Parabolic cylindrical coordinates are related to Cartesian coordinates by , ,

PHYSICS · Multi-part · Fermi University Physics · Mathematical Methods for Physics · Vector calculus and field geometry · Tensors and differential geometry

Invariance, Ch Differential forms and the exterior derivative

Problem

Parabolic cylindrical coordinates (u,v,z)(u,v,z) are related to Cartesian coordinates by x=12(u2−v2)x=\tfrac12(u^2-v^2), y=uvy=uv, z=zz=z, with u∈Ru\in\mathbb R, v≥0v\ge0. ((a)) Compute dxdx and dydy in terms of du,dvdu,dv and show that dx∧dy=(u2+v2) du∧dvdx\wedge dy=(u^2+v^2)\,du\wedge dv; hence write dx∧dy∧dzdx\wedge dy\wedge dz in these coordinates. ((b)) The line element in these coordinates is ds2=(u2+v2)(du2+dv2)+dz2ds^2=(u^2+v^2)(du^2+dv^2)+dz^2 (held from the curvilinear-coordinates course), so the orthonormal coframe is e1=h due^1=h\,du, e2=h dve^2=h\,dv, e3=dze^3=dz with h=u2+v2h=\sqrt{u^2+v^2}. Using the Hodge star of R3\mathbb R^3, derive ∇2f=⋆d⋆df\nabla^2f={\star}d{\star}df for a function f(u,v,z)f(u,v,z) in terms of ∂u,∂v,∂z\partial_u,\partial_v,\partial_z and hh. ((c)) Apply the result to f=ln⁡(u2+v2)f=\ln(u^2+v^2), show that it is harmonic everywhere except at u=v=0u=v=0, and explain why this must be so, given that ln⁡(u2+v2)=ln⁡(2r)\ln(u^2+v^2)=\ln(2r) where r=x2+y2r=\sqrt{x^2+y^2}.

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