%------------------------------------------------------------------------------
%$Author: saulius $
%$Date: 2006-10-25 08:43:54 +0300 (Wed, 25 Oct 2006) $ 
%$Revision: 146 $
%$URL: svn+ssh://kolibris.ibt.lt/home/saulius/svn-repositories/RA-mokslo-kontekste/difrakcijos-teorija-lt.tex $
%------------------------------------------------------------------------------

\documentclass{chaksem}
%% \documentclass[a4paper,landscape]{slides}
%% \documentclass[a4]{seminar}
\usepackage{amsmath}
\usepackage{amssymb}
%% \usepackage{amsfonts}
\usepackage{amsbsy}
\usepackage[utf8x]{inputenc}
\usepackage[T1]{fontenc}
\usepackage{graphicx}
\usepackage{xcolor}
\pagestyle{empty}

%% \newcommand{\vvec}[1]{\vec{#1}}
%% \newcommand{\vvec}[1]{\vec{\boldsymbol{#1}}}
\renewcommand{\vec}[1]{\boldsymbol{#1}}
\newcommand{\vvec}[1]{\boldsymbol{#1}}
\newcommand{\uvec}[1]{\boldsymbol{\hat{#1}}}

\begin{document}

%------------------------------------------------------------------------------

\begin{slide}

\begin{center}
  \Large\color{gray}
  Tai, ką Jūs visada norėjote žinoti apie \\
  rentgenostruktūrinę analizę

  \bigskip
  \large ... bet nedrįsote paklausti
\end{center}

\end{slide}

%------------------------------------------------------------------------------

\begin{slide}

\heading{Difrakcijos eksperimentas}

\begin{center}
  % For some reason, the inclusion of the '.ps.gz' file causes TeX
  % stack size to be exceeded on Ubuntu-20.04:
  %% \includegraphics[width=5cm,angle=-90]{photos/beamlines/img_1082_crop.ps.gz}
  \includegraphics[width=5cm]{photos/beamlines/img_1082_crop.ps}
\end{center}

\end{slide}

%------------------------------------------------------------------------------

\begin{slide}

\heading{Difrakcijos eksperimento schema}

\begin{center}
\includegraphics[width=9cm]{exp-setup.ps}
\end{center}

\end{slide}

%------------------------------------------------------------------------------

\begin{slide}

\heading{Sin banga}

\begin{center}
\includegraphics[width=5cm,angle=-90]{drawings/sin-wave.ps}
\end{center}

\begin{equation}
y = E \sin( kx - \omega t - \varphi )
\end{equation}

\end{slide}

%------------------------------------------------------------------------------

\begin{slide}

\heading{Sin bangos anatomija}

\begin{center}
\input{drawings/sin-wave-annotated.pstex_t}
\end{center}

\begin{minipage}{0.44\textwidth}
\begin{align}
  y = E \sin( kx - \omega t - \varphi ) \\
  \notag
\end{align}
\end{minipage}
%
\begin{minipage}{0.44\textwidth}
\begin{align}
  k \underset{\mathit{def}}{=} & \frac{2\pi}{\lambda}
  \\
  \omega \underset{\mathit{def}}{=} & \frac{2\pi}{T}
\end{align}
\end{minipage}

\end{slide}

%------------------------------------------------------------------------------

\begin{slide}

\heading{Plokščios monochromatinės bangos erdvėje}

Plokščia monochromatinė banga erdvėje gali būti aprašyta vectoriniu
pavidalu:

\begin{center}
\input{drawings/flat-wave.pstex_t}
\end{center}

\begin{equation}
    E( \vec{r}, t ) = E_0 \sin ( \vec{k} \vec{r} - \omega t - \varphi )
\end{equation}

\end{slide}

%------------------------------------------------------------------------------

\begin{slide}

\heading{Dviejų bangų superpozicija}

\bigskip
\centerline{Elektromagnetinių bangų amplitudės sumuojasi:}

\begin{equation}
  E = E_1 + E_2
\end{equation}

\end{slide}

%------------------------------------------------------------------------------

\begin{slide}

\heading{El.\ m.\ bangų sklaidymas elektronu}

\centerline{Thomsonas parodė, kad:}

\begin{center}
\input{drawings/electron.pstex_t}
\end{center}

\begin{minipage}{0.63\textwidth}
\begin{gather}
  E_1( \vec{R}, t ) = E_1 \sin( \vec{k'}\vec{R} - \omega t -
  \Delta\varphi ) \\
  E_1 = E_0 \frac{1}{R} \frac{e^2}{mc^2} \sin \phi
  = E_0 \frac{A}{R}
\end{gather}
\end{minipage}
%
\begin{minipage}{0.35\textwidth}
\begin{align*}
  m_{p^+} &\approx 1800 \cdot m_{e^-} \\
  I_1     &\approx 2\% \cdot I_0
\end{align*}
\end{minipage}

