Three short derivations extend results built for the multipole expansion.

PHYSICS · Multi-part · Fermi University Physics · Mathematical Methods for Physics · Electricity, Magnetism & Electromagnetism · Electrostatic fields and potential

The Boundary-Value Problem, Ch The multipole expansion of the potential

Problem

Three short derivations extend results built for the multipole expansion. ((a)) Carry the binomial expansion of (1−2tcos⁡γ+t2)−1/2(1-2t\cos\gamma+t^2)^{-1/2} one order further than the t2t^2 term already found, using (1+u)−1/2=1−12u+38u2−516u3+⋯(1+u)^{-1/2} = 1 - \tfrac12u + \tfrac38u^2 - \tfrac{5}{16}u^3 + \cdots with u=−2tcos⁡γ+t2u = -2t\cos\gamma + t^2. Collect the coefficient of t3t^3 and show it equals P3(cos⁡γ)=12(5cos⁡3γ−3cos⁡γ)P_3(\cos\gamma) = \tfrac12(5\cos^3\gamma - 3\cos\gamma). Check the four-term series numerically at t=0.1t=0.1, γ=0\gamma=0 against the exact value 1/(1−t)1/(1-t). ((b)) A point dipole p=pz^\mathbf{p} = p\hat{\mathbf{z}} sits at the origin. Starting from E=14πε0r3[3(p⋅r^)r^−p]\mathbf{E} = \dfrac{1}{4\pi\varepsilon_0r^3}\left[3(\mathbf{p}\cdot\hat{\mathbf{r}})\hat{\mathbf{r}} - \mathbf{p}\right] (valid away from the origin), and using z^=cos⁡θ r^−sin⁡θ θ^\hat{\mathbf{z}} = \cos\theta\,\hat{\mathbf{r}} - \sin\theta\,\hat{\boldsymbol\theta}, find the spherical components Er(θ)E_r(\theta), Eθ(θ)E_\theta(\theta) and Eϕ(θ)E_\phi(\theta). Confirm that θ=0\theta=0 reproduces the on-axis value 2p/4πε0r32p/4\pi\varepsilon_0r^3 and that θ=π/2\theta=\pi/2 reproduces the equatorial value p/4πε0r3p/4\pi\varepsilon_0r^3 directed against p\mathbf{p}. ((c)) Starting from the ℓ=2\ell=2 term of the multipole expansion, V2=14πε0r3∫r′2P2(cos⁡γ) ρ dτ′V_2 = \dfrac{1}{4\pi\varepsilon_0r^3}\displaystyle\int r'^2P_2(\cos\gamma)\,\rho\,d\tau', and using r′cos⁡γ=r^ixi′r'\cos\gamma = \hat r_ix_i' and r^ir^i=1\hat r_i\hat r_i=1, show step by step in index notation that V2=18πε0r3∑ijQijr^ir^j,Qij=∫(3xi′xj′−r′2δij)ρ dτ′.V_2 = \frac{1}{8\pi\varepsilon_0r^3}\sum_{ij}Q_{ij}\hat r_i\hat r_j, \qquad Q_{ij} = \int(3x_i'x_j' - r'^2\delta_{ij})\rho\,d\tau'. Then, for a distribution that is axially symmetric about zz (so Qxx=QyyQ_{xx}=Q_{yy} by symmetry and the trace condition fixes Qxx=Qyy=−12QzzQ_{xx}=Q_{yy}=-\tfrac12Q_{zz}), show that V2V_2 reduces to Qzz8πε0r3P2(cos⁡θ)\dfrac{Q_{zz}}{8\pi\varepsilon_0r^3}P_2(\cos\theta).

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