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)

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