高等电动力学的一个证明

命题:

对平面波\(\displaystyle E=\hat{e}_{x} E_{0} e^{ikz} =\hat{e}_{x} E_{0} e^{ikr\cos \theta }\)按矢量球波函数进行展开。

其中,\(\displaystyle \hat{e}_{x} =\sin \theta \cos \phi \hat{e}_{r} +\cos \theta \cos \phi \hat{e}_{\theta } -\sin \phi \hat{e}_{\phi }\)

证明:

矢量球波函数: \[\begin{gathered} \vec{M}_{emn} =\frac{-m}{\sin \theta }\sin( m\phi ) P_{n}^{m}(\cos \theta ) z_{n}( kr)\hat{e}_{\theta } -\cos( m\phi )\frac{\mathrm{d} P_{n}^{m}(\cos \theta )}{\mathrm{d} \theta } z_{n}( kr)\hat{e}_{\phi }\\ \vec{M}_{omn} =\frac{m}{\sin \theta }\cos( m\phi ) P_{n}^{m}(\cos \theta ) z_{n}( kr)\hat{e}_{\theta } -\sin( m\phi )\frac{\mathrm{d} P_{n}^{m}(\cos \theta )}{\mathrm{d} \theta } z_{n}( kr)\hat{e}_{\phi }\\ \vec{N}_{emn} =\frac{z_{n}( kr)}{kr}\cos( m\phi ) n( n+1) P_{n}^{m}(\cos \theta )\hat{e}_{r}\\ +\cos( m\phi )\frac{\mathrm{d} P_{n}^{m}(\cos \theta )}{\mathrm{d} \theta }\frac{1}{kr}\frac{\mathrm{d}}{\mathrm{d}( kr)}[( kr) z_{n}( kr)]\hat{e}_{\theta }\\ -m\sin( m\phi )\frac{P_{n}^{m}(\cos \theta )}{\sin \theta }\frac{1}{kr}\frac{\mathrm{d}}{\mathrm{d}( kr)}[( kr) z_{n}( kr)]\hat{e}_{\phi }\\ \vec{N}_{omn} =\frac{z_{n}( kr)}{kr}\sin( m\phi ) n( n+1) P_{n}^{m}(\cos \theta )\hat{e}_{r}\\ +\sin( m\phi )\frac{\mathrm{d} P_{n}^{m}(\cos \theta )}{\mathrm{d} \theta }\frac{1}{kr}\frac{\mathrm{d}}{\mathrm{d}( kr)}[( kr) z_{n}( kr)]\hat{e}_{\theta }\\ +m\cos( m\phi )\frac{P_{n}^{m}(\cos \theta )}{\sin \theta }\frac{1}{kr}\frac{\mathrm{d}}{\mathrm{d}( kr)}[( kr) z_{n}( kr)]\hat{e}_{\phi }\end{gathered}\] 其中,\(\displaystyle z_{n}\)可以是四个球贝塞尔函数\(\displaystyle \left\{j_{n} ,y_{n} ,h_{n}^{( 1)} ,h_{n}^{( 2)}\right\}\)之中的任意一个。

对于平面波: \[E=\hat{e}_{x} E_{0} e^{ikr\cos \theta } =\hat{e}_{x} =E_{0} e^{ikr\cos \theta }(\sin \theta \cos \phi \hat{e}_{r} +\cos \theta \cos \phi \hat{e}_{\theta } -\sin \phi \hat{e}_{\phi })\] 由矢量球波函数的各分量特点,首先考虑使用\(\displaystyle \vec{N}_{emn} 、\vec{N}_{omn}\)组合出\(\displaystyle \hat{e}_{r}\)分量。

考虑\(\displaystyle \phi\)的变化规律,则\(\displaystyle m=1\) \[\begin{gathered} N_{e1n}^{( r)} =\frac{z_{n}( kr)}{kr}\cos \phi n( n+1) P_{n}^{1}(\cos \theta )\\ N_{o1n}^{( r)} =\frac{z_{n}( kr)}{kr}\sin \phi n( n+1) P_{n}^{1}(\cos \theta )\end{gathered}\] 考虑\(\displaystyle \phi\)的对称性,则在展开中只存在\(\displaystyle N_{e1n}^{( r)}\)项,\(\displaystyle A_{o1n} =0\)(同理可得\(\displaystyle B_{e1n} =0\))。

