& \rho\dot{\bf  u}\ddot{\bf  u}  +  \frac{1}{2}\dot{\rho}\dot{u}^2
%+ \sigma_{\alpha\beta}a_i^{\alpha}a_k^{\beta}
%\frac{\partial\dot{u}_i}{\partial x_k}  - \dot{\bfu}\nabla E_0
+ \frac{\partial}{\partial t}\angg{ \eps\varphi}
 + m\ddot{\bf  u} {\bf
a}_{\alpha}\angg{\varphi\frac{\partial\eps}{\partial  k_{\alpha}}}
 \nonumber\\ & + & m\dot{\bf u}\dot{\bf a}_{\alpha}
 \angg{\varphi\frac{\partial\eps}{\partial k_{\alpha}}}
  +  m\dot{\bf  u}{\bf a}_ {\alpha}
\angg{\varphi\frac{\partial\dot{\eps}}{\partial
k_{\alpha}} }  +m\dot{\bf  u}{\bf a}_ {\alpha}
\angg{\dot{\varphi}\frac{\partial\eps}{\partial
k_{\alpha}} }  + \dot{W} .\end{aligned}$$ The time derivative of the elastic energy $E_0(g^{\alpha\beta})$ can be taken in the following way. $${\dot E}_0 = \frac{\p E_0}{\p g^{\alpha\beta}}{\dot g}^{\alpha\beta} =
-\sigma_{\alpha \beta} {\dot a}^\alpha_i a^\beta_i
\ee            %57
where
\begin{equation}
\sigma_{\alpha \beta} = -2 \frac{\p E_0}{\p g^{\alpha\beta}} .
\ee         %58
   Substituting $\,{\dot a}^\alpha_i\,$ from the evolution equation (10)
in (57) yields
\begin{equation}
{\dot E}_0 = \sigma_{\alpha \beta} a^\beta_i \left(
\frac{\p a^\alpha_i}{\p x_k} {\dot u}_k +
a^\alpha_k \frac {\p {\dot u}_k} {\p x_i} \right) .
\ee         %59
On the other hand
\begin{equation}
\nabla_k E_0 =  \frac{\p E_0}{\p g^{\alpha\beta}} \nabla g^{\alpha\beta} =
-\sigma_{\alpha \beta}  a^\beta_i \frac {\partial a^\alpha_i}{\p x_k} .
\ee         %60

It follows from (59) and (60) that
\begin{equation}
{\dot E}_0 =
 \sigma_{\alpha\beta}a_i^{\alpha}a_k^{\beta}
\frac{\partial {\dot u}_i}{\partial x_k}  - \dot{\bf u}\nabla E_0 .
\ee          %61
\par
The time derivative of the full momentum (52) gives
\begin{equation}
0 = -{\bf{\dot J}}+ {\dot \rho}{\dot {\bf u}} + \rho\ddot{\bf u} +
m\dot{\bf a}_ 