$i$ [@Butcher; @CA]. Basically, the idea is to write the particle current, which is given by with $e=1$, as well as the energy current, which is also given by with $e=1$ but with an extra $E$ in the integrand, and then to invoke the first law of thermodynamics $\delta Q_i = dE_i - \mu_i
  dN_i$. One finds $$\label{Current Heat}
J_i = \frac{1}{h} \sum_{j} \int_{0}^{\infty} \left[t_{ji}(E) f_i(E) - t_{ij}(E) f_j(E)\right] (E-\mu_i) dE~.$$

[As we shall see in the linear response analysis, if one considers $\delta \mu_i / e$ and $\delta T_i / T$ as the thermodynamic forces, then $I_i$ and $J_i$ turn out to be the right currents to obtain the Onsager relations.]{.nodecor}

Although the expressions – for the electric and heat currents have a direct physical interpretation in terms of $t_{ij}(E)$, we prefer to work in terms of the quantities $$\Gamma_{ij}(E) = M_i
\delta_{ij} - t_{ij}(E)~,$$ where $\delta_{ij}$ denotes the Kronecker delta. For this, we use the relations to rewrite the integrand in as follows: $$\begin{aligned}
\sum_{j} \left[f_i(E) t_{ji}(E) - f_j(E) t_{ij}(E)\right] &=& f_i(E) M_i - \sum_j f_j(E) t_{ij}(E) \nonumber\\
&=& \sum_j f_j(E) \underbrace{\left(M_i \delta_{ij} - t_{ij}(E)\right)}_{= \
  \Gamma_{ij}(E)}~.\end{aligned}$$ Hence, one can rewrite the currents in the following form ($i\in 