那么命题变为考虑\(\displaystyle A_{emn}\)(记为\(\displaystyle A_{n}\)),使得 \[E_{0} e^{ikr\cos \theta }\sin \theta =\sum _{n=1}^{\infty } A_{n}\frac{z_{n}( kr)}{kr} n( n+1) P_{n}^{1}(\cos \theta )\]\(\displaystyle z_{n} =j_{n}\),则命题变为: \[E_{0} e^{ikr\cos \theta }\sin \theta =\sum _{n=1}^{\infty } A_{n} n( n+1)\frac{j_{n}( kr)}{kr} P_{n}^{1}(\cos \theta )\] 由连带勒让德函数正交关系: \[\begin{gathered} \int _{-1}^{1} P_{l}^{m}( x) P_{k}^{m}( x)\mathrm{d} x=\frac{( l+m) !}{( l-m) !}\frac{2}{2l+1} \delta _{kl}\\ \int _{-1}^{1} P_{l}^{1}( x) P_{k}^{1}( x)\mathrm{d} x=\frac{2( l+1) l}{2l+1} \delta _{kl}\end{gathered}\] 对(1)式乘\(\displaystyle P_{l}^{1}(\cos \theta )\)并对\(\displaystyle \cos \theta\)求积分,可得: \[\begin{gathered} \int _{-1}^{1} E_{0} e^{ikr\cos \theta }\sin \theta P_{l}^{1}(\cos \theta )\mathrm{d}(\cos \theta ) =\sum _{n=1}^{\infty } A_{n} n( n+1)\frac{j_{n}( kr)}{kr}\int _{-1}^{1} P_{n}^{1}(\cos \theta ) P_{l}^{1}(\cos \theta )\mathrm{d}(\cos \theta ) \notag\\ =\sum _{n=1}^{\infty } A_{n} n( n+1)\frac{j_{n}( kr)}{kr}\frac{2( l+1) l}{2l+1} \delta _{ln}\\ =A_{l} l( l+1)\frac{j_{l}( kr)}{kr}\frac{2( l+1) l}{2l+1} \notag\end{gathered}\] 继续计算需要得到\(\displaystyle K_{n}( x) =\frac{j_{n}( x)}{x}\)的正交关系: \[\int _{0}^{\infty } K_{m}( x) K_{n}( x)\mathrm{d} x=-\frac{8\sin\left(\frac{1}{2} \pi (m-n)\right)}{(m-n-2)(m-n)(m-n+2)(m+n-1)(m+n+1)(m+n+3)}\] \(\displaystyle K_{n}( x)\)不是正交的,但有: \[\int _{0}^{\infty } K_{n}( x) K_{n}( x)\mathrm{d} x=\frac{\pi }{8n^{3} +12n^{2} -2n-3} =\frac{\pi }{8\left( n+\frac{3}{2}\right)\left( n-\frac{1}{2}\right)\left( n+\frac{1}{2}\right)}\] 这不影响我们的结果,(2)两边乘\(\displaystyle K_{l}( kr)\),对\(\displaystyle kr\)求积分: \[\begin{gathered} \int _{0}^{\infty }\int _{-1}^{1} E_{0} e^{ikr\cos \theta }\sin \theta P_{l}^{1}(\cos \theta )\frac{j_{l}( kr)}{kr}\mathrm{d}(\cos \theta )\mathrm{d}( kr) =\int _{0}^{\infty } A_{l} l( l+1)\frac{2( l+1) l}{2l+1} K_{l}^{2}( kr)\mathrm{d}( kr)\\ =A_{l} l( l+1)\frac{( l+1) l}{2l+1}\frac{\pi }{4\left( l+\frac{3}{2}\right)\left( l-\frac{1}{2}\right)\left( l+\frac{1}{2}\right)}\end{gathered}\] 可得: \[A_{l} =\frac{2l+1}{( l+1)^{2} l^{2}}\frac{4\left( l+\frac{3}{2}\right)\left( l-\frac{1}{2}\right)\left( l+\frac{1}{2}\right)}{\pi }\int _{0}^{\infty }\int _{-1}^{1} E_{0} e^{ikr\cos \theta }\sin \theta P_{l}^{1}(\cos \theta )\frac{j_{l}( kr)}{kr}\mathrm{d}(\cos \theta )\mathrm{d}( kr)\]

