Why the unitary vectors in polar coordinates are defined as such? I read in this post that says $\mathbf{e_r}$ and $\mathbf{e_\theta}$ are defined in the direction in which the coordinates increases.
But, for example, in cartesian coordinates you can do this:
$$
\mathbf v = a\mathbf i + b\mathbf j
$$
where $a$, $b$, represent the coordinates in the $x$ and $y$ axes respectively.
So you could guess that in polar coordinates:
$$
\mathbf v = r\mathbf{e_r} + \theta\mathbf{e_\theta}
$$
where, $r$ is the radius (or distance to the origin) and $\theta$ is the angle with the horizontal axes. But as you might know, this do not work.
I suppose this only works in cartesians coordinates because it is an "exceptional case". But then, why are the basis defined as this:
$$
\mathbf{e_r} = (\cos{\theta}, \sin{\theta}) \\
\mathbf{e_\theta} = (-\sin{\theta}, \cos{\theta})
$$
It loses some intuitive meaning (contrary to cartesian coordinates). Why are they defined like so?
 A: Let consider a position vector in cartesian coordinates
$$P(t) = x(t) \mathbf i+y(t) \mathbf j$$
then consider its velocity in cartesian coordinates
$$v(t) = \dot P(t) = \dot x(t) \mathbf i+\dot y(t) \mathbf j$$
In polar coordinates, the position vector is
$$P(t) = r(t)\cos \left(\theta(t)\right) \mathbf i+r(t)\sin\left(\theta(t)\right) \mathbf j$$
and its velocity is
$$v(t) = \dot P(t) = 
\left[\dot r(t) \cos\left(\theta(t)\right)-r(t)\dot \theta(t)\sin\left(\theta(t)\right)\right] \mathbf i
+\left[\dot r(t) \sin\left(\theta(t)\right)+r(t)\dot \theta(t)\cos\left(\theta(t)\right)\right] \mathbf j=$$
$$=\dot r(t)\left[\cos\left(\theta(t)\right)\mathbf i+\sin\left(\theta(t)\right)\mathbf j\right]+r(t)\dot \theta(t)\left[-\sin\left(\theta(t)\right)\mathbf i+\cos\left(\theta(t)\right)\mathbf j\right]=$$
$$=\dot r(t)\mathbf{e_r}+r(t)\dot \theta(t)\mathbf{e_\theta}$$
with

*

*$\mathbf{e_r} = \left[\cos\left(\theta(t)\right), \sin\left(\theta(t)\right)\right]$ which is parallel to $r(t)$

*$\mathbf{e_\theta} =   \left[-\sin\left(\theta(t)\right), \cos\left(\theta(t)\right)\right]$ which is orthogonal to $r(t)$
and as we can see such unit vectors are really useful to describe derivatives or gradients in polar coordinates.


A: The position vector in polar coordinates is $\vec r=r\hat r$ where $r$ is the distance from the origin along the direction specified by the unit vector $\hat r$. We don't use $\hat\theta$ for the position because it is not needed.
The unit vector $\hat\theta$ comes into play when we derive the velocity and acceleration because $$\frac{d\hat r}{d\theta}=\hat\theta$$ and $$\frac{d\hat\theta}{d\theta}=-\hat r$$
which are easily verified by looking at the definitions of these unit vectors.
A: $
\newcommand\PD[2]{\frac{\partial#1}{\partial#2}}
\newcommand\R{\mathbb R}
$
Point Functions and Coordinate Functions
A system of coordinates $x^1, x^2$ can be specified in two ways:

*

*As a function $\mathbf p(x^1, x^2)$ taking coordinates to points. We will call this the point function of a coordinate system.

*As functions $X^1(\mathbf q), X^2(\mathbf q)$ taking points to coordinates. We will call these the coordinate functions of a coordinate system.

