Kinda reasonable form of 1D in 3D Ewald.
Former-commit-id: eb5c54026cff0c2889d32cd2c6288e7be2cbb3e9
This commit is contained in:
Marek Nečada
parent
cae5cee97d
commit
299bb6fc03
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@ -162,6 +162,25 @@
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\end_inset
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\end_layout
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\begin_layout Section
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General formula
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\end_layout
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\begin_layout Standard
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We need to find the expansion coefficient
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\end_layout
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\begin_layout Standard
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\begin_inset Formula
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\begin{equation}
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\tau_{l'}^{m'}\left(\vect s,\vect k\right)=\frac{i}{\kappa j_{l'}\left(\kappa\left|\vect r\right|\right)}\int\ud\Omega_{\vect r}\,G_{\Lambda}^{(\kappa)}\left(\vect s+\vect r,\vect k\right)\ushD{l'}{m'}\left(\uvec r\right).\label{eq:tau extraction formula}
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\end{equation}
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\end_inset
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\end_layout
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\begin_layout Standard
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@ -354,8 +373,153 @@ e^{-\left|\vect s_{\bot}+\vect r_{\bot}\right|^{2}\kappa^{2}\gamma_{\vect K}^{2}
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hence
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\begin_inset Formula
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\begin{align*}
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\int\ud\Omega_{\vect r}\,G_{\Lambda}^{(1;\kappa)}\left(\vect s+\vect r\right)\ushD{l'}{m'}\left(\uvec r\right) & =-\frac{1}{2\pi\mathcal{A}}\int\ud\Omega_{\vect r}\,\ushD{l'}{m'}\left(\uvec r\right)\sum_{\vect K\in\Lambda^{*}}e^{i\vect K\cdot\vect s}\sum_{lm}4\pi i^{l}j_{l}\left(\left|\vect K\right|\left|\vect r\right|\right)\ushD lm\left(\uvec r\right)\ush lm\left(\uvec K\right)\sum_{n=0}^{\infty}\frac{1}{n!}\left(-\frac{\left(\left|\vect r_{\bot}\right|^{2}+2\vect r_{\bot}\cdot\vect s_{\bot}\right)\kappa^{2}\gamma_{\vect K}^{2}}{4}\right)^{n}\underbrace{\int_{\kappa^{2}\gamma_{\vect K}^{2}/4\eta^{2}}^{\infty\exp\left(i\pi/2\right)}e^{-\tau}e^{-\left|\vect s_{\bot}\right|^{2}\kappa^{2}\gamma_{\vect K}^{2}/4\tau}\tau^{-1-n}\ud\tau}_{\Delta_{n+1/2}}\\
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& =-\frac{1}{2\pi\mathcal{A}}\sum_{\vect K\in\Lambda^{*}}e^{i\vect K\cdot\vect s}\sum_{lm}4\pi i^{l}j_{l}\left(\left|\vect K\right|\left|\vect r\right|\right)\ush lm\left(\uvec K\right)\sum_{n=0}^{\infty}\frac{\Delta_{n+1/2}}{n!}\int\ud\Omega_{\vect r}\,\ushD{l'}{m'}\left(\uvec r\right)\ushD lm\left(\uvec r\right)\left(-\frac{\left(\left|\vect r_{\bot}\right|^{2}+2\vect r_{\bot}\cdot\vect s_{\bot}\right)\kappa^{2}\gamma_{\vect K}^{2}}{4}\right)^{n}\\
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& =-\frac{1}{2\pi\mathcal{A}}\sum_{\vect K\in\Lambda^{*}}e^{i\vect K\cdot\vect s}\sum_{lm}4\pi i^{l}j_{l}\left(\left|\vect K\right|\left|\vect r\right|\right)\ush lm\left(\uvec K\right)\sum_{n=0}^{\infty}\frac{\left(-1\right)^{n}}{n!}\Delta_{n+1/2}\left(\frac{\kappa\gamma_{\vect K}}{2}\right)^{2n}\sum_{k=0}^{n}\int\ud\Omega_{\vect r}\,\ushD{l'}{m'}\left(\uvec r\right)\ushD lm\left(\uvec r\right)\left|\vect r_{\bot}\right|^{2(n-k)}\left(2\vect r_{\bot}\cdot\vect s_{\bot}\right)^{k}
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\end{align*}
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\end_inset
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If we label
