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I am wondering how to actually determine the gradient of a vector in cylindrical coordinates. I have seen a lot of websites that just say what the general form is but I cannot seem to understand how they got there.

The vector in cylindrical coordinates that I am going to use so everyone can follow along is going to be $\vec{V}=V_{r}\hat{r}+V_{\theta}\hat{\theta}+V_{z}\hat{z}$

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  • $\begingroup$ Did you mean "the gradient of an scalar function" ? $\endgroup$
    – Mikasa
    Aug 21, 2013 at 15:30
  • $\begingroup$ Yes that's what I meant. Sorry. The function I would be dealing with is still the one above $\endgroup$ Aug 21, 2013 at 15:44
  • $\begingroup$ As far as I have seen, gradient of a scalar function is well-known not a vector. for a vector we can speak about its "div" or "curl". $\endgroup$
    – Mikasa
    Aug 21, 2013 at 15:48
  • $\begingroup$ I am working on a problem where I am trying to find the divergence of the vector in cylindrical coordinates but I need to find its gradient in order to do that. I have found the general form of the gradient online but I would like to understand how that was produced instead of copying it $\endgroup$ Aug 21, 2013 at 15:50
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    $\begingroup$ The gradient of a vector is defined in tensor calculus : $\nabla x_i e_i = e_j \frac{\partial}{\partial x_j} x_i e_i = \delta_{ij}$. $\endgroup$
    – Doctor Dan
    Aug 21, 2013 at 16:08

2 Answers 2

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On any Riemannian manifold (not necessarily curved), the gradient of a function is the metric dual of the exterior derivative.

The exterior derivative relative to any coordinate system of a function is just

$$ \mathrm{d}f = \partial_{x^1} f \mathrm{d}x^1 + \partial_{x^2} f \mathrm{d}x^2 + \cdots + \partial_{x^k} f \mathrm{d}x^k $$

What we need to do, then, is to "hit it with the inverse metric".

In cylindrical coordinates, the metric is

$$ \mathrm{d}r^2 + r^2 \mathrm{d}\theta^2 + \mathrm{d}z^2 $$

which we can write as the matrix $\mathrm{diag}(1, r^2, 1)$. Inverting the matrix gives $\mathrm{diag}(1, r^{-2}, 1)$ and so the inverse metric is

$$ \hat{r}^2 + r^{-2} \hat{\theta}^2 + \hat{z}^2 $$

So applying the inverse metric to the differential form $\mathrm{d}f$ we get

$$ \nabla f = \partial_r f \hat{r} + r^{-2} \partial_\theta f \hat{\theta} + \partial_z f \hat{z} $$

the coefficients I note are computed by taking the inverse metric as a matrix, and multiplying to it the exterior derivative as a column vector:

$$ \begin{pmatrix} 1 \\ & r^{-2} \\ & & 1\end{pmatrix} \begin{pmatrix} \partial_r f \\ \partial_\theta f \\ \partial_z f \end{pmatrix} $$

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  • $\begingroup$ Shouldn't it be $1/r$ instead of $1/r^2$? $\endgroup$
    – rivendell
    Feb 14, 2018 at 15:41
  • $\begingroup$ Depends on what you use as the basis. I am using the coordinate basis $\{\partial_r, \partial_\theta, \partial_z\}$ (in the above denoted by $\hat{r}, \hat{\theta}, \hat{z}$) to decompose tangent vectors. Alternatively you can use a normalized basis consisting of $\{\partial_r, r^{-1} \partial_\theta, \partial_z\}$ in which case multiplication by $1/r$ would be correct. $\endgroup$ Feb 14, 2018 at 16:27
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I've derived it in the easiest way. Note each term in the gradient tensor is described in tensor notation: $$\nabla \vec v_{ij}=\nabla_j\vec v \cdot e_i$$ Where $\nabla_j$ means jth component of del operator. Apply this to each term in gradient tensor as below.

