James Casey

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

Meccanica (2011) 46: 471–476
DOI 10.1007/s11012-010-9296-x
B R I E F N OT E
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
curve, a kinematical decomposition, discovered by
Siacci, expresses the acceleration vector as the sum
of two special oblique components in the osculating
plane to the curve. A new proof of Siacci’s theorem is
presented.
Keywords Siacci · Space curve · Kinematics ·
Classical mechanics · Central forces
1 Introduction
For planar motions of a particle, a resolution of the
acceleration vector, due to Siacci [1], is well known
and is particularly useful for problems in which angu-
lar momentum is constant [2–6]. This resolution com-
prises a radial component and a tangential component,
which, in general, are not orthogonal to each other.1
1Whittaker [2, p. 21] gives a geometrical proof of Siacci’s theo-
rem in the plane. Grossman [3] has a more modern proof (but the
accompanying diagram is misdrawn). Francesco Siacci (1839–
1907) held professorships in ballistics at the military academy
in Genoa, and in mechanics at the Universities of Turin and
Naples [7]. He played an active role in the movement for unifi-
cation of Italy, and in his later years was elected a Senator for
life. A street in Rome is named in his honor.
J. Casey ()
Department of Mechanical Engineering, University
of California, Berkeley, CA 94720, USA
e-mail: jcasey@me.berkeley.edu
For spatial motions of a particle, Siacci succeeded
in showing that a similar resolution can be obtained [8].
In this case, the two components are in the instanta-
neous osculating plane to the path of the particle. One
is tangent to the path, while the other is directed to-
wards the point where the perpendicular from a fixed
origin meets the osculating plane. Although Siacci’s
formulas are quite remarkable, his statement of the
theorem is not altogether precise and his proof is cum-
bersome.2
The purpose of this note is to present a proof of
Siacci’s theorem in space that is based on the Serret-
Frenet formulae. Siacci’s theorem in the plane is in-
cluded as a corollary.
2 Preliminaries
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
in E3, and let x be the position vector of P at time t .
Let C be the oriented curve traced out by P, and let s
be the arc length of C corresponding to time t .
Let {et , en, eb} be a right-handed orthonormal basis
consisting of the unit tangent vector
et = x′ = dx
ds
, (1)
2Whittaker [2, p. 24, Problem 12] repeats Siacci’s statement of
the theorem.
472 Meccanica (2011) 46: 471–476
the unit normal vector en, and the unit binormal vec-
tor eb . Recall the Serret-Frenet formulae
e′t = κen, e′n = −κet + τeb, e′b = −τen, (2)
where κ and τ are the curvature and torsion functions.
For κ = 0, ρ = 1/κ is the radius of curvature of C .
The velocity v and acceleration a of P at time t are
given by
v = x˙ = vet , v = s˙, a = v˙ = v˙et + κv2en, (3)
where the superposed dot denotes d/dt . Along C , v
may be expressed as a function of s, and the accelera-
tion may then be written as
a = v dv
ds
et + κv2en, (4)
which always lies in the osculating plane Π to C at P.
The angular velocity vector of the Serret-Frenet basis
is
ω = s˙(τet + κeb) (5)
and lies in the rectifying plane to C . Clearly,
e˙t = ω × et , e˙n = ω × en, e˙b = ω × eb. (6)
The angular momentum of P about O is
HO = x × mv = x × mvet (7)
and lies in the plane perpendicular to et (i.e., the nor-
mal plane to C ).
3 Siacci’s theorem
Let the position vector of P be resolved on the Serret-
Frenet basis:
x = qet − pen + beb, (8)
where
q = x · et , −p = x · en, b = x · eb. (9)
Also, define a vector r in the osculating plane Π by
r = −pen + qet . (10)
Note that
r2 = r · r = p2 + q2. (11)
The foregoing resolution of x is indicated in Fig. 1.
