International Journal of Engineering Science 49 (2011) 1388–1396

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International Journal of Engineering Science

journal homepage: www.elsevier .com/locate / i jengsci

On the representation of rigid body rotational dynamics in Hertzianconfiguration space

James CaseyDepartment of Mechanical Engineering, 6125 Etcheverry Hall, University of California, Berkeley, CA 94720-1740, USA

a r t i c l e i n f o

Article history:Received 12 October 2010Received in revised form 24 February 2011Accepted 24 March 2011Available online 19 April 2011

In memory of Professor A.C. Eringen and hislifelong dedication to engineering science.

Keywords:Rotational dynamicsConfiguration spaceLagrange’s equations

0020-7225/$ - see front matter � 2011 Elsevier Ltddoi:10.1016/j.ijengsci.2011.03.012

E-mail address: jcasey@me.berkeley.edu1 A similar procedure, influenced by the work of He

Lagrange’s equations for a system of particles. Additi

a b s t r a c t

In a previous paper by the author, a geometrical procedure was presented for derivingLagrange’s equations for a rigid body. The rigid body was represented by an abstractparticle moving in a 12-dimensional Euclidean space, called Hertzian configuration space,the metric of which is determined by the radius of gyration of the body. The present paperfocuses on the representation of the underlying rotational dynamics in Hertzian space.

� 2011 Elsevier Ltd. All rights reserved.

1. Introduction

Each configuration of a rigid body B rotating about a fixed point O is represented by a proper orthogonal second-ordertensor Q. Hence, each configuration can be represented geometrically by a point P belonging to a fixed 9-dimensional Euclid-ean space, E9. Three parameters (e.g., Euler angles) suffice to determine Q, and therefore P lies in a 3-dimensional manifoldM embedded in E9. One method (Casey, 1995) for deriving Lagrange’s equations for B is to transform Euler’s second law (thebalance of angular momentum) into an equation in E9, choose an appropriate inner product on E9, and then calculate thecovariant components of the latter equation, using a basis that lies in the tangent space to M at P. It transpires that ifthe distance function on E9 is taken to be equal to the radius of gyration of B about O, one can represent B by an abstractparticle Z moving inM and having the same mass m and the same kinetic energy T as B.1 The corresponding inner productis denoted by h i0 and involves the mass m of B and the Euler tensor E0 of B about O in a fixed reference configuration of B (seeEq. (3.1)). When this inner product is assigned to E9, we obtain Hertzian configuration space, denoted by hE9i0. The purpose of thepresent paper is to investigate the nature of the equations in Hertzian configuration space that describe the dynamics of a rigidbody rotating about a fixed point. A basic dynamical Eq. (2.20) is obtained. It is equivalent to two component Eq. (3.36)1,2. Thecomponent (3.36)1 is equivalent to Euler’s second law and lies in a 3-dimensional subspace D of Hertzian space. The component(3.36)2 involves the Möbius tensor of the external force system and the reactive stresses that must exist in the rigid body. Thiscomponent lies in a 6-dimensional subspace N of Hertzian space, see Fig. 1. The component (3.36)1 is also equivalent toLagrange’s equations.

. All rights reserved.

rtz, 1900, Synge, 1927, Synge, 1936, and Synge and Schild, 1949, was employed by Casey, 1994 to deriveonal applications are treated in Casey and O’Reilly, 2006, Casey, 2004, and Casey, 2007.

Fig. 1. A schematic representation of the dynamics of the abstract particle Z moving in the configuration manifoldM embedded in 9-dimensional Hertzianspace hE9i0.

J. Casey / International Journal of Engineering Science 49 (2011) 1388–1396 1389

Standard notation of vector and tensor algebra is employed and the usual convention of summation over repeated indicesis understood.

