Deriving Vorticity Transport in Index Notation

Definitions and Useful Tools #︎

Notation #︎

The vorticity transport equation can simply be calculated by taking the curl of the conservation of momentum evolution equations. See my earlier post going over expressing curl in index summation notation. In summary, the curl of a vector $a_j$ can be expressed as:

$$ \nabla \times a_j = b_k \ \Rightarrow \ \varepsilon_{ijk} \partial_i a_j = b_k $$

where $\varepsilon_{ijk}$ is the Levi-Civita symbol.

Note that I’ll be using shorthand to express the differential operator, $\partial_\phi$, where the index $\phi$ is the index of a spacial variable except for $t$, which represents a time derivative:

$$ \frac{\partial}{\partial x_i} = \partial_i \quad \mathrm{and} \quad \frac{\partial}{\partial t} = \partial_t$$

For double derivatives, a superscript will be used:

$$ \frac{\partial}{\partial x_i} \frac{\partial}{\partial x_i} = \frac{\partial^2}{\partial x_i^2} = \partial_i^2 $$

Math Properties #︎

Some useful properties of the Levi-Civita symbol that will be used are that it is commutative in multiplication and, since it is invariant of space and time, it can be brought in or outside a differential operator operator like a constant:

$$ \varepsilon_{ijk}\frac{\partial}{\partial x_i} (\phi ) = \frac{\partial}{\partial x_i} (\varepsilon_{ijk} \phi ) $$

Using the product identity of permutation tensors , you can also convert permutation tensors into Kronecker deltas, $\delta_{ij}$:

$$ \varepsilon_{ijk} \varepsilon_{imn} = \delta_{jm}\delta_{kn} - \delta_{jn}\delta_{km} $$

Hint: To convert from some arbitrary pair of Levi-Civita symbols that share one index, first rearrange their indices such that the shared index is the first for both. Then, the ordering of the Kronecker delta indices is $ \delta_{a_2\ b_2} \delta_{a_3\ b_3} - \delta_{a_2\ b_3} \delta_{a_3\ b_2}$ , where $a_i$ and $b_i$ represent the indices of the different Levi-Civita symbols and their subscripts represent which index is placed in that Kronecker delta.

Speaking of Kronecker deltas, they have the handy effect of changing indices:

$$ \delta_{ij} a_i = a_j $$

Differential operators are order invariant:

$$ \partial_i (\partial_j (u_k)) = \partial_j (\partial_i (u_k))$$

I will be dropping the use of parentheses for the differential operator, but note that it is an operator and is not commutative: $\partial_i u_j \neq u_j \partial_i$

With that taken care of, onto the derivation!

Curl of Momentum Evolution #︎

The index notation form of the incompressible momentum evolution (or conservation of momentum equations) is:

$$ \partial_t u_i + u_j \partial_j u_i = - \tfrac{1}{\rho} \partial_i p + \nu \partial_j^2 u_i $$

Vorticity, $\omega_k$, is given as the curl of velocity, or:

$$ \nabla \times u_j = \omega_k \ \Rightarrow \ \varepsilon_{ijk} \partial_i u_j = \omega_k $$

To get vorticity evolution, we can take the curl of the momentum transport equations:

$$ \nabla \times [\partial_t u_i + u_j \partial_j u_i = - \tfrac{1}{\rho} \partial_i p + \nu \partial_j^2 u_i ]$$

In index notation, this is the equivalent of multiplying by the Levi-Civita symbol and a corresponding differential operator:

$$ \Rightarrow \varepsilon_{k\ell i} \partial_\ell [\partial_t u_i + u_j \partial_j u_i = - \tfrac{1}{\rho} \partial_i p + \nu \partial_j^2 u_i ]$$

Distributing this across the terms, we get:

$$ \begin{align} \underbrace{\varepsilon_{k\ell i} \partial_\ell \partial_t u_i}_\text{Temporal Term} + \underbrace{\varepsilon_{k\ell i} \partial_\ell u_j \partial_j u_i}_\text{Advection Term} & = \underbrace{- \varepsilon_{k\ell i} \partial_\ell \tfrac{1}{\rho} \partial_i p}_\text{Pressure Term} + \underbrace{\varepsilon_{k\ell i} \partial_\ell \nu \partial_j^2 u_i}_\text{Viscous Term} \\
\Rightarrow \quad \mathbb{T} + \mathbb{A} & = \mathbb{P} + \mathbb{V} \end{align}$$

We’ll treat the named terms individually, then put them back together. I’ll leave the advection term for last since it’s more involved than the other three.