后半部分积分: \[\begin{gathered} \int _{0}^{\infty }\int _{-1}^{1} E_{0} e^{ikr\cos \theta }\sin \theta P_{l}^{1}(\cos \theta )\frac{j_{l}( kr)}{kr}\mathrm{d}(\cos \theta )\mathrm{d}( kr)\\ =E_{0}\int _{0}^{\infty }\frac{j_{l}( kr)}{kr}\left[\int _{-1}^{1} e^{ikr\cos \theta }\sin \theta P_{l}^{1}(\cos \theta )\mathrm{d}(\cos \theta )\right]\mathrm{d}( kr)\end{gathered}\]\(\displaystyle P_{l}^{1}(\cos \theta ) =-\left( 1-\cos^{2} \theta \right)^{1/2}\frac{\mathrm{d} P_{l}(\cos \theta )}{\mathrm{d}(\cos \theta )}\),\(\displaystyle \theta \in ( 0,\pi )\),可得: \[\begin{gathered} \int _{-1}^{1} e^{ikr\cos \theta }\sin \theta P_{l}^{1}(\cos \theta )\mathrm{d}(\cos \theta ) =\int _{-1}^{1} e^{ikr\cos \theta }\left(\sin^{2} \theta \right)\frac{\mathrm{d} P_{l}(\cos \theta )}{\mathrm{d}(\cos \theta )}\mathrm{d}(\cos \theta )\\ =-\int _{-1}^{1} e^{ikrx}\left( x^{2} -1\right)\mathrm{d} P_{l}( x)\\ =-\int _{-1}^{1} e^{ikrx} l[ xP_{l} (x)-P_{l-1} (x)]\mathrm{d} x\end{gathered}\] mathematica进行计算符合命题。

\(\displaystyle l\) 1 2 3 4 5
\(\displaystyle ( 4)\) \(\displaystyle \frac{4}{15} \pi\) \(\displaystyle -\frac{4i}{35} \pi\) \(\displaystyle \frac{8}{105} \pi\)$ $ \(\displaystyle \frac{40i}{693} \pi\) \(\displaystyle -\frac{20}{429} \pi\)

根据 \[\begin{gathered} j_{n}( x) =\frac{i^{-n}}{2}\int _{0}^{\pi } e^{ix\cos \theta } P_{n}(\cos \theta )\sin \theta \mathrm{d} \theta \\ =\frac{i^{-n}}{2}\int _{0}^{\pi } e^{ix\cos \theta } P_{n}(\cos \theta )\mathrm{d}\cos \theta \\ =\frac{i^{-n}}{2}\int _{1}^{-1} e^{ixy} P_{n}( y)\mathrm{d} y\end{gathered}\] \[\begin{gathered} -\int _{0}^{\infty }\frac{j_{l}( kr)}{kr}\left[\int _{-1}^{1} e^{ikrx} l[ xP_{l} (x)-P_{l-1} (x)]\mathrm{d} x\right]\mathrm{d}( kr)\\ =-\int _{0}^{\infty }\frac{j_{l}( y)}{y} l\left[\int _{-1}^{1} e^{ixy}[ xP_{l} (x)-P_{l-1} (x)]\mathrm{d} x\right]\mathrm{d} y\end{gathered}\] 根据\(\displaystyle xP_{l}( x) =\frac{1}{2l+1}[( l+1) P_{l+1}( x) +lP_{l-1}( x)]\),上式化为: \[\begin{gathered} =-\int _{0}^{\infty }\frac{j_{l}( y)}{y} l\left\{\int _{-1}^{1} e^{ixy}\left[\frac{1}{2l+1}[( l+1) P_{l+1}( x) +lP_{l-1}( x)] -P_{l-1} (x)\right]\mathrm{d} x\right\}\mathrm{d} y\\ =-\int _{0}^{\infty }\frac{j_{l}( y)}{y} l\left[\int _{-1}^{1}\frac{( l+1) e^{ixy}}{2l+1}[ P_{l+1} (x)-P_{l-1} (x)]\mathrm{d} x\right]\mathrm{d} y\\ =-\int _{0}^{\infty }\frac{j_{l}( y)}{y}\frac{l( l+1)}{2l+1}\left[\int _{-1}^{1} e^{ixy}[ P_{l+1} (x)-P_{l-1} (x)]\mathrm{d} x\right]\mathrm{d} y\\ =-\int _{0}^{\infty }\frac{j_{l}( y)}{y}\frac{l( l+1)}{2l+1}\frac{2}{i^{-n-1}}[ j_{l+1}( y) +j_{l-1}( y)]\mathrm{d} y\\ =-i^{l+1}\frac{l( l+1)}{\left( l+\frac{1}{2}\right)}\frac{\pi }{4\left( l+\frac{3}{2}\right)\left( l-\frac{1}{2}\right)}\\ =-i^{l+1}\frac{l( l+1) \pi }{4\left( l+\frac{3}{2}\right)\left( l-\frac{1}{2}\right)\left( l+\frac{1}{2}\right)}\end{gathered}\]