The vector beb is the position vector of the perpendic-
ular OB from O to the osculating plane, and r is the
position vector of P relative to B . BZ is the perpen-
dicular from B to the tangent line SPT , so that the
position vector of Z relative to B is −pen. The posi-
tion vector of P relative to Z is qet .
Substituting (8) in (1) and invoking (2)1,2,3, we ob-
tain
et = q ′et + qκen − p′en − p(−κet + τeb) + b′eb
+ b(−τen), (12)
which yields
q ′ = 1 − κp, p′ = κq − τb, b′ = τp. (13)
It follows from (11) and (13)1,2,3 that
rr ′ = q − τpb = q − bb′. (14)
With the aid of (8), the angular momentum vector
in (7) takes the form
HO = m(ben + peb)v. (15)
Let
h = pv, w = bv. (16)
The two components, h and w, of angular momentum
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,
respectively.3
Siacci’s idea is to resolve the acceleration vector
a in (4) along the tangential direction SPT and the
radial direction BP in the osculating plane. To do this,
we need to be able to express en in terms of r and et ,
which in view of (10), is possible if and only if p =
0. We may ensure that p never vanishes by making
the physical assumption that the binormal component
of angular momentum is nonzero. Then, h = pv = 0.
Hence, in view of (11), r = 0 and we may define a unit
vector er by
er = 1
r
r. (17)
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
tangent line to C at P and Π is the osculating plane at P . OB is
the perpendicular, from an arbitrary fixed origin O in space,
to Π . BZ is perpendicular to SPT , and r is the position vec-
tor of P relative to B
Equation (10) then yields the desired resolution
en = 1
p
(−rer + qet ). (18)
Substituting (18) in (4), we obtain the fundamental
form of Siacci’s resolution of the acceleration vector:
a = −κv
2r
p
er +
(
v
dv
ds
+ κv
2q
p
)
et = Srer + Stet .
(19)
In general, the unit vectors er and et are not orthogonal
to each other and the two components of a in (19) are
not equal to the orthogonal projections of a onto er
and et . We call Sr and St , respectively, the radial and
tangential Siacci components of the acceleration.
Invoking (16)1 with p = 0, we may express the ra-
dial Siacci component as
Sr = −κrh
2
p3
. (20)
The tangential Siacci component may be put into a
number of different forms. First note that q satisfies
the equations
κq = p′ + τb, q = rr ′ + τpb = rr ′ + bb′, (21)
which follow from (13)2 and (14). With the aid of
(21)1, (16)1 and (13)3, we find that
St = 12 (v
2)′ + κv
2q
p
= 1
2
(v2)′ + v
2p′
p
+ τv
2b
p
= (h
2)′
2p2
+ τbh
2
p3
= 1
2p2
{
(h2)′ + h
2
p2
(b2)′
}
. (22)
Likewise, employing (21)2, we obtain
St = 12 (v
2)′ + κv2
{
(r2)′
2p
+ τb
}
. (23)
474 Meccanica (2011) 46: 471–476
If instead, we use the second expression for q in (21)2,
we may deduce that
St = 12 (v
2)′ + κv
2
2p
(r2 + b2)′
= 1
2
(v2)′ + κv
2
2p
(x · x)′, (24)
where (10) and (8) have also been utilized.
Summarizing the above derivation, we state
Siacci’s Theorem Let a particle P of mass m(> 0)
travel along a curve C in space, and suppose that the
binormal component of its angular momentum vector
never vanishes. Then, the acceleration vector of P may
be expressed as the sum of two oblique components,
the Siacci components, as in (19). One Siacci compo-
nent lies along the tangent to C , while the other one is
directed from P towards the point where the perpen-
dicular from an arbitrary fixed origin meets the oscu-
lating plane. The radial component may also be writ-
ten as in (20), and alternative expressions for the tan-
gential component are given by (22), (23), and (24).
