 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2018/paper-2/6/a/solution

Split the Cartesian second-rank tensor into symmetric and antisymmetric parts:

S_{ij} = \frac{T _{ij} + T _{ji}}{2}, A_{ij} = \frac{T _{ij} - T _{ji}}{2} .

u_{k} = \frac{1}{2} ϵ_{kij} T_{ij}

. The symmetric part contributes zero to this contraction. The epsilon-delta identity then gives

ϵ_{ijk} u_{k} = \frac{1}{2} ϵ_{ijk} ϵ_{k l m} T_{l m} = \frac{1}{2} (δ_{i l} δ_{jm} - δ_{im} δ_{j l}) T_{l m} = \frac{T _{ij} - T _{ji}}{2} = A_{ij} .

Therefore

T_{ij} = S_{ij} + ϵ_{ijk} u_{k}, S_{ij} = \frac{T _{ij} + T _{ji}}{2}, u_{k} = \frac{1}{2} ϵ_{kij} T_{ij} .

The symmetric part and the three independent antisymmetric components determine these quantities uniquely. Under proper rotations

u

is a vector; under reflections it is an axial vector.

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