# Conservation of probability in phase space flow

In J.Binney's notes on classical mechanics, under the section 'Liouville's theorem', he states that (paraphrasing):

the conservation of probability requires that $\frac{df}{dt} = 0.$

where $f$ is the phase space density.

I'm not sure why the probability is conserved. If we fix some initial phase space region $V_0$, which in time $t$ will become $V_t$, the probability of the system occupying that region is given by $$P_t = \int_{V_t} dxdp \ f(x,p;t).$$ But why is this constant in time? I know that the phase space volume is conserved, so at least the volume of the region won't change, but its shape will. What allows me to conclude that the function $f$ evolves in time in such a way that $P_t$ in fact remains constant?

• The answer to your question is the proof of Liouville's theorem. – Robin Ekman Jan 17 '15 at 12:21
• also time-evolution is linked to $\{H,f\}$, since the Hamiltonian is the infinitesimal generator – Phoenix87 Jan 17 '15 at 12:24
• @RobinEkman well, the proof I've seen (damtp.cam.ac.uk/user/tong/dynamics/four.pdf page 88) is not very illuminating to me, that 'version' of the theorem is that the volume is conserved. But I don't know how to prove the probability is, as well. It goes on to say "...since the volume elements $dxdp$ are preserved, we have $df/dt =0$". I don't see why. – Spine Feast Jan 17 '15 at 12:33
• @DepeHb did you see my answer? Liouville's theorem is Gauss like - see my answer. It's not that volume is conserved, it's that the content of the volume, if not constant, then there is a flux. – Sofia Jan 17 '15 at 12:46
• Well, by volume I don't mean a particular region of phase space, but rather its..well..volume, as in, measure, as it travels around in phase space. Volume in this sense is conserved. – Spine Feast Jan 17 '15 at 12:51

## 1 Answer

No, no, you miss something. Let me show you in a simple way. Since your density function $f$ is a function of $x, p$ and $t$, its total derivative by time looks as follows,

(1) $\frac {df}{dt} = \frac {∂f}{∂t} + \frac{∂f}{∂x} \frac {dx}{dt} + \frac{∂f}{∂p} \frac {dp}{dt}$.

Do you agree? This is what says Liouville's theorem, i.e. that if the left-hand-side (LHS) is zero, then the right-hand-side (RHS) is zero.

Now let me integrate (1) on both sides

(2) $\frac {d}{dt} \int_{V_t} f(x, p; t) dx dp = \frac {∂}{∂t} \int_{V_t} f(x, p; t) dx dp + \int_{V_t} \frac{∂f(x, p; t)}{∂x} \frac {dx}{dt} dx dp + \int_{V_t} \frac{∂f(x, p; t)}{∂p} \frac {d p}{dt} dx dp$

$= \frac {∂}{∂t} \int_{V_t} f(x, p; t) dx dp + \int_{V_t} \nabla f(x, p, t) \ \vec v \ dx dp$

Here, you apply a variant of the Divergence theorem, in fact, the energy conservation: the change of the total probability in a volume, due to transport with velocity $\vec v$ is equal to the total flux through the surface of the volume

$\int_{V_t} (\nabla f) \ \vec v \ dpdx = \int_{S_t} f \vec v$ • $\vec n \ dS$,

In all, we have similarity with Gauss' theorem : nothing "disappears". If the amount of energy confined to some volume decreases, then there is some outwards flux. Alternatively, if it increases, then some inwards flux exists there.

(I notice that you speak of a volume $V_t$ and I went with that, but in 1 dimension you have a surface, and my $S_t$ is then a contour - i.e. the flux is through the contour of the surface.)