# How to determine gradient of vector in cylindrical coordinates?

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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Did you mean "the gradient of an scalar function" ? – Babak S. Aug 21 '13 at 15:30
Yes that's what I meant. Sorry. The function I would be dealing with is still the one above – Greg Harrington Aug 21 '13 at 15:44
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". – Babak S. Aug 21 '13 at 15:48
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 – Greg Harrington Aug 21 '13 at 15:50
The gradient of a vector is defined in tensor calculus : $\nabla x_i e_i = e_j \frac{\partial}{\partial x_j} x_i e_i = \delta_{ij}$. – Doctor Dan Aug 21 '13 at 16:08

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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