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In section 5.3.2 of the book Advanced Classical Electromagnetism by Robert Wald, in deriving the multipole expansion for the retarded solution of electromagnetic field in presence of charge-current distribution, it was asserted that $$\nabla \cdot\left(h_{\ell}^{(1)}\left(\frac{\omega r}{c}\right) \boldsymbol{r} \times \nabla Y_{\ell m}\right)=0$$ and $$\left[\nabla^{2}+\frac{\omega^{2}}{c^{2}}\right]\left(h_{\ell}^{(1)}\left(\frac{\omega r}{c}\right) \boldsymbol{r} \times \nabla Y_{\ell m}\right)=0$$ where $h_{\ell}^{(1)}$ is the Hankel function of the first kind, and $\boldsymbol{r} \times \nabla Y_{\ell m}$ is vector spherical harmonics.

The first equation is a property of vector spherical harmonics, as I saw on Wikipedia, but I am not able to derive this property myself. I also tried to prove the second equation by brute force, using the identity $\nabla^2(\boldsymbol{r} \times \nabla Y_{\ell m})=-\frac{l(l+1)}{r^2}(\boldsymbol{r} \times \nabla Y_{\ell m})$, but the terms does not cancel at the end. I wonder if someone can explain how can theses two identities be shown, or suggest some literature I can look into.

Sofvar
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1 Answers1

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A good overview over identities and definition can be found on en.wikipedia.org/wiki/Vector_spherical_harmonics. In the notation of wikipedia we are dealing with identities $$ \begin{align} \nabla\cdot(f(r)\, \mathbf{\Phi}_{lm}) &= 0,\tag{1}\\ \nabla^2(f(r)\, \mathbf{\Phi}_{lm}) &= \Delta(f(r)\, \mathbf{\Phi}_{lm}) = \left(\frac{1}{r^2}\frac{\partial}{\partial r}r^2\frac{\partial f(r)}{\partial r}\right) \mathbf{\Phi}_{lm} +f(r)\Delta\mathbf{\Phi}_{lm}\\ &= \left(\frac{1}{r^2}\frac{\partial}{\partial r}r^2\frac{\partial f(r)}{\partial r}\right) \mathbf{\Phi}_{lm} - \frac{1}{r^2}l(l+1)f(r)\mathbf{\Phi}_{lm},\tag{2} \end{align} $$

for $$ \mathbf{\Phi}_{lm}=\mathbf{r}\times\nabla Y_{lm}. $$

The first identity is simple to prove using the definition $$ Y_{lm}\equiv\sqrt{\frac{2 l+1}{4 \pi }} \sqrt{\frac{(l-m)!}{(l+m)!}} P_l^m(\cos \theta)\mathrm{e}^{\mathrm{i} m \phi }\equiv c_{lm} P_l^m(\cos \theta)\mathrm{e}^{\mkern1mu\mathrm{i}\mkern1mum\mkern1mu\phi }\tag{3} $$ and the relations for the gradient and divergence in spherical coordinates: $$ \begin{align} \nabla\cdot(f(r)\, \mathbf{\Phi}_{lm})&=\nabla\cdot(f(r)\,\mathbf{r}\times\nabla Y_{lm}) \tag{4.1}\\[.75em] &= \nabla\cdot(f(r)\,\mathbf{r}\times \frac{c_{lm}\mathrm{e}^{\mkern1mu\mathrm{i}\mkern1mum\mkern1mu\phi }}{r}(0,-\sin\theta (P_l^{m})'(\cos\theta),\mathrm{i}\,m\csc\theta P_l^m(\cos\theta))^T ) \tag{4.2}\\[.75em] &=\nabla\cdot(c_{lm}f(r){\mathrm{e}^{\mkern1mu\mathrm{i}\mkern1mum\mkern1mu\phi }}(0,-\mathrm{i}\,m\csc\theta P_l^m(\cos\theta),-\sin\theta (P_l^{m})'(\cos\theta))^T ) \tag{4.3}\\[.75em] &=0+ \mathrm{i}\,m \,c_{lm}f(r)(P_l^{m})'(\cos\theta)\mathrm{e}^{\mkern1mu\mathrm{i}\mkern1mum\mkern1mu\phi }(1-1)=0\qquad\square,\tag{4.4} \end{align} $$ where we used the notation $(a,b,c)^T\equiv a \hat{r}+b \hat{\theta}+c \hat{\phi}$.

Proofing the second identity (2) is a bit more involved before we start with an auxillary computation of $\Delta\mathbf{\Phi}_{lm}$ we recall two properties of Associated Legendre polynomials: $$ \begin{align} (P_l^{m})'(\cos\theta)&=(l-m+1) \csc \theta P_{l+1}^m(\cos \theta )-(l+1) \cot \theta P_l^m(\cos \theta),\tag{5.1}\\ (1 + l - m)P_{l+1}^m(\cos \theta)&=(2 l+1) \cos \theta P_l^m(\cos \theta)-(l+m) P_{l-1}^m(\cos \theta). \tag{5.2} \end{align} $$ We now first explicitly compute $\mathbf{\Phi}_{lm}$ using Eq. (5.1): $$ \mathbf{\Phi}_{lm} = -c_{lm}\mathrm{e}^{\mkern1mu\mathrm{i}\mkern1mum\mkern1mu\phi }\csc\theta \,(0,\mathrm{i}\, m P_l^m(\cos \theta),(l+1) \cos \theta P_l^m(\cos \theta )-(l-m+1) P_{l+1}^m(\cos\theta ))^T. \tag{6} $$ Using the Laplacian in spherical coordinates and Eq. (5.2) repeatedly for $l+3$ and $l+2$ we recover the identity $$ \Delta \mathbf{\Phi}_{lm} = -\frac{1}{r^2}l(l+1) \mathbf{\Phi}_{lm},\tag{7} $$ which proves Eq. (2) after using the chain rule $\square$.

To apply Eq. (2) to $f(r)=h_l^{(1)}(\frac{\omega}{c}r)$ we look up (see e.g. dlmf.nist.gov/10.6) an identity for the derivative of the Hankel function $$ (h_l^{(1)})'(z)=\frac{l}{z}h_l^{(1)}(z)-h_{l+1}^{(1)}(z) \tag{8} $$ and using the chain and product rule with Eq. (8) we compute $$ \frac{1}{r^2}\frac{\partial}{\partial r}r^2\frac{\partial}{\partial r}h_l^{(1)}\left(\frac{\omega}{c}r\right)=\left(\frac{l (l+1)}{r^2}-\frac{\omega^2}{c^2}\right)h_l^{(1)}\left(\frac{\omega}{c}r\right), \tag{9} $$ which together with Eq. (2) proves $$ \left(\nabla^2+\frac{\omega^2}{c^2}\right)\left(h_l^{(1)}\left(\frac{\omega}{c}r\right)\, \mathbf{\Phi}_{lm}\right) = 0 \qquad \square.\tag{10} $$

N0va
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