I have a probably real stupid question, but I still can not wrap my head around it.
The change in inner energy per unit volume $v$ of the actual state of a stressed crystal is given by the well known formula
$$
de_v=\sigma_{ij}d\epsilon_{ij} \label{eq1} \tag{1}
$$
If the crystal is only homogeneously deformed, this formula should simplify to
$$
de_v=\sigma d\epsilon \label{eq2} \tag{2}
$$
With $\sigma=\sigma_{kk}$ and $\epsilon=\epsilon_{kk}$ (using Einsteins summation convention).
I know how to derive this formula using virtual displacements, but I wondered if I can also derive this formula from the general formula for internal energy with volume work performed by the system.
So, I start with the mother of all thermodynamical equations for a single phase system
$$
e=-vp+TS+\mu N \label{eq3} \tag{3}
$$
$v$ being the actual volume and $p$ the hydrostatical pressure, $e$ the internal energy, $T$ temperature, $S$ entropy, $\mu$ chemical potential and $N$ the number of particles. The Gibbs-Duhem Relation for a process reads
$$
-vdp+SdT+Nd\mu=0 \label{eq4} \tag{4}
$$
After deviding \eqref{eq4} by the actual volume of the system $v$, we end up with
$$
-dp+s_v dT+\rho d\mu = 0 \label{eq5} \tag{5}
$$
Here the subscript $v$ means the value per volume and $\rho = N/v$.
I also devide equation \eqref{eq3} by the systems actual volume $v$ and calculate the total differential of the resulting equation.
$$
de_v = -dp + s_v dT + Tds_v+\rho d\mu + \mu d\rho \label{eq6} \tag{6}
$$
Using \eqref{eq5}, I immediately obtain
$$
de_v = T ds_v + \mu d\rho \label{eq7} \tag{7}
$$
So, the internal energy density seems to be independent of the deformation of the crystal. There is no term that could be understood as resembling the $de_v=\sigma d\epsilon$ term from equation \eqref{eq2}.
What am I missing????!!!!