Metric tensor: parabolic coordinates

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 5; Problem 5.7.

Another example of a non-rectangular 2-d curved coordinate system. This time we have a parabolic bowl with equation

$\displaystyle z=br^{2} \ \ \ \ \ (1)$

where ${r^{2}=x^{2}+y^{2}}$ and ${b}$ is a constant. We use the two coordinates ${r}$ (as defined here) and the azimuthal angle ${\phi}$.

Using the same technique as with the sphere, we consider infinitesimal displacements along the constant curves. The curve of constant ${\phi}$ is a parabola, while the curve of constant ${r}$ is a circle at height ${z=br^{2}}$. 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

$\displaystyle d\mathbf{s}=dr\mathbf{e}_{r}+d\phi\mathbf{e}_{\phi} \ \ \ \ \ (2)$

and our job is to find the lengths of the two basis vectors.

Consider first a displacement along ${\mathbf{e}_{r}}$. As ${r}$ increases, we move a distance up the side of the parabolic bowl. This displacement consists of a horizontal increment of size ${dr}$ and a vertical increment of size ${dz=2brdr}$. By Pythagoras, the total displacement is ${ds=\sqrt{1+\left(2br\right)^{2}}dr}$. The length of ${\mathbf{e}_{r}}$ is therefore ${\sqrt{1+\left(2br\right)^{2}}}$.

A displacement along ${\mathbf{e}_{\phi}}$ is a displacement around a circle of constant ${r}$, so here ${ds=rd\phi}$, giving the length of ${\mathbf{e}_{\phi}}$ as ${r}$. The metric is then

$\displaystyle g_{ij}=\left[\begin{array}{cc} 1+\left(2br\right)^{2} & 0\\ 0 & r^{2} \end{array}\right] \ \ \ \ \ (3)$