2. Equations in physical space

Consider a rigid 3-dimensional continuum B of mass m (>0) rotating about the origin O of a Newtonian frame of reference.Let E3 be the Euclidean point space of the frame, and let the vector space V3 be its translation space. Choose a fixed right-handed orthonormal basis {EK, K = 1,2,3} for V3. Let LðV3;V3Þ be the vector space of linear mappings that take V3 into V3,i.e., the set of (second-order) tensors. The sets of symmetric and skew-symmetric tensors are denoted by Sym and Skw respec-tively. If B is any tensor, we have the decomposition B ¼ Bsym þ Bskw;Bsym ¼ 1

2 ðBþ BTÞ 2 Sym;Bskw ¼ 12 ðB� BTÞ 2 Skw, where BT

is the transpose of B. Bskw possesses an axial vector b such that Bskwc = b � c for every vector c. For any two vectors a,b belong-ing to V3, the tensor product a � b is defined by (a � b)c = a b � c for all c. LðV3;V3Þ becomes a Euclidean space when the usualinner product is employed: B � C = tr(BTC) = tr(BCT), for any tensors B, C. The tensor products EK � EL form an orthonormal ba-sis for LðV3;V3Þ : ðEK � ELÞ � ðEM � ENÞ ¼ dKMdLN , where dKM is the Kronecker delta. Let E9 be a Euclidean point space with trans-lation space LðV3;V3Þ and let O be the origin of E9.

Let X be any particle belonging to B, and let X and x, respectively, denote the position vectors of X in a fixed occupiablereference configuration of B and in the current configuration at time t. Let q be the mass density of B (not necessarily uni-form). Every rigid motion v of B that leaves O fixed is of the form

x ¼ vðX; tÞ ¼ QX; ð2:1Þ

where Q is a proper orthogonal tensor-valued function of t : QTQ = QQT = I, detQ = 1. It is clear that the configuration of B attime t can be represented by a point P belonging to a 3-dimensional manifoldM embedded in E9 and lying in an 8-dimen-sional sphere. (Many other tensors, including all the improper orthogonal ones, belong to this sphere.) The position vector ofP relative to O is Q.

The velocity field of B at time t has referential and spatial descriptions _x ¼ @v@t¼ _QX; _x ¼ Xx, respectively, where

X ¼ _QQ T ¼ �XT is the angular velocity tensor. Let x denote the angular velocity vector (Xc = x � c for all c). The acceler-ation field has the descriptions

€x ¼ €Q X ¼ Ax; A ¼ €QQ T ¼ _XþX2; ð2:2Þ

where A is the spatial acceleration gradient. Note that Asym ¼ X2; Askw ¼ _X.Let us choose rectangular Cartesian coordinates XK on E3, and then employ XK as convected coordinates on B. Thus,

X = XKEK, x = xKEK = XKeK, where we have set eK = QEK. The three vectors eK depend on t and form a right-handed orthonormalbasis for V3 for all t; they are tangent to the convected coordinate (straight) lines. The rotation tensor may be expressed asQ = QI = Q(EK � EK) = eK � EK. It is obvious that _eK ¼ _QEK ¼ XeK ¼ x� eK .

Suppose that u = u(x, t) and B = B(x, t) are, respectively, a vector field and a tensor field defined on the current configura-tion of B. We may represent these fields in the forms u = uKeK, B = BKLeK � eL. We may utilize the motion v to pull back u andB to the reference configuration to produce a vector field u0 and a tensor field B0 as follows:

u0 ¼ uK EK ¼ Q T u ¼ vHu; B0 ¼ BKLEK � EL ¼ Q T BQ ¼ vHB: ð2:3Þ

In general, u0 and B0 will vary with time. The push forward operation vw is the inverse of vw:

u ¼ vH

u0 ¼ Qu0; B ¼ vH

B0 ¼ QB0Q T : ð2:4Þ

We observe that vwQ = Q = vwQ.

1390 J. Casey / International Journal of Engineering Science 49 (2011) 1388–1396

The material derivatives of u and B may be expressed as

2 Forpushingderivati

3 Thi

_u ¼ u�þXu; _B ¼ B

�þXB� BX; ð2:5Þ

where

u�¼ _uK eK ; B

�¼ _BKLeK � eL ð2:6Þ

are the corotational rates of u and B. These rates may be written as Lie derivatives:2

u�¼ £u ¼ v

H

_vHu� �

¼ Q _u0; B�¼ £B ¼ v

H

_vHB� �

¼ Q _B0Q T : ð2:7Þ

We note that £I ¼ I�¼ 0, and £Q ¼ Q

�¼ v

H

_Q ¼ Q _QQ T ¼ QX.If the components uK are constants, then the pull back u0 is independent of time and _u0 ¼ 0; u

�¼ 0. We then say that the

field u is corotational.3 Likewise, if BKL are constants, then _B0 ¼ 0;B�¼ 0, and we say that B is corotational. It is obvious that for

all the motions (2.1), eK ¼ vH

EK ; £eK ¼ e�

K ¼ 0; eK � eL ¼ vHðEK � ELÞ; £ðeK � eLÞ ¼ eK � eL

�¼ 0.

