## Fluid Flows and Infinite-Dimensional Manifolds (Part 2)

### Or: ideal fluids—dry water?

guest post by Tim van Beek

Last time in this series, we set the stage by explaining infinite dimensional manifolds. Then we looked at a simple example: the inviscid Burgers equation. We saw this was the equation for geodesics in the diffeomorphism group of the circle.

Now let’s look at a more interesting example! It will still be a simplified model of fluid flow: it will describe an ideal fluid that is incompressible. I’ll start by explaining these concepts. We will then see how the equation of motion for ideal incompressible fluids can be interpreted as a geodesic equation.

En route I will also repeat some stuff from classical vector analysis, mostly for my own sake. The last time I seriously had to calculate with it was when I attended a class on “classical electrodynamics”, which was almost 15 years ago!

When we delve into differential geometry, it is always a good idea to look both at the “coordinate free” formulation using abstract concepts like differential forms, and also at the “classical vector analysis” part, that is best for calculating stuff once suitable coordinates have been chosen. Our fluid flows will take place in a smooth, orientable, compact, $n$-dimensional Riemannian manifold $M,$ possibly with a smooth boundary $\partial M.$

I will frequently think of $M$ as an open set in $\mathbb{R}^2$ or $\mathbb{R}^3,$ so I will use the globally defined coordinate chart of Euclidean coordinates on $\mathbb{R}^n$ denoted by $x, y$ (and $z,$ if needed) without further warning.

Before we continue: Last time our reader “nick” pointed out a blog post by Terence Tao about the same topic as ours, but—as could be expected—assuming a little bit more of a mathematical background: The Euler-Arnold equation. If you are into math, you might like to take a look at it.

So, let us start with the first important concept: the ‘ideal fluid’.

### What is an ideal fluid?

When you are a small parcel in a fluid flow, you will feel two kinds of forces:

external forces like gravity that are there whether or not your fellow fluid parcels surround you or are absent,

internal forces that come from your interaction with the other fluid parcels.

If there is friction between you and other fluid parcels, for example, then there will be a force slowing down faster parcels and speeding up slower parcels. This is called viscosity. I already explained it back in the post Eddy Who? High viscosity means that there is a lot of friction: think of honey.

The presence of viscosity leads to shear stress whenever there are differences in the velocities of nearby fluid parcels. These lead to the formation of eddies and therefore to turbulence. This complicates matters considerably! For this reason, sometimes people like to simplify matters and to assume that the fluid flow that they consider has zero viscosity. This leads us to the physics definition of an ideal fluid:

An ideal fluid (as physicists say) is a fluid with zero viscosity.

As you can guess, I have also a mathematical definition in store for you:

An ideal fluid (as mathematicians say) is a fluid with the following property: For any motion of the fluid there is a (real valued) function $p(x, t)$ called the pressure such that if $S$ is a surface in the fluid with a chosen unit normal $n,$ the force of stress exerted across the surface $S$ per unit area at $x \in S$ at time $t$ is $p(x,t) n.$

This implies that there is no force acting tangentially to the surface $S$:

This picture is from

• Alexandre Chorin and Jerrold E. Marsden, A Mathematical Introduction to Fluid Mechanics, 3rd edition, Springer, New York 1993.

An ideal fluid cannot form eddies by itself without the help of external forces, nor can eddies vanish once they are present. So this simplification exclude a lot of very interesting phenomena, including everything that is usually associated with the term ‘turbulence’. But it is a necessary simplification for describing fluid flow using geodesic equations, because something moving along a geodesic doesn’t lose energy due to friction! So we will have to stick with it for now.

Historically, ideal fluids were almost exclusively studied during the 19th century, because the mathematics of viscous fluids seemed to be too hard—which it still is, although there has been a lot of progress. T his led to a schism of theoretical hydrodynamics and engineering hydrodynamics, because engineers had to handle effects like turbulence that ideal fluids cannot model. A very problematic aspect is that no body with a subsonic velocity feels any drag force in an ideal fluid. This is known as D’Alembert’s paradox. This means that one cannot find out anything about optimal design of ships or aircrafts or cars using ideal fluids as a model. This situation was overcome by the invention of ‘boundary layer techniques’ by the physicist Ludwig Prandtl at the beginning of the 20th century.

John von Neumann is cited by Richard Feynman in his physics lectures as having said that ideal fluids are like “dry water”, because they are so unlike real water. This is what the subtitle of this post alludes to. I don’t think this is quite fair to say. Along these lines one could say that quantum mechanics is the theory of stagnant light, because it does not include relativistic effects like quantum field theory does. Of course every mathematical model is always just an approximation to a Gedankenexperiment. And ideal fluids still have their role to play.

