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Mechanics~polar coordinate & radial and transverse component

  1. Jul 10, 2013 #1
    http://www.answers.com/topic/radial-and-transverse-components

    From the above link,
    1) e θ is a unit vector perpendicular to r in the direction of increasing θ.
    Where is the direction of increasing θ? Is that a circle? θ Increase from 0 to 2∏.then eθ moves in a circle? direction always changes?
    2) dr/dθ= eθ , and eθ is a unit vector, this is being defined ?why do we know that dr/dθ is the unit vector that is perpendicular to the radius? Do we have anything to prove it?
    Please teach, thanks
     
  2. jcsd
  3. Jul 10, 2013 #2

    vanhees71

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    The cartesian coordinates of the position vector in the plane (taken out the origin) can be described in terms of polar coordinates [itex](r,\theta[/itex] by
    [tex]\vec{r}=\begin{pmatrix}x \\ y \end{pmatrix} = r \begin{pmatrix} \cos \theta \\ \sin \theta \end{pmatrix}.[/tex]
    The polar-coordinate lines define the basis vectors of polar coordinates
    [tex]\vec{b}_r=\frac{\partial \vec{r}}{\partial r}=\begin{pmatrix} \cos \theta \\ \sin \theta \end{pmatrix}, \quad \vec{b}_{\vartheta}=\frac{\partial \vec{r}}{\partial \theta}=r \begin{pmatrix} -\sin \theta \\ \cos \theta \end{pmatrix}.[/tex]
    You can easily check that these two vectors are perpendicular to each other
    [tex]\vec{b}_r \cdot \vec{b}_{\theta}=0.[/tex]
    Usually for such orthogonal curved coordinates one introduces the normalized vectors along the coordinate lines. The lengths of the basis vectors are [itex]|\vec{b}_r|=1[/itex] and [itex]|\vec{b}_{\theta}|=r[/itex]. Thus the normalized basis vectors are given by
    [tex]\vec{e}_r=\vec{b}_r=\begin{pmatrix} \cos \theta \\ \sin \theta \end{pmatrix}, \quad \vec{e}_{\theta}=\frac{1}{r} \vec{b}_{\theta} = \begin{pmatrix}-\sin \theta \\ \cos \theta \end{pmatrix}.[/tex]
     
  4. Jul 13, 2013 #3
    understand already ,thanks :smile:
     
  5. Jul 13, 2013 #4

    BruceW

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    I think generally, with polar coordinates, if you are ever unsure about something, you can substitute Cartesian coordinates, and see why it works. A good way to check.
     
  6. Jul 14, 2013 #5
    Cartesian and polar coordinate.
    ok, thanks for advice.
     
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