# Metric tensor: spherical coordinates

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 5; Problem 5.6.

The non-rectangular coordinate systems (semi-log and sinusoidal) we’ve looked at so far have all been flat, so it’s time to look at one in curved space. We’ll use the surface of a sphere, but rather than the usual spherical coordinates we’ll use a slight variation. We keep the azimuthal angle ${\phi}$ but use as the second coordinate the quantity ${r}$ which is the distance along the surface of the sphere measured from the north pole. If the radius of the sphere is ${R}$, then in terms of normal spherical coordinates, ${r=R\theta}$.

Curves of constant ${\phi}$ are the usual lines of longitude, while curves of constant ${r}$ are lines of latitude. The tangents to the two curves at a given point are always perpendicular, so the metric ${g_{ij}}$ will be diagonal. To find the diagonal components, consider an infinitesimal displacement ${d\mathbf{s}}$. We have

$\displaystyle d\mathbf{s}=dr\mathbf{e}_{r}+d\phi\mathbf{e}_{\phi} \ \ \ \ \ (1)$

and our job is to find the two basis vectors.

The displacement along ${\mathbf{e}_{r}}$ is just ${dr=Rd\theta}$, so ${\mathbf{e}_{r}}$ is a unit vector. A displacement along ${\mathbf{e}_{\phi}}$ depends on the radius of the constant ${r}$ curve. In spherical coordinates, this is ${R\sin\theta}$, so in our new coordinate system we get the displacement as ${R\sin\theta d\phi=R\sin\frac{r}{R}d\phi}$. Therefore the magnitude of ${\mathbf{e}_{\phi}}$ is ${R\sin\frac{r}{R}}$. The metric tensor is thus

$\displaystyle g_{ij}=\left[\begin{array}{cc} 1 & 0\\ 0 & \left(R\sin\frac{r}{R}\right)^{2} \end{array}\right] \ \ \ \ \ (2)$