supposed to be the corresponding quasimomentum.
However, $\tilde \Phi \, $ does not depend on the bare mass, $m$, of the
quasiparticle
and therefore does not take into account any inertial effects (e.g. the
 Stewart-Tolman
effect in metals, centrifugal forces in rotating bodies etc).
\par
If $\eps ({\bf k}, u_{ik})$ is considered as quasiparticle energy then
 the Galilean transformation applies as a consequence of
the requirement that energy density $\int \eps d^3 k$
(as a macroscopic quantity) has to obey Galilean transformations. This yields
\begin{equation}
\tilde {\cal E} = \eps_0 ({\bf k}) + \lambda_{ik}({\bf k})u_{ik} + m\dot{\bf u}
\frac{\p \eps_0}{\p {\bf k}}.
\ee
\par
It has been shown by many authors that such a transformation is incompatible
with the Boltzmann equation. That is why some artificial ways have been used
the most successful being that of Landau (cf the footnote in Ref.
\cite{KontUFN}). He
has suggested that (in order to take into account the noninertial properties of
 the lattice frame) one has first to add to the dispersion law (1)
the term $\dx -m\dot{\bf u}\frac{\p \eps}{\p {\bf k}}$, setting
\begin{equation}
\eps({\bf k}, u_{ik}) = \eps_0({\bf k}) + \lambda_{ik}({\bf k})u_{ik}
 -m {\dot {\bf u}}\f{\p \eps}{\p {\bf k}}.
\ee
 and then to 