## Fluid Flows and Infinite-Dimensional Manifolds I

### Or: waves that take the shortest path through infinity

guest post by Tim van Beek

Water waves can do a lot of things that light waves cannot, like “breaking”:

In mathematical models this difference shows up through the kind of partial differential equation (PDE) that models the waves:

• light waves are modelled by linear equations while

• water waves are modelled by nonlinear equations.

Physicists like to point out that linear equations model things that do not interact, while nonlinear equations model things that interact with each other. In quantum field theory, people speak of “free fields” versus “interacting fields”.

Some nonlinear PDE that describe fluid flows turn out to also describe geodesics on infinite-dimensional Riemannian manifolds. This fascinating observation is due to the Russian mathematician Vladimir Arnold. In this blog post I would like to talk a little bit about the concepts involved and show you a little toy example.

#### Fluid Flow modelled by Diffeomorphisms

The Euler viewpoint on fluids is that a fluid is made of tiny “packages” or “particles”. The fluid flow is described by specifying where each package or particle is at a given time $t$. When we start at some time $t_0$ on a given manifold $M$, the flow of every fluid package is described by a path on $M$ parametrized by time, and for every time $t > t_0$ there is a diffeomorphism $g^t : tiM \to M$ defined by the requirement that it maps the initial position $x$ of each fluid package to its position $g^t(x)$ at time $t$:

This picture is taken from the book

• V.I. Arnold and B.A. Khesin, Topological Methods in Hydrodynamics, Springer, Berlin, 1998. (Review at Zentralblatt Mathematik.)

We will take as a model of the domain of the fluid flow a compact Riemannian manifold $M$. A fluid flow, as pictured above, is then a path in the diffeomorphism group $\mathrm{Diff}(M)$. In order to apply geometric concepts in this situation, we will have to turn $\mathrm{Diff}(M)$ or some closed subgroup of it into a manifold, which will be infinite dimensional.

The curvature of such a manifold can provide a great deal about the stability of fluid flows: On a manifold with negative curvature geodesics will diverge from each other. If we can model fluid flows as geodesics in a Riemannian manifold and calculate the curvature, we could try to infer a bound on weather forecasts (in fact, that is what Vladimir Arnold did!): The solution that you calculate is one geodesic. But if you take into account errors with determining your starting point (involving the measurement of the state of the flow at the given start time), what you are actually looking at is a bunch of geodesics starting in a neighborhood of your starting point. If they diverge fast, that means that measurement errors make your result unreliable fast.

If you never thought about manifolds in infinite dimensions, you may feel a little bit insecure as to how the concepts that you know from differential geometry can be generalized from finite dimensions. At least I felt this way when I first read about it. But it turns out that the part of the theory one needs to know in order to understand Arnold’s insight is not that scary, so I will talk a little bit about it next.

#### What you should know about infinite-dimensional manifolds

The basic strategy when handling finite-dimensional, smooth, real manifolds is that you have a complicated manifold $M$, but also locally for every point $p \in M$ a neighborhood $U$ and an isomorphism (a “chart”) of $U$ to an open subset of the vastly simpler space $\mathbb{R}^n$, the “model space”. These isomorphisms can be used to transport concepts from $\mathbb{R}^n$ to $M$. In infinite dimensions it is however not that clear what kind of model space $E$ should be taken in place of $\mathbb{R}^n$. What structure should $E$ have?

Since we would like to differentiate, we should for example be able to define the derivative of a curve in $E$:

$\gamma: \mathbb{R} \to E$

If we write down the usual formula for a derivative

$\gamma'(t_0) := \lim_{t \to 0} \frac{1}{t} (\gamma(t_0 +t) - \gamma(t_0))$

we see that to make sense of this we need to be able to add elements, have a scalar multiplication, and a topology such that the algebraic operations are continuous. Sets $E$ with this structure are called topological vector spaces.

A curve that has a first derivative, second derivative, third derivative… and so on at every point is called a smooth curve, just as in the finite dimensional case.