LetR be the region in E3 which is occupied by the rigid body B at time t and let @R be the boundary ofR. Let dv and da bevolume and area elements, respectively. Let b be the body force field per unit mass acting on B at time t and let t be thetraction field acting on the boundary of B. Let T = TT be the Cauchy stress tensor.

Euler’s second law may be stated as

Z@R

x� tdaþZRqx� bdv ¼ _h ¼

ZRqx� €xdv ; ð2:8Þ

where h is the angular momentum of B about O. The Euler tensor of B at time t, taken about O, is defined by

E ¼ZRqx� xdv ð2:9Þ

and is a symmetric positive definite tensor. Let E0 be its value in the reference configuration of B. It follows from (2.9), (2.1),and (2.4)2 that

E ¼ QE0Q T ¼ vH

E0: ð2:10Þ

Since E0 is a constant tensor, E is corotational:

E ¼ EKLeK � eL; E0 ¼ EKLEK � EL; EKL ¼ const;

£E ¼ E�¼ 0; _E ¼ XE� EX:

ð2:11Þ

The inertia tensor about O is defined by J = (trE)I � E. It too is corotational.The angular momentum of B about O is given in vector form by h = Jx and, equivalently, in tensor form by

bH ¼ 2ðXEÞskw ¼ XEþ EX; ð2:12Þ

where bH is the skew-symmetric tensor whose axial vector is h (i.e., bHc ¼ h� c for all c). Let m be the resultant external tor-que acting on B at time t, and let cM be the corresponding skew-symmetric torque tensor. Euler’s second law may be writtenin vector and tensor forms as

m ¼ _h ¼ J _xþx� Jx; cM ¼ _bH ¼ 2ðAEÞskw ¼ 2 €Q E0Q T� �

skw: ð2:13Þ

Since each term in (2.13)2 is skew-symmetric, this tensor equation furnishes three linearly independent componentequations.

The kinetic energy T of B at time t is given by the expressions

T ¼ 12x � Jx ¼ 1

2tr XEXT� �

¼ 12

tr _Q E0_Q T

� �¼ 1

2_QKLELM

_Q KM: ð2:14Þ

For a rigid continuum, useful additional global information can be obtained by utilizing Signorini’s mean stress theorem.Thus, define the mean stress T and the Möbius tensor M of the external force system by

T ¼ 1V

ZR

Tdv; M ¼Z@R

t� xdaþZRqb� xdv; ð2:15Þ

deformable continua, the Lie derivative of a spatial field is obtained by pulling it back to a fixed reference configuration, differentiating materially, andthe result forward to the current configuration (see e.g., (Casey & Papadopoulos, 2002)). In the present case, the motions are all rigid, and the Lie

ves become corotational rates.s is a special case of ‘‘material transport’’ of a field (Casey & Papadopoulos, 2002).

J. Casey / International Journal of Engineering Science 49 (2011) 1388–1396 1391

where V is the volume of B. Then, it can be shown (Casey, 2004; Casey, 2007) that the differential equations that describe thebalance of linear momentum (Cauchy’s first law) imply that4

4 It ispoint o(2.16), tand the

5 Forin Case

M� VT ¼ AE ¼ €Q E0Q T : ð2:16Þ

The skew-symmetric and symmetric parts of (2.16) are

12cM ¼ ðAEÞskw ¼ €QE0Q T

� �skw; Msym � VT ¼ ðAEÞsym ¼ €Q E0Q T

� �sym: ð2:17Þ

Eq. (2.17)1 is equivalent to Euler’s second law, while (2.17)2 supplies new global information involving the mean reactivestress that is required to support the motion of the rigid body.

