An Introduction to the Deviatoric Tensor Decomposition in Three Dimensions and its Multipole Representation
aa r X i v : . [ m a t h . G M ] S e p A N I NTRODUCTION TO THE D EVIATORIC T ENSOR D EC OMPOSITION IN T HR EE D IMENSIONS AND ITS M ULTIPOLE R EPR ESENTATION
Chiara Hergl
Department of Computer ScienceLeipzig UniversityGermany
Thomas Nagel
Geotechnical InstituteTechnische Universität Bergakademie FreibergGermany
Gerik Scheuermann
Department of Computer ScienceLeipzig UniversityGermanySeptember 25, 2020 A BSTRACT
The analysis and visualization of tensor fields is a very challenging task. Besides the cases of zeroth-and first-order tensors, most techniques focus on symmetric second-order tensors. Only a few worksconcern totally symmetric tensors of higher-order. Work on other tensors of higher-order than twois exceptionally rare. We believe that one major reason for this gap is the lack of knowledge aboutsuitable tensor decompositions for the general higher-order tensors. We focus here on three dimen-sions as most applications are concerned with three-dimensional space. A lot of work on symmetricsecond-order tensors uses the spectral decomposition. The work on totally symmetric higher-ordertensors deals frequently with a decomposition based on spherical harmonics. These decompositionsdo not directly apply to general tensors of higher-order in three dimensions. However, another optionavailable is the deviatoric decomposition for such tensors, splitting them into deviators. Togetherwith the multipole representation of deviators, it allows to describe any tensor in three dimensionsuniquely by a set of directions and non-negative scalars. The specific appeal of this methodology isits general applicability, opening up a potentially general route to tensor interpretation. The underly-ing concepts, however, are not broadly understood in the engineering community. In this article, wetherefore gather information about this decomposition from a range of literature sources. The goalis to collect and prepare the material for further analysis and give other researchers the chance towork in this direction. This article wants to stimulate the use of this decomposition and the searchfor interpretation of this unique algebraic property. A first step in this direction is given by a detailedanalysis of the multipole representation of symmetric second-order three-dimensional tensors. K eywords Tensor · Higher-order · Deviatoric Decomposition · Multipole Decomposition · Stiffness Tensor
Analysis and visualization of tensor fields focuses mainly on symmetric second-order fields like mechanical stress andstrain. There is also work on general second-order tensor fields like the velocity gradient in fluid flow. But there aremany more tensor fields used in natural sciences and engineering to describe the behaviour of matter and various fields.However, there is only very limited work on their analysis and visualization. In the life sciences, there is some workon the analysis and visualization of higher-order tensors, especially higher-order diffusion tensors. However, thesetensors are totally symmetric and of even-order. If this property does not hold, literature gets exceptionally sparse.
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25, 2020A look at most of this work [1–4] reveals that analysis and visualization of symmetric tensor fields nearly alwaysuses the eigendecomposition, i.e. eigenvalues and eigenvectors. For general tensors of second-order, authors splitthem into the asymmetric part which is interpreted as vector or rotation and a symmetric part which again is analyzedusing eigenvalues and eigenvectors most often. The work on totally symmetric tensors of higher even order uses adecomposition based on spherical harmonics in most cases. Unfortunately, these decompositions do not work forgeneral tensors of higher order in three dimensions. We therefore suspect that the lack of work on the analysis andvisualization of other tensors is caused by missing knowledge on tensor decompositions of these tensors.The good news is that there is such a decomposition, namely the deviatoric decomposition, allowing to split eachtensor in three dimensions into a set of deviators, i.e. traceless totally symmetric tensors. Furthermore, these deviatorscan be uniquely represented by a finite set of directions and a non-negative scalar. The generality of this methodsrenders it quite appealing for study in tensor analysis and visualization. Because this decomposition is not very wellestablished and has gone largely unnoticed in the applied sciences, this work gives an overview of the so calleddeviatoric decomposition. The deviatoric decomposition is an orthogonal decomposition of a tensor of arbitrary-orderup to dimension three. The deviatoric decomposition of a second-order tensor is well known. It is given by