Maybe I will tell you more about this in a follow-up post, but before this one gets too long, let us move on to our second topic: incompressible fluids and ‘volume preserving’ diffeomorphisms.

### What is an incompressible fluid flow?

If you are a parcel of an incompressible fluid, this means that your volume does not change over time. But your shape may, so if you start out as a sphere, after some time you may end up as an ellipsoid. Let’s make this mathematically precise.

But first note, that “incompressible” in the sense above means that the density of a given fluid parcel does not change over time. It does not mean that the density of the whole fluid is everywhere the same. A fluid like that is actually called homogeneous. So we have two different notions:

incompressible means that the volume of an infinitesimal fluid parcel does not change as it moves along the fluid flow,

homogeneous means that the density at a given time is everywhere the same, that is: constant in space.

This distinction is important, but for now we will study fluid flows that are both homogeneous and incompressible.

Let us see how we can make the notion of “incompressible” mathematically precise:

Remember from the last post: The flow of each fluid parcel is described by a path on $M$ parametrized by time, so that for every time $t \ge t_0$ there is a diffeomorphism

$g^t : M \to M$

defined by the requirement that it maps the initial position $x$ of each fluid parcel to its position $g^t(x)$ at time $t$:

Now let’s assume our fluid flow is incompressible. What does that mean for the diffeomorphisms that describe the flow? Assuming that we have a volume form $\mu$ on $M,$ these diffeomorphisms must conserve it:

$\mathrm{SDiff}(M) := \{ f \in \mathrm{Diff}(M): f^* \mu = \mu \}$

For people who need a reminder of the concepts involved (which includes me), here it is:

Remember that $M$ is a smooth orientable Riemannian manifold of dimension $n.$ A volume form $\mu$ is a $n$-form that vanishes nowhere. In $\mathbb{R}^3$ with Cartesian coordinates $x, y, z$ the canonical example would be

$\mu = d x \wedge d y \wedge d z$

The dual basis of $d x, d y, d z$ is denoted by $\partial_x, \partial_y, \partial_z$ in our example.

Given two manifolds $M, N$ and a differentiable map $f: M \to N,$ we can pull back a differential form $\mu$ on $N$ to one on $M$ via

$f^{*} \mu_p (v_1, ..., v_n) = \mu_{f(p)} (d f(v_1), ..., d f(v_n))$

For the übernerds out there: remember that we see the group of diffeomorpisms $\mathrm{Diff}(M)$ as a Fréchet Lie group modelled on the Fréchet space of vector fields on $M,$ $\mathrm{Vec}(M).$ For those who would like to read more about this concept, try this:

• Karl-Hermann Neeb, Monastir Summer School: infinite-dimensional Lie groups.

$\mathrm{SDiff}(M)$ is clearly a subgroup of $\mathrm{Diff}(M).$ It is less obvious, but true, that it is a closed subgroup and therefore itself a Lie group. What about its Lie algebra? For a vector field to give a flow that’s volume preserving, it must have zero divergence. So, the vector fields that form the tangent space $T_{\mathrm{id}} \mathrm{SDiff}(M)$ consist of all smooth vector fields $V$ with zero divergence:

$\mathrm{div}(V) = 0$

These vector fields form a vector space we denote by $\mathrm{SVec}(M).$ Remember $T_{\mathrm{id}}$ stands for the tangent space at the identity element of the group $\mathrm{SDiff}(M),$ which is the identity diffeomorphism $\mathrm{id}$ of $M.$ The tangent space at the identity of a Lie group is a Lie algebra, so $\mathrm{SVec}(M)$ is a Lie algebra.

I will need a little refresher about the definition of divergence. Then I will point you to a proof of the claim above, namely that zero-divergence vector fields form the Lie algebra of volume preserving diffeomorphism. This may seem obvious on an intuitive level, if you ever learned that the zero-divergence vector fields have ‘no sinks and no sources’, for example in a course on classical electromagnetism.

So, what is the divergence, again? You’ve probably seen it somewhere if you’ve survived reading this so far, but you may not have seen it in full generality.

The divergence of a vector field $V$ with respect to a volume form $\mu$ is the unique scalar function $\mathrm{div}(V)$ such that:

$\mathrm{div}(V)\, \mu = d (i_V \mu)$

Here, $i_X$ is the contraction with $X.$ Contraction means that you feed the vector $X$ in the first slot of the differential form $\mu,$ and therefore reduce the function $\mu$ of $n$ vector fields to one of $n-1$ vector fields.

