Why in polar coordinate, $(r,\theta )$ doesn't mean $re_r+\theta e_\theta $?

Question

1) I asked a question here to compute length of set written in polar form. I'm very confuse about something : In cartesian, when I write $(x,y)$ this mean $xe_x+ye_y$. In $\{e_u,e_v\}$, when I right $(a,b)$ this mean $ae_u+be_v$. I really don't get why in $\{e_r,e_\theta \}$, the vector $(r,\theta )$ is not $re_r+\theta e_\theta $.

2) Also, if a curve is given by it's polar coordinate as $(r(\varphi ),\varphi )$, then it's graph is given by $$\{r(\varphi )e_r\mid \varphi \in I\}.$$

I have the impression that all curve in polar form can be written as $(r(\varphi ),\varphi )$. If not, what would be for example a curve s.t. the graph is given by $$\{r_1(\varphi )e_r+r_2(\varphi )e_\varphi \mid \varphi \in I\} \ \ ?$$ Can such a graph exist ? If yes, what would be his "polar coordinate" ?

Answer 1

There are many types of coordinate systems. The way the coordinates define the location of a point depends on the coordinate system.

Some, but not all, coordinate systems use fixed basis vectors, where $(x_1 , x_2 , \dots , x_n)$ corresponds to $\sum_{i=1}^{n} x_i \vec{e}_i$. So, it is just common; not a property of coordinate systems in general.

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In two dimensions, the correspondence between Cartesian coordinates $(x, y)$ and polar coordinates $(r, \varphi)$ is (usually defined as) $$\bbox{\begin{cases} x = r \cos \varphi \\ y = r \sin \varphi \\ \end{cases}} \quad \iff \quad \bbox{\begin{cases} r = \sqrt{x^2 + y^2} \\ \varphi = \operatorname{atan2}(y , x) \\ \end{cases}}$$ where $\operatorname{atan2}(y , x)$ is the two-argument form of arctangent; like $\arctan(y / x)$, except that it takes the quadrant (signs of $x$ and $y$) into account also, and therefore can yield any angle.

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If you have a two-dimensional skewed coordinate system, where the basis vectors are $$\bbox{\vec{e}_u = \left [ \begin{matrix} x_u \\ y_u \end{matrix} \right ]}, \quad \bbox{\vec{e}_v = \left [ \begin{matrix} x_v \\ y_v \end{matrix} \right ]}$$ but not necessarily perpendicular to each other (like they are in Cartesian coordinate systems), with the origin of the skewed coodinate system at $(x_0 , y_0)$ in Cartesian coordinates, the correspondence to Cartesian coordinates is $$\bbox{\begin{cases} x = x_0 + u x_u + v x_v \\ y = y_0 + u y_u + v y_v \\ \end{cases}} \quad \iff \quad \bbox{\begin{cases} \displaystyle u = \frac{ x_v ( y - y_0 ) - y_v ( x - x_0 ) }{ x_v y_u - x_u y_v } \\ \displaystyle v = \frac{ y_u ( x - x_0 ) - x_u ( y - y_0 ) }{ x_v y_u - x_u y_v } \end{cases}}$$ This works as long as $x_v y_u - x_u y_v \ne 0$, i.e. the basis vectors $\vec{e}_u$ and $\vec{e}_v$ are not parallel.

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You can define your own coordinate systems when needed. For example, you could use a two-dimensional coordinate system $(\varphi, d)$, $$\bbox{\begin{cases} x = x_0 + (r + d) \cos \varphi \\ y = y_0 + (r + d) \sin \varphi \\ \end{cases}} \quad \iff \quad \bbox{\begin{cases} d = \sqrt{ ( x - x_0 )^2 + ( y - y_0 )^2 } - r \\ \varphi = \operatorname{atan2}( y - y_0 , x - x_0 ) \\ \end{cases}}$$ where $d$ is the distance from the $r$-radius circle centered at $(x_0 , y_0)$, and $\varphi$ is the angle. So, very similar to polar coordinates, except that $d$ is relative to a fixed radius $r$. (You could use this for example if you had a large circular racing track, so $\varphi$ would identify your location along the track, and $d$ deviance from the track midpoint.)