thetensor’s symmetric part and a vector, representing its asymmetric part. These deviators again can be represented bythe symmetrization of a tensor product of first-order tensors (vectors) which are called multipoles. The deviatoricdecomposition and the multipole representation are unique, so each tensor can be represented by a unique set offirst-order tensors, i.e. vectors.The underlying calculations go back to Maxwell [5] and were summarized by Backus [6] who also provided furtherinformation and some mathematical background. Zou et al. [7] gave explicit formulae to calculate the multipoles of atensor.The goal of this article is to give a summary of the multipole decomposition to enable further analysis of this repre-sentation. Until now, the meaning of the individual deviators is, to the best of our knowledge, not known for caseshigher than order two. This paper should enable the analysis of higher-order tensors using the deviatoric decomposi-tion within a single framework. To gain a first impression we analyze the second-order multipole decomposition forsymmetric tensors in more detail and discuss a connection between the multipoles and the eigenvectors.The paper is organized in the following way. First, a general introduction to the required tensor algebra is given. Then,the mathematical background of the deviatoric decomposition and the multipole representation is presented. Becauseof their prevalence and familiarity, the second- and fourth-order decompositions are described in more detail to explainthe calculations. The final chapter describes a more or less well known application of the multipole decomposition:The multipoles of a stiffness tensor can be used to calculate the anisotropy type of the underlying material. However,as stated earlier, the exact meaning of the various deviators is, to the best of our knowledge, not known. In furtherworks we want to change this. Identifying the meaning of such abstract object requires interpretations from a rangeof different fields which can all contribute unique insights. Therefore, we want to provide other researchers with thenecessary tools and concepts to analyze the meaning of the deviators. Furthermore, we hope to find more applicationsfor the deviatoric decomposition and the multipole representation to learn more about their potential meaning in thelight of their generality.
The reader may consider the dimension n to be three in most cases. To ease the description without loss of generality,we will assume an orthonormal basis at all times, so there will be no distinction between co- and contravariant tensorsand all indices will be lower indices.We denote the n -dimensional Euclidean vector space as V n . Its scalar product is a bilinear mapping of two vectors x , y to a real number and is denoted as x · y . We assume that we are given an orthonormal basis { e , . . . , e n } , i.e. e i · e j = δ i j with the Kronecker delta δ i j . Because no distinction between co- and contravariant tensors is required, wedefine an n -dimensional tensor of order q as multilinear map of q vectors to the real numbers T : ( V n ) q → R . (1)As a multilinear map, the tensor can also be described by its coefficients with respect to a fixed orthonormal basis of V n , say { e i } T ( e i , . . . , e i q ) = T i ,..., i q (2)Therefore, a zeroth-order tensor can be represented as a scalar, a first-order tensor as a vector of dimension n and asecond-order tensor as an n × n -matrix. Higher-order tensors can be represented as arrays of order q . For example, thecoefficients of a three-dimensional fourth-order tensor can be represented as 3 × × × EPTEMBER
25, 2020the tensor. If there is another orthonormal basis { f , . . . , f n } of V n expressed in the basis { e i } as f j = ∑ ni = c i j e i then T is represented by the numbers T ′ j j ... j q = n ∑ i = . . . n ∑ i q = c i j . . . c i q j q T i ... i q (3)with respect to the basis { f j } . In the following, we describe our tensors always as coefficient arrays of order q , i.e. asrepresented with respect to the basis { e i } . This is also the representation in our (and probably any typical) visualizationapplication.Let S , T be tensors of order q over V n , and a , b ∈ R . Then, the set of all q -arrays, with the scalar multiplication andthe tensor addition a S + b T = aS i i ... i q + bT i i ... i q (4)create a n q -dimensional vector space V qn representing all n -dimensional tensors of order q .An often used tensor operation is the tensor product or outer product . The tensor product of a q th-order tensor A and a r th-order tensor B results in a ( q + r ) th-order tensor C as follows C i ... i q j ... j r = A ⊗ B = A i ... i q B j ... j r (5)Mathematically, this creates an algebra of all tensors of arbitrary order over V n .Another tensor operation we use is called tensor contraction . The tensor