A uniform cube of mass and side has

PHYSICS · Multi-part · Fermi University Physics · Mathematical Methods for Physics · Mechanics & Dynamical Systems · Rotation, rigid bodies and accelerated frames

Invariance, Ch General bases, the metric and tensors in spacetime

Problem

A uniform cube of mass M=1.5 kgM = 1.5\,\mathrm{kg} and side L=0.20 mL = 0.20\,\mathrm{m} has, about its centre and referred to orthonormal axes e1,e2,e3\mathbf{e}_1,\mathbf{e}_2,\mathbf{e}_3 along its edges, the inertia tensor Iij=16ML2 δijI_{ij} = \tfrac16 ML^2\,\delta_{ij}, a result held from the mechanics course. A new, non-orthogonal basis is defined by e1′=e1\mathbf{e}'_1 = \mathbf{e}_1, e2′=e1+e2\mathbf{e}'_2 = \mathbf{e}_1+\mathbf{e}_2, e3′=e3\mathbf{e}'_3 = \mathbf{e}_3, so that ei′=Ajiej\mathbf{e}'_i = A^j{}_i\mathbf{e}_j with A11=1A^1{}_1=1, A21=0A^2{}_1=0, A12=1A^1{}_2=1, A22=1A^2{}_2=1, A33=1A^3{}_3=1, all other entries zero. ((a)) Using the covariant transformation law Iij′=AkiAljIklI'_{ij} = A^k{}_iA^l{}_jI_{kl}, find I11′I'_{11}, I12′I'_{12} and I22′I'_{22} (the 3333 component is unaffected). ((b)) Find the metric gij′=ei′⋅ej′g'_{ij} = \mathbf{e}'_i\cdot\mathbf{e}'_j of the new basis, invert its 11–22 block, and use g′ijg'^{ij} to raise both indices, I′ij=g′ikg′jlIkl′I'^{ij} = g'^{ik}g'^{jl}I'_{kl}, obtaining I′11I'^{11}, I′12I'^{12} and I′22I'^{22}.

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