Confusion with Dot Product in Polar Coordinates with the Metric Tensor

[MrBillyShears]

Alright, so I was reading up on tensors and such with non-Cartesian coordinate systems all day but now I'm a bit tired an confused so you'll have to forgive me if it's a stupid question. So to express the dot product in some coordinate system, it's:
$g(\vec{A}\,,\vec{B})=A^aB^bg_{ab}$
And, if we're dealing with polar coordinates, then the metric is:
$g_{ab}=\begin{pmatrix}1&0\\0&r^2\end{pmatrix}$
Alright, so the dot product is:
$A^1B^1+(r)^2A^2B^2$
But which r? I know I'm probably only confused because I'm so tired right now, but both A and B have r's, do't they? Which r is used to compute this?
Remember, I'm just a student at this, so don't get too technical in the response, and sorry for any typos.
Thanks!

 

[WWGD]

In my experience, both, r, theta are given as functions.

 

[MrBillyShears]

I'm trying to figure this out on my own. Perhaps no one understands what I'm asking. I'm asking if the metric tensor I listed above is valid to use in the dot product of two different vectors. Because the basis vectors are different at different points,
$\vec{e}_{a'}=\Lambda^b{}_{a'}\vec{e}_{b}$
where
$\Lambda^b{}_{a'}=\begin{pmatrix}cos{Θ}&sin{Θ}\\-rsin{Θ}&rcos{Θ}\end{pmatrix}$
So, if there are different basis vectors for different points, you can't use the same metric tensor as you would to say, transform a polar vector to a polar covector. Is this correct?

 

[Fredrik]

The metric $g$ is a tensor field of type (2,0), so it associates a tensor $g_{\vec r}$ of type (2,0) with each $\vec r\in\mathbb R^2$. This $g_{\vec r}$ is an inner product on $T_{\vec r}\mathbb R^2$, the tangent space of $\mathbb R^2$ at $\vec r$. You have determined the components (in the polar coordinate system) of $g_{\vec r}$ for an arbitrary $\vec r$. Your r is the absolute value of this $\vec r$.

If you want a mental image, imagine a second copy of $\mathbb R^2$ with its (0,0) point attached to the point $\vec r$. The A and B that you feed into $g_{\vec r}$ are vectors in that space, so if you visualize them as arrows, those arrows should start at $\vec r$.

 

[MrBillyShears]

Ok, thanks. It's becoming much clearer to me now.

 

[JonnyMaddox]

You can calculate the metric like this $g_{ij} = (\partial_{i}, \partial_{j})$ for your parametrization i.e. for cylindrical coordinates. Then if you want to convert a vector in this coordinate system from a contravariant form into a covariant form you have to contract it with this metric tensor like $A_{i}=g_{ij}A^{j}$, the solution to this system of equations is your covariant vector in cylindrical coordinates. I'm not sure this helps...