# 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.