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The curl of an arbitrary vector, $\vec{A}$ is The curl of an arbitrary vector $\vec{A}$ in spherical coordinates

\begin{align*} \nabla \times \vec{A} &= \frac{1}{r^{2}\sin{\theta}}\left| \begin{array}{ccc} \hat{r} & r\hat{\theta} & r\sin{\theta}\hat{\phi}\\ \dfrac{\partial}{\partial r} & \dfrac{\partial}{\partial \theta} & \dfrac{\partial}{\partial \phi} \\ A_{r} & rA_{\theta} & r\sin{\theta}A_{\phi} \end{array} \right| \\ &=\frac{\hat{r}}{r\sin{\theta}}\bigg[ \frac{\partial}{\partial\theta}(A_{\phi}\sin{\theta})-\frac{\partial A_{\theta}}{\partial \phi}\bigg] +\frac{\hat{\theta}}{r\sin{\theta}}\bigg[\frac{\partial A_{r}}{\partial\phi}-\sin{\theta}\frac{\partial}{\partial r}(rA_{\phi}) \bigg]+\frac{\hat{\phi}}{r}\bigg[\frac{\partial}{\partial r}(rA_{\theta})-\frac{\partial A_{r}}{\partial\theta} \bigg] \end{align*}

Can I simply let $\nabla = E$ and $\vec{A} = \vec{B}$ to say that the cross product of $\vec{E}$ and $\vec{B}^{*}$ in spherical coordinates \begin{align*} \vec{E} \times \vec{B}^{*}&= \frac{\hat{r}}{r\sin{\theta}}\big( E_{\theta}B_{\phi}^{*}\sin{\theta} - E_{\phi}B_{\theta}^{*} \big) +\frac{\hat{\theta}}{r\sin{\theta}}(E_{\phi}B_{r}^{*}-r \sin{\theta}E_{r}B_{\phi}^{*})+\frac{\hat{\phi}}{r}(rE_{r}B_{\theta}^{*}-E_{\theta}B_{r}^{*}) \\ \end{align*}

Paul Childs
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Joshua Burrow
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    You already got the answer, I'd just like to state you cannot use the formula for curl for cross product, because usually in vector calculus, we do not use the coordinate basis vectors ($\partial\mathbf{r}/\partial u^ i$, where $u^i$ is a coordinate), but their normalized versions (which are less "natural" in regards to the coordinate system), while the operators $\partial/\partial r, \partial/\partial\theta$ etc... refer to the coordinate system itself, and the extra factors are to compensate for this. – Bence Racskó Oct 12 '15 at 07:09
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    Might [math.se] be better suited for this math question? – Kyle Kanos Oct 12 '15 at 09:54

1 Answers1

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The cross product in spherical coordinates is given by the rule,

$$ \hat{\phi} \times \hat{r} = \hat{\theta},$$

$$ \hat{\theta} \times \hat{\phi} = \hat{r},$$

$$ \hat{r} \times \hat{\theta} = \hat{\phi},$$

this would result in the determinant,

$$ \vec{A} \times \vec{B} = \left| \begin{array}{ccc} \ \hat{r} & \hat{\theta} & \hat{\phi} \\ A_r & A_\theta & A_\phi \\ B_r & B_\theta & B_\phi \\ \end{array}\right|$$

This rule can be verified by writing these unit vectors in Cartesian coordinates.

The scale factors are only present in the determinant for the curl. This has to do with the definition of the curl and its use of length and area.

Chappers
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Spencer
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    This is a completely false and misleading answer. You can't expect to do linear algebra with a curvilinear coordinate system. E.g. using this determinant, a simple cross product of the x and y unit vectors would give an r of pi^2 / 4 instead of 1. – Paul Childs Nov 16 '18 at 03:47
  • You miss the point. The coordinate system is not linear. The basis can be localised but not globalised in a comparable manner (as needed by the cross product) without applying a connection. For your determinant, if A=x, B=y then A_{\theta}=B_{\theta}={\pi}/4, A_{\phi}=0, B_{\phi}={\pi}/4 therefore by your reasoning the radius of AxB is {\pi}^2/16 – Paul Childs Nov 28 '18 at 02:50
  • Yes and A_{\phi} = 0 as sin{phi} = 0 as \phi = 0. And what good is expressing things in sin {phi} when you have two vectors each with different angles. – Paul Childs Nov 30 '18 at 00:42
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    You can read through this to get familiar with the basics. In particular take note of the cross product relationship between $e_r, e_\theta,$ and $e_\phi$ on page 3.

    https://www2.ph.ed.ac.uk/~mevans/mp2h/VTF/lecture23.pdf

     – Spencer Dec 13 '18 at 00:27
  • @PaulChilds, I allowed my temper to get the better of me in our interactions over the last couple of months and I apologize for that.

    I do not change my position, that you have misunderstood the meaning of the notation here, but I do not doubt that this is a good faith misunderstanding. You can examine the link provided in the previous comment for further clarification. I do not have the time to write a full treatment myself here in the comments.

     – Spencer Dec 14 '18 at 01:08