# Transforming Metric from (position-dependent) Polar Coordinates to (position-independent) Cartesian Coordinates

The following example is taken form Relativity, Gravitation and Cosmology, Ta-Pei Cheng, 2nd edition, page 89.

Fig. 1

$$ds^2=(dx^1, dx^2) \begin{pmatrix}g_{11} & g_{12} \\ g_{21} & g_{22}\end{pmatrix}\begin{pmatrix}dx^1\\ dx^2\end{pmatrix} \tag{1}$$

Here are further examples of metric tensors for two dimensional surfaces, calculated by using the fact the any surface in the small can be approximated by a plane having Cartesian coordinates. We also take this occasion to discuss the possibility of using the metric tensor to determine whether a surface is curved or not. For a flat surface, we can find a set of coordinates so that the metric tensor is position-independent; this is not the case in a curved surface as $$g_{ab} = \mathbf{e}_a\cdot\mathbf{e}_b$$ in a curved space must change from point to point.

1. A plane surface with Cartesian coordinates For the coordinates $$(x^1,x^2)=(x,y)$$, we have the infinitesimal length $$ds^2 = dx^2 + dy^2$$ Fig. 1(a). Comparing this to the general expression in (1), we see that the metric must be

$$g_{ab} = \begin{pmatrix} 1 && 0 \\ 0 && 1 \end{pmatrix} \tag{2}$$
which is of course position-independent. This is possible only if the space is not curved.

1. A plane surface with polar coordinates For the coordinates $$(x^1, x^2)=(r,\phi)$$, we have the infinitesimal length $$ds^2 = (dr)^2 + (rd\phi)^2$$, Fig. 1(b), thus according to (1), a metric

$$g_{ab}=\begin{pmatrix}1 & 0 \\ 0 & r^2\end{pmatrix} \tag{3}$$

which is position dependent! But we can find a coordinate transformation $$(x^1, x^2) \to (x'^1, x'^2)$$ so that the metric in the new coordinates is position independent, $$g_{ab}'=\delta_{ab}$$. Of course, the new coordinates are just the Cartesian coordinates $$(x'^1, x'^2)=(x,y)$$:

$$x = r\cos\phi, \quad y=r\sin\phi \tag{4}$$

Now while the whole thing makes intuitively sense I fail to use the given transformation in 2. to transform (3) into a position independent metric.

What exactly would be the correct transformation of $$ds^2 = (dr)^2 + (rd\phi)^2$$?

The metric tensor is just like any other tensor. Its components transform as $$\bar{g}_{ij} = \frac{\partial x^k}{\partial \bar{x}^i} \frac{\partial x^l}{\partial \bar{x}^j} g_{kl}$$ Where $$\bar{x}$$ and $$x$$ are the new and old coordinates respectively. Tensors themselves always remain the same regardless of the choice of coordinates. It's the components that depend on the choice of coordinates.