# A Bit Confused About Polar Basis Vectors

1. Aug 22, 2014

### MrBillyShears

Let me say from the beginning I'm not talking about the non-coordinate unit vectors for polar coordinates. I'm talking about basis vectors. Let me just ask it as boldly as possible: how does one use these basis vectors in order to describe a vector? I know they are different at every point, so which point do you use? Is it completely arbitrary? Why is there different basis vectors at every point? And, I am new with this kind of stuff, so try to keep it as simple as possible in your explanation.
And also, if so, if you pick $\vec{r}$ to describe your vectors, would the "tails" of your vectors come from the origin, or from your point $\vec{r}$? And, if someone could give an example with numbers, that would be great.

Last edited: Aug 22, 2014
2. Aug 22, 2014

### Blazejr

To really understand this concept properly one needs to learn some basic introduction to manifolds and tangent spaces. But let's first clarify on the underlying idea, without going into details. Those vectors you are refering to are in a way "glued to a point", and if you are going to imagine them as arrows, they would come from points in space, not always from origin. We call them vectors tangent to a point. Space of all vectors tangent to a given point is obviously linear space. Space of all such vectors (at any point) is not, because it's meaningless to add vectors with tails at different points! Therefore, for every point we have a tangent space. You can have different basis in those spaces (from now on I assume that we chose one point). We can have, for example, basis related to cartesian cordinates. Those basis vectors are of unit length and point in direction of coordinate axes. You can decompose any vector in terms of those basis vectors and get cartesian components. Make no mistake though - those components are not coordinates of a vector. Vector is not a point in the original space. Now you can also have basis related to polar coordinates. One vector pointing in the direction of growing $r$ etc., three vectors, each parallel to each other. There is important difference between those and cartesian vectors. If you go to different point in space. There direction of growing r is different! At point $(3,0,0)$ in euclidean space cartesian coordinates of r vector would be just $(1,0,0)$. However, at point $(-3,0,0)$ has cartesian components $(-1,0,0)$. Both point radially outwards and are of unit length, but what it means to point radially outward clearly depends on where you are.

3. Aug 22, 2014

### WWGD

One issue is that, outside of R^n , there is rarely a natural isomorphism between vector spaces at different points. The differential quotient then takes tangent vectors at different points, and, like blazejr said, this difference --the whole expression-- is not well-defined. To "well-define" it , one uses connections, which are choices of vector-space isomorphisms between the tangent spaces.

4. Aug 22, 2014

### Fredrik

Staff Emeritus
I don't understand what you're saying here. "Non-coordinate unit vectors" sounds like something that has nothing to do with the coordinate system. "for polar coordinates" sounds like the exact opposite. Then the second sentence suggests that the vectors you were talking about in the first sentence aren't basis vectors. Every linearly independent set with two vectors is a basis.

I'm still assuming that you're talking about these guys:
\begin{align}
&\hat r=(\cos\varphi,\sin\varphi)\\
&\hat\varphi =(-\sin\varphi,\cos\varphi)
\end{align} For all $r,\varphi$, the vectors above are the ones that the polar coordidinate system associates with the tangent space at the point $(r\cos\varphi,r\sin\varphi)$. Since that tangent space is an identical copy of the vector space that your $\vec r$ is an element of, you can also use these vectors as a basis for that space.

Same way you use any other basis.

A point on a particle's trajectory at which you intend to calculate something, like that particle's centripetal acceleration.

They're defined by
\begin{align}
\hat r=\frac{\frac{d\vec r}{dr}}{\left|\frac{d\vec r}{dr}\right|},\qquad \hat\varphi=\frac{\frac{d\vec r}{d\varphi}}{\left|\frac{d\vec r}{d\varphi}\right|}
\end{align} The right-hand sides have different values at each point. If you don't find this useful, then you can use some other basis. But you will certainly find these bases useful when you describe circular motion, where the particle's position and velocity are respectively $r\hat r$ and $r\dot\varphi\hat\varphi$, at every point.

The position vector should be drawn from the origin, but the velocity vector from the particle's location. It makes sense to think of the position vector $\vec r$ as an element of your original copy of $\mathbb R^2$ ("the configuration space"), and the velocity vector as an element of a different copy of $\mathbb R^2$ ("the tangent space at $\vec r$") with its origin attached to the point $\vec r$. When you're dealing with a significantly less trivial manifold than $\mathbb R^2$, you pretty much have to think this way.

Last edited: Aug 22, 2014