When we use our standard example $M = \mathbb{R}^3,$ we of course write a vector field as

$V = V_x \partial_x + V_y\partial_y + V_z \partial_z$

where $V_x, V_y$ and $V_z$ are smooth real-valued functions. The divergence of $V$ is then

$\mathrm{div}(V) = \partial_x V_x + \partial_y V_y + \partial_z V_z$

which we get if we plug in the expression for $V$ into the formula $d(i_V \mu).$

So, how does one see that ‘zero divergence’ of a vector field is equivalent to ‘volume preserving’ for the flow it generates?

If we write

$\phi(t) = (x(t), y(t), z(t))$

for the path of a fluid particle and $u$ for its velocity, then of course we have:

$\displaystyle{ \frac{d \phi}{d t} = u }$

For a scalar function $f(t, x(t), y(t), z(t))$ we get

$\displaystyle{ \frac{d f}{d t} = \frac{\partial f}{\partial t} + u \cdot \mathrm{grad}(f) }$

Here $\cdot$ is the inner product. The latter part is often written with the help of the nabla operator $\nabla$ as

$u \cdot \mathrm{grad}(f) = u \cdot \nabla \; f$

This is really just a handy short notation, there is no mystery behind it: it’s just like how we write the divergence as $\mathrm{div}(X) = \nabla \cdot X$ and the curl as $\mathrm{curl}(X) = \nabla \times X.$

The operator

$D_t = \partial_t + u \cdot \nabla$

appears so often that it has its own name: it is called the material derivative.

Why ‘material’? Because if we follow a little bit of material—what we’re calling a parcel of fluid—something about it can change with time for two different reasons. First, this quantity can explicitly depend on time: that’s what the first term, $\partial_t,$ is about. Second, this quantity can depend on where you are, so it changes as the parcel moves: that’s what $u \cdot \nabla$ is about.

Now suppose we have a little parcel of fluid. We’ve been talking about it intuitively, but mathematically we can describe it at time zero as an open set $W_0$ in our manifold. After a time $t,$ it will be mapped by the fluid flow $g^t$ to

$W_t := g^t (W_0)$

This describes how our parcel moves. We define the fluid to be incompressible if the volume of $W_t$ for all choices of $W_0$ is constant, that is:

$\displaystyle{ 0 = \frac{d}{d t} \int_{W_t} d \mu }$

If we write $J^t$ for the Jacobian determinant of $g^t,$ then we have

$\displaystyle{ 0 = \frac{d}{d t} \int_{W_t} d \mu = \frac{d}{d t} \int_{W_0} J^t d \mu }$

So in a first step we get that a fluid flow is incompressible iff the Jacobian determinant $J$ is $1$ for all times, which is true iff $g^t$ is volume preserving.

It is not that hard to show by a direct calculation that

$\displaystyle{ \left. \partial_t J\right|_{t=0} = \mathrm{div}(u) J }$

If you don’t want to do it yourself, you can look it up in a book that I already mentioned:

• Alexandre Chorin and Jerrold E. Marsden, A Mathematical Introduction to Fluid Mechanics, 3rd edition, Springer-Verlag, New York 1993.

This is the connection between ‘volume preserving’ and ‘zero divergence’! Inserting this into our equation of incompressibility, we finally get:

$\begin{array}{ccl} 0 &=& \displaystyle{ \frac{d}{d t} \int_{W_t} d \mu } \\ \\ &=& \displaystyle{\frac{d}{d t} \int_{W_0} J^t d \mu } \\ \\ &=& \displaystyle{\int_{W_0} \mathrm{div}(u) J d \mu } \end{array}$

which is true for all open sets $W_0$ iff $\mathrm{div}(u) = 0.$ The equation of continuity for a fluid flow is:

$\displaystyle{ \frac{\partial \rho}{\partial t} + \mathrm{div}(\rho u) = 0 }$

This says that mass is conserved. Written with the material derivative it is:

$\displaystyle{ \frac{D \rho}{D t} + \rho \, \mathrm{div}(u) = 0 }$

So, since we’re assuming $\mathrm{div}(u) = 0,$ we get

$\displaystyle{ \frac{D \rho}{D t} = 0 }$

which is what we intuitively expect, namely that the density is constant for a fluid parcel following the fluid flow.

### Euler’s equation for the ideal incompressible fluid

The equation of motion for an ideal incompressible fluid is Euler’s equation:

$\partial_t u + (u \cdot \nabla) u = - \nabla p$

$p$ is the pressure function mentioned in the mathematical definition of an ideal fluid above. As I already mentioned, to be precise I should say that we also assume that the fluid is homogeneous. This means that the density $\rho$ is constant both in space and time and therefore can be cancelled from the equation of motion.

