Isotropic invariants of traceless symmetric tensors of orders three and four

Pergamon

ISOTROPIC

Int. J. Engng Sci. Vol. 35, No. 15, pp. 1457-1462, 1997

~) 1997 Elsevier Science Limited. All rights reserved Printed in Great Britain PH: S0020-7225(97)00048-7 0020-7225/97 $17.00+0.00

INVARIANTS OF TRACELESS SYMMETRIC OF ORDERS THREE AND FOUR

TENSORS

G. E SMITH Lehigh University, Bethlehem, PA, USA

G. BAO Johns Hopkins University, Baltimore, MD, USA Abstrael--We determine integrity bases for functions of a traceless symmetric third-order tensor A0k and for functions of a traceless symmetric fourth-order tensor A~ikj which are invariant under the threedimensional proper orthogonai group R3. Integrity bases are also given for the full orthogonal group O3. {~) 1997 Elsevier Science Ltd.

1. I N T R O D U C T I O N We consider the problem of determining an integrity basis for functions of a three-dimensional traceless symmetric tensor Aij...i, which are invariant under the three-dimensional proper orthogonal group Ra for the cases where n = 3 and n = 4. We may refer to these invariants as isotropic invariants. It has been observed by Spencer [1] and Littlewood [2] that the problem of determining an integrity basis for functions of a three-dimensional traceless symmetric tensor A,-,...i~ which are invariant under the three-dimensional proper orthogonal group R 3 is essentially the same as the problem of determining an integrity basis for functions of a two-dimensional symmetric tensor a,-...~2, which are invariant under the two-dimensional unitary unimodular group/-/2. We may determine (see Murnaghan [3], p. 328) an expression

air ... i2,, =

C i I ,..

i2nJl...j,,Ajl..4,

(1.1)

such that a function l(ai,...J which is invariant under U2 yields a function

J(A,,...,.) = I(C,,...,2,,j,...jAj,...j,,)

(1.2)

which is invariant under R3. Thus, given an integrity basis Ik(a~,...i2,), (k = 1,...,p), for functions of the two-dimensional symmetric tensor a~,...~2,' which are invariant under /-/2, we may list an integrity basis Jk(A~,...i,,) = Ik(C~,...~,j,...j, Aj,...j,,), (k = 1,...,p) for functions of the three-dimensional traceless symmetric tensor Ai,...~,,which are invariant under R3. We observe that an integrity basis for functions of the two-dimensional symmetric secondorder tensor a~j which are invariant under/./2 is given (see Grace and Young [4], p. 86) by

I(aij) = aria22 - a22

(1.3)

Let all = A1 +

i A 2 , at2 =

A3, a22

= -- A 1 + i A 2

(1.4)

define the relation % = C~jkAk of the form of equation (1.1) for the case n = 1. With equations (1.2)-(1.4), we have J(A,) = I ( C , i , A , ) = (A, + iA2)( - A t + iA2) - A 2 = - A.,A,

(1.5)

The invariant A,A; forms an integrity basis for functions of the three-dimensional vector Ai which are invariant under R 3. This of course means that any polynomial function of the vector A~ which is invariant under R3 can be expressed as a polynomial in the invariant A,A~. 1457

1458

G.F.

SMITH

and G. BAO

Integrity bases for functions of two-dimensional symmetric tensors of orders six and eight which are invariant under U2 are available in the literature. We may in principle employ these results together with equation (1.2) to generate integrity bases for functions of traceless symmetric tensors A~,...~,which are invariant under R3 for the cases n = 3 and n = 4. The algebra required leads us to consider an alternative procedure. We observe that, given the integrity basis for functions of two-dimensional symmetric tensors ai~...i2, ' (n = 3 or 4) which are invariant under /-/2, we may immediately list the number and degrees of the integrity basis elements for functions of three-dimensional traceless symmetric tensors Ai,...~, (n = 3 or 4) which are invariant under R 3. Given this information, we employ a method due to Sylvester [5] in order to obtain the desired integrity bases. 2. T R A C E L E S S S Y M M E T R I C T H I R D - O R D E R T E N S O R Aij, The tensor Aijk is said to be symmetric if Aijk = Aiki = Ajki = Aj~k = Akij = Ak~

(2.1)

The tensor A~jk is said to be traceless if A l l k = O,

Aiji = O, Airy = 0

(2.2)

