How to obtain the determinant of the Curl in cylindrical coordinates?

[SebastianRM]

TL;DR

Hey guys, I wrote my determinant for directions r, $\theta$, z. For vectors of the same cylindrical space. So, I proceed to calculate the determinant, but its form is far different from what shown online. How can I derive the proper way.

I have a vector in cylindrical Coordinates:
$$\vec{V} = \left < 0 ,V_{\theta},0 \right> $$
where $V_\theta = V(r,t)$.

The Del operator in $\{r,\theta,z\} is: \vec{\nabla} = \left< \frac{\partial}{\partial r}, \frac{1}{r}\frac{\partial}{\partial \theta}, \frac{\partial}{\partial z} \right>$

I tried to obtain the curl as follows:
$$\vec{\nabla} \times \vec{V} = \begin{vmatrix} \hat{r}\ & \hat{\theta} & \hat{z} \\ \partial/\partial r & (1/r)\partial/\partial \theta & \partial/\partial z \\ 0 & V_{\theta} & 0 \end{vmatrix} = \left< -\partial V_{\theta}/\partial z, 0 , \partial V_\theta/\partial r \right> = \frac{\partial V_\theta}{\partial r} \hat{z} $$

However i have seen very different determinants for the curl, and I am not sure why my approach is incorrect. How can I derive the proper determinant?

Your time and answers are very appreciated.

 

[Tony Hau]

Summary:: Hey guys, I wrote my determinant for directions r, $\theta$, z. For vectors of the same cylindrical space. So, I proceed to calculate the determinant, but its form is far different from what shown online. How can I derive the proper way.

I have a vector in cylindrical Coordinates:
$$\vec{V} = \left < 0 ,V_{\theta},0 \right> $$
where $V_\theta = V(r,t)$.

The Del operator in $\{r,\theta,z\} is: \vec{\nabla} = \left< \frac{\partial}{\partial r}, \frac{1}{r}\frac{\partial}{\partial \theta}, \frac{\partial}{\partial z} \right>$

I tried to obtain the curl as follows:
$$\vec{\nabla} \times \vec{V} = \begin{vmatrix} \hat{r}\ & \hat{\theta} & \hat{z} \\ \partial/\partial r & (1/r)\partial/\partial \theta & \partial/\partial z \\ 0 & V_{\theta} & 0 \end{vmatrix} = \left< -\partial V_{\theta}/\partial z, 0 , \partial V_\theta/\partial r \right> = \frac{\partial V_\theta}{\partial r} \hat{z} $$

The resultant vector would be:
$$<

However i have seen very different determinants for the curl, and I am not sure why my approach is incorrect. How can I derive the proper determinant?

Your time and answers are very appreciated.

The general formula for cylindrical coordinate is as follows:
$$\vec{\nabla} \times \vec{V} = \frac{1}{r}\begin{vmatrix} \hat{r}\ & r\hat{\theta} & \hat{z} \\ \partial/\partial r & \partial/\partial \theta & \partial/\partial z \\ V_{r} & V_{\theta} & V_{z} \end{vmatrix} $$
Because $V_r$ and $V_z$ are zero, the resultant vector would be:
$$\frac{1}{r} < -\frac{\partial (rV_{\theta})}{\partial z}\hat r ,0,\frac{\partial (rV_{\theta})}{\partial r} \hat z >$$

 

[SebastianRM]

I was trying to derive the determinant itself, I know that is the correct form. I was using the definition of a cross product to do the curl; however, the curl is not really a cross product of vectors in an orthonormal basis.

 

[Tony Hau]

I was trying to derive the determinant itself, I know that is the correct form. I was using the definition of a cross product to do the curl; however, the curl is not really a cross product of vectors in an orthonormal basis.

The del operator is not a vector that crosses with vectors, although it resembles the property of vectors. I am sorry I am not an expert who can explain clearly to you about it, but here is a derive of the curl in other coordinates:https://en.wikipedia.org/wiki/Del_in_cylindrical_and_spherical_coordinates

The definition of curl is: $$\nabla \times \vec A = \lim_{dS \rightarrow 0}\frac{\int \vec A \cdot dl}{\iint dS}$$, which is shrinking the path integral of $\vec A$ over a path $dl$ enclosing an infinitely small area $dS$

I suppose you need to apply the Jacobian to transform the $dl$ element from Cartesian coordinate to other coordinates in deriving the curl for other coordinates.