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\begin_inset Formula $\left|\vect r_{\bot}\right|\left|\vect s_{\bot}\right|\cos\varphi\equiv\vect r_{\bot}\cdot\vect s_{\bot}$
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\end_inset
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, we have
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\begin_inset Formula
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\[
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\int\ud\Omega_{\vect r}\,G_{\Lambda}^{(1;\kappa)}\left(\vect s+\vect r\right)\ushD{l'}{m'}\left(\uvec r\right)=-\frac{1}{2\pi\mathcal{A}}\int\ud\Omega_{\vect r}\,\ushD{l'}{m'}\left(\uvec r\right)\frac{1}{2\pi\mathcal{A}}\sum_{\vect K\in\Lambda^{*}}e^{i\vect K\cdot\vect s}\sum_{lm}4\pi i^{l}j_{l}\left(\left|\vect K\right|\left|\vect r\right|\right)\ushD lm\left(\uvec r\right)\ush lm\left(\uvec K\right)\sum_{n=0}^{\infty}\frac{1}{n!}\left(-\frac{\left(\left|\vect r_{\bot}\right|^{2}+2\vect r_{\bot}\cdot\vect s_{\bot}\right)\kappa^{2}\gamma_{\vect K}^{2}}{4}\right)^{n}\underbrace{\int_{\kappa^{2}\gamma_{\vect K}^{2}/4\eta^{2}}^{\infty\exp\left(i\pi/2\right)}e^{-\tau}e^{-\left|\vect s_{\bot}\right|^{2}\kappa^{2}\gamma_{\vect K}^{2}/4\tau}\tau^{-1-n}\ud\tau}_{\Delta_{n+1/2}}
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\int\ud\Omega_{\vect r}\,G_{\Lambda}^{(1;\kappa)}\left(\vect s+\vect r\right)\ushD{l'}{m'}\left(\uvec r\right)=-\frac{1}{2\pi\mathcal{A}}\sum_{\vect K\in\Lambda^{*}}e^{i\vect K\cdot\vect s}\sum_{lm}4\pi i^{l}j_{l}\left(\left|\vect K\right|\left|\vect r\right|\right)\ush lm\left(\uvec K\right)\sum_{n=0}^{\infty}\frac{\left(-1\right)^{n}}{n!}\Delta_{n+1/2}\left(\frac{\kappa\gamma_{\vect K}}{2}\right)^{2n}\sum_{k=0}^{n}\left(2\left|\vect s_{\bot}\right|\right)^{k}\int\ud\Omega_{\vect r}\,\ushD{l'}{m'}\left(\uvec r\right)\ushD lm\left(\uvec r\right)\left|\vect r_{\bot}\right|^{2n-k}\left(\cos\varphi\right)^{k}
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\]
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\end_inset
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and if we label
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\begin_inset Formula $\left|\vect r\right|\sin\theta\equiv\left|\vect r_{\bot}\right|$
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\end_inset
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\begin_inset Formula
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\[
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\int\ud\Omega_{\vect r}\,G_{\Lambda}^{(1;\kappa)}\left(\vect s+\vect r\right)\ushD{l'}{m'}\left(\uvec r\right)=-\frac{1}{2\pi\mathcal{A}}\sum_{\vect K\in\Lambda^{*}}e^{i\vect K\cdot\vect s}\sum_{lm}4\pi i^{l}j_{l}\left(\left|\vect K\right|\left|\vect r\right|\right)\ush lm\left(\uvec K\right)\sum_{n=0}^{\infty}\frac{\left(-1\right)^{n}}{n!}\Delta_{n+1/2}\left(\frac{\kappa\gamma_{\vect K}}{2}\right)^{2n}\sum_{k=0}^{n}\left|\vect r\right|^{2n-k}\left(2\left|\vect s_{\bot}\right|\right)^{k}\int\ud\Omega_{\vect r}\,\ushD{l'}{m'}\left(\uvec r\right)\ushD lm\left(\uvec r\right)\left(\sin\theta\right)^{2n-k}\left(\cos\varphi\right)^{k}
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\]
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\end_inset
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Now let's put the RHS into
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\begin_inset CommandInset ref
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LatexCommand eqref
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reference "eq:tau extraction formula"
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plural "false"
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caps "false"
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noprefix "false"
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\end_inset
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and try eliminating some sum by taking the limit
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\begin_inset Formula $\left|\vect r\right|\to0$
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\end_inset
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.