These are inverses of each other, in that we want the following to be true:
$$
  \mathbf p(X^1(\mathbf q), X^2(\mathbf q)) = \mathbf q,
\tag{1}
$$$$
  X^1(\mathbf p(x^1, x^2)) = x^1,\quad X^2(\mathbf p(x^2, x^2)) = x^2.
\tag{2}
$$
In addition to Cartesian and polar coordinates,
for the sake of example we will also consider a simple
non-orthogonal coordinate system.
We have
$$
  \mathbf p_{xy}(x, y) = x\mathbf i + y\mathbf j,
\tag{Cartesian}
$$$$
  \mathbf p_{r\theta}(r,\theta) = r(\cos\theta)\mathbf i + r(\sin\theta)\mathbf j,
\tag{Polar}
$$$$
  \mathbf p_{uv}(u, v) = (u^3 + v)\mathbf i + v\mathbf j,
\tag{NonOrtho}
$$
and
$$
  X(\mathbf q) = \mathbf q\cdot\mathbf i,\quad
  Y(\mathbf q) = \mathbf q\cdot\mathbf j,
\tag{Cartesian}
$$$$
  R(\mathbf q) = |\mathbf q|,\quad
  \cos\Theta(\mathbf q) = \frac{\mathbf q\cdot\mathbf i}{|\mathbf q|},
\tag{Polar}
$$$$
  \mathbf U(\mathbf q) = \sqrt[3]{\mathbf q\cdot(\mathbf i - \mathbf j)},\quad
  \mathbf V(\mathbf q) = \mathbf q\cdot\mathbf j.
\tag{NonOrthog}
$$
Equations (1) and (2) allow us to turn any function on points $f(\mathbf q)$
into a function on coordinates:
$$
  f(\mathbf p(x^1, x^2))
$$
and any function on coordinates $g(x^1, x^2)$ into a function on points:
$$
  g(X^1(\mathbf q), X^2(\mathbf q)).
$$
We tend to prefer functions on points,
since that is more geometrically meaningful,
but I will freely switch between both perspectives.
Basis from Coordinates
Conceptually, vectors are differences or changes in points;
if you have two points $\mathbf q_1, \mathbf q_2 \in \R^2$,
then $\mathbf q_1 - \mathbf q_2$ is exactly the vector
pointing from $\mathbf q_2$ to $\mathbf q_1$.
Given a coordinate function $\mathbf p(x^1, x^2)$,
this idea motivates the definition of the vector basis
associated to the coordinate system $x^1, x^2$
as the "infinitesimal change" between points when $x^1$ and $x^2$ vary:
$$
  \mathbf e_1 = \PD{\mathbf p}{x^1},\quad \mathbf e_2 = \PD{\mathbf p}{x^2}.
$$
For example we may compute
$$
  \mathbf e_x = \mathbf i,\quad
  \mathbf e_y = \mathbf j,
\tag{Cartesian}
$$$$
  \mathbf e_r = (\cos\theta)\mathbf i + (\sin\theta)\mathbf j,\quad
  \mathbf e_\theta = -r(\sin\theta)\mathbf i + r(\cos\theta)\mathbf j,
\tag{Polar}
$$$$
  \mathbf e_u = 3u^2\mathbf i,\quad
  \mathbf e_v = \mathbf i + \mathbf j.
\tag{NonOrtho}
$$
Note that $\mathbf e_r$ and $\mathbf e_\theta$ are orthogonal,
and $\mathbf e_u$ and $\mathbf e_v$ are not orthogonal.
Though I've left it implicit,
it's important to note here that $\mathbf e_1$ and $\mathbf e_2$
are functions of $x^1, x^2$ and hence also functions on points;
that is to say that we've actually assigned a basis to each point,
and each such basis may be and usually is distinct from those at other points.
Reciprocal Basis
For any vector basis $\mathbf e_1, \mathbf e_3$,
there is a unique reciprocal basis $\mathbf e^1, \mathbf e^2$ such that
$$
  \mathbf e^1\cdot\mathbf e_1 = \mathbf e^2\cdot\mathbf e_2 = 1,\quad
  \mathbf e^1\cdot\mathbf e_2 = \mathbf e^2\cdot\mathbf e_1 = 0.
\tag{3}
$$
If $\mathbf e_1, \mathbf e_2$ are implicitly functions (of points or coordinates),
