# Taking curl of Euler equation

Consider an inviscid incompressible ﬂow. Euler’s equation can be written as $$\frac{\partial \textbf u}{\partial t} + \textbf ω × \textbf u = −\textbf∇\bigg( \frac pρ + \frac 12 \textbf u^2 + V \bigg)$$ where the vorticity $\textbf ω = \textbf ∇ × \textbf u$. By taking the curl of this equation and using the vector identity $\textbf ∇ × (\textbf a × \textbf b) = (\textbf b \cdot \textbf ∇)\textbf a − (\textbf a\cdot \textbf∇)\textbf b + \textbf a(\textbf∇ \cdot \textbf b) − \textbf b(\textbf∇ \cdot \textbf a)$ show that $$\frac{D \textbf ω}{Dt} = (\textbf ω \cdot \textbf ∇)\textbf u$$ This is the vorticity equation.

I am very stuck on this. Does taking the curl mean we have to do this:$$\textbf ∇ \times \bigg( \frac{\partial \textbf u}{\partial t} + \textbf ω × \textbf u \bigg)= −\textbf ∇ \times \bigg( \textbf∇\bigg( \frac pρ + \frac 12 \textbf u^2 + V \bigg) \bigg)$$ and the LHS becomes $$\textbf ∇ \times \frac{\partial \textbf u}{\partial t}+ \textbf ∇ \times (\textbf ω × \textbf u )$$ and then use the identity? I was not sure if we can just kind off expand the LHS like how I did. But even still, after using the identity, it gets me nowhere.

You can directly apply that formula to the LHS: $$\nabla\times(\omega\times u)=(u\nabla)\omega-(\omega\nabla)u+\omega(\nabla u)-u(\nabla \omega)$$ The last two terms are zero. First one because of incompressibility, $\nabla v=0$. Second because a divergence of a curl is zero (vorticity is the curl of velocity).

What's left:

$$\frac{\partial \omega}{\partial t}+(u\nabla)\omega-(\omega\nabla)u=0$$ First two terms are the total derivative, $\frac{D\omega}{Dt}$ and you get the equation you are looking for.

Explanation of the total derivative:

$$\frac{D \Box}{Dt}=\frac{\partial\Box}{\partial t}+ \frac{\partial\Box}{\partial x}\frac{d x}{d t}+ \frac{\partial\Box}{\partial y}\frac{d y}{d t}+ \frac{\partial\Box}{\partial z}\frac{d z}{d t}=\frac{\partial\Box}{\partial t}+( {\bf u}\cdot \nabla)\Box$$ This relation is universal, no matter what physical quantity and what dimension (scalar, vector,...). It's simply the chain rule for the time derivative. It simply describes the advection effect: the time derivative of whatever quantity you are describing, but moving with the fluid (imagine floating in the fluid and measuring the quantity along the stream). This is why the velocity appears in the equation: the second term is the derivative caused by change of the quantity along the streamline. We recognized the $dx/dt$ terms as velocity components and observed that the notation can be shortened by using the nabla operator.

You can put $\bf u$ instead of the box $\Box$, but notice that the other $\bf u$ that's already in there has a different role (it's how much you're carried downstream).

• So my LHS expansion was correct? May 8, 2016 at 15:06
• Yes, the curl simply distributes over addition (and in this case, you can also exchange curl and time derivative in the first term). And RHS vanishes because curl(grad)=0. May 8, 2016 at 15:07
• Order of derivatives isn't important, just exchange and recognize the definition of vorticity $$\nabla\times \frac{\partial \bf u}{\partial t}= \frac{\partial (\nabla\times\bf u)}{\partial t}$$ May 8, 2016 at 16:43
• Yep, that's exactly what I wrote down, only I made the fraction line longer. May 8, 2016 at 16:57
• I added the explanation for that. May 8, 2016 at 20:31