Mathematical Methods for Physics and Engineering : A Comprehensive Guide

(Darren Dugan) #1

TENSORS


(iii) referred to these axes as coordinate axes, the inertia tensor is diagonal
with diagonal entriesλ 1 ,λ 2 ,λ 3.

Two further examples of physical quantities represented by second-order tensors

are magnetic susceptibility and electrical conductivity. In the first case we have


(in standard notation)


Mi=χijHj, (26.43)

and in the second case


ji=σijEj. (26.44)

HereMis the magnetic moment per unit volume andjthe current density


(current per unit perpendicular area). In both cases we have on the left-hand side


a vector and on the right-hand side the contraction of a set of quantities with


another vector. Each set of quantities must therefore form the components of a


second-order tensor.


For isotropic mediaM∝Handj∝E, but for anisotropic materials such as

crystals the susceptibility and conductivity may be different along different crystal


axes, makingχijandσijgeneral second-order tensors, although they are usually


symmetric.


The electrical conductivityσin a crystal is measured by an observer to have components
as shown:

[σij]=



1



√^20


231


011



. (26.45)


Show that there is one direction in the crystal along which no current can flow. Does the
current flow equally easily in the two perpendicular directions?

The current density in the crystal is given byji=σijEj,whereσij,relativetotheobserver’s
coordinate system, is given by (26.45). Since [σij] is a symmetric matrix, it possesses
three mutually perpendicular eigenvectors (or principal axes) with respect to which the
conductivity tensor is diagonal, with diagonal entriesλ 1 ,λ 2 ,λ 3 , the eigenvalues of [σij].
As discussed in chapter 8, the eigenvalues of [σij] are given by|σ−λI|= 0. Thus we
require ∣
∣∣

∣∣


1 −λ


√^20


23 −λ 1
011 −λ


∣∣



∣∣=0,


from which we find
(1−λ)[(3−λ)(1−λ)−1]−2(1−λ)=0.


This simplifies to giveλ=0, 1 ,4 so that, with respect to its principal axes, the conductivity
tensor has componentsσ′ijgiven by


[σ′ij]=



400


010


000



.


Sincej′i=σ′ijEj′, we see immediately that along one of the principal axes there is no current
flow and along the two perpendicular directions the current flows are not equal.

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