So $E$ should at least be a topological vector space. We can, of course, put more structure on $E$ to make it “more similar” to $\mathbb{R}^n$, and choose as model space in ascending order of generality:

1) A Hilbert space, which has an inner product,

2) a Banach space that does not have a inner product, but a norm,

3) a Fréchet space that does not have a norm, but a metric,

4) a general topological vector space that need not be metrizable.

People talk accordingly of Hilbert, Banach and Fréchet manifolds. Since the space $C^{\infty}(\mathbb{R}^n)$ consisting of smooth maps from $\mathbb{R}^n$ to $\mathbb{R}$ is not a Banach space but a Fréchet space, we should not expect that we can model diffeomorphism groups on Banach spaces, but on Fréchet spaces. So we will use the concept of Fréchet manifolds.

But if you are interested in a more general theory using locally convex topological vector spaces as model spaces, you can look it up here:

• Andreas Kriegl and Peter W. Michor, The Convenient Setting of Global Analysis, American Mathematical Society, Providence Rhode Island, 1999.

Note that Kriegl and Michor use a different definition of “smooth function of Fréchet spaces” than we will below.

If you learn functional analysis, you will most likely start with operators on Hilbert spaces. One could say that the theory of topological vector spaces is about abstracting away as much structure from a Hilbert space and look what structure you need for important theorems to still hold true, like the open mapping/closed graph theorem. If you would like to learn more about this, my favorite book is this one:

• Francois Treves, Topological vector Spaces, Distributions and Kernels, Dover Publications, 2006.

Since we replace the model space $\mathbb{R}^n$ with a Fréchet space $E$, there will be certain things that won’t work out as easily as for the finite dimensional $\mathbb{R}^n$, or not at all.

It is nevertheless possible to define both integrals and differentials that behave much in the expected way. You can find a nice exposition of how this can be done in this paper:

• Richard S. Hamilton, The inverse function theorem of Nash and Moser, Bulletin of the American Mathematical Society 7 (1982), pages 65-222.

The story starts with the definition of the directional derivative that can be done just as in finite dimensions:

Let $F$ and $G$ be Fréchet spaces, $U \subseteq F$ open and $P: U \to G$ a continuous map. The derivative of $P$ at the point $f \in U$ in the direction $h \in F$ is the map

$D P: U \times F \to G$

given by:

$D P(f) h = \lim_{t \to 0} \frac{1}{t} ( P(f + t h) - P(f))$

A simple but nontrivial example is the operator

$P: C^{\infty}[a, b] \to C^{\infty}[a, b]$

given by:

$P(f) = f f'$

with the derivative

$D P(f) h = f'h + f h'$

It is possible to define higher derivatives and also prove that the chain rule holds, so that we can define that a function between Fréchet spaces is smooth if it has derivatives at every point of all orders. The definition of a smooth Fréchet manifold is then straightforward: you can copy the usual definition of a smooth manifold word for word, replacing $\mathbb{R}^n$ by some Fréchet space.

With tangent vector s, you may remember that there are several different ways to define them in the finite dimensional case, which turn out to be equivalent. Since there are situations in infinite dimensions where these definitions turn out to not be equivalent, I will be explicit and define tangent vector s in the “kinematic way”:

The (kinematic) tangent vector space $T_p M$ of a Fréchet manifold $M$ at a point $p$ consists of all pairs $(p, c'(0))$ where $c$ is a smooth curve

$c: \mathbb{R} \to M \; \textrm{\; with\; } c(0) = p$

With this definition, the set of pairs $(p, c'(0)), p \in M$ forms a Fréchet manifold, the tangent bundle $T M$, just as in finite dimensions.

The first serious (more or less) problem we encounter is the definition of the cotangent bundle: $\mathbb{R}^n$ is isomorphic to its dual vector space. This is still true for every Hilbert space (this is known as the Riesz representation theorem). It fails already for Banach spaces: The dual space will still be a Banach space, but a Banach space does not need to be isomorphic to its dual, or even the dual of its dual (though the latter situation happens quite often, and such Banach spaces are called reflexive).

