# What is the cross product in spherical coordinates?

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*}

• 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 7:09
• Might Mathematics be better suited for this math question? – Kyle Kanos Oct 12 '15 at 9:54

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.

• 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 3:47
• 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 2:50
• 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 0:42
• I find your defence by attacking my understanding arrogant and completely offensive. – Paul Childs Nov 30 '18 at 0:50
• There's a world of difference between criticising the answer and criticising the person. The former allows for correction, refinement, elaboration and the general improvement of an answer; the latter is just plain Trollish. You haven't even made an attempt to address my counter example. – Paul Childs Dec 10 '18 at 0:29