### continued with writing lecture: derived B matrix

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 ... ... @@ -17,9 +17,92 @@ \begin{document} Useful formulas, T. Weber, tweber@ill.fr, July 13, 2018. % ------------------------------------------------------------------------------------------------------------------------------------ \section{Fractional Coordinates} \subsection*{Basic Properties} From the cosine theorem we get: \left< a | b \right > = ab \cos \gamma, \label{ab} \left< a | c \right > = ac \cos \beta, \label{ac} \left< b | c \right > = bc \cos \alpha. \label{bc} \subsection*{Basis Vectors} We first choose $\left| a \right>$ along $x$, \boxed{ \left| a \right> = \left( \begin{array}{c} a_1 = a \\ 0 \\ 0 \end{array} \right), } \label{avec} $\left| b \right>$ in the $xy$ plane, \left| b \right> = \left( \begin{array}{c} b_1 \\ b_2 \\ 0 \end{array} \right), and $\left| c \right>$ in general: \left| c \right> = \left( \begin{array}{c} c_1 \\ c_2 \\ c_3 \end{array} \right). Inserting $\left| a \right>$ and $\left| b \right>$ into Eq. \ref{ab} gives: \left< a | b \right > = a_1 b_1 = ab \cos \gamma, b_1 = b \cos \gamma. Using the cross product between $\left| a \right>$ and $\left| b \right>$, we get: \left\Vert \left| a \right> \times \left| b \right> \right\Vert = \left\Vert \left( \begin{array}{c} 0 \\ 0 \\ a_1 b_2 \end{array} \right) \right\Vert = ab \sin \gamma, \label{crossab} b_2 = b \sin \gamma, \boxed{ \left| b \right> = \left( \begin{array}{c} b \cos \gamma \\ b \sin \gamma \\ 0 \end{array} \right). } \label{bvec} Inserting $\left| a \right>$ and $\left| c \right>$ into Eq. \ref{ac} gives: \left< a | c \right > = a_1 c_1 = ac \cos \beta, c_1 = c \cos \beta. Inserting $\left| b \right>$ and $\left| c \right>$ into Eq. \ref{bc} gives: \left< b | c \right > = b_1 c_1 + b_2 c_2 = bc \cos \alpha, b \cos \gamma \cdot c \cos \beta + b \sin \gamma \cdot c_2 = bc \cos \alpha, c_2 = \frac{c \cos \alpha - c \cos \gamma \cos \beta}{\sin \gamma}. The last component, $c_3$, can be obtained from the vector length normalisation, $\left< c | c \right> = c^2$: \left< c | c \right > = c_1^2 + c_2^2 + c_3^2 = c^2, c_3^2 = c^2 - c_1^2 - c_2^2, c_3^2 = c^2 \left[1 - \cos^2 \beta - \left(\frac{\cos \alpha - \cos \gamma \cos \beta}{\sin \gamma} \right)^2 \right], \boxed{ \left| c \right> = \left( \begin{array}{c} c \cdot \cos \beta \\ c \cdot \frac{\cos \alpha - \cos \gamma \cos \beta}{\sin \gamma} \\ c \cdot \sqrt{ 1 - \cos^2 \beta - \left(\frac{\cos \alpha - \cos \gamma \cos \beta}{\sin \gamma} \right)^2 } \end{array} \right). } \label{avec} The crystallographic $A$ matrix, which transforms real-space fractional to lab coordinates (A), is formed with the basis vectors in its columns: A = \left( \begin{array}{ccc} \left| a \right> & \left| b \right> & \left| c \right> \end{array} \right). The $B$ matrix, which transforms reciprocal-space relative lattice units (rlu) to lab coordinates (1/A), is: B = 2 \pi A^{-t}, where $-t$ denotes the transposed inverse. % ------------------------------------------------------------------------------------------------------------------------------------ \section{Scattering Triangle} \begin{center} \includegraphics[width = 0.2 \textwidth]{triangle} ... ...
 ... ... @@ -131,6 +131,10 @@ def get_A(lattice, angles): (cs[0]-cs[1]*cs[2]) / s2, \ (np.sqrt(1. - np.dot(cs,cs) + 2.*cs[0]*cs[1]*cs[2])) / s2]) # testing equality with own derivation #print((np.sqrt(1. - np.dot(cs,cs) + 2.*cs[0]*cs[1]*cs[2])) / s2) #print(np.sqrt(1. - cs[1]*cs[1] - ((cs[0] - cs[2]*cs[1])/s2)**2.)) # the real-space basis vectors form the columns of the A matrix return np.transpose(np.array([a, b, c])) ... ...
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