3.2 The volumes of the unit cells in direct and reciprocal space are inverse

Next: 3.3 Calculation of the reciprocal lattice Up: 3. Reciprocal Space and Dual Space Previous: 3.1 Definition

Let V be the volume of the unit cell. In a change of coordinate:

$\begin{displaymath} \textbf{e}_i \times \textbf{e}^{\prime}_j B^j_i \end{displaymath}$ (3.9)

we have

$\begin{displaymath} V = V^{\prime}\Delta(B)\end{displaymath}$

where $\Delta(B)$ is the determinant built on B^j_i. In the same way, we may write:

$\begin{displaymath} {\Delta}g_{ij} = {\Delta}g^{\prime}_{ij}{\Delta}\end{displaymath}$ (3.10)

Let us now assume that the base e_j is orthonormal. There comes:

$\begin{displaymath} V = {\Delta}(B), \quad {\Delta}g_{ij} = {\Delta}(B)^2 \end{displaymath}$ (3.11)

We have then demonstrated the following general result:

$\begin{displaymath} {\Delta}g_{ij} = V^2 \end{displaymath}$ (3.12)

From (3.6) we know that

$\begin{displaymath} {\Delta}g_{ij} \cdot {\Delta}g^{ij} = 1\end{displaymath}$ (3.13)

It is easy to show the following relation, equivalent to (3.12):

$\begin{displaymath} {\Delta}g^{ij} = V^{*2}\end{displaymath}$ (3.14)

where V^* is the volume of the unit cell in reciprocal space.

From (3.12), (3.13) and (3.14), we obtain finally:

$\begin{displaymath} V \cdot V^* = 1\end{displaymath}$

Next: 3.3 Calculation of the reciprocal lattice Up: 3. Reciprocal Space and Dual Space Previous: 3.1 Definition