contraction is the summation over a deter-mined number of indices. The single contraction C = A · B is the summation of two tensors A and B over one index C i i ... i n − j ... j m = A · B = n ∑ k = A i i ... i n − k B k j ... j m = A i i ... i n − k B k j ... j m (6)where the implicit summation over a repeated index, i.e. A i i ... i n − k B k j ... j m , is called Einstein summation convention.The double contraction A : B is analogously defined as C i i ... i n − j ... j m = A : B = n ∑ k = n ∑ l = A i i ... i n − kl B kl j ... j m = A i i ... i n − kl B kl j ... j m (7)One can define a trace tr i , j for each index pair i , j , but here we will define the general trace tr () of a tensor as thefollowing sum about the first two indices tr ( T ) = tr , ( T ) = n ∑ s = D ssi i ... i n (8)which generalizes the notion well-known for second-order tensors. Therefore, a tensor of order higher than two hasmore than one trace. The determinant of a second-order tensor T is defined asdet ( T ) = T T T + T T T + T T T (9) − T T − T T − T T Let S q be the q th -order symmetric group of all q ! permutations π on the integers ( , , . . . , q ) . Let T be a three-dimensional tensor of order q . Then the permutation of the tensor π T can be defined as ( π T ) i π ... i π q = T i ... i q (10)We call a tensor T totally symmetric if π T = T for all π ∈ S q . So, every symmetric second-order tensor is totallysymmetric. A tensor of order 2 n can have the so called minor symmetry , which means it is invariant to indexpermutations in the sets { , } and { n − , n } of index positions. These even-order tensors can also have the majorsymmetry , which describes the invariance of a permutation of the index positions { , . . . , n } with { n + , . . ., n } . Fora fourth-order three-dimensional tensor T the minor symmetries and the major symmetry are given by T i jkl = T jikl = T i jlk , T i jkl = T kli j . (11)It is important for the following discussion that total symmetry implies more symmetry than minor and major sym-metries alone. This becomes clear if one realizes that a totally symmetric three-dimensional fourth-order tensor has3 EPTEMBER
25, 202015 independent coefficients, i.e. dim ( S ) =
15. In contrast, a three-dimensional fourth-order tensor with minor andmajor symmetries has 21 independent coefficients. Based on these definitions, we define the totally symmetric parts T of a general tensor as s T = q ! ∑ π ∈S q π T (12)We call the remainder a T = T − s T (13)the asymmetric part of T . Obviously, the asymmetric part of any tensor has no totally symmetric part which createsa vector space of all asymmetric tensors which is actually orthogonal to the totally symmetric tensor space, see [6].The symmetric and traceless part of a tensor T will be described by ⌊ T ⌋ . For the tensor product n ⊗ m it is ⌊ n ⊗ m ⌋ = ( n ⊗ m + m ⊗ n ) − ( n · m ) I . This work focuses on so called deviators . In the visualization literature and classical mechanics texts, a deviatoris defined as a traceless tensor of second-order. Here, we extend this definition following Zou et al. [8] to generalthree-dimensional q th-order deviators. A deviator D ( q ) is a three-dimensional tensor of arbitrary order q that is totallysymmetric and traceless, i.e. D i i i ... i q = D i i i ... i q = · · · = D i q i i ... i , tr (cid:16) D ( q ) (cid:17) = The best known decomposition for n × n tensors is the spectral decomposition. Such a tensor T can be described bythree n -dimensional eigenvectors v i and three eigenvalues λ i T = ∑ i = λ i v i ⊗ v i (15)For symmetric tensors the results are real.General tensors of order higher than two can not be decomposed using this decomposition. There are some specialcases where a generalization of the spectral decomposition exists. One of these special cases is for example a fourth-order tensor T with the two minor symmetries and the major symmetry. By choosing a suitably defined tensor basisexploiting these symmetries, this tensor can be mapped onto a symmetric second-order six-dimensional tensor usingthe Mandel or Kelvin mapping [9–11] as follows: K mn ( T i jkl ) = T T T √ T √ T √ T T T T √ T √ T √ T T T T √ T √ T √ T √ T √ T √ T T T T √ T √ T √ T T T T √ T √ T √ T T T T . (16)This tensor can be decomposed by the known second-order spectral decomposition with the eigenvalues Λ i . The resultsare six first-order six-dimensional tensors. These can be mapped using the inverse Kelvin mapping onto second-orderthree-dimensional tensors M i . These are the eigentensors of the fourth-order tensor T , which can be represented by T = ∑ i = Λ i M i ⊗ M i . (17) There is a well known relation between totally symmetric tensors and spherical harmonics. This section will describethis connection, following the representation in Backus’ [6] paper.Let T be a three-dimensional tensor of order q . We consider the polynomial T ( r ) = T i ... i q r i . . . r i q (18)4 EPTEMBER
