0
$\begingroup$

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?

$\endgroup$

1 Answer 1

0
$\begingroup$

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.
$\endgroup$
4
  • $\begingroup$ 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. $\endgroup$
    – Aria
    Commented Dec 11, 2023 at 19:33
  • $\begingroup$ 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. $\endgroup$
    – Kurt G.
    Commented Dec 11, 2023 at 19:46
  • $\begingroup$ My question is regarding the phi component of the curl of the same vector field calculation comparison in two different coordinates systems $\endgroup$
    – Aria
    Commented Dec 11, 2023 at 21:44
  • $\begingroup$ I got that from the beginning. See updated answer. $\endgroup$
    – Kurt G.
    Commented Dec 12, 2023 at 8:10

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .