# general definition of curl

I am studying about 2-dimensional Euler equation's fluid vorticity, and I want to know how to calculate it.

$$\omega = ∇\times u$$ if $$\omega$$ is a fluid vorticity and u is the velocity vector of the fluid.

I tried to calculate it using polar coordinates and since $$∇=(\frac{\partial}{\partial r},\frac{1}{r}\frac{\partial}{\partial \theta})$$, it becomes $$∇\times u=(\frac{\partial}{\partial r},\frac{1}{r}\frac{\partial}{\partial \theta},\frac{\partial}{\partial z})\times(u_r,u_\theta,0)=(0,0,\partial_r u_\theta - \frac{1}{r}\partial_\theta u_r)$$.

But my book says it should be $$\omega = \frac{1}{r}\partial_r (ru_\theta)-\frac{1}{r}\partial_\theta u_r$$.

I think this difference is from the general definition of curl. When I studied divergence in polar coordinate, I saw that the general definition of dot product is $$\textrm{div(F)}=\frac{1}{\rho}\frac{\partial(\rho F^i)}{\partial x^i},$$ where $$\rho=\sqrt{\det(g)}$$ and g is an metric tensor.

So similarly, I want to know what is the general definition of curl, and how to evaluate exact fluid vorticity in that case. Does anyone help me?

The dot and cross product notation only works easily with cartesian coordinates. Otherwise, you need some work to get the right expressions, because $$\nabla$$ is a certain differential operator, it is NOT a vector by itself. Since you're asking about the curl, we have to restrict ourselves to $$3$$-dimensional space. First let us fix some notation,

• Let $$g=dx\otimes dx+dy\otimes dy+dz\otimes dz$$ denote the standard metric tensor on $$\Bbb{R}^3$$.

• Let us suppose that we have orthogonal coordinates $$(x^1,x^2,x^3)$$ (not necessarily cartesian), which means $$g_{ij}=0$$ if $$i\neq j$$ (see next bullet point).

• Let $$g_{ij}$$ be the $$i,j$$ component of the Euclidean metric tensor relative to these coordinates: $$g_{ij}:=g\left(\frac{\partial}{\partial x^i},\frac{\partial}{\partial x^j}\right)$$, and let $$h_i:=\sqrt{g_{ii}}$$. The $$h_i$$'s are usually what's called the scale factors (of the metric tensor relative to the coordinates $$(x^1,x^2,x^3)$$).

• Let $$\mathbf{e}_i$$ denote the $$i^{th}$$ unit vector (in differential geometry notation, we have $$\frac{\partial}{\partial x^i}= \sqrt{g_{ii}}\mathbf{e}_i = h_i\mathbf{e}_i$$).

• Finally, let $$F^1,F^2,F^3$$ denote the components of the vector field $$\mathbf{F}$$ relative to the basis $$\{\mathbf{e}_1,\mathbf{e}_2,\mathbf{e}_3\}$$, i.e $$\mathbf{F}=F^1\mathbf{e}_1 + F^2\mathbf{e}_2 + F^3\mathbf{e}_3$$.

Then, we have: \begin{align} \nabla \times \mathbf{F} &=\frac{1}{h_1h_2h_3}\det \begin{pmatrix} h_1\mathbf{e}_1 & h_2\mathbf{e}_2 & h_3\mathbf{e}_3\\ \frac{\partial}{\partial x^1}&\frac{\partial}{\partial x^2}&\frac{\partial}{\partial x^3}\\ h_1F^1& h_2F^2 & h_3F^3 \end{pmatrix}\\\\ &=\frac{1}{h_2h_3}\left[\frac{\partial(h_3F^3)}{\partial x^2}-\frac{\partial(h_2F^2)}{\partial x^3}\right]\mathbf{e}_1\\ &+ \frac{1}{h_1h_3}\left[\frac{\partial(h_1F^1)}{\partial x^3}-\frac{\partial(h_3F^3)}{\partial x^1}\right]\mathbf{e}_2\\ &+ \frac{1}{h_1h_2}\left[\frac{\partial(h_2F^2)}{\partial x^1}-\frac{\partial(h_1F^1)}{\partial x^2}\right]\mathbf{e}_3 \end{align} The way I calculated this is using the exterior calculus: $$\nabla \times \mathbf{F}:= g^{\sharp}\star d [g^{\flat}(\mathbf{F})]$$.

If you wish, here are a set of lecture notes I found online which deals with this question (albeit with a bit of handwaving... which can be made more rigorous if you really want). If I remember correctly, Griffiths' electrodynamics book has an appendix which addresses these questions, and the presentation there is similar to the notes I linked to.

• thank you so much! That was exactly I wanted:) can I ask a question more? I want to know how to call that # and flat operator to search it. Jan 23, 2021 at 5:20
• @yunjoo $g^{\flat}$ is read as "gee-flat" and $g^{\sharp}$ is read as "gee-sharp". These are called the musical isomorphisms (with respect to $g$); what they do is set up an isomorphism between the the various tangent and cotangent spaces of the underlying manifold. Jan 23, 2021 at 5:25