 Solution

For the positive-frequency branch write $s=1+\delta$, $\delta >0$, and $Q=k^2{V}_e^2/(3{\omega }_{\mathrm{p}e}^2)={\kappa }^2/3$. The degenerate-plasma dielectric response is equivalent toAt large $Q$, a logarithmic divergence is needed to balance the right-hand side; hence $\delta \ll 1$. Keeping the constant as well as the large term givesMore precisely the relative correction is $O((1+Q)e^{-2(1+Q)})$. The additive constant $-2$ in the exponent fixes the leading prefactor $2e^{-2}$ and should not be discarded.

The negative-frequency branch follows from the even real dielectric function outside the occupied interval. ThusUsing the physical Thomas-Fermi screening length instead gives $\delta \sim 2e^{-2-2(k{\lambda }_{\mathrm{TF}}{)}^2}$. This is degenerate-plasma zero-point sound: the phase velocity lies exponentially close to, but strictly beyond, the edge of the Fermi sphere for every finite $k$ in this approximation. The limit is a property of the supplied classical-trajectory kinetic theory; applying it at microscopic wavelengths would require checking quantum recoil, which that equation omits.