Vector Calculus | Part IA, 2006

Define what is meant by an isotropic tensor. By considering a rotation of a second rank isotropic tensor BijB_{i j} by 9090^{\circ} about the zz-axis, show that its components must satisfy B11=B22B_{11}=B_{22} and B13=B31=B23=B32=0B_{13}=B_{31}=B_{23}=B_{32}=0. Now consider a second and different rotation to show that BijB_{i j} must be a multiple of the Kronecker delta, δij\delta_{i j}.

Suppose that a homogeneous but anisotropic crystal has the conductivity tensor

σij=αδij+γninj\sigma_{i j}=\alpha \delta_{i j}+\gamma n_{i} n_{j}

where α,γ\alpha, \gamma are real constants and the nin_{i} are the components of a constant unit vector n\mathbf{n} (nn=1)(\mathbf{n} \cdot \mathbf{n}=1). The electric current density J\mathbf{J} is then given in components by

Ji=σijEjJ_{i}=\sigma_{i j} E_{j}

where EjE_{j} are the components of the electric field E\mathbf{E}. Show that

(i) if α0\alpha \neq 0 and γ0\gamma \neq 0, then there is a plane such that if E\mathbf{E} lies in this plane, then E\mathbf{E} and J\mathbf{J} must be parallel, and

(ii) if γα\gamma \neq-\alpha and α0\alpha \neq 0, then E0\mathbf{E} \neq 0 implies J0\mathbf{J} \neq 0.

If Dij=ϵijknkD_{i j}=\epsilon_{i j k} n_{k}, find the value of γ\gamma such that

σijDjkDkm=σim\sigma_{i j} D_{j k} D_{k m}=-\sigma_{i m}

Typos? Please submit corrections to this page on GitHub.