- #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.
$$\vec{V} = \left < 0 ,V_{\theta},0 \right> $$
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.