综上\[A_{e1n} =A_{n} =\frac{2n+1}{( n+1)^{2} n^{2}}\frac{4\left( n+\frac{3}{2}\right)\left( n-\frac{1}{2}\right)\left( n+\frac{1}{2}\right)}{\pi } *\left( i^{l+1}\frac{n( n+1) \pi }{4\left( n+\frac{3}{2}\right)\left( n-\frac{1}{2}\right)\left( n+\frac{1}{2}\right)}\right) =- E_{0} i^{n+1}\frac{2n+1}{n( n+1)}\] 此时, \[E=\sum _{n=1}^{\infty } -E_{0} i^{n+1}\frac{2n+1}{n( n+1)} N_{e1n}^{( 1)} +B_{o1n} M_{o1n}^{( 1)} E_{0} =E_{0} e^{ikr\cos \theta }(\sin \theta \cos \phi \hat{e}_{r} +\cos \theta \cos \phi \hat{e}_{\theta } -\sin \phi \hat{e}_{\phi })\] 接下来需要考虑\(\displaystyle \phi 、\theta\)方向,此时没有正交关系,将十分难以进行计算,于此补上正交关系证明\[\begin{gathered} \int _{0}^{2\pi }\int _{0}^{\pi } M_{em'n'} \cdotp M_{omn}\sin \theta \mathrm{d} \theta \mathrm{d} \phi \\ =\int _{0}^{2\pi }\int _{0}^{\pi }\left(\frac{-m}{\sin \theta }\sin( m'\phi ) P_{n'}^{m'}(\cos \theta ) z_{n'}( kr)\hat{e}_{\theta } -\cos( m'\phi )\frac{\mathrm{d} P_{n'}^{m'}(\cos \theta )}{\mathrm{d} \theta } z_{n'}( kr)\hat{e}_{\phi }\right) \cdotp \notag\\ \left(\frac{m}{\sin \theta }\cos( m\phi ) P_{n}^{m}(\cos \theta ) z_{n}( kr)\hat{e}_{\theta } -\sin( m\phi )\frac{\mathrm{d} P_{n}^{m}(\cos \theta )}{\mathrm{d} \theta } z_{n}( kr)\hat{e}_{\phi }\right)\sin \theta \mathrm{d} \theta \mathrm{d} \phi \notag\\ =\int _{0}^{2\pi }\int _{0}^{\pi }\left(\frac{-m}{\sin \theta }\sin( m'\phi ) P_{n'}^{m'}(\cos \theta ) z_{n'}( kr)\right)\left(\frac{m}{\sin \theta }\cos( m\phi ) P_{n}^{m}(\cos \theta ) z_{n}( kr)\right)\hat{e}_{\theta } \notag\\ +\left(\cos( m'\phi )\frac{\mathrm{d} P_{n'}^{m'}(\cos \theta )}{\mathrm{d} \theta } z_{n'}( kr)\right)\left(\sin( m\phi )\frac{\mathrm{d} P_{n}^{m}(\cos \theta )}{\mathrm{d} \theta } z_{n}( kr)\right)\hat{e}_{\phi }\sin \theta \mathrm{d} \theta \mathrm{d} \phi \notag\\ =\int _{0}^{\pi }\left[\int _{0}^{2\pi }\cos( m\phi )\sin( m'\phi )\mathrm{d} \phi \right]\left(\frac{-m}{\sin \theta } P_{n'}^{m'}(\cos \theta ) z_{n'}( kr)\right)\left(\frac{m}{\sin \theta } P_{n}^{m}(\cos \theta ) z_{n}( kr)\right)\hat{e}_{\theta } \notag\\ +\left[\int _{0}^{2\pi }\cos( m'\phi )\sin( m\phi )\mathrm{d} \phi \right]\left(\frac{\mathrm{d} P_{n'}^{m'}(\cos \theta )}{\mathrm{d} \theta } z_{n'}( kr)\right)\left(\frac{\mathrm{d} P_{n}^{m}(\cos \theta )}{\mathrm{d} \theta } z_{n}( kr)\right)\hat{e}_{\phi }\sin \theta \mathrm{d} \theta \notag\end{gathered}\] 对于任意\(\displaystyle m,m'\)时, \[\int _{0}^{2\pi }\cos( m'\phi )\sin( m\phi )\mathrm{d} \phi =0\] 故(4)式对任意\(\displaystyle m,m',n,n'\)都等于零。

