1
$\begingroup$

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:

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

and

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

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.

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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 $A_y$.

$\endgroup$
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.