Polar coordinates: Orthonormal basis

Question

In polar coordinates we have $r= c(\hat{r})$, where $c$ is the distance of a point from origin, and $r$ is the position vector.

So, what is the use of $\hat{\theta}$ especially given that it is always orthonormal to $\hat{r}$ ?

As an extension: What does the vector $r = a(\hat{r}) + b(\hat{\theta})$ mean at all? Does this mean that $b$ is the magnitude of "radians" the position vector makes in the $x$-$y$ plane? But this sounds absurd.

Furthermore, given that the $\hat{\theta}$ and the $\hat{r}$ vectors are always "moving", how do the coordinate axes of the $\hat{\theta}$ and the $\hat{r}$ plane look like?

Answer 1

You are only thinking about position vectors - vectors that represent the position of a point relative to the origin. For this type of vector you are correct - the position $\vec r$ of a point is

$\vec r = |\vec r| \hat r = r \hat r$

and so a position vector has no component in the $\hat \theta$ direction.

However, position vectors are not the only type of vector. We can, for example, express the velocity or acceleration of an object as a vector. Or we can represent forces as vectors. All of these other types of vectors may have non-zero components in the $\hat \theta$ direction.

As a concrete example, consider the general velocity vector

$\vec v = \dot {\vec r} = \dot r \hat r + r \dot {\hat r} = \dot r \hat r + r \dot \theta \hat \theta = \dot r \hat r + r \omega \hat \theta$

In general, the $\hat \theta$ component of the velocity vector will not be zero. Indeed, if the object is moving in a circle then $\dot r$ is zero and its velocity vector $r \omega \hat \theta$ is entirely in the $\hat \theta$ direction.

Answer 2

You are making a gigantic mistake.

In flat space, we have a rather important convenience that we implicitly assume is available, namely that we can identify a position with a vector, hence position vector.

Position vector $\vec r=\left|\vec r\right|\hat{\vec r}$ defines both its magnitude and its unit direction vector. It is obvious that this is never going to have a $\hat{\vec\theta}$ component.

But other vectors can have. A general other kind of vector may be expressed as $\vec v=a\hat{\vec r}+b\hat{\vec\theta}$ and be understood.