同理,对任意\(\displaystyle m,m',n,n'\)成立(提取\(\displaystyle \phi\)方向即得) \[\begin{gathered} \int _{0}^{\infty }\int _{0}^{2\pi }\int _{0}^{\pi } M_{em'n'} \cdotp N_{emn}\sin \theta \mathrm{d} \theta \mathrm{d} \phi \mathrm{d} r=0\\ \int _{0}^{\infty }\int _{0}^{2\pi }\int _{0}^{\pi } N_{om'n'} \cdotp M_{omn}\sin \theta \mathrm{d} \theta \mathrm{d} \phi \mathrm{d} r=0\\ \int _{0}^{\infty }\int _{0}^{2\pi }\int _{0}^{\pi } N_{em'n'} \cdotp N_{omn}\sin \theta \mathrm{d} \theta \mathrm{d} \phi \mathrm{d} r=0\end{gathered}\] \[\begin{gathered} \int _{0}^{\infty }\int _{0}^{2\pi }\int _{0}^{\pi } M_{om'n'} \cdotp N_{emn}\sin \theta \mathrm{d} \theta \mathrm{d} \phi \mathrm{d} r\\ =\int _{0}^{\infty }\int _{0}^{2\pi }\int _{0}^{\pi }\{\left(\frac{m'}{\sin \theta }\cos( m'\phi ) P_{n'}^{m'}(\cos \theta ) z_{n'}( kr)\right)\left(\cos( m\phi )\frac{\mathrm{d} P_{n}^{m}(\cos \theta )}{\mathrm{d} \theta }\frac{1}{kr}\frac{\mathrm{d}}{\mathrm{d}( kr)}[( kr) z_{n}( kr)]\right) \notag\\ +\left( -\sin( m\phi )\frac{\mathrm{d} P_{n}^{m}(\cos \theta )}{\mathrm{d} \theta } z_{n}( kr)\right)\left( -m\sin( m\phi )\frac{P_{n}^{m}(\cos \theta )}{\sin \theta }\frac{1}{kr}\frac{\mathrm{d}}{\mathrm{d}( kr)}[( kr) z_{n}( kr)]\right)\}\sin \theta \mathrm{d} \theta \mathrm{d} \phi \mathrm{d} r \notag\end{gathered}\]\(\displaystyle m'\neq m\)时,上式等于零,当\(\displaystyle m=m'\)时,上式化为: \[\begin{gathered} =m\pi \int _{0}^{\infty }\int _{0}^{\pi }\{\left(\frac{1}{\sin \theta } P_{n‘}^{m}(\cos \theta ) z_{n}( kr)\right)\left(\frac{\mathrm{d} P_{n}^{m}(\cos \theta )}{\mathrm{d} \theta }\frac{1}{kr}\frac{\mathrm{d}}{\mathrm{d}( kr)}[( kr) z_{n}( kr)]\right)\\ +\left(\sin( m\phi )\frac{\mathrm{d} P_{n'}^{m}(\cos \theta )}{\mathrm{d} \theta } z_{n'}( kr)\right)\left(\frac{P_{n}^{m}(\cos \theta )}{\sin \theta }\frac{1}{kr}\frac{\mathrm{d}}{\mathrm{d}( kr)}[( kr) z_{n}( kr)]\right)\}\sin \theta \mathrm{d} \theta \mathrm{d} r\\ =m\pi \int _{0}^{\infty }\int _{0}^{\pi }\{\left(\frac{1}{\sin \theta } P_{n‘}^{m}(\cos \theta ) z_{n}( kr)\right)\left(\frac{\mathrm{d} P_{n}^{m}(\cos \theta )}{\mathrm{d} \theta }\frac{1}{kr}\frac{\mathrm{d}}{\mathrm{d}( kr)}[( kr) z_{n}( kr)]\right)\\ +\left(\sin( m\phi )\frac{\mathrm{d} P_{n'}^{m}(\cos \theta )}{\mathrm{d} \theta } z_{n'}( kr)\right)\left(\frac{P_{n}^{m}(\cos \theta )}{\sin \theta }\frac{1}{kr}\frac{\mathrm{d}}{\mathrm{d}( kr)}[( kr) z_{n}( kr)]\right)\}\sin \theta \mathrm{d} \theta \mathrm{d} r\end{gathered}\] 由于 \[\begin{gathered} \int _{0}^{\pi }\frac{1}{\sin \theta } P_{n‘}^{m}(\cos \theta )\frac{\mathrm{d} P_{n}^{m}(\cos \theta )}{\mathrm{d} \theta }\sin \theta \mathrm{d} \theta \\ =\int _{0}^{\pi } P_{n‘}^{m}(\cos \theta )\frac{\mathrm{d} P_{n}^{m}(\cos \theta )}{\mathrm{d} \theta }\mathrm{d} \theta \\ =\int _{-1}^{1} P_{n'}^{m}( x)\frac{\mathrm{d} P_{n}^{m}( x)}{\mathrm{d} x}\mathrm{d} x\\ =\int _{-1}^{1} P_{n'}^{m}( x)\mathrm{d}\left[ P_{n}^{m}( x)\right]\\ =P_{n'}^{m}( x) P_{n}^{m}( x) |_{-1}^{1} -\int _{-1}^{1} P_{n}^{m}( x)\mathrm{d}\left[ P_{n'}^{m}( x)\right]\\ =-\int _{-1}^{1} P_{n}^{m}( x)\mathrm{d}\left[ P_{n'}^{m}( x)\right]\end{gathered}\] 故(5)式恒等于0。

