= Solution
For the <Bogoliubov spectrum>, define $\epsilon_k=\hbar^2k^2/(2m)$ and $E_k=\sqrt{\epsilon_k(\epsilon_k+2\mu_0)}$. For repulsive interactions and $\mu_0>0$,
$$
E_k\simeq\hbar c_sk\quad(k\to0),\qquad
c_s=\sqrt{\mu_0/m},\qquad
E_k\simeq\epsilon_k+\mu_0\quad(\epsilon_k\gg\mu_0).
$$
Thus the low-momentum excitations are gapless <phonons> and the high-momentum branch approaches free-particle curvature. The crossover occurs when kinetic and interaction <energy> are comparable. The <Landau criterion> follows because an obstacle moving at <velocity> $v$ can supply <energy> $\mathbf v\cdot\mathbf p$ to an excitation. Here
$$
\frac{E_k}{\hbar k}=\sqrt{\frac{\mu_0}{m}+\frac{\hbar^2k^2}{4m^2}},\qquad
\boxed{v_c=\inf_{k>0}\frac{E_k}{\hbar k}=c_s.}
$$
This is the homogeneous <phonon> kinematic threshold, not a guarantee that every perturbation dissipates immediately above it.
\Image[/past-exam-of-the-atomic-and-optical-physics-course-of-the-university-of-cambridge/2013/paper-1-aop-bogoliubov-stirring.png]
{title=Bogoliubov phonon-to-particle crossover and the local stirring threshold during slow atom loss}
In two dimensions, $n=|\psi|^2$ has units of inverse <area> and $\hbar^2/m$ has units of <energy> times <area>. Therefore \b[$\tilde g$ is dimensionless]. Neglecting the kinetic term in the <Gross–Pitaevskii equation>, with positive coupling, gives
$$
\boxed{n_0=\frac{m\mu_0}{\hbar^2\tilde g},\qquad
R^2=\frac{2\mu_0}{m\omega_r^2},\qquad
n(r)=n_0(1-r^2/R^2)\quad(0\leq r\leq R),}
$$
with $n=0$ outside the radius in the <Thomas–Fermi approximation for a condensate>. Multiplication gives
$$
\boxed{n_0R^2=\frac2{\tilde g}\left(\frac{\mu_0}{\hbar\omega_r}\right)^2.}
$$
The sharp edge is an approximation; the neglected gradient term rounds it over a <healing length> scale.
Using the <local sound-speed criterion for circular condensate stirring>, the moving beam has <speed> $v=\Omega D$. The local interaction <chemical potential> is $\mu_0-V(D)=m\omega_r^2(R^2-D^2)/2$, so
$$
c(D)=\frac{\omega_r}{\sqrt2}\sqrt{R^2-D^2}.
$$
The onset equality is $\Omega D=c(D)$ and the permitted excitation condition is
$$
\boxed{\frac DR>\frac{\omega_r}{\sqrt{\omega_r^2+2\Omega^2}}.}
$$
This uses the beam <speed> as the local relative-flow scale. A finite hole can distort the surrounding flow or nucleate <vortices>, so the result is the requested local <phonon> estimate, not a universal necessary condition for all experimental stirring.
For the <loss-driven onset of condensate stirring>, first integrate the density:
$$
N=2\pi n_0\int_0^Rr(1-r^2/R^2)\,dr
=\frac\pi2n_0R^2
=\frac\pi{\tilde g}\left(\frac{\mu_0}{\hbar\omega_r}\right)^2.
$$
At fixed coupling and trap <frequency>, $\mu_0\propto N^{1/2}$ and $R\propto N^{1/4}$. Hence $R(t)=R(0)e^{-t/(4\tau)}$ while $D$ stays fixed. For $\Omega=\omega_r/\sqrt2$, threshold requires $D/R=1/\sqrt2$. Starting at $D/R(0)=1/2$,
$$
\frac12e^{t_c/(4\tau)}=\frac1{\sqrt2},\qquad
\boxed{t_c=2\tau\ln2\simeq1.386\tau.}
$$
At onset, $N=N_0/4$, $R=R(0)/\sqrt2$ and the beam is still inside the cloud. Excitations become kinematically allowed just beyond this equality, provided the stipulated adiabatic regime of the <Thomas–Fermi approximation for a condensate> remains valid.
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