If $M$ has a nonempty (smooth) boundary $\partial M,$ the equation is supplemented by the boundary condition that $u$ is tangential to $\partial M.$

How can we turn this equation into a geodesic equation on $\mathrm{SDiff}(M)$? Our strategy will be the same as last time when we handled the diffeomorphism group of the circle. We will define the necessary gadgets of differential geometry on $\mathrm{SDiff}(M)$ using the already existing ones on $M.$ First we define them on $T_{\mathrm{id}}\mathrm{SDiff}(M).$ Then, for any diffeomorphism $\phi \in \mathrm{SDiff}(M),$ we use right translation by $\phi$ to define them on $T_{\phi}\mathrm{SDiff}(M).$ After that, we can use the version of the abstract version of the geodesic equation for right invariant metrics to calculate the explicit differential equation behind it.

Let us start with defining right invariant vector fields on $\mathrm{SDiff}(M).$ A right invariant vector field $U$ is a vector field such that there is a $u \in \mathrm{SVec}(M)$ such that $U_{\phi} = u \circ \phi.$ In the following, we restrict ourselves to right invariant vector fields only.

We define the usual $L^2$ inner product of vector fields $u, v$ on $M$ just as last time:

$\displaystyle{ \langle u, v \rangle = \int_M \langle u_x, v_x \rangle \; d \mu (x) }$

The inner product used on the right is of course the one on $M.$

For two right invariant vector fields $U, V$ with $U_{\phi} = u \circ \phi$ and $V_{\phi} = v \circ \phi,$ we define the inner product on $T_{\phi}\mathrm{SDiff}(M)$ by

$\langle U, V \rangle_{\phi} = \langle u, v \rangle$

This definition induces a right invariant metric on $\mathrm{SDiff}(M).$ Note that it is right invariant because we are only considering volume preserving diffeomorphisms. It is not right invariant on the larger group of all diffeomorphims $\mathrm{Diff}(M)$!

For an incompressible ideal fluid without external fields the only kind of energy one has to consider is the kinetic energy. The inner product that we use is actually proportional to the kinetic energy of the whole fluid flow at a fixed time. So geodesics with respect to the induced metric will correspond to Hamilton’s extremal principle. In fact it is possible to formulate all this in the language of Hamiltonian systems, but I will stop here and return to the quest of calculating the geodesic equation.

Last but not least, we define the following right invariant connection:

$\nabla_{U_{\phi}} V_{\phi} = (\nabla_{u} v) \circ \phi$

Here $\nabla$ on the right is the connection on $M$—sorry, this is not quite the same as the $\nabla$ we’d been using earlier! But in $\mathbb{R}^3$ or Euclidean space of any other dimension, $\nabla_u v$ is just another name for $(u \cdot \nabla) v,$ so don’t get scared.

Remember from last time that the geodesic equation says

$\nabla_u u = 0$

where $u$ is the velocity vector of our geodesic, say

$\displaystyle{ u(t) = \frac{d}{d t} \gamma(t) }$

where $\gamma$ is the curve describing our geodesic. We saw that for a right-invariant metric on a Lie group, this equation says

$\partial_t u = \mathrm{ad}^*_u u$

where the coadjoint operator $\mathrm{ad}^*$ is defined by

$\langle \mathrm{ad}^*_u v, w \rangle = \langle v, \mathrm{ad}_u w \rangle = \langle v, [u, w] \rangle$

For simplicity, let us specialize to $\mathbb{R}^3,$ or an open set in there. What can we say about the right hand side of the above equation in this case? First, we have the vector identity

$\nabla \times (u \times w) = - [u, w] + u \; \nabla \cdot w - w \; \nabla \cdot u$

Since we are talking about divergence-free vector fields, we actually have

$[u, w] = - \nabla \times (u \times w)$

Also note that for a scalar function $f$ and the divergence-free vector field $u$ we have

$\begin{array}{ccl} \langle u, \nabla f \rangle &=& \int_M \langle u(x), \nabla f(x) \rangle \; d \mu (x) \\ \\ &=& \int_M \nabla \cdot (f(x) u(x)) \; d \mu (x) \\ \\ &=& \int_{\partial M} f(x) \; \langle u, n \rangle \; d S (x) \\ \\ &=& 0 \end{array}$

The last term is zero because of our boundary condition that says that the velocity field $u$ is tangent to $\partial M.$

