How to show $\operatorname{curl}\operatorname{curl}(e_r) = 0$

Question

$\DeclareMathOperator{curl}{curl}$I want to figure out how to calculate $\text{curl}(e_r$). Where $e_r$ is a base vector for the Spherical co-ordinate system.

Taking $e_r = (\sin\theta \cos\phi)i+(\sin\theta \sin\phi)j+(\cos\theta)k$

and I tried taking the $\text{curl}(e_r)$ as follows,

$\text{curl }e_r=\begin{vmatrix}i & j & k\\ \frac{\partial}{\partial r} & \frac{\partial}{\partial \theta} & \frac{\partial}{\partial \phi} \\ (\sin\theta \cos\phi) & (\sin\theta \sin\phi) & (\cos\theta)\end{vmatrix}$

And then took the $\operatorname{curl}$ again for the output of the above. But this gives a non-zero value.
I would like to know whether the above steps and the values I am using for $e_r$ are correct. If not can you please tell me what I am doing wrong.

Answer 1

You provide $e_r$ in Cartesian coordinates, but the second row of your determinant features derivatives for spherical coordinates.

Try again with the second row using a gradient in Cartesian coordinates $\partial/\partial x_i$ or change to a spherical version of the curl, but then formulate $e_r$ in spherical coordinates as well.

Derivation in Cartesian Coordinates

In Cartesian coordinates we can write $$ e_r = r/\lVert r\rVert \\ \partial_i ((e_r)_j) = \partial_i (x_j/\lVert r \rVert) = (\delta_{ij} \lVert r \rVert - x_j(1/2)(1/\lVert r \rVert)(2x_i))/\lVert r \rVert^2 = \delta_{ij}/\lVert r \rVert - x_i x_j/\lVert r \rVert^3 $$ This gives $$ \DeclareMathOperator{curl}{curl} (\curl e_r)_1 = \{(-x_2 x_3)-(-x_3 x_2)\}/\lVert r \rVert^3 = 0 $$ Only different indices contribute to the curl, so the first term has its Kronecker delta vanish. The second term is symmetric, one can exchange the indices, so this vanishes as well.

The curl of $e_r$ vanishes already, and so would a second application.

Example in Spherical Coordinates

The curl operator in spherical coordinates is:

$$ \curl A = \frac{1}{r\sin\theta}\left({\partial \over \partial \theta} \left( A_\varphi\sin\theta \right) - {\partial A_\theta \over \partial \varphi}\right) e_r + \frac 1 r \left({1 \over \sin\theta}{\partial A_r \over \partial \varphi} - {\partial \over \partial r} \left( r A_\varphi \right) \right) e_\theta + \frac 1 r \left({\partial \over \partial r} \left( r A_\theta \right) - {\partial A_r \over \partial \theta}\right) e_\varphi $$ For $A = e_r$ we have $(A_r, A_\varphi, A_\theta) = (1,0,0)$

This gives $\curl e_r = 0$.

For $A = e_\varphi$ we have $(A_r, A_\varphi, A_\theta) = (0,1,0)$

This gives $\curl e_\varphi = \frac{1}{r}\cot \theta \, e_r - \frac{1}{r} e_\theta$.

For $A = e_\theta$ we have $(A_r, A_\varphi, A_\theta) = (0,0,1)$.

This gives $\curl e_\theta = (1/r) e_\varphi$ and $\curl \curl e_\theta = \frac{1}{r \sin \theta}(\partial_\theta((1/r)\sin\theta)e_r $

Answer 2

The unit vector $e_r$ is the is nothing but the normalised vector radius: letting $$ r=\left(\begin{array}[c] \ x=\rho\cos\phi\sin\theta\\ y=\rho\sin\phi\sin\theta\\ z=\rho\cos\theta \end{array}\right) $$ then $$ e_r=\frac{r}{|r|},\text{ where }|r|=\sqrt{x^2+y^2+z^2}=\rho. $$ Now, consider $V(r)=|r|$ and compute its gradient $$ \nabla V(r) =\left(\begin{array}[c] \ \partial_x|r|\\ \partial_y|r|\\ \partial_z|r| \end{array} \right)= \frac{1}{|r|}\left(\begin{array}[c] \ x\\ y\\ z \end{array} \right)=\frac{r}{|r|}=e_r. $$ The curl of a gradient is zero, so: $$ \nabla\times e_{r}=\nabla\times\nabla V(r)=0. $$

In the comments, the OP asks for some guidance in computing $\nabla \times e_{\theta},$ as well, where I assume $$ e_\theta=\left( \begin{array}[c] \ \cos\phi \cos\theta\\ \sin\phi \cos\theta\\ -\sin\theta \end{array} \right). $$ Consider the path $\Gamma$ made as follows: for $0<\alpha<\pi/2$ and $\beta$ small, start from $(\rho=a,\theta=\alpha,\varphi=0)$ and move to $(a,\alpha+\beta,0)$ along an arc of fixed $\rho=a$, then go to $(a-\varepsilon,\alpha+\beta,0)$ along the radial segment of fixed angles, go back up to $(a-\varepsilon, \alpha,0)$ along the arc of radius $a-\varepsilon$ and finally return to the starting point moving radially. Since $e_\theta$ is a unit vector which is tangential to the above arcs and orthogonal to the above segments, the line integral $$ \oint_\Gamma e_\theta\cdot dl= \beta\varepsilon; $$ since, if $\Gamma=\partial \Sigma$, where $\Sigma$ is the surface enclosed in the path $\Gamma$, $$ \oint_\Gamma e_\theta\cdot dl = \int_\Sigma (\nabla\times e_{\theta})\cdot da $$ we conclude that $\nabla\times e_{\theta}\neq0$. Furthermore, since geometrically $\nabla\times e_{\theta}=|\nabla\times e_{\theta}|e_{\phi}$ and the area element of $\Sigma$ is precisely $e_{\phi}\rho d\theta d\rho$, so that $$ \beta\varepsilon=\int_\Sigma (\nabla\times e_{\theta})\cdot da=\int_\alpha^{\alpha+\beta}d\theta \int_{a-\varepsilon}^a \rho d\rho |\nabla\times e_{\theta}| $$ and the mean over the surface $\Sigma$ of area $\beta(a^2-(a-\varepsilon)^2)/2$ is given by $$ \frac{2}{\beta(a^2-(a-\varepsilon)^2)}\beta\varepsilon = \frac{2}{\beta(a^2-(a-\varepsilon)^2)}\int_\alpha^{\alpha+\pi/2}d\theta \int_b^a \rho d\rho |\nabla\times e_{\theta}|. $$ Letting $\varepsilon,\beta\to0$, the right-hand side approaches $|\nabla\times e_{\theta}|$ computed at $(a,\alpha,0)$ by continuity and hence $$ |\nabla\times e_{\theta}|=\frac{1}{\rho}, $$ where I have renamed $a=\rho$, to make the expression general, and finally $$ \nabla\times e_{\theta}=\frac{1}{\rho}e_{\phi}. $$