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

  • #1
SebastianRM
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TL;DR 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:
$$\vec{\nabla} \times \vec{V} = \begin{vmatrix} \hat{r}\ & \hat{\theta} & \hat{z} \\ \partial/\partial r & (1/r)\partial/\partial \theta & \partial/\partial z \\ 0 & V_{\theta} & 0 \end{vmatrix} = \left< -\partial V_{\theta}/\partial z, 0 , \partial V_\theta/\partial r \right> = \frac{\partial V_\theta}{\partial r} \hat{z} $$

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.
 
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  • #2
SebastianRM said:
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:
$$\frac{1}{r} < -\frac{\partial (rV_{\theta})}{\partial z}\hat r ,0,\frac{\partial (rV_{\theta})}{\partial r} \hat z >$$
 
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  • #3
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.
 
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  • #4
SebastianRM said:
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.
 
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