n_B}{2\pi \hbar^3 k_F^2} \right ) \left ( \frac{2\pi e^2}{\kappa} \right )^2
\frac{2\pi}{s^2},
\label{eq26}$$ $$I^{(B)}_{q} = \left ( \frac{m n_B}{2\pi \hbar^3 k_F^2} \right ) \left ( \frac{2\pi e^2}{\kappa} \right )^2
\frac{2\pi}{s^2},
\label{eq27}$$ with $\tau_{t,q}^{-1} = I_{t,q}^{(R)} + I_{t,q}^{(B)}$. Equations (\[eq24\]) – (\[eq27\]) provide us with approximate analytical expressions for the contributions $I_{t,q}^{(R)}$ and $I_{t,q}^{(B)}$ by the remote and the background scattering respectively to the transport and quantum scattering rates. We note that in these asymptotic limits ($a_R \gg 1$ and $s \gg 1$) $$\frac{I_t^{(B)}}{I_t^{(R)}} = 3 \pi a_R^3 \frac{n_B}{n_R},
\label{eq29}$$ and $$\frac{I_q^{(R)}}{I_q^{(B)}} = \frac{1}{2\pi a_R}\frac{n_R}{n_B}.
\label{eq30}$$ In addition, $${I_q^{(R)}}/{I_t^{(R)}} = 3 a_R^2/2,
\label{eq31}$$ $${I_q^{(B)}}/{I_t^{(B)}} = 1.
\label{eq32}$$ We note, therefore, that $I_q^{(R)} \gg I_t^{(R)}$ (since $a_R \gg 1$) as is already well established, and that $I_t^{(B)} > I_t^{(R)}$ if $3\pi a_R^3 > n_R/n_B$, whereas $I_q^{(R)} > I_q^{(B)}$ if $n_R/n_B >
2 \pi a_R$. Consistency demands that $3\pi a_R^3 \gg 2 \pi a_R$, i.e., $a_R^2 \gg 2/3$, which is guaranteed since $a_R \gg 1$.

It is thus possible for the $B$-scatterers to dominate $\tau_t^{-1}$, i.e., $I_t^{(B)} \gg I_t^{(R)}$, and $R$-scatterers to dominate $\tau_q^{-1}$, i.e., $I_q^{(R)} \gg I_q^{(B)}$ if the following conditions are both satisfied $$\begin{aligned}
3 \pi a_R^3 & \gg & n_R/n_B, 