So, now I am ready to formulate my claim that

$\mathrm{ad}^*_u v = - (\nabla \times v) \times u + \nabla f$

for some yet undetermined scalar function $f.$ This can be verified by a direct calculation:

$\begin{array}{ccl} \langle \mathrm{ad}^*_u v, w \rangle &=& \langle v, \mathrm{ad}_u w \rangle \\ \\ &=& \langle v, [u, w] \rangle \\ \\ &=& \int_M \langle v_x, [u, w]_x \rangle \;d\mu(x) \\ \\ &=& - \int_M \langle v_x, (\nabla \times (u \times w))_x \rangle \;d \mu(x) \end{array}$

What next? We can use the following 3 dimensional version of Green’s theorem for the curl operator:

$\int_M ( \langle \nabla \times a, b \rangle - \langle a, \nabla \times b \rangle ) d \mu = \int_{\partial M} \langle a \times b, n \rangle d S$

That is, the curl operator is symmetric when acting on vector fields that have no component that is tangent to $\partial M.$ Note that I deliberately forgot to talk about function spaces that our vector fields need to belong to and the regularity assumptions on the domain $M$ and its boundary, because this is a blog post and not a math lecture. But the operators we use on vector fields obviously depend on such assumptions.

If you are interested in how to extend the symmetric curl operator to a self-adjoint operator, for example, you could look it up here:

• R. Hiptmair, P. R. Kotiuga, S. Tordeux, Self-adjoint curl operators.

Since our vector fields are supposed to be tangent to $\partial M,$ we have that the boundary term in our case is

$\int_{\partial M} \langle u_x \times w_x \times v_x, n \rangle \; dS = 0$

because $u_x \times w_x$ is normal, and therefore $u_x \times w_x \times v_x$ is tangent to $\partial M,$ so its inner product with the normal vector $n$ is zero.

So we can shift the curl operator from right to left like this:

$\begin{array}{ccl} - \int_M \langle v_x, (\nabla \times (u \times w))_x \rangle \;d \mu(x) &=& - \int_M \langle (\nabla \times v)_x, (u \times w)_x \rangle \;d \mu(x) \\ \\ &=& - \int_M \langle (\nabla \times v)_x \times u_x, w_x \rangle \;d \mu(x) \end{array}$

In the last step we used the cyclicity of the relation of the vector product and the volume spanned by three vectors:

$\langle a \times b, c \rangle = \mu(a, b, c) = \mu (c, a, b) = \langle c \times a, b \rangle$

This verifies the claim, since the part $\nabla f$ does not contribute, as stated above.

And now, yet another vector identity comes to our rescue:

$(\nabla \times v) \times u = (u \cdot \nabla) v - u_k \nabla v_k$

So, we finally end up with this:

$\begin{array}{ccl} \mathrm{ad}^*_u u &=& - (u \cdot \nabla) u - u_k \nabla u_k + \nabla f \\ \\ &=& - (u \cdot \nabla) u + \nabla g \end{array}$

for some function $g.$ Why? Since the middle term $u_k \nabla u_k = \frac{1}{2} \nabla u^2$ is actually a gradient, we can absorb this summand and $\nabla f$ into one summand with a new function, $\nabla g.$

Thanks to this formula we derived, the abstract and elegant equation for a geodesic on any Lie group

$\partial_t u = \mathrm{ad}^*_u u$

becomes, in this special case

$\partial_t u = - (u \cdot \nabla) u + \nabla g$

If we can convince ourselves that $-g$ is the pressure $p$ of our fluid, we get Euler’s equation:

$\partial_t u + (u \cdot \nabla) u = - \nabla p$

Wow! Starting with abstract stuff about infinite-dimensional Lie groups, we’ve almost managed to derive Euler’s equation as the geodesic equation on $\mathrm{SDiff}(M)$! We’re not quite done: we still need to talk about the role of the function $g,$ and why it’s minus the pressure. But that will have to wait for another post.

### 31 Responses to Fluid Flows and Infinite-Dimensional Manifolds (Part 2)

1. Giuseppe Tortorella says:

Dear Tim van Beek, I like even this your second post on the subject. For what it is worth, I would point out some possible typos hoping to be useful.

1) The measure form $\mu$ is missing in the lhs of the equation defining the divergence of $V$ w.r.t. to $\mu$.

