For a vector field $X$, the divergence in coordinates is given by $\nabla\cdot X=\sum_n\frac{X^i}{\partial x^i}$. In polar coordinates, the metric is $\begin{bmatrix}1 & 0\\ 0 & r^2\end{bmatrix}$, and so $\frac{1}{\sqrt{g(\frac{\partial}{\partial r},\frac{\partial}{\partial r})}}\frac{\partial}{\partial r}=\frac{\partial}{\partial r}$ and $\frac{1}{\sqrt{g(\frac{\partial}{\partial\theta},\frac{\partial}{\partial\theta})}}\frac{\partial}{\partial\theta}=\frac{1}{r}\frac{\partial}{\partial\theta}$ are unit vectors. Then for $X=X_{r}\frac{\partial}{\partial r}+X_{\theta}\frac{\partial}{r\partial\theta}$, $\nabla\cdot X=\frac{\partial X_r}{\partial r}+\frac{\partial}{\partial\theta}\frac{X_{\theta}}{r}=\frac{\partial X_r}{\partial r}+\frac{1}{r}\frac{\partial X_{\theta}}{\partial\theta}$. But this disagrees with the usual formula given in vector calculus books. Does anyone see the error?
Differential Geometry – Divergence in Polar Coordinates
differential-geometryvector-spaces
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Best Answer
$\DeclareMathOperator\div{div}$The formula for $\nabla\cdot X$ is incorrect. The notation with the 'usual' dot product is misleading. Properly it is: $$\div F = \frac 1\rho\frac{\partial(\rho F^i)}{\partial x^i}$$ where $\rho=\sqrt{\det g}$ is the coefficient of the differential volume element $dV=\rho\, dx^1\wedge\ldots \wedge dx^n$, meaning $\rho$ is also the Jacobian determinant, and where $F^i$ are the components of $F$ with respect to an unnormalized basis.
In polar coordinates we have $\rho=\sqrt{\det g}=r$, and: $$\div X = \frac 1r \frac{\partial(r X^r)}{\partial r} + \frac 1r\frac{\partial(r X^\theta)}{\partial \theta}$$
In the usual normalized coordinates $X=\hat X^{r}\frac{\partial}{\partial r} + \hat X^{\theta}\frac 1r\frac{\partial}{\partial\theta}$ this becomes: $$\div X = \frac 1r \frac{\partial(r \hat X^{r})}{\partial r} + \frac 1r\frac{\partial \hat X^{\theta}}{\partial \theta}$$ which agrees with the usual formula given in calculus books.