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An unconstrained natural state is a stress-free homogeneous deformation gradient. For $J>0$, its equivalent conditions are $DW(F)=0$ and $J^{-1}DW(F)F^T=0$. An interior differentiable minimum of the strain energy density is a natural state, although a stationary state need not minimize the energy. Constraints require allowing their reaction stress; the unconstrained equivalence must then be modified. Under strict Baker-Ericksen inequalities, an isotropic natural state is a dilation times a proper rotation.

The radial gradient $F=D\mathbf{y}$ is symmetric positive definite. If it belongs to a well $SO(3)U_i$, write $F=Q{U}_i$. Then $F^{T}F=U_i^2$, while positive symmetry makes $F$ itself the unique positive square root of an operator of $F^{T}F$. Hence $F=U_i$. Likewise a symmetric positive-definite matrix in $SO(3)$ must be $I$. ThereforeFor each radius, the tangential stretch $r(R)/R$ must belong to the finite set of eigenvalues of these matrices. The ratio is continuous on the connected radial interval, so it is one constant $v>0$. Thus $r(R)=vR$ and differentiation gives $r^{\prime }=v$ as well. HenceThis proof uses finiteness and positive definiteness of the wells, not isotropy or the Baker-Ericksen inequalities. It also covers $R_1=0$: the punctured three-dimensional ball remains connected, and no condition at the excluded center was used.

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 $F=R\mathrm{diag}(v_1,v_2,v_3)Q$ 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, soRight 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: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 $f(R)=r(R)/R$, $R=|\mathbf{x}|$, $\mathbf{\theta }=\mathbf{x}/R$. Since ${\partial }_{j}R={\theta }_j$,This symmetric tensor has radial eigenvalue $r^{\prime }$ 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.

Consider an orientation-preserving classical radial deformation of an elastic solid on a connected spherical annulus. If its nonnegative energy is zero everywhere and the material is isotropic with strict Baker-Ericksen inequalities, every gradient is a natural state, so $r^{\prime }=r/R$. Thus $(r/R{)}^{\prime }=0$ and $y=vx$.

A separate rigidity argument applies when the zero wells are $SO(3)\cup {\bigcup }_{i=1}^{N}SO(3)U_i$, with finitely many positive definite symmetric $U_i$. The radial gradient $F$ is positive symmetric; if $F=Q{U}_i$, then $F^2=U_i^2$, so uniqueness of the positive square root gives $F=U_i$. The rotation well similarly gives $F=I$. Hence the continuous ratio $r(R)/R$ takes values in a finite set of eigenvalues and must be constant on the connected radial interval. Differentiating $r=vR$ finishes the proof. This version needs neither material isotropy nor Baker-Ericksen inequalities and does not require an assumed continuity of $r^{\prime }$ beyond the radial-map regularity used for the ratio.