I'm following along with these notes, and at a certain point it talks about change of basis to go from polar to Cartesian coordinates and vice versa. It gives the following relations:

$$\begin{pmatrix} A_r \\ A_\theta \end{pmatrix} = \begin{pmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{pmatrix} \begin{pmatrix} A_x \\ A_y \end{pmatrix}$$


$$\begin{pmatrix} A_x \\ A_y \end{pmatrix} = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix} \begin{pmatrix} A_r \\ A_\theta \end{pmatrix}$$

I was struggling to figure out how these were arrived at, and then I noticed what is possibly a mistake. In (1), shouldn't it read $$A_r=A_x+A_y$$

Is this a mistake, or am I making a wrong assumption somewhere?

I'm kinda stuck here, and would appreciate some inputs on this. Thanks.

  • $\begingroup$ I edited the MathJax to reflect the matrix notation, so you can look at the new source to see how it's done. (Though it's totally fine to write the equations separately.) $\endgroup$ – David Z Dec 7 '14 at 12:03

This properly belongs on math.se, but to properly derive these you need to remember that we can write a vector in terms of basis vectors. The vector $\vec{A}$ is unchanged, but it is just expressed as a different linear combination: $$\vec{A} = A_x \hat {x} + A_y \hat{y} = A_r \hat{r} + A_\theta\hat{\theta} $$.

Because you can write $\hat{r}$ as a linear combination of $\hat x$ and $\hat y$, i.e., $\hat {r} = \frac{x}{r} \hat {x} + \frac{y}{r} \hat y = \cos\theta \hat{x} + \sin\theta \hat{y}$, and similarly for $\hat{\theta} = -\sin\theta \hat{x} + \cos\theta \hat y$, you can solve for the $A_r$ and $A_\theta$ in terms of $A_x$ and $A_y$.

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    $\begingroup$ I don't understand the $A_\theta\hat{\theta}$ term. Shouldn't it vanish? After all, a vector starting from the origin only has magnitude in the radial direction, right? $\endgroup$ – Joebevo Dec 8 '14 at 2:52
  • $\begingroup$ True, the polar basis vectors aren't defined at the origin, but $A$ is better thought of as being some displacement vector away from the origin. Said differently, $A$ lives in a vector space defined by the basis vectors. $\endgroup$ – lionelbrits Dec 8 '14 at 10:03
  • $\begingroup$ I understand that it isn't defined at the origin. However, a displacement vector $A$ that starts at the origin only has a radial component, when we consider it in polar coordinates. $\endgroup$ – Joebevo Dec 8 '14 at 10:14
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    $\begingroup$ I think the source of confusion is that the notes use the vector $\vec{A}$ as a position vector in the beginning, but later it refers to it as a "generic vector $\vec{A}$. A good example would be the velocity vector. In that case the "origin" for the vector is at the point specified by $r,\theta$. It is not at $r=0$. $\endgroup$ – lionelbrits Dec 8 '14 at 15:12

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