= Solution
Define the local <mean velocity> and <momentum flux> by
$$
\boxed{\rho\mathbf u=\int\mathbf v f\,d^3v,\qquad
\Pi_{ij}=\int v_iv_j f\,d^3v.}
$$
Here $\rho=\int f\,d^3v$ and $\mathbf u$ is defined where $\rho>0$. The <pressure tensor> separates random motion from the bulk <velocity>:
$$
\Pi_{ij}=\rho u_i u_j+P_{ij},\qquad
P_{ij}=\int(v_i-u_i)(v_j-u_j)f\,d^3v.
$$
Integrating the <mean-field alignment equation> over <velocity space> removes the force <divergence> and gives the <continuity equation>
$$
\boxed{\partial_t\rho+\nabla_{\mathbf x}\cdot(\rho\mathbf u)=0.}
$$
Its first <velocity moment> follows by multiplying by $v_i$ and using $\int v_i\partial_{v_j}(fL_j)\,d^3v=-\int fL_i\,d^3v$:
$$
\partial_t(\rho u_i)+\partial_{x_j}\Pi_{ij}=\int fL_i[f]\,d^3v.
$$
Inserting the <mean-field alignment equation>'s force and integrating each particle <velocity> separately gives
$$
\boxed{\partial_t(\rho\mathbf u)+\nabla_{\mathbf x}\cdot\boldsymbol\Pi
=\int K(\mathbf x,\mathbf x_\star)\bigl[\mathbf u(\mathbf x_\star,t)-\mathbf u(\mathbf x,t)\bigr]\rho(\mathbf x_\star,t)\rho(\mathbf x,t)\,d^3x_\star.}
$$
The right-hand side exchanges <momentum> between positions and aligns their bulk <velocities>. Its spatial <integral> vanishes by exchange <symmetry>, recovering total <momentum conservation>. No closure for the <pressure tensor> is needed for these exact <velocity moment> identities.
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