# Curl is zero when I have radial symmetry?

I'm trying to understand why, when we have radial symmetry of a vector quantity, the curl of this quantity is zero. For example, I have a sphere. I know that the polarization of the sphere, which is just a vector quantity, has radial symmetry. Does this means that the curl of this quantity, the curl of the polarization, is zero? Why?

• Well, $\vec{\nabla}\times f(r)\vec{r}=0$ because $f(r)\vec{r}=\vec{\nabla}g(r)$ with $g'=rf$.
– J.G.
Commented Jul 19, 2022 at 14:53

Just rephrasing the comment from @J.G, the spherical symmetry means that curl of a vector with a spherical symmetry $$\vec{a} = f(r) \vec{e_r}$$ will be: $$$$\mbox{curl} \, \vec{a} = \nabla \times \vec{a} = \nabla_r \times f(r) \vec{e_r}$$$$ The nabla operator. It is easy to show that $$f(r) \vec{e_r} = \nabla_r g(r) = \nabla g(r)$$ and finally $$$$\mbox{curl} \, \vec{a} = \nabla \times \nabla g(r)$$$$ and the operator $$\nabla \times \nabla$$ is known to be zero

• This answer doesn't make sense. You wrote that $\mbox{curl}\; \vec a = \nabla g$ and you also wrote that $\mbox{curl}\; \vec a = \nabla \times \nabla g$.
– hft
Commented Jul 19, 2022 at 23:11
• If you are paraphrasing J.G.'s comment, I think you want to write $f(r)\vec e_r = \nabla g(r)$ not $\nabla \times f(r)\vec e_r = \nabla g(r)$
– hft
Commented Jul 19, 2022 at 23:15
• Yes that is true, thank you for the remark. I have corrected. Commented Jul 20, 2022 at 7:45

I'm trying to understand why, when we have radial symmetry of a vector quantity, the curl of this quantity is zero.

Assuming, "radial symmetry" means you are looking at a field of the form: $$\vec F = f(r) \vec r,$$

Then you have $$\left(\nabla \times \vec F\right)_ i = \epsilon_{ijk}\nabla_j \left(f(r) \vec r_k\right)$$ $$=\epsilon_{ijk}\left(f'(r)\frac{r_j r_k}{r}+f(r)\delta_{jk}\right)$$ $$=0\;,$$ since both $$r_j r_k$$ and $$\delta_{jk}$$ are symmetric and $$\epsilon_{ijk}$$ is antisymmetric.