# Why the form factors on the green function of relativistic two-component liquid model for superfluid have this form?

I am working in trying to understand Appendix A of arXiv:0906.4810. On this appendix, the authors set up an external velocity field perturbating a superfluid, described by a two-component model. This field will couple to the momentum density $T^{0 i}$ and the perturbation on it can be estimated using linear-response theory by $\delta T^{0i}(0,\vec{k}) = U_j(0,\vec{k}) G^{ij}(0,\vec{k})$. They then argue that due to the rotational symmetry $$G^{ij}(0,\vec{k}) = \frac{k^i k^j}{\vec{k}^2} G^\parallel(\vec{k}^2) + \left(\delta^{ij} - \frac{k^i k^j}{\vec{k}^2}\right) G^\bot(\vec{k}^2)\,.$$

The argument follows stating that if $\vec{U} \parallel \vec{k}$, both components will answer, but if $\vec{U} \bot \vec{k}$ only the viscous component will answer. He then jumps to state that

$$\lim_{\vec{k}\to 0} G^\bot(\vec{k}^2) = -(sT + \mu \rho_n) \quad \text{and} \quad \lim_{\vec{k}\to 0} G^\parallel(\vec{k}^2) = -(sT + \mu \rho)\,,$$ where $\rho_n$ is the density of the normal component and $\rho$ is the total density.

I can see why is a good idea to decompose the Green function in components parallel and perpendicular to $\vec{k}$ and understands why only the viscous component answers to the perturbation when $\vec{U} \bot \vec{k}$. However, I cannot see why the expressions for the form factors $G^\bot$ and $G^\parallel$ on the limit $\vec{k} \to 0$ are the ones the authors state. Why they are true? Also, why the minus sign? I expected that the liquid would gain momentum on the direction of $\vec{U}$, not on the direction of $-\vec{U}$.