With Fréchet spaces things are even a little bit worse, because the dual of a Fréchet space (which is not a Banach space) is not even a Fréchet space! Since I did not know that and could not find a reference, I asked about this on Mathoverflow here and promptly got an answer. Mathoverflow is a really amazing platform for this kind of question!

So, if we naively define the cotangent space as in finite dimensions by taking the dual space of every tangent space, then the cotangent bundle won’t be a Fréchet manifold.

We will therefore have to be careful with the definition of differential forms for Fréchet manifolds and define it explicitly:

A differential form (a one form) $\alpha$ is a smooth map

$\alpha: T M \to \mathbb{R}$

where $T M$ is the tangent bundle, such that $\alpha$ restricts to a linear map on every tangent space $T_p M$.

Another pitfall is that theorems from multivariable calculus may fail in Fréchet spaces, like the existence and uniqueness theorem of Picard-Lindelöf for ordinary differential equations. Things are much easier in Banach spaces: If you take a closer look at multivariable calculus, you will notice that a lot of definitions and theorems actually make use of the Banach space structure of $\mathbb{R}^n$ only, so that a lot generalizes straight forward to infinite dimensional Banach spaces. But that is less so for Fréchet spaces.

By now you should feel reasonably comfortable with the notion of a Fréchet manifold, so let us talk about the kind of gadget that Arnold used to describe fluid flows: diffeomorphism groups that are both infinite-dimensional Riemannian manifolds and Lie groups.

### The geodesic equation for an invariant metric

If $M$ is both a Riemannian manifold and a Lie group, it is possible to define the concept of left or right invariant metric. A left or right invariant metric $d$ on $M$ is one that does not change if we multiply the arguments with a group element:

A metric $d$ is left invariant iff for all $g, h_1, h_2 \in G$:

$d (h_1, h_2) = d(g h_1, g h_2)$

Similarly, $d$ is right invariant iff:

$d(h_1, h_2) = d(h_1 g, h_2 g)$

How does one get a one-sided invariant metric?

Here is one possibility: If you take a Lie group $M$ off the shelf, you get two automorphisms for free, namely the left and right multiplication by a group element $g$:

$L_g, R_g: M \to M$

given by:

$L_g(h) := g h$

$R_g(h) := h g$

Pictorially speaking, you can use the differentials of these to transport vector s from the Lie algebra $\mathfrak{m}$ of $M$ – which is the tangent space at the identity of the group, $T_\mathrm{id}M$ – to any other tangent space $T_g M$. If you can define an inner product on the Lie algebra, you can use this trick to transport the inner product to all the other tangent spaces by left or right multiplication, which will get you a left or right invariant metric.

To be more precise, for every tangent vectors $U, V$ of a tangent space $T_{g} M$ there are unique vectors $X, Y$ that are mapped to $U, V$ by the differential of the right multiplication $R_g$, that is

$d R_g X = U \textrm{\; and \;} d R_g Y = V$

and we can define the inner product of $U$ and $V$ to have the value of that of $X$ and $Y$:

$\langle U, V \rangle := \langle X, Y \rangle$

This works for the left multiplication $L_g$, too, of course.

For a one-sided invariant metric, the geodesic equation looks somewhat simpler than for general metrics. Let us take a look at that:

On a Riemannian manifold $M$ with tangent bundle $T M$ there is a unique connection, the Levi-Civita connection, with the following properties for vector fields $X, Y, Z \in T M$:

$Z \langle X, Y \rangle = \langle \nabla_Z X, Y \rangle + \langle X, \nabla_Z Y \rangle \textrm{\; (metric compatibility)}$

$\nabla_X Y - \nabla_Y X = [X, Y] \textrm{\; (torsion freeness)}$

If we combine both formulas we get

$2 \langle \nabla_X Y, Z \rangle = X \langle Y, Z \rangle + Y \langle Z, X \rangle - Z \langle X, Y \rangle + \langle [X, Y], Z \rangle - \langle [Y, Z], X \rangle + \langle [Z, X], Y \rangle$

If the inner products are constant along every flow, i.e. the metric is (left or right) invariant, then the first three terms on the right hand side vanish, so that we get

