For the plane in polar coordinates $(r,\theta)$ with metric $$ds^2=dr^2+r^2d\theta^2,$$
the curl on a vector field $v^a\partial_a$ is given by the rank-2 antisymmetric tensor $\nabla_av_b-\nabla_bv_a$. This tensor has only one independent & non-zero component:
$$\nabla_rv_\theta-\nabla_\theta v_r=\partial_rv_\theta – \Gamma_{r\theta}^av_a -\partial_\theta v_r +\Gamma^a_{\theta r}v^a=\partial_rv_\theta-\partial_\theta v_r$$
since $\Gamma^a_{\theta r} = \Gamma^a_{r \theta }$ for Christoffel symbols. Expressed in contravariant vector components, $$\nabla_rv_\theta-\nabla_\theta v_r=\partial_r (g_{\theta a}v^a)-\partial_\theta (g_{rb}v^b)=\partial_r(r^2v^\theta)-\partial_\theta(v^r)=2rv^\theta+r^2\partial_rv^\theta-\partial_\theta v^r$$
I am trying to compare this to the formula for curl in vector calculus:
$$\nabla\times \textbf{v} = \frac{1}{r}\left[\frac{\partial}{\partial r}(rv^\theta) – \frac{\partial}{\partial\theta}v^r\right] \vec{\hat{z}}=\left[\frac{1}{r}v^\theta+\partial_rv^\theta-\frac{1}{r}\partial_\theta v^r\right]\vec{\hat{z}}$$, where $\vec{\hat{z}}$ is the unit vector pointing out of the plane.
The difference in the two formulas is because in manifolds we used the coordinate basis $\partial_\theta$ while in vector calculus we used the normalized coordinate basis $\frac{1}{r}\partial_\theta$. In the coordinate basis, $\textbf{v}=v^r\partial_r+v^\theta\partial_\theta$ while in the normalized coordinate basis, $\textbf{v}=v^r\partial_r+v^\theta\frac{1}{r}\partial_\theta$.
Thus to convert between the 'manifold' formula and the 'vector calculus' formula, we just have to rescale the $v^\theta$ component of the vector.
From the 'manifold' formula, replace $v^\theta\rightarrow \frac{v^\theta}{r}$ to get the 'vector calculus' formula:
$$\nabla_rv_\theta-\nabla_\theta v_r=2rv^\theta+r^2\partial_rv^\theta-\partial_\theta v^r \rightarrow2r\frac{v^\theta}{r}+r^2\partial_r\frac{v^\theta}{r}-\partial_\theta v^r =v^\theta+r\partial_rv^\theta -\partial_\theta v^r$$
However, this answer larger than the correct formula by a factor of $r$. Why is that so?
Best Answer
The curl of $v$ is given by $$\mathrm{curl}(v) = E^{ab}\nabla_av_b$$ where $E^{ab}$ is the Levi-Civita tensor. Not the Levi-Civita symbol $\epsilon^{ij}$. The Levi-Civita symbols $\epsilon^{ij}$ and $\epsilon_{ij}$ satisfy $$\epsilon_{ij}=\epsilon^{ij} = \begin{cases} 1 & (i,j)=(1,2) \\ -1 & (i,j)=(2,1) \\ 0 & \text{else} \end{cases}$$ However, the Levi-Civita tensors $E^{ab}$ and $E_{ab}$ have components $$E_{ij} = \sqrt{\det g}\;\epsilon_{ij}$$ $$E^{ij} = \frac{1}{\sqrt{\det g}}\epsilon^{ij}.$$ That is where the $\frac{1}{r}$ factor comes from, since for polar plane coordinates you have $\sqrt{\det g} = r$ and hence \begin{align} \mathrm{curl}(v) &= E^{ab}\nabla_av_b \\ &= E^{ij}\partial_iv_j \\ &= \frac{1}{r}(\partial_1v_2-\partial_2v_1) \\ &= \frac{1}{r}(2rv^2+r^2\partial_1v^2-\partial_2 v^1), \end{align} as we wanted.