A crystal's dielectric response along its two principal axes ( ) is the linear map

PHYSICS · Multi-part · Fermi University Physics · Mathematical Methods for Physics · Electricity, Magnetism & Electromagnetism · Electric fields in matter

Mathematical Methods for Physics, Ch Vector spaces, bases, linear maps and dual vectors

Problem

A crystal's dielectric response along its two principal axes (1,21,2) is the linear map E↦P/ϵ0=XE\mathbf{E}\mapsto\mathbf{P}/\epsilon_0 = \mathsf{X}\mathbf{E}, where E=(E1,E2)\mathbf{E} = (E_1, E_2) is the electric field and P\mathbf{P} the polarisation. In the crystal-axis basis X=diag(χ1,χ2)\mathsf{X} = \mathrm{diag}(\chi_1, \chi_2) with χ1=2.0\chi_1 = 2.0 and χ2=5.0\chi_2 = 5.0 (dimensionless susceptibilities). A slab is cut so that its faces define lab axes rotated by θ=π/4 rad\theta = \pi/4\,\mathrm{rad} from the crystal axes, related by e^1′=cos⁡θ e^1+sin⁡θ e^2\hat{\mathbf{e}}_1' = \cos\theta\,\hat{\mathbf{e}}_1 + \sin\theta\,\hat{\mathbf{e}}_2, e^2′=−sin⁡θ e^1+cos⁡θ e^2\hat{\mathbf{e}}_2' = -\sin\theta\,\hat{\mathbf{e}}_1 + \cos\theta\,\hat{\mathbf{e}}_2. The dielectric is linear and lossless, and the field applied is uniform. ((a)) Write the matrix X\mathsf{X} of the map E↦P/ϵ0\mathbf{E}\mapsto\mathbf{P}/\epsilon_0 in the crystal-axis basis. ((b)) With R=(cos⁡θ−sin⁡θsin⁡θcos⁡θ)\mathsf{R} = \begin{pmatrix}\cos\theta & -\sin\theta\\ \sin\theta & \cos\theta\end{pmatrix} as the change-of-basis matrix from crystal axes to lab axes, compute X′=R−1XR\mathsf{X}' = \mathsf{R}^{-1}\mathsf{X}\mathsf{R}, the matrix of the same map in the lab-axis basis. ((c)) Confirm tr X=tr X′\mathrm{tr}\,\mathsf{X} = \mathrm{tr}\,\mathsf{X}' and det⁡X=det⁡X′\det\mathsf{X} = \det\mathsf{X}'. ((d)) In the crystal-axis basis the potential is φ(x1,x2)=−E0x1\varphi(x_1, x_2) = -E_0x_1 with E0=1.0×106 V m−1E_0 = 1.0\times10^6\,\mathrm{V\,m^{-1}}, so E=−∇φ=(E0,0)\mathbf{E} = -\nabla\varphi = (E_0, 0). For a displacement δx1=2.0 nm\delta x_1 = 2.0\,\mathrm{nm} along axis 11 (with 1 nm=10−9 m1\,\mathrm{nm} = 10^{-9}\,\mathrm{m}), compute the change in potential δφ=(∇φ)⋅δx\delta\varphi = (\nabla\varphi)\cdot\delta\mathbf{x}. ((e)) Re-express the calculation with axis-11 distance measured in nanometres, x1′=x1/(1 nm)x_1' = x_1/(1\,\mathrm{nm}): state the rescaled gradient component ∂φ/∂x1′\partial\varphi/\partial x_1' and the rescaled displacement δx1′\delta x_1', and confirm that δφ\delta\varphi from part (d) is recovered. ((f)) State which quantity from parts (a)-(c) is the same in both bases despite the matrix entries changing, and what is unchanged in parts (d)-(e) despite the gradient's and displacement's numbers both changing.

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