Mechanics~polar coordinate & radial and transverse component

1. Jul 10, 2013

Outrageous

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. Jul 10, 2013

vanhees71

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

3. Jul 13, 2013

Outrageous

understand already ,thanks

4. Jul 13, 2013

BruceW

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.

5. Jul 14, 2013

Outrageous

Cartesian and polar coordinate.
ok, thanks for advice.

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