\end{slide}

%------------------------------------------------------------------------------

\begin{slide}

\heading{El.\ m.\ bangų sklaidymas pavyzdžiu}

\begin{center}
\input{drawings/scattering.pstex_t}
\end{center}

\begin{equation}
  E_\Sigma =
  E_1 \sin( \vec{k}\vec{r_1} + \vec{k'}\vec{R_1} - \omega t - \Delta\varphi ) +
  E_2 \sin( \vec{k}\vec{r_2} + \vec{k'}\vec{R_2} - \omega t - \Delta\varphi )
\end{equation}

\end{slide}

%------------------------------------------------------------------------------

\begin{slide}

\heading{Kompleksinės eksponentės}

Skaičiavimams patogiau naudoti ne sinusus ir kosinusus, o kompleksines
eksponentes

\begin{equation}
    e^{ix} = \cos x + i \sin x
\end{equation}

Susitarkime vietoj $\sin$ rašyti kompleksinę eksponentę, atsimindami,
kad mūsų ``tikroji'' banga yra šios eksponentės menamoji dalis. Arba
realioji. Jeigu mums tai iš viso svarbu...

\end{slide}

%------------------------------------------------------------------------------

\begin{slide}

\heading{Prisiminimai: kompleksiniai skaičiai ...}

\bigskip
\begin{center}
\input{drawings/complex-numbers.pstex_t}
\end{center}

\begin{align*}
  z   &= a + ib & |z|   = \sqrt{a^2 + b^2} & \qquad \tan\varphi = \frac{b}{a} \\
  z^* &= a - ib & |z|^2 = z \cdot z^*      & 
\end{align*}

\end{slide}

%------------------------------------------------------------------------------

\begin{slide}

\heading{Prisiminimai: kompleksinių skaičių vektorinis pavidalas ...}

\bigskip
\begin{center}
\input{drawings/complex-numbers-as-vectors.pstex_t}
\end{center}

\begin{gather}
  z = z_1 + z_2 = a + i b = (a_1 + a_2) + i (b_1 + b_2) \\
  z_1 z_2 = (a_1 + i b_1) (a_2 + i b_2) 
  = (a_1 a_2 - b_1 b_2) + i (a_1 b_2 + a_2 b_1)
\end{gather}

\end{slide}

%------------------------------------------------------------------------------

\begin{slide}

\heading{Oilerio (Euler) formulės įrodymo eskizas...}

\begin{gather*}
%
\begin{split}
    e^{ix} = \cos x + i \sin x
\end{split} \\[12pt]
%
\boxed{
\small
\begin{align*}
%
    \sin x &= x - \frac{1}{3!}x^3 + \frac{1}{5!}x^5 - \dots 
    + (-1)^k \frac{1}{(2k+1)!} x^{2k+1} + \dots \\
%
    \color{red}
    i \sin x &
    \color{red} 
    = ix + \frac{1}{3!}(ix)^3 + \frac{1}{5!}(ix)^5 + \dots 
    + \underset{=1}{\underbrace{(-1)^k i^{-2k-1} i}}
      \frac{1}{(2k+1)!} (ix)^{2k+1} + \dots \\
%
    \color{blue}
    \cos x &= 1 - \frac{1}{2!}x^2 + \frac{1}{4!}x^4 - \dots 
    + (-1)^k \frac{1}{(2k)!} x^{2k} + \dots \\
           &
    \color{blue}
    = 1 + \frac{1}{2!}(ix)^2 + \frac{1}{4!}(ix)^4 + \dots
    + \underset{=1}{\underbrace{(-1)^k i^{-2k}}} 
      \frac{1}{(2k)!} (ix)^{2k} + \dots \\
%
    e^{x}  &= 1 + \frac{1}{1!} x +  \frac{1}{2!} x^2 + \dots
    + \frac{1}{k!} x^k + \dots \\
%
    e^{ix} &= \color{blue} 1 + \color{red} ix
    + \color{blue} \frac{1}{2!} (ix)^2
    + \color{red}  \frac{1}{3!} (ix)^3
    + \color{blue} \frac{1}{4!} (ix)^4 + \dots
    + \normalcolor  \frac{1}{k!} (ix)^k + \dots
\end{align*}
}
\end{gather*}