Corollary 1 Suppose that the motion of P is confined
to a fixed plane, not necessarily containing the origin
O , and that the component of angular momentum that
is perpendicular to the plane never vanishes. Then, the
acceleration of P can be expressed in the form (19),
where er is the radial basis vector of a polar coordi-
nate system in the plane. The tangential Siacci compo-
nent reduces to
St = (h
2)′
2p2
, (25)
or equivalently to
St = 12
{
(v2)′ + κv
2
p
(r2)′
}
. (26)
Proof In the planar case, τ = 0, eb is constant and or-
thogonal to the plane, and b is constant. Also, the com-
ponent mpveb of angular momentum in (15) is orthog-
onal to the plane, and, by assumption, never vanishes.
By virtue of (8), (10) and (11) the position vector of P
is
x = rer + beb. (27)
The expression (19) now holds in the plane of the
motion. Equations (25) and (26) follow immediately
from (22) and (23). 
It may be noted that in the planar case, (14) and
(13)2 yield
rr ′ = q, p′ = κrr ′. (28)
Corollary 2 Suppose that the fixed plane in Corol-
lary 1 passes through O, and that the angular momen-
tum of P never vanishes. Then, Siacci’s resolution (19)
holds, with St being given by (25) and (26).
Proof This is a special case of Corollary 1 with
b = 0. 
4 Illustrative examples
Example 1: Motion along a helix
Let the curve C be a circular helix, along which the
particle P is traveling. The helix lies on a cylinder of
radius A.
Using a cylindrical coordinate system (R, θ, z), let
the position vector of P be given by
x = AeR + zk, z = Bθ, (29)
where eR = cos θ i + sin θ j, eθ = − sin θ i + cos θ j,
{i, j,k} is a fixed right-handed orthonormal basis, and
A(> 0) and B( = 0) are constants. Let ω = θ˙ . The ve-
locity and acceleration vectors of P are
v = ω(Aeθ + Bk),
a = −Aω2eR + ω˙(Aeθ + Bk).
(30)
Let C = √A2 + B2. The speed of P and its time deriv-
ative are
v = s˙ = Cω, v dv
ds
= Cω˙. (31)
The arc length increases linearly with θ .
The Serret-Frenet basis for the helix is
et = cosα eθ + sinα k, en = −eR,
eb = − sinα eθ + cosα k,
(32)
Meccanica (2011) 46: 471–476 475
where the helix angle α is given by tanα = B/A. The
curvature and torsion are constant:
κ = A
C2
= cos
2 α
A
, τ = B
C2
= κ tanα. (33)
The angular velocity of the Serret-Frenet basis is
ω = ωk. (34)
The components of x on the Serret-Frenet basis
may be calculated from (9)1,2,3, (29)1, and (32)1,2,3:
q = z sinα, p = A, b = z cosα. (35)
The two components of the angular momentum per
unit mass, given by (16)1,2, are
h = ACω, w = Azω. (36)
Applying Siacci’s theorem in space, we find that
Sr = −rω2 = −ω2
√
A2 + z2 sin2 α,
St = Cω˙ + ω2z sinα.
(37)
Example 2: Central forces
Suppose that the hypotheses of Corollary 2 are satis-
fied. Suppose also that the angular momentum of P is
a nonzero constant and that the radial component Sr of
acceleration is a function of r :
Sr = −f (r). (38)
The tangential Siacci component of acceleration in
(25) now vanishes. From (20) and (38), it follows that
κrh2
p3
= f (r). (39)
It is shown in the Appendix that at every point of an or-
bit, except possibly at its apses, the curvature is given
by the formula in (50). Thus, for any portion of the
orbit not containing an apse, (39) may be written as a
first-order differential equation4
h2
p3
dp
dr
= f (r). (40)
4The key formula (40) is attributed to Abraham de Moivre, al-
though a similar result may have been previously known to
Newton. De Moivre communicated it in a letter to Johann
Bernoulli in 1705; Bernoulli responded with a proof in 1706,
which he later published. For more details, see the historical
study [10].