$2 \langle \nabla_X Y, Z \rangle = \langle [X, Y], Z \rangle - \langle [Y, Z], X \rangle + \langle [Z, X], Y \rangle$

This latter formula can be written in a more succinct way if we introduce the coadjoint operator. Remeber the adjoint operator defined to be

$\mathrm{ad}_X Z = [X, Z]$

With the help of the inner product we can define the adjoint of the adjoint operator:

$\langle \mathrm{ad}^*_X Y, Z \rangle := \langle Y, \mathrm{ad}_X Z \rangle = \langle Y, [X, Z] \rangle$

Beware! We’re using the word ‘adjoint’ in two completely different ways here, both of which are very common in math. One way is to use ‘adjoint’ for the operation of taking a Lie bracket: $\mathrm{ad}_X Z = [X,Z]$. Another is to use ‘adjoint’ for the linear map $T: W \to V$ coming from a linear map between inner product spaces $T: V \to W$ given by $\langle T^* w, v \rangle =$ $\langle w, T v \rangle.$ Please don’t blame me for this.

Then the formula above for the covariant derivative can be written as

$2 \langle \nabla_X Y, Z \rangle = \langle \mathrm{ad}_X Y, Z \rangle - \langle \mathrm{ad}^*_Y X, Z \rangle - \langle \mathrm{ad}^*_X Y, Z \rangle$

Since the inner product is nondegenerate, we can eliminate $Z$ and get

$2 \nabla_X Y = \mathrm{ad}_X Y - \mathrm{ad}^*_X Y - \mathrm{ad}^*_Y X$

A geodesic curve is one whose tangent vector $X$ is transported parallel to itself. That is, we have

$\nabla_X X = 0$

Using the formula for the covariant derivative for an invariant metric above we get

$\nabla_X X = - \mathrm{ad}^*_X X = 0$

as a reformulation of the geodesic equation.

For time dependent dynamical systems, we have the time axis as an additional dimension and every vector field has $\partial_t$ as an additional summand. So, in this case we get as the geodesic equation (again, for an invariant metric):

$\nabla_X X = \partial_t X - \mathrm{ad}^*_X X = 0$

### A simple example: the circle

As a simple example we will look at the circle $S^1$ and its diffeomorphism group $\mathrm{Diff} S^1.$ The Lie algebra $\mathrm{Vect}(S^1)$ of $\mathrm{Diff} S^1$ can be identified with the space of all vector fields on $S^1.$ If we sloppily identify $S^1$ with $\mathbb{R}/\mathbb{Z}$ with coordinate $x$, then we can write for vector fields $X = u(x) \partial_x$ and $Y = v(x) \partial_x$ the commutator

$[X, Y] = (u v_x - u_x v) \partial_x$

where $u_x$ is short for the derivative:

$\displaystyle{ u_x := \frac{d u}{d x} }$

And of course we have an inner product via

$\langle X, Y \rangle = \int_{S^1} u(x) v(x) d x$

which we can use to define either a left or a right invariant metric on $\mathrm{Diff} S^1$, by transporting it via left or right multiplication to every tangent space.

Let us evaluate the geodesic equation for this example. We have to calculate the effect of the coadjoint operator:

$\langle \mathrm{ad}^*_X Y, Z \rangle := \langle Y, \mathrm{ad}_X Z \rangle = \langle Y, [X, Z] \rangle$

If we write for the vector fields $X = u(x) \partial_x$, $Y = v(x) \partial_x$ and $Z = w(x) \partial_x$, this results in

$\langle \mathrm{ad}^*_X Y, Z \rangle = \int_{S^1} v (u w_x - u_x w) d x = - \int_{S^1} (u v_x + 2 u_x v) w d x$

where the last step employs integration by parts and uses the periodic boundary condition $f(x + 1) = f(x)$ for the involved functions.

