Past exam of the solid mechanics course of the University of Oxford / 2010 / 1

A homogeneous hyperelastic material has a stored energy per reference volume $W(F)$ depending on the local deformation gradient but not explicitly on the material point. Its first Piola-Kirchhoff response isA superposed rigid rotation changes the observer's description without changing the energy of the deformation. Material frame indifference is thereforeThis left-rotation invariance is distinct from a material isotropy assumption on right rotations. Differentiate it in an arbitrary matrix direction $H$:The Frobenius inner product then gives $Q^T\hat{T}(QF)=\hat{T}(F)$, proving $\hat{T}(QF)=Q\hat{T}(F)$.

For the converse, that stress covariance makes the derivative in $F$ of $W(QF)-W(F)$ zero. The positive-determinant matrices form a connected set, as also follows from polar decomposition, so this difference is a constant $c(Q)$ independent of $F$. Composition gives $c(Q_1{Q}_2)=c(Q_1)+c(Q_2)$. For fixed $F$, continuity of $W$ and compactness of $\mathrm{SO}(3)$ make $c$ bounded. But $c(Q^n)=nc(Q)$ for every positive integer, forcing $c(Q)=0$. Thus stress covariance and frame indifference of the energy are equivalent. Compactness removes the possible additive energy constant in the converse.

For small displacement gradients write $F=I+H$. Taylor expansion around the stress-free reference state givesThe linear term vanishes because $DW(I)=0$. Drop the irrelevant constant and retain the quadratic term:The major symmetry $\boxed{C_{i\alpha j\beta }=C_{j\beta i\alpha }}$ follows from equality of mixed second derivatives. For a skew matrix $K$, stress covariance along $e^{tK}$ givesDifferentiation at zero yields $D\hat{T}(I)[K]=C[K]=0$. Thus $C_{i\alpha j\beta }{K}_{j\beta }=0$ for every skew matrix, which forces the input minor symmetry $C_{i\alpha j\beta }=C_{i\alpha \beta j}$. Combining it with the major symmetry gives the output minor symmetry as well:These identify reference and spatial indices at $F=I$. Decompose $H=e+K$ into symmetric and skew parts. Since $C[K]=0$ and the major symmetry also eliminates the other cross term,Thus infinitesimal rotations cost no energy in the quadratic approximation.

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.