Geodesic equation: paraboloid

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 8; Problem 8.7.

This is an example of the geodesic equation in a 2-d curved space (with two space dimensions). 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}. We’ve already looked at this example and found that the metric is

\displaystyle  ds^{2}=\left(1+\left(2br\right)^{2}\right)dr^{2}+r^{2}d\phi^{2} \ \ \ \ \ (2)

Using the geodesic equation for the two coordinates, we get

\displaystyle   \frac{d}{ds}\left(\left(1+\left(2br\right)^{2}\right)\frac{dr}{ds}\right)-\frac{1}{2}8b^{2}r\left(\frac{dr}{ds}\right)^{2}-\frac{1}{2}2r\left(\frac{d\phi}{ds}\right)^{2} \displaystyle  = \displaystyle  0\ \ \ \ \ (3)
\displaystyle  \frac{d}{ds}\left(r^{2}\frac{d\phi}{ds}\right) \displaystyle  = \displaystyle  0 \ \ \ \ \ (4)

The second equation can be integrated once to give

\displaystyle  \frac{d\phi}{ds}=\frac{c}{r^{2}} \ \ \ \ \ (5)

where {c} is a constant. Substituting this into the first equation and working out the first term, we get

\displaystyle  4b^{2}r\left(\frac{dr}{ds}\right)^{2}+\left(1+\left(2br\right)^{2}\right)\frac{d^{2}r}{ds^{2}}-\frac{c^{2}}{r^{3}}=0 \ \ \ \ \ (6)

Integrating these equations directly isn’t easy, but if we look at the special case of a curve defined by the condition {\phi=k} for some constant {k} (that is, a curve starting at the bottom of the bowl and going directly up the side of the bowl), then {d\phi=0} and from the metric:

\displaystyle   \frac{dr}{ds} \displaystyle  = \displaystyle  \frac{1}{\sqrt{1+\left(2br\right)^{2}}}\ \ \ \ \ (7)
\displaystyle  \frac{d^{2}r}{ds^{2}} \displaystyle  = \displaystyle  -\frac{4b^{2}r}{\left(1+\left(2br\right)^{2}\right)^{3/2}}\frac{dr}{ds}\ \ \ \ \ (8)
\displaystyle  \displaystyle  = \displaystyle  -\frac{4b^{2}r}{\left(1+\left(2br\right)^{2}\right)^{2}} \ \ \ \ \ (9)

Inserting this into the above, we get

\displaystyle  \frac{4b^{2}r}{1+\left(2br\right)^{2}}-\left(1+\left(2br\right)^{2}\right)\frac{4b^{2}r}{\left(1+\left(2br\right)^{2}\right)^{2}}-\frac{c^{2}}{r^{3}}=-\frac{c^{2}}{r^{3}} \ \ \ \ \ (10)

This will be equal to zero if {c=0}, which also satisfies the condition {\frac{d\phi}{ds}=\frac{c}{r^{2}}=0}. Thus these radial curves are in fact geodesics.

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