= Solution
Under an orthogonal change of <Cartesian coordinates> with $x_i'=R_{ia}x_a$, a rank-$n$ <Cartesian tensor> transforms as $T'_{i_1\cdots i_n}=R_{i_1j_1}\cdots R_{i_nj_n}T_{j_1\cdots j_n}$, summing repeated indices. An <isotropic tensor> is invariant under every rotation. For an <isotropic fourth-rank tensor>, half-turns about the coordinate axes eliminate components unless each coordinate label occurs an even number of times. Coordinate rotations make the remaining two-pair patterns depend on three constants $c_{1122}=\alpha$, $c_{1212}=\beta$, $c_{1221}=\gamma$. Invariance after rotating the first axis to $(e_1+e_2)/\sqrt2$ gives $c_{1111}=\alpha+\beta+\gamma$. Hence
$$
c_{ijkl}=\alpha\delta_{ij}\delta_{kl}+\beta\delta_{ik}\delta_{jl}+\gamma\delta_{il}\delta_{jk}.
$$
This is a direct proof of the rank-four isotropic decomposition, without assuming any prior interchange symmetry of the four indices. Each displayed delta pairing is itself rotation invariant.
For symmetric <strain>, contraction gives $\sigma_{ij}=\alpha e_{kk}\delta_{ij}+(\beta+\gamma)e_{ij}$. Comparing with the <Lamé parameters> yields
$$
\boxed{\lambda=\alpha,\qquad\mu=\frac{\beta+\gamma}2.}
$$
Write $p=e_{kk}/3$ and $d_{ij}=e_{ij}-p\delta_{ij}$, so $d_{kk}=0$. Since $e_{ij}e_{ij}=3p^2+d_{ij}d_{ij}$, the <strain energy density> becomes
$$
E=\frac\lambda2(e_{kk})^2+\mu e_{ij}e_{ij}
=\frac12\left(\lambda+\frac{2\mu}3\right)(e_{kk})^2+\mu d_{ij}d_{ij}.
$$
The two terms describe volumetric and shear deformation, respectively. Thus <nonnegative isotropic elastic strain energy> is equivalent to
$$
\boxed{\mu\geq0,\qquad\lambda\geq-\frac{2\mu}3.}
$$
Necessity also follows by taking a nonzero traceless <strain> and a pure dilation separately. The combination $K=\lambda+2\mu/3$ is the <bulk modulus>; strict inequalities give positive energy for every nonzero <strain>.
Back to article page
