(\dot{\bf u}\nabla) {\bf a}_{\alpha} -
({\bf a}_{\alpha}\nabla)
\dot{\bf u} = 0$$ $$\dot{\bf a}^{\alpha} + \nabla ({\bf a}^{\alpha}\dot{\bf u}) = 0.$$ In the notation used the metrical tensors in the real and reciprocal space are, respectively: $$g_{\alf \beta} = \ba_\alf \ba_\beta, \qquad g^{\alf \beta} = \ba^\alf
\ba^\beta.
\ee            %11

    Then the lattice cell volume equals $g^{1/2}$, where
$g = \det\, g_{\alf \beta} $.
\par
    The relations between the components of the metrical tensors and the
deformation tensor $\, u_{ik}\,$ follow from the expression for the interval:
\begin{equation}
ds^2 = g_{\alf \beta} dN^\alf dN^\beta .
\ee          %12

The squared interval between the same points in the undeformed crystal is
$$
ds_0^2 = \,\stackrel{\circ}{g}_{\alf \beta} dN^\alf dN^\beta .
$$
        Hence
\begin{equation}
ds^2 - ds_0^2 = 2w_{\alf \beta} dN^\alf dN^\beta
\ee            %13
where
\begin{equation}
w_{\alf \beta} = \f{1}{2}(g_{\alf \beta} - \,\stackrel{\circ}{g}_{\alf \beta})
\ee             %14
plays the role of the deformation tensor in our notation. The invariance of the
interval yields:
\begin{equation}
w_{\alf \beta}dN^\alf dN^\beta = w_{ik}dx^i dx^k .
\ee              %15

    Taking into account relations (7) and (8) one obtains
\begin{equation}
w_{ik} = w_{\alf \beta} a_i^\alf a_k^\beta .
\ee              %16
    In order to obtain the components $u_{ik}$  of the tensor of small
deformations (2) as 