 Solution

Under an orthogonal change of Cartesian coordinates with $x_i^{\prime }=R_{ia}{x}_a$, a rank-$n$ Cartesian tensor transforms as $T_{i_1\cdots i_n}^{\prime }=R_{i_1{j}_1}\cdots R_{i_n{j}_n}{T}_{j_1\cdots j_n}$, 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 $(e_1+e_2)/\sqrt{2}$ gives $c_{1111}=\alpha +\beta +\gamma$. HenceThis 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 yieldsWrite $p=e_{kk}/3$ 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 becomesThe two terms describe volumetric and shear deformation, respectively. Thus nonnegative isotropic elastic strain energy is equivalent toNecessity 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.