2)In the equation just before the reference to Chorin and Marsden, I think that some time dependence is missing, in fact the equation for the time derivative of the Jacobian determinant $J^t$ of the flow $g^t$ of the vector field $u$ should be

$\partial_t J^t=\textrm{div}(u)\circ g^t .J^t,$

or equivalently

$\partial_t|_{t=0}J^t=\textrm{div}(u).$

3) In the third and fourth line of the equation for $\langle\textrm{ad}^\ast_u v,w\rangle$ we should replace $[u_x,w_x]$ with $[u,w]_x,$ and $\nabla\times (u_x\times w_x)$ with $(\nabla\times (u\times w))_x$.

Bye.

• Tim van Beek says:

Thanks for the corrections – I myself cannot change the blog post, only John can. But fortunately, I don’t think that the errors are very disrupting to the flow of reading.

• John Baez says:

I want to fix all the errors here (and in every blog post on Azimuth), and in this case it looks easy to do. I was just waiting for Tim to give the okay. Thanks, Giuseppe!

• John Baez says:

Okay, I’ve made the changes Giuseppe suggested—maybe someone can check to see if I did them correctly.

I’ve also changed

$\begin{array}{ccl} - \int_M \langle v_x, \nabla \times (u_x \times w_x) \rangle \;d \mu(x) &=& - \int_M \langle \nabla \times v_x, u_x \times w_x \rangle \;d \mu(x) \\ \\ &=& - \int_M \langle (\nabla \times v_x) \times u_x, w_x \rangle \;d \mu(x) \end{array}$

to

$\begin{array}{ccl} - \int_M \langle v_x, (\nabla \times (u \times w))_x \rangle \;d \mu(x) &=& - \int_M \langle (\nabla \times v)_x, (u \times w)_x \rangle \;d \mu(x) \\ \\ &=& - \int_M \langle (\nabla \times v)_x \times u_x, w_x \rangle \;d \mu(x) \end{array}$

• Tim van Beek says:

Looks good to me :-)

2. John Baez says:

What do you think we should do next in this series of posts, Tim? I guess one of your goals is to explain how the flow of an ideal fluid can be chaotic? That would be nice. It’s probably less useful, but I also (as you know) feel it’s amusing to study the Burgers equation in a bit more detail, relating it to a ‘gas of noninteracting particles’.

Moving to something more speculative: I wonder if there’s any way to incorporate dissipation into these infinite-dimensional Lie group models. If instead of working with an infinite-dimensional Lie group we were studying SO(3), we’d be studying a spinning object. This is well-understood… and I can’t imagine nobody has ever tried incorporating dissipation into that model. It seems there should be some reasonably elegant way to do it. And then one could try to generalize to other Lie groups.

There’s a pretty nice geometric formalism for studying systems with friction—see this.

• Giuseppe Tortorella says:

Dear John Baez, I have read the fragment from the previous post you refer to, I have a question: setting the “Generic” framework there is the requirement $\{H,F\}=[S,F]=0,$ for every function $F,$ but in order to get the conservation of energy and the increasing of entropy, is not the requirement $\{H,S\}=[S,H]=0$ enough? In the context of the Kahler manifold, does it mean the conservation of $S$ along the hamiltonian and the gradient flow of $H$? and is it again too much strong?

• John Baez says:

$\{H, F \} = [S,F] = 0$

for every function $F$ seems absurd to me now! What’s the use of a Hamiltonian whose Poisson bracket with every observable vanishes???

It’s possible that a virtual black hole destroyed some of my neurons when I was originally typing that article and that Oettinger’s framework merely demands

$\{H, S \} = [S,H] = 0$

I’ll need to reread his book. It’s a bit annoying that nobody ever pointed out this problem before! Thanks for catching it.

• Giuseppe Tortorella says:

Dear John Baez

Please excuse me, I made a mistake typing the requirement, I should have written $[H,F]=\{S,F\}=0,$ for every function $F.$ They are equations (1.4) and (1.5) in Oettinger’s book. And however on page 17 they are considered indeed as stronger formulations of the energy conservation and of the second law of thermodynamics, because these laws could be ensured just by assuming $[H,S]=\{H,S\}=0$. I should read more deeply to understand what is the rationale of these strong formulations.

• Eugene says:

“What’s the use of a Hamiltonian whose Poisson bracket with every observable vanishes???”

Such functions are called Casimirs. They are conserved quantities for whatever dynamics there is around.

• John Baez says:

Giuseppe wrote:

Please excuse me, I made a mistake typing the requirement, I should have written

$[H,F]=\{S,F\}=0$

for every function $F$.

Oh, good, that makes more sense. I haven’t thought about this for a long time, but I know that once it made sense to me. This makes more sense. But you’re right that it’s still stronger than necessary.

Eugene wrote:

Such functions are called Casimirs. They are conserved quantities for whatever dynamics there is around.