25, 2020where r = ( r , r , r ) are coordinates with respect to our orthonormal basis ( e , e , e ) . All monomials of this polyno-mial have the degree q , so T ( r ) is a homogeneous polynomial of degree q . It is called the polynomial generated by T . Since the products of the r i j are products of real numbers and therefore commutative, we have T ( r ) = S ( r ) if S isthe totally symmetric part of T . More specifically, it is S i ... i q = q ! ∂ i . . . ∂ i q [ T ( r )] (19)Therefore, two different totally symmetric tensors generate different polynomials, but two tensors with the same totallysymmetric part generate the same polynomial. Let P q be the linear space of all homogeneous polynomials of degree q in three dimensions. Then, we have an isomorphism between P q and S q , the space of the totally symmetric tensorsof order q .Harmonic polynomials P ( q ) ( r ) in P q are called spherical harmonics . The well-known decomposition of homoge-neous polynomials, interpreted as functions on the sphere, by spherical harmonics states the following: For everypolynomial in P q , there are spherical harmonics H ( q ) ( r ) , H ( q − ) ( r ) , . . . such that P ( q ) ( r ) = H ( q ) ( r ) + r H ( q − ) ( r ) + . . . (20)Applying the three-dimensional Laplacian, it follows that H ( q ) is unique. This allows a nice characterization of devia-tors. Let H be a q th -order totally symmetric tensor that generates a harmonical polynomial H ( r ) . It follows ∆ H ( r ) = H j ji ... i q r i . . . r i q = S oforder q , we may use the spherical harmonic decomposition of S ( r ) to compute tensors H ( q ) using (20). By comparingthe coefficients, we can derive that S = H ( q ) + s ( I H ( q − ) ) + . . . (22)where I is the identity tensor and s () denotes the totally symmetric part of a tensor as before. Therefore, any totallysymmetric tensor in three dimensions is uniquely described by a series of deviators H i , which will be described in thefollowing by D i , of order q , q − , . . . . As mentioned before, the connection between totally symmetric tensors and spherical harmonics is well known. Thus,each totally symmetric tensor can be described by deviators. Also, every tensor of arbitrary order up to dimensionthree can be described by deviators uniquely. Backus [6] described this so-called deviatoric decomposition .It is well known that every tensor T of order q can be decomposed into a totally symmetric s T and an asymmetric part a T T = s T + a T The deviatoric decomposition of the totally symmetric part is described in the previous section using the sphericalharmonics. The deviatoric decomposition of the asymmetric part is far less well-known. The gap is bridged by aunique isomorphism from the totally symmetric tensor space S p into the asymmetric tensor space A q of order q φ ( S p ) → A q . Through Schur’s Lemma [12] it is known that the space of the isormorphisms uniquely determines the isomorphism.These totally symmetric tensors can then be decomposed into deviators. Thus, eventually this isomorphism allows thedecomposition of the asymmetric part of the tensor into deviators.
We complete our description by decomposing deviators even further as tensor products of vectors which are called multipoles . This result was first described by Maxwell [5], who found it as an elegant geometric description ofspherical harmonics. Using a theorem by Sylvester [13], Maxwell stated in our terms that for any three-dimensionaldeviator D of order q , there are q unit real vectors n , . . . , n q called multipoles and a real non-negative number a suchthat D = a ⌊ n ⊗ · · · ⊗ n q ⌋ (23)5 EPTEMBER
25, 2020Figure 1: Each tensor T of every order up to dimension three can be decomposed into a set of vectors (of differentlength or a set of vectors and scalars). First, it must be decomposed into its symmetric part sT and its asymmetric part aT . For the asymmetric part an isomorphism can be used to represent this part also with a symmetric tensor s ˆ T . Thesesymmetric tensors can be decomposed in deviators D j . Each of these deviators can be witten as the symmetrization ofa set of vectors, called multipoles, multiplied by a scalar.The unit vectors n i are uniquely defined by D up to an even number of sign changes, and the number a is unique.The major algorithmic task is to actually compute the multipoles of all deviators in the deviatoric decomposition of agiven tensor. To identify multipoles of a q th -order deviator, Zou et al. [14] define a polynomial of degree 2 q over thecomplex numbers by a q , + q ∑ r = s q ! q ! ( q + r ) ! ( q − r ) ! (cid:0) x r a q , r + ( − ) r x − r a q , r (cid:1) = . (24)We do not explain the background of this polynomial here, but refer the reader interested in such detail to the originalpaper by Zou et al [15]. There, one