Here we discus how material line elements evolve in a fluid and further we show that in an inviscid fluid vorticity lines move with the flow, that is vorticity lines move as if they were material lines.

## How material line elements move in a fluid?¶

Initially at time $t$ our material line element is given as:

$$\delta\boldsymbol{\ell} = \boldsymbol{x}_B-\boldsymbol{x}_A. \tag{1}$$

We are interested to find the rate of change of $\delta\boldsymbol{\ell}$ as we follow it with the flow. This is Lagrangian thinking.

In a Lagrangian framework we follow the fluid parcels as they move with the flow and compute in this manner the rate of change of any quantity we are interested along with the flow. In an Eulerian framework, stay put at a fixed point in space and we compute the rate of change of a quantity as fluid is being advected. We can always translate form one framework to the other.

For our calculation here, it is much easier to compute the rate of change of $\delta\boldsymbol{\ell}$ in the Lagrangian framework. We follow points $A$ and $B$ for time $\delta \tau$. At $t=t+\delta t$, the points $A$ and $B$ have moved to points $A'$ and $B'$ respectively. Thus, our material line at $t=t+\delta\tau$ is:

\begin{align*} \delta\boldsymbol{\ell}' & = \boldsymbol{x}_{B'}-\boldsymbol{x}_{A'} \\ & = \big\{\boldsymbol{x}_B + \delta\tau\,\boldsymbol{u}(\boldsymbol{x}_B,t) + \mathcal{O}\big[(\delta\tau)^2\big]\big\} -\big\{\boldsymbol{x}_A + \delta\tau\,\boldsymbol{u}(\boldsymbol{x}_A,t) + \mathcal{O}\big[(\delta\tau)^2\big]\big\}\\ & = \delta\boldsymbol{\ell} + \delta\tau\big[\boldsymbol{u}(\boldsymbol{x}_B,t) - \boldsymbol{u}(\boldsymbol{x}_A,t)\big] + \mathcal{O}\big[(\delta\tau)^2\big]. \tag{2} \end{align*}

But, since points $A$ and $B$ are very close to each other we can write:

$$\boldsymbol{u}(\boldsymbol{x}_B,t) = \boldsymbol{u}(\boldsymbol{x}_A,t) + (\underbrace{\boldsymbol{x}_B-\boldsymbol{x}_A}_{=\delta\boldsymbol{\ell}})\boldsymbol{\cdot}\boldsymbol{\nabla}\boldsymbol{u}(\boldsymbol{x}_A,t) + \mathcal{O}\big[(\delta\boldsymbol{\ell})^2\big], \tag{3}$$

which implies:

\begin{align*} \delta\boldsymbol{\ell}' & = \delta\boldsymbol{\ell} + \delta\tau\,\delta\boldsymbol{\ell}\boldsymbol{\cdot}\boldsymbol{\nabla}\boldsymbol{u}(\boldsymbol{x}_A,t) + \mathcal{O}\big[(\delta\tau)^2, \delta \tau\,(\delta\boldsymbol{\ell})^2\big]. \tag{4} \end{align*}

From (4) we can compute the Lagrangian rate of change of the infinitesimal material element $\delta\boldsymbol{\ell}$. That is the rate of change $\delta\boldsymbol{\ell}$ as we follow it with the flow:

$$\left.\frac{\mathrm{d}}{\mathrm{d}t}\delta\boldsymbol{\ell}\right|_{\mathrm{L}} = \lim_{\delta\tau\to0} \frac{\delta\boldsymbol{\ell}' - \delta\boldsymbol{\ell}}{\delta\tau} = \delta\boldsymbol{\ell}\boldsymbol{\cdot}\boldsymbol{\nabla}\boldsymbol{u}(\boldsymbol{x}_A,t) + \mathcal{O}\big[(\delta\boldsymbol{\ell})^2\big]. \tag{5}$$

Subscript $\mathrm{L}$ in equation (5) denotes that the rate of change is computed in the Lagrangian framework.

For infinitesimal line elements we can neglect terms of order $(\delta\boldsymbol{\ell})^2$ and thus (5) simplifies to:

$$\left.\frac{\mathrm{d}}{\mathrm{d}t}\delta\boldsymbol{\ell} \right|_{\mathrm{L}}= \delta\boldsymbol{\ell} \boldsymbol{\cdot}\boldsymbol{\nabla}\boldsymbol{u}, \tag{6}$$

with both $\delta\boldsymbol{\ell}$ and $\boldsymbol{u}$ evaluated at $\boldsymbol{x},t$.