If f (r) is integrable, we obtain
1
2
h2
p2
+
∫
f (r)dr = const., (41)
which is equivalent to energy conservation for the par-
ticle P .
Additional material on central forces may be found
in [2–6].
Acknowledgements I would like to thank Professor Charles
E. Smith, of Oregon State University, and Professor Nathaniel
Grossman of the University of California, Los Angeles, for their
comments on a draft of this note.
Open Access This article is distributed under the terms of the
Creative Commons Attribution Noncommercial License which
permits any noncommercial use, distribution, and reproduction
in any medium, provided the original author(s) and source are
credited.
Appendix: The radial-perpendicular
representation of planar curves
In this appendix, some useful background material on
the radial-perpendicular, or (r,p), representation of
planar curves is provided.
Consider a particle P moving along a curve C in a
fixed plane through O (Fig. 2). {i, j} is a fixed ortho-
normal basis and k = i × j. Choose polar coordinates
and let {er , eθ } be the associated basis. Clearly,
x = r = rer , der
dθ
= eθ , deθ
dθ
= −er . (42)
Suppose that C is specified in polar form r = r(θ). The
unit tangent vector to C is5
et = dr
ds
=
(
dr
dθ
er + reθ
)
dθ
ds
. (43)
Hence,[
r2 +
(
dr
dθ
)2](
dθ
ds
)2
= 1. (44)
Observing that the triangle OZP in Fig. 2 is right-
angled, and ∠OPZ = β , we note that et can also be
expressed as
et = cosβ er + sinβ eθ = 1
r
(qer + peθ ). (45)
5In the plane (Fig. 2), et may also be expressed as et = cosφ i +
sinφ 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
p = r2 dθ
ds
, q = r dr
dθ
dθ
ds
. (46)
By virtue of (46)1 and (44),
1
p2
= 1
r2
+
(
1
r2
dr
dθ
)2
. (47)
The relation (28)2 furnishes
dp
ds
= κr dr
dθ
dθ
ds
. (48)
If an (r,p) representation of a curve is given, (48) be-
comes
dp
dr
dr
dθ
= κr dr
dθ
. (49)
For all points at which dr/dθ does not vanish, (49)
produces a first-order differential equation for curva-
ture:
κ = 1
r
dp
dr
(
dr
dθ
= 0
)
. (50)
At an apse of an orbit, the radius vector is perpendic-
ular to the tangent, and hence p = r , q = 0, β = π/2;
further, in view of (46)2, (47) and (48), dr/ds = 0,
dr/dθ = 0, and dp/ds = 0 at an apse. The limits of
both sides of (50) must be examined, as the apse is
approached, to check whether the equality in (50) still
holds at the apse.
References
1. Siacci F (1879) Moto per una linea piana. Atti R Accad Sci
Torino 14:750–760
2. Whittaker ET (1937) A treatise on the analytical dynamics
of particles and rigid bodies, 4th edn. Cambridge University
Press, Cambridge. Dover, New York (1944)
3. Grossman N (1996) The sheer joy of celestial mechanics.
Birkhäuser, Basel
4. Love AEH (1921) Theoretical mechanics, 3rd edn. Cam-
bridge University Press, Cambridge
5. Ramsey AS (1933) Dynamics. Cambridge University
Press, Cambridge
6. Lamb H (1923) Dynamics, 2nd edn. Cambridge University
Press, Cambridge
7. http://www.torinoscienza.it/accademia/personaggi,
accessed 20 October 2008
8. Siacci F (1879) Moto per una linea gobba. Atti R Accad Sci
Torino 14:946–951
9. Casey J (2007) Areal velocity and angular momentum for
non-planar problems in particle mechanics. Am J Phys
75:677–685
10. Guicciardini N (1995) Johann Bernoulli, John Keill and the
inverse problem of central forces. Ann Sci 52:537–575


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