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A dead-load traction is prescribed force per unit reference area whose vector is independent of the deformation. It is imposed through $T_{R}N=t_R$ on the reference boundary. It does not rotate or rescale with the current surface as a follower load would.

Elastic incompressibility imposes $\det F=1$ and adds a scalar Lagrange multiplier to the constitutive law. With the pressure sign convention used here,The pressure field is determined by equilibrium and boundary conditions, rather than by the strain energy density alone.

For the diagonal homogeneous deformation, $F=\mathrm{diag}(v_1,v_2,v_3)$ and $v_1{v}_2{v}_3=1$. Isotropy makes $DW(F)$ diagonal here, with entries ${\Phi }_i$, soThe six face conditions are exactly $T_R{e}_i=\tau e_i$. Thus ${\Phi }_i-p/v_i=\tau$ for each $i$, orNecessity follows from the boundary conditions. Conversely these equations give a constant pressure and constant stress satisfying those conditions, while $\mathrm{Div}{T}_R=0$ because the stress is constant and there is no body force. This proves sufficiency as well.

For the particular energy, ${\Phi }_i=\mu (v_i-v_i^{-2})$. Put $S=v_1+v_2+v_3$ and $Q=v_1{v}_2+v_1{v}_3+v_2{v}_3$. Elastic incompressibility gives $v_i^{-1}=v_j{v}_k$, henceEquality for every pair is therefore equivalent to $(\mu S-\tau )(v_i-v_j)=0$. If all stretches are equal, their positive product one gives $v_1=v_2=v_3=1$, with $p=-\tau$. Otherwise $S=\tau /\mu$. Conversely any positive triple with this sum and product one solves the equations, with $p=-\mu Q$. Thus the complete solutions within the stated homogeneous diagonal ansatz are the unit triple and the stated sum-product family.

By the arithmetic-geometric mean inequality, $S\ge 3$, with equality only at the unit triple. Consequently $\tau \le 3\mu$ admits only that triple, including equality, when the two descriptions coincide.

To prove infinitely many for $s=\tau /\mu >3$, choose $v_1=t$ near one and setAt $t=1$ the discriminant is $(s-1{)}^2-4>0$; it stays positive on an interval about one, and both roots remain positive. Every such $t$ gives a different admissible triple. HenceThe degeneracy has an energy explanation: under the product constraint the strain energy density is $\mu S^2/2$, and the total potential under dead loading is $\mu S^2/2-\tau S$. It depends on the three stretches only through their sum. Its minimum over attainable $S\ge 3$ moves from $S=3$ to the whole level set $S=\tau /\mu$ when the load exceeds $3\mu$.

As a function of the full deformation gradient, use $e=\mathrm{sym}(F-I)$. Let $Q=\mathrm{diag}(1,-1,-1)$, a proper rotation. Assumed full frame indifference would give $\tilde{W}(Q)=\tilde{W}(I)=0$. For $F_s=I+s(Q-I)$, its infinitesimal strain is $s\mathrm{sym}(Q-I)$. Quadratic homogeneity therefore givesChoose $s=1/4$. ThenThe homogeneous deformation $y(X)=F_{s}X$ 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.

A deformation maps a material point in the reference body to its current position, $y:\Omega \to {\mathbb{R}}^3$, with deformation gradient $F=Dy$. A homogeneous deformation has constant gradient and is affine, $y(X)=FX+b$.

If $\det F>0$, the matrix $F^{T}F$ is symmetric positive definite. The matrix-square-root theorem says that any such matrix has a unique symmetric positive definite square root, obtained by taking positive square roots in its orthogonal spectral decomposition. SetThen $R^{T}R=U^{-1}{F}^{T}F{U}^{-1}=I$ and $\det R=\det F/\det U>0$, so $R\in \mathrm{SO}(3)$ and $F=RU$. If also $F=\tilde{R}\tilde{U}$, then $F^{T}F={\tilde{U}}^2$; uniqueness of the positive square root gives $\tilde{U}=U$ and then $\tilde{R}=R$. This proves existence and uniqueness of polar decomposition.

The principal stretches are the positive eigenvalues of $U$, equivalently the singular values of $F$ or square roots of the eigenvalues of $F^{T}F$.