Intuition about conservation of volume in phase space (Liouville's theorem)

I'm reading a text that says "Conservative dynamics in physical systems requires that volumes are exactly preserved". I'm assuming this means volumes in phase space, since this seems to me what Liouville's theorem is all about.

I have trouble understanding the intuition behind this. Suppose we take a spherical volume of point particles around the origin (in real space, not phase space). And suppose each of these has a momentum that points radially outwards. Then as that system evolves over time, the momentum of these particles would not change (assuming there are no forces acting on them), but their locations in real space would get ever further from the origin. Note that not all particles have the same absolute momentum, so some of them would move very slow, staying near the origin. In other words we have a gradually expanding ball filled with particles. If we look at this same system of particles in phase space, then nothing should happen to the coordinates of their momentums (since no forces are there to change it). But their locations are expanding outwards. Doesn't this then increase the volume in phase space? Clearly I'm interpreting something wrong, but I'm not sure where to look.

For clarity, I'm assuming a world of classical mechanics.

• Try one dimension first. You’ll see it. Commented Jul 4, 2021 at 11:29

Not a demonstration, but maybe something more visual than the other correct answers.

Instead of being in 2D like you suggest, let's try to see what happens in 1D when the particles move "radially" outward of $$x=0$$. Green balls go to the left at a constant momentum $$-p$$ and red balls go to the right at the same absolute momentum $$p$$ (I suppose they all have the same absolute velocity. The argument is way easier to explain this way... Even though you are asking for different momenta...)

In the phase space, this configuration gives at two different times:

The momentum doesn't change since there is no force (so same $$y$$ value whatever $$t$$ is) and the distance between two balls of the same color (in the real space) will be the same at any time (the distance between my 2 green balls et my 2 red balls is $$1$$ in my awful figure).

I also tried to draw the surface area (ie volume in $$d>2$$) in gray at time $$t=0$$ and at a later time $$t=a$$.

Now, the area of these parallelograms (the "volume" bounded by our 4 balls at $$t=0$$ or $$t=a$$) is of course the height of the parallelogram times the lenght of its base. It's trivial to see that on the figure, both parallelogram have then the same area. So the volume bounded by the 4 balls is constant although in real space, the balls move in opposite directions.

edit

It also works in the case of different momenta (even though, it's kind of harder to see (I changed the position of 2 balls so that the calculation of the area is easier)). We could have taken a different absolute momentum for each particle, but we would lose the symetry and thus, it would be hard to calculate the area (for example, taking one momentum equal to 0 like you suggest would make the conservation of the area hard to see visualy).

Yes, I plotted points but we're not talking about the volume of these points, we are talking about the volume bounded by a cloud of points (so 3 points are required).

We can also view this area as a density of probability, so even if in classical mechanics a particle is Dirac delta (it takes no volume) in the phase space, it's not really what the Liouville theorem is dealing with (at first at least). For example, we can say that we don't really know where our particle (with 0 volume) is, but we are sure that it's located in a tiny square (or parallelogram) of area $$A$$ in phase space with an uniform probability. Then, what we can say is, after some time, the particle will still be located inside a surface of area $$A$$ (but not necessarily a square).

This explanation is not very general, but kind of useful.

Now, we talk about conservation of volume because $$dpdx$$ at time $$t$$ is equal to $$dxdp$$ at time $$t'>t$$. This is equivalent to the assertion I mentioned above, namely: the probability distribution is constant along (so following the fluid) a classical trajectory in phase space. Where $$dx$$ and $$dp$$ are the horizontal and vertical lenghts of your fluid element (or your cloud of points, since these are the same).

At the end of the day, all you want is to define a volume, be it a probability density with your fluid elements mentionned in your comments, or an area bounded by points like in my figures, and show that this volume in conserved. The probability density/fluid approach is better understood using $$\frac{d\rho}{dt}=0$$ (see the wiki page), while the volume approach is better understood by seeing that $$dx(t)dp(t)=dx(t+dt)dp(t+dt)$$ because of the egality of the mixed derivative of the Hamiltonian. But they are both equivalent.

Still talking about your second comment, here is the "fluid approach" (my first figures take the "volume bounded by every points" approach, but I repeat both are equivalent)

So now, as you proposed, we take 2 fluids elements instead of 1 to describe 4 particles (it could be troublesome to do so but, let's assume nothing too serious happens at $$t>0$$).

What is important here, is that, at $$t = 0$$ you don't consider that you perfectly know the position of your particles, you say: well, i've got 2 particles inside the top ellipse and 2 inside the bottom ellipse, but I don't know where (you define a probability density $$\rho$$). Then, you proceed to move every points inside these volumes according to their corresponding equations of motion. This gives a new volume/area (on the figure these are the outermost ellipses at $$t=2$$), these new clouds of points/clouds of particles/densities of probability/fluid elements (containing 2 particles each) have the same area that what we started with. Note that in this case, we do not care about the volume between the fluid elements!

I feel like I went all over the place, so it might no be very understandable but I wanted to give a lot of examples.