So we get for the coadjoint operator

$\mathrm{ad}^*_X Y = - (u v_x + 2 u_x v) \partial_x$

Finally, the geodesic equation

$\partial_t X + \nabla_X X = 0$

turns out to be

$u_t + 3 u u_x = 0$

A similar equation,

$u_t + u u_x = 0$

is known as the Hopf equation or inviscid Burgers’ equation. It looks simple, but its solutions can produce behaviour that looks like turbulence, so it is interesting in its own right.

If we take a somewhat more sophisticated diffeomorphism group, we can get slightly more complicated and therefore more interesting partial differential equations like the Korteweg-de Vries equation. But since this post is quite long already, that topic will have to wait for another post!

### 28 Responses to Fluid Flows and Infinite-Dimensional Manifolds I

1. mathenaetor says:

Reblogged this on mathenaetor.

2. James Griffin says:

Thank you for the post, I’ve been meaning to brush up my differential geometry, I now look forward to the next post. I do have a couple of questions:

1) I don’t quite understand the Levi-Cevita connection part. Is there a term missing when you combine the two formulae? Also I don’t see how to make the next jump. Is this compatibility of the inner product with the Lie bracket?

2) In your final example, what generality are we working in? When we integrate around the circle, is that an inner product, does it define a norm, or is it just the metric of a Fréchet manifold? I wish I had the time to investigate myself! Also is there any kind of physical system associated to this example?

• John Baez says:

Also is there any kind of physical system associated to this example?

Yes, near the end of this post Tim is deriving the inviscid Burgers equation in the case where space is a circle. The same idea works in higher dimensions, for example 2 or 3. The idea is that we have a pressure-free, zero-viscosity gas of particles each of which move along a straight line (or geodesic) at constant speed. Thus, the convective derivative of the velocity vector field is zero:

$\displaystyle{ \frac{d}{d t} u + u \cdot \nabla u = 0 }$

I hope you (and indeed everyone in the universe) knows about the convective derivative of the velocity vector field, which is the left-hand side of the above equation. It’s the time derivative of the velocity vector field as seen by a particle moving along with that velocity—that’s what adds that second term! Does everyone still learn about this in school?

There’s a funny factor of 3 in Tim’s version of this equation, which I find puzzling, but we can make it go away by doing a simple transformation:

$v = \frac{1}{3} u$

One fun thing about the inviscid Burgers equation is that it develops ‘shock waves’ when particles with different velocities crash into each other. Here we see the shock wave developing as the faster particles move into a region of zero-velocity particles and speed of the particles becomes a discontinuous function:

This is the beginning of a longer story…

• James Griffin says:

Thank you, that’s really helped. So perhaps the most natural solutions for this equation aren’t continuous at all, they’re delta or Green’s functions. Steps travel around the circle with velocity proportional to their height.

Obviously you can’t just add any two solutions to these differential equations; but from the physical description you can take the unions of the two sets of particles, but I suppose we are assuming that they don’t ‘overlap’. I feel that there should be an algebraic analogue of this by which you can combine any two solutions, but perhaps you have to re-parametrize either the circle or time to make it work.

• Correct me if I’m wrong, but doesn’t the formation of a shock just correspond to the geodesic hitting the boundary (I think $\mathrm{Diff}(S^1)$ is noncompact and has a boundary)? So the shock would simply be an endpoint after which the solution is no longer defined (at least in the differential geometric setting).

• also, how do TeX here?!

• John Baez says:

James Griffin wrote:

Obviously you can’t just add any two solutions to these differential equations; but from the physical description you can take the unions of the two sets of particles, but I suppose we are assuming that they don’t ‘overlap’.

Hmm, that’s interesting! Your somewhat ‘algebraic’ outlook on this problem sounds worth pursuing. The awkwardness of assuming two sets don’t overlap seems related to the existence of shocks, since shocks happen when they start to overlap.

It might be possible to enlarge $\mathrm{Diff}(S^1)$ to the space of smooth maps from the circle to itself and extend the solutions of the equations beyond shocks that way. A map that’s not 1-1 would have several particles at the same point, typically with different velocities.

There’s been a lot of work on this stuff already, most of which I don’t know.