We may solve (2.16) for €Q :

€Q ¼ M� VT� �

QE�10 ¼ Msym � VT

� �QE�1

0 þ12cMQE�1

0 : ð2:18Þ

If we introduce an abstract force tensor U in Hertzian space by

U ¼ m M� VT� �

QE�10 ; ð2:19Þ

then (2.18) takes the suggestive form

U ¼ m €Q : ð2:20Þ

This is the full dynamical equation for the abstract particle Z moving in Hertzian space. It is equivalent to the two equationsin (2.17).

3. Dynamics in Hertzian space

The configuration of B in physical space at time t is uniquely determined by the tensor Q, which belongs to the vectorspace LðV3;V3Þ and corresponds to a point P in the point space E9. Define an alternative inner product on LðV3;V3Þ by

hB;Ci0 ¼1m

trðBE0CTÞ; ð3:1Þ

where B, C are any tensors. We observe that hEK � EL;EM � ENi0 ¼ 1m dKMELN ¼ const. Let hE9i0 denote the Euclidean point space

that is obtained from E9 when the usual inner product ‘‘�’’ is replaced by (3.1). The new ‘‘distance’’, dðO;PÞ, from O to P, isgiven by

d2 ¼ hQ ;Q i0 ¼1m

trE ¼ 1m

trE0 ¼ k2; ð3:2Þ

where the constant k is the radius of gyration of B about O. Thus, P now lies in an 8-dimensional sphere of radius k and cen-ter O in hE9i0. We call hE9i0 the Hertzian configuration space of B, or simply Hertzian space.5 We note thath _Q ;Q i0 ¼ trðXEÞ ¼ X � E ¼ 0.

In terms of the inner product (3.1), the kinetic energy of B in (2.14) may be expressed as

T ¼ 12

mh _Q ; _Q i0: ð3:3Þ

Thus, the abstract particle Z, of mass m, may be regarded as having the same kinetic energy as B as it moves through Hertzianspace with velocity _Q .

The orthogonality condition places six restrictions on the nine components QKL of Q. Consequently, three parameters suf-fice to determine Q (e.g., three Euler angles). Thus, the point P lies in a 3-dimensional manifold M embedded in Hertzianspace. M is called the configuration manifold of B. Let nc (c = 1,2,3) be arbitrary coordinates on M (or ‘‘generalizedcoordinates’’):

Q ¼ Q ðncÞ ðc ¼ 1;2;3Þ: ð3:4Þ

The three tensors

Ac ¼@Q@nc ¼

@Q MN

@nc EM � EN ð3:5Þ

worth mentioning that, even for a rigid continuum, the standard Cauchy tetrahedron argument establishes the existence of the Cauchy tensor at eachf the body. It is true that the stress in a rigid continuum is indeterminate (i.e., not specified by a constitutive equation). Nonetheless, as indicated inhe mean stress plays a mechanical role. In the static case, the mean stress can be calculated from the body force field, the traction field on the boundary,volume of the body.

a system of particles, Hertz, 1900 employed a Riemannian metric based on the moment of inertia of the system. The inner product (3.1) was introducedy, 1995.

1392 J. Casey / International Journal of Engineering Science 49 (2011) 1388–1396

are tangent to the coordinate curves on M at the generic point P. They form a basis for the 3-dimensional tangent vectorspace toM at P, which is denoted by T PM. Let N be any tensor that is h i0-orthogonal to T PM; N belongs to the 6-dimen-sional subspace N which is the orthogonal complement of T PM when the inner product (3.1) is used. We note that

@Ac

@nk ¼@Ak

@nc : ð3:6Þ

It is obvious that eK = QEK are functions of nc. Clearly,

@eK

@nc ¼ AcEK ¼ XceK ; ð3:7Þ

where we have set

Xc ¼ AcQ T : ð3:8Þ

Since eK � eK = I, it follows with the help of (3.7) that

Xc þXTc ¼ 0; ð3:9Þ

so that each Xc is a skew-symmetric tensor. Let xc be the axial vector of Xc.The following formulae may be recorded:

Ac ¼@eK

@nc � EK ; Xc ¼@eK

@nc � eK : ð3:10Þ

Taking the time derivative of Q, and noting (3.5), we obtain

_Q ¼ _ncAc; ð3:11Þ

which resolves the velocity of the abstract particle Z into components in the tangent space T PM. It then follows that theangular velocity tensor and vector may be expressed as

X ¼ _ncXc; x ¼ _ncxc: ð3:12Þ

Clearly,

@ _Q@ _nc¼ Ac;

@X

@ _nc¼ Xc;

@x

@ _nc¼ xc: ð3:13Þ

Lemma 3.1.