\end{slide}

%------------------------------------------------------------------------------

\begin{slide}

\heading{Kompleksinių skaičių užrašymas eksponentėmis}

\bigskip
\begin{center}
\input{drawings/complex-numbers-as-exponents.pstex_t}
\end{center}

\begin{gather*}
  \begin{align*}
    |z| &= \sqrt{a^2 + b^2} &
    a &= |z|\cos\varphi &
    b &= |z|\sin\varphi
  \end{align*} \\
  z = a + ib = |z|\cos\varphi + i |z|\sin\varphi 
  = |z|(\cos\varphi + i \sin\varphi)
  = |z|e^{i\varphi} 
\end{gather*}

\end{slide}

%------------------------------------------------------------------------------

\begin{slide}

\heading{Naudingos eksponenčių savybės}

\begin{equation}
    e^a e^b = e^{a+b}
\end{equation}

\begin{equation}
    e^{ix} e^{i\varphi} = e^{i(x+\varphi)}
\end{equation}

\end{slide}

%------------------------------------------------------------------------------

\begin{slide}

\heading{Fazės postūmis}

\bigskip
\begin{center}
\input{drawings/phase-shift.pstex_t}
\end{center}

\begin{equation}
  z_1 = ze^{i\varphi} = |z|e^{i\varphi_z}e^{i\varphi} =
  |z|e^{i(\varphi_z + \varphi)}
\end{equation}

\end{slide}

%------------------------------------------------------------------------------

\begin{slide}

\heading{Dviejų eksponenčių sudėtis}

\begin{align}
    e^{ix} + e^{iy}
    &= e^{ix} (1 + e^{i(y-x)})
    \notag \\
    &= e^{i\textcolor{blue}{x}} e^{i\frac{y-\textcolor{blue}{x}}{2}}
      (e^{-i\frac{y-x}{2}} + e^{i\frac{y-x}{2}})
    \notag \\
    &= e^{i2\frac{\textcolor{blue}{x}}{2}} e^{i\frac{y-\textcolor{blue}{x}}{2}}
    (\cos\frac{y-x}{2} \textcolor{red}{- i \sin\frac{y-x}{2}} + 
     \cos\frac{y-x}{2} \textcolor{red}{+ i \sin\frac{y-x}{2}} )
     \notag \\
    &= e^{i\frac{x+y}{2}} \cdot 2\cos\frac{y-x}{2}
\end{align}

\end{slide}

%------------------------------------------------------------------------------

\begin{slide}

\heading{Amplitudės kvadratas}

\begin{align}
    \intertext{let}
    z &= |z| e^{i\varphi} \, \text{,} \\
    \intertext{then}
    z \cdot z^*
    &= |z| e^{i\varphi} \cdot |z| e^{-i\varphi} \notag \\
    &= |z|^2
    \underset{=1}{
      \underbrace{
	e^{\overset{\scriptscriptstyle =0}{
	    \overbrace{\scriptstyle i(\varphi-\varphi)}}
	}
      }
    }
    \notag \\
    &= |z|^2
\end{align}

\end{slide}

%------------------------------------------------------------------------------

\begin{slide}

\heading{Bangos vaizdavimas eksponente}

\begin{equation}
    E \sin( \vec{k}\vec{r} - \omega t ) =
    \text{\rm Im}(E e^{i( \vec{k}\vec{r} - \omega t )})
\end{equation}

\begin{align}
    E \sin( \vec{k}\vec{r} - \omega t -\varphi ) &=
    \text{\rm Im}(E e^{i( \vec{k}\vec{r} - \omega t - \varphi )}) \\
    &=
    \text{\rm Im}(E  e^{-i \varphi } e^{i( \vec{k}\vec{r} - \omega t)})
\end{align}

\begin{align}
    E_1 \sin( \vec{k}\vec{r} - \omega t -\varphi_1 ) & + \notag \\
    E_2 \sin( \vec{k}\vec{r} - \omega t -\varphi_2 )
    &=
    \text{\rm Im}(
    E_1 e^{i( \vec{k}\vec{r} - \omega t - \varphi_1 )} +
    E_2 e^{i( \vec{k}\vec{r} - \omega t - \varphi_2 )}) \notag \\
    &=
    \text{\rm Im}(
    (E_1 e^{-i \varphi_1} + E_2 e^{-i \varphi_2})
    e^{i( \vec{k}\vec{r} - \omega t )}
    ) \\
\end{align}

