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?


1 Answer 1


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.

  • $\begingroup$ 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. $\endgroup$
    – yunjoo
    Jan 23, 2021 at 5:20
  • $\begingroup$ @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. $\endgroup$
    – peek-a-boo
    Jan 23, 2021 at 5:25

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