Cross product direction in Spherical coordiantes

Question

The cross product of a vector $\vec{m}$ with $\hat{r}$ gives $m\sin\theta \hat{\phi}$ as claimed by the author.

This would probably be trivial but a lot of these subtle technicalities were not encountered in my first year multivariate course. This is slowing down my progression considerably in Physics.

I want to know how $\hat{\phi}$ is determined.

Edit:

I'd like to add that the direction due to the above cross product must be perpendicular to both $\vec{m}$ and $\hat{r}$. Clearly, It must lie on the xy-plane for it to be mutually perpendicular to $\vec{m}$ and $\hat{r}$. But the $\phi$ is elusive.

Answer 1

Let the unit vectors in spherical coordinates be $\hat{\rho},\hat{\theta},\hat{\phi}$.

Note that $\vec{m}=m \hat{z}$, that $\hat{z}=\cos\theta \hat{\rho}-\sin\theta \hat{\theta}$ and that $\hat{\theta}\times\hat{\rho}=-\hat{\phi}$.

So the cross-product is $$\vec{m}\times \hat{\rho}=m(\cos\theta \hat{\rho}-\sin\theta \hat{\theta})\times\hat{\rho}=m\sin\theta \hat{\phi}.$$

Answer 2

Compare the spherical coordinates and Cartesian coordinates: $$\vec{n}(r, \varphi, \theta) = \begin{cases} x = r \cos(\varphi) \sin(\theta) \\ y = r \sin(\varphi) \sin(\theta) \\ z = r \cos(\theta) \end{cases}$$

By definition, the spherical coordinate system unit vectors are $$\hat{r} = \begin{cases} x = \cos(\varphi) \sin(\theta) \\ y = \sin(\varphi) \sin(\theta) \\ z = \cos(\theta) \end{cases}, \; \hat{\varphi} = \begin{cases} x = -\sin(\varphi) \\ y = \cos(\varphi) \\ z = 0 \end{cases}, \; \hat{\theta} = \begin{cases} x = \cos(\varphi) \cos(\theta) \\ y = \sin(\varphi) \cos(\theta) \\ z = -\sin(\theta) \end{cases}$$

If $$\vec{m} = \begin{cases} x = 0 \\ y = 0 \\ z = m \end{cases}$$ then $\vec{m} \times \hat{r}$ in Cartesian coordinates is $$\vec{m} \times \hat{r} = \begin{cases} x = -m \sin(\varphi) \sin(\theta) \\ y = m \cos(\varphi) \sin(\theta) \\ z = 0 \end{cases}$$ and $m \sin(\theta) \hat{\varphi}$ is $$m \sin(\theta) \hat{\varphi} = \begin{cases} x = -m \sin(\theta) \sin(\varphi) \\ y = m \sin(\theta) \cos(\varphi) \\ z = 0 \end{cases}$$ The two are obviously equal.