Covariant coordinates – dual or reciprocal space

Authier, A.

doi:10.1107/97809553602060000628

International
Tables for
Crystallography
Volume D
Physical properties of crystals
Edited by A. Authier

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International Tables for Crystallography (2006). Vol. D. ch. 1.1, pp. 6-7

Section 1.1.2.4. Covariant coordinates – dual or reciprocal space

A. Authier^a ^*

^a Institut de Minéralogie et de la Physique des Milieux Condensés, Bâtiment 7, 140 rue de Lourmel, 75015 Paris, France
Correspondence e-mail: aauthier@wanadoo.fr

1.1.2.4. Covariant coordinates – dual or reciprocal space

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1.1.2.4.1. Covariant coordinates

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Using the developments (1.1.2.1) and (1.1.2.5), the scalar products of a vector x and of the basis vectors $[{\bf e}_{i}]$ can be written $[x_{i} = {\bf x} \cdot {\bf e}_{i} = x\hskip 1pt^{j}{\bf e}_{j} \cdot {\bf e}_{i} = x\hskip 1pt^{j}g_{ij}. \eqno(1.1.2.9)]$ The n quantities $[x_{i}]$ are called covariant components, and we shall see the reason for this a little later. The relations (1.1.2.9) can be considered as a system of equations of which the components $[x\hskip 1pt^{j}]$ are the unknowns. One can solve it since $[\Delta (g_{ij}) \neq 0]$ (see the end of Section 1.1.2.2). It follows that $[x\hskip 1pt^{j} = x_{i}g^{ij} \eqno(1.1.2.10)]$ with $[g^{ij}g_{jk} = \delta_{k}^{i}. \eqno(1.1.2.11)]$

The table of the $[g^{ij}]$ 's is the inverse of the table of the $[g_{ij}]$ 's. Let us now take up the development of x with respect to the basis $[{\bf e}_{i}]$ : $[{\bf x} = x^{i}{\bf e}_{i}.]$

Let us replace $[x^{i}]$ by the expression (1.1.2.10): $[{\bf x} = x_{j}g^{ij}{\bf e}_{i}, \eqno(1.1.2.12)]$ and let us introduce the set of n vectors $[{\bf e}\hskip 1pt^{j} = g^{ij}{\bf e}_{i} \eqno(1.1.2.13)]$ which span the space $[E^{n}\,\,(j = 1, \ldots, n)]$ . This set of n vectors forms a basis since (1.1.2.12) can be written with the aid of (1.1.2.13) as $[{\bf x} = x_{j}{\bf e}\hskip 1pt^{j}. \eqno(1.1.2.14)]$

The $[x_{j}]$ 's are the components of x in the basis $[{\bf e}\hskip 1pt^{j}]$ . This basis is called the dual basis. By using (1.1.2.11) and (1.1.2.13), one can show in the same way that $[{\bf e}_{j} = g_{ij}{\bf e}\hskip 1pt^{j}. \eqno(1.1.2.15)]$

It can be shown that the basis vectors $[{\bf e}\hskip 1pt^{j}]$ transform in a change of basis like the components $[x\hskip 1pt^{j}]$ of the physical space. They are therefore contravariant. In a similar way, the components $[x_{j}]$ of a vector x with respect to the basis $[{\bf e}\hskip 1pt^{j}]$ transform in a change of basis like the basis vectors in direct space, $[{\bf e}_{j}]$ ; they are therefore covariant: $[\left. \matrix{{\bf e}\hskip 1pt^{j} = B\hskip1pt_{k}^{j} {\bf e}'^{k}\semi &{\bf e}'^{k} = A_{j}^{k} {\bf e}\hskip 1pt^{j}\cr x_{i} = A\hskip1pt_{i}^{j} x'_{j}\semi &x'_{j} = B_{j}^{i}x_{i}.\cr}\right\} \eqno(1.1.2.16)]$

1.1.2.4.2. Reciprocal space

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Let us take the scalar products of a covariant vector $[{\bf e}_i]$ and a contravariant vector $[{\bf e}\hskip 1pt^{j}]$ : $[{\bf e}_{i} \cdot {\bf e}\hskip 1pt^{j} = {\bf e}_{i} \cdot g\hskip 1pt^{jk}{\bf e}_{k} = {\bf e}_{i} \cdot {\bf e}_{k} g\hskip 1pt^{jk} = g_{ik}g\hskip 1pt^{jk} = \delta\hskip1pt_{i}^{j}]$ [using expressions (1.1.2.5), (1.1.2.11) and (1.1.2.13)].