• John Baez says:

Heiki Arponen wrote:

Correct me if I’m wrong, but doesn’t the formation of a shock just correspond to the geodesic hitting the boundary (I think $\mathrm{Diff}(S^1)$ is noncompact and has a boundary)?

I get what you mean, and it’s a good intuition, but technically it’s a bit slippery.

First of all, it’s hard for to make sense of the idea that a topological group has a ‘boundary’. In a group every point is just like every other (thanks to left or right translation symmetry), so it doesn’t seem to make sense to say one is ‘on the boundary’. We can try to imagine some sort of ideal boundary added on to $\mathrm{Diff}(S^1)$ consisting of points that describe shocks. But I haven’t worked out the details. Maybe someone has.

Second, on a lesser note, $\mathrm{Diff}(S^1)$ isn’t even locally compact, because it’s infinite-dimensional.

The safest thing to say for now is that some geodesics in $\mathrm{Diff}(S^1)$ ‘run off to infinity in a finite amount of time’.

This can’t happen for geodesics on a finite-dimensional Lie group equipped with some left-invariant metric, even a noncompact Lie group.

So the shock would simply be an endpoint after which the solution is no longer defined (at least in the differential geometric setting).

Note the shock isn’t actually in $\mathrm{Diff}(S^1)$, so to make sense of this ‘endpoint’ we’d need to add some ‘boundary points’ to this group.

also, how do TeX here?!

On this blog, and all WordPress blogs, you need to type

$latex x = y$

to get the equation

$x = y$

The ‘latex’ must occur directly after the dollar sign, with no space in between. There must be a space after it. And double dollar signs don’t work, at least on my blog.

• Thanks for the reply John, that makes a lot of sense! I meant with a “boundary” a similar concept from finite dimensional noncompact symmetric spaces, where indeed the geodesic distance to the boundary is infinite, unlike in $\mathrm{Diff}(S^1)$ (insofar as a concept of a boundary makes sense here).

Again, this is kind of a stretch, but I guess $\mathrm{Diff}(S^1)$ is isomorphic to the group of conformal maps on the upper half plane (preserving the real axis), so that might provide the “analytic continuation”?

• John Baez says:

Heikki wrote:

Again, this is kind of a stretch, but I guess $\mathrm{Diff}(S^1)$ is isomorphic to the group of conformal maps on the upper half plane (preserving the real axis) [….]

Interesting idea. It needs to be polished a bit:

There’s a problem with how a rotation of $S^1$ corresponds to a transformation that maps $\infty$ to a finite real number.

Also, we’d better work with real-analytic diffeomorphisms of $S^1$, not just smooth ones, if we want something like this.

I think it’s a bit easier to work with the unit disk than the upper half-plane, and try to show any real-analytic diffeomorphism of the unit circle extends to an analytic map from the unit disk to itself. Is this true? If it’s true, clearly the extension is unique.

[…] so that might provide the “analytic continuation”?

What I really want is a larger space such that when a solution of the inviscid Burgers’ equation develops a shock, it ceases to be an element of $\mathrm{Diff}(S^1)$ but still lies in this larger space.

I think the obvious choice is the space of smooth maps from the circle to itself. Using the ‘particle picture’ of the inviscid Burgers’ equation, a shock occurs when two or more particles hit, and this means we get a smooth map that’s no longer one-to-one.

Someone should have studied this already! Time to do a little poking around in the literature…

• Willie Wong says:

For question (1a) Yes, the Koszul formula is missing one term. See e.g. http://en.wikipedia.org/wiki/Fundamental_theorem_of_Riemannian_geometry For question (1b) that is the compatibility of the choice of the vector fields $X,Y,Z$. In other words, it is postulated that $X\langle Y,Z\rangle$ etc. are 0. In the case of the Lie group, the assumption probably is that the vector fields $X,Y,Z$ is given by transporting tangent vectors at the identity with the group action.

• Tim van Beek says:

Sorry for being absent, I have been terribly busy yesterday and am today.
(1a) Willie is right, but the missing term is recovered in the next line :-)

(1b) Willie is right again. I should have been clearer: The step is actually that we assume that we look at vector fields such that the scalar product is constant along all flows generated by all fields. This happens with an invariant metric and invariant vector fields.