An integrity basis for functions of a two-dimensional symmetric sixth-order tensor a~,..~, which are invariant under U2 is comprised (see Grace and Young [4], p. 156) of five invariants 12, I4, I6, Ito and 115 of degrees 2, 4, 6, 10 and 15 respectively in the ai,...~. An integrity basis for functions of the traceless symmetric tensor Aijk which are invariant under R 3 is then comprised of five invariants of degrees 2, 4, 6, 10 and 15 in the A,.~k.We make the conjecture that the functions Kp(A~jk), (p=2, 4, 6, 10, 15), listed below form an integrity basis for functions of Aijk invariant under R 3. Let Bij = AipqAjpq, C~ = AijkBjk

(2.3)

and K 2 -~-

AijkAijk,

K 4 ~--

nijsij,

K6 =

Cifi, Klo "-- mijkfifjC k, K 1 5

=

~,ijkfi8jpCpAkqrfqfr

(2.4)

where e~jk is the alternating symbol defined by I~ eijk =

if ijk = 123,231,312; 1

if ijk = 132,213,321;

(2.5)

otherwise It is obvious that the Kp defined by equation (2.4) are invariant under R 3. We list below the sets of monomial invariants of degrees 2, 4, 6, 10 and 15 which are obtained from products of the Kp. •

.

2

.

3

.

.

5

3

2

2

2.K2, 4:K2,K4, 6:K2,K2Ka, K6,10.K2,K2K4,K2K6,K2K4,K4K6,K10; 15:K15

(2.6)

Consider the set of invariants (equation (2.6)) 3 of degree six. It may be the case that K 3, K2K4 and I(6 are linearly dependent, i.e. there are constants a, /3, y not all zero such that a K 3 +/3K2K4 + TK6 ~ 0. Suppose this is so with a # 0,/3 # 0, y = 0. Then we would have a syzygy aK 3 +/3K2K4 = 0 and would require two invariants K6 and K~ (say) of degree six as elements of the integrity basis. Suppose that aK 3 +/3K2K4 + yK6 -- 0 holds with 7 # 0 so that K6=a'K3+/3'K2K4. Then K6 is said to be reducible and we must then determine a different invariant K~ of degree six to serve as an element of the integrity basis• Suppose that K 3, K2K4, K6 are linearly independent, i.e. a K 3 +/3K2K4 + yK6 -- 0 holds only if a =/3 = 3' = 0. We may then rule out the existence of a syzygy and the possibility that K6 is redundant. We then conclude that K6 is an element of an integrity basis. Thus, in order to establish that K2, K4, K6, Klo, K15 form an integrity basis, we must show that the invariants comprising the set of invariants of degree p in equation (2.6) are linearly independent for the cases p = 2, 4, 6, 10, 15. In order to

Isotropic invariants of traceless symmetric tensors of orders three and four

1459

accomplish this, we consider the traceless symmetric third-order tensor Aq, whose components are given by All 1 = a, A122 := - a, A133 = 0, All 2 = b, A222 = - b, A233 = 0, All a = c, A223 = - c, A333 = 0, g

A123 = 0; b 2 = - a 2 - --2-c 2 2

(2.7)

We note that AI22=A212=A221 .... since Auk is symmetric. With equations (2.4) and (2.7), we have K2 = 0, K4 = - 6c 4, K6 = - 24c6,K10 = 16c6(3234 + 4832c 2 + 9c4), Kls = - 2048abc9(a 2 - b2) 2

(2.g) Suppose that t]he invariants (equation (2.6))2 of degree four satisfy a K 2 + ilK4 -~ 0

(2.9)

Let the Aijk be given by equation (2.7) so that K2 = 0, K4 # 0. Then fl = 0 and hence a K 2 ~" 0 so that a = 0. Thus, if equation (2.9) holds, we must have a = fl = 0 so that K~ and K4 are linearly independent. Suppose that t]~e invariants (equation (2.6))3 of degree six satisfy a K 3 + flK2K4 + TK6 ~ 0

(2.10)

Let the Aiik be given by equation (2.7) so that K 2 = 0, K6 # 0. Then y = 0, KE(aK 2 + ilK4) -- 0 and hence a K 2 + i l K 4 - 0. We have seen that K 2 and K4 are linearly independent so that a = fl = 0. Thus, if equation (2.10) holds, we must have a = fl = y = 0 and the invariants (equation (2.6))3 are linearly independent. Suppose that tl~e invariants (equation (2.6))4 of degree ten satisfy aKl0 + flK4K6 + K2(TK 4 + 6K22K4 + eK2K6 + vK42) - 0

(2.11)

Let the Auk be given by equation (2.7) so that K2 = 0 and equation (2.11) then yields ac6(3234 + 48a2c 2 + 9C4) + 9tiC I° = 0