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We have
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\begin_inset Formula $j_{l}\left(\left|\vect K\right|\left|\vect r\right|\right)\sim\left(\left|\vect K\right|\left|\vect r\right|\right)^{l}/\left(2l+1\right)!!$
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\end_inset
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; the denominator from
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\begin_inset CommandInset ref
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LatexCommand eqref
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reference "eq:tau extraction formula"
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plural "false"
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caps "false"
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noprefix "false"
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\end_inset
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behaves like
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\begin_inset Formula $j_{l'}\left(\kappa\left|\vect r\right|\right)\sim\left(\kappa\left|\vect r\right|\right)^{l'}/\left(2l'+1\right)!!.$
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\end_inset
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The leading terms are hence those with
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\begin_inset Formula $\left|\vect r\right|^{l-l'+2n-k}$
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\end_inset
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.
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So
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\begin_inset Formula
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\[
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\tau_{l'}^{m'}\left(\vect s,\vect k\right)=\frac{-i}{2\pi\mathcal{A}\kappa^{1+l'}}\left(2l'+1\right)!!\sum_{\vect K\in\Lambda^{*}}e^{i\vect K\cdot\vect s}\sum_{lm}4\pi i^{l}\frac{\left|\vect K\right|^{l}}{\left(2l+1\right)!!}\ush lm\left(\uvec K\right)\sum_{n=0}^{\infty}\frac{\left(-1\right)^{n}}{n!}\Delta_{n+1/2}\left(\frac{\kappa\gamma_{\vect K}}{2}\right)^{2n}\sum_{k=0}^{n}\delta_{l'-l,2n-k}\left(2\left|\vect s_{\bot}\right|\right)^{k}\int\ud\Omega_{\vect r}\,\ushD{l'}{m'}\left(\uvec r\right)\ushD lm\left(\uvec r\right)\left(\sin\theta\right)^{l'-l}\left(\cos\varphi\right)^{k}.
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\]
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\end_inset
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Let's now focus on rearranging the sums; we have
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\begin_inset Formula
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\[
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S(l')\equiv\sum_{l=0}^{\infty}\sum_{n=0}^{\infty}\sum_{k=0}^{n}\delta_{l'-l,2n-k}f(l',l,n,k)=\sum_{l=0}^{\infty}\sum_{n=0}^{\infty}\sum_{k=0}^{n}\delta_{l'-l,2n-k}f(l',l,n,2n-l'+l)
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\]
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\end_inset
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We have
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\begin_inset Formula $0\le k\le n$
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\end_inset
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, hence
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\begin_inset Formula $0\le2n-l'+l\le n$
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\end_inset
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, hence
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\begin_inset Formula $-2n\le-l'+l\le-n$
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\end_inset
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, hence also
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\begin_inset Formula $l'-2n\le l\le l'-n$
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\end_inset
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, which gives the opportunity to swap the
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\begin_inset Formula $l,n$
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\end_inset
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sums and the
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\begin_inset Formula $l$
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\end_inset
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-sum becomes finite; so also consuming
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\begin_inset Formula $\sum_{k=0}^{n}\delta_{l'-l,2n-k}$
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\end_inset
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we get
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\begin_inset Formula
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\[
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S(l')=\sum_{n=0}^{\infty}\sum_{l=\max(0,l'-2n)}^{l'-n}f(l',l,n,2n-l'+l).
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\]
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\end_inset
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Finally, we see that the interval of valid
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\begin_inset Formula $l$
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\end_inset
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becomes empty when
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\begin_inset Formula $l'-n<0$
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\end_inset
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, i.e.
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\begin_inset Formula $n>l'$
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\end_inset
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; so we get a finite sum
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\begin_inset Formula
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\[
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S(l')=\sum_{n=0}^{l'}\sum_{l=\max(0,l'-2n)}^{l'-n}f(l',l,n,2n-l'+l).
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\]
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\end_inset
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Applying rearrangement,
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\begin_inset Formula
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\[
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\tau_{l'}^{m'}\left(\vect s,\vect k\right)=\frac{-i}{2\pi\mathcal{A}\kappa^{1+l'}}\left(2l'+1\right)!!\sum_{\vect K\in\Lambda^{*}}e^{i\vect K\cdot\vect s}\sum_{n=0}^{l'}\frac{\left(-1\right)^{n}}{n!}\Delta_{n+1/2}\left(\frac{\kappa\gamma_{\vect K}}{2}\right)^{2n}\sum_{l=\max\left(0,l'-2n\right)}^{l'-n}4\pi i^{l}\left(2\left|\vect s_{\bot}\right|\right)^{2n-l'+l}\frac{\left|\vect K\right|^{l}}{\left(2l+1\right)!!}\sum_{m=-l}^{l}\ush lm\left(\uvec K\right)\int\ud\Omega_{\vect r}\,\ushD{l'}{m'}\left(\uvec r\right)\ushD lm\left(\uvec r\right)\left(\sin\theta\right)^{l'-l}\left(\cos\varphi\right)^{2n-l'+l}.