3. X says:

I respectfully submit that Arnold, in all his glory, is not “the” Russian mathematician. :-)

4. Excellent post! I just happened to be getting to know these things in the course of my work, so it’s doubly interesting. I wish all math textbooks were written this clearly :)

Now only if there was a way of describing viscous flows as geodesic flows on Diff(M)… Maybe some stochasticity would give rise to viscosity?

• Tim van Beek says:

Interesting question, but unfortunately I have no idea. I’m still working on an understanding of ideal incompressible flows. That’s what Arnold actually worked on.

• WebHubTelescope says:

Stochasticity contributes to the wave energy spectrum of typical open-ocean waves.

First, we make a maximum entropy estimation of the energy of a one-dimensional propagating wave driven by a prevailing wind direction. The mean energy of the wave is related to the wave height by the square of the height, H. This makes sense because a taller wave needs a broader base to support that height, leading to a scaled pseudo-triangular shape, as shown in this figure.

Since the area of such a scaled triangle goes as H^2, the MaxEnt cumulative probability is:

$P(H) = e^{-a H^2}$

where a is related to the mean energy of an ensemble of waves. This relationship is empirically observed from measurements of ocean wave heights over a sufficient time period. However, we can proceed further and try to derive the dispersion results of wave frequency, which is the very common oceanography measure. So we consider — based on the energy stored in a specific wave — the time, t, it will take to drop a height, H, by the Newton’s law relation:

$t^2 \sim H$

and since t goes as 1/f, then we can create a new PDF from the height cumulative as follows:

$p(f) df = \frac{dP(H)}{dH} \frac{dH}{df} df$

where

$H \sim \frac{1}{f^2}$

$\frac{dH}{df} \sim -\frac{1}{f^3}$

then

$p(f) \sim \frac{1}{f^5} e^{-\frac{c}{f^4}}$

which is just the Pierson-Moskowitz wave spectra that oceanographers have observed for years (developed first in 1964, variations of this include the Bretschneider and ITTC wave spectra).

One can actually test this by visiting one of the real-time buoy measuring stations, this one off of San Diego.
http://cdip.ucsd.edu/?nav=historic&sub=data&units=metric&tz=UTC&pub=public&map_stati=1,2,3&stn=167&stream=p1&xyrmo=201201&xitem=product25

A typical spectrum with a fit looks like this

This is an interesting derivation that I have not seen anywhere before and it is one of those cases that first-order physics (Newton’s Law plus MaxEnt) works pretty effectively to describe real-world behavior for a seemingly complex phenomena.

5. Nick says:

Excellent article, Tim!

I have a few naive questions:

Besides the correspondence of the fluid parcel pathlines and the curvature of the manifold, is there any other information we can get from the manifold that will tell us about the nature of the fluid flow?

For instance, if we were ignorant about the nature of the solutions to the inviscid Burger equation, are there any properties of $Diff(S^1)$ that would alert us to the fact that the characteristics of the equation governing the geodesics cross?

Can you assign a corresponding geodesic equation to the dynamical equation of any inviscid fluid flow? Is there then necessarily a manifold that corresponds with this geodesic equation?

Although most of this is out of my league, I find the connection interesting and beautiful and in my ignorance I am trying to find places where this alternative way of looking at fluids would yield new information that would not be apparent, or would be very difficult to deduce, in the classical description.

Nick

• Tim van Beek says:

Those are not naive questions, at least not from where I am standing.

Besides the correspondence of the fluid parcel pathlines and the curvature of the manifold, is there any other information we can get from the manifold that will tell us about the nature of the fluid flow?

For instance, if we were ignorant about the nature of the solutions to the inviscid Burger equation, are there any properties of $\mathrm{Diff}(S^1)$ that would alert us to the fact that the characteristics of the equation governing the geodesics cross?

Both John and I plan to write a follow-up to this post where we will address this (although that is easier announced than done, of course). A relevant paper about shock waves and how they can be described using the viewpoint of manifolds is this:

• B. Khesina and G. Misiolek, Shock waves for the Burgers equation and curvatures of diffeomorphism groups.

