when $\gamma_{\mu\nu}=0,\ \forall\mu,\nu$, the FPE (\[eq:4\]) becomes the Liouville equation initially derived by @condiff_brownian_1966[@condiff_brownian_1966; @condiff_transport_1969; @dahler_kinetic_1963] for molecular fluids consisting of non-spherical particles. In this case one usually invokes a Kirkwood-type time smoothing technique[@condiff_brownian_1966; @kirkwood_statistical_1946], rather than a local-equilibrium approximation involving a free-energy functional (see Section \[sec:extended-unified-DDFT\]).

We now define the reduced distributions obtained by integration of the phase-space PDF, $$\begin{aligned}
 {f^{(n)}}({\boldsymbol{\mathfrak{r}}}^n,{\boldsymbol{\mathfrak{p}}}^n;t)=\frac{N!}{(N-n)!} \int d{\boldsymbol{\mathfrak{r}}}^{N-n}d{\boldsymbol{\mathfrak{p}}}^{N-n}\,{f^{(N)}}({\boldsymbol{\mathfrak{r}}}^N,{\boldsymbol{\mathfrak{p}}}^N;t)
\label{eq:7}\end{aligned}$$ where $d{\boldsymbol{\mathfrak{r}}}^{N-n}\equiv d{\boldsymbol{\mathfrak{r}}}_{n+1}\dots d{\boldsymbol{\mathfrak{r}}}_N$, with an analogous expression for $d{\boldsymbol{\mathfrak{p}}}$, and the corresponding configuration-space distribution, $$\begin{aligned}
{\rho^{(n)}}({\boldsymbol{\mathfrak{r}}}^n;t) = \int d{\boldsymbol{\mathfrak{p}}}^n\,{f^{(n)}}({\boldsymbol{\mathfrak{r}}}^n,{\boldsymbol{\mathfrak{p}}}^n;t).
 \label{eq:8}\end{aligned}$$ We denote the domains of integration by $\mathfrak{V}=\mathcal{V}\times\mathcal{S}$, with $\mathcal{V}=\mathbb{R}^3$ and $\mathcal{S}=[0,2\pi)\times[0,\pi)\times[0,2\pi)$, and so $\int d{\boldsymbol{\mathfrak{r}}}\equiv \int_{\mathcal{V}}d\mathcal{V}\int_{\mathcal{S}}d\mathcal{S}$ and $\int d{\boldsymbol{\mathfrak{p}}}\equiv \int_{\mathcal{V}}d\mathcal{V}\int_{\mathcal{V}}d\mathcal{V}$, where $$\begin{aligned}
 \int_{\mathcal{V}}d\mathcal{V}=& \int_{0}^\infty dx \int_{0}^\infty dy \int_{0}^\infty dz \label{eq:9}\\
 \int_{\mathcal{S}}d\mathcal{S}=&\int_{0}^{2\pi}d\theta \int_{0}^{\pi}d\phi\, \sin(\phi)\int_{0}^{2\pi} d\chi.
 \notag\end{aligned}$$ For ease of notation $\rho^{(1)}$ is replaced with $\rho$, also known as the one-body density. We now take successive momentum moments of Equation (\[eq:4\]) starting at zeroth order and integrating over all but one-particle position and momentum, i.e. $N \int \text{(\ref{eq:4}) } d{\boldsymbol{\mathfrak{r}}}^{N-1} d{\boldsymbol{\mathfrak{p}}}^{N}$, resulting in the continuity equation $$\begin{aligned}
 \partial_t\rho({\boldsymbol{\mathfrak{r}}}_1;t) + {\boldsymbol{\nabla}}_{{\boldsymbol{\mathfrak{r}}}_1}\cdot{\boldsymbol{\mathfrak{j}}}({\boldsymbol{\mathfrak{r}}}_1;t) = 0,
 \label{eq:10}\end{aligned}$$ with the flux defined as ${\boldsymbol{\mathfrak{j}}}({\boldsymbol{\mathfrak{r}}}_1;t)=\int d{\boldsymbol{\mathfrak{p}}}\,({\boldsymbol{\mathfrak{m}}}^{-1}{\boldsymbol{\mathfrak{p}}}_1)f^{(1)}({\boldsymbol{\mathfrak{r}}}_1,{\boldsymbol{\mathfrak{p}}}_1;t)$.

To obtain the first-momentum moment equation we use $N\int ({\boldsymbol{\mathfrak{m}}}^{-1}{\boldsymbol{\mathfrak{p}}}_1)\otimes \text{(\ref{eq:4}) }d{\boldsymbol{\mathfrak{r}}}^{N-1}d{\boldsymbol{\mathfrak{p}}}^N$ giving $$\begin{aligned}
 \partial_t{\boldsymbol{\mathfrak{j}}}({\boldsymbol{\mathfrak{r}}}_1;t)&+{\boldsymbol{\nabla}}_{{\boldsymbol{\mathfrak{r}}}_1}\cdot\int 