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\]
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\end_inset
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@ -363,6 +527,107 @@ hence
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\end_layout
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\begin_layout Section
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Z-aligned lattice
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\end_layout
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\begin_layout Standard
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Now we set some conventions: let the lattice lie on the
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\begin_inset Formula $z$
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\end_inset
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axis, so that
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\begin_inset Formula $\vect s_{\bot},\vect r_{\bot}$
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\end_inset
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lie in the
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\begin_inset Formula $xy$
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\end_inset
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-plane.
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\begin_inset Note Note
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status open
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\begin_layout Plain Layout
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(TODO check the meaning of
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\begin_inset Formula $\vect k$
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\end_inset
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and possible additional phase factor.)
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\end_layout
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\end_inset
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If we write
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\begin_inset Formula $\vect s_{\bot}=\uvec x\left|\vect s_{\bot}\right|\cos\Phi+\uvec y\left|\vect s_{\bot}\right|\sin\Phi$
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\end_inset
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,
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\begin_inset Formula $\vect r_{\bot}=\uvec x\left|\vect r_{\bot}\right|\cos\phi+\uvec y\left|\vect r_{\bot}\right|\sin\phi=\uvec x\left|\vect r\right|\sin\theta\cos\phi+\uvec y\left|\vect r\right|\sin\theta\sin\phi$
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\end_inset
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, we have
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\begin_inset Formula $\varphi=\phi-\Phi$
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\end_inset
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.
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Also, in this convention
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\begin_inset Formula $\ush lm\left(\uvec K\right)=0$
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\end_inset
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for
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\begin_inset Formula $m\ne0$
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\end_inset
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, so
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\begin_inset Formula
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\[
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\tau_{l'}^{m'}\left(\vect s,\vect k\right)=\frac{-i}{2\pi\mathcal{A}\kappa^{1+l'}}\left(2l'+1\right)!!\sum_{\vect K\in\Lambda^{*}}e^{i\vect K\cdot\vect s}\sum_{n=0}^{l'}\frac{\left(-1\right)^{n}}{n!}\Delta_{n+1/2}\left(\frac{\kappa\gamma_{\vect K}}{2}\right)^{2n}\sum_{l=\max\left(0,l'-2n\right)}^{l'-n}4\pi i^{l}\left(2\left|\vect s_{\bot}\right|\right)^{2n-l'+l}\frac{\left|\vect K\right|^{l}}{\left(2l+1\right)!!}\ush l0\left(\uvec K\right)\underbrace{\int\ud\Omega_{\vect r}\,\ushD{l'}{m'}\left(\uvec r\right)\ushD l0\left(\uvec r\right)\left(\sin\theta\right)^{l'-l}\left(\cos\varphi\right)^{2n-l'+l}}_{\equiv A_{l',l,n,m'}}.
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\]
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\end_inset
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Let's also fix the (dual) spherical harmonics for now,
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\begin_inset Formula
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\[
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\ushD lm\left(\uvec r\right)=\lambda'_{lm}e^{-im\phi}P_{l}^{-m}\left(\cos\theta\right);
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\]
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\end_inset
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the angular integral then becomes (we also use
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\begin_inset Formula $e^{-im'\phi}=e^{im'\Phi}e^{-im'\varphi}$
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\end_inset
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)
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\begin_inset Formula
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\begin{align*}
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A_{l',l,n,m'} & \equiv\int\ud\Omega_{\vect r}\,\ushD{l'}{m'}\left(\uvec r\right)\ushD l0\left(\uvec r\right)\left(\sin\theta\right)^{l'-l}\left(\cos\varphi\right)^{2n-l'+l}\\
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& =\lambda'_{l'm'}\lambda'_{l0}e^{im'\Phi}\int_{0}^{\pi}\ud\theta\,\sin\theta P_{l'}^{-m'}\left(\cos\theta\right)P_{l}^{0}\left(\cos\theta\right)\left(\sin\theta\right)^{l'-l}\int_{0}^{2\pi}\ud\varphi\,e^{-im'\varphi}\left(\cos\varphi\right)^{2n-l'+l}.