\end{slide}

%------------------------------------------------------------------------------

\begin{slide}

\heading{Vėl pavyzdžio sklaidymas}
\subheading{eksponentinis pavidalas}

\begin{center}
\input{drawings/scattering-exp.pstex_t}
\end{center}

\begin{equation}
  E_\Sigma =
  E_1 e^{i( \vec{k}\vec{r_1} + \vec{k'}\vec{R_1} - \omega t - \Delta\varphi )} +
  E_2 e^{i( \vec{k}\vec{r_2} + \vec{k'}\vec{R_2} - \omega t - \Delta\varphi )}
\end{equation}

\end{slide}

%------------------------------------------------------------------------------

\begin{slide}

\heading{Dviejų monochromatinių bangų sudėtis}

\begin{align}
  E_1(\vvec{R_1},t)
  &= E_1 e^{i(\vec{k}\vec{r_1} +
              \vec{k'}\vec{R_1} - \omega t - \Delta\varphi)} \notag \\
  &= \underset{\underset{\mathit{def}}{=}E'_0}{
       \underbrace{
         E_0 \frac{A}{R}
         e^{-i\Delta\varphi}
	 e^{-i\omega t}
       }
     }
     e^{i(\vec{k}\vec{r_1} + \vec{k'}\vec{R_1})} \notag \\
  &= E'_0 e^{i(\vec{k}\vec{r_1} + \vec{k'}\vec{R_1})} \\
  E_2(\vvec{R_2},t)
  &= E'_0 e^{i(\vec{k}\vec{r_2} + \vec{k'}\vec{R_2})}
\end{align}

\begin{equation}
\begin{split}
  E_\Sigma
  &=
  E'_0 e^{i( \vec{k}\vec{r_1} + \vec{k'}\vec{R_1})} +
  E'_0 e^{i( \vec{k}\vec{r_2} + \vec{k'}\vec{R_2})} \\
  &=
  E'_0 [
    e^{i( \vec{k}\vec{r_1} + \vec{k'}\vec{R_1})} +
    e^{i( \vec{k}\vec{r_2} + \vec{k'}\vec{R_2})}
  ] \\
  &=
  E'_0 e^{i( \vec{k}\vec{r_1} + \vec{k'}\vec{R_1})}
  [
    1 + 
    e^{i( \vec{k}\vec{r_2} - \vec{k}\vec{r_1} +
          \vec{k'}\vec{R_2} - \vec{k'}\vec{R_1}
        )
      }
  ]
\end{split}
\end{equation}

\end{slide}

%------------------------------------------------------------------------------

\begin{slide}

\heading{Dviejų monochromatinių bangų sudėtis (2)}

\begin{equation}
\begin{split}
  E_\Sigma
  &=
  \underset{=E''_0}{
    \underbrace{
      E'_0 e^{i( \vec{k}\vec{r_1} + \vec{k'}\vec{R_1})}
    }
  }
  [
    1 + 
    e^{i( \vec{k}\vec{r_2} - \vec{k}\vec{r_1} +
          \vec{k'}\vec{R_2} - \vec{k'}\vec{R_1}
        )
      }
  ] \\
  &=
  E''_0
  [
    1 + e^{i( \vec{k}\Delta\vec{r} + \vec{k'}\Delta\vec{R} ) }
  ]
\end{split}
\end{equation}

\end{slide}

%------------------------------------------------------------------------------

\begin{slide}

\heading{$\Delta\vec{r}$ projekcija}

\begin{center}
\input{drawings/projection-of-delta-k.pstex_t}
\end{center}

\begin{gather*}
\begin{align*}
\vec{\hat{k'}} =& \frac{\vec{k'}}{k'} &
\Delta\vec{\vec{R}} =& -\vec{\hat{k'}} ( \Delta\vec{r} \cdot \vec{\hat{k'}} )
\end{align*} \\
%
\vec{k'}\Delta\vec{R} =
-\vec{k'}\cdot\frac{\vec{k'}}{k'}(\Delta\vec{r}\cdot\frac{\vec{k'}}{k'}) =
-\frac{(\vec{k'}\cdot\vec{k'})}{k'^2}(\Delta\vec{r}\cdot\vec{k'}) =
-\vec{k'}\Delta\vec{r}
\end{gather*}

