# Mechanics~polar coordinate & radial and transverse component

1. Jul 10, 2013

### Outrageous

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?

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

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.