Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2015 / Paper 2 / 6

Preserving Cartesian lengths and Euclidean inner products requires an orthogonal matrix:Its determinant is $\pm 1$. Relative to an originally right-handed frame, $\det M=+1$ gives a right-handed frame and $\det M=-1$ a left-handed frame. More generally, this sign determines whether the two frames have the same or opposite orientation.

The tensor component transformation law isFor a tensor field, the components are compared at the same physical point, whose coordinates satisfy $x=M^T{x}^{\prime }$.

Define the symmetric tensor and antisymmetric tensor parts by $S=(T+T^T)/2$ and $A=(T-T^T)/2$. The transformed transpose satisfies $(T^{\prime }{)}^T=(MT{M}^T{)}^T=M{T}^T{M}^T$. Thereforeand similarlyBoth obey the tensor component transformation law. Thus the symmetric and antisymmetric parts are themselves second-order tensors. Taking a transpose and forming linear combinations commute with this orthogonal change of frame.

Let $r=\sqrt{x_1^2+x_2^2}$. Averaging each pair of transposed entries gives the symmetric tensorFor $r>0$, the upper block is the outer product of $(x_1,x_2)$ with itself. It sends that vector to $r^2$ times itself and annihilates its perpendicular. Thus its orthonormal principal axes and corresponding eigenvalues areTaking these vectors as the rows of the change-of-components matrix givesThis matrix is orthogonal with determinant one. These are the principal axes of a planar rank-one tensor block.

The matrix is a pointwise frame transformation: because its angle varies with position, it does not represent one fixed Cartesian rotation diagonalizing the entire field everywhere. On the axis $r=0$, the symmetric tensor is zero; all principal values coincide and any orthonormal axes may be chosen. The displayed formula for $M$ is therefore used only off that axis.

The scalar, symmetric-traceless and axial tensor decomposition fixes the scalar part by the trace:Subtracting this isotropic part from the symmetric tensor in part (c) givesIts trace is zero. The antisymmetric tensor part isAdding $PI+\hat{S}+\hat{A}$ reconstructs every entry of $F$.

Subtracting a scalar multiple of the identity preserves every eigenvector: if $Sv=\lambda v$, then $(S-PI)v=(\lambda -P)v$. Thus the principal axes are the same as in part (c), with shifted principal valuesThe two tensors are coaxial symmetric tensors. At $r=0$ both vanish, so neither specifies unique axes.