$\nabla\vec v_{rr}=\nabla_r\vec v\cdot\hat r=\frac{\partial\vec v}{\partial r}\cdot\hat r=\left(\frac{\partial v_r}{\partial r}\hat r+\frac{\partial v_\theta}{\partial r}\hat\theta+\frac{\partial v_z}{\partial r}\hat z\right)\cdot\hat r=\frac{\partial v_r}{\partial r}$ $\nabla\vec v_{r\theta}=\nabla_{\theta }\vec v\cdot \hat r=\frac{\partial \vec v}{r\partial \theta }\cdot \hat{r}=\frac{1}{r}\left(\frac{\partial {v}_r}{\partial \theta }\hat{r}+{v}_r\hat{\theta }+\frac{\partial {v}_{\theta }}{\partial \theta }\hat{\theta }-{v}_{\theta }\hat{r}+\frac{\partial {v}_z}{\partial \theta }\hat{z}\right)\cdot \hat{r}=\frac{\partial {v}_r}{r\partial \theta }-\frac{{v}_{\theta }}{r}$ $\nabla\vec v_{rz}=\nabla_z\vec v\cdot\hat r=\frac{\partial\vec v}{\partial z}\cdot\hat r=\left(\frac{\partial v_r}{\partial z}\hat r+\frac{\partial v_\theta}{\partial z}\hat\theta+\frac{\partial v_z}{\partial z\hat z}\right)\cdot\hat r=\frac{\partial v_r}{\partial z}$ $\nabla\vec v_{\theta r}=\nabla_r\vec v\cdot\hat\theta=\frac{\partial\vec v}{\partial r}\cdot\hat\theta=\left(\frac{\partial v_r}{\partial r}\hat r+\frac{\partial v_\theta}{\partial r}\hat\theta+\frac{\partial v_z}{\partial r}\hat z\right)\cdot\hat\theta=\frac{\partial v_\theta}{\partial r}$ $\nabla\vec v_{\theta\theta}=\nabla_\theta\vec v\cdot\hat\theta=\frac{\partial \vec v}{r\partial\theta}\cdot\hat\theta=\frac{1}{r}\left(\frac{\partial v_r}{\partial\theta}\hat r+v_r\hat\theta+\frac{\partial v_\theta}{\partial\theta}\hat\theta-v_\theta\hat r+\frac{\partial v_z}{\partial\theta}\hat z\right)\cdot\hat\theta=\frac{\partial v_\theta}{r\partial\theta}+\frac{v_r}{r}$ $\nabla\vec v_{\theta z}=\nabla_{z}\vec v\cdot\hat\theta=\frac{\partial\vec v}{\partial z}\cdot\hat\theta=\left(\frac{\partial v_r}{\partial z}\hat r+\frac{\partial v_\theta}{\partial z}\hat\theta+\frac{\partial v_z}{\partial z}\hat z\right)\cdot\hat\theta=\frac{\partial v_\theta}{\partial z}$ $\nabla\vec v_{zr}=\nabla_r\vec v\cdot\hat z=\frac{\partial\vec v}{\partial r}\cdot\hat z=\left(\frac{\partial v_r}{\partial r}\hat r+\frac{\partial v_\theta}{\partial r}\hat\theta+\frac{\partial v_z}{\partial r}\hat z\right)\cdot\hat z=\frac{\partial v_z}{\partial r}$ $\nabla\vec v_{z\theta}=\nabla_\theta\vec v\cdot\hat z=\frac{\partial\vec v}{r\partial\theta}\cdot\hat z=\frac{1}{r}\left(\frac{\partial v_r}{\partial\theta }\hat r+v_r\hat\theta+\frac{\partial v_\theta}{\partial\theta}\hat\theta-v_\theta\hat r+\frac{\partial v_z}{\partial\theta}\hat z\right)\cdot \hat z=\frac{\partial v_z}{r\partial\theta}$ $\nabla\vec v_{zz}=\nabla_z\vec v\cdot\hat z=\frac{\partial\vec v}{\partial z}\cdot\hat z=\left(\frac{\partial v_r}{\partial z}\hat r+\frac{\partial v_\theta}{\partial z}\hat\theta+\frac{\partial v_z}{\partial z}\hat z\right)\cdot\hat z=\frac{\partial v_z}{\partial z}$

Finally, we've obtained the full terms of tensor. $$\nabla\vec v=\begin{bmatrix}\nabla\vec v_{rr}&\nabla\vec v_{r\theta}&\nabla\vec v_{rz}\\\nabla\vec v_{\theta r}&\nabla\vec v_{\theta \theta}&\nabla\vec v_{\theta z}\\\nabla\vec v_{zr}&\nabla\vec v_{z\theta}&\nabla\vec v_{zz}\end{bmatrix}=\begin{bmatrix}\frac{\partial v_r}{\partial r}&\frac{\partial v_r}{r\partial\theta}-\frac{v_\theta}{r}&\frac{\partial v_r}{\partial z}\\\frac{\partial v_\theta}{\partial r}&\frac{\partial v_\theta}{r\partial\theta}+\frac{v_r}{r}&\frac{\partial v_\theta}{\partial z}\\\frac{\partial v_z}{\partial r}&\frac{\partial v_z}{r\partial\theta}&\frac{\partial v_z}{\partial z}\end{bmatrix}$$ Refer to my site for more information : https://blog.naver.com/richscskia/221744783715

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    $\begingroup$ While this link may answer the question, it is better to include the essential parts of the answer here and provide the link for reference. Link-only answers can become invalid if the linked page changes. - From Review $\endgroup$
    – MafPrivate
    Dec 11, 2019 at 14:24
  • $\begingroup$ I've added the essential parts of it to address your comment and leave the link just for reference. Thank you for advice. $\endgroup$
    – user733991
    Dec 21, 2019 at 13:08

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