(a) Relative to the inner product (3.1), a tensor N is h i0-orthogonal to T PM if and only if it is of the form

N ¼ SQE�10 ; ð3:14Þ

where S is a symmetric tensor.(b) If B is any tensor, and we employ the decomposition

BQE�10 ¼ BsymQE�1

0 þ BskwQE�10 ; ð3:15Þ

then, the component BsymQE�10 is h i0-orthogonal to T PM, and

hBQE�10 ;Aci0 ¼ hBskwQE�1

0 ;Aci0: ð3:16Þ

(c) A tensor R belongs to T PM if and only if it is of the form

R ¼WQ ; ð3:17Þ

where W is a skew-symmetric tensor.

Proof.

(a) A tensor N is h i0-orthogonal to T PM if and only if hN,Aci0 = 0, or equivalently,

trðNE0Q TXcÞ ¼ 0: ð3:18Þ

Suppose that N is h i0-orthogonal to T PM. It then follows from (3.18) and (3.12)1 that

tr NE0Q TX� �

¼ _nctr NE0Q TXc

� �¼ 0: ð3:19Þ

J. Casey / International Journal of Engineering Science 49 (2011) 1388–1396 1393

Since X can be any skew-symmetric tensor, and X is independent of NE0QT, we deduce from (3.19) that (NE0QT)skw

must vanish, and hence that NE0QT is a symmetric tensor S (=(NE0QT)sym). Therefore, N is of the form (3.14). Con-versely, let S be any symmetric tensor, and define N by (3.14). Then tr (NE0QTXc) = tr (SXc) = 0 and (3.18) is satisfied.Therefore, N is h i0-orthogonal to T PM.

(b) Since Bsym is a symmetric tensor, it follows directly from (a) that the tensor BsymQE�10 is h i0-orthogonal to T PM. Hence,

6 A malong so(3.17) m

7 As a

hBsymQE�10 ;Aci0 ¼ 0: ð3:20Þ

Taking an inner product of both sides of (3.15) with Ac, and employing (3.20), we immediately arrive at (3.16).(c) Suppose that R 2 T PM. Then,

hR;Ni0 ¼ 0 ð3:21Þ

for all N of the form (3.14). Hence, in view of (3.1), tr(RQTST) = 0 for every symmetric tensor S. Therefore, (RQT)sym = 0,and RQT must be a skew-symmetric tensor W (=(RQT)skw). Therefore, (3.17) holds.6 Conversely, let W be any skew-symmetric tensor and define R by (3.17). With the use of (3.14) and (3.1), it may be readily seen that (3.21) then holdsfor all N. Therefore R 2 T PM. h

The following remarks may be made:

(i) Choosing S = I in (3.14), we find that QE�10 is h i0-orthogonal to T PM. Similarly, choosing S = QE0QT = E, we see that Q is

hi0-orthogonal to T PM.(ii) If S is symmetric, so is the tensor S ¼ Q T SQ , and conversely. Hence, N in (3.14) may also be expressed as

N ¼ QSE�10 ; ð3:22Þ

(and in other ways as well). Similarly, W is skew-symmetric if and only if the tensor W ¼ Q T WQ is skew-symmetric,and R in (3.17) can be written equivalently as

R ¼ QW: ð3:23Þ

Since Q�¼ QX, it follows from (3.23) that Q

�belongs to T PM. (It can also be easily checked that hQ

�;Ni0 ¼ 0.)

