Why polar coordinates r basis vector is constant at large r?

Question

The Christoffel symbols for polar coordinates r base vector are:

$$\Gamma_{r\theta}^r=0$$

$$\Gamma_{r\theta}^\theta=\frac{1}{r}$$

I understand intuitively why the first one is true: the rate of change of r basis vector in radial direction is zero.

I have difficulty understanding why the second one is true. It says that at large r, the rate of change of r basis vector in azimuthal direction is close to zero. My intuition tells me that the rate of change in azimuthal direction should be bigger (ie. proportional to r).

To illustrate my confusion, in the diagram above the rate of change of the smaller r basis vector (yellow) is represented by the green arrow. The rate of change of the bigger r basis vector (red) is represented by the blue arrow. So $\Gamma_{r\theta}^\theta$ should be proportional to r.

What am I missing?

Answer 1

\begin{equation} \begin{cases} x=r\cos\theta\\ y=r\sin\theta \end{cases} \ \ \ \text{and}\ \ \ \begin{cases} r=\sqrt{x^2+y^2}\\ \theta=\arctan(\frac{y}{x}) \end{cases} \end{equation} Calculating the symbol, $$\Gamma^\theta_{r\theta}=\frac{\partial^2x}{\partial r\partial\theta}\frac{\partial \theta}{\partial x}+\frac{\partial^2y}{\partial r\partial\theta}\frac{\partial \theta}{\partial y}=\left(-\sin\theta\frac{-y}{x^2+y^2}\right)+\left(\cos\theta\frac{x}{x^2+y^2}\right)=\dots=\frac{1}{r}$$ For more see https://physics.stackexchange.com/a/567731/331865

Answer 2

First, the vectors should be based at the origin. So the change from a unit vector at angle $\theta$ to a unit vector at angle $\theta+\Delta\theta$ has length approximately $\Delta\theta$, independent of $r$. But at the point $(r,\theta)$, the $\theta$ basis vector has length $r$, and so we need a coefficient of $1/r$. This is the Christoffel symbol you have. But the net vector $\frac 1r \frac{\partial}{\partial\theta}$ is a unit vector, certainly not a vector going to $0$ as $r\to\infty$.