From 890dbc3e76fc205f4db86d13b8c15bdf865efb5e Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?Marek=20Ne=C4=8Dada?= Date: Thu, 3 Aug 2017 13:09:54 +0000 Subject: [PATCH] Continue writing on Ewald summation. Former-commit-id: a01eb920f0e49ce30fd29b8694a614a4af9993b1 --- ewald.lyx | 216 +++++++++++++++++++++++++++++++++++++++++++++++++----- 1 file changed, 199 insertions(+), 17 deletions(-) diff --git a/ewald.lyx b/ewald.lyx index d9c1a60..acba36c 100644 --- a/ewald.lyx +++ b/ewald.lyx @@ -51,7 +51,7 @@ \output_sync 0 \bibtex_command default \index_command default -\paperfontsize default +\paperfontsize 10 \spacing single \use_hyperref true \pdf_title "Accelerating lattice mode calculations with T-matrix method" @@ -65,8 +65,8 @@ \pdf_colorlinks false \pdf_backref false \pdf_pdfusetitle true -\papersize default -\use_geometry false +\papersize a5paper +\use_geometry true \use_package amsmath 1 \use_package amssymb 1 \use_package cancel 1 @@ -90,6 +90,10 @@ \shortcut idx \color #008000 \end_index +\leftmargin 2cm +\topmargin 2cm +\rightmargin 2cm +\bottommargin 2cm \secnumdepth 3 \tocdepth 3 \paragraph_separation indent @@ -234,26 +238,85 @@ Now suppose that the scatterers constitute an infinite lattice Due to the periodicity, we can write \begin_inset Formula $S_{\vect aα\leftarrow\vect bβ}=S_{α\leftarrowβ}(\vect b-\vect a)$ +\end_inset + + and +\begin_inset Formula $T_{\vect aα}=T_{\alpha}$ \end_inset . In order to find lattice modes, we search for solutions with zero RHS \begin_inset Formula \[ -\sum_{\vect bβ}(\delta_{\vect{ab}}\delta_{αβ}-T_{\vect aα}S_{\vect aα\leftarrow\vect bβ})A_{\vect bβ}=0 +\sum_{\vect bβ}(\delta_{\vect{ab}}\delta_{αβ}-T_{α}S_{\vect aα\leftarrow\vect bβ})A_{\vect bβ}=0 \] \end_inset and we assume periodic solution -\begin_inset Formula $A_{\vect b\alpha}(\vect k)=A_{\vect a\alpha}e^{i\vect k\cdot\vect r_{\vect b-\vect a}}$ +\begin_inset Formula $A_{\vect b\beta}(\vect k)=A_{\vect a\beta}e^{i\vect k\cdot\vect r_{\vect b-\vect a}}$ \end_inset -. +, yielding +\begin_inset Formula +\begin{eqnarray*} +\sum_{\vect bβ}(\delta_{\vect{ab}}\delta_{αβ}-T_{α}S_{\vect aα\leftarrow\vect bβ})A_{\vect a\beta}\left(\vect k\right)e^{i\vect k\cdot\vect r_{\vect b-\vect a}} & = & 0,\\ +\sum_{\vect bβ}(\delta_{\vect{0b}}\delta_{αβ}-T_{α}S_{\vect 0α\leftarrow\vect bβ})A_{\vect 0\beta}\left(\vect k\right)e^{i\vect k\cdot\vect r_{\vect b}} & = & 0,\\ +\sum_{β}(\delta_{αβ}-T_{α}\underbrace{\sum_{\vect b}S_{\vect 0α\leftarrow\vect bβ}e^{i\vect k\cdot\vect r_{\vect b}}}_{W_{\alpha\beta}(\vect k)})A_{\vect 0\beta}\left(\vect k\right) & = & 0,\\ +A_{\vect 0\alpha}\left(\vect k\right)-T_{α}\sum_{\beta}W_{\alpha\beta}\left(\vect k\right)A_{\vect 0\beta}\left(\vect k\right) & = & 0. +\end{eqnarray*} + +\end_inset + +Therefore, in order to solve the modes, we need to compute the +\begin_inset Quotes eld +\end_inset + +lattice Fourier transform +\begin_inset Quotes erd +\end_inset + + of the translation operator, +\begin_inset Formula +\begin{equation} +W_{\alpha\beta}(\vect k)\equiv\sum_{\vect b}S_{\vect 0α\leftarrow\vect bβ}e^{i\vect k\cdot\vect r_{\vect b}}.\label{eq:W definition} +\end{equation} + +\end_inset + + \end_layout \begin_layout Section -Multidimensional Dirac comb +Computing the Fourier sum of the translation operator +\end_layout + +\begin_layout Standard +The problem evaluating +\begin_inset CommandInset ref +LatexCommand eqref +reference "eq:W definition" + +\end_inset + + is the asymptotic behaviour of the translation operator, +\begin_inset Formula $S_{\vect 0α\leftarrow\vect bβ}\sim\left|\vect r_{\vect b}\right|^{-1}e^{ik_{0}\left|\vect r_{\vect b}\right|}$ +\end_inset + + that makes the convergence of the sum quite problematic for any +\begin_inset Formula $d>1$ +\end_inset + +-dimensional lattice. + In electrostatics, one can solve this with problem with Ewald summation. + Its basic idea is that if what asymptoticaly decays poorly in the direct + space, will perhaps decay fast in the Fourier