I would like to make sense of the transformation of the differentials in polar coordinates (to fix the ideas). To be more precise, the "right" way to find the transform for the differential and the gradient is :

$$(x,y) = (r \cos\theta, r \sin\theta)$$

$$(\text{d}x \; \text{d}y) = \left(\begin{array}{cc}\cos \theta & -r \sin \theta \\ \sin \theta & r \cos \theta\end{array}\right) (\text{d}r \; \text{d}\theta)$$

and $\left(\begin{array}{cc} \partial_x \\ \partial_y \end{array}\right) = \left(\left(\begin{array}{cc}\cos \theta & -r \sin \theta \\ \sin \theta & r \cos \theta\end{array}\right)^T\right)^{-1} \left(\begin{array}{cc} \partial_r \\ \partial_\theta \end{array}\right) = \left(\begin{array}{cc}\cos \theta & -\frac{\sin \theta}{r} \\ \sin \theta & \frac{\cos \theta}{r}\end{array}\right) \left(\begin{array}{cc} \partial_r \\ \partial_\theta \end{array}\right)$

I guess I have these formula because the differential is covariant, and the gradient is contravariant (although I don't really feel why).

The way I would guess the formula for the gradient is e.g. :

$$\partial_x = \partial_{r \cos \theta} = \frac{1}{\cos\theta}\partial_r - \frac{\sin \theta}{r}\partial_\theta$$

which admittedly looks odd (and wrong), but more straightforward than the first formula...

Could someone explain what is behind this ? Is there a correct "one liner" trick to remember the expression for $\partial _x$ or is it doomed to fail due to the matrix inversion above ?


1 Answer 1


Just remember the chain rule for partial derivatives:

$$ \frac\partial{\partial x} = \frac{\partial r}{\partial x}\frac\partial{\partial r} + \frac{\partial \theta}{\partial x}\frac{\partial}{\partial \theta}.$$

The matrix formula is just this along with the fact that $$\frac{\partial(r,\theta)}{\partial(x,y)} = \left(\frac{\partial(x,y)}{\partial(r,\theta)}\right)^{-1}$$ - this is probably easier to compute despite the matrix inversion simply because $(r(x,y),\theta(x,y))$ are much uglier to differentiate.

I don't really know any tricks to simplify this. You're never going to find one that works using just the expression for $x(r,\theta)$ - unlike $dx$, the coordinate basis frame $\partial_x$ can change if you change the coordinate $y$ while keeping $x$ fixed.


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