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\end{align*}
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\end_inset
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The asimuthal integral evaluates to
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\begin_inset Formula
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\[
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\int_{0}^{2\pi}\ud\varphi\,e^{-im'\varphi}\left(\cos\varphi\right)^{2n-l'+l}=\pi\delta_{\left|m'\right|,2n-l'+l}
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\]
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\end_inset
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(note that
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\begin_inset Formula $2n-l'+l\ge0$
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\end_inset
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as it's the former index
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\begin_inset Formula $k$
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\end_inset
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).
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That eliminates one of the two remaining (finite) sums.
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\end_layout
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\begin_layout Standard
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\begin_inset Note Note
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status open
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@ -386,6 +651,10 @@ BTW:
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\end_layout
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\begin_layout Standard
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\begin_inset Note Note
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status open
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\begin_layout Plain Layout
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Now we set the conventions: let the lattice lie on the
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\begin_inset Formula $z$
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\end_inset
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@ -418,6 +687,38 @@ Now we set the conventions: let the lattice lie on the
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\end_inset
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\end_layout
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\begin_layout Plain Layout
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Also, in this convention
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\begin_inset Formula $\ush lm\left(\uvec K\right)=0$
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\end_inset
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for
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\begin_inset Formula $m\ne0$
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\end_inset
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, so
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\end_layout
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\begin_layout Plain Layout
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\begin_inset Formula
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\begin{align*}
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\int\ud\Omega_{\vect r}\,G_{\Lambda}^{(1;\kappa)}\left(\vect s+\vect r\right)\ushD{l'}{m'}\left(\uvec r\right) & =-\frac{1}{2\pi\mathcal{A}}\sum_{\vect K\in\Lambda^{*}}e^{i\vect K\cdot\vect s}\sum_{lm}4\pi i^{l}j_{l}\left(\left|\vect K\right|\left|\vect r\right|\right)\ush l0\left(\uvec K\right)\times\\
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& \quad\times\int\ud\Omega_{\vect r}\,\ushD{l'}{m'}\left(\uvec r\right)\ushD l0\left(\uvec r\right)\sum_{n=0}^{\infty}\Delta_{n+1/2}\frac{1}{n!}\left(-\frac{\left(\left|\vect r_{\bot}\right|^{2}+2\vect r_{\bot}\cdot\vect s_{\bot}\right)\kappa^{2}\gamma_{\vect K}^{2}}{4}\right)^{n}.
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\end{align*}
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\end_inset
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Let's also fix the spherical harmonics for now,
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\begin_inset Formula
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\[
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\ushD lm\left(\uvec r\right)=\lambda'_{lm}e^{-im\phi}P_{l}^{-m}\left(\cos\theta\right)
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\]
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\end_inset
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Also, in this convention
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\begin_inset Formula $\ush lm\left(\uvec K\right)=0$
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\end_inset
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@ -437,7 +738,7 @@ Also, in this convention
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\end_layout
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\begin_layout Standard
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\begin_layout Plain Layout
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Let's also fix the spherical harmonics for now,
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\begin_inset Formula
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\[
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@ -449,7 +750,7 @@ Let's also fix the spherical harmonics for now,
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\end_layout
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\begin_layout Standard
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\begin_layout Plain Layout
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The angular integral (assuming it can be separated from the rest like this)
|
||||
is
|
||||
\begin_inset Formula
|
||||
|
|
@ -462,7 +763,7 @@ I_{l'}^{m'}\equiv\int\ud\Omega_{\vect r}\,\ushD{l'}{m'}\left(\uvec r\right)e^{-\
|
|||
|
||||
\end_layout
|
||||
|
||||
\begin_layout Standard
|
||||
\begin_layout Plain Layout
|
||||
Let's further extract the azimuthal part
|
||||
\begin_inset Formula $\left(w\equiv2r_{\bot}s_{\bot}\kappa^{2}\gamma_{\vect K}^{2}/4\tau\right)$
|
||||
\end_inset
|
||||
|
|
@ -556,7 +857,7 @@ Althought it's not superobvious, this sum is symmetric w.r.t.
|
|||
.
|
||||
\end_layout
|
||||
|
||||
\begin_layout Standard
|
||||
\begin_layout Plain Layout
|
||||
Let's do the polar integration next:
|
||||
\begin_inset Formula $r_{\bot}=r\sin\theta$
|
||||
\end_inset
|
||||
|
|
@ -616,6 +917,11 @@ so we can fix
|
|||
\end_inset
|
||||
|
||||
|
||||
\end_layout
|
||||
|
||||
\end_inset
|
||||
|
||||
|
||||
\begin_inset Formula $ $
|
||||
\end_inset
|
||||
|
||||
|
|
|
|||
Loading…
Reference in New Issue