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Suppose we want to write the vector: $$\vec{A}=4\hat{x} +3\hat{y}$$ in polar coordinates $(\rho, \phi)$, by making use of the identities: $$ \hat{x}=\cos\phi\,\hat{\rho} -\sin{\phi}\,\hat{\phi}, \quad \hat{y}=\sin\phi\,\hat{\rho} +\cos{\phi}\,\hat{\phi} $$ which simply yield: $$ \vec{A}= \left(4\cos{\phi}+3\sin{\phi}\right)\hat{\rho} + \left(-4\sin{\phi}+3\cos{\phi}\right)\hat{\phi} $$ Now expressing $\vec{A}$ as above does not make much sense to me because $\vec{A}$'s orientation is fixed so there should appear no free parameter such as the angle $\phi$. If $\phi$ indeed represents the orientation of the given vector then: $$ \cos\phi=\frac{A_x}{||A||}=\frac{4}{5}, \quad \sin\phi=\frac{A_y}{||A||}=\frac{3}{5} $$ If I plug these in $\vec{A}$, I end up with: $$A=5\hat{\rho}$$ By using the same reasoning above, a generic vector's, $\vec{A}=A_x \hat{x} + A_y\hat{y}$, representation in polar coordinates appears to be $$\vec{A}=||A|| \hat{\rho}, \quad ||A||=\sqrt{A_x^2+A_y^2}$$

But on the other hand, a generic vector in polar coordinates is given in the form: $$ \vec{A}=A_{\rho}\hat{\rho} + A_{\phi}\hat{\phi} $$ If my steps of converting $\vec{A}$ from cartesian to polar coordinates is correct, then why does the $\hat{\phi}$ component seems to disappear ?

EDIT

Thank you all for your answers. I need to clear up few more things. In one of the answers below it is stated the representation (although I have seen such notation on the web): $$ \vec{A}=A_{\rho}\hat{\rho} + A_{\phi}\hat{\phi} $$ with constant coefficients is wrong. Is this also true for the representation of say: $$ \vec{A}=4\hat{x} +3\hat{y} + 2\hat{z} $$ in cylindrical and spherical coordinates ? If so, then the proper conversion of $\vec{A}$ must read $$ \vec{A}=5\hat{\rho} + 2\hat{z}, \quad \rho=\sqrt{x^2 + y^2} $$ in cylindrical coordinates and $$ \vec{A}=\sqrt{29}\hat{r}, \quad r=\sqrt{x^2+y^2+z^2} $$ in spherical coordinates. Therefore the representation of a generic vector with arbitrary but constant coefficients: $$ \vec{A}= A_{\rho} \hat{\rho} + A_{\phi}\hat{\phi} + A_{z}\hat{z}\\ \vec{A}= A_{r} \hat{r} + A_{\phi}\hat{\phi} + A_{\theta}\hat{\theta} $$ is wrong. Is this conclusion correct ?

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    $\begingroup$ No question is ever dumb! $\endgroup$ – xX A C E Xx Dec 17 '20 at 0:26
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    $\begingroup$ You've aligned your polar coordinate frame so that $\hat{\rho}$ is pointing in the same direction as $A$, so that's why there is no $\hat{\phi}$ component. $\endgroup$ – Bungo Dec 17 '20 at 1:20
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    $\begingroup$ Your last conclusion is correct. Given a curvilinear coordinate system $\xi_1,...,\xi_n$ with unit vectors $\boldsymbol{\epsilon}_1,...,\boldsymbol{\epsilon}_n$ a vector will be in general some linear combination of these basis vectors: $$\mathbf{v}=\sum_{i=1}^n f_i(\xi_1,...,\xi_n) \boldsymbol{\epsilon}_i$$ In a lot of coordinate systems, many of these coefficients will be zero - this is precisely what makes them so useful. $\endgroup$ – K.defaoite Dec 17 '20 at 13:56
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But on the other hand, a generic vector in polar coordinates is given in the form: $$ \vec{A}=A_{\rho}\hat{\rho} + A_{\phi}\hat{\phi} $$

The formula $$\underline{v}=v_r \hat{\underline{r}}+v_\theta \hat{\underline{\theta}}$$ Is wrong. The radius unit vector is defined such that the position vector $\underline{\mathrm{r}}$ can be written as $$\underline{\mathrm{r}}=r~\hat{\underline{r}}$$ That's what makes polar coordinates so useful. Sometimes we only care about things that point in the direction of the position vector, making the theta component ignorable.

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  • $\begingroup$ I have edited the question a bit. Could you take a look at it ? $\endgroup$ – user91411 Dec 17 '20 at 11:15
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Probably I'm not understanding your notation or your question properly, but if what you want is to express the point $(4,3)$ in the Cartesian plane in polar coordinates, I'll use the identities $r=\sqrt{x^2+y^2}$ and $\tan(\theta)=\frac{y}{x}$ Substituting your values, the answer will be $\left(5, \arctan\left(\frac{3}{4}\right)\right)$

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Cartesian coordinates have some very convenient properties. The vector that expresses the distance and direction from point $P = (p_x,p_y)$ to point $Q = (q_x,q_y)$ is simply $(q_x - p_x)\hat x + (q_y - p_y)\hat y$; that is, the vector's $x$ coordinate is simply the difference of the two points' $x$ coordinates, and similarly with $y$ coordinates.

This works exactly the same no matter where you are in the plane.

We don't have this convenience with polar coordinates. You cannot just take the polar coordinates of $P$ and $Q$, find the difference in $\rho$ coordinates, and use this as the $\rho$ coordinate of the vector between the points. It doesn't work for the $\phi$ coordinate either.

Usually when someone uses $\hat\rho$ and $\hat\phi$ as unit vectors like $\hat x$ and $\hat y$, what they are actually doing is (in at least some sense) looking at the grid of the polar coordinate system at some point in the plane and building a Cartesian coordinate system with an origin at that point and axes aligned with the radial and tangential directions. That's why you can construct a vector with components like $A_{\rho}\hat{\rho} + A_{\phi}\hat{\phi}$; it is actually a vector in a locally defined Cartesian coordinate system where $\hat\rho$ is the unit vector pointing along one axis and $\hat\rho$ is the unit vector pointing along the other axis.

It turns out we can define the same locally defined coordinates for any point along the same radial line out from the origin--that is, for all points with a given $\phi.$ But if you take a point whose $\phi$-coordinate is different you have a different local coordinate system. That's where the "free" $\phi$ in your vector expression comes from.

That "free" $\phi$ is not really free; it is a function of the point on which you decided to construct the locally-defined coordinate system. It expresses the angle between the $\hat x$-$\hat y$ coordinate axes and the $\hat\rho$-$\hat\phi$ coordinate axes for those two local Cartesian coordinate systems at that point, which you must know in order to correctly convert the coordinates of a given vector between those two coordinate systems.

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