# Confusion with Dot Product in Polar Coordinates with the Metric Tensor

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

Gold Member
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
Staff Emeritus
Gold Member
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##.

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

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...