This chapter's key result for an azimuthally symmetric potential,

PHYSICS · Multi-part · Fermi University Physics · Mathematical Methods for Physics · Partial differential equations and boundary-value problems · Special functions and orthogonal polynomials

Conditions of Validity, Ch Separation of variables as a chain of Sturm-Liouville problems

Problem

This chapter's key result for an azimuthally symmetric potential, u=∑ℓ(Aℓrℓ+Bℓr−ℓ−1)Pℓ(cos⁡θ)u=\sum_\ell(A_\ell r^\ell+B_\ell r^{-\ell-1})P_\ell(\cos\theta), quotes the two radial powers without deriving them. ((a)) Laplace's equation ∇2u=0\nabla^2u=0, restricted to azimuthal symmetry (∂u/∂ϕ=0\partial u/\partial\phi=0), reads, from the general spherical Laplacian, ∇2u=1r2∂∂r ⁣(r2∂u∂r)+1r2sin⁡θ∂∂θ ⁣(sin⁡θ∂u∂θ)=0.\nabla^2u=\frac{1}{r^2}\frac{\partial}{\partial r}\!\left(r^2\frac{\partial u}{\partial r}\right)+\frac{1}{r^2\sin\theta}\frac{\partial}{\partial\theta}\!\left(\sin\theta\frac{\partial u}{\partial\theta}\right)=0. Substitute u=R(r)Θ(θ)u=R(r)\Theta(\theta) and use that the polar-angle equation already separated in this chapter fixes the angular separation constant at ℓ(ℓ+1)\ell(\ell+1), to show that RR satisfies r2R′′+2rR′−ℓ(ℓ+1)R=0,r^2R''+2rR'-\ell(\ell+1)R=0, and confirm that R=rℓR=r^\ell and R=r−ℓ−1R=r^{-\ell-1} both solve it. ((b)) A physical dipole has a point charge q=3.0 nCq=3.0\,\mathrm{nC} at z=+dz=+d and −q-q at z=−dz=-d on the polar axis, with d=4.0 cmd=4.0\,\mathrm{cm}. Using the moment integral Qℓ=∫ρ r′ℓPℓ(cos⁡θ′) dV′Q_\ell=\int\rho\,r'^\ell P_\ell(\cos\theta')\,\mathrm dV' quoted earlier in this chapter, applied to the two point charges, find Q0Q_0, Q1Q_1 and Q2Q_2, and use the leading surviving term of the key result to estimate the potential on the axis at r=0.30 mr=0.30\,\mathrm m. Check the estimate against the exact potential of the two point charges added directly.

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