A. J. M. Spencer - Continuum Mechanics

Suppose that T and D are second-order tensors, such that the components of T are functions of the components of D , thus

T = T(D)

Then if

(A.1)

for all orthogonal tensors M, we say that T ( D ) is an isotropic tensor function of D. We consider the case in which T and D are symmetric tensors, and denote

(A.2)

Theorem . T is an isotropic tensor function of D if and only if

(A.3)

where , , are scalar functions of tr D , tr D 2 and tr D 3 .

Proof (a) Sufficiency . Since M is orthogonal, tr D = tr , tr D 2 = tr 2, and tr D 3 = tr 3. Hence , and are unchanged if D ij are replaced by ij.

Assume (A.3) holds. Then from (A.2),

(b) Necessity . Assume that (A.1) is satisfied, and choose the x i coordinate system so that the coordinate axes are the principal axes of D . Then, in these coordinates,

(A.4)

and

(A.5)

Choose

Then

(A.6)

(A.7)

However, (A.1) and (A.6) require that = T ij. Hence T 12 = 0, T 13 = 0. Similarly, by another choice of M , it can be shown that T 23=0. Thus if (D j ) is a diagonal matrix, so is ( T ij); that is, D and T have the same principal axes. Therefore we can now write

(A.8)

Next choose

Then

and so (A.1) gives

(A.9)

Hence T 1, T 2 and T 3 can be expressed in terms of the single function F ( D 1, D 2, D 3) as

(A.10)

Finally, choose

Then

and then (A.1) gives

(A.11)

Now the equations

(A.12)

have solutions for , and as functions of D 1 , D 2 and D 3. Also, because F ( D 1, D 2, D 3) has the symmetry expressed by (A.11), equations (A.12) are unaltered if any pair of D 1 D 2 and D 3 are interchanged. Hence , and are symmetric functions of D 1 D 2 and D 3. It follows from a theorem in the theory of symmetric functions that , and can be expressed as functions of

(A.13)

Also, from (A.10) and (A.12),

which, with (A.13), is equivalent to (A.3).