• I really like the visuals, thx for this! But what if, on each side, we take two points with different momentum. For example one stationary particle and one with momentum p. In that case the parallelogram would expand, no? Commented Jul 4, 2021 at 14:35
• Looking at the answer by Andrew Steane perhaps this can also be solved by taking a fluid element of length dx instead of 2 particles. If we again assume that all parts of the fluid element travel at the same velocity, then the volume in phase space for 1 fluid element does indeed stay constant. If we take 2 fluid elements (instead of 2 particles) on each side, then the volume in phase space of each fluid element stays the same. But the volume in between them does change. Commented Jul 4, 2021 at 14:43
• The more I think about it, the more I'm starting to think that the fundamental problem with my question is that I'm thinking about point particles. Each of these particles has only 1 location and 1 momentum, and therefore takes up no volume in phase space. Any countable number of such points would still take up 0 volume. It's only when considering things like fluid elements that we start taking up any volume in phase space. What do you think? Is this the answer? Commented Jul 4, 2021 at 14:48
• i added something at the end of my answer (but not sure it will be very useful) Commented Jul 4, 2021 at 15:14
• Yes! I think the visuals plus the comparison between the particle case plus fluid element case is what finally made it click for me. Thank you. Commented Jul 4, 2021 at 18:24

Let's consider one particle in one-dimensional space; the general case just adds more terms. The theorem states the phase space density satisfies $$\rho_t+p\rho_q=0$$ (I've dropped the last term as you conserve $$p$$, and assumed unit mass so $$p=\dot{q}$$). This has general solution $$\rho=f(q-pt)$$, showing the blob (fluid element, fluid parcel, whatever term you prefer) in phase space drifts right at speed $$p$$, but doesn't deform or otherwise change volume. (Note also this is an advective derivative $$D_t\rho=0$$, which has the same "moves but doesn't reshape" intuition in fluid mechanics.)

1. Well, Liouville's theorem can be formulated as Hamiltonian vector fields in phase space are divergencefree.

Let us try to unpack the above statement. Consider an infinitesimal fluid parcel of $$m$$ points in the $$2n$$-dimensional phase space (i.e. imagine that we repeat the same experiment $$m$$ times with slightly different initial conditions). Then the fluid parcel has constant volume under time evolution.

2. Consider for simplicity a free particle. In $$n=1$$ dimension, the trajectories in the 2D phase are then horizontal lines, which are clearly divergencefree despite the position coordinate changes over time.

A free particle in higher dimensions $$n>1$$ can be viewed as a Cartesian product of $$n$$ 2D phase spaces. The time evolution is still divergencefree.

• I don't really see how this counters my example. Suppose indeed we take such a fluid parcel and place it at the origin. Let's say the first of the m points has no momentum. Then for each of the other m-1 points, let's indeed take slightly different initial conditions, namely a very small but nonzero momentum in a random direction. These m points would slowly scatter through space. So I don't understand how you can immediately conclude "Then the fluid parcel has constant volume under time evolution". In this example, could you point to what I'm getting wrong intuitively? Commented Jul 4, 2021 at 12:42
• I updated the answer. Commented Jul 4, 2021 at 13:43

It may help your intuition if instead of the concept of 'particles' you think in terms of the concept of 'fluid element'. A fluid element is a small volume of fluid, small enough such that all of its contents have the same velocity, and whose mass is conserved as the fluid evolves. If you like you could then regard such a fluid element as containing some fixed number of particles. Anyway as your example evolves you will find that each such fluid element gets drawn out into a line which becomes thinner as it becomes longer. This is easiest to see for motion in one dimension in the first instance (giving $$(x,p_x)$$ phase space) and then you can either try to picture it in more dimensions, or just trust the algebra.

• You state that all contents of the fluid element have the same velocity. To me this would imply that its shape is fixed. However, you also state that "each such fluid element gets drawn out into a line which becomes thinner as it becomes longer". To me these two statements cannot both be true. Or am I misunderstanding something? The idea of these fluid elements did give me some intuition to solve my original problem however. I will make another comment to explain how I understand it now. Could you check if my reasoning is correct? Thanks very much in advance! Commented Jul 4, 2021 at 12:51
• New attempt at understanding: In my original example, let's say we have a sphere of fluid elements centered around the origin at t=0. All elements touch, so the entire sphere is filled. Each of these elements has some (random) constant momentum that points radially outwards. As time evolves these elements will start to move outwards. However, because they move at different speeds, cracks will start to form between them. I.e. the sphere containing all elements will get bigger, but it will no longer be entirely filled up. Is this a correct way of solving my problem intuitively? Commented Jul 4, 2021 at 12:56
• @Willem better to let the fluid elements remain continuous with one another, without cracks between them, so that the volume of phase space filled by the system is equal to the total volume of the fluid elements. My statement about all particles in a fluid element having the same velocity has to be modified a little, I now see. It must be that their velocities are almost the same, differing by an amount that tends to zero as the size of the fluid element does. The idea is to allow the fluid to have continuity. Commented Jul 4, 2021 at 15:53