= Solution
{c}
The shifted <Fermi circle> satisfies $|\mathbf p-p_0\hat{\mathbf x}|=p_F$. Its radial boundary at polar angle $\theta$ is the positive solution of $p^2-2pp_0\cos\theta+p_0^2=p_F^2$. Thus, for $p_0<p_F$,
$$
p_F+\nu(\theta)=p_0\cos\theta+\sqrt{p_F^2-p_0^2\sin^2\theta},
$$
and its leading <Taylor expansion> is
$$
\boxed{\nu(\theta)=p_0\cos\theta+O(p_0^2/p_F).}
$$
Forward-pointing <momenta> gain occupied states and backward-pointing <momenta> lose them. With the paper's explicitly reversed <Heaviside step function> convention, $\Theta'(x)=-\delta(x)$; consequently
$$
\delta n(\mathbf p)=\nu(\theta_{\mathbf p})\delta(p-p_F)+O(p_0^2),
$$
with a positive leading sign.
For the later quadratic <energy> calculation it is useful to distinguish the first-order shape from exact <particle number> conservation. The <particle-conserving displacement of a Fermi circle> has
$$
\nu(\theta)=p_0\cos\theta-\frac{p_0^2}{2p_F}\sin^2\theta+O(p_0^4/p_F^3),\qquad
\delta N=C\int_0^{2\pi}\left[p_F\nu+\frac{\nu^2}{2}\right]d\theta=0.
$$
The second-order inward correction cancels the apparent area increase one would obtain by treating $\nu=p_0\cos\theta$ as exact.
\Image[/past-exam-of-the-quantum-matter-course-of-the-university-of-oxford/2023-boosted-fermi-circle.png]
{title=Translation of a <Fermi circle> and its radial displacement}
{description=Original geometric construction with $p_0/p_F=0.25$. The <Fermi sea> keeps its area under translation; the first-order radial approximation differs from the exact boundary at second order.}
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