continued with formulas

doc/formulas/formulas.tex

@@ -8,6 +8,8 @@
 \documentclass{article}
 
 \usepackage{amsmath}
+\usepackage{tensor}
+\usepackage{bm}
 \usepackage{graphicx}
 
 \usepackage[a4paper]{geometry}
@@ -51,7 +53,8 @@ Angle $\psi$ between $\left| k_i \right>$ and $\left| Q \right>$, in units of \A
 
 
 \subsubsection*{Angle $\xi$}
-Angle $\xi$ between $\left| Q \right>$ and orientation vector $\left| a \right>$ (i.e. $ax$, $ay$, $az$), in units of rlu; $g_{ij} = \left| b_i \left> \right< b_j \right|$ is the covariant metric of the reciprocal lattice with basis $\left| b_i \right>$:
+% $g_{ij} = \left< \bm{b}_i | \bm{b}_j \right> = \tensor{B}{_{ik}} \tensor{B}{^{k}_{j}}$
+Angle $\xi$ between $\left| Q \right>$ and orientation vector $\left| a \right>$ (i.e. $ax$, $ay$, $az$), in units of rlu; $g_{ij} = \left< \bm{b}_i | \bm{b}_j \right>$ is the covariant metric of the reciprocal lattice with basis vectors $\left| \bm{b}_i \right>$, which form the columns of the crystallographic $B$ matrix:
 \begin{equation} \xi = \sigma_{\mathrm{side}} \cdot \arccos \left( \frac{ \left< Q | a \right> }{ \sqrt{\left< Q | Q \right>} \sqrt{\left< a | a \right>} } \right) \end{equation}
 \begin{equation} \boxed{ \xi = \sigma_{\mathrm{side}} \cdot \arccos \left( \frac{ Q^i g_{ij} a^j }{ \sqrt{Q^i g_{ij} Q^j} \sqrt{a^i g_{ij} a^j} } \right) } \end{equation}
 

tools/tascalc/tascalc.py

@@ -36,6 +36,7 @@ def rotate(_axis, vec, phi):
 
 
 # get metric from crystal B matrix
+# basis vectors are in the columns of B, i.e. the second index
 def get_metric(B):
 	return np.einsum("ij,ik -> jk", B, B)
 
@@ -130,10 +131,12 @@ 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])
 
+	# the real-space basis vectors form the columns of the A matrix
 	return np.transpose(np.array([a, b, c]))
 
 
 # crystallographic B matrix converting rlu to 1/A
+# the reciprocal-space basis vectors form the columns of the B matrix
 def get_B(lattice, angles):
 	A = get_A(lattice, angles)
 	B = 2.*np.pi * np.transpose(la.inv(A))