# Geodesic equation: 2-d polar coordinates

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 8; Problem 8.3.

Another example of using the geodesic equation to calculate geodesics, this time in flat, 2-d space. We know that geodesics here are straight lines, but let’s prove this using polar coordinates.

First, consider some arbitrary straight line. Draw a perpendicular from the origin to the line and call this distance ${b}$. The polar angle from the ${x}$ axis to this perpendicular we define as ${\alpha}$. The point where the perpendicular intersects the line marks the zero point for path length, so ${s=0}$ there.

Starting with the perpendicular, we increase ${\theta}$ and draw the radius vector from the origin to the line at this angle. This vector has length ${r}$ and intersects the line a distance ${s}$ along from ${b}$. The vector, the perpendicular, and the line segment of length ${s}$ make a right-angled triangle, so ${r^{2}=s^{2}+b^{2}}$, and the angle between the perpendicular and the vector is ${\arctan\frac{s}{b}}$, making the polar angle between the ${x}$ axis and the vector ${\theta=\alpha+\arctan\frac{s}{b}}$. The parametric equations for the straight line in polar coordinates are then

 $\displaystyle r^{2}$ $\displaystyle =$ $\displaystyle s^{2}+b^{2}\ \ \ \ \ (1)$ $\displaystyle \theta$ $\displaystyle =$ $\displaystyle \alpha+\arctan\frac{s}{b} \ \ \ \ \ (2)$

We now need to show that the geodesics given by the geodesic equation have the same form. The geodesic equation is

$\displaystyle \frac{d}{d\tau}\left(g_{aj}\frac{dx^{j}}{d\tau}\right)-\frac{1}{2}\frac{\partial g_{ij}}{\partial x^{a}}\frac{dx^{i}}{d\tau}\frac{dx^{j}}{d\tau}=0 \ \ \ \ \ (3)$

The metric tensor in polar coordinates is

$\displaystyle g_{ij}=\left[\begin{array}{cc} 1 & 0\\ 0 & r^{2} \end{array}\right] \ \ \ \ \ (4)$

The geodesic equation gives us

 $\displaystyle \frac{d^{2}r}{ds^{2}}-r\left(\frac{d\theta}{ds}\right)^{2}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (5)$ $\displaystyle \frac{d}{ds}\left(r^{2}\frac{d\theta}{ds}\right)$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (6)$

The second equation can be integrated once to give

 $\displaystyle r^{2}\frac{d\theta}{ds}$ $\displaystyle =$ $\displaystyle k\ \ \ \ \ (7)$ $\displaystyle \frac{d\theta}{ds}$ $\displaystyle =$ $\displaystyle \frac{k}{r^{2}} \ \ \ \ \ (8)$

for some constant ${k}$. Substituting this into the first equation gives

 $\displaystyle \frac{d^{2}r}{ds^{2}}$ $\displaystyle =$ $\displaystyle r\left(\frac{d\theta}{ds}\right)^{2}\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{k^{2}}{r^{3}} \ \ \ \ \ (10)$

Using the condition

$\displaystyle g_{ij}\frac{dx^{i}}{ds}\frac{dx^{j}}{ds}=\left(\frac{ds}{ds}\right)^{2}=+1 \ \ \ \ \ (11)$

we get

 $\displaystyle \left(\frac{dr}{ds}\right)^{2}+r^{2}\left(\frac{d\theta}{ds}\right)^{2}$ $\displaystyle =$ $\displaystyle 1\ \ \ \ \ (12)$ $\displaystyle \frac{dr}{ds}$ $\displaystyle =$ $\displaystyle \pm\sqrt{1-\frac{k^{2}}{r^{2}}} \ \ \ \ \ (13)$

This is the first integral of the second derivative above, as we can verify:

 $\displaystyle \frac{d^{2}r}{ds^{2}}$ $\displaystyle =$ $\displaystyle \pm\frac{1}{2}\left(1-\frac{k^{2}}{r^{2}}\right)^{-1/2}\left(2\frac{k^{2}}{r^{3}}\right)\frac{dr}{ds}\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{k^{2}}{r^{3}} \ \ \ \ \ (15)$

We can now integrate the first derivative to get

 $\displaystyle \pm\int\left(1-\frac{k^{2}}{r^{2}}\right)^{-1/2}dr$ $\displaystyle =$ $\displaystyle \int ds\ \ \ \ \ (16)$ $\displaystyle \pm\sqrt{r^{2}-k^{2}}$ $\displaystyle =$ $\displaystyle s+s_{0} \ \ \ \ \ (17)$

If we take the constant ${k=b}$ and ${s_{0}=0}$, then ${r^{2}=s^{2}+b^{2}}$ which agrees with the radial equation for the straight line above. If we then substitute this into ${d\theta/ds}$, we get

 $\displaystyle \frac{d\theta}{ds}$ $\displaystyle =$ $\displaystyle \frac{k}{r^{2}}\ \ \ \ \ (18)$ $\displaystyle \theta$ $\displaystyle =$ $\displaystyle \int\frac{k}{k^{2}+s^{2}}ds\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \arctan\frac{s}{k}+\alpha \ \ \ \ \ (20)$

where we can identify the constant of integration with the angle ${\alpha}$ defined above. This matches the ${\theta}$ equation for the straight line above. Thus the geodesic equation does indeed generate straight lines for the geodesics.