(iii) With the aid of (3.1) and (3.14), we note that

m BskwQE�10 ;N

D E0¼ trfBskwE�1Sg; ð3:24Þ

this need not vanish, since the product E�1S is not necessarily symmetric.7 Thus, in general, the tensor BskwQE�10 has a

component that is h i0-orthogonal to T PM. (For an inertially spherical rigid body, E0 = mk2I (k = const. > 0), and BskwQE�10

belongs to T PM.)As was mentioned in (i), both Q and QE�1

0 are h i0-orthogonal to T PM. Setting S = E and S = I successively in (3.24), it isobvious that hBskwQE�1

0 ;Q i0 ¼ 1m trðBskwÞ ¼ 0; hBskwQE�1

0 ;QE�10 i0 ¼ 1

m trðBskwE�1Þ ¼ 0.(iv) The subspaces Sym and Skw are orthogonal complements in LðV3;V3Þ when the usual inner product is employed. N is

the orthogonal complement of T PM relative to the inner product (3.1). It is a 6-dimensional subspace of LðV3;V3Þ. Wemay rephrase the results (3.14) and (3.22) as

N ¼ ðSymÞQE�10 ¼ Q ðSymÞE�1

0 : ð3:25Þ

Let D be the image of Skw under post-multiplication by QE�10 :

D ¼ ðSkwÞQE�10 : ð3:26Þ

Let D1 and D2 be any tensors belonging to D. Then, there exist skew-symmetric tensors W1, W2 such thatD1 ¼W1QE�1

0 , D2 ¼W2QE�10 . Since W1 + W2 is also skew-symmetric, ðW1 þW2ÞQE�1

0 belongs to D, and henceD1 + D2 belongs to D. Likewise, for any real number a, aW1 is skew-symmetric, ðaW1ÞQE�1

0 belongs to D, and henceaD1 belongs to D. Therefore, D is a subspace of LðV3;V3Þ. To find the dimension of D, let {W1, W2, W3} be a basisfor Skw, and let DK ¼WK QE�1

0 ðK ¼ 1;2;3Þ. Clearly, aKDK = 0 implies that aKWK = 0. Since {WK} is a basis, aK = 0. There-fore, D1, D2, D3 are linearly independent. Furthermore, if D is any tensor in D, there exists a skew-symmetric tensor Wsuch that D ¼WQE�1

0 . Representing W as W = aKWK, we see that D = aKDK, and hence D1, D2, D3 generate D. Thus, {DK}is a basis for D, and D is 3-dimensional.

ore primitive proof of (3.17) that does not employ the inner product (3.1) may be given. Thus, every element of T PM may be regarded as the velocityme curve inM and must be of the form _Q ¼ XQ , where X is skew-symmetric. For any given value of Q, X can be prescribed independently, and henceust hold.

counterexample, consider the trace of0 1 0�1 0 0

0 0 0

0@ 1A l1 0 00 l2 00 0 l3

0@ 1A 0 1 01 0 00 0 0

0@ 1A; where l1, l2, l3 are arbitrary positive numbers.

1394 J. Casey / International Journal of Engineering Science 49 (2011) 1388–1396

If C is any tensor, we may express C as C ¼ BQE�10 , with B = CE0QT. Then, C ¼ BsymQE�1

0 þ BskwQE�10 . Thus, C can be un-

iquely resolved into a component in N and a component in D. In other words, the direct sum of N and D is LðV3;V3Þ:

D�N ¼ LðV3;V3Þ: ð3:27Þ

(v) The results (3.17) and (3.23) can be rephrased as

T PM ¼ ðSkwÞQ ¼ Q ðSkwÞ: ð3:28Þ

Additionally, (3.26) can then be rewritten as

D ¼ ðT PMÞE�10 : ð3:29Þ

(vi) The three skew-symmetric tensors Xc, given by (3.8) form a basis for Skw. For, suppose that bcXc = 0 for real numbersb1, b2, b3. Then, in view of (3.8), bcAc = 0. Since {AK} form a basis for T PM, it follows that bc = 0. Therefore, X1, X2, X3

are linearly independent elements of Skw. Since Skw has only three dimensions, the three tensors Xc must form abasis for Skw.

