Geodesic equation: 2-d polar coordinates

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 {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} $latex \displaystyle
= &fg=000000&bg=eedbbd$
\displaystyle s^{2}+b^{2}
\displaystyle \theta \displaystyle = \displaystyle \alpha+\arctan\frac{s}{b}

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 \ \ \ \ \ (1)


The metric tensor in polar coordinates is

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

The geodesic equation gives us

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

The second equation can be integrated once to give

\displaystyle r^{2}\frac{d\theta}{ds} \displaystyle = \displaystyle k
\displaystyle \frac{d\theta}{ds} \displaystyle = \displaystyle \frac{k}{r^{2}}

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}
\displaystyle \displaystyle = \displaystyle \frac{k^{2}}{r^{3}}

Using the condition

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

we get

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

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}
\displaystyle \displaystyle = \displaystyle \frac{k^{2}}{r^{3}}

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
\displaystyle \pm\sqrt{r^{2}-k^{2}} \displaystyle = \displaystyle s+s_{0}

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}}
\displaystyle \theta \displaystyle = \displaystyle \int\frac{k}{k^{2}+s^{2}}ds
\displaystyle \displaystyle = \displaystyle \arctan\frac{s}{k}+\alpha

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.

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