Mathematical Methods for Physics and Engineering - Matematica.NET

TENSORS(iii) referred to these axes as coordinate axes, the inertia tensor is diagonalwith diagonal entries λ 1 ,λ 2 ,λ 3 .Two further examples of physical quantities represented by second-order tensorsare magnetic susceptibility and electrical conductivity. In the first case we have(in standard notation)M i = χ ij H j , (26.43)and in the second casej i = σ ij E j . (26.44)Here M is the magnetic moment per unit volume and j the current density(current per unit perpendicular area). In both cases we have on the left-hand sidea vector and on the right-hand side the contraction of a set of quantities withanother vector. Each set of quantities must therefore form the components of asecond-order tensor.For isotropic media M ∝ H and j ∝ E, but for anisotropic materials such ascrystals the susceptibility and conductivity may be different along different crystalaxes, making χ ij and σ ij general second-order tensors, although they are usuallysymmetric.◮The electrical conductivity σ in a crystal is measured by an observer to have componentsas shown:⎛[σ ij ]= ⎝ 1 √ ⎞√ 2 02 3 1 ⎠ . (26.45)0 1 1Show that there is one direction in the crystal along which no current can flow. Does thecurrent flow equally easily in the two perpendicular directions?The current density in the crystal is given by j i = σ ij E j ,whereσ ij ,relativetotheobserver’scoordinate system, is given by (26.45). Since [σ ij ] is a symmetric matrix, it possessesthree mutually perpendicular eigenvectors (or principal axes) with respect to which theconductivity tensor is diagonal, with diagonal entries λ 1 ,λ 2 ,λ 3 , the eigenvalues of [σ ij ].As discussed in chapter 8, the eigenvalues of [σ ij ] are given by |σ − λI| = 0. Thus werequire∣ √ ∣∣∣∣∣ 1√− λ 2 02 3− λ 10 1 1− λ ∣ =0,from which we find(1 − λ)[(3 − λ)(1 − λ) − 1] − 2(1 − λ) =0.This simplifies to give λ =0, 1, 4 so that, with respect to its principal axes, the conductivitytensor has components σ ij ′ given by ⎛[σ ij] ′ = ⎝ 4 0 0⎞0 1 0 ⎠ .0 0 0Since j ′ i = σ ′ ijE ′ j, we see immediately that along one of the principal axes there is no currentflow and along the two perpendicular directions the current flows are not equal. ◭952