\end{slide}

%------------------------------------------------------------------------------

\begin{slide}

\enlargethispage{1\baselineskip}

\heading{Dviejų monochromatinių bangų sudėtis (3)}

\begin{center}
\input{drawings/definition-of-s.pstex_t}
\end{center}

\begin{equation}
\begin{split}
  E_\Sigma
  &=
  E''_0
  [
    1 + e^{i( \vec{k}\Delta\vec{r} - \vec{k'}\Delta\vec{r} ) }
  ] \\
  &=
  E''_0
  [
    1 + e^{i( \Delta\vec{r} ( \vec{k} - \vec{k'} )) }
  ] \\
  &= E''_0 [ 1 + e^{i\Delta\vec{r}\vec{s}} ] \\
  &=
  E''_0 [ 1
    \textcolor{gray}{\cdot e^{i\Delta\vec{r_0}\vec{s}}} +
    e^{i\Delta\vec{r}\vec{s}}
  ]
\end{split}
\end{equation}

\end{slide}

%------------------------------------------------------------------------------

\begin{slide}

\heading{Dviejų monochromatinių bangų sudėtis (4)}

\begin{center}
\input{drawings/small-volumes.pstex_t}
\end{center}

\begin{equation}
\begin{split}
  E_{1+2}
  &=
  E''_0 [
    n_1 \textcolor{gray}{e^{i\vec{r_0}\vec{s}}} +
    n_2 e^{i\vec{r}\vec{s}}
  ] \\
  E_\Sigma
  &=
  E''_0 \sum_{k=1}^N n_k e^{i\vec{r}_k\vec{s}} \\
  &=
  E''_0 \sum_{k=1}^N \rho_k \Delta V_k e^{i\vec{r}_k\vec{s}} \\
\end{split}
\end{equation}

\end{slide}

%------------------------------------------------------------------------------

\begin{slide}

\heading{Furje (Fourier) transformacija}

\begin{center}
\input{drawings/Fourier-arguments.pstex_t}
\end{center}

\begin{align}
  E_\Sigma
  &=
  \lim_{
    \substack{
      \scriptscriptstyle N \rightarrow \infty \\
      \scriptscriptstyle \Delta V_k \rightarrow 0
    }
  }
  E''_0 \sum_{k=1}^N \rho_k \Delta V_k e^{i\vec{r}_k\vec{s}} \\
  &=
  E''_0 \int_{V} \rho(\vec{r}) e^{i\vec{r}\vec{s}} dV
\end{align}

\end{slide}

%------------------------------------------------------------------------------

\begin{slide}

\enlargethispage{2\baselineskip}

\heading{Furje transformacija (2)}

\begin{center}
\input{drawings/Fourier-arguments.pstex_t}
\end{center}

\begin{equation}
  \vec{S} = \frac{\vec{s}}{2\pi}
\end{equation}

\begin{align}
  E_\Sigma
  &=
  E''_0 \int_{V} \rho(\vec{r}) e^{2\pi i\vec{r}\vec{S}} dV
\end{align}

\end{slide}

%------------------------------------------------------------------------------

\begin{slide}

\heading{Struktūriniai faktoriai}

\begin{align}
  E_\Sigma
  &=
  E''_0 \underset{F(\vec{S})}{
    \underbrace{
      \int_{V} \rho(\vec{r}) e^{2\pi i\vec{r}\vec{S}} dV
    }
  }
\end{align}

\begin{gather}
  \boxed{
    \mathcal{F}[\rho()](\vec{S}) =
    F(\vec{S}) = \int_{V} \rho(\vec{r}) e^{2\pi i\vec{r}\vec{S}} dV
  }
\end{gather}

\end{slide}

%------------------------------------------------------------------------------

\begin{slide}

\heading{Atvirkštinė Furje transformacija}

\begin{align}
  \intertext{let}
  F(\vec{S}) &= \int_{-\infty}^{+\infty} \rho(\vec{r})
  e^{2\pi i\vec{r}\vec{S}} dV \\
  \intertext{then}
  \rho(\vec{r}) &= \int_{-\infty}^{+\infty} F(\vec{S})
  e^{-2\pi i\vec{r}\vec{S}} d\vec{S}
\end{align}

