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We can compute the divergence of a vector filed $X$ on $\mathbb{R}^2$ expressed in polar coordinates $(r,\theta)$ in two ways: the first one is the classical formula

$$\text{div}(X)=\frac{1}{r}\frac{\partial(rX^r)}{\partial r}+\frac{1}{r}\frac{\partial X^\theta}{\partial\theta},$$

and the second one is the formula given by the Riemannian definition of the divergence (here $(x^1,x^2)=(r,\theta)$)

$$\text{div}(X)=\frac{\partial X^i}{\partial x^i}+\Gamma_{ij}^iX^j=\frac{\partial X^r}{\partial r}+\frac{\partial X^\theta}{\partial\theta}+\frac{1}{r}X^r.$$

The two expression are not the same: the term $\frac{\partial X^\theta}{\partial\theta}$ is rescaled differently. Why? I suppose that it has something to do with some kind of renormalisation.

For the Christoffel symbols I looked here.

Duca_Conte
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  • This is not the Riemannian formula for divergence with which I'm familiar. Where did you get this? Ultimately, you should use the metric to convert $X$ to a $1$-form $\omega$, apply the Hodge star operator to get an $(n-1)$-form $\star\omega$, apply $d$, and then $d{\star}\omega = \text{div},X,dV$. The Christoffel symbols aren't involved. – Ted Shifrin Nov 03 '19 at 18:51
  • In my geometry class, but you can find it here for instance: https://math.stackexchange.com/questions/137573/divergence-of-a-vector-field-on-a-manifold – Duca_Conte Nov 03 '19 at 18:53
  • Got it: It's the trace of the covariant derivative. – Ted Shifrin Nov 03 '19 at 19:27

1 Answers1

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The issue is that the first formula comes from taking components of $X$ with respect to the orthonormal basis $e_r,e_\theta$, whereas the differential geometry formula is in terms of components with respect to the coordinate basis $\partial/\partial r, \partial/\partial\theta$. In particular, the unit vector $e_\theta = \frac1r\partial/\partial\theta$. So the $X^\theta$ appearing in the two formulas are different.

Ted Shifrin
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