Scalar product invariance The question is somehow silly, but I can't seem to find a way out right now.
Consider the vector $v\in\mathbb R^2$, expressed in Cartesian coordinates as $(1,1)$ and in polar coordinates as $(\sqrt 2, \pi/4)$. The length of the vector should be the same regardless of the coordinate system, right?
In Cartesian coordinates the metric tensor is the identity, so that the norm squared is just $1+1=2$; in polar coordinates $g = diag(1, r^2)$, so that the norm squared is $2 + 4\cdot (\pi^2/16)$. This looks just wrong.
One also easily checks that vectors which are orthogonal with the ordinary metric are not orthogonal anymore in this new metric.
I would then conclude that I just cannot use the metric tensor like that - but it looks like that's what I'm doing all the time in GR... What's going on? What am I missing?
 A: The point is that the length of your vector is given by
\begin{equation}
s=\int_0^a \sqrt{g_{11} u'^{2}+g_{22} v'^{2}} dt
\end{equation}
In cartesian coordinate the length vector (1,1) can be computed by considering the parametric representation of a straight line going through the origin forming an angle of 45°. Thus $(u(t),v(t))=(t,t)$. The integral must be evaluated between 0 and 1.By doing that we have
\begin{equation}
s=\int_0^1 \sqrt{1 +1} dt=\sqrt{1 +1}
\end{equation}
When considering the Polar coordinate the same straight line is given by the parametric representation  $(u(t),v(t))=(t \sqrt 2 ,\pi/4)$, where u(t) represent r, and v(t) the angle $\theta$. Then
\begin{equation}
s=\int_0^1 \sqrt{1 \cdot (\sqrt 2)^2 + 0 \cdot (t \sqrt 2)^2 }  dt=\sqrt{1 +1}
\end{equation}
A: Let's play it according to the rules of tensor calculus.
Introduce polar coordinates and calculate the local base
vectors of that system:
$$
\left. \begin{matrix} x = r\cos(\phi) \\
y = r\sin(\phi) \end{matrix} \right\} \quad \Longrightarrow \quad
\left\{ \begin{matrix} \vec{c}_r =
\left[ \begin{matrix} \partial x/\partial r\\
\partial y/\partial r\end{matrix} \right]
= \left[\begin{matrix}\cos(\phi)\\ \sin(\phi)\end{matrix}\right] \\
\vec{c}_\phi =
\left[ \begin{matrix} \partial x/\partial \phi\\
\partial y/\partial \phi\end{matrix} \right]
= r \left[\begin{matrix}-\sin(\phi)\\ \cos(\phi)\end{matrix}\right] 
\end{matrix} \right.
$$
Next calculate the metric tensor:
$$
\left[ \begin{matrix} g_{rr} & g_{r\phi} \\ g_{\phi r} & g_{\phi\phi}
\end{matrix} \right] =
\left[ \begin{matrix} \left(\vec{c}_r\cdot\vec{c}_r\right) &
\left(\vec{c}_r\cdot\vec{c}_\phi\right)\\
\left(\vec{c}_r\cdot\vec{c}_\phi\right) &
\left(\vec{c}_\phi\cdot\vec{c}_\phi\right) \end{matrix}\right] =
\left[ \begin{matrix} 1 & 0 \\ 0 & r^2 \end{matrix}\right]
$$
Then all we can say is something about the infinitesimal length:
$$
ds^2 = \left[ \begin{matrix} dr & d\phi \end{matrix} \right] 
\left[ \begin{matrix} 1 & 0 \\ 0 & r^2 \end{matrix}\right]
\left[ \begin{matrix} dr \\ d\phi \end{matrix} \right]
\quad \Longrightarrow \quad
ds^2 = dx^2 + dy^2 = dr^2 + r^2 d\phi^2
$$
It is shown in the answer by Upax how to integrate this to obtain
a finite length. Substitute $r=t\sqrt{2}$ and $\phi=\pi/4$ into:
$$
s = \int_0^1 \sqrt{\left(\frac{dr}{dt}\right)^2+r^2\left(\frac{d\phi}{dt}\right)^2} dt =
\int_0^1 \sqrt{\left(\sqrt{2}\right)^2+\left(t\sqrt{2}\right)^2\cdot 0}\; dt = \sqrt{2}
$$
