I was reading my professor's slides (a sort of Introduction to Statistical Mechanics) and unfortunately, they do not seem to be as clear as I'd want them to be, therefore I've come here for help.

After some introductory statements, a probability density function $\rho(\mathbf{q,p},t), \mathbf{q,p} = q_1,....,q_n,p_1,...,p_n$ is defined as follows (I'll drop the vector signs when obvious):

An ensemble is represented by a distribution of several points in the phase space $\Gamma$, usually a continuous distribution. The ensemble can be described by a probability function $\rho(\mathbf{q,p})$, defined in such a way that the quantity: $$\rho d\mathcal{\mathbf{V}}, d\mathbf{V} = d\mathbf{q}d\mathbf{p}$$ tell us "how much" the points representative of the system, which at a given time $t$ are contained in the infinitesimal volume, contribute.

Then, after some derivations, the continuity equation is presented as a consequence of the necessary existence of a conservation law that regulates $\rho$: $$\frac{\partial \rho(q,p,t)}{\partial t} = -\nabla \cdot (\rho \mathbf{v}) \tag{1}$$ where $\mathbf{v}$ " compactly denotes the time-derivatives of the variables associates with $\rho$"

It is then said that, from $(1)$, one easily obtains the following equation: $$ -\frac{\partial \rho(q,p,t)}{\partial t} = \sum_i \left[ \frac{\partial ( \rho \dot{q_i})}{\partial q_i}+ \frac{\partial (\rho \dot{p_i} )}{\partial p_i} \right]$$

I was unsuccessful in computing this final result, mainly because I do not know how to interpret the divergence operator when applied to quantities depending on generalized coordinates and conjugated momenta, and I'm not sure if $\mathbf{v}$ can be seen as a sort of $\{\dot{\mathbf{q}}, \dot{\mathbf{p}}\}$. I'd gladly accept any help in deriving it.

  • 1
    $\begingroup$ There appears to be a typo on the right side of the last equation. Derivatives should be taken with respect to $q_i$ and $p_i$, and not with respect to $t$. $\endgroup$
    – Gec
    Oct 7, 2023 at 20:32
  • $\begingroup$ Yes, Thanks for making me notice the typo $\endgroup$ Oct 7, 2023 at 21:12

1 Answer 1


There is no real derivation going on, it's just a matter of interpreting equation (1), as you rightly notice. Maybe you'd have to go back to where your professor introduced the divergence in phase space or the velocity $\vec{v}$ in (1), check definitions, and provide more info here if you would like a more detailed answer than the following:

Are you familiar with the continuity equation in fluid dynamics or electrodynamics, $$ \partial_t \rho + \nabla \cdot (\rho \vec{v}) = 0\,, $$ which would typically be used in a space with coordinates $\vec{x} = (x_1, x_2, x_3)$? Write it out in components, noting that the particle velocity is $\vec{v} = \dot{\vec{x}}$, thus $v_i = \dot{x}_i$, and you get $$ \partial_t \rho + \sum_i \frac{\partial}{\partial x_i}\left(\rho \dot{x}_i\right)\,. $$ Now, if instead of in those coordinates you are using a continuity equation in a space with coordinates $(\vec{q},\vec{p}) = (q_1,...,q_n,p_1,...,p_n)$ where the velocity is $\vec{v} = (\dot{q}_1,...,\dot{q}_n,\dot{p}_1,...,\dot{p}_n)$ you get... You see the analogy?

The divergence operator is just $$ \nabla \cdot \vec{A} = \sum_i \frac{\partial}{\partial x_i} A_i $$ and in phase space nothing has changed except the coordinate names.

  • $\begingroup$ It was much simpler than expected.I just needed to consider generalized velocities. And yes I’ve already encountered in electrodynamics, in fact the way I derived the formula for the statistical case is basically identical to the one I learned in Griffiths’s textbook, done in the 2n dimensional phase space $\Gamma$ (I asked the professor if this was viable and the answer was positive) $\endgroup$ Oct 8, 2023 at 12:57

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