Individual Terms #︎

Temporal Term $\mathbb{T}$ #︎

$$\mathbb{T} = \varepsilon_{k\ell i} \partial_\ell \partial_t u_i $$

Using the order invariance of derivatives and moving $\varepsilon_{k\ell i}$ inside the derivative operators, we can get the following:

$$ \Rightarrow \quad \partial_t \varepsilon_{k\ell i} \partial_\ell u_i $$

Since we know that $\omega_k = \varepsilon_{k\ell i} \partial_\ell u_i$, then we can write:

$$\mathbb{T} = \partial_t \omega_k $$

Pressure Term $\mathbb{P}$ #︎

$$ \mathbb{P} = - \varepsilon_{k\ell i} \partial_\ell \tfrac{1}{\rho} \partial_i p $$

First we can move the density term out of the derivatives:

$$ \Rightarrow \quad - \tfrac{1}{\rho} \varepsilon_{k\ell i} \partial_\ell \partial_i p $$

The next step can go one of two ways. First you can simply use the fact that the curl of a gradient of a scalar equals zero ($\nabla \times (\partial_i \phi) = \mathbf{0}$). Or, you can be like me and want to prove that it is zero. I’ll probably do the former here, and put the latter in a separate post. Using the first method, we get that:

$$ \mathbb{P} = \mathbf{0} $$

Viscous Term $\mathbb{V}$ #︎

$$ \mathbb{V} = \varepsilon_{k\ell i} \partial_\ell \nu \partial_j^2 u_i $$

This step is almost identical to the Temporal Term; rearrange the terms such that you have the curl of velocity in order to get vorticity:

$$ \Rightarrow \quad \nu \partial_j^2 \varepsilon_{k\ell i} \partial_\ell u_i $$

$$ \Rightarrow \quad \mathbb{V} = \nu \partial_j^2 \omega_k $$

Advection Term $\mathbb{A}$ #︎

$$ \mathbb{A} = \varepsilon_{k\ell i} \partial_\ell u_j \partial_j u_i $$

This is quite a bit more tricky than the other three terms. Note that the derivatives are operators, so this maybe more explicitly written as:

$$ \mathbb{A} = \varepsilon_{k\ell i} \partial_\ell (u_j \partial_j (u_i) ) $$

First, we’ll use a specialized property/rule:

$$ u_j \partial_j u_i = \partial_i (\tfrac{1}{2} u_j u_j ) + (\nabla \times u_n) \times u_q $$

This is proved in the Appendix A. Converting the right term into index notation we get:

$$ u_j \partial_j u_i = \partial_i (\tfrac{1}{2} u_j u_j ) + \varepsilon_{ijq} u_q (\varepsilon_{jmn} \partial_m u_n) $$

Notice that the term in parentheses on the right is already the curl of velocity, so we’ll go ahead and turn that into vorticity.

$$ u_j \partial_j u_i = \partial_i (\tfrac{1}{2} u_j u_j ) + \varepsilon_{ijq} \omega_j u_q $$

Plugging this back into our expression for $\mathbb{A}$, we get:

$$\mathbb{A} = \varepsilon_{k\ell i} \partial_\ell [\partial_i (\tfrac{1}{2} u_j u_j ) + \varepsilon_{ijq} \omega_j u_q ] $$

Distributing this through, we get:

$$\Rightarrow \ \varepsilon_{k\ell i} \partial_\ell [\partial_i (\tfrac{1}{2} u_j u_j ) ] + \varepsilon_{k\ell i} \partial_\ell [ \varepsilon_{ijq} \omega_j u_q ] $$

For the lefthand term, note that $u_j u_j$ is just a scalar. Therefore, the left expression can be surmised as the curl of the gradient of a scalar and it is then equal to zero. This leaves us with:

$$\Rightarrow \ \mathbb{A} = \varepsilon_{k\ell i} \partial_\ell [ \varepsilon_{ijq} \omega_j u_q ] $$

Now let’s move $\varepsilon_{ijq}$ outside the differential and change the indices of $\varepsilon_{k\ell i} \Rightarrow \varepsilon_{ik\ell}$: :

$$\Rightarrow \ \mathbb{A} = \varepsilon_{ik\ell} \varepsilon_{ijq} \partial_\ell \omega_j u_q$$

Since we have two Levi-Civita symbols, we can utilize the permutation tensor’s product identity shown in the Math Properties section to turn this into an expression of Kronecker deltas.

