vertical axis, they also tend to diverge to plus or minus infinity. However, Svintsov et al. expression [@svintsov18] (with a correction of a sign error in the denominator), $$\begin{aligned}
\Pi^r_{(k_x, k_y \!=\! 0)}(\omega, v_1) &=& 
-\frac{2}{\pi} \frac{e^2 \mu}{(\hbar v_F)^2} \frac{1}{(1-s\beta)^2}\Bigg(\qquad\qquad\qquad \\
 &&\qquad \sqrt{1-\beta^2}  - \frac{s-\beta}{\sqrt{s^2-1}}\Bigg), \\
\beta &=& {\rm sgn}(k_x) \frac{v_1}{v_F}, \quad s = \frac{\hbar \omega + i\eta}{\hbar |k_x| v_F},\end{aligned}$$ agrees very well with a full lattice model calculation [@jia-huei17]. When the chemical potential is much larger than $k_b T$ and at $\omega > v_F k \to 0$, we have an excellent approximation for the equilibrium polarization $$\Pi^r_{\bf k}(\omega) \approx \frac{e^2 \mu}{\pi} \frac{k^2}{(\hbar \omega + i \eta)^2}.$$ Doppler shift of this expression in the small $v_1$ limit gives $\Delta \Pi^r = \Pi^r(v_1) - \Pi^r(0) \approx
2 D$, while Svintsov et al. expression in this limit is $-D$, where $D= (e^2 \mu/\pi)\hbar v_1 [k/(\hbar \omega + i \eta)]^3$. Both the sign and magnitude are different, contrary to the claim in ref. .

Possible connection to Coulomb drag\[app-c\]
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Since we expect that the drift velocity $v_1$ is a small quantity, we can make linear, or rather, quadratic response calculations. Taylor expanding in variable $\delta \omega = - k_x v_1$ of the 