\displaystyle
\frac{d\rho}{dt}
&amp; = &amp;
\displaystyle
- \frac{\partial H}{\partial \theta }
&amp; = &amp;
\displaystyle
- \frac{\partial T}{\partial \theta}
- \frac{\partial V}{\partial \theta}.
\end{array}
\]</span></p>
<p>With the momentum density being independent of the target density,
i.e., <span class="math inline">\(p(\rho | \theta) = p(\rho)\)</span>, the first term in the
momentum time derivative, <span class="math inline">\({\partial T} / {\partial \theta}\)</span> is
zero, yielding the pair 