= Solution
Put $z_i=(\mathbf x_i,\mathbf v_i)$ and $dz_i=d^3x_i\,d^3v_i$. Normalize the symmetric $N$-particle <phase-space distribution function> $D_N$ to one. Use factorial normalization for its one-particle marginal and <two-particle distribution function>:
$$
f(z_1)=N\int D_N\,dz_2\cdots dz_N,\qquad f_2(z_1,z_2)=N(N-1)\int D_N\,dz_3\cdots dz_N.
$$
Thus $\int f\,dz=N$ and $\int f_2(z_1,z_2)\,dz_2=(N-1)f(z_1)$. For unit particle <mass>, $f$ also carries <mass density>. These conventions explain why no extra factor $N-1$ occurs in the result.
Separate the <Hamiltonian> as $H=H_1+H_{\mathrm{rest}}+H_{\mathrm{int}}$, where
$$
H_1=\frac12|\mathbf v_1|^2,\qquad H_{\mathrm{int}}=\sum_{j=2}^N\phi(|\mathbf x_1-\mathbf x_j|),
$$
and $H_{\mathrm{rest}}$ contains all terms involving only particles $2,\ldots,N$. The stated <Poisson bracket> gives <Hamilton's equations>
$$
\dot{\mathbf x}_i=\mathbf v_i,\qquad \dot{\mathbf v}_i=-\sum_{j\ne i}\nabla_{\mathbf x_i}\phi(|\mathbf x_i-\mathbf x_j|).
$$
Conservation of probability along the <Hamiltonian flow> gives the <Hamiltonian Liouville equation>
$$
\partial_tD_N+\sum_{i=1}^N\left[\mathbf v_i\cdot\nabla_{\mathbf x_i}D_N-\sum_{j\ne i}\nabla_{\mathbf x_i}\phi_{ij}\cdot\nabla_{\mathbf v_i}D_N\right]=0.
$$
Assume decay of the required <velocity moments> and either decay or periodic <boundary conditions> in position. Integrating over $z_2,\ldots,z_N$ eliminates the $H_{\mathrm{rest}}$ terms by <integration by parts>. The terms in $H_{\mathrm{int}}$ differentiating an integrated <velocity> also vanish. Indistinguishability makes the remaining $N-1$ terms equal, so
$$
\begin{aligned}
\partial_tf+\mathbf v_1\cdot\nabla_{\mathbf x_1}f
&=N\sum_{j=2}^N\int\nabla_{\mathbf x_1}\phi_{1j}\cdot\nabla_{\mathbf v_1}D_N\,dz_2\cdots dz_N\\
&=\int dz_2\,\nabla_{\mathbf x_1}\phi_{12}\cdot\nabla_{\mathbf v_1}f_2.
\end{aligned}
$$
Relabeling the two particles gives the first <BBGKY hierarchy> equation:
$$
\boxed{\partial_tf+\mathbf v\cdot\nabla_{\mathbf x}f=\int d^3x_\star\,d^3v_\star\,\nabla_{\mathbf v}f_2\cdot\nabla_{\mathbf x}\phi(|\mathbf x-\mathbf x_\star|).}
$$
This is exact for a differentiable <pair potential>; no independent-particle closure has been made.
Back to article page