(2.12)

for all values of a and c. We set a = 0 and a = c in turn in equation (2.12) to obtain a + fl = 0, 893 + 9fl = 0 which implies that a =/3 = 0. With a = fl = 0, equation (2.11) yields K 2 ( y K 3 + ~KEK4 + eK6) + vK 2 - 0

(2.13)

Let the Aijk be given by equation (2.7) so that K2=0, K 4 # 0 . Then v = 0 , KE(yK23+6KEK4+eK6)---0 and hence yK23+SK2K4+eK6-=0. We have seen above that K23, KEK4 and K6 are, linearly independent so that y=~$=e =0. Thus, if equation (2.11) holds, we must have a =/~ . . . . . v = 0 and the invariants (equation (2.6))4 of degree ten are linearly independent. We may then conclude that the K2, I(4, K6, Kio and K15 given by equation (2.4) form an integrity basis for functions of the traceless symmetric t e n s o r Aqk which are invariant under Ra. A function F(A~,) which is invariant under the proper orthogonal group R3 and which is also invariant under the central inversion C = II - 60 II will be invariant under the full orthogonal group O3. Application of C to the Ao., will change the sign of each of the Aok. Thus, the K2, K4, K6, Klo defined by equation (2.4) which are of even degree in the Aqk remain invariant under C and the function K~5 which is of odd degree in the AUk will change sign under C. We then see that K2, I(4, K6, }[10 and K~5 will form an integrity basis for functions of Ae, invariant under 03. It may be shown that K25 is expressible as a polynomial in K2, K4, K6, Kto. Thus, an integrity basis for functions of a traceless symmetric third-order tensor A~k which are invariant under 03 is formed by the K2, K4, K6, K~o defined by equation (2.4).

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G.F. SMITH and G. BAO

3. T R A C E L E S S S Y M M E T R I C F O U R T H - O R D E R

T E N S O R A0k ~

The tensor Aokt is said to be symmetric if

Aim = Apqrs

(3.1)

holds for the 4[ cases where (p,q,r,s) is any permutation of (i,j,k,l). The tensor A~m is said to be traceless if

Aiikt = O, Aim = O, Aqki = 0 ....

(3.2)

An integrity basis for functions of a two-dimensional symmetric eighth-order tensor a~,...i, which are invariant under U2 is comprised (see Sylvester [5], von Gall [6]) of nine invariants I2, I3, It, I5, I6, 17, Is, I9, I10 of degrees 2, 3, 4, 5, 6, 7, 8, 9, 10 respectively in the ai,...i,. An integrity basis for functions of the traceless symmetric fourth-order tensor Aijkt which are invariant under R3 is then comprised of nine invariants of degrees 2, 3, 4, 5, 6, 7, 8, 9, 10 in the Aqkt. We proceed by listing nine functions K2(Aok3 ..... K~o(Aok3, each of which is invariant under R3 and which are chosen upon inspection of the I2(a~,...~,),..., Ito(ai~...i,). We then demonstrate that the K2..... Kto form an integrity basis. Let (3.3)

Cijkl = Aijmnaktmn, Bij = Aipqralpqr

and K2 = aqklaqkh K3 = Cijklaijkl, K4 = BijBij, K5 = ei]AiiklBkh K6 = BoCijklekl, K7 = B2i]AijklBkh

Ks = B2CjmBk,, K9 = B2.rAqktB2,, Kl0 = B2CimB 2,

(3.4)

The Kp defined by equation (3.4) are obviously invariant under R 3. It has been shown by Smith [7] that there are 1, 1, 2, 2, 4, 4, 7, 8, 12 linearly independent functions of degrees 2, 3 .... , 10 in the Aqkt which are invariant under R 3. We list below the nine sets of invariants of degrees 2 ..... 10 respectively which are given by monomials in the products of the invariants K2 ..... K~o. We establish that the K2..... K10 form an integrity basis upon showing that the invariants of degree p in the sets listed below are linearly independent for the cases p = 2 .... ,10. .

.

.

.

.

2

.

.

.

.

3

2

.

2

4

2

2

2.K2, 3.K3,4.K2,K4, 5.K2K3,K5,6.K2,K2K4,K3,K6, 7:K2K3,K2Ks, K3K4,K7;8:K2,K2K4,K2K3, 2

. .

3

2

3

.

.