The metric coefficients for M are defined by

Aab ¼ hAa;Abi0 ð3:30Þ

and the square of the length of the line element of M is

ds2 ¼ hdQ ;dQ i0 ¼ h _Q dt; _Q dti0 ¼ Aabdnadnb: ð3:31Þ

The configuration manifold M is Riemannian.The kinetic energy of Z may be expressed as

T ¼ 12

mAab_na _nb; ð3:32Þ

where (3.3), (3.11), and (3.30) have been employed. In view of (3.31) and (3.33),

ds2 ¼ 2Tm

dt2: ð3:33Þ

We will consider in a moment a special oblique decomposition of €Q . But, we also recall that €Q can always be decomposedinto a unique piece €Q k that lies in T PM and a unique piece €Q? that is h i0-orthogonal to T PM:

€Q ¼ €Q k þ €Q?; h €Q?;Aai0 ¼ 0: ð3:34Þ

Theorem 3.1.

(a) For the abstract particle Z moving in Hertzian space hE9i0, the tensor €Q admits the decomposition

€Q ¼ €Q E0Q T� �

symQE�1

0 þ €Q E0Q T� �

skwQE�1

0 ; ð3:35Þ

the first component of which lies in the 6-dimensional subspace N , and the second component of which lies in the 3-dimen-sional subspace D (see Fig. 1).

(b) The dynamical Eq. (2.20) is equivalent to the two component equations

12cMQE�1

0 ¼ €QE0Q T� �

skwQE�1

0 ; Msym � VT� �

QE�10 ¼ €QE0Q T

� �sym

QE�10 : ð3:36Þ

Each term in (3.36)1 is a tensor belonging to the subspace D, while each term in (3.36)2 is a tensor belonging to the subspaceN . Also, (3.36)1 is equivalent to Euler’s second law.

(c) Eq. (3.36)1 is equivalent to

12cMQ ¼ €Q E0Q T

� �skw

Q ; ð3:37Þ

in which each term is a tensor belonging to the tangent space T PM.(d) Eq. (2.20) implies the three covariant component equations

hm €Q ;Aci0 ¼ Mc; ð3:38Þ

where

Mc ¼ hU;Aci0: ð3:39Þ

J. Casey / International Journal of Engineering Science 49 (2011) 1388–1396 1395

(e) The following identities hold:

m €Q ;Ac

D E0¼ m €QE0Q T

� �skw

QE�10 ;Ac

D E0¼ m €Q k;Ac

D E0¼ €Q E0Q T� �

skw�Xc¼ðAEÞskw �Xc¼ðAEÞ �Xc¼ J _xþx� Jxð Þ �xc:

ð3:40Þ

(f) Other expressions for Mc are

Mc ¼12

mcMQE�10 ;Ac

� �0¼ 1

2cM �Xc ¼m �xc: ð3:41Þ

(g) The three component equations in (3.38) are equivalent to the tensor Eq. (2.13)2 [and hence also to (2.17)1 and (3.36)1].

Proof.

(a) For any €Q , we may write €Q ¼ €QE0Q T� �

QE�10 , which readily yields (3.35). By Lemma 3.1 (b), the component

ð €QE0Q TÞsymQE�10 lies in the subspace N which is the hi0-orthogonal complement of T PM, while in view of (3.26),

the component €QE0Q T� �

skwQE�1

0 lies in the subspace D.(b) The dynamical Eq. (2.20) is equivalent to the two equations in (2.17). Since E0 is invertible and Q is orthogonal, it is

obvious that (3.36)1 is equivalent to (2.17)1. By virtue of (3.26), each term in (3.36)1 represents a tensor lying in thesubspace D. Similarly, (3.36)2 is equivalent to (2.17)2. By virtue of (3.25)1, each term in (3.36)2 is a tensor belonging toN . Further, (2.17)1 is equivalent to Euler’s second law, and hence, so also is (3.36)1.

