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

International Tables for Crystallography (2013). Vol. D, ch. 1.1, pp. 6-7

Section Covariant coordinates – dual or reciprocal space

A. Authiera*

aInstitut de Minéralogie et de Physique des Milieux Condensés, 4 Place Jussieu, 75005 Paris, France
Correspondence e-mail: Covariant coordinates – dual or reciprocal space

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Using the developments ([link] and ([link], 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(]The n quantities [x_{i}] are called covariant components, and we shall see the reason for this a little later. The relations ([link] 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[link]). It follows that [x\hskip 1pt^{j} = x_{i}g^{ij} \eqno(]with [g^{ij}g_{jk} = \delta_{k}^{i}. \eqno(]

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

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 ([link] and ([link], one can show in the same way that [{\bf e}_{j} = g_{ij}{\bf e}\hskip 1pt^{j}. \eqno(]

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(] 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 ([link], ([link] and ([link]].

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}]. Properties of the metric tensor

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In a change of basis, following ([link] and ([link], 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(]Let us now consider the scalar products, [{\bf e}^{i} \cdot {\bf e}\hskip 1pt^{j}], of two contravariant basis vectors. Using ([link] and ([link], it can be shown that [{\bf e}^{i} \cdot {\bf e}\hskip 1pt^{j} = g^{ij}. \eqno(]

In a change of basis, following ([link], 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(]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, &(\cr}]where [\Delta (B_{j}^{i})] is the determinant associated with the transformation matrix between the two bases. From ([link] and ([link], 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(]

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.

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