\end{slide}

%------------------------------------------------------------------------------

\begin{slide}

\heading{Furje transformacijos savybės}

\begin{itemize}

\item Tiesiškumas

\begin{equation}
  \mathcal{F}[\alpha\rho_1 + \beta\rho_2] =
  \alpha\mathcal{F}[\rho_1] + \beta\mathcal{F}[\rho_2]
\end{equation}

\item Postūmis

\begin{equation}
  \mathcal{F}[\rho(\vec{r}-\Delta\vec{r})] =
  e^{2\pi i (\Delta\vec{r}\vec{S})}\, \mathcal{F}[\rho(\vvec{r})]
\end{equation}

\item Posūkis

\begin{align}
    \vvec{r}' &= ||R||\vvec{r} \notag \\
    \rho'(\vvec{r}) & \underset{\mathit{def}}{=}
    \rho(\vvec{r}') = \rho(||R||\vvec{r}) \notag \\
    \mathcal{F}[\rho(||R||\vvec{r})] &= F(||R||\vvec{S})
\end{align}

\end{itemize}

\end{slide}

%------------------------------------------------------------------------------

\begin{slide}

\heading{Dirako (Dirac) delta funkcija}

\begin{center}
\input{drawings/Diracs-delta-function.pstex_t}
\end{center}

\begin{align}
  \int_{T_1} \delta(x) f(x) dx
  &\underset{\mathit{def}}{=} 0
  \\
  \int_{T_2} \delta(x) f(x) dx
  &\underset{\mathit{def}}{=} f(0)
  \\
  \int_{-\infty}^{+\infty} \delta(x) f(x) dx
  &\underset{\mathit{def}}{=} f(0)
\end{align}

\end{slide}

%------------------------------------------------------------------------------

\begin{slide}

\heading{Delta funkcijos savybės}

\begin{gather}
  \delta(-x) = \delta(x) \\
  \int_{-\infty}^{+\infty} \delta(x) dx
  = 1 \\
  \int_{-\infty}^{+\infty} \delta(x-x_0) f(x) dx
  = f(x_0) \\
  \mathcal{F}[\delta] =
  \int_{-\infty}^{+\infty}
  \delta(x) e^{2\pi i x S} dx
  = e^0 = 1 \\
  \mathcal{F}^{-1}[1] =
  \int_{-\infty}^{+\infty} e^{-2\pi i x S} dS
  = \delta(x) \\
  \mathcal{F}[1] =
  \int_{-\infty}^{+\infty} e^{2\pi i x S} dx
  = \delta(S)
\end{gather}

\end{slide}

%------------------------------------------------------------------------------

\begin{slide}

\heading{Konvoliucija (sąsūka)}

\begin{equation}
  (f * g)(x) \underset{\mathit{def}}{=}
  \int_{-\infty}^{+\infty} f(v) g(x-v) dv
\end{equation}

\end{slide}

%------------------------------------------------------------------------------

\begin{slide}

\heading{Konvoliucija su delta funkcija}

\begin{center}
\input{drawings/convolution.pstex_t}
\end{center}

\begin{equation}
  f(x) * \delta(x-\Delta x) =
  \int_{-\infty}^{+\infty} f(v) \delta(x-v-\Delta x) dv =
  f(x-\Delta x)
\end{equation}

\end{slide}

%------------------------------------------------------------------------------

\begin{slide}

\heading{Konvoliucijos Furje transformacija}

\begin{align}
  \mathcal{F}[f * g]
  &=
  \int_{-\infty}^{+\infty} 
  \bigg( \int_{-\infty}^{+\infty} f(v) g(x-v) dv \bigg) \,
  e^{2\pi i S x} dx \notag \\
  &=
  \int_{-\infty}^{+\infty} 
  \int_{-\infty}^{+\infty} f(v) g(x-v)
  e^{2\pi i S (x-v)} e^{2\pi i S v} dvdx \notag \\
  &=
  \int_{-\infty}^{+\infty} f(v) e^{2\pi i S v} dv \cdot \notag \\
  & \int_{-\infty}^{+\infty} g(x-v) e^{2\pi i S (x-v)} d(x-v)
  \\
  &=
  \mathcal{F}[f] \cdot \mathcal{F}[g]
\end{align}