How can we tranform the above to our usual beloved Eulerian frame? The time-derivative $\mathrm{d}/\mathrm{d}t\big|_{\mathrm{L}}$ becomes in the Eulerian framework the material derivative:

$$\big(\partial_t + \boldsymbol{u}\boldsymbol{\cdot}\boldsymbol{\nabla}\big) \delta\boldsymbol{\ell} = \delta\boldsymbol{\ell} \boldsymbol{\cdot}\boldsymbol{\nabla}\boldsymbol{u}. \tag{7}$$

## Vorticity equation for an inviscid fluid¶

Let's take the simplest case of a fluid with homogeneous density, $\rho=\rho_m=\textrm{const}$. Then the mass conservation equation $\partial_t\rho + \boldsymbol{\nabla} \boldsymbol{\cdot} (\rho\,\boldsymbol{u}) = 0$ simplifies to

$$\boldsymbol{\nabla} \boldsymbol{\cdot} \boldsymbol{u} = 0. \tag{8}$$

The momentum equatio n for a fluid under the influence of gravity is then:

$$\partial_t \boldsymbol{u} + \boldsymbol{u} \boldsymbol{\cdot} \boldsymbol{\nabla} \boldsymbol{u} = -\frac{\boldsymbol{\nabla}p}{\rho_m} + \rho_m g\widehat{\boldsymbol{z}}.\tag{9}$$

Note that because (i) gravity is a conservative force (i.e., it derives from a potential) and (ii) $\rho_m=\textrm{const.}$, we can write the right-hand-side of the above as

$$-\boldsymbol{\nabla} \Big( \frac{p}{\rho_m} - \rho_m g z \Big).\tag{10}$$

The vorticity $\boldsymbol{\omega}$ of the fluid is defined as:

$$\boldsymbol{\omega} =\boldsymbol{\nabla} \times \boldsymbol{u}.\tag{11}$$

We would like to find how the vorticity evolves. To do so we take the curl of (9) and use (11) to get:

$$\partial_t \boldsymbol{\omega} + \boldsymbol{\nabla}\times\big(\boldsymbol{u} \boldsymbol{\cdot} \boldsymbol{\nabla} \boldsymbol{u}\big) = 0.\tag{12}$$

The term $\boldsymbol{\nabla}\times\big(\boldsymbol{u} \boldsymbol{\cdot} \boldsymbol{\nabla} \boldsymbol{u}\big)$ involves the flow field $\boldsymbol{u}$ as well as spatial derivatives of $\boldsymbol{u}$. After doing some (rather elaborate) vector calculus and use of equation (8) we can rewrite

$$\boldsymbol{\nabla}\times\big(\boldsymbol{u} \boldsymbol{\cdot} \boldsymbol{\nabla} \boldsymbol{u}\big) = \boldsymbol{u}\boldsymbol{\cdot}\boldsymbol{\nabla} \boldsymbol{\omega} - \boldsymbol{\omega}\boldsymbol{\cdot}\boldsymbol{\nabla}\,\boldsymbol{u}.\tag{13}$$

(The steps to obtain (13) are described below. But if you find those difficult or confusing then just take equation (13) as an identity that holds whenever (8) is true.)

Therefore, the vorticity equation for a homogeneous inviscid fluid becomes:

$$\partial_t \boldsymbol{\omega} + \boldsymbol{u}\boldsymbol{\cdot}\boldsymbol{\nabla} \boldsymbol{\omega} = \boldsymbol{\omega}\boldsymbol{\cdot}\boldsymbol{\nabla}\,\boldsymbol{u},\tag{14}$$

or more compactly written,

$$\frac{\mathrm{D} \boldsymbol{\omega}}{\mathrm{D}t} = \boldsymbol{\omega}\boldsymbol{\cdot}\boldsymbol{\nabla}\,\boldsymbol{u}.\tag{15}$$

Exercise: Assume the flow $\boldsymbol{u}$ is two-dimensional. What does that imply about the direction of $\boldsymbol{\omega}$? In this case, what does equation (15) imply?

## Vorticity lines move with the flow in an inviscid fluid¶

The resemblance of equations (7) and (14) demonstrates the above. Vorticity lines and material lines evolve in the same way in an inviscid fluid.

This is an extremely powerful and useful result.

Exercise: Does this argument follow on in a viscous fluid? Consider a fluid with viscosity $\nu$. Does equation (7) change? Does equation (14) change?

### Steps to derive equation (13)¶

First let's bring this vector identity to our short-term memory:

$$\boldsymbol{a} \times (\boldsymbol{b} \times \boldsymbol{c}) = (\boldsymbol{a}\boldsymbol{\cdot}\boldsymbol{c})\,\boldsymbol{b} - (\boldsymbol{a}\boldsymbol{\cdot}\boldsymbol{b})\,\boldsymbol{c}.\tag{16}$$

Does the identity above make sense? Definitely the vector on the left-hand-side will be lying on the plane which is (i) normal to $\boldsymbol{a}$ and (ii) normal to $(\boldsymbol{b}\boldsymbol{\times}\boldsymbol{c})$. But (ii) implies that $\boldsymbol{a} \times (\boldsymbol{b} \times \boldsymbol{c})$ will by lying on the plane that is spanned by vectors $\boldsymbol{b}$ and $\boldsymbol{c}$ and this is exactly what (16) means.