can see that the polynomial is based on the construction of an orthogonal basisof the corresponding space of deviators for each x ∈ C and each order q . The coefficients a q , r , r = , . . . , q of thepolynomial are the coefficients of the deviator expressed in terms of the basis. The idea is that a root of the polynomialselects a basis that allows one to read off the multipoles. The coefficients a q , r for an explicit example are givenin subsubsection 8.1.2. However, the roots of the above polynomial come in pairs as x − r is a root if x r is a root.Therefore, we actually get 2 · q different roots, but we need only one member of each pair to compute the multipoles.The multipoles of our deviators are found via x r = e i ϕ r tan θ r n r = e cos ( θ r ) + ( e cos ϕ r + e sin ϕ r ) sin θ r r = , . . . . (26) Based on the general considerations above, we now study the important special case of a second-order tensor to applythe method on familiar ground. A general second-order tensor can be decomposed into the zeroth-order deviator d ,the first-order deviator d and the second-order deviator D by T i j = dI i j + ε i jk · d k + D i j , d = T ii , d i = ε i jk T jk (27) εεε describes the third-order permutation tensor and I the second-order identity tensor. If T is symmetric the antisym-metric part represented by d becomes zero. In the second-order three-dimensional symmetric tensor case, the tensor T can be decomposed by the spectral decom-position into three vectors v i and three scalars λ i T = ∑ i = λ i v i ⊗ v i (28)6 EPTEMBER
25, 2020or by the deviatoric decomposition into two vectors m and m and two scalars d and a T = d · I + a ⌊ m ⊗ m ⌋ (29)This suggests there is a connection between the two decompositions. This connection is analyzed in the sequel. Onthe one hand, there are three orthonormal eigenvectors and three real eigenvalues that describe such a tensor. Onthe other hand, there are two multipoles m and m that describe the second-order deviator of this tensor. Togetherwith the isotropic part, i.e. trace times identity, this is the deviatoric decomposition of this tensor. However, becauseeigendecomposition, deviatoric decomposition and multipole decomposition of deviators are all unique, there must bea connection. This will be elaborated on in this section.There are three different cases for the position of two multipoles. Case 1.
The scalar a from Eq. (23) equals zero or the multipoles are given by m = m = ( , , ) T . This is the case,if and only if the tensor has a triple eigenvalue, i.e. is an isotropic or spherical tensor. Case 2.
The multipoles m and m are identical, if and only if the tensor has a double eigenvalue. Then, the multipolesare given by the eigenvector according to the eigenvalue with the largest absolute value. Proof idea.
Assume m = m = : m . Then T = tr ( T ) I + a ⌊ m ⊗ m ⌋ (30)We know that adding a multiple of the identity to a tensor does not change the eigenvectors of a tensor and doesnot change the multiplicity of the eigenvalues, so we define m · m = : b . Because the multipoles are the same, thesymmetrization of the tensor product is the tensor product itself and reads a ⌊ n ⊗ n ⌋ = a m m m m m m m m m m m m m m m | {z } = : M − ab I (31)The eigenvalues of M are given by λ = , λ = , λ = a · ( m + m + m ) . (32)Thus, the tensor has a double eigenvalue.Assume now, the tensor is degenerated and s is the double eigenvalue. Zheng [16] states that a degenerate second-orderthree-dimensional tensor can be represented by T = s I ± V · V T (33)where V is defined by the eigenvector to the single eigenvalue by e = V / k V k . It follows VV T = e k V k · ( e k V k ) T = k V k ( e · e T ) = k V k e e e e e e e e e e e e e e e e (34)We can derive m = m = e . Therefore, the multipoles collapse into one vector and equal the eigenvector correspond-ing to the eigenvalue with the largest absolute value, which is in this case the single eigenvalue. Thus, the assumptionis true. Case 3.
When the multipoles m and m are different, a connection between the eigenvectors and eigenvalues of thesymmetric part and the multipoles can be found. The eigenvector deduced by the, according to the absolute value,largest eigenvalue and the, sorted by absolute value, medium eigenvalue are the bisecting lines of the two multipoles.The eigenvector associated to the eigenvalue with largest absolute value is the bisecting line which has the smallerangle to the multipoles. In other words, the closer two of the eigenvalues become, the closer the multipoles biastowards the direction of the largest (absolute) eigenvalue. In the eigenvector system where | λ | > | λ | > | λ | , themultipoles are given by m = ± q λ + λ a q − λ − λ a m = ± q λ + λ a − q − λ − λ a (35)7 EPTEMBER
25, 2020The angle between the first multipole and the eigenvector e is given byarccos q λ + λ a q λ − λ a . (36) Proof idea.
