Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge 2002 Paper 2 6 c Solution Created 2026-10-18 Updated 2026-10-20
Contract the given isotropic fourth-rank tensor with the strain array. The stress tensor isWrite , where is symmetric and traceless. Then and . Substitution givesBy uniqueness of the scalar, symmetric-traceless and axial tensor decomposition,This is isotropic elasticity on tensor components: the three tensor sectors do not mix. If the strain is symmetric, , hence and . No equality between and is needed for that conclusion.
Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge 2008 Paper 2 6 Solution Created 2026-10-18 Updated 2026-10-20
Under an orthogonal change of Cartesian coordinates with , a rank- Cartesian tensor transforms as , summing repeated indices. An isotropic tensor is invariant under every rotation. For an isotropic fourth-rank tensor, half-turns about the coordinate axes eliminate components unless each coordinate label occurs an even number of times. Coordinate rotations make the remaining two-pair patterns depend on three constants , , . Invariance after rotating the first axis to gives . HenceThis is a direct proof of the rank-four isotropic decomposition, without assuming any prior interchange symmetry of the four indices. Each displayed delta pairing is itself rotation invariant.
For symmetric strain, contraction gives . Comparing with the Lamé parameters yieldsWrite and , so . Since , the strain energy density becomesThe two terms describe volumetric and shear deformation, respectively. Thus nonnegative isotropic elastic strain energy is equivalent toNecessity also follows by taking a nonzero traceless strain and a pure dilation separately. The combination is the bulk modulus; strict inequalities give positive energy for every nonzero strain.