= Solution
The <material symmetry group> consists of invertible admissible reference transformations $H$ for which $W(FH)=W(F)$ for every admissible $F$. Its rotational part lies in $SO(3)$; a <determinant>-one reference convention restricts the full group to the corresponding special linear group. This right action is distinct from the spatial left action $W(QF)=W(F)$ of <material frame indifference>.
The proper cube-rotation group has $\boxed{24}$ elements. Choose the image of one outward face normal in six ways, then the image of an adjacent perpendicular normal in four ways; orientation fixes the third direction. No reflections are included. For an isotropic material every proper rotation belongs to the <material symmetry group>.
Combine <material isotropy> with the physical frame-indifference condition. The <singular value decomposition> may be written $F=R\operatorname{diag}(v_1,v_2,v_3)Q$ with $R,Q\in SO(3)$ because $\det F>0$. Left and right rotational invariance remove $R,Q$. Signed permutation rotations permit any permutation of the three diagonal stretches, so
$$
\boxed{W(F)=\Phi(v_1,v_2,v_3),\qquad\Phi\text{ symmetric in its three arguments}.}
$$
Right invariance alone, without the usual <material frame indifference> assumption, would not suffice to remove spatial orientation.
The strict <Baker-Ericksen inequalities> say that larger distinct principal stretches carry larger principal Cauchy stresses:
$$
\boxed{(\sigma_i-\sigma_j)(v_i-v_j)>0\quad\text{if }v_i\ne v_j,\qquad
\sigma_i=J^{-1}v_i\Phi_i.}
$$
A <natural state of a hyperelastic material> has zero stress. If two stretches differed, their equal zero stresses would contradict the strict inequality. Thus all three equal some $v>0$, and <polar decomposition> gives $\boxed{F=vQ}$, $Q\in SO(3)$. A weak non-strict inequality alone would not prove this conclusion, so its strict convention is explicit here.
For the radial deformation put $f(R)=r(R)/R$, $R=|\mathbf x|$, $\boldsymbol\theta=\mathbf x/R$. Since $\partial_jR=\theta_j$,
$$
D\mathbf y=fI+Rf'\boldsymbol\theta\otimes\boldsymbol\theta
=\boxed{\frac rR I+\left(r'-\frac rR\right)\boldsymbol\theta\otimes\boldsymbol\theta.}
$$
This symmetric tensor has radial <eigenvalue> $r'$ and the repeated tangential <eigenvalue> $r/R$. Both are positive in the open annulus under the stated monotonicity/nonnegativity conditions. Therefore these are its principal stretches. The two cases below establish <radial zero-energy rigidity> by different arguments.
Back to article page