Identity (16) is often referred to as the "$a$-$b$-$c$ identity". We can derive (16) easily using index notation:

\begin{align*} [ \boldsymbol{a} \times (\boldsymbol{b} \times \boldsymbol{c}) ]_i &= \epsilon_{ijk} a_j [ \boldsymbol{b} \times \boldsymbol{c} ]_k \\ &= \epsilon_{ijk} a_j \epsilon_{klm} b_l c_m \\ &= \big(\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl}\big) a_j b_l c_m \\ &= a_j b_i c_j - a_j b_j c_i \\ &= (\boldsymbol{a}\boldsymbol{\cdot}\boldsymbol{c}) b_i - (\boldsymbol{a}\boldsymbol{\cdot}\boldsymbol{b}) c_i \end{align*}

(We can also derive it using geometry if we consider two cases: first take $\boldsymbol{a}$ to be perpendicular to $\boldsymbol{b}$ and then $\boldsymbol{a}$ to be perpendicular to $\boldsymbol{c}$.)

So can we use the $a$-$b$-$c$ identity to compute $\boldsymbol{u} \times \boldsymbol{\omega}$?

Yes, but we have to be careful on where the $\boldsymbol{\nabla}$ acts on since the differentiation does not commute with the vectos (i.e., $f\partial_x g \ne \partial_x (fg)$). For example, we have $\boldsymbol{a}\mapsto\boldsymbol{u}$, $\boldsymbol{b}\mapsto\boldsymbol{\nabla}$, and $\boldsymbol{c}\mapsto\boldsymbol{u}$. The term $(\boldsymbol{a}\boldsymbol{\cdot}\boldsymbol{b})\,\boldsymbol{c}$ is rather trivial:

\begin{align*} (\boldsymbol{a}\boldsymbol{\cdot}\boldsymbol{b})\,\boldsymbol{c} \mapsto (\boldsymbol{u}\boldsymbol{\cdot}\boldsymbol{\nabla})\,\boldsymbol{u},\tag{17} \end{align*}

but the term $(\boldsymbol{a}\boldsymbol{\cdot}\boldsymbol{c})\,\boldsymbol{b}$ is tricky. If we just replace everything we have

\begin{align*} (\boldsymbol{a}\boldsymbol{\cdot}\boldsymbol{c})\,\boldsymbol{b} \mapsto (\boldsymbol{u}\boldsymbol{\cdot}\boldsymbol{u})\,\boldsymbol{\nabla}.\tag{18} \end{align*}

But (18) makes no sense since $\boldsymbol{\nabla}$ is left acting on thin air that is on its right! What we should have written is

\begin{align*} (\boldsymbol{a}\boldsymbol{\cdot}\boldsymbol{c})\,\boldsymbol{b} \mapsto \boldsymbol{\nabla} (\boldsymbol{u}\boldsymbol{\cdot}\boldsymbol{u}).\tag{19} \end{align*}

But wait again! Now $\boldsymbol{\nabla}$ acts on both vectors $\boldsymbol{u}$ (while initially it was acting only on one of them)! This is wrong... The correct way to do that is

\begin{align*} (\boldsymbol{a}\boldsymbol{\cdot}\boldsymbol{c})\,\boldsymbol{b} \mapsto \tfrac1{2} \boldsymbol{\nabla} (\boldsymbol{u}\boldsymbol{\cdot}\boldsymbol{u}),\tag{20} \end{align*}

which gives:

\begin{align*} \boldsymbol{u} \times \boldsymbol{\omega} = \boldsymbol{\nabla}\big(\tfrac1{2}\left\|\boldsymbol{u}\right\|^2\big) - (\boldsymbol{u}\boldsymbol{\cdot}\boldsymbol{\nabla})\,\boldsymbol{u}.\tag{21} \end{align*}

Rather mind boggling, right?!

I find it much easier to do these kind of calculations with index notation. Then everything is much clearer; we don't have to remember e.g. on which fields the differential operators were acting on. We just have to be careful with putting parenthesis at the right places.

\begin{align*} [\boldsymbol{u} \times \boldsymbol{\omega}]_i &= \big[ \boldsymbol{u} \times (\boldsymbol{\nabla} \times \boldsymbol{u}) \big]_i \\ &= \epsilon_{ijk} u_j [ \boldsymbol{\nabla} \times \boldsymbol{u} ]_k \\ &= \epsilon_{ijk} u_j \epsilon_{klm} \partial_{l}\, u_m \\ &= \big(\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl}\big) u_j \partial_{l}\, u_m \\ &= u_j \partial_i u_j - u_j \partial_j u_i \\ &= \partial_i \big(\tfrac1{2}\boldsymbol{u}\boldsymbol{\cdot}\boldsymbol{u}\big) - (\boldsymbol{u}\boldsymbol{\cdot}\boldsymbol{\nabla}) u_i. \end{align*}

Same as (21) in a straightforward manner! (Index notation and Einstein summation are great tools once you get the grips out of them.)

