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Confused with polar vector

  1. Feb 19, 2014 #1
    It is known that the vector in polar coordinate system can be expressed as [tex]\mathbf{r}=r\hat{r}[/tex]. In this formula, we don't see [tex]\hat{\theta}[/tex] appear.
    But after the derivation yielding speed, [tex]\mathbf{v}=\dot{r}\hat{r}+r\dot{\theta}\hat{\theta}[/tex]. Where does theta come from? And how to define its magnitude and direction?
     
    Last edited: Feb 19, 2014
  2. jcsd
  3. Feb 19, 2014 #2
    The [itex]\dot{θ}[/itex] appears because of the differentiation of the unit vector [itex]\hat{\underline{r}}[/itex].
     
  4. Feb 19, 2014 #3

    sophiecentaur

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    The angle will turn up if the position vector is changing direction (if motion is not radial).
    The direction of the unit theta vector will be normal to the unit r vector.

    It can sometimes be difficult to see any point in using anything other than cartesian co ordinates - until you come upon a suitable problem, when it suddenly makes good sense.
     
  5. Feb 19, 2014 #4

    vanhees71

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    You see everything most easily by expressing all vectors in cartesian coordinates. For polar coordinates you have
    [tex]\vec{r}=r \cos \theta \vec{e}_1 + r \sin \theta \vec{e}_2.[/tex]
    Here, [itex]\vec{e}_j[/itex] are a Cartesian basis, i.e., two fixed orthonormalized vectors in the plane.
    Then you get
    [tex]\hat{r}=\cos \theta \vec{e}_1 + \sin \theta \vec{e}_2, \quad \hat{\theta}=-\sin \theta \vec{e}_1+\cos \theta \vec{e}_2.[/tex]
    The derivatives of the polar unit-basis vectors thus are
    [tex]\partial_r \hat{r}=0, \quad \partial_{\theta} \hat{r}=\hat{\theta}, \quad \partial_r \hat{\theta}=0, \quad \partial_{\theta} \hat{\theta}=-\hat{r}.[/tex]
    From this you get
    [tex]\partial_r \vec{r}=\hat{r}, \quad \partial_{\theta} \vec{r}=r \hat{\theta}[/tex]
    and thus, using the product and chain rule
    [tex]\vec{v}=\frac{\mathrm{d}}{\mathrm{d}t} \vec{r}=\dot{r} \hat{r} + r \dot{\theta} \hat{\theta}.[/tex]
     
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