5

3

2

2

K2K6,K3Ks, K4,Ks,9.K2K3,K2K 5,K2K3K4,K2KT,K3,K3K6,K4K5,K9,1o.K2,K2K4,K2K3,

2 2 2 2 K2K6,K2K3Ks, K2K4, K2Ks,K3K4, K3K7,K4K6,Ks,K 10 (3.5)

We follow Sylvester [5] and set 1

Amt-16

2'

1--6 ' A m 3 = ~ ( b - 3 d ) ,

1

i

A,,22 = - 1-6 '

Ai222 -- i

.

16

d

c

An23 = -~- ( - b + d),

'

A1223= -

(b + d),

1

AI333 = - ~ ,

c A2233=

2

c

A2222 = - ~ - + -~-,

AH33 = -~-,

A1233 = - i 2 ' i A2223= -ff (b + 3d),

d '

A2333--"

-

iT'

A3333=0

(3.6)

Isotropic invariants of traceless symmetric tensors of orders three and four

1461

With equations (3.4) and (3.6), we have 3 5 25 KE = 0, K3 = - -ff (bd - 3c2), K , = - ~- cd2,K5 = -64 d4'

K6 = ~

Ks = ~

1 (6c 4 -

1(14 -4- b

4bc2d + 4b2d2),K7 = 5 - ~

1( 70c2d4 - --23 Kl0 =

)

- 40c3d 2 + 30bcd 3 ,

25 b2c3 + L4 b3cd - - - bd 5 ,K9 = 5 2

__

_ lOb2cd 4 + 3 b4c2 _ 1 b5 d 8 8

b2d2(4c 2 - bd),

)

(3.7)

The algebra involved in obtaining equation (3.7) from equations (3.4) and (3.6) is extensive and it would be reasonable to question the accuracy of the results of equation (3.7). We have checked equation (3.7) by choosing values of b, c, d in equation (3.6) so as to obtain values for the Aijkl; we then substituted those values in equation (3.4) and employed a computer program to calculate the values of K2..... K10. Using the same values of b, c, d, we computed the values of K2,..., Klo from equation (3.7). The results coincided for various choices of b, c, d which strongly indicates that the algebra leading to equation (3.7) is correct. Suppose that the invariants (equation (3.5)) 3 of degree four satisfy aK22 +/3K4 =- 0

(3.8)

Let the A~kl be given by equation (3.6). We see that /3 = 0 since K2 = 0 and K4 ¢ 0. Then aK~ -~ 0 and hence a = 0. Thus, if equation (3.8) holds, we must have a = 13 = 0 so that Kzz and K4 are linearly independent. The same argument shows that K2 K3 and Ks are linearly independent. Suppose that the invariants (equation (3.5))5 of degree six satisfy aK~ + flK2K4 + yK32+ 8K6 --- 0.

(3.9)

Let the A~/kt be given by equation (3.6) with b d - 3c 2= 0 so that K2 = K3 = 0. Then ¢5= 0 in equation (3.9) and we have a K 3 +/3K2K4 + yK~ = 0. Let the Ai/kt be given by equation (3.6) but with b d - 3 c 2 ¢ O so that K2=0, K 3 ¢ 0 . Then y = 0 , K 2 ( a K 2 + f l K 4 ) ~ 0 and hence aK~ +/3K4 =- 0. We have seen above that K~ and K4 are linearly independent so that a =/3 = 0. Thus, if equation (3.9) holds, we must have a =/3 = y = ~ = 0 so that the invariants (equation (3.5))s of degree six are linearly independent. We may show in exactly the same manner that the invariants (equation (3.5))6 of degree seven are linearly independent. Suppose that the invariants (equation (3.5))7 of degree eight satisfy trlK ~ + ot2K2K4 + oL3K2K2 + ot4K2K6 + asK3K5 + ot6K2 + a r K s ~ 0

(3.10)

Let the Ai~kl be given by equation (3.6) with d = 0 so that K2 = K4 = Ks = 0. Then 0 ' 7 : 0 and we have a l K 4 + ... + a6K42 - 0. Let the Aiyk/ be given by equation (3.6) with b d - 3 c 3 = 0 so that K2 = K3 = 0, K4 ~ 0. Then 0,6 = 0 and otlK ~ + ... + asK3K5 -- 0. Let the A~/k~be given by equation (3.6) with b d - 3 c 2 ~ O so that K2=0, K 3 ~ 0 , Ksv~0. Then 0,5=0, K2(0,1K~+a2K2K4+ a3 K2 + t~4K6) -- 0 and hence a l K 3 + o~2K2K4+ aaK~ + ot4K6 ~ 0. Since the invariants (equation (3.5))5 of degree six are linearly independent, we have 0,1 = 0"2= O~3=0"4 = 0 . Thus, if equation (3.10) holds, we must have a~=0 (i= 1 ..... 7) so that the invariants (equation (3.5)) 7 of degree eight are linearly independent. Suppose that the invariants (equation (3.5))8 of degree nine satisfy ot,K23K3 + a2K2K5 + aaK2K3K4 + 0"4K2K7 + asK33 + a6K3K6 + aTK4K5 + asK9 -= 0

