Divergence in cartesian coordinates conflicts with spherical divergence.

Question

Consider the vector $$ \vec{a}=f(r)\hat{r}+r\hat{\theta}+r\hat{\phi} $$ Where $\vec{a}$ is a vector in spherical coordinates and $f(r)$ is a function of $r$.
Let's calculate $\nabla\cdot(\vec{a}\times\vec{r})$, where $\vec{r}$ is the position vector.
$$ \nabla\cdot(\vec{a}\times\vec{r})=\vec{r}\cdot(\nabla\times\vec{a})-\vec{a}\cdot(\nabla\times\vec{r}) $$ $\nabla\times\vec{r}=0$. Thus, by using divergence formula in spherical coordinates, I concluded that $$ \nabla\cdot(\vec{a}\times\vec{r})=r\cot(\theta) $$ But, the same calculation in cartesian coordinates lead to $$ \nabla\cdot(\vec{a}\times\vec{r})=0 $$ Why is this quantity different in these coordinates?

Answer 1

I think you are using the wrong identity .

$\nabla \cdot (\vec{a} \times \vec{r}) = \frac{\partial }{\partial x_i}(\epsilon_{ijk}a_jr_k)=\epsilon_{ijk}(\partial a_j/\partial x_i)r_k + \epsilon_{ijk} ( \partial r_k/ \partial x_i)a_j= \vec{r}\cdot (\nabla \times \vec{a} )-\vec{a} \cdot (\nabla \times \vec{r}) $

$\nabla \cdot (\vec{a} \times \vec{r}) = \vec{r}\cdot (\nabla \times \vec{a} )-\vec{a} \cdot (\nabla \times \vec{r})$

$\vec{a}=f(r)\hat{r} + r \hat{\theta}+r \hat{\phi}$

$\vec{r}=r\hat{r}$

$\vec{a}\times \vec{r}= -r^2\hat{\phi}+r^2\hat{\theta}$

$\nabla \cdot \vec{E} = \frac{1}{r^2}\frac{\partial }{\partial r}(r^2 E_r)+ \frac{1}{r \sin \phi }\frac{\partial E_\theta}{\partial \theta}+\frac{1}{r \sin \phi}\frac{\partial }{\partial \phi}(\sin \phi E\phi)$

Since the $r$ component is $0$ and the other components only have radial dependence, the divergence is $0$.

Further $\nabla (r^2/2)=\vec{r}$ and for any scalar $f$, $\nabla \times \nabla f=0$ in any coordinate system.

It follows that $\nabla \times \vec{r}=0. $

$\nabla \times \vec{E} = \frac{1}{r \sin \phi}(\frac{\partial }{\partial \phi }(A_\theta \sin \phi)-\frac{\partial A_\phi}{\partial \theta })\hat{r}+ \frac{1}{r}(\frac{1}{\sin \phi} \frac{\partial A_r}{\partial \theta}-\frac{\partial }{\partial r}(rA_\theta))\hat{\theta}+\frac{1}{r}(\frac{\partial }{\partial r}(rA_\phi)-\frac{A_r}{\partial \phi})\hat{\phi}$

Since the components of $\vec{a}$ are only dependent on $r$, the derivatives for the r-component of the curl are 0, so the r coordinate of the curl is 0.

$\nabla \times \vec{a}= 0 \hat{r} -2\hat{\theta}+2\hat{\phi}$

So $\vec{r} \cdot (\nabla \times \vec{a}) = 0. $

Combining this we have $\nabla \cdot (\vec{a} \times \vec{r})=0$