Quadratic Invariants of the Elasticity Tensor

Abstract

We study the quadratic invariants of the elasticity tensor in the framework of its unique irreducible decomposition. The key point is that this decomposition generates the direct sum reduction of the elasticity tensor space. The corresponding subspaces are completely independent and even orthogonal relative to the Euclidean (Frobenius) scalar product. We construct a basis set of seven quadratic invariants that emerge in a natural and systematic way. Moreover, the completeness of this basis and the independence of the basis tensors follow immediately from the direct sum representation of the elasticity tensor space. We define the Cauchy factor of an anisotropic material as a dimensionless measure of a closeness to a pure Cauchy material and a similar isotropic factor is as a measure for a closeness of an anisotropic material to its isotropic prototype. For cubic crystals, these factors are explicitly displayed and cubic crystal average of an arbitrary elastic material is derived.

Appendix: Calculating the relations between invariants

1.1 A.1 The first Ting invariant

For the first invariant of Ting, we write

$$ B_{1}=C^{ijkl}C_{ijkl}=\sum _{I=1}^{5}{}^{(I)} C^{ijkl} \sum _{J=1}^{5}{}^{(J)} C_{ijkl} . $$

(148)

Due to the orthogonality of the decomposition, we are left with

$$ B_{1}={}^{(1)} S^{2}+{}^{(2)} S^{2}+{}^{(3)} S^{2}+ {}^{(1)} A^{2}+{}^{(2)} A^{2}. $$

(149)

We calculate step by step

$$\begin{aligned} {}^{(1)} S^{2} =&\frac{1}{225}S^{2} \bigl(g^{ij}g^{kl}+g^{ik}g^{jl}+g^{il}g^{jk} \bigr)^{2}=\frac{1}{5} S^{2}=\frac{1}{5}Z_{1} , \end{aligned}$$

(150)

$$\begin{aligned} {}^{(2)} S^{2} =&\frac{1}{49} \bigl(P^{ij}g^{kl}+P^{ik}g^{jl}+P^{il}g^{jk}+P^{jk}g^{il} +P^{jl}g^{ik}+P^{kl}g^{ij} \bigr)^{2}=\frac{6}{7} P^{ij}P_{ij} \\ =& \frac{6}{7}Z_{4} , \end{aligned}$$

(151)

$$\begin{aligned} {}^{(3)} S^{2} =&R^{ijkl}R_{ijkl}=Z_{7} , \end{aligned}$$

(152)

and

$$\begin{aligned} {}^{(1)} A^{2} =&\frac{1}{36}A^{2} \bigl(2g^{ij}g^{kl}-g^{il}g^{jk}-g^{ik}g^{jl} \bigr)^{2}=A^{2}=Z_{3}, \end{aligned}$$

(153)

$$\begin{aligned} {}^{(2)} A^{2} =&\frac{1}{4} \bigl( \epsilon^{ikm}\epsilon^{jln}+ \epsilon^{ilm} \epsilon^{jkn} \bigr) (\epsilon_{ikp}\epsilon_{jlq}+ \epsilon_{ilp}\epsilon_{jkq} )Q_{mn}Q^{pq}=12Q^{mn}Q_{mn} \\ =& 12Z_{6} . \end{aligned}$$

(154)

Consequently, the first invariant of Ting reads

$$ B_{1}=\frac{1}{5} Z_{1}+Z_{3}+ \frac{6}{7}Z_{4}+12Z_{6}+Z_{7} . $$

(155)

1.2 A.2 The second Ting invariant

For the second invariant of Ting, $B_{2}$, we first calculate the trace

$$ C^{i}{}_{ikl}=g^{ij}C_{ijkl}=g^{ij} \bigl(S_{ijkl}+ {}^{(1)} A_{ijkl}+{}^{(2)} A_{ijkl} \bigr) . $$

(156)

Here,

$$\begin{aligned} g^{ij}S_{ijkl} =&S_{kl}=P_{kl}+ \frac{1}{3} Sg_{kl} , \end{aligned}$$

(157)

$$\begin{aligned} g^{ij}{}^{(1)} A_{ijkl} =&g^{ij} (g_{ik}Q_{jl}+g_{jk}Q_{il}+g_{il}Q_{jk}+ g_{jl}Q_{ik}-2g_{kl}Q_{ij}-2g_{ij}Q_{kl} ) \\ =& -2Q_{kl} , \end{aligned}$$

(158)

and

$$ g^{ij} {}^{(2)} A_{ijkl}= \frac{1}{6} Ag^{ij} (2g_{ij}g_{kl}-g_{il}g_{jk}-g_{ik}g_{jl} )=\frac{2}{3}Ag_{kl} . $$

(159)

Hence,

$$ C^{i}{}_{ikl}=P_{kl}-2Q_{kl} +\frac{1}{3} ( S+2A )g_{kl} . $$

(160)

Consequently, the second invariant of Ting reads

$$ B_{2}=C^{i}{}_{ikl}C_{j}{}^{jkl}=P_{kl}P^{kl}-4P_{kl}Q^{kl}+4Q_{kl}Q^{kl}+ \frac{1}{3} (S+2A )^{2} , $$

(161)

or

$$ B_{2}=\frac{1}{3}Z_{1} + \frac{4}{3}Z_{2}+\frac{4}{3}Z_{3} + Z_{4}-4Z_{5}+4Z_{6} . $$

(162)