(3.11)

Let the Aqkt be given by equation (3.6) with bd - 3c 2 = 0, d = 1 so that K2 = K3 = 0. Then, with

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G . F . SMITH and G. BAO

equation (3.11), we have 200a7c-9otsc6=O for all c. This implies that a T = a S = 0 . Next, let equation (3.6) hold with bd = 1. This yields (81a5 + 6ot6)c 4 - (54a5 + 4Ot6)C2 q- (9ot5 + 4ot6) = 0

(3.12)

for all c. The coefficients of c °, c 2, c n in equation (3.12) must then all equal zero and hence or5 = a6 = 0. Also, K2(OtlK2K3 + a2K2K5 + ot3K3K4 + ot4K7) ~ 0 and hence otlK2K3 + ot2K2K5+ Ot3K3K4+Ot4KT-=0. Since the invariants (equation (3.5)) 6 of degree seven are linearly independent, or1 = ot2 = or3 = a4 = 0. Thus, if equation (3.11) holds, we must have ai = 0 (i = 1..... 8) so that the invariants (equation (3.5))8 of degree nine are linearly independent. Suppose that the invariants (equation (3.5)) 9 of degree ten satisfy 2 2 ot~K5 + ot2K3K4 + ot3K2K3 + ce4K2K6 + OtsK2K3K5 + aeK2K] + otvK2K8 + otsKEK4 + o t 9 K a K 7 + OtloKaK6+ a l l K ] + ot~2K~o--0

(3.13)

Let the A~jk~ be given by equation (3.6) with b = 3c 2, d = 1 so that K2 = K3 = 0 and equation (3.13) then yields

243ot~2c~°+ (9600ot~0 + 720ot12)c 5 - 5000otll = 0

(3.14)

for all c. Each of the coefficients in equation (3.14) must equal zero so that ot~o=a~ =aa~2=0 and otlK25+... + o t 9 K 3 K 7 ~ 0 . Let the Aij, t be given by equation (3.6) with b = l , d=O. Then, K2 = K4 = 0, K3K7 ~ 0 so that ot9 = 0 and a~K~ + ,.. + otsK2K4- 0. Let equation (3.6) hold so that K 2 = 0 , K23K4~0. Then ors=0, K 2 ( o t l K ~ + . . . + avK8)~0 and hence otlK~+otzKzZK4 + asKzK 2 + ot4KzK6+ otsK3K5 + a6K] + otTK8~ 0. Since the invariants (equation (3.5))7 of degree eight are linearly independent, we must have oti= 0 (i = 1 ..... 7). Thus, if equation (3.13) holds, we must have oti= 0 (i = 1..... 12) so that the invariants (equation (3.5))9 of degree ten are linearly independent. We may conclude that the invariants K2.... , K~o given by equation (3.4) form an integrity basis for functions of the three-dimensional traceless symmetric fourth-order tensor Aij,t which are invariant under R 3. Since the Aij,~ are unaltered under the central inversion transformation C = I] - 6q II, the restrictions imposed on a function of the A6k t by the requirement of invariance under the proper orthogonal group R3 and by the requirement of invariance under the full orthogonal group O3 are the same. Hence, the K2..... K~o defined by equation (3.4) also form an integrity basis for functions of the traceless symmetric fourth-order tensor Ao, ~which are invariant under O3.

REFERENCES 1. 2. 3. 4. 5. 6. 7.

Spencer, A. J. M. Mathematika 1970, 17, 275 Littlewood, D. E. Proc. London Math. Soc. 1947, 50, 349 Murnaghan, E D., The Theory of Group Representations, Johns Hopkins, Baltimore, 1938. Grace, J. H. and Young, A., The Algebra oflnvariants, Cambridge University Press, London, 1903. Sylvester, J. J. Amer. J. Math. 1881, 4, 62 Von Gall, Math. Ann., 1881, 17, 139. Smith, G. E, Constitutive Equations forAnisotropic and Isotropic Materials, North Holland, Amsterdam, 1994.

(Received 15 January 1996; accepted 5 February 1997)