The relation we obtain, $[{\bf e}_{i} \cdot {\bf e}\hskip1pt^{j} = \delta\hskip1pt_{i}^{j}]$ , is identical to the relations defining the reciprocal lattice in crystallography; the reciprocal basis then is identical to the dual basis $[{\bf e}^{i}]$ .

1.1.2.4.3. Properties of the metric tensor

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In a change of basis, following (1.1.2.3) and (1.1.2.5), the $[g_{ij}]$ 's transform according to $[\left. \matrix{g_{ij} \ = A_{i}^{k} A_{j}^{m} g'_{km}\cr g'_{ij} \ = B_{i}^{k} B_{j}^{m} g_{km}.\cr}\right\} \eqno(1.1.2.17)]$ Let us now consider the scalar products, $[{\bf e}^{i} \cdot {\bf e}\hskip 1pt^{j}]$ , of two contravariant basis vectors. Using (1.1.2.11) and (1.1.2.13), it can be shown that $[{\bf e}^{i} \cdot {\bf e}\hskip 1pt^{j} = g^{ij}. \eqno(1.1.2.18)]$

In a change of basis, following (1.1.2.16), the $[g^{ij}]$ 's transform according to $[\left. \matrix{g^{ij} \ = B\hskip1pt_{k}^{i} B\hskip1pt_{m}^{j} g'^{km}\cr g'^{ij} \ = A_{k}^{i} A\hskip1pt_{m}^{j} g^{km}.\cr}\right\} \eqno(1.1.2.19)]$ The volumes V ′ and V of the cells built on the basis vectors $[{\bf e}'_{i}]$ and $[{\bf e}_{i}]$ , respectively, are given by the triple scalar products of these two sets of basis vectors and are related by $[\eqalignno{V' &= ({\bf e}'_{1}, {\bf e}'_{2}, {\bf e}'_{3}) &\cr &= \Delta (B_{j}^{i}) ({\bf e}_{1}, {\bf e}_{2}, {\bf e}_{3}) &\cr & = \Delta (B_{j}^{i}) V, &(1.1.2.20)\cr}]$ where $[\Delta (B_{j}^{i})]$ is the determinant associated with the transformation matrix between the two bases. From (1.1.2.17) and (1.1.2.20), we can write $[\Delta (g'_{ij}) = \Delta (B_{i}^{k}) \Delta (B_{j}^{m}) \Delta(g_{km}).]$

If the basis $[{\bf e}_{i}]$ is orthonormal, $[\Delta (g_{km})]$ and V are equal to one, $[\Delta (B_{j})]$ is equal to the volume V ′ of the cell built on the basis vectors $[{\bf e}'_{i}]$ and $[\Delta (g'_{ij}) = V'^{2}.]$ This relation is actually general and one can remove the prime index: $[\Delta (g_{ij}) = V^{2}. \eqno(1.1.2.21)]$

In the same way, we have for the corresponding reciprocal basis $[\Delta (g^{ij}) = V^{*2},]$ where $[V^{*}]$ is the volume of the reciprocal cell. Since the tables of the $[g_{ij}]$ 's and of the $[g^{ij}]$ 's are inverse, so are their determinants, and therefore the volumes of the unit cells of the direct and reciprocal spaces are also inverse, which is a very well known result in crystallography.

References

International Tables for Crystallography (2006). Vol. D. ch. 1.1, pp. 6-7