The first step towards getting equation (13) is to use (21) to rewrite the advection term as:

$$\boldsymbol{u}\boldsymbol{\cdot}\boldsymbol{\nabla} \boldsymbol{u} = \boldsymbol{\nabla}\big(\tfrac1{2}\left\|\boldsymbol{u}\right\|^2\big) + \boldsymbol{\omega} \times \boldsymbol{u}.\tag{22}$$

Next we want to compute the curl of (22). The curl of $\boldsymbol{\nabla}\big(\tfrac1{2}\left\|\boldsymbol{u}\right\|^2\big)$ vanishes; that's good. The real challenge is the term $\boldsymbol{\nabla}\times (\boldsymbol{\omega} \times \boldsymbol{u} )$.

If we use again the $a$-$b$-$c$ identity without much thinking, we get:

\begin{align*} \boldsymbol{\nabla}\times (\boldsymbol{\omega} \times \boldsymbol{u} ) & = (\boldsymbol{\nabla}\boldsymbol{\cdot}\boldsymbol{u})\,\boldsymbol{\omega} - (\boldsymbol{\nabla}\boldsymbol{\cdot}\boldsymbol{\omega})\,\boldsymbol{u}.\tag{23} \end{align*}

Wrong! Why? Same reason as before; the $\boldsymbol{\nabla}$ operator on the left-hand-side now acts on both $\boldsymbol{\omega}$ and $\boldsymbol{u}$ and on the right-hand-side it only acts on one of them.

All these is very confusing (for me) so I'll do it with index notation to be safe!

\begin{align*} [\boldsymbol{\nabla}\times (\boldsymbol{\omega} \times \boldsymbol{u} )]_i &= \epsilon_{ijk} \partial_j [ \boldsymbol{\omega} \times \boldsymbol{u} ]_k \\ &= \epsilon_{ijk} \partial_j \epsilon_{klm} \omega_l\, u_m \\ &= \big(\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl}\big) \partial_j \big(\omega_l \, u_m \big) \\ &= \partial_j (\omega_i \, u_j ) - \partial_j (\omega_j \, u_i ) \\ &= u_j \partial_j \omega_i + \omega_i \partial_j u_j - u_i \partial_j \omega_j - \omega_j \partial_j u_i \\ &= (\boldsymbol{u}\boldsymbol{\cdot}\boldsymbol{\nabla})\omega_i + \omega_i (\boldsymbol{\nabla}\boldsymbol{\cdot}\boldsymbol{u}) - u_i (\boldsymbol{\nabla}\boldsymbol{\cdot}\boldsymbol{\omega}) - (\boldsymbol{\omega}\boldsymbol{\cdot}\boldsymbol{\nabla}) u_i. \tag{24} \end{align*}

But, $\boldsymbol{\nabla}\boldsymbol{\cdot}\boldsymbol{\omega}=0$ (identically) and also for a homogeneous fluid we have $\boldsymbol{\nabla}\boldsymbol{\cdot}\boldsymbol{u}=0$ (see equation (1)). Thus, (24) simplifies to equation (13):

$$\boldsymbol{\nabla}\times (\boldsymbol{\omega} \times \boldsymbol{u} ) = \boldsymbol{u}\boldsymbol{\cdot}\boldsymbol{\nabla} \boldsymbol{\omega} - \boldsymbol{\omega}\boldsymbol{\cdot}\boldsymbol{\nabla}\,\boldsymbol{u}.$$

Exercise: Try generalizing the vorticity conservation for an inviscid fluid with varying density. Prove that for a barotropic fluid, that is, for a fluid for which $\boldsymbol{\nabla}\rho\times\boldsymbol{\nabla}p=0$ then:

$$\frac{\mathrm{D}}{\mathrm{D}t}\Big(\frac{\boldsymbol{\omega}}{\rho}\Big) = \Big(\frac{\boldsymbol{\omega}}{\rho}\Big)\boldsymbol{\cdot}\boldsymbol{\nabla}\,\boldsymbol{u}.\tag{25}$$

(Hint: You will have to revisit the mass conservation equation. Now, the mass conservation equation does not imply $\boldsymbol{\nabla}\boldsymbol{\cdot}\boldsymbol{u}=0$.)