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Let $\vec{v}\in\mathbb{R}^2$ be a vector in cartesian coordinates that can be written as the linear combination of its vector components and the unit base vectors $(\vec{e}_1,\vec{e}_2)$, e.g.: $$ \vec{v}=x\,\vec{e}_1+y\,\vec{e}_2 \,.$$

My understanding of coordinate transforms is still pretty limited, but I understand there are some important distinctions to be made:

  1. The vector itself is different from the vector components
  2. If we change to a different coordinate system, the vector components have to change
  3. If we change to a different coordinate system, the base vectors have to change as well

So, the linear combination in polar coordinates $$ \vec{v}=r\,\vec{e}_r+\theta\,\vec{e}_\theta \,.$$ should yield the same vector as before. To see if that's true, I chose $\vec{v}=(x,y)^T=(1,1)^T$.


First, let's compute the unit base vectors $(\vec{e}_r,\vec{e}_\theta)^T$ in polar coordinates. We have $$ \vec{v}=\begin{pmatrix} x \\ y \end{pmatrix}=\begin{pmatrix} r\,\cos\theta \\ r\,\sin\theta \end{pmatrix} \,.$$ With that, the unit base vectors in polar coordinates become $$ \vec{e}_r = \frac{1}{h_r}\frac{\partial \vec{v}}{\partial r} = \begin{pmatrix}\cos\theta \\ \sin\theta \end{pmatrix}\quad , \quad \vec{e}_\theta = \frac{1}{h_\theta}\frac{\partial \vec{v}}{\partial \theta} = \frac{1}{r}\begin{pmatrix} -r\sin\theta \\ r\,\cos\theta \end{pmatrix} \,.$$

I did not want to "silently" apply the scale factors $h_r=1$, $h_\theta=r$ to make it more clear how to get from the base vectors to the unit base vectors.

The vector components in polar coordinates are $$ \begin{pmatrix} r \\ \theta \end{pmatrix} = \begin{pmatrix} \sqrt{x^2+y^2} \\ \arctan{y/x} \end{pmatrix} = \begin{pmatrix} \sqrt{2} \\ \pi/4 \end{pmatrix} $$

Now, all that is left is to plug in all that we have and we should get the vector that we started with: \begin{align} \vec{v} = r\,\vec{e}_r + \theta\,\vec{e}_\theta &= \sqrt{2}\begin{pmatrix} \cos\pi/4 \\ \sin\pi/4 \end{pmatrix} + \frac{\pi}{4}\begin{pmatrix} -\sin\pi/4 \\ \cos\pi/4 \end{pmatrix}= \\[8pt] &=\begin{pmatrix} 1 \\ 1 \end{pmatrix} + \frac{\pi}{4\sqrt{2}}\begin{pmatrix} -1 \\ 1 \end{pmatrix} \end{align} which is not the same vector that we began with in cartesian coordinates. What did I miss?

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You are confusing coordinates of a point with components of a vector. It so happens that in flat space points can be represented with position vectors, but generally you need to keep vectors and points distinct. A point is a location and a vector is a direction from a given location. All of this to say that the expression $r\vec e_r + \theta\vec e_\theta$ makes no sense.

Remember that $\vec e_r$ and $\vec e_\theta$ depend on $r, \theta$; this is because they give a basis to specify directions (vectors) from the location (point) with coordinates $(r,\theta)$.

It so happens (in what could be considered a coincidence) that the position vector of a point with coordinates $(r,\theta)$ is $r\vec e_r(r,\theta)$.

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  • $\begingroup$ "All of this to say that the expression $r\,\vec{e}_r+\theta\,\vec{e}_\theta$ makes no sense." OK, then how would a vector (its linear combination) in polar coordinates has to look like? $\endgroup$ Commented Mar 15, 2023 at 22:59
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    $\begingroup$ @user358572 We have to choose a point to base our vectors at. If this point has coordinates $(r,\theta)$ and we take the vector $v = a\vec e_x + b\vec e_y$ then what we need is $$a\vec e_x + b\vec e_y = v = a'\vec e_r(r,\theta) + b'\vec e_\theta(r,\theta) = (a'\cos\theta-b'\sin\theta)\vec e_x + (a'\sin\theta + b'\cos\theta)\vec e_y$$ hence we get a system of two linear equations $$\begin{pmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{pmatrix}\begin{pmatrix}a'\\b'\end{pmatrix}=\begin{pmatrix}a\\b\end{pmatrix}.$$ $\endgroup$ Commented Mar 16, 2023 at 0:50
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    $\begingroup$ The matrix is a rotation matrix and easily invertible; we find $$a' = a\cos\theta + b\sin\theta,\quad b' = -a\sin\theta + b\cos\theta.$$ Thus $$a\vec e_x + b\vec e_y = v = (a\cos\theta + b\sin\theta)\vec e_r(r,\theta) + (-a\sin\theta + b\cos\theta)\vec e_\theta(r,\theta). $$ I'm not sure if that answers you question. $\endgroup$ Commented Mar 16, 2023 at 0:50
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Welcome to MSE. Your confusion probably stems from the fact that you are assuming the new coordinates of $v$ to be $(\sqrt 2,\pi/4)$. Note that with $r = \sqrt 2$ and $\theta = \pi/4$, then $$e_r = \frac{\sqrt 2}{2}v \quad \Longleftrightarrow \quad v = \sqrt 2 e_r.$$ So $v = e_1+e_2 = \sqrt 2 e_r$, meaning that in terms of the standard basis $v$ is $(1,1)$, whereas in the new basis it is $(\sqrt 2,0)$ (and not $(\sqrt 2,\pi/4)$).

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