# Does laplacian, curl and gradient commute in spherical coordinate?

I just cannot get my head around this seeming contraction:

First, we expect gradient, curl and lapalcian etc. to commute in $$R^n$$ as $$R^n$$ is flat space, even when we use spherical coordinate. However, by inspecting the formula for laplacian and gradient etc. in spherical corrdiantes, it seems that they do not commute. For example, take a function $$f(r)$$ which only depends on $$r$$, then \begin{align} \Delta \nabla f &= \frac{1}{r^2}\partial_r r^2 \partial_r \partial_r f \boldsymbol{e}_r\\ &=\frac{1}{r^2} (r^2 \partial_r^3 + 2r \partial_r^2) f \boldsymbol{e}_r \end{align} while \begin{align} \nabla \Delta f &= \partial_r (\frac{1}{r^2} \partial_r r^2 \partial_r f )\boldsymbol{e}_r\\ &=\partial_r (\partial_r^2 + \frac{2}{r}\partial_r) f \boldsymbol{e}_r \\ &= (f\partial_r^3 +\frac{2}{r}\partial_r^2-\frac{2}{r^2}\partial_r)f \boldsymbol{e}_r \end{align} and the two differ by a term of $$-\frac{2}{r^2}\partial_r$$! Similar things happen to laplacian and curl.

• I'm pretty sure they do commute, your calculation should have an error. Can you show the calculation? Commented Feb 4, 2021 at 8:48
• All these operations are coordinate-independent (although they do depend on the metric, which is flat in Euclidean spaces), so if they commute in some coordinate system, they commute in any. Commented Feb 4, 2021 at 8:52
• @CalvinKhor Thanks! I just edited to show the calculation. Commented Feb 4, 2021 at 8:57
• @BowenZhao The result of either $\Delta \nabla f$ or $\nabla \Delta f$ should be a vector, but in your calculations it is a scalar. Commented Feb 4, 2021 at 9:00
• @lisyarus You are right but since only r-dependence presents here i guess one can just add the e_r directional vector? Commented Feb 4, 2021 at 9:06

The gradient of a radial scalar function $$f=f(r)$$ in spherical coordinates is $$\nabla f(r,\theta)= f'(r)\hat {\mathbf r}(\theta)$$ Note that this is a vector. The vector Laplacian, in general is a slight mouthful- from trusty wikipedia if $$\mathbf F = A_r \hat{\mathbf r} + A_\theta \hat{\boldsymbol \theta} + A_\varphi \hat{\boldsymbol \varphi}$$, $$\Delta \mathbf F = \left(\Delta A_{r}-\frac{2 A_{r}}{r^{2}}-\frac{2}{r^{2} \sin \theta} \frac{\partial\left(A_{\theta} \sin \theta\right)}{\partial \theta}-\frac{2}{r^{2} \sin \theta} \frac{\partial A_{\varphi}}{\partial \varphi}\right) \hat{\mathbf{r}} \\+\left(\Delta A_{\theta}-\frac{A_{\theta}}{r^{2} \sin ^{2} \theta}+\frac{2}{r^{2}} \frac{\partial A_{r}}{\partial \theta}-\frac{2 \cos \theta}{r^{2} \sin ^{2} \theta} \frac{\partial A_{\varphi}}{\partial \varphi}\right) \hat{\boldsymbol{\theta}}\\+\left(\Delta A_{\varphi}-\frac{A_{\varphi}}{r^{2} \sin ^{2} \theta}+\frac{2}{r^{2} \sin \theta} \frac{\partial A_{r}}{\partial \varphi}+\frac{2 \cos \theta}{r^{2} \sin ^{2} \theta} \frac{\partial A_{\theta}}{\partial \varphi}\right) \hat{\boldsymbol{\varphi}}$$ In our case $$A_\theta = A_\varphi = 0$$, and $$A_r$$ is a function of $$r$$, leading to the simplification $$\Delta \nabla f = \left(\Delta f'(r)-\frac{2 f'(r)}{r^{2}}\right) \hat{\mathbf{r}}(\theta) = \left( f'''(r)+\frac{2 f''(r)}r -\frac{2 f'(r)}{r^{2}}\right) \hat{\mathbf{r}}(\theta)$$ where we used the scalar Laplacian $$\Delta g(r)=\frac1{r^2}(r^2 g'(r))' = g''(r) + \frac{2}{r}g'(r)$$.
Lets start from the other side: the scalar Laplacian of $$f=f(r)$$ is as before, the radial scalar function $$\Delta f(r)=\frac1{r^2}(r^2 f'(r))' = f''(r) + \frac{2}{r}f'(r)$$ And now take the gradient: $$\nabla \Delta f (r)=\left(f'''(r)+\left(\frac2r f'(r)\right)' \right)\hat {\mathbf r}(\theta)$$ Upon expanding $$\left(\frac2r f'(r)\right)'=\frac2rf''(r) - \frac{2}{r^2} f'(r)$$, we again obtain the same result. Maths is consistent for another day...!