On Euler: the vorticity vector is not invariant, but vorticity over density is

Question

I've been reading Frankel's Geometry of Physics but I'm struggling to understand a section devoted to "Additional problems on fluid flow" (Sec. 4.3c in my edition).

Consider a fluid flow in $\mathbb R^3$ with density $\rho=\rho(t,x)$ and velocity vector ${\bf v}={\bf v} (t,x)$. Let $vol^3$ be the (standard?) volume form in $\mathbb R^3$.

By using the metric structure, one can define the velocity covector $\nu$ associated with ${\bf v}$ and define then the vorticity 2-form $\omega^2 := d\nu$ (the two is obviously not a square, but it is the notation adopted in the book for 2-forms).

It is easy to show (and I think I did) that the vorticity form is invariant under the flow (Frankel attributes this to Helmholtz: it easily follows from Euler's equation). This means that $$ L_{{\bf v}+\frac{\partial}{\partial t}} \omega^2 = 0, $$ where $L_{{\bf v}+\frac{\partial}{\partial t}}$ denotes the Lie derivative w.r.t. the space-time vector field ${\bf v}+\frac{\partial}{\partial t}$.

The author now makes the following observation:

Can you clarify the first three lines of this remark, please?

1. I do not really know what it means that a vector field is invariant (in the text, the definition of invariant under the flow is given only for forms);
2. maybe this is not best practice, but let's write everything in coordinates: $\nu$ is the 1-form whose coordinates are $v^i$. In turn, $\omega^2$ is the two form whose components make up $\text{curl} u$. How is it possible that $d\nu$ is invariant and its components (i.e. $\text{curl} u$) is not?
3. Do you understand where the density comes from? What's its role? What's the idea behind the sentence "the mass form $\rho vol^3$ is conserved... the vector $\omega/\rho$ should be conserved".

Answer 1

Small caveat, might be best to precise that you are considering an Eulerian fluid (no viscosity etc), which is not the default choice for a physicist.

For your first question, the vector field $\omega$ would be considered invariant if it has a zero Lie derivative along $v$ (Lie bracket of two vector fields). In usual physicist's notation, this would lead to: $$ \mathcal L_v\omega := \partial_t\omega+(v\cdot \nabla)\omega -(\omega\cdot \nabla)v = 0 $$ This equation is true only when the fluid is incompressible. If not, you'd be missing the term $-\omega (\nabla\cdot u)$ on the RHS.

For the second question, the issue is explained in the quote. Since $dv = i_\omega vol^3$, because $vol^3$ is not conserved for incompressible fluids. In usual notation: $$ \mathcal L_v vol^3 = (\nabla\cdot v)vol^3 $$ so the conservation of $vol^3$ is equivalent to the compressibility of the fluid. From the coordinate point of view, say in cartesian coordinates, you can see this as well as: $$ dv = \omega_x dy\wedge dz+\omega_y dz\wedge dx+\omega_z dx\wedge dy $$ When you apply the Lie derivative, you take the derivative, you apply it to the components of $\omega$, but to the the two forms $dx\wedge dy ...$ as well. Both terms given by Leibniz' rule are non zero, but cancel out when you add them to get the conservation of $dv$.

For the third question, the idea is to cancel out the pesky $vol^3$ factor. Formally, $dv=i_\omega vol^3$ and $\rho vol^3$ both have the common factor of the $vol^3$. By dividing them, you would get a vector field that would be automatically invariant since it is constructed from conserved quantities. Using physicist's notation: $$ \begin{align} \mathcal L_v \rho vol^3 &= 0 &&\to &\partial_t\rho+\nabla\cdot(\rho v) &= 0 \\ \mathcal L_v i_\omega vol^3 &= 0 &&\to &\partial_t\omega+(v\cdot \nabla)\omega -(\omega\cdot \nabla)v+\omega (\nabla\cdot u) &= 0 \\ \mathcal L_v \frac{\omega}{\rho} &= 0 &&\to & \partial_t\left(\frac{\omega}{\rho}\right)+(v\cdot \nabla)\left(\frac{\omega}{\rho}\right)-\left(\left(\frac{\omega}{\rho}\right)\cdot \nabla\right)v &= 0 \end{align} $$ and you can check that the third equation can be deduced from the first two. This is where you realise the power k-forms since it would be hard to guess that the ratio is conserved just from staring at the equations written with $\nabla$.

Hope this helps.