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Confusion with Dot Product in Polar Coordinates with the Metric Tensor

  1. Aug 21, 2014 #1

    MrBillyShears

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    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:
    [itex]g(\vec{A}\,,\vec{B})=A^aB^bg_{ab}[/itex]
    And, if we're dealing with polar coordinates, then the metric is:
    [itex]g_{ab}=\begin{pmatrix}1&0\\0&r^2\end{pmatrix}[/itex]
    Alright, so the dot product is:
    [itex]A^1B^1+(r)^2A^2B^2[/itex]
    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!
     
  2. jcsd
  3. Aug 21, 2014 #2

    WWGD

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    In my experience, both, r, theta are given as functions.
     
  4. Aug 21, 2014 #3

    MrBillyShears

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    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,
    [itex]\vec{e}_{a'}=\Lambda^b{}_{a'}\vec{e}_{b}[/itex]
    where
    [itex]\Lambda^b{}_{a'}=\begin{pmatrix}cos{Θ}&sin{Θ}\\-rsin{Θ}&rcos{Θ}\end{pmatrix}[/itex]
    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?
     
  5. Aug 21, 2014 #4

    Fredrik

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    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##.
     
    Last edited: Aug 21, 2014
  6. Aug 21, 2014 #5

    MrBillyShears

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    Ok, thanks. It's becoming much clearer to me now.
     
  7. Aug 22, 2014 #6
    You can calculate the metric like this [itex]g_{ij} = (\partial_{i}, \partial_{j})[/itex] 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 [itex]A_{i}=g_{ij}A^{j}[/itex], the solution to this system of equations is your covariant vector in cylindrical coordinates. I'm not sure this helps...
     
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