then the reciprocal basis is as well.
We may in fact compute them by taking the gradient
of the coordinate functions:
$$
  \mathbf e^1 = \nabla_{\mathbf q}X^1(\mathbf q),\quad
  \mathbf e^2 = \nabla_{\mathbf q}X^2(\mathbf q).
$$
In fact, this is in a sense a definition of the gradient,
but I will not get into that here.
Returning to our examples, we can compute
$$
  \mathbf e^x = \mathbf i,\quad
  \mathbf e^y = \mathbf j,
\tag{Cartesian}
$$$$
  \mathbf e^r = \frac{\mathbf q}{|\mathbf q|} = \mathbf e_r,\quad
  \mathbf e^\theta = \frac1{|\mathbf q|^2}\mathbf e_\theta
\tag{Polar}
$$$$
  \mathbf e^u = \frac{\mathbf i-\mathbf j}{3[U(\mathbf q)]^2},\quad
  \mathbf e^v = \mathbf j.
\tag{NonOrthog}
$$
$e^\theta$ is easiest to compute using the definition (3),
rather than trying to take a gradient.
Vector "Coordinates"
The definition (3) of the reciprocal vectors
immediately gives us for any vector $\mathbf v$
$$
  \mathbf v
     = (\mathbf e^1\cdot\mathbf v)\mathbf e_1 + (\mathbf e^2\cdot\mathbf v)\mathbf e_2
     = (\mathbf e_1\cdot\mathbf v)\mathbf e^1 + (\mathbf e_2\cdot\mathbf v)\mathbf e^2.
$$
Say that $\mathbf v$ is expanded at the point $\mathbf q$
if we evaluate the above implicit functions at the point $\mathbf q$.
This gives the expansion of a vector in the corresponding basis,
what might be called the "coordinates" of the vector,
but these "coordinates" are not the same thing
as the point coordinates of our coordinate system.
We will instead call them the component magnitudes of a vector.
In our examples, we make the implicit functional dependences explicit,
consider the specific case where $\mathbf v$
is thought of as a position vector,
and expand the vector $\mathbf v$ at the point it represents:
$$
  \mathbf e^x(\mathbf v)\cdot\mathbf v = X(\mathbf v),\quad
  \mathbf e^y(\mathbf v)\cdot\mathbf v = Y(\mathbf v),
\tag{Cartesian}
$$$$
  \mathbf e^r(\mathbf v)\cdot\mathbf v = R(\mathbf v),\quad
  \mathbf e^\theta(\mathbf v)\cdot\mathbf v = 0,
\tag{Polar}
$$$$
  \mathbf e^u(\mathbf v)\cdot\mathbf v = \frac13U(\mathbf v),\quad
  \mathbf e^v(\mathbf v)\cdot\mathbf v = V(\mathbf v),
\tag{NonOrthog}
$$
yielding
$$
  \mathbf v = X(\mathbf v)\mathbf e_x(\mathbf v) + Y(\mathbf v)\mathbf e_y(\mathbf v),
\tag{Cartesian}
$$$$
  \mathbf v = R(\mathbf v)\mathbf e_r(\mathbf v),
\tag{Polar}
$$$$
  \mathbf v = \frac13U(\mathbf v)\mathbf e_u(\mathbf v) + V(\mathbf v)\mathbf e_v.
\tag{NonOrthog}
$$
Component magnitudes are not coordinates,
even if they may be coincidentally for some coordinate systems.
Even then, question of component magnitudes coinciding with coordinates
only makes sense for position vectors expanded around the point they represent;
this is a very particular situation.
The moral should be that vectors and points are distinct as geometric entities;
really, it is a side-effect of modeling flat space with a vector space
that points may be identified with vectors.
Another way to say this is that, considered as a manifold,
the tangent space of e.g. $\R^2$ at a point is just $\R^2$;
this should be considered a very special case.