$$ \varepsilon_{ik\ell} \varepsilon_{ijq} = \delta_{kj}\delta_{\ell q} - \delta_{kq}\delta_{\ell j} $$

which gives us

$$ \mathbb{A} = (\delta_{kj}\delta_{\ell q} - \delta_{kq}\delta_{\ell j} ) \partial_\ell \omega_j u_q $$

Using the Kronecker deltas to change indices yields:

$$ \mathbb{A} = \partial_q \omega_k u_q - \partial_j \omega_j u_k $$

Next, use the product rule to expand the derivatives:

$$ \Rightarrow \ (u_q \partial_q \omega_k + \omega_k \partial_q u_q) - ( u_k \partial_j \omega_j + \omega_j \partial_j u_k)$$

By incompressibility $\partial_q u_q =0$ (ie. the divergence of velocity is zero). Also, $\partial_j \omega_j$ also equals zero, which is proven in Appendix B. Substituting in we are left with:

$$ (u_q \partial_q \omega_k + 0) - ( 0 + \omega_j \partial_j u_k) $$

And thus we finally get the final form of the advection term:

$$ \Rightarrow \ \mathbb{A} = \underbrace{u_q \partial_q \omega_k}_\text{Vorticity Advection} - \underbrace{\omega_j \partial_j u_k}_\text{Vorticity Stretching} $$

Putting It All Together #︎

When we combine the terms together, we finally get the full vorticity transport equation:

$$ \underbrace{\partial_t \omega_k}_\mathbb{T} + \underbrace{u_q \partial_q \omega_k - \omega_j \partial_j u_k}_\mathbb{A} = \underbrace{0}_\mathbb{P} + \underbrace{\nu \partial_j^2 \omega_k}_\mathbb{V} $$

Appendix #︎

Appendix A #︎

Prove: $$ u_j \partial_j u_i = \partial_i (\tfrac{1}{2} u_j u_j ) + (\nabla \times u_n) \times u_q $$

Starting with the righthand side, put in index notation:

$$\partial_i (\tfrac{1}{2} u_j u_j ) + \varepsilon_{ijq} u_q (\varepsilon_{jmn} \partial_m u_n) $$

Take the right term and rearrange it: $$ … + \varepsilon_{jqi} \varepsilon_{jmn} u_q \partial_m u_n $$

Convert the product of Levi-Civita symbols to Kronecker deltas:

$$ … + (\delta_{qm} \delta_{in} - \delta_{qn}\delta_{im} ) u_q \partial_m u_n $$

$$ \Rightarrow \ … + u_m \partial_m u_i - u_n \partial_i u_n $$

Change dummy indices in each term to match (for aesthetics):

$$ u_j \partial_j u_i = \tfrac{1}{2} \partial_i (u_j u_j) + u_j \partial_j u_i - u_j \partial_i u_j $$

Using product rule, we have $\partial_i (u_j u_j) = u_j \partial_i u_j + u_j \partial_i u_j = 2 u_j \partial_i u_j $. Substituting this in yields:

$$ u_j \partial_i u_j + u_j \partial_j u_i - u_j \partial_i u_j $$

$$\Rightarrow u_j \partial_j u_i $$

Appendix B #︎

Prove: $$\partial_j \omega_j = 0$$

Revert vorticity back to it’s original definition:

$$\partial_j \varepsilon_{jik} \partial_i u_k \Rightarrow \varepsilon_{jik} \partial_j \partial_i u_k $$

Since the permutation tensor is antisymmetric in its indices, $\varepsilon_{jik} = -\varepsilon_{ijk}$.

$$\varepsilon_{jik} \partial_j \partial_i u_k = - \varepsilon_{ijk} \partial_j \partial_i u_k $$

Since all the indices are dummy indices, they maybe changed/switched arbitrarily. Let’s switch the $j$ and $i$ indices:

$$\Rightarrow \ \varepsilon_{jik} \partial_j \partial_i u_k = - \varepsilon_{jik} \partial_i \partial_j u_k $$

Additionally, the derivatives are order invariant, so we can rearrange them:

$$\varepsilon_{jik} \partial_j \partial_i u_k = - \varepsilon_{jik} \partial_j \partial_i u_k $$

This is only possible if $\varepsilon_{jik} \partial_j \partial_i u_k = 0$. Thus:

$$ \varepsilon_{jik} \partial_j \partial_i u_k = \partial_j \omega_j = 0$$

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James Wright
James Wright
PhD Student, Aerospace Engineering

I’m interested in CFD, turbulence modeling, motorsports, and disc golf.