0
$\begingroup$

The unit vectors in polar coordinates describe the directions of increasing $\rho$ and $\phi$:

$$ \begin{align*} \hat \rho &= \cos(\phi) \hat i + \sin(\phi) \hat j \\ \hat \phi &= -\sin(\phi) \hat i + \cos(\phi) \hat j \\ \end{align*} $$

Note the above formulas give the polar unit vectors in-terms of the cartesian unit vectors $\hat i$ and $\hat j$. This somewhat bothers me, since it doesn't seem like there should be anything "fundamental" about cartesian coordinates. Why can't I start with polar coordinates and describe the cartesian unit vectors in terms of $\hat \rho$ and $\hat \phi$?

$$ \begin{align*} \hat i &= ?\ \hat \rho\ + ?\ \hat \phi \\ \hat j &= ?\ \hat \rho\ + ?\ \hat \phi \\ \end{align*} $$

I think that, given $\hat \rho$ and $\hat \phi$, (both of which depend on $\phi$ - or where I am in the plane), I can rotate clockwise by $\phi$ to obtain $\hat i$ and $\hat j$. This rotation is given (in the standard basis) by the matrix: $$R(\phi) = \begin{pmatrix} \cos \phi & \sin \phi \\ -\sin \phi & \cos \phi \end{pmatrix}.$$ So, for example, to get $\hat i$, I want to rotate $\hat \rho$ by $\phi$ radians. But if I want to apply $R(\phi)$ to $\hat \rho$, I need to express $\hat \rho$ in the standard basis again, which is what I'm trying to avoid.

$\endgroup$

1 Answer 1

1
$\begingroup$

Note that $$\begin{bmatrix} \hat{\rho} \\ \hat{\phi} \end{bmatrix}=R(\phi)\begin{bmatrix} \hat{i} \\ \hat{j}\end{bmatrix}$$ Now multiply by $(R(\phi))^{-1}$.

$\endgroup$

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.