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2-dimension polar basis vectors

  1. Mar 3, 2014 #1
    I'd like to understand why i cannot seem to be able to define unit polar basis vectors. Let me explain:

    We have our usual polar coordinates relation to Cartesian:

    x = r cosθ ; y = r sinθ

    if I define [itex]\hat{e_{r}}[/itex], [itex]\hat{e_{\vartheta}}[/itex] as the polar basis vectors, then they should be contravariant, meaning that they can be obtained from [itex]\hat{u_{x}}[/itex], [itex]\hat{u_{y}}[/itex] as:

    [itex]\hat{e_{r}}[/itex] = [itex]\delta x/ \delta r\ \hat{u_{x}} + \delta y / \delta r \ \hat{u_{y}}[/itex] = cosθ [itex]\hat{u_{x}}[/itex] + sin θ [itex]\hat{u_{y}}[/itex]

    and
    [itex]\hat{e_{\vartheta}}[/itex] = [itex]\delta x/ \delta \vartheta \ \hat{u_{x}} + \delta y / \delta \vartheta \ \hat{u_{y}} [/itex] = -r sinθ [itex]\hat{u_{x}}[/itex] + r cosθ [itex]\hat{u_{y}}[/itex]

    which implies that |[itex]\hat{e_{\vartheta}}[/itex]| = r, rather than being a unit vector as usually considered.

    Is this right?
     
  2. jcsd
  3. Mar 3, 2014 #2
    Yes. This is correct. The coordinate basis vectors are not (necessarily) unit vectors. Also, they should be considered covariant. That is,
    [tex]d\vec{r}=\vec{e}_rdr+\vec{e}_θdθ[/tex]
    where dr and dθ are considered contravariant.

    Chet
     
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