Right, but a framework that demands that the Hamiltonian be a Casimir would be pretty silly, because such Hamiltonian generate trivial dynamics. That’s what I was expostulating about.

• John Baez says:

Giuseppe: I think it’s worthwhile to look for physically interesting examples where we start with a Kähler manifold, define symmetric and antisymmetric brackets as I proposed back there, and then find physically plausible examples of energy and entropy functions obeying

$\{H, S \} = [H,S] = 0$

I’ll try to do this…

• Tim van Beek says:

One post will be about the relation of solutions of Euler’s equation (no viscosity) to the Navier-Stokes equation for incompressible fluids (viscosity > 0). Then we should of course apply concepts from Riemannian geometry to fluid flows (we have only set up the framework, but haven’t seen any applications of it).
Therefore, one post should be about the geodesic deviation equations and what one can say about the stability of fluid flows using it.
A math application is of course the proof of existence and smoothness of Euler’s equation, but I don’t plan to go into it. Even if it would seem that Arnold (or Ebin and Marsden, I’m not quite sure, actually) did pioneering work here. But that’s just math, as an engineer I’m not that interested in the existence&smoothness proofs, I simply assume that those are settled :-)
Then one post will be about the Burger’s equation as the simplest example of a nonlinear equation, whose solutions can develop shock waves. But I’m of course taking suggestions :-)

• Tim van Beek says:

John wrote:

I wonder if there’s any way to incorporate dissipation into these infinite-dimensional Lie group models.

Arnold and Khesin have a chapter about this in their book “Topological Methods in Hydrodynamics” which I don’t quite understand. I also remember that Terry Tao mentions this in his blog post. I could try to ask about it on mathoverflow, but I think I should do some basic research first :-) (basic research being googling in our days).

• John Baez says:

Okay, let’s try to learn about this. If you find out any more nice references, let us know here! I think Öttinger uses fluid flow as an example of his general theory of dissipative dynamics, so there could be some fun math going on here.

3. Eugene says:

Sorry. I did answer a rhetorical question, didn’t I? I kind of realized it the second after I hit return. Nobody studies the dynamics of Casimirs. But perhaps you can publish a paper on their dynamics in “Chaos, solitons and fractals” :)

• John Baez says:

I did answer a rhetorical question, didn’t I?

You can claim it was a rhetorical answer.

4. […] In Part 2 of this series, I told you what ideal incompressible fluids are […]

5. yoshi says:

Hello !

I have a bit of misanderstanding about the nature of the vector field u in the Euler equation : is it a vector field on R^3 or a vector field on the infinite dimensional group of volume preserving diffeos ? In part 1 of the lecture, we did an identification between S^(1) and its group of diffeomorphisms, but here…well its not that obvious…

• John Baez says:

For any compact manifold $M$, the Lie algebra of the diffeomorphism group $\mathrm{Diff}(M)$ can be identified with the Lie algebra of vector fields on $M$, since we can exponentiate any vector field and get a 1-parameter group of diffeomorphisms (called a ‘flow’). Furthermore, an element of a Lie algebra can always be identified with a left-invariant vector field on the corresponding Lie group (see Prop. 7.1 here for the basic idea).

So, we can freely switch between talking about:

• a vector field on $M$

• an element in the Lie algebra of $\mathrm{Diff}(M)$

• a left-invariant vector field on $\mathrm{Diff}(M)$

The same is true for divergence-free vector fields and the group of volume-preserving diffeomorphisms, and that’s what Tim is using here.

• Tim van Beek says:

It can be a little bit confusing at first :-)

I’ll rephrase what John said, HTH.

We start with a finite dimensional real manifold M. Then we choose a vector field u on M. This is the vector field that is used in the Euler equation. It describes the velocity of every point at a fixed time, which we can set to $t = 0$. So picture every point of M at time $t=0$ as a little packet of fluid that flows with time, with $u_x$ being its velocity.

Now we switch perspective and talk about the diffeomorphisms of M as a Lie group. Here, u is not a vector field, but is re-interpreted as an element of the Lie algebra, i.e. of the tangential space at the group identity. The group identity is the diffeomorphism that is the identity of M, sending each point to itself. The tangential space of it consists of all vector fields of M.

So these are the two different ways of interpreting u.

• yoshi says:

thank you for these answers ! please keep doing the good work

6. Pedro says:

Can you go a bit deeper in the connection between group theory, Noether’s theorem and its connection with perfect fluids?

• John Baez says:

What do you want to know? The symmetry of the perfect fluid under the group of volume-preserving diffeomorphisms should, by Noether’s theorem, give infinitely many conserved quantities. Did Tim say what these conserved quantities are? I forget.

