# Challenging Susskind's uncompressible flow in Liouville’s Theorem

1. Oct 19, 2013

### VRT

Challenging Susskind's "uncompressible flow" in Liouville’s Theorem

It’s in Susskind’s 7th lection on the classical mechanics when he’s approaching Liouville’s Theorem he’s using "uncompressible flow" term to express the idea that divergence of flow in phase space is 0 – as much of flow is entering a volume the same amount of flow is leaving the volume. The idea is very simple but I didn’t understand why he called the flow "uncompressible” and based his explanation on "uncompressibility” of the flow. Now I know (please correct me if I’m wrong here) that by "uncompressible flow" he meant exactly div=0 but I feel that "uncompressible” and “div=0” do not follow one from another.

1) It’s obvious that in some places of the phase space lines with the same H will go closer to each other than in other places what is exactly “compression” in my book (but div=0).

2) Even compressible flows can have div=0. Imaging a cube volume with a flow going into it from 5 sides with some speed v and going out of the volume from 6th side with the same speed v but compressed 5 times denser – div still will be 0 despite it’s compressible flow.

Please help me understand whether "uncompressible flow" was a poor choice of words or "uncompressibility" is really related to div=0 (at least for Liouville’s Theorem).

Thank you.

2. Oct 20, 2013

### mikeph

I've always thought incompressible flow and div(v) = 0 were equivalent statements. Incompressible is equivalent to saying the volume of a fluid element doesn't change over time, and this is only true if the integral of the normal component of velocity through the surface enclosing the element is zero, i.e. if one side moves to the left quickly and the other side moves to the right quickly then the top and bottom MUST move inwards to preserve its volume. The divergence theorem makes this statement equivalent to div(v) = 0.

A compressible flow can have div(v) = 0 if you carefully engineer the correct situation.

3. Oct 20, 2013

### Jano L.

The rate of change of the volume of liquid body $V$ is given by the integral of velocity over its boundary $\Sigma$:

$$\frac{dV}{dt} = \oint_\Sigma \mathbf V \cdot d\boldsymbol \Sigma.$$

Using the Gauss theorem, this is equal to integral of divergence of $\mathbf V$ over the volume of the domain:

$$\frac{dV}{dt} =\int_V \text{div} \mathbf V dV.$$

If $\text{div} \mathbf V = 0$, the rate of change is 0 and such flow is called incompressible.

4. Oct 20, 2013

### Q_Goest

For what it's worth, I'm pretty sure he means "incompressible flow". I've never heard of the phrase "uncompressible flow" and if he actually said that, I suspect it was just a mistake.

There are online notes that mention incompressible flow. The lecture you refer to is also online but I haven't watched the YouTube video. It sounds very interesting though.

Notes: http://www.lecture-notes.co.uk/susskind/classical-mechanics/lecture-7/liouvilles-theorem/

Last edited by a moderator: Sep 25, 2014
5. Oct 20, 2013

### Staff: Mentor

6. Oct 20, 2013

### VRT

I see what you all are saying - it probably comes to the terminology. Maybe he said "incompressible" but not "uncompressible" - most likely I've changed it - sorry for the confusion. Susskind proved that div=0 and therefore concluded that volume is preserved (I think I didn't said it this way in the first post) - this part is the puzzle for me.

What about my 2nd example? Is it also considered "incompressible flow" (div of the flow is 0 here) despite it got 5 times denser?

That wouldn't be too hard with a gas - just cool it down in this volume.

---------
By saying "liquid body" you already presumed that it's practically "incompressible". In general case V should be considered not as 'volume' but as 'number of elements of the flow'='volume'*'density of the flow'. Div of this value would be 0 despite div of the volume is not 0.

7. Oct 20, 2013

### mikeph

How can div(v) be zero when you have a velocity flux of 5 entering the cube and 1 exiting? This is the divergence theorem, if the integral of the normal component of a field over a closed surface is nonzero then the volume integral of the divergence cannot be zero inside the cube. That means div(v) must be nonzero somewhere inside your cube.

All you have achieved by specifying the density is 5 times greater is the conservation of mass, which is a law rather than an approximation of incompressibility. If the density increases then the fluid has by definition been compressed.

8. Oct 20, 2013

### VRT

Number of elements of the flow entering the volume is the same as number of elements of the flow leaving the volume - that make div = 0, doesn't it?

9. Oct 20, 2013

### mikeph

If you're talking about elements of constant mass then the number entering is the same as the number leaving, but you explicitly stated that the density increases by 5 which means the volume of each element must be reduced by a factor of 5. Incompressible means elements maintain their volume, not their mass (which should always be conserved).

10. Oct 20, 2013

### VRT

I appreciate you time, mikeph.

Just to prevent going in circles I'll make a step back. There are 3 statements here: if (1) div = 0 then (2) flow is imcompressible and (3) volume is preserved.

I understand that (2) and (3) are equivalent to each other. I do not understand how (2) and (3) follow from (1). I tried to show that in my case when (1) was true, (3) was not true.

Do you think div in my case was not 0? Then what it was?

Thank you.

11. Oct 20, 2013

### lurflurf

This is a common point of confusion. Nothing is really uncompressible, meaning the density is unaffected by pressure. Water and oil are often assumed to be incompressible, but that is more an assumption that the pressure gradient is negligible. We know water and oil can be compressed and that the compression is important to account for in some situations. On the other hand we often think of gasses as compressible, but many practical situations involve the flow of gases with small pressure gradients. Gasses can be treated as incompressible for these purposes. The choice to treat a flow as compressible of incompressible does not depend on how compressible the fluid is, it depends on how much compression is happening.

