# A paradox on curl equations in cylindrical and spherical coordinates

Let $$\mathbf{A}=\sin(\theta)\hat{\phi}$$ be an azimuthal vector field in either cylindrical (cylindrical radial, azimuthal, vertical)=$$(\rho,\phi,z)$$ or spherical (spherical radial, colatitude, azimuthal)= $$(r,\theta,\phi)$$ coordinates where $$\phi=\tan^{-1}(y/x)$$ and $$\theta=\cos^{-1}(z/r)$$ for the corresponding Cartesian coordinates $$(x,y,z)$$ transformation.

As we see $$\mathbf{A}$$ has no radial or colatitude component in spherical coordinates: $$A_r=A_\theta=0$$.

Also $$\mathbf{A}$$ has no radial or vertical component in cylindrical coordinates: $$A_\rho=A_z=0$$.

The $$\phi-$$ component of $$\nabla\times(A_\phi\hat{\phi})$$ in spherical coordinates is: $$\hat{\phi}\cdot\nabla\times(A_\phi\hat{\phi})=\frac{1}{r}\left[\frac{\partial}{\partial r}(rA_\phi)-\frac{\partial A_r}{\partial\theta}\right]=\frac{1}{r}\sin\theta$$

However the $$\phi$$-component of curl of $$\mathbf{A}$$ is zero in cylindrical component. Because: $$\hat{\phi}\cdot\nabla\times(A_\phi\hat{\phi})=\frac{\partial A_\rho}{\partial z}-\frac{\partial A_z}{\partial \rho}=0$$

Therefore using cylindrical and spherical coordinates the $$\phi$$-component of curl of $$\mathbf{A}$$ gets two different values. What causes such a paradox happen and how to resolve the apparent contradiction?

The vector field $$\mathbf{A}=\sin(\theta)\,\boldsymbol{\hat{\phi}}$$ you start with is in spherical coordinates. Otherwise, what is $$\theta\,?$$

It is not the case that the curl of his field only has a $$\boldsymbol{\hat{\phi}}$$-component.

The curl of this field has the two other components: \begin{align} \nabla\times\mathbf{A}&= \frac{1}{r\sin\theta}\Big(\frac{\partial}{\partial\theta}(A_\phi\sin\theta)-\underbrace{\frac{\partial A_\theta}{\partial\phi}}_{\textstyle 0}\Big)\boldsymbol{\hat{r}}+ \underbrace{\frac1r\Big(\underbrace{\frac1{\sin\theta}\frac{\partial A_r}{\partial\phi}}_{0}-\frac{\partial}{\partial r}(rA_\phi)\Big)}_{\textstyle-\frac1rA_\phi}\boldsymbol{\hat\theta}\\ &+\frac1r\underbrace{\Big(\frac{\partial}{\partial r}(rA_\theta)-\frac{\partial A_r}{\partial\theta}\Big)}_{\textstyle 0}\boldsymbol{\hat\phi}\tag1 \end{align} In cylindrical coordinates this field must be written as $$\tag2 \mathbf{A}=\sin\left(\arccos\left(\frac zr\right)\right)\,\boldsymbol{\hat{\phi}}=\underbrace{\sin\left(\arccos\left(\frac{z}{\sqrt{z^2+\rho^2}}\right)\right)}_{\textstyle A_\phi}\,\boldsymbol{\hat{\phi}}\,.$$ I wish you a lot of fun calculating the curl in these coordinates.

The $$\boldsymbol{\hat{\phi}}$$-component is the easisest. It is zero in both coordinate systems. When you calculated in spherical you had a typo $$\tag3 \hat{\phi}\cdot\nabla\times(A_\phi\hat{\phi})=\frac{1}{r}\left[\frac{\partial}{\partial r}(rA_{\color{red}{\phi}})-\frac{\partial A_r}{\partial\theta}\right]=\frac{1}{r}\sin\theta$$ which led to the wrong result. The correct expression is (1) and gives zero.

• Your calculation of the $$\boldsymbol{\hat{\phi}}$$-component in cylindrical coordinates looks correct.
• My question is regarding the $\phi$ component of the curl and got two different results in spherical and cylindrical coordinates although the same result is expected.
– Aria
Commented Dec 11, 2023 at 19:33
• To rephrase what I wrote in that answer: that curl has no $\phi$ component. I spent a lot of time going through Wikipedia to make that clear. Commented Dec 11, 2023 at 19:46
• My question is regarding the phi component of the curl of the same vector field calculation comparison in two different coordinates systems
– Aria
Commented Dec 11, 2023 at 21:44
• I got that from the beginning. See updated answer. Commented Dec 12, 2023 at 8:10