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$(1)$ Compute the gradient of polar basis vectors, \begin{equation} \tilde{\nabla} e_\rho=\frac{1}{\rho} \widetilde{e}_\rho \otimes e_\theta \text { and } \tilde{\nabla} e_\theta=\frac{1}{\rho} \tilde{e}^\rho \otimes e_\theta-\tilde{\rho} \tilde{e}^\theta \otimes e_\rho \end{equation} $(2)$ What are the gradients of the vectors $e_\rho$ and $e_\theta$ on a cylinder? (Hints: A cylinder is a surface of constant $\rho$ in cylindrical coordinates in $\mathbb{R}$)

I know,

\begin{align} [g_{ij}] = \begin{pmatrix} g_{rr} & g_{r\theta}\\ g_{\theta r} & g_{\theta \theta} \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & r^2 \end{pmatrix} \end{align}

\begin{align} [g^{ij}] = \begin{pmatrix} g^{rr} & g^{r\theta}\\ g^{\theta r} & g^{\theta \theta} \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1/r^2 \end{pmatrix} \end{align}

and gradient in polar coordinates,

\begin{align} \text{grad}(f) &= g^{rr}\dfrac{\partial f}{\partial r}\dfrac{\partial }{\partial r} + g^{\theta \theta}\dfrac{\partial f}{\partial \theta}\dfrac{\partial }{\partial \theta} \\ &= \dfrac{\partial f}{\partial r}\dfrac{\partial }{\partial r} + \dfrac{1}{r^2}\dfrac{\partial f}{\partial \theta}\dfrac{\partial }{\partial \theta} \tag{$**$} \end{align}

But I didn't see from these information how can I compute the given one? From here I knew that there are two kind of basis. Covariant basis and orthonormal basis (which we mostly familiar with linear algebra). From the question context I couldn't understand how to interpret which basis was used to compute the gradient.

Any help will be appreciated. TIA.

As @Paul Sinclair suggested, I need to use gradient for vector field. but what I get was scalar gradient in my book only.

I have the following definitions in my book.

$$ \begin{align} \text{grad}(\phi) &= \phi_{,i} = \frac{\partial \phi}{\partial x^{i}}\\ \text{div}(A^{i}) &= A^{i}_{,i} = \frac{1}{\sqrt g}\frac{\partial }{\partial x^{k}}(\sqrt g A^k)\\ \text{curl}(A_i) &= A_{i,j} - A_{j,i} \end{align} $$

Where I see any of them can't help me to get the solution. And I was wondering why $\text{curl}$ operator only defined for covariant tensor? Where $\text{div}$ defined for both covariant and contravarint tensor $\text{div}(A^{i})=\text{div}(A_{i})$

What bother me a lot that from vector calculus we taught the operations grad (which maps functions to vector fields), curl (which maps vector fields to vector fields), and div (which maps vector fields to functions). But in tensor calculus we are talking about gradient of vector field, isn't that seem wrong?

falamiw
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  • If your cylinder is in $\Bbb R^3$, then $e_{\rho}$ is not tangent to it but normal. $e_z$ is. – Didier Nov 08 '22 at 07:39
  • Notice that the formula for gradient that you have is for the gradient of a function, not a vector field. You need to back up and see how your text defines the gradient of a vector field. – Paul Sinclair Nov 08 '22 at 20:34
  • I have add all the definition of the book @PaulSinclair. Could you redirect me any resource where I get the definition of gradient of a vector field? I search in google like here but could get some useful to solve the problem in tensor calculus context. – falamiw Nov 09 '22 at 17:53
  • See here for the definition of gradient of a tensor field (and divergence and Laplacian). You're going to have to compute covariant derivatives here, so you need the Christoffel symbols in polar coordinates (a not-so-fun, but standard exercise). Also, what are the tildes and superscripts on the $e$'s? – peek-a-boo Nov 10 '22 at 17:19
  • If my calculation are correct then the only non-zero Christoffel symbols for second kind are $\Gamma^r_{\theta\theta}=-r,\Gamma^\theta_{r\theta}\stackrel{\textrm{symmetry}}{=}\Gamma^r_{\theta r}=\frac1r$. Now, how to proceed further ? This was my previous year question on Introductory tensor calculus course. I couldn't know what was those tildes mean. @peek-a-boo And thanks for your quick response – falamiw Nov 10 '22 at 18:02
  • Can you say something on this, "...But in tensor calculus we are talking about gradient of vector field, isn't that seem wrong?" @peek-a-boo – falamiw Nov 10 '22 at 18:03
  • it’s not wrong, it just requires a more general definition, and this is what I have provided in the link. Ok, now that you have the Christoffel symbols, recall that $\nabla_{e_i}e_j=\Gamma^k_{ij}e_k$, sum over $k$. So, you can compute $\nabla_{e_{\rho}}e_{\rho}$ etc. To get the gradient as a vector, you then use the metric to convert this $(1,1)$ tensor field to a $(2,0)$ tensor field (but I’d rather leave $\nabla e_j$ in its natural state as a $(1,1)$ tensor field). – peek-a-boo Nov 10 '22 at 21:24

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