Tensors and coordinate transformations I was wondering if there's a difference between the terms "coordinate transformation" and "change of coordinates" from the point of view of tensors.
Here's what I mean. If we have a cartesian coordinate system $X$ and we rotate it by some angle $\theta$, we have a new coordinate system $X'$ where we can get the components $v'^i$ of any vector $\vec V$ in terms of its components $v^i$ in $X$. Since $V$ is a tensor, its components in $X'$ & $X$ are related by $$V'^i = \frac{\partial x'^i}{\partial x^j}V^j$$
Now consider a vector $A$ in cartesian coordinates and the same vectors in polar coordinates. Are the components of $A$ in polar coordinates related to the components in Cartesian coordinates via the same equation given above?
When write down the equations, it gives me that the components are functions of the position. For example, $$A^r = cos(\theta)A^x + sin(\theta)A^y$$
And I don't really understand what that means. Why are the components of a constant vector, that's one and the same in any coordinate representation, depend on a coordinate?
I worded it as best as I could, yet it still sounds confusing.
 A: A lot of confusion seems to come from mixing up vectors (and tensors) with vector (or tensor) fields. In a lot of literature the word field is omitted for brevity.
In two dimensions the coordinate transformation between cartesian and polar coordinates is
$$
\left(\begin{matrix}x\\y\end{matrix}\right)\mapsto\left(\begin{matrix}r\\\varphi\end{matrix}\right)=\left(\begin{matrix}\sqrt{x^2+y^2}\\\arctan(\frac{y}{x})\end{matrix}\right)\,.
$$
The back transformation is
$$
\left(\begin{matrix}r\\\varphi\end{matrix}\right)\mapsto\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}r\cos\varphi\\r\sin\varphi\end{matrix}\right)\,.
$$
These maps are one to one when restricted to $M=\mathbb R^2\setminus\{0\}\,.$
So far the above maps only express points on the manifold in different languages. Recall that a tensor field on the manifold is a map from $M$ into the tensor product of a selection of (co-)tangential spaces. For example, a contravariant vector field is
$$\tag{1}
V=v^x\partial_x+v^y\partial_y
$$
where $v^x,v^y$ are functions on $M$ and $\partial_x,\partial_y$ are a basis of $T_pM\simeq\mathbb R^2\,.$ The $x,y$-notation indicates that $V$ is expressed in cartesian coordinates. To express $V$ in polar coordinates we use the chain rule:
\begin{align}
\partial_x&=\frac{\partial r}{\partial x}\partial_r+\frac{\partial \varphi}{\partial x}\partial_\varphi=\frac{x}{\sqrt{x^2+y^2}}\partial_r-\frac{y}{x^2+y^2}\partial_\varphi=\cos\varphi\,\partial_r-\frac{\sin\varphi}{r}\,\partial_\varphi\,,\\
\partial_y&=\frac{\partial r}{\partial y}\partial_r+\frac{\partial \varphi}{\partial y}\partial_\varphi=\frac{y}{\sqrt{x^2+y^2}}\partial_r+\frac{x}{x^2+y^2}\partial_\varphi=\sin\varphi\,\partial_r+\frac{\cos\varphi}{r}\,\partial_\varphi\,.
\end{align}
Plugging this into (1) gives us the vector field in polar coordinates:
$$\tag{2}
V=\underbrace{(v^x\cos\varphi\,+v^y\sin\varphi)}_{\displaystyle =:v^r}\,\partial_r+\underbrace{\frac{-v^x\sin\varphi+v^y\cos\varphi}{r}}_{\displaystyle=:v^\varphi}\,\partial_\varphi\,.
$$
To shed some light on the concept of $\partial_x,\partial_y$ being vector fields recall that, for a real valued function $f$ on $M$ and a curve $\gamma$ passing through $p=\gamma(0)\,,$ the chain rule gives for the derivative of $f$ along the tangent of the curve
\begin{align}
\frac{d}{dt}\Bigg|_{t=0}f(\gamma(t))&=\partial_xf(\gamma(0))\,\gamma_1'(0)+\partial_yf(\gamma(0))\,\gamma_2'(0)\\
&=\partial_xf(p)\,\gamma_1'(0)+\partial_yf(p)\,\gamma_2'(0)\\
%&=\langle\nabla f(p),\gamma'(0)\rangle\,.
\end{align}
If we remove the test function $f$ from this relationship we get the RHS as
$$
\gamma_1'(0)\,\partial_x|_p+\gamma_2'(0)\,\partial_y|_p\,.
$$
This shows that $\gamma_1'(0),\gamma_2'(0)$ are the components of a general tangential vector in $T_pM$, and $\partial_x|_p,\partial_y|_p$ are the basis vectors.
See also Tangent Space.