12. Oct 20, 2013

### mikeph

I think Jano L's proof is sufficient to show their equivalence.

In your particular case div(v) couldn't have been zero EVERYWHERE. It could be zero almost everywhere, but if the integral of a function is nonzero in a region (which we know from the divergence theorem), then that function cannot possible be zero everywhere. That would be like saying "the sum of n zeros equals 1". If the sum equals one, then one of the terms must be nonzero. If the total flux is positive then the divergence of v somewhere must be positive.

So the non-preservation of volume of a mass element in the fluid necessarily implies that div(v) =/= 0.

Conversely, if div(v) =/= 0 then it is still possible that the velocity flux in/out of one particular closed surface is zero if you chose the right surface, (it could be negative somewhere and positive somewhere else to balance out), but I can always chose a different closed surface where the velocity flux will not be zero, which means there is some compression going on. In other words, div(v) =/= 0 also implies that the fluid is compressible.

Are you familiar with the divergence theorem? This is really the central idea behind all this.

13. Oct 20, 2013

### Jano L.

The term liquid was unfortunate, better is "fluid", as this includes both liquids and gases.

My post #3 does not assume the flow is incompressible; on the contrary, the formula for dV/dt is general for any continuous flow.

From 1) follows 3), see the derivation in the post #3.

14. Oct 20, 2013

### Staff: Mentor

Your second example is wrong. In your second example div(v) ≠ 0.

Say the velocity is a vector field $\mathbf{V}=v_x \mathbf{i} + v_y \mathbf{j}+ v_z \mathbf{k}$ then the divergence would be $$\nabla \cdot \mathbf{V} = \frac{\partial}{\partial x} v_x +\frac{\partial}{\partial y} v_y +\frac{\partial}{\partial z} v_z$$

Now, if we choose units such that the cube is a unit length, and if we take a first difference approximation then $\nabla \cdot \mathbf{V} \approx (v_{x+}-v_{x-})+(v_{y+}-v_{y-})+(v_{z+}-v_{z-})$ where the + and - signs identify the positive and negative faces of the cube respectively. So then, if we choose units such that the speed of the flow is 1, the divergence would be approximately -4, not 0.

For any fluid, the conservation of mass requires that $\nabla \cdot (\rho \mathbf{v}) = 0$. In your example, that condition does hold. But that is a different condition than $\nabla \cdot \mathbf{v} = 0$. Conservation of mass implies the first, but to get the second requires that $\rho$ is constant (in addition to the conservation of mass). Such a flow is called incompressible precisely because the density is constant, which is not the case in your 2nd example.

Last edited: Oct 20, 2013
15. Oct 20, 2013

### VRT

Thank you for help with terminology. $\rho \mathbf{v}$ is what I considered as "flow" (and insisted that its div = 0), not just $\mathbf{v}$. Switch to $\mathbf{v}$ from $\rho \mathbf{v}$ doesn't go easy on me (imagine somebody proposed to use speed in places of momentum in mechanics)...

Back to Liouville’s theorem - it seems then that $\rho$ in the phase space (if there is such a thing) is constant everywhere even though lines with the same H can go further from and closer to each other (what was my analogy of variable $\rho$). Is that the case? But then why would we even need this theorem? If $\rho$ is constant then the flow is incompressible by definition. So here I started vary $\rho$...

Last edited: Oct 20, 2013
16. Oct 20, 2013

### Staff: Mentor

I don't know what you mean by "switch". If you know that k is a non-zero constant then ka=0 implies a=0. There isn't any "switch", just dividing out a constant on both sides of an equation.

If momentum is zero then you certainly can divide both sides by the (constant) mass and get speed is zero also.

The theorem is what shows that ρ is constant. If the phase space equivalent of div(v) were not zero then the phase space equivalent of ρ would not be constant.

Last edited: Oct 20, 2013
17. Oct 21, 2013

### VRT

Beautiful! That's what I didn't know before you said it. Susskind's description is completely different. According to him the theorem states "flows in phase space are incompressible (volume preserved)", what also "means div(H)=0" (see http://www.lecture-notes.co.uk/susskind/classical-mechanics/lecture-7/liouvilles-theorem/). No "density" nor "distribution" was mentioned. He stopped by proving div=0. My point was that if you allow variable density/distribution than div=0 cannot define "incompressibility".

Thanks!

18. Oct 21, 2013

### Jano L.

In the beginning, he writes $\nabla H = 0$. The left-hand side is gradient, not divergence since $H$ has no indices.

However, this equation is wrong, the gradient of $H$ in the phase space is not generally zero vector. This seems like an error created in transcript.

Later he never returns to it. What he really proves in the end is

$$\sum_k \frac{\partial \dot{q}_k}{\partial q_k} + \frac{\partial \dot{p}_k}{\partial p_k} = 0,$$

which follows directly from the Hamilton equations.

He proved the above, which can be written also as $\text{div} ~\mathbf V = 0$. This implies the flow is incompressible, see #3 and the divergence theorem:

http://en.wikipedia.org/wiki/Divergence_theorem

Variable density is allowed in the Susskind notes. In fact nothing special is assumed about density $\rho$ (as you noticed, he did not introduce it at all.)

Last edited by a moderator: May 6, 2017
19. Oct 21, 2013

### Staff: Mentor

Any vector field F for which div(F)=0 can be considered analogous to the velocity field of an incompressible flow. Geometrically, if you start with a fixed volume, and the flow evolves over time according to F then you wind up with a volume that is the same as what you started with.