= Solution
The <Green-Lagrange strain tensor> has <eigenvalues> $(\lambda_j^2-1)/2$. Its <Frobenius norm> and trace are invariant under the orthogonal principal-axis change, so
$$
\boxed{W(F)=\frac\mu4\sum_{j=1}^3(\lambda_j^2-1)^2+\frac\lambda8\left(\sum_{j=1}^3\lambda_j^2-3\right)^2.}
$$
For the homogeneous dilation $F=\rho I$, put $K=\mu+3\lambda/2$. Then $E=(\rho^2-1)I/2$. Differentiating the energy through $E$ gives the <second Piola-Kirchhoff stress tensor> $2\mu E+\lambda\operatorname{tr}(E)I=K(\rho^2-1)I$, and multiplication by $F$ gives the <first Piola-Kirchhoff stress tensor>. Thus
$$
\boxed{T=\rho K(\rho^2-1)I,\qquad S=J^{-1}TF^T=\frac{K(\rho^2-1)}\rho I.}
$$
The normals of the reference and deformed spheres point in the same radial direction. As $\rho\downarrow0$, the reference <traction> magnitude $|T N|$ tends to zero, while the current <traction> magnitude $|S n|$ tends to infinity, provided $K>0$, as for physically stable <Lamé parameters>. The negative signs of the <stress> coefficients for $0<\rho<1$ indicate compression.
There is no contradiction: the current <area> is $\rho^2$ times the reference <area>, so the same <force> divided by the vanishing current <area> has the larger <stress>. Indeed $T N=\rho^2S n$. The limiting map at $\rho=0$ is not itself an admissible <deformation in continuum mechanics>; the statement concerns the approach through positive $\rho$. A degenerate choice $K=0$ would not have the asserted growth without bound.
Back to article page
