# What does $\frac{d\rho}{dt} =0$ but $\frac{\partial \rho}{\partial t} \neq 0$ mean intuitively?

For concreteness, let's say that $$\rho(q,p,t)$$ describes the probability density in phase space.

On a superficial level both $$\frac{d\rho}{dt}$$ and $$\frac{\partial \rho}{\partial t}$$ tells us how $$\rho$$ changes as time passes on.

But for $$\frac{\partial \rho}{\partial t}$$ we keep $$q=q(t)$$ and $$p=p(t)$$ fixed and only measure how $$\rho$$ changes "itself". So in a sense, for $$\frac{\partial \rho}{\partial t}$$ we stay at one particular location $$(q,p)$$ and observe how $$\rho$$ then changes at this particular location as time passes on. Alternatively, in order to calculate $$\frac{\partial \rho}{\partial t}$$ we can imagine that phase space remains frozen and we then observe if $$\rho$$ still changes.

In contrast, for $$\frac{d\rho}{dt}$$ we take into account that $$q=q(t)$$ and $$p=p(t)$$ get dragged along as time passes on, too. (Phase space, in contrast to spacetime, is not a static stage.)

So in a sense, for $$\frac{\partial \rho}{\partial t}$$ we consider $$\rho$$ as some kind of background field and observe how it changes at each particular point. But for $$\frac{d\rho}{dt}$$, $$\rho$$ is an active agent which also gets dragged along in phase space through the Hamiltonian vector field because the phase space points themselves get moved around. (This is discussed nicely, for example, here, here and here.)

Formulated differently, $$\frac{d\rho}{dt}$$ captures the total change which is due to two effects: the change in $$\rho$$ itself as measured by $$\frac{\partial \rho}{\partial t}$$ and the change which is due to phase space points getting moved around.

But then, how can we understand a situation for which $$\frac{d\rho}{dt} =0$$ and $$\frac{\partial \rho}{\partial t} \neq 0$$ specifically for a situation in which $$\rho$$ describes a probability density in phase space.? (We have such a situation for Classical Mechanics systems. This is what Liouville's equation tells us.)

• I don't really understand your question. You explain what each derivative means, so how are you confused on this situation? It seems like you have a good grasp on everything already. – Aaron Stevens Mar 6 at 12:50
• Perhaps the same way you understand the Lagrangian comoving of your fleet as it drifts down the river? The density of boats may remain the same, while the shape of your fleet cluster changes. – Cosmas Zachos Mar 6 at 16:02
• @CosmasZachos Thanks, the boat analogy is really helpful! – JakobH Mar 7 at 12:14

Not sure whether this is what you are looking for, but here goes.

$$\partial\rho/\partial t=0$$ is the condition for thermodynamic equilibrium, which is of course separate from anything that comes directly out of the equations of motion. When the equilibrium condition is met, the net inflow to any small subvolume in phase space is equal to the net outflow. Hence the distribution $$\rho(q,p,t)$$ becomes $$\rho_{\text{eq}}(q,p)$$. Throughout the whole process of reaching equilibrium, from some arbitrary initial distribution $$\rho(q,p,0)$$, the Liouville equation is obeyed: $$d\rho/dt=0$$.

So, for example, imagine a dilute gas, initially confined to one half of a container; then the constraint is released and the gas expands to fill the whole volume. The initial distribution is clearly non-equilibrium: let's say the Maxwell-Boltzmann distribution of momenta, but non-uniform distribution of positions with respect to the whole volume. We are confident that it will evolve in time, and eventually become a uniform distribution in positions (while the Maxwell-Boltzmann distribution remains unaltered). This happens naturally, as the atoms all move at different velocities and get out of phase with each other. So, $$\rho(q,p,t)$$ in a small region around fixed values of $$q$$ and $$p$$, will change in time. Still, throughout the process, if you follow the flow in phase space, it will always satisfy $$d\rho/dt=0$$ (going along with the flow).

I'm not trying to confuse the density in phase space with the average density of particles per unit volume in 3D, of course, but the latter can be expressed in terms of the former, by summing over all the atoms and integrating over the momenta, and gives a physical indication of whether the phase space distribution is changing or not.

In Classical Mechanics (unlike in Quantum Mechanics) there is nothing which mixes probabilities around.

In particular, this means if we start with an initial state $$A$$ which occurs with a probability of $$70$$% at $$t=t_0$$, at some later point in time $$T=t_1$$ the corresponding final state $$\tilde A$$ will occurs with a probability of $$70$$%, too.

This means, if we start with a fixed set of possible initial states and specific probabilities for each of them, the corresponding final states will each occur with exactly the same probabilities. Mathematically:

\begin{align} P(A,t_0) &= 70 \% \notag \\ P(B,t_0) &= 10 \% \notag \\ P(C,t_0) &= 5 \% \notag \\ & \ \ \vdots \tag {1} \end{align} And at a later point in time \begin{align} P(\tilde A,t_1) &= 70 \% \notag \\ P(\tilde B,t_1) &= 10 \% \notag \\ P(\tilde C,t_1) &= 5 \% \notag \\ \tag {2} & \ \ \vdots \, , \end{align} where $$\tilde A, \tilde B, \tilde C$$ are the phase space points we get by evolving our initial points $$A,B,C$$ using Hamilton's equations. Most importantly, this implies \begin{align} P(A,t_1) &\neq 70 \% \notag \\ P(B,t_1) &\neq 10 \% \notag \\ P(C,t_1) &\neq 5 \% \notag \\\tag {3} & \ \ \vdots \end{align}

And this is exactly what $$\frac{d\rho}{dt} =0$$ but $$\frac{\partial \rho}{\partial t} \neq 0$$ means in the context of probability densities.

The statements from above in terms of the phase space probability density read $$\begin{equation} (1) \neq (3) \quad \leftrightarrow \quad \rho(q,p,t_0) \neq \rho(q,p,t_1) \end{equation}$$ because, for example, state $$A\equiv q_A,p_A$$, will have moved to a new location $$\tilde{A}\equiv Q_A,P_A$$. In terms of derivatives, this means $$\frac{\partial \rho}{\partial t} \neq 0$$.

But in contrast, we have $$\begin{equation} (1) = (2) \quad \leftrightarrow \quad \rho(q,p,t_0) = \rho(Q,P,t_1) \end{equation}$$ because the probabilities remain attached to their points as they move through phase space.

Here \begin{align} Q&=q + \epsilon \{H,q\} \notag \\ P&= p + \epsilon \{H,p\}\, \notag . \end{align}

The state described by $$A\equiv q_A,p_A$$ has now evolved into the state described by $$\tilde A \equiv Q_A,P_A$$. Therefore, the probability to observe this state at $$t_1$$ is exactly the same as the probability to observe the state corresponding to $$A$$ at $$t_0$$. And in terms of derivatives, this means $$\frac{d \rho}{d t} = 0$$.

Take note that this does not imply that our uncertainties are constant. Even if the initial states are close together (=small uncertainties), the final states can spread around wildly in phase space (=large uncertainties). (This is known as filamentation.)