space. + I use the same idea here, but things will be somehow harder than in electrostat +ics. +\end_layout + +\begin_layout Section +(Appendix) Multidimensional Dirac comb \end_layout \begin_layout Subsection @@ -286,9 +349,9 @@ Fourier series representation \begin_layout Standard \begin_inset Formula -\[ -Ш_{T}(t)=\sum_{n=-\infty}^{\infty}e^{2\pi int/T} -\] +\begin{equation} +Ш_{T}(t)=\sum_{n=-\infty}^{\infty}e^{2\pi int/T}\label{eq:1D Dirac comb Fourier series} +\end{equation} \end_inset @@ -313,9 +376,9 @@ With unitary ordinary frequency Ft., i.e. we have \begin_inset Formula -\[ -\uoft{Ш_{T}}(f)=\frac{1}{T}Ш_{\frac{1}{T}}(f)=\sum_{n=-\infty}^{\infty}e^{-i2\pi fnT} -\] +\begin{equation} +\uoft{Ш_{T}}(f)=\frac{1}{T}Ш_{\frac{1}{T}}(f)=\sum_{n=-\infty}^{\infty}e^{-i2\pi fnT}\label{eq:1D Dirac comb Ft ordinary freq} +\end{equation} \end_inset @@ -439,16 +502,41 @@ From the scaling property of delta function, \end_inset + +\end_layout + +\begin_layout Standard +From the book: +\end_layout + +\begin_layout Standard +\begin_inset Formula +\[ +\dc A(\vect x)=\frac{1}{\left|\det A\right|}\dc{}^{(d)}\left(A^{-1}\vect x\right) +\] + +\end_inset + + +\begin_inset Note Note +status open + +\begin_layout Plain Layout Applying both sides to a test function that is one at the origin, we get \begin_inset Formula $c_{\basis u}=\prod_{i=1}^{d}\left\Vert \rec{\vect u_{i}}\right\Vert $ \end_inset -, and hence + SRSLY?, and hence \begin_inset Formula -\[ -\dc{\basis u}(\vect x)=\prod_{i=1}^{d}\left\Vert \rec{\vect u_{i}}\right\Vert \dc{}\left(\vect x\cdot\rec{\vect u_{i}}\right). -\] +\begin{equation} +\dc{\basis u}(\vect x)=\prod_{i=1}^{d}\left\Vert \rec{\vect u_{i}}\right\Vert \dc{}\left(\vect x\cdot\rec{\vect u_{i}}\right).\label{eq:Dirac comb factorisation} +\end{equation} + +\end_inset + + +\end_layout \end_inset @@ -459,9 +547,103 @@ Applying both sides to a test function that is one at the origin, we get Fourier series representation \end_layout +\begin_layout Standard +\begin_inset Note Note +status open + +\begin_layout Plain Layout +Utilising the Fourier series for 1D Dirac comb +\begin_inset CommandInset ref +LatexCommand eqref +reference "eq:1D Dirac comb Fourier series" + +\end_inset + + and the factorisation +\begin_inset CommandInset ref +LatexCommand eqref +reference "eq:Dirac comb factorisation" + +\end_inset + +, we get +\begin_inset Formula +\begin{eqnarray*} +\dc{\basis u}(\vect x) & = & \prod_{j=1}^{d}\left\Vert \rec{\vect u_{j}}\right\Vert \sum_{n_{j}=-\infty}^{\infty}e^{2\pi in_{i}\vect x\cdot\rec{\vect u_{i}}}\\ + & = & \left(\prod_{j=1}^{d}\left\Vert \rec{\vect u_{j}}\right\Vert \right)\sum_{\vect n\in\mathbb{Z}^{d}}e^{2\pi i\vect x\cdot\sum_{k=1}^{d}n_{k}\rec{\vect u_{k}}}. +\end{eqnarray*} + +\end_inset + + +\end_layout + +\end_inset + + +\end_layout + \begin_layout Subsubsection Fourier transform \end_layout +\begin_layout Standard +From the book https://see.stanford.edu/materials/lsoftaee261/chap8.pdf +\end_layout + +\begin_layout Standard +\begin_inset Formula +\[ +\uoft{\dc A}\left(\vect{\xi}\right)=\dc{}^{(d)}\left(A^{T}\vect{\xi}\right). +\] + +\end_inset + + +\end_layout + +\begin_layout Standard +\begin_inset Note Note +status open + +\begin_layout Plain Layout +So, from the stretch theorem +\begin_inset Formula $\uoft{(f(A\vect x))}=\frac{1}{\left|\det A\right|}\uoft{f\left(A^{-T}\vect{\xi}\right)}=\left|\det A^{-T}\right|\uoft{f\left(A^{-T}\vect{\xi}\right)}$ +\end_inset + + +\end_layout + +\begin_layout Plain Layout +From +\begin_inset CommandInset ref +LatexCommand eqref +reference "eq:Dirac comb factorisation" + +\end_inset + + and +\begin_inset CommandInset ref +LatexCommand eqref +reference "eq:1D Dirac comb Ft ordinary freq" + +\end_inset + + +\begin_inset Formula +\[ +\uoft{\dc{\basis u}}(\vect{\xi})=\prod_{i=1}^{d}\left\Vert \rec{\vect u_{i}}\right\Vert \dc{}\left(\vect x\cdot\rec{\vect u_{i}}\right). +\] + +\end_inset + + +\end_layout + +\end_inset + + +\end_layout + \end_body \end_document