Vorticity equation in index notation (curl of Navier-Stokes equation)

I am trying to derive the vorticity equation and I got stuck when trying to prove the following relation using index notation: $$ {\rm curl}((\textbf{u}\cdot\nabla)\mathbf{u}) = (\mathbf{u}\cdot\nabla)\pmb\omega - ( \pmb\omega \cdot\nabla)\mathbf{u} $$ considering that the fluid is incompressible $\nabla\cdot\mathbf{u} = 0 $, $\pmb \omega = {\rm curl}(\mathbf{u})$ and that $\nabla \cdot \pmb \omega = 0.$

Here follows what I've done so far: $$ (\textbf{u}\cdot\nabla) \mathbf{u} = u_m\frac{\partial u_i}{\partial x_m} \mathbf{e}_i = a_i \mathbf{e}_i \\ {\rm curl}(\mathbf{a}) = \epsilon_{ijk} \frac{\partial a_k}{\partial x_j} \mathbf{e}_i = \epsilon_{ijk} \frac{\partial}{\partial x_j}\left( u_m\frac{\partial u_k}{\partial x_m} \right) \mathbf{e}_i = \\ = \epsilon_{ijk}\frac{\partial u_m}{\partial x_j}\frac{\partial u_k}{\partial x_m} \mathbf{e}_i + \epsilon_{ijk}u_m \frac{\partial^2u_k}{\partial x_j \partial x_m} \mathbf{e}_i \\ $$

the second term $\epsilon_{ijk}u_m \frac{\partial^2u_k}{\partial x_j \partial x_m} \mathbf{e}_i$ seems to be the first term "$(\mathbf{u}\cdot\nabla)\pmb\omega$" from the forementioned identity. Does anyone have an idea how to get the second term?

Solution 1:

The trick is the following:

$$ \epsilon_{ijk} \frac{\partial u_m}{\partial x_j} \frac{\partial u_m}{\partial x_k} = 0 $$

by antisymmetry.

So you can rewrite

$$ \epsilon_{ijk} \frac{\partial u_m}{\partial x_j} \frac{\partial u_k}{\partial x_m} = \epsilon_{ijk} \frac{\partial u_m}{\partial x_j}\left( \frac{\partial u_k}{\partial x_m} - \frac{\partial u_m}{\partial x_k} \right) $$

Note that the term in the parentheses is something like $\pm\epsilon_{kml} \omega_l$

Lastly use the product property for Levi-Civita symbols

$$ \epsilon_{ijk}\epsilon_{lmk} = \delta_{il}\delta_{jm} - \delta_{im}\delta_{jl} $$

Solution 2:

Not a direct answer to the original post's question, but I have done a full write-up on deriving the vorticity transport equations. I go into more detail in my post, but I've copied the general gist of the derivation below:

Derivation

Incompressible conservation of momentum equations:

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

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

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

Distributing this across the terms, we get:

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

Temporal Term $\mathbb{T}$

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

Pressure Term $\mathbb{P}$

Since the curl of the gradient of a scalar is 0, $\mathbb{P} = 0$.

Viscous Term $\mathbb{V}$

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

Advection Term $\mathbb{A}$

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

Using:

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

get:

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

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:

$$\Rightarrow \ \mathbb{A} = \varepsilon_{k\ell i} \partial_\ell ( \varepsilon_{ijq} \omega_j u_q ) \Rightarrow \ \varepsilon_{ik\ell} \varepsilon_{ijq} \partial_\ell \omega_j u_q$$

Plug:

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

into previous expression:

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

$$ \Rightarrow \ \mathbb{A} = \partial_q \omega_k u_q - \partial_j \omega_j u_k \Rightarrow \ (u_q \partial_q \omega_k + \omega_k \partial_q u_q) - ( u_k \partial_j \omega_j + \omega_j \partial_j u_k)$$

By incompressibility $\partial_q u_q =0$. Also, $\partial_j \omega_j$ also equals zero:

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

Putting It All Together

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