• Tim van Beek says:

Actually I am not aware of any results in this direction.

In case I pick up this topic again, I would also rather go deeper into the calculation/estimation of the curvature and what that implies for the stability of geodesics, instead of talking about how one can formulate the system in the terms of Hamiltonian dynamics etc.

• John Baez says:

Okay. But it’s really tempting to figure out the conserved quantities associated to the symmetries in the problem we’re talking about—the incompressible frictionless fluid. Let’s try!

For starters, let’s remember (since it’s been half a year!) that this problem is all about geodesic motion on a certain infinite-dimensional Lie group with a right-invariant metric. So, the symmetry group is precisely that Lie group. It acts on itself by right translations.

Next, I’ll remember from my own education that this is a fancy version of a much simpler problem: a rotating object, like a tennis racket, floating at rest in intergalactic space. There the symmetry group is the rotation group $SO(3).$ And there the conserved quantities are the 3 components of angular momentum: $J_x, J_y$ and $J_z.$ 3-dimensional Lie group, 3 conserved quantities.

So, we need to generalize this, and if we were smart enough we could guess the correct answer without doing no calculations with pencil and paper.

We could guess that angular momentum is related to rotation symmetry simply because they both have to do with rotating things… and the numbers match: 3-dimensional Lie group, 3 conserved quantities.

It’s a bit harder to do this guessing for our problem here, because our Lie group is infinite-dimensional. But we know right away we’re looking for infinitely many conserved quantities.

My first guess is this. Take the volume of a little parcel of fluid. Since the fluid is incompressible, this doesn’t change with time. So, we get a conservation law! And we get infinitely many of these conservation laws, since there are infinitely many little parcels to look at.

We need to polish this up a bit to get a rigorous formulation of these conservation laws as

$\displaystyle{ \frac{d}{d t} \;\mathrm{something} = 0}$

But I think Tim already did this in his blog article! He wrote this condition for the incompressibility of a little parcel of fluid:

$\displaystyle{ \frac{d}{d t} \int_{W_t} d \mu = \frac{d}{d t} \int_{W_0} J^t d \mu = 0}$

So my guess is that these conservation laws, and linear combinations of these, are what we want. To test this guess, we could think a bit harder and try to see a correspondence between such linear combinations and infinitesimal symmetries—which for this problem are divergence-free vector fields.

Actually, I think Tim gave us a clue here, too, since he wrote the equation

$\displaystyle{ \left. \partial_t J\right|_{t=0} = \mathrm{div}(u) J }$

7. jayprich says:

The equivalence of divergence free vector fields to conservation of volume is surely a tautology and not what Noether’s theorem is about. Decomposing a flow into the family of solutions of the PDE is very dependent on symmetries in the boundary conditions. Noether is about local symmetries in the tensor parameterised by time when viewed in coordinates that follow the flow itself. Here is a link to a paper that discussed fluid/particle “re-labelling symmetries” deriving from Noether the conservation of advanced physical macro properties, (advection of vortices along surfaces of constant entropy) : Nikhil Subhash Padhye, Ph.D. The University of Texas at Austin,1998 http://w3fusion.ph.utexas.edu/ifs/ifsreports/825_padhye.pdf
this paper also contains nice references to work on conservation laws from this symmetry dating back to 1962.

• John Baez says:

Jayprich wrote:

The equivalence of divergence free vector fields to conservation of volume is surely a tautology and not what Noether’s theorem is about.

You don’t need Noether’s theorem to see that divergence-free vector fields lead to conservation of volume. However, conservation of the volume of each little bit of the fluid is an infinite set of conservation laws, so Noether’s theorem (used in the reverse direction) gives an infinite collection of symmetries, and these would be interesting to think about.

• jayprich says:

Understood. It seems Ertel’s “Potential Vorticity” PV Theorem, mentioned in that paper, characterises these symmetries as a conservation. PV being carried with the flow (materially conserved) then so is any local function of PV alone in particular the local energy, inertia and angular momenta. ( “The conservation of potential enstrophy-the second moment of potential vorticity-is thought to be especially important because it prevents the spurious cascade of energy into high wavenumbers.” – http://coaps.fsu.edu/pub/eric_back/OCP5930/Papers/GeneralMethodsForConservation.non5_5_R01.pdf )

8. phillip says:

How can I prove that the ideal fluid bracket is a Poisson bracket?

• John Baez says:

If you’re trying to check the Jacobi identity, you can probably use its relation to the Lie bracket of vector fields.

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