所以,对任意\(\displaystyle m,m',n,n'\)成立 \[\int _{0}^{\infty }\int _{0}^{2\pi }\int _{0}^{\pi } M_{om'n'} \cdotp N_{emn}\sin \theta \mathrm{d} \theta \mathrm{d} \phi \mathrm{d} r=0\] 同理可得: \[\int _{0}^{\infty }\int _{0}^{2\pi }\int _{0}^{\pi } N_{om'n'} \cdotp M_{emn}\sin \theta \mathrm{d} \theta \mathrm{d} \phi \mathrm{d} r=0\] \[\begin{gathered} E=\sum _{n=1}^{\infty }\left( -E_{0} i^{n+1}\frac{2n+1}{n( n+1)} N_{e1n}^{( 1)} +B_{o1n} M_{o1n}^{( 1)} E_{0}\right)\\ =E_{0} e^{ikr\cos \theta }(\sin \theta \cos \phi \hat{e}_{r} +\cos \theta \cos \phi \hat{e}_{\theta } -\sin \phi \hat{e}_{\phi })\end{gathered}\] \[\vec{M}_{o1n} =\frac{1}{\sin \theta }\cos( \phi ) P_{n}^{1}(\cos \theta ) z_{n}( kr)\hat{e}_{\theta } -\sin( \phi )\frac{\mathrm{d} P_{n}^{1}(\cos \theta )}{\mathrm{d} \theta } z_{n}( kr)\hat{e}_{\phi }\] \[\begin{gathered} \int _{0}^{2\pi }\int _{0}^{\pi } M_{o1n} \cdotp M_{o1n}\sin \theta \mathrm{d} \theta \mathrm{d} \phi \notag\\ =\int _{0}^{2\pi }\int _{0}^{\pi }\left[\left(\frac{1}{\sin \theta }\cos( \phi ) P_{n}^{1}(\cos \theta ) z_{n}( kr)\right)^{2} +\left(\sin( \phi )\frac{\mathrm{d} P_{n}^{1}(\cos \theta )}{\mathrm{d} \theta } z_{n}( kr)\right)^{2}\right]\sin \theta \mathrm{d} \theta \mathrm{d} \phi \notag\\ =\pi ( z_{n}( kr))^{2}\int _{0}^{\pi }\left[\left(\frac{1}{\sin \theta } P_{n}^{1}(\cos \theta )\right)^{2} +\left(\frac{\mathrm{d} P_{n}^{1}(\cos \theta )}{\mathrm{d} \theta }\right)^{2}\right]\sin \theta \mathrm{d} \theta \notag\\ =\pi ( z_{n}( kr))^{2}\int _{0}^{\pi } -\left[\left(\frac{1}{\sin \theta } P_{n}^{1}(\cos \theta )\right)^{2} +\left(\frac{\mathrm{d} P_{n}^{1}(\cos \theta )}{\mathrm{d}\cos \theta }\frac{\mathrm{d}\cos \theta }{\mathrm{d} \theta }\right)^{2}\right]\mathrm{d}(\cos \theta ) \notag\\ =\pi ( z_{n}( kr))^{2}\int _{-1}^{1}\left[\left(\frac{\mathrm{d} P_{n}( x)}{\mathrm{d} x}\right)^{2} +\left(\frac{\mathrm{d} P_{n}^{1}( x)}{\mathrm{d} x}\right)^{2}\left( 1-x^{2}\right)\right]\mathrm{d} x\end{gathered}\]

