Confused with polar vector

1. Feb 19, 2014

Gzyousikai

It is known that the vector in polar coordinate system can be expressed as $$\mathbf{r}=r\hat{r}$$. In this formula, we don't see $$\hat{\theta}$$ appear.
But after the derivation yielding speed, $$\mathbf{v}=\dot{r}\hat{r}+r\dot{\theta}\hat{\theta}$$. Where does theta come from? And how to define its magnitude and direction?

Last edited: Feb 19, 2014
2. Feb 19, 2014

grzz

The $\dot{θ}$ appears because of the differentiation of the unit vector $\hat{\underline{r}}$.

3. Feb 19, 2014

sophiecentaur

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.

4. Feb 19, 2014

vanhees71

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