds = 0 at an apse. The limits of both sides of (50) must be examined, as the apse is dθ dr dθ p = r2 ,q= r . (46) approached, to check whether the equality in (50) still ds dθ ds holds at the apse. By virtue of (46)1 and (44), 1 1 1 dr 2 = + . (47) References p2 r2 r2 dθ

The relation (28)2 furnishes 1. Siacci F (1879) Moto per una linea piana. Atti R Accad Sci Torino 14:750–760 dp dr dθ 2. Whittaker ET (1937) A treatise on the analytical dynamics = κr . (48) ds dθ ds of particles and rigid bodies, 4th edn. Cambridge University Press, Cambridge. Dover, New York (1944) If an (r, p) representation of a curve is given, (48) be- 3. Grossman N (1996) The sheer joy of celestial mechanics. comes Birkhäuser, Basel 4. Love AEH (1921) Theoretical mechanics, 3rd edn. Cam- dp dr dr bridge University Press, Cambridge = κr . (49) dr dθ dθ 5. Ramsey AS (1933) Dynamics. Cambridge University Press, Cambridge For all points at which dr/dθ does not vanish, (49) 6. Lamb H (1923) Dynamics, 2nd edn. Cambridge University produces a ﬁrst-order differential equation for curva- Press, Cambridge ture: 7. http://www.torinoscienza.it/accademia/personaggi, accessed 20 October 2008 1 dp dr 8. Siacci F (1879) Moto per una linea gobba. Atti R Accad Sci κ = = 0 . (50) Torino 14:946–951 r dr dθ 9. Casey J (2007) Areal velocity and angular momentum for non-planar problems in particle mechanics. Am J Phys At an apse of an orbit, the radius vector is perpendic- 75:677–685 = = = ular to the tangent, and hence p r, q 0, β π/2; 10. Guicciardini N (1995) Johann Bernoulli, John Keill and the further, in view of (46)2,(47) and (48), dr/ds = 0, inverse problem of central forces. Ann Sci 52:537–575