framework, we can write $$f = f^0 - \frac{df^0}{d \epsilon} \Phi \approx f^0(\epsilon - \Phi),$$ here $\Phi \equiv \Phi_{\bf k}$ is mode ${\bf k}$ dependent, and $f^0 = 1/\bigl[\exp((\epsilon - \mu)/(k_bT)) + 1\bigr]$ is the equilibrium Fermi distribution at temperature $T$ and chemical potential $\mu$.

The effect of the current drift is to introduce anisotropy to the problem, thus we expect $\Phi$ should be angle and magnitude dependent, $\Phi_{\bf k} = \Phi(\theta, k)$, here $\theta$ is the angle between ${\bf k}$ and the drift velocity ${\bf v}_1$ and $k = | {\bf k}|$ is the magnitude of the wavevector. We make a Legendre polynomial expansion of the angular dependence and keep only the lowest non-trivial term, i.e., we write, $\Phi \propto \cos(\theta)$. It is convenient to assume $$\label{eqphik}
\Phi_{\bf k} =  \hbar k_x v_1 =  \hbar v_1 k \cos\theta.$$ If $\Phi$ does take this linear dependence on $k_x$, the nonequilibrium distribution might be transformed back to the equilibrium one of $f^0$ by a change of reference frame. That is, $$\label{noneq-eq}
\Pi^{>,<,r}_{{\bf k}, \rm noneq}(\omega) \approx \Pi^{>,<,r}_{{\bf k}, \rm eq}(\omega - k_x v_1).$$ In steady state, the fluctuation-dissipation like relation for photon self-energy can be obtained from Eq. (\[noneq-eq\]) even if 