\end{slide}

%------------------------------------------------------------------------------

\begin{slide}

\heading{Gardelės funkcija}

\begin{center}
\input{drawings/lattice-function.pstex_t}
\end{center}

\begin{equation}
  L(x) = \sum_{n=-\infty}^{+\infty} \delta(x - an)
\end{equation}

\end{slide}

%------------------------------------------------------------------------------

\begin{slide}

\heading{Kristalo aprašymas}

\begin{center}
\input{drawings/crystal-function.pstex_t}
\end{center}

\begin{equation}
  f * L = \sum_{n=-\infty}^{+\infty} f(x - an)
\end{equation}

\end{slide}

%------------------------------------------------------------------------------

\begin{slide}

\heading{Kristalo el.\ tankio Furje transformacija}

\begin{center}
\input{drawings/crystal-function.pstex_t}
\end{center}

\begin{align}
  \mathcal{F}[f * L]
  &= \mathcal{F}[f] \cdot \mathcal{F}[L] \\
  &= \mathcal{F}[ \sum_{n=-\infty}^{+\infty} f(x - an) ] \\
  &= \sum_{n=-\infty}^{+\infty} \mathcal{F}[f(x - an)]
\end{align}

\end{slide}

%------------------------------------------------------------------------------

\begin{slide}

\heading{Kristalo el.\ tankio Furje transformacija (2)}

\begin{align}
  \mathcal{F}[f * L]
  &= \mathcal{F}[f] \cdot \mathcal{F}[L] \\
  &= F(S) \cdot \mathcal{F}[L] \\
  &= \mathcal{F}[ \sum_{n=-\infty}^{+\infty} f(x - an) ] \\
  &= \sum_{n=-\infty}^{+\infty} \mathcal{F}[f(x - an)] \\
  &= \sum_{n=-\infty}^{+\infty} F(S) e^{2\pi i S an} \\
  &= F(S) \sum_{n=-\infty}^{+\infty} e^{2\pi i S an}
\end{align}

\end{slide}

%------------------------------------------------------------------------------

\begin{slide}

\heading{Kristalo el.\ tankio Furje transformacija (3)}

\begin{align}
  \sum_{n=-\infty}^{\infty} f(x - an)
  &\underset{\mathit{def}}{=} S(x) \\
  &= \sum_{n=-\infty}^{\infty} c_n e^{-2\pi i \cdot n \frac{x}{a}}
\end{align}

\begin{align}
  c_n 
  &=
  \frac{1}{a} \int_{-a/2}^{a/2} S(x) e^{2\pi i \cdot n
  \frac{x}{a}} dx \\
  &=
  \frac{1}{a} \int_{-a/2}^{a/2} 
  \sum_{n=-\infty}^{\infty} f(x - an)
  e^{2\pi i \cdot n
  \frac{x}{a}} dx \\
  &=
  \frac{1}{a}
  \sum_{n=-\infty}^{\infty}
  \int_{-a/2}^{a/2} 
  f(x - an)
  e^{2\pi i \cdot n
  \frac{x}{a}} dx \\
  &=
  \frac{1}{a}
  \int_{-\infty}^{\infty} 
  f(x)
  e^{2\pi i \cdot x
  \frac{n}{a}} dx \\
  &= \frac{1}{a} F(n/a)
\end{align}

\end{slide}

%------------------------------------------------------------------------------

\begin{slide}

\heading{Kristalo el.\ tankio Furje transformacija (4)}

\begin{align}
  c_n &= \frac{1}{a} F(n/a)
\end{align}

\begin{align}
  \sum_{n=-\infty}^{\infty} f(x - an)
  &= \sum_{n=-\infty}^{\infty} c_n e^{-2\pi i \cdot n \frac{x}{a}} \\
  &= \frac{1}{a} \sum_{n=-\infty}^{\infty} F(n/a)
  e^{-2\pi i \cdot n \frac{x}{a}}
\end{align}

\begin{align}
  \sum_{n=-\infty}^{\infty} \delta(x - an)
  = \frac{1}{a} \sum_{n=-\infty}^{\infty} e^{-2\pi i \cdot n \frac{x}{a}}
\end{align}