Can you assign a corresponding geodesic equation to the dynamical equation of any inviscid fluid flow? Is there then necessarily a manifold that corresponds with this geodesic equation?

I have no idea, but I guess that we’d have to make “any” in “any inviscid fluid flow” more precise in order to get an answer.

6. I recently developed a non-breaking-wave derivation here:
http://theoilconundrum.blogspot.com/2012/01/wave-energy-spectrum.html
The math may seem a bit pedantic, but I like pedantic. Nothing but Newton’s law, MaxEnt principle, and inverting probability densities are involved.

I believe the key to describing all these physical phenomenon in practice is to assume disorder. I would appreciate getting some feedback on the approach, especially in regards to finding an alternate derivation for wave spectra. Every oceanography text I have flipped through just states the Pierson-Moskowitz wave spectra as a given without providing a first-principles derivation.

I think that is kind of common in the earth and geosciences disciplines and is why I appreciate the Azimuth — as the understanding always seems to come first here.

7. James Griffin says:

Interesting discussions. I’d just like to make three points:
1) It’s sort of obvious but I only just noticed that if you integrate u around the circle then you get a conserved quantity (w.r.t. time). Both the physical and geodesic interpretations are quite natural.

2) If you make sure that the above integral is zero then a step function describes a periodic geodesic (just integrate by hand starting at the identity, or constant map). Having said that I haven’t done this rigorously, I could be mistaken.

2) One could try to do the differential geometry of a function which takes multiple values (multiple ‘particles’ with differing velocities) by replacing $Diff(S^1)$ with a category of spans. When composing one has to be careful taking the pullbacks (I think it’s possible to keep our pullbacks being closed 1-manifolds if not a single circle). The Lie algebra looks like the tensor product of two copies of the Lie algebra above, however there is an issue with the invariant metric, because it isn’t both left and right invariant I think you have to change the Lie algebra slightly. You also need to change the metric a little and make sure that everything you do is invariant with respect to a diagonal copy of $Diff(S^1)$ which corresponds to equivalence of spans, or reparametrisation of our curve.

Having said all this I’m not convinced that what you get behaves nicely. I’ve so far failed to fill in all the details.

8. nick says:

A few more observations:

It seems clear that there will be a Hamiltonian structure associated with the geodesics of the manifold, since these flow lines are inherently minimizing a quantity.

From these will we be able to deduce relevant conservation laws of the fluid? Also, I would expect the Hamiltonian structure to yield additional information about the dynamic equations of the geodesics, but precisely what I’m unsure of.

Since speculation gets us nowhere, I’ll start to scour the internet for relevant resources.

9. Tim van Beek says:

Thanks for the input, I will have a look at it. I wrote a stub on the Azimuth Projekt, analytical hydrodynamics, with three references to textbooks that I use to fill me in. All three of them seem to be quite useful to me, and you can find some answers to your questions there.

10. […] 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. […]

11. pedro says:

Hello,

I’m quite puzzled by the burgers equation… If we dealing with this group can we identify the $g_{t}(x) = e^{ix}$ then the eulerian velocity $v = \dot{g} g^{-1} = i$ so the velocity is only imaginary?

• John Baez says:

I don’t know where you’re getting that $i$ in your $e^{ix}$. There are no complex numbers in Tim’s post or in the Burgers equation. Physicists doing quantum mechanics like to insert $i$ in the definition of Lie algebras, so that group elements look like $e^{ix}$ where $x$ is a Lie algebra element. But mathematicians know this is unnecessary, and in contexts like the Burgers equation this it would be very confusing. If $x$ is a vector field on a compact manifold, we can exponentiate it and get a diffeomorphism $e^{x}.$

12. pedro says:

Hello,

I’m quite puzzled by the burgers equation… If we dealing with this group can we identify the $g_{t}(x) = e^{ix}$ then the eulerian velocity $v = \dot{g} g^{-1} = i$ so the velocity is only imaginary?