Let | λ | > | λ | > | λ | , where λ i are the eigenvalues of the deviator. Because the deviator is traceless, λ = − ( λ + λ ) holds. The idea for proving the other cases is analogous. Assume, the eigenvectors are given by thecoordinate axes e , e , e . Then the deviator D is given by D = diag ( λ , λ , λ ) This tensor can also be written as D = a q λ + λ a − q − λ + λ a
00 0 0 − a (cid:18) λ + λ a − − λ − λ a (cid:19) I (37) Using (35) it follows D = a · ⌊ m ⊗ m ⌋ Then, the rotation invariance must be proven and the result follows through the equivalence of each step.This connection underlines the importance of the multipoles. We do not claim that the multipole decompositionis better than the eigendecomposition. We just want to demonstrate that in the symmetric second-order case, theinformation in the eigendecomposition can also be found in the multipole decomposition. Thus, an analysis of higher-order tensors by using the deviatoric decomposition appears reasonable. With this connection, we want to emphasizethe close connection between the two decompositions to show the importance of multipoles, in particular as they allowa generalization to tensors not amenable to an eigendecomposition.
The deviatoric decomposition described above can be used for every tensor of any order up to dimension three. Backus[6] gave a decomposition of a general fourth-order three-dimensional tensor. It can be decomposed into a fourth-orderdeviator D , three third-order deviators D ( i ) , six second-order deviators D ( i ) , six first-order deviators d ( i ) and three8 EPTEMBER
25, 2020zeroth-order ones d ( i ) . The tensor is then given by T i jkl = D i jkl + ε ikm D ( ) jlm + ε jkm D ( ) ilm + ε ilm D ( ) jkm + ε jlm D ( ) ikm + D ( ) i jm ε mkl + D ( ) klm ε mi j + δ i j D ( ) kl + δ kl D ( ) i j + δ ik D ( ) jl + δ jl D ( ) ik + δ il D ( ) jk + δ jk D ( ) il + δ i j D ( ) kl + δ kl D ( ) i j − δ ik D ( ) jl − δ jl D ( ) ik − δ il D ( ) jk − δ jk D ( ) il + δ i j D ( ) kl − δ kl D ( ) i j + D ( ) ik δ jl + D ( ) jk δ il − D ( ) il δ jk − D ( ) jl δ ik + D ( ) ki δ l j + D ( ) li δ k j − D ( ) k j δ l j − D ( ) l j δ k j + D ( ) ik δ jl − D ( ) il δ jk + ( ε ikm δ jl + ε jkm δ il + ε ilm δ jk + ε jlm δ ik ) d ( ) m + d ( ) i ε jkl + d ( ) j ε ikl − δ i j ε klm d ( ) m + d ( ) k ε li j + d ( ) l ε ki j − δ kl ε i jm d ( ) m + δ i j ε klm d ( ) m + δ kl ε i jm d ( ) m + ε i jk d ( ) l − ε i jl d ( ) k − ε kli d ( ) j + ε kl j d ( ) i + d ( ) ( δ i j δ kl + δ ik δ jl + δ il δ jk )+ d ( ) ( δ i j δ kl − δ ik δ jl − δ il δ jk )+ d ( ) ( δ ik δ jl − δ il δ jk ) (38)This decomposition is very complex, but for tensors with some kinds of symmetries, some of the deviators vanish. The best known application of the deviatoric decomposition is the calculation of symmetries of materials described bythe stiffness tensor. To understand the background of this calculation we give a short introduction to the tensor.We describe the stress at a point of the deformed material by the symmetric second-order Cauchy stress tensor σσσ andthe strain by εεε σσσ = σ σ σ σ σ σ σ σ σ ! , εεε = ε ε ε ε ε ε ε ε ε ! (39)The eigenvectors of σσσ are called principal stress directions and the eigenvalues principal stresses.The stiffness tensor describes the linear mapping from strain increments into stress increments which can be describedby the Hooke’s law d σσσ = C :d εεε . (40)It is a fourth-order three-dimensional tensor and can be represented as 3 × × × ψ from which stressesare derived by differentiation with respect to a work-conjugate symmetric deformation measure, the stiffness tensor hasthe two minor symmetries and the major symmetry. Under these conditions, the number of independent coordinatesreduces from 81 to 21. 9 EPTEMBER
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The deviatoric decomposition of the stiffness tensor can for example be used to compute all symmetry planes for allsymmetry classes. Hergl et al. [17] used this fact to visualize the symmetries of the stiffness tensor by designing aglyph. Since our stiffness tensor is not totally symmetric, we also need a decomposition of the asymmetric part whichwas the major achievement of Backus in his paper. Let C be a three-dimensional fourth-order tensor with major andminor symmetry. Then, the totally symmetric and the asymmetric parts are given by S i jkl = (cid:0) C i jkl + C ikl j + C il jk (cid:1) , A i jkl = C i jkl − C ikl j − C il jk (41)The trick, to represent the antisymmetric