\end{slide}

%------------------------------------------------------------------------------

\begin{slide}

\heading{Gardelės Furje transformacija}

\begin{align}
  \sum_{n=-\infty}^{\infty} \delta(x - an)
  = \frac{1}{a} \sum_{n=-\infty}^{\infty} e^{2\pi i \cdot n \frac{x}{a}}
\end{align}

\begin{align}
  \sum_{n=-\infty}^{\infty} \delta(x - n/a)
  = a \sum_{n=-\infty}^{\infty} e^{-2\pi i \cdot n a x}
\end{align}

\begin{align}
  \mathcal{F}[L] &= \mathcal{F}[
    \sum_{n=-\infty}^{\infty} \delta(x - an)
  ] \\
  &=
  \int_{-\infty}^{\infty}
  \sum_{n=-\infty}^{\infty} \delta(x - an)
  e^{2\pi i x S} dx \\
  &=
  \sum_{n=-\infty}^{\infty} 
  \int_{-\infty}^{\infty}
  \delta(x - an)
  e^{2\pi i x S} dx \\
  &=
  \sum_{n=-\infty}^{\infty} 
  e^{2\pi i a n S} \\
  &=
  \frac{1}{a} \sum_{n=-\infty}^{\infty} \delta(S - n/a)
\end{align}

\end{slide}

%------------------------------------------------------------------------------

\begin{slide}

\heading{Atvirkštinė gardelė, vienmatė}

\begin{center}
\input{drawings/crystal-function.pstex_t}
\end{center}

\begin{equation}
  L(x) = \sum_{n=-\infty}^{+\infty} \delta(x - an)
\end{equation}

\begin{align}
  \mathcal{F}[f * L]
  &= \mathcal{F}[f] \cdot \mathcal{F}[L] \\
  &= F(S) \cdot \frac{1}{a} \sum_{n=-\infty}^{\infty} \delta(S - n/a)
\end{align}

\end{slide}

%------------------------------------------------------------------------------

\begin{slide}

\heading{Atvirkštinė gardelė, trimatė}

%% \begin{center}
%% \input{drawings/crystal-function.pstex_t}
%% \end{center}

\begin{equation}
  L(x) = \sum_{m,n,p=-\infty}^{+\infty}
  \delta(\vec{r} - \vec{a}m - \vec{b}n - \vec{c}p)
\end{equation}

\begin{equation}
    L^*(\vvec{S}) = {\cal{F}}[L] = 
    \frac{1}{V} \sum_{h,k,l=-\infty}^{+\infty}
    \delta(\vvec{S}-h\vvec{a}^*-k\vvec{b}^*-l\vvec{c}^*)
    \label{lattice_transform_3}
\end{equation}

\begin{align}
    \vvec{a}^* &= \frac{[\vvec{b}\times\vvec{c}]}{V}, \qquad
    \vvec{b}^*  = \frac{[\vvec{c}\times\vvec{a}]}{V}, \qquad
    \vvec{c}^*  = \frac{[\vvec{a}\times\vvec{b}]}{V} \\
    V &= (\vvec{a}\vvec{b}\vvec{c})
       = (\vvec{a}\cdot[\vvec{b}\times\vvec{c}])
    \label{reciprocal_vectors}
\end{align}

\end{slide}

%------------------------------------------------------------------------------

\begin{slide}

\heading{Lauės (Laue) sąlygos}

\begin{align}
    {\cal{F}}[\rho_{\mathit{cryst}}] &=
        {\cal{F}}[\rho * L] =
        {\cal{F}}[\rho] \cdot {\cal{F}}[L] \\
     &= F(\vvec{S}) \cdot L^\star(\vvec{S}) \\
     &= F(\vvec{S}) \cdot \frac{1}{V} \sum_{h,k,l=-\infty}^{+\infty}
                          \delta(\vvec{S}-h\vvec{a}^*-k\vvec{b}^*-l\vvec{c}^*)
\end{align}

Matome, kad periodinio kristalo Furje transformacija nelygi nuliui tik
tam tikroms vektoriaus $\vvec{S}$ reikšmėms

\begin{align}
    (\vvec{S}\cdot\vvec{a}) &= h, \qquad
    (\vvec{S}\cdot\vvec{b})  = k, \qquad
    (\vvec{S}\cdot\vvec{c})  = l \label{von_Laue_conditions} \\
    h,k,l & \in {\mathbb{Z}}
\end{align}

\end{slide}

%------------------------------------------------------------------------------

\begin{slide}

\heading{Evaldo (Ewald) konstrukcija ir Evaldo sfera}

\begin{center}
\input{drawings/Ewald-sphere.pstex_t}
\end{center}

\end{slide}

%------------------------------------------------------------------------------

\end{document}