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I've done a lot of algebra in polar coordinates, but now some basic stuff is confusing me. A general vector in Cartesian coordinates can be written as $\mathbf{v} = x \hat{x} + y\hat{y}$. And in polar coordinates it is $r \hat{r} + \phi \hat{\phi}$ where $\phi$ is the angle made by the vector with x axis. Now I want to write the vectors say $\hat{x} + \hat{y}$ and $\hat{x} - \hat{y}$ in polar coordinates. How do i do that? will they be $\sqrt{2} \hat{r} + \pi/4\hat{\phi}$ and $\sqrt{2} \hat{r} - \pi/4\hat{\phi}$? I got confused because a vector field which goes radially outward will look like $\mathbf{V} = r\hat{r}$. And so polar coordinates both the mentioned vectors have the same representation. Am i missing something here? I understand that the basis vectors for polar coordinates at each point in 2D space are different, but shouldnt a vector described anywhere on the 2d plane, be uniquely described by some r and theta?

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    $\begingroup$ Would you clarify your question? V=rr is a vector field. but x + y is a vector. so x + y is one specific vector in V = rr vector field. a vector and a vector field can both uniquely described in polar coordinate. $\endgroup$ Dec 8 '20 at 20:04
  • $\begingroup$ oh....i'll try. Take the vector field V=rr for example. at the point (0,3) this vector field will have a vector pointing in the drection of y axis and of the magnitude 3, which in caretsian coordiantes is 0x + 3y. But also at that point the vector field has the value 3r. so does the vector 0x+3y correspond to the vector 3r? or the vector 3r + $\pi/2 \hat{\phi}$? I would think the latter because just the vector 3r doesnt specify the direction of the vector. But then should a radially outward pointing vector field look like $r \hat{r} + \phi \hat{\phi}$? $\endgroup$ Dec 10 '20 at 7:10
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You will have to calculate the corresponding values of $r$ and $\phi$, yes.

$r$ is found with Pythagoras' Theorem and $\phi$ with one of the trigonometric relations for right-angled triangles. I'm not sure where your numbers $\sqrt 2$ and $\pi/4$ come from because $r$ and $\phi$ will depend on $x$ and $y$. Unless your $x$ and $y$ are fixed in your scenario of course. In general, these equations might do:

$$r=\sqrt{x^2+y^2}$$ $$\phi=\arctan{\left(\frac yx\right)}$$

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  • $\begingroup$ the vectors I have chosen are (1,1) and (1,-1) hence $\sqrt(2)$ and $\pi/4$. unlike a general vector which is (x,y). Even so, how do you explain the vector field $r \hat{r}$? Or maybe more concretely, where does the vector $3 \hat{r}$ point? if it's (3,0) then shouldn't the vector field $r \hat{r}$ be only defined on x-axis? $\endgroup$ Dec 8 '20 at 13:14

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