= Solution
As a function of the full <deformation gradient>, use $e=\operatorname{sym}(F-I)$. Let $Q=\operatorname{diag}(1,-1,-1)$, a proper rotation. Assumed full frame indifference would give $\widetilde W(Q)=\widetilde W(I)=0$. For $F_s=I+s(Q-I)$, its infinitesimal strain is $s\operatorname{sym}(Q-I)$. Quadratic homogeneity therefore gives
$$
\widetilde W(F_s)=s^2\widetilde W(Q)=0.
$$
Choose $s=1/4$. Then
$$
\boxed{F_s=\operatorname{diag}(1,1/2,1/2),\qquad
\det F_s=1/4\ne1,\qquad\widetilde W(F_s)=0.}
$$
The <homogeneous deformation> $y(X)=F_sX$ changes <volume> but has zero energy. This illustrates why a nondegenerate small-strain quadratic energy cannot also be imposed as a fully frame-indifferent finite-<deformation> energy.
Back to article page