part by deviators, is to define an isomorphism between the second-ordertotally symmetric tensors S and the fourth-order asymmetric tensors A . Let R be a totally symmetric tensor of ordertwo in three dimensions. We define the isomorphism φ by φ ( R ) i jkl = δ i j R kl + δ kl R i j − δ ik R jl − δ jl R ik − δ il R jk − δ jk R il (42)This isomorphism is the only possible linear isomorphism that is rotation invariant for any rigid rotation which wasproven by Backus using the classification of all isotropic tensors by Weyl [18]. Using (42), the second-order tensor R is given by R i j = ( n − ) A i jll − δ i j ( n − )( n − ) A kkll (43)and the fourth-order three-dimensional tensor C can be represented by the deviatoric decomposition C = D + s ( ID ) + s ( II d ) + φ ( ˆ D ) + φ ( I ˆ d ) (44)where R = D + I ˆ d is the deviatoric decomposition of the totally symmetric tensor R . Therefore, the stiffness tensorcan be uniquely decomposed into one fourth-order deviator D , two second-order deviators D and D and two zeroth-order deviators d and ˆ d .Finally, let us shortly denote the complete deviatoric decomposition of the stiffness tensor C in terms of its coefficientswhich allows to implement it even without understanding the theory in this section. The two zeroth-order deviatorsare called Lamé coefficients in engineering and can be computed as λ = d = ( C iikk − C ikik ) , µ = ˆ d = ( C ikik − C iikk ) . (45)The two second-order deviators can be calculated by D i j = (cid:18) C kk j j − C kkll δ i j (cid:19) − (cid:18) C kiki − C klkl δ i j (cid:19) , (46)ˆ D i j = (cid:18) C kik j − C klkl δ i j (cid:19) − (cid:18) C kki j − C kkll δ i j (cid:19) (47)This allows to compute the fourth-order deviator by removing the other parts of the deviatoric decomposition D i jkl = C i jkl − (cid:0) λδ i j δ kl + µ ( δ ik δ jl + δ il δ jk ) + δ i j D kl + δ kl D i j + (cid:0) δ ik ˆ D jl + δ il ˆ D jk + δ jk ˆ D il + δ jl ˆ D ik (cid:1)(cid:1) (48)with the Kronecker delta δ i j .As a side comment, one can also use this decomposition to calculate Young’s modulus E ( d ) in a specified direction d . Basically, the Young’s modulus describes the stiffness upon uniaxial stretching in this direction. Böhlke andBrüggemann [19] calculated it by 1 E ( d ) = E RI + D · d ⊗ d + d ⊗ d · D : ( d ⊗ d ) (49)where E RI = / ( µ + λ ) is used to normalize the directional dependent quantities.10 EPTEMBER
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Starting with a stiffness tensor given by its 21 coefficients in arbitrary Cartesian coordinates, the deviatoric decom-position (44) and the multipole representation (23) of the stiffness tensor can be used to calculate the position of thesymmetry planes of any anisotropic material.To calculate the multipoles the equation (24) is used. For the second order deviators D = D and ˆ D = D , we set q = a k , = − r ( D k + D k ) , a k , = D k − iD k a k , = ( D k − D k ) − iD k (50)where k = ,
2. For the fourth-order deviator D , we use q = a , = r ( D + D + D ) a , = r ( − D − D + i ( D + D )) a , = √ ( D − D + i ( D + D )) a , = √ ( D − D − i ( D − D )) a , = D + D − D + i ( D − D ) (51)We need the complex roots of these three polynomials. In the second-order case, we can use the known formula forquadratic equations to solve for its four roots. In the fourth-order case, we need to use a numerical method to findthe eight roots. We use Laguerre’s method to calculate the roots given by Press et al. [20]. The multipoles can becalculated by using (25) and (26). One of the useful applications of the multipole decomposition is to compute the anisotropy type of the stiffness tensor.For that purpose, the symmetry planes of each deviator must be calculated. The intersection of these are the symmetryplanes of the stiffness tensor and determine the anisotropy type.A general stiffness tensor describes a rather complicated relation between stress and strain despite its linear form dueto its incremental definition. In practice, most engineering materials exhibit symmetries that simplify this relationfurther. Of special interest are materials showing symmetry change under load.In this context, a plane symmetry means that the elastic behavior, i.e. the stress-strain relation does not change undera reflection at this plane. The simplest materials are isotropic . In this case, any plane is a symmetry plane and therelation between strain and stress is the same for all directions. In this case, one needs