(6)式前半部分结果为: \[\begin{gathered} \int _{-1}^{1}\left(\frac{\mathrm{d} P_{n}( x)}{\mathrm{d} x}\right)^{2}\mathrm{d} x\\ =\int _{-1}^{1}\frac{\left( P_{n}^{1}( x)\right)^{2}}{1-x^{2}}\mathrm{d} x\\ =n( n+1)\end{gathered}\] (6)式后半部分变为: \[\begin{gathered} \int _{-1}^{1}\left(\frac{\mathrm{d} P_{n}^{1}( x)}{\mathrm{d} x}\right)^{2}\left( 1-x^{2}\right)\mathrm{d} x \notag\\ =\frac{1}{( 2n+1)^{2}}\int _{-1}^{1}\frac{\left[( n+1)^{2} P_{n-1}^{1}( x) -n^{2} P_{n+1}^{1}( x)\right]^{2}}{1-x^{2}}\mathrm{d} x \notag\\ =\frac{-2( n+1)^{2} n^{2}}{( 2n+1)^{2}}\int _{-1}^{1}\frac{P_{n+1}^{1}( x) P_{n-1}^{1}( x)}{1-x^{2}}\mathrm{d} x+\frac{( n+1)^{4}( n-1) n}{( 2n+1)^{2}} +\frac{n^{4}( n+2)( n+1)}{( 2n+1)^{2}}\end{gathered}\] 由于 \[\begin{gathered} \int _{-1}^{1}\frac{P_{n+1}^{1}( x) P_{n-1}^{1}( x)}{1-x^{2}}\mathrm{d} x \notag\\ =\int _{-1}^{1}\left(\frac{\mathrm{d} P_{n+1}( x)}{\mathrm{d} x}\right)\left(\frac{\mathrm{d} P_{n-1}( x)}{\mathrm{d} x}\right) \ \mathrm{d} x \notag\\ =\int _{-1}^{1}\frac{\mathrm{d} P_{n-1}( x)}{\mathrm{d} x}\left[( 2n+1) P_{n}( x) +\frac{\mathrm{d} P_{n-1}( x)}{\mathrm{d} x}\right]\mathrm{d} x \notag\\ =( n-1) n+( 2n+1)\int _{-1}^{1}\frac{\mathrm{d} P_{n-1}( x)}{\mathrm{d} x} P_{n}( x)\mathrm{d} x\end{gathered}\] 由于 \[\begin{gathered} \int _{-1}^{1}\frac{\mathrm{d} P_{n-1}( x)}{\mathrm{d} x} P_{n}( x)\mathrm{d} x \notag\\ =\int _{-1}^{1}\frac{n[ P_{n-1}( x) -xP_{n}( x)]}{1-x^{2}} P_{n}( x)\mathrm{d} x \notag\\ =\int _{-1}^{1}\frac{n\left[ P_{n-1}( x) -\frac{1}{2n+1}[( n+1) P_{n+1}( x) +nP_{n-1}( x)]\right]}{1-x^{2}} P_{n}( x)\mathrm{d} x \notag\\ =\int _{-1}^{1}\frac{n( n+1)[ P_{n-1}( x) -P_{n+1}( x)]}{( 2n+1)\left( 1-x^{2}\right)} P_{n}( x)\mathrm{d} x \notag\\ =\frac{n( n+1)}{2n+1}\int _{-1}^{1}\left[ P_{n-1}^{1}( x) -P_{n+1}^{1}( x)\right] P_{n}^{1}( x)\mathrm{d} x \notag\\ =0\end{gathered}\]\[( 8) =n( n-1)\]\[( 7) =\frac{n( n+1)\left( 2n^{2} -1\right)}{2n+1}\] (6)式化为: \[\begin{gathered} =\pi ( z_{n}( kr))^{2} n( n+1)\left[\frac{2n^{2} +2n}{2n+1}\right]\\ =\pi ( z_{n}( kr))^{2}\frac{2n^{2}( n+1)^{2}}{2n+1}\end{gathered}\] \[\vec{M}_{o1n} =\frac{1}{\sin \theta }\cos( \phi ) P_{n}^{1}(\cos \theta ) z_{n}( kr)\hat{e}_{\theta } -\sin( \phi )\frac{\mathrm{d} P_{n}^{1}(\cos \theta )}{\mathrm{d} \theta } z_{n}( kr)\hat{e}_{\phi }\] \[\begin{gathered} E=\hat{e}_{x} E_{0} e^{ikr\cos \theta } =\hat{e}_{x} =E_{0} e^{ikr\cos \theta }(\sin \theta \cos \phi \hat{e}_{r} +\cos \theta \cos \phi \hat{e}_{\theta } -\sin \phi \hat{e}_{\phi })\\ \int _{0}^{2\pi }\int _{0}^{\pi } E\cdotp M_{o1n}\sin \theta \mathrm{d} \theta \mathrm{d} \phi \\ =E_{0}\int _{0}^{2\pi }\int _{0}^{\pi }\{\left[\frac{1}{\sin \theta }\cos( \phi ) P_{n}^{1}(\cos \theta ) z_{n}( kr)\right]\left[ E_{0} e^{ikr\cos \theta }\cos \theta \cos \phi \right] +\\ \left[\sin( \phi )\frac{\mathrm{d} P_{n}^{1}(\cos \theta )}{\mathrm{d} \theta } z_{n}( kr)\right]\left[ E_{0} e^{ikr\cos \theta }\sin \phi \right]\}\sin \theta \mathrm{d} \theta \mathrm{d} \phi \\ =\pi z_{n}( kr) E_{0}\int _{0}^{\pi } -\left[ E_{0} e^{ikr\cos \theta }\right]\left\{\frac{1}{\sin \theta } P_{n}^{1}(\cos \theta )\cos \theta -\frac{\mathrm{d} P_{n}^{1}(\cos \theta )}{\mathrm{d}(\cos \theta )}\sin \theta \right\}\mathrm{d}(\cos \theta )\\ =\pi z_{n}( kr) E_{0}\int _{-1}^{1}\left[ E_{0} e^{ikrx}\right]\left\{\frac{1}{\sqrt{1-x^{2}}} xP_{n}^{1}( x) -\frac{\mathrm{d} P_{n}^{1}( x)}{\mathrm{d} x}\sqrt{1-x^{2}}\right\}\mathrm{d} x\\ =\pi z_{n}( kr) E_{0}\int _{-1}^{1} E_{0} e^{ikrx}\left\{x\frac{\mathrm{d} P_{n}( x)}{\mathrm{d} x} -\frac{1}{2n+1}\left[( n+1)^{2}\frac{\mathrm{d} P_{n-1}( x)}{\mathrm{d} x} -n^{2}\frac{\mathrm{d} P_{n+1}( x)}{\mathrm{d} x}\right]\right\}\mathrm{d} x\\ =\pi z_{n}( kr) E_{0}\int _{-1}^{1} E_{0} e^{ikrx}\{\left[\frac{\mathrm{d} P_{n+1}( x)}{\mathrm{d} x} -( n+1) P_{n}( x)\right]\\ -\frac{1}{2n+1}\left[( n+1)^{2}\frac{\mathrm{d} P_{n-1}( x)}{\mathrm{d} x} -n^{2}\frac{\mathrm{d} P_{n+1}( x)}{\mathrm{d} x}\right]\}\mathrm{d} x\\ =\pi z_{n}( kr) E_{0}\int _{-1}^{1} E_{0} e^{ikrx}\left[\frac{( n+1)^{2}}{2n+1}\frac{\mathrm{d} P_{n+1}( x)}{\mathrm{d} x} -( n+1) P_{n}( x) -\frac{( n+1)^{2}}{2n+1}\frac{\mathrm{d} P_{n-1}( x)}{\mathrm{d} x}\right]\mathrm{d} x\\ =\pi z_{n}( kr) E_{0}\int _{-1}^{1} E_{0} e^{ikrx}\left[\frac{( n+1)^{2}}{2n+1}\frac{\mathrm{d} P_{n+1}( x)}{\mathrm{d} x} -( n+1) P_{n}( x) -\frac{( n+1)^{2}}{2n+1}\frac{\mathrm{d} P_{n-1}( x)}{\mathrm{d} x}\right]\mathrm{d} x\\ =\pi z_{n}( kr) E_{0}\int _{-1}^{1} E_{0} e^{ikrx}\left[ -( n+1) P_{n}( x) +n^{2} P_{n}( x) +( 2n+1) P_{n}( x)\right]\mathrm{d} x\\ =\pi z_{n}( kr) E_{0}\int _{-1}^{1} E_{0} e^{ikrx}[ n( n+1) P_{n}( x)]\mathrm{d} x\\ =2\pi ( j_{n}( kr))^{2} i^{n} n( n+1)\end{gathered}\] 所以可以得到: \[B_{o1n} =\frac{2\pi ( j_{n}( kr))^{2} i^{n} n( n+1)}{\pi ( j_{n}( kr))^{2}\frac{2n^{2}( n+1)^{2}}{2n+1}} =i^{n}\frac{2n+1}{n( n+1)}\] (4)和(10)为所证展开系数。

\[\vec{E} =\sum _{n=1}^{\infty } i^{n}\frac{2n+1}{n( n+1)}\left( M_{o1n}^{( 1)} -iN_{e1n}^{( 2)}\right)\]