(c) Clearly, (3.37) is equivalent to (3.36)1. In view of (3.28)1, each term in (3.37) represents a tensor belonging to T PM.(d) Taking the inner product (3.1) of both sides of (2.19) with the basis vectors Ac 2 T PM, and defining the covariant com-

ponents of U by (3.39), we arrive at once at (3.38).(e) Using the decomposition (3.35) and noting that the component belonging to N satisfies

€Q E0Q T� �

symQE�1

0 ;Ac

� �0¼ 0; ð3:42Þ

we obtain (3.40)1. The second equality in (3.40) follows immediately from (3.34)1,2. With the aid of (3.1) and (3.8), it is clearthat

m €QE0Q T� �

skwQE�1

0 ;Ac

D E0¼ tr €QE0Q T

� �skw

XTc

n o¼ €QE0Q T� �

skw�Xc: ð3:43Þ

Thus, the third equality in (3.40) holds. The fourth equality then follows easily with the use of (2.2)2 and (2.10)1. Then,observing that (AE)sym�Xc = 0, we deduce the fifth equality. To obtain the sixth equality, recall that J _xþx� Jx is the axialvector of 2(AE)skw and xc is the axial vector of Xc.

(f) It is evident from (2.18), (2.19) and (3.39) that

Mc ¼12

mcMQE�10 ;Ac

� �0þm Msym � VT

� �QE�1

0 ;Ac

D E0: ð3:44Þ

But the last term vanishes because ðMsym � VTÞQE�10 belongs toN . Therefore, (3.41)1 holds. The second equality in (3.41) may

be obtained using the same steps by which (3.43) was deduced. The third inequality in (3.41) follows from the fact that m isthe axial vector of cM and xc is the axial vector of Xc.

(g) Eq. (2.13)2 is equivalent to (2.17)1, which in turn is equivalent to (3.36)1. Therefore (2.13)2 is equivalent to (3.36)1.Taking the inner product h i0 of each side of (3.36)1 with Ac, and utilizing (3.41)1 and (3.40)1, we arrive at (3.38). Conversely,suppose that (3.38) holds. It then follows from (3.40)3 and (3.41)2 that

AEð Þskw �Xc ¼12cM �Xc: ð3:45Þ

But, recall from Remark 3.1 (vi) that {Xc} is a basis for Skw. Hence, if W is any skew-symmetric tensor, we may express W asW = WcXc. It then follows from (3.45) that

ðAEÞskw �12cM� �W ¼ 0 ð3:46Þ

for all W 2 Skw. Since the terms in braces are independent of W, they must vanish, thereby yielding the desired result(2.13)2. h

Theorem 3.2.

m €Q ;Ac

D E0¼ d

dt@T

@ _nc

�� @T@nc : ð3:47Þ

1396 J. Casey / International Journal of Engineering Science 49 (2011) 1388–1396

Proof (See (Casey, 1995)). It follows from (3.11), (3.5) and (3.6) that

@ _Q@nk ¼ _nc @Ac

@nk ¼ _nc @Ak

@nc ¼ _Ak: ð3:48Þ

From (3.3), we obtain

@T@nc ¼ m _Q ;

@ _Q@nc

* +0

;@T

@ _nc¼ m _Q ;

@ _Q@ _nc

* +0

: ð3:49Þ

Hence, in view of (3.48) and (3.13),

@T@nc ¼ m _Q ; _Ac

D E0;

@T

@ _nc¼ m _Q ;Ac

D E0: ð3:50Þ

It follows from (3.50)2 and the constancy of E0 that

ddt

@T

@ _nc

�¼ m €Q ;Ac

D E0þm _Q ; _Ac

D E0: ð3:51Þ

The desired result (3.47) follows at once from (3.51) and (3.50)1. h

Theorem 3.3. An equivalent form of (2.13)2, which represents Euler’s second law for a rigid body rotating about a fixed point, isgiven by Lagrange’s equations

ddt

@T

@ _nc

�� @T@nc ¼ Mc: ð3:52Þ

Proof. By virtue of (3.47), the Eqs. (3.52) are equivalent to (3.38). But, in part (g) of Theorem 3.1, it was shown that the threeequations in (3.38) are equivalent to (2.13)2. Therefore, (3.52) is equivalent to (2.13)2. h

References

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Casey, J. (2004). Pseudo-rigid continua: Basic theory and a geometrical derivation of Lagrange’s equations. Proceedings of the Royal Society of London, A 460,2021–2049.

Casey, J. (2007]). A new definition of a pseudo-rigid continuum. Note di Matematica, 27, 43–53.Casey, J., & O’Reilly, O. M. (2006]). Geometrical derivation of Lagrange’s equations for a system of rigid bodies. Mathematics and Mechanics of Solids., 11,

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