3
$\begingroup$

I was asked to calculate the curl of the vector field $\theta^2\bf e_\theta$ in spherical coordinates. If we use the typical formula for curl in curvilinear coordinates: $$ \nabla\times \bf F = \left|\begin{matrix}h_1\hat{\bf e_1} & h_2\hat{\bf e_2} & h_3\hat{\bf e_3} \\ \displaystyle\frac{\partial}{\partial q_1} & \displaystyle\frac{\partial}{\partial q_2} & \displaystyle\frac{\partial}{\partial q_3} \\ h_1F_1 & h_2F_2 & h_3F_3\end{matrix}\right| $$ where, in spherical coordinates, $(q_1,q_2,q_3) = (r,\theta,\phi)$, and $h_1 = 1$, $h_2 = r$, $h_3 = r\sin\theta$, then we get $$ \nabla\times(\theta^2{\bf e}_\theta) = r\theta^2\sin\theta\bf e_\phi $$

I wanted to check this, so I thought I'd try converting the original vector field into Cartesian coordinates, taking the curl in Cartesian coordinates, and converting back. Using $$ \theta = \arctan\left({\frac{\sqrt{x^2+y^2}}{z}}\right) \\ {\bf e_\theta} = \frac{(x\hat{\bf i} + y\hat{\bf j})z - (x^2+y^2)\hat{\bf k}}{\sqrt{x^2+y^2+z^2}\sqrt{x^2+y^2}} $$ we get $$ \theta^2{\bf e}_\theta = \frac{\arctan^2\left(\frac{\sqrt{x^2+y^2}}{z}\right)}{\sqrt{x^2+y^2}\sqrt{x^2+y^2+z^2}}\left(xz,yz,-(x^2+y^2)\right) $$

After some tedious calculation, this (somewhat remarkably) comes out to $$ \nabla\times{\bf F} = \frac{\arctan^2\left(\frac{\sqrt{x^2+y^2}}{z}\right)}{\sqrt{x^2+y^2}\sqrt{x^2+y^2+z^2}}(-y{\bf\hat{i}}+x{\bf\hat{j}}) $$ This takes some work to convert back to polar coordinates, but without converting the unit vectors back immediately we have $$ \nabla\times{\bf F} = \frac{\theta^2}{r}(-\sin\phi{\bf\hat{i}} + \cos\phi{\bf\hat{j}}) $$ After substituting $$ {\bf\hat{i}} = \sin\theta\cos\phi{\bf\hat{r}}+\cos\theta\cos\phi\hat{\theta}-\sin\phi\hat{\phi} \\ {\bf\hat{j}} = \sin\theta\sin\phi{\bf\hat{r}} + \cos\theta\sin\phi\hat{\theta} + \cos\phi\hat{\phi} $$ into the above formula, we get $$ \frac{\theta^2}r{\bf e}_\theta $$ which is... not the same.

So clearly, my intuition is wrong, or my assumption that these two "curls" in different coordinate systems are the same (since, assuming I didn't make any mistakes in my derivation, this would imply that "spherical curl" is somehow fundamentally different from "Cartesian curl"), or I made a mistake somewhere. Or something else.

Which is it? Can someone guide me in the right direction please?

$\endgroup$
1
$\begingroup$

I see that you miss a factor in your first formula for curl. It should read $$\nabla\times \mathbf F = \frac{1}{h_1 h_2 h_3} \left|\begin{matrix}h_1\hat{\bf e_1} & h_2\hat{\bf e_2} & h_3\hat{\bf e_3} \\ \displaystyle\frac{\partial}{\partial q_1} & \displaystyle\frac{\partial}{\partial q_2} & \displaystyle\frac{\partial}{\partial q_3} \\ h_1F_1 & h_2F_2 & h_3F_3\end{matrix}\right|$$

Thus you get $$\nabla\times\mathbf F = \frac{1}{r^2 \sin\theta} \, r\theta^2\sin\theta \, \mathbf{e}_\phi = \frac{\theta^2}{r} \mathbf{e}_\phi$$

But now I noticed that there is still a difference. In your second attempt, via Cartesian coordinates, you get $\mathbf{e}_\theta$ instead of $\mathbf{e}_\phi$. I believe $\mathbf{e}_\phi$ to be the correct since that's orthogonal to the field.

And now I see that you make some error when you convert $-\sin\phi{\bf\hat{i}} + \cos\phi{\bf\hat{j}}$ back. It should be $\mathbf{e}_\phi$, not $\mathbf{e}_\theta$.

$\endgroup$
  • $\begingroup$ Ah I see, I honestly can't believe I forgot the factor of $\frac1{h_1h_2h_3}$! As for the second error, I'll take another look at it, but I used wolfram alpha to calculate it so I was fairly certain at the time, but I may have substituted something wrong... $\endgroup$ – user3002473 Feb 24 '18 at 14:40
  • $\begingroup$ Actually, though I haven't found the exact error yet, I do see that $$ {\bf e}_\phi = -\sin\phi\hat{\bf i} + \cos\phi\hat{\bf j} $$ which can be found here: en.wikipedia.org/wiki/… I'm not sure why substituting the given formulas for $\hat{\bf i}$ and $\hat{\bf j}$ found there and simplifying proved so error-prone, I'll look at it again. $\endgroup$ – user3002473 Feb 24 '18 at 14:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.