Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 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 . The polar angle from the axis to this perpendicular we define as . The point where the perpendicular intersects the line marks the zero point for path length, so there.
Starting with the perpendicular, we increase and draw the radius vector from the origin to the line at this angle. This vector has length and intersects the line a distance along from . The vector, the perpendicular, and the line segment of length make a right-angled triangle, so , and the angle between the perpendicular and the vector is , making the polar angle between the axis and the vector . The parametric equations for the straight line in polar coordinates are then
We now need to show that the geodesics given by the geodesic equation have the same form. The geodesic equation is
The metric tensor in polar coordinates is
The geodesic equation gives us
The second equation can be integrated once to give
for some constant . Substituting this into the first equation gives
Using the condition
This is the first integral of the second derivative above, as we can verify:
We can now integrate the first derivative to get
If we take the constant and , then which agrees with the radial equation for the straight line above. If we then substitute this into , we get
where we can identify the constant of integration with the angle defined above. This matches the equation for the straight line above. Thus the geodesic equation does indeed generate straight lines for the geodesics.