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)

2 thoughts on “Metric tensor: spherical coordinates

  1. Pingback: Metric tensor: parabolic coordinates | Physics tutorials

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