 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2021/paper-2/6/a/i/solution

Under the specified passive change of orthonormal basis, each Cartesian tensor index acquires one copy of

L

. Thus a tensor of order

n

S_{a_{1} \dots a_{n}}^{'} = L_{a_{1} b_{1}} \dots L_{a_{n} b_{n}} S_{b_{1} \dots b_{n}} .

Repeated old-basis indices are summed from one to three. Orthogonality of the new basis gives

L L^{T} = I

. Taking determinants yields

(det L)^{2} = 1

, hence

det L = + 1 or - 1

. The positive case preserves orientation and the negative case reverses it.

 New to topics? Read the docs here!