1.3 A.3 The first Ahmad invariant

For the first invariant of Ahmad, $B_{3}$, we need the trace

$$ C^{j}{}_{kjl}=g^{ij}C_{ikjl}=g^{ij} \bigl(S_{ikjl}+ {}^{(1)} A_{ikjl}+{}^{(2)} A_{ikjl} \bigr) . $$

(163)

Here,

$$\begin{aligned} g^{ij}S_{ikjl} =&S_{kl}=P_{kl}+ \frac{1}{3} Sg_{kl} , \end{aligned}$$

(164)

$$\begin{aligned} g^{ij} {}^{(1)} A_{ikjl} =& \frac{1}{6} Ag^{ij} (2g_{ik}g_{jl}-g_{il}g_{jk}-g_{ij}g_{kl} )=-\frac{1}{3}Ag_{kl} , \end{aligned}$$

(165)

and

$$\begin{aligned} g^{ij} {}^{(2)} A_{ikjl} =&g^{ij} (g_{ij}Q_{kl}+g_{jk}Q_{il}+g_{il}Q_{jk}+ g_{kl}Q_{ij}-2g_{jl}Q_{ik}-2g_{ik}Q_{jl} ) \\ =& Q_{kl} . \end{aligned}$$

(166)

Consequently,

$$ C^{j}{}_{kjl}=P_{kl}+Q_{kl}+ \frac{1}{3} g_{kl}(S-A) . $$

(167)

Hence using (160) and (167) we get

$$\begin{aligned} B_{3} =& \biggl(P_{kl}+Q_{kl}+ \frac{1}{3} g_{kl}(S-A) \biggr) \biggl(P^{kl}-2Q^{kl} +\frac{1}{3} ( S+2A )g^{kl} \biggr) \\ =&P_{kl}P^{kl}-P_{kl}Q^{kl}-2Q_{kl}Q^{kl}+ \frac{1}{3}(S+2A) (S-A) , \end{aligned}$$

(168)

or

$$ B_{3}=\frac{1}{3}Z_{1} + \frac{1}{3}Z_{2}-\frac{2}{3}Z_{3} + Z_{4}-Z_{5}-2Z_{6} . $$

(169)

1.4 A.4 The second Ahmad invariant

This invariant is obtained by the use of the formula (167),

$$\begin{aligned} B_{3} =& \biggl(P^{kl}+Q^{kl}+ \frac{1}{3} g^{kl}(S-A) \biggr) \biggl(P_{kl}+Q_{kl}+ \frac{1}{3} g_{kl}(S-A) \biggr) \\ =&P_{kl}P^{kl}+2P_{kl}Q^{kl}+Q_{kl}Q^{kl}+ \frac{1}{3}(S-A)^{2} , \end{aligned}$$

(170)

or

$$ B_{4}=\frac{1}{3}Z_{1} - \frac{2}{3}Z_{2}+\frac{1}{3}Z_{3} + Z_{4}+2Z_{5}+Z_{6} . $$

(171)

1.5 A.5 The Norris invariant

We put the invariant of Norris first in the form

$$ B_{5}=C^{ijkl}C_{ikjl}= \bigl(S^{ijkl}+A^{ijkl} \bigr) (S_{ikjl}+A_{ikjl} )= S^{ijkl}S_{ijkl}+A^{ijkl}A_{ikjl} . $$

(172)

Using (150,151,152), we have

$$ S^{ijkl}S_{ikjl}=S^{ijkl}S_{ijkl}= \frac{1}{5} Z_{1}+\frac{6}{7} Z_{4}+Z_{7} . $$

(173)

We observe

$$\begin{aligned} A^{ijkl}A_{ikjl} =& \bigl({}^{(1)} A^{ijkl}+{}^{(2)} A^{ijkl} \bigr) \bigl({}^{(1)} A_{ikjl}+{}^{(2)} A_{ikjl} \bigr) \\ =&{}^{(1)} A_{ijkl} {}^{(1)} A^{ikjl}+{}^{(2)} A_{ijkl} {}^{(2)} A^{ikjl} . \end{aligned}$$

(174)

Thus we find

$$\begin{aligned} {}^{(1)} A_{ijkl} {}^{(1)} A^{ikjl} =&\frac{1}{36}A^{2} (2g_{ij}g_{kl}-g_{il}g_{jk}-g_{ik}g_{jl} ) \bigl(2g^{ik}g^{jl}-g^{il}g^{jk}-g^{ij}g^{kl} \bigr) \\ =&-\frac{1}{2}A^{2} , \end{aligned}$$

(175)

$$\begin{aligned} {}^{(2)} A_{ijkl} {}^{(2)} A^{ikjl} =& (g_{ik}Q_{jl}+g_{jk}Q_{il}+g_{il}Q_{jk}+ g_{jl}Q_{ik}-2g_{kl}Q_{ij}-2g_{ij}Q_{kl} ) \\ &{}\times\bigl(g^{ij}Q^{kl}+g^{jk}Q^{il}+g^{il}Q^{jk}+ g^{kl}Q^{ij}-2g^{jl}Q^{ik}-2g^{ik}Q^{jl} \bigr) \\ =&-6Q_{ij}Q^{ij} . \end{aligned}$$

(176)

Consequently,

$$ B_{5}=C^{ijkl}C_{ikjl}= \frac{1}{5} Z_{1}-\frac{1}{2}Z_{3}+ \frac{6}{7} Z_{4} -6Z_{6}+Z_{7} . $$

(177)