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There are very few limitations on what you can use as a coordinate system, and that is why the expectation of always being able to use basis vectors to convert the coordinates to Cartesian coordinates is wrong.

Answer 2

Fundamentally, a vector is a geoemtric object whos coordinates transform in a specific way. Suppose you have a vector pointing from the origin to the point (1,1). Now lets set upa new coordinate system by rotating the the x and ycoordinate axes counter-clockwise $45^\circ$ about the z axis. According to the new coordinate axes, this point is at $(\sqrt{2},0)$. Even though there are different coordinate representations, the vector is considered the same direction, $\vec{v}=x\hat{i}+y\hat{j}=x'\hat{i'}+y'\hat{j'}$, where $x'$ and $y'$ are the coordinates relative to the new basis vectors. Some basis vectors change from point to point introducing the behavior you mention.

Let $\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}$ Then:

$$\frac{\partial\vec{r}}{\partial x}=\hat{i}$$ $$\frac{\partial\vec{r}}{\partial y}=\hat{j}$$ $$\frac{\partial\vec{r}}{\partial z}=\hat{k}$$

It's no coincidence that you get the standard cartesian basis vectors.

Take an additional partial derivative of these with respect to any other coordinate, including x, e.g. $\frac{\partial^2 \vec{r}}{\partial x\partial y}$,or $\frac{\partial^2 \vec{r}}{\partial x^2}$, you get zero. This is because the cartesian basis vectors are constant.

More generally, the basis vectors in some coordinate system at a specific point is $\frac{\partial \vec{r}}{\partial \mu}$, where $\mu$ is the new coordinate.

So in your specific case: $$\vec{r}=x\hat{i}+y\hat{j}=r\cos{\theta}\hat{i}+r\sin{\theta}\hat{j}$$

So (keeping in mind the cartesian vectors are constant at every point): $$\frac{\partial \vec{r}}{\partial r}=\cos{\theta}\hat{i}+\sin{\theta}\hat{j}=\frac{x}{r}\hat{i}+\frac{y}{r}\hat{j}=\hat{e_r}$$

$$\frac{\partial \vec{r}}{\partial \theta}=-r\sin{\theta}\hat{i}+r\cos{\theta}\hat{j}=-y\hat{i}+x\hat{j}$$

Notice the second vector is not a unit vector, but has magnitude $r$. This could still be used as a basis vector and actually makes some advanced concepts easier to manipulate, but for now, divide through by r to get the usual basis vector for polar coordinates:

$$\hat{e_\theta}=-\sin{\theta}\hat{i}+\cos{\theta}\hat{j}=\frac{-y}{r}\hat{i}+\frac{x}{r}\hat {j}$$

What is $r\hat{e_r}$? What is $r\hat{e_\theta}$?

Basis vectors aren't simply a pair of vectors. They are an entire Vector Field. Given a position in coordinate system $(\mu, \alpha,\theta)$ The basis vectors at that coordinate point could be different from what they are elsewhere. So they have a non-zero derivative.

$\frac{\partial \hat{e_r}}{\partial \theta}=\hat{e_\theta}$.

Again, a non zero derivative of the coordinate basis implies a different basis at different coordinates. Every point in space has its own basis vectors.

The basis vectors in polar coordinates at a given point consists of:

$\hat{e_r}=$ vector pointing from origin to point, divided by how far from the orin. In other words, it's a unit vector representing the direction from origin to the point.

$\hat{e_\theta}=$ unit vector perpendicular to $\hat{e_r}$ and pointing counter clockwise.

Having unit magnitude, the basis vectors represent direction of fast change of the associated coordinate. For $r$ this is directly away from the origin. For $\theta$ this is as would be expected if $r$ is held constant while going around a circle.