only the two Lamé coefficientsin Eq. (45) to describe the relation. In a linear setting, λ enters the scalar relation between a uniform compressionand isostatic pressure, and µ describes the scalar relation between any volume-preserving (i.e. isochoric or deviatoric)strain and isochoric stress with the same direction.In the general case, engineers distinguish the following different classes of symmetries: isotropic , transversallyisotropic , cubic , tetragonal , orthotropic , monoclinic , trigonal and triclinic materials. Of course, these characteri-zations hold only pointwise if the material exhibits different symmetry types at different positions. In the following,we give a short characterization of all classes. A more comprehensive treatment can be found in textbooks on solidmechanics such as [21].A second-order deviator can either be isotropic, transversally isotropic or orthogonal. The set of symmetry planenormals MP [ D ] in these three cases is given by MP [ D ] = all directions if A = { n , m ϕ , ϕ ∈ [ , π ) } if A = , n = n = nN , N , N if A = , n = n (52) EPTEMBER
25, 2020Figure 2: Anisotropy Classes of the stiffness tensor. Left: The symmetry planes of the material in this point. Right:The symmetries of the multipoles (yellow: fourth-order, blue: second-order).where the infinite set { m ϕ } contains all vectors orthogonal to n , and N , N , N are given by N = n + n | n + n | , N = n × n | n × n | , N = N × N (53)The set MP [ D ] can be calculated equivalently using the multipoles n , n .The set MP [ D ] is a bit more complicated. We follow the description in Zou et al. [14]. If the deviator D is transversallyisotropic the normals are all the same n = n = n = n = n . The symmetry plane normals can be calculated likethese in the second-order transversally isotropic deviator case.If the deviator has cubic symmetry, the cube is given by a right-hand coordinate system { m , m , n } such that the axisdirection set { n , n , n , n } is given by (cid:26) n + m + m √ , n − m + m √ , n + m − m √ , − n + m + m √ (cid:27) (54)and the nine symmetry planes are the three coordinate planes and the six planes created by rotating in each of thecoordinate planes by an angle of ± π /
4. If the deviator is tetragonal, the four multipoles are the result of a rotationat 90 ◦ around some axis. The symmetry plane normals of D are given by the vector orthogonal to all multipoles, themultipoles themselves and the vectors that lie with an angle of 45 ◦ between two of the multipoles. If the deviator istrigonal, one normal is orthogonal to the other three, while the other three are coplanar. The coplanar ones result froma rotation around the fourth multipole at 120 ◦ . The three coplanar multipoles are the symmetry plane normals. If thedeviator is orthogonal, the multipoles result from a rotation of one of the following sets S W = { n ( θ , ϕ ) , n ( θ , π − ϕ ) , n ( θ , π + ϕ ) , n ( θ , π − ϕ ) } ; S U = { n ( π / , ϕ ) , n ( π / , π − ϕ ) } or { n ( θ , ) , n ( θ , π ) } or { n ( θ , π / ) , n ( θ , π / ) } ; S V = { e } or { e } or { e } (55)12 EPTEMBER
25, 2020and the symmetry plane normals equal the coordinate surface normals. If all multipoles lie on a plane or there exists aplane that is their mid-separating surface, the normal to this plane is the symmetry plane normal. A triclinic deviatorhas four arbitrary multipoles not fulfilling any of the above relations and there is no symmetry plane. Finally, thesymmetry plane normals of the stiffness tensor are given by the intersection of the symmetry plane normals of threedeviators. MP [ C ] = MP [ D ] ∩ MP [ ˆ D ] ∩ MP [ D ] (56) Every tensor of arbitrary order up to dimension three can be described uniquely by a set of vectors and scalars. Thisis the main statement of this work. This decomposition is not new, but it is neither well known nor well understood.Thus, this general concept has not found its way into natural or engineering sciences and its physical interpretationremains unclear in many areas. As a step towards obtaining such interpretations, we gave a connection between themultipole decomposition and the spectral decomposition of a symmetric three-dimensional second-order tensor. Weare sure, there are other interesting facts about this decomposition and we want